Print Page - TT's Maths Thread

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: TrueTears on November 12, 2009, 05:01:58 pm

Title: TT's Maths Thread
Post by: TrueTears on November 12, 2009, 05:01:58 pm

Well yeah, just started doing some reading now, got a few questions, help would be appreciated!

First one:

Let a and b be two integers, not both zero. Then $\{xa + yb | x, y \in Z\}$ is the set of all integral multiples of $GCD(a, b)$ . In particular; $GCD(a, b)$ is the smallest positive number in this set.

What exactly does this theorem mean? I just read the proof, but can't seem to figure out the meaning of it haha

Title: Re: TT's Maths Thread
Post by: dcc on November 12, 2009, 05:08:25 pm

Quote from: TrueTears on November 12, 2009, 05:01:58 pm

Let a and b be two integers, not both zero. Then $\{xa + yb | x, y \in Z\}$ is the set of all integral multiples of $GCD(a, b)$ . In particular; $GCD(a, b)$ is the smallest positive number in this set.

What exactly does this theorem mean? I just read the proof, but can't seem to figure out the meaning of it haha

http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity

Title: Re: TT's Maths Thread
Post by: TrueTears on November 12, 2009, 05:32:07 pm

Just a bit stuck in the middle of my proof by contradiction.

1. Show that $p^2 = 2$ is not satisfied by any rational p.

Let $p = \frac{m}{n}$ where $m,n \in Z$ and let $m,n$ both not be even.

Thus $m^2 = 2n^2$

This implies the RHS is even no matter what. (Let $N = n^2$ , since n is an integer then $n^2$ must also be an integer, thus $2N$ is even)

Now $m^2 = \mbox{some even number}$ which implies that n is also even. However we said that $m$ is non-even, thus this is a contradiction.

Now how do you go about showing n is even?

Title: Re: TT's Maths Thread
Post by: /0 on November 12, 2009, 05:34:54 pm

Are you using Zuckerman's number theory book?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 12, 2009, 05:36:43 pm

Quote from: dcc on November 12, 2009, 05:08:25 pm

Quote from: TrueTears on November 12, 2009, 05:01:58 pm
Let a and b be two integers, not both zero. Then $\{xa + yb | x, y \in Z\}$ is the set of all integral multiples of $GCD(a, b)$ . In particular; $GCD(a, b)$ is the smallest positive number in this set.

What exactly does this theorem mean? I just read the proof, but can't seem to figure out the meaning of it haha

http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity

Thanks.

Quote from: /0 on November 12, 2009, 05:34:54 pm

Are you using Zuckerman's number theory book?

Nah Art and Craft of Problem Solving. I'm reading Number Theory with Algebra first (Cause the book says so lol)

Title: Re: TT's Maths Thread
Post by: Ahmad on November 12, 2009, 05:45:16 pm

I'll explain the significance of the theorem, and hopefully that will give you a better idea of what it means and why it's useful.

In basic number theory we tend to use facts which we learn to intuitively accept without proof such as every positive integer being uniquely factorised into prime factors i.e. 15 = 3*5. But being particular as they are, mathematicians require that this be proved. To prove such a result one naturally asks the question, under what assumptions? What can I assume is true about the integers which I can use without proof to derive results such as this.

The key property of the integers which we assume without proof is that any non-empty subset of the positive integers, (such as {5, 100} or {x such that x is a prime number}), has a smallest element in it (5 and 2 respectively). And naturally we give it a hopefully suggestive name:

The Well Ordering Property - Every non-empty subset of the positive integers has a smallest element.

This property seems very obvious right? What isn't quite obvious is how we derive so much using such a simple axiom! But lo and behold this statement is logically equivalent to mathematical induction, meaning they imply each other, I'll leave this as an exercise. So now we have induction too, great!

Divisibility is easily defined, 20 is divisible by 2 because there's an integer which when multiplied by 2 gives 20, with the appropriate generalisation. Primes are also easily defined as positive integers divisible only by 1 and itself, and noting that we normally exclude 1 from being a prime.

Now what we want to do is show that we can factorise any positive integer into primes. If you think about how we normally factor an integer a natural idea which may occur to you is that we can start off with an integer n > 1, if it's prime then we're done, we've written it as a product of primes, otherwise we can write n = pq, with p,q > 1, and furthermore since both p and q are smaller than n we can use an inductive argument to show that each of them can be written as a product of primes, then n = pq can be written as a product of primes. (We can think of 1 as a product of primes, namely as the empty product to get the induction started).

We have the result that a positive integer can be written as a product of primes, but there's a subtlety, how do we know that 15 = 3 * 5 and that the only way to write 15 as a product of primes is as 3*5? In other words, how do we know that the factorisation is unique! How do we go about doing this? Well if 15 = 3 * 5 = p1 * p2 * ... * pn for some primes p1.., then we want to conclude that one of those is a 3, say p1 = 3, then we could divide off by 3 and get 5 = p2 * ... * pn, and we can repeat the same argument to show that one of these primes say p2 must be equal to 5, and dividing off by 5 gives 1 = p3*...*pn, from which we conclude that n must've been 2, and so this statement is really 1 = 1. Then we'd have shown that p1 = 3 and p2 = 5, and so the factorisation was really the same!

But I've been a little sly, because I didn't say how I concluded that some pm, say p1, must be equal to 3. This is in fact the key step to proving unique factorisation, and it is named after Euclid. Euclid's Lemma: if a prime p divides a*b, then p divides a or p divides b. This is where your result comes in, in order to prove Euclid's lemma we first prove a very very useful result called Bezout's identity: there exists integers x and y such that ax + by = gcd(a, b), with gcd(a,b) denoting the greatest common divisor of a and b. It's not obvious why this is useful, or how this might be useful, and unfortunately I don't have the time to motivate it (I'm supposed to be studying for an exam I have tomorrow morning). I'm assuming you read the proof of Bezout's identity, the key step being the well ordering theorem (which can be used to prove the division algorithm with remainder which we all know from grade 3).

Once we have Bezout's identity we're ready to prove Euclid's lemma, which as a reminder states that if a prime p divides ab then p divides a or p divides b. Proof:
If p divides a then we're done! So assume that p doesn't divide a, so that gcd(p, a) = 1, then by Bezout's identity we have ax + py = 1 for some integers x and y. Multiplying both sides by b, gives abx + pby = b, but we know that p divides ab, so that ab = k*p, for some integer k which we can substitute into our equation, giving kpx + pby = b, or p(kx + by) = b, which shows that p must divide b and we're done! This completes our proof of Euclid's lemma and hence our proof of unique factorisation. So we have proved the Fundamental Theorem of Arithmetic that each positive integer can be written uniquely as a product of primes. :)

(Sorry the ending is kind of not motivated very well, I'm kinda strapped for time atm, also apologise for any typos/errors).

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 12, 2009, 05:55:57 pm

m is even right, meaning that:

$m=2k$ for some k

therefore:

$4k^2=2n^2$
$n^2=2k^2$ so n is even too.

====================================================================================

Another proof that I prefer is to show that p is not a natural number, hence $p=\frac{a}{b}$ hence
$ \frac{a^2}{b^2}=2 $

now we can assume a and b have no common factor, so they are composed of different primes, but $a^2$ is composed of the same primes as $a$ and same for $b$ . (only the powers double in the prime factorization), hence the fraction cannot cancel to a whole number. This easily generalises to higher powers.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 12, 2009, 05:57:33 pm

hahahaha what an easy step and I missed it!

thanks kamil!

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 12, 2009, 06:30:08 pm

Yeah this is a common approach in textbooks. But I quite like Gauss' proof of the Fundamental theorem of arithmetic, where he proves Euclid's lemma without even mentioning Bezout's identity :D . In my spare moments where I have fantasized about being a teacher of number theory, I have always wanted to show both approaches to my students :D

Title: Re: TT's Maths Thread
Post by: TrueTears on November 12, 2009, 06:41:56 pm

Amazing Ahmad, thank you so much!

Also:

1. Prove that for $n \in N$

$\left \lfloor \sqrt{n} + \sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{4n+2} \right \rfloor$

Where do I start with this?

2. The function $f : Z \to Z$ satisfies $f(n) = n - 3$ if $n \ge 1000$ and $f(n) = f(f(n+5))$ if $n<1000$ . Find $f(84)$

Title: Re: TT's Maths Thread
Post by: Ahmad on November 12, 2009, 06:44:19 pm

Quote from: kamil9876 on November 12, 2009, 06:30:08 pm

Yeah this is a common approach in textbooks. But I quite like Gauss' proof of the Fundamental theorem of arithmetic, where he proves Euclid's lemma without even mentioning Bezout's identity :D . In my spare moments where I have fantasized about being a teacher of number theory, I have always wanted to show both approaches to my students :D

I'm not sure if I know this one. I'd be delighted if you could enlighten us! :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 12, 2009, 07:45:46 pm

Quote from: TrueTears on November 12, 2009, 06:41:56 pm

Amazing Ahmad, thank you so much!

Also:

1. Prove that for $n \in N$

$\left \lfloor \sqrt{n} + \sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{4n+2} \right \rfloor$

Where do I start with this?

2. The function $f : Z \to Z$ satisfies $f(n) = n - 3$ if $n \ge 1000$ and $f(n) = f(f(n+5))$ if $n<1000$ . Find $f(84)$

I think I got Q 2...

I tried the following:

Consider $f(999)$

$f(999) = f(f(999+5)) = f(f(1004)) = f(1001) = 998$

Now as we move further down to say $f(990)$

$f(990) = f(f(995)) = f(f(f(1000))) = f(f(997)) = f(f(f(1002))) = f(f(999)) = f(f(f(1004))) = f(f(1001)) = f(998) = f(f(1003)) = f(1000) = 997$

Now a pattern can be seen:

As long as the starting number is even then $f(n) = f(f(n+5))$ can be expressed in the form of $f(f((X+5) + 2k))$ where X is the number which is even and $k \in Z$

We find that $(X+5) + 2k = 1003$ and IFF $k \in Z$ , then $X$ is an even number.

Which means every even n, namely $f(n)$ , that is below 1000 can be simplified to merely f(1003), which equals 997.

Thus $f(84) = 997$ .

I did a few more trials on paper, and generalised that every odd n below 1000 equals 998 :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 12, 2009, 07:59:04 pm

Quote

I'm not sure if I know this one. I'd be delighted if you could enlighten us!

Ok so here it is, uses some simple results from modular arithmetic.

lemma: if $a$ and $b$ are two numbers less than a prime $p$ , then $ab$ is not a multiple of $p$ .

suppose this is false, then we can find numbers $a_1, a_2....$ such that each, $a_i b$ is a multiply of $p$ . Assume $a$ is the smallest of these.

$p$ is a prime hence it is between two multiples of $a$ :

am<p<a(m+1)
0<p-am<a

however (p-am)b=pb-amb=pb-kpm=p(b-km)

hence a multiple of p, so we have found a number(p-am) that is smaller than a that satisfies that condition, a contradiction.

(If fermat could be bothered doing more than writing conjectures in margins then he probably would've used infinite descent here, i reckon)

edit: oh yeah... and now see if you can use this lemma to prove Euclid's lemma.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 12, 2009, 08:48:30 pm

By division algorithm a = p*k1 + m, b = p*k2 + n, p | ab implies p | mn, with both m and n less than p, which implies at least one of m, n is 0, and therefore a multiple of p. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 12, 2009, 09:11:30 pm

Quote from: TrueTears on November 12, 2009, 06:41:56 pm

Amazing Ahmad, thank you so much!

Also:

1. Prove that for $n \in N$

$\left \lfloor \sqrt{n} + \sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{4n+2} \right \rfloor$

Where do I start with this?

2. The function $f : Z \to Z$ satisfies $f(n) = n - 3$ if $n \ge 1000$ and $f(n) = f(f(n+5))$ if $n<1000$ . Find $f(84)$

Just random brainstorming for Q 1...

Consider first the perfect squares: $\sqrt{1}, \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}...$ $= 1, 2, 3, 4, 5...$

It can be seen that the largest difference between the perfect squares is 1.

Thus for $\sqrt{n}$ and $\sqrt{n+1}$ the difference between them will be smaller than 1.

This implies that $\left \lfloor \sqrt{n} \right \rfloor = \left \lfloor \sqrt{n+1} \right \rfloor$ IFF n+1 is NOT a perfect square.

$RHS = \left \lfloor \sqrt{4(n+\frac{1}{2})} \right \rfloor = \left \lfloor 2\sqrt{(n+\frac{1}{2})} \right \rfloor = \left \lfloor \sqrt{(n+\frac{1}{2})} + \sqrt{(n+\frac{1}{2})} \right \rfloor$

kk I'm leaving this question for now, coming back to it later... brain just can't break through it rofl

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 12:06:53 am

Another one:

The polynomial $1 - x + x^2 - x^3 + . . . + x^{16} - x^{17}$ may be written in the form $a_0 + a_1y + a_2y^2 + a_3y^3 + . . . + a_{16}y^{16} + a_{17}y^{17}$ , where $y = x + 1$ and the $a_i$ 's are constants. Find the value of $a_2$ .

My working:

So immediately one can see this is a terminating geometric sum

thus, $1 - x + x^2 - x^3 + . . . + x^{16} - x^{17} = \frac{(x^{18} - 1)}{-x-1}$

But $y = x+1 \implies x = y - 1$

Subbing this in yields: $\frac{((y-1)^{18} - 1)}{-(y-1)-1} = \frac{(y-1)^{18}-1}{-y} = \frac{1}{-y}(y-1)^{18} + \frac{1}{y}$

Where to go from here? I know binomial expansion is somehow involved but how...?

Title: Re: TT's Maths Thread
Post by: dejan91 on November 13, 2009, 12:16:26 am

So this is 'real' maths? Different. Definitely different. :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 12:18:37 am

Quote from: TrueTears on November 13, 2009, 12:06:53 am

$\frac{(y-1)^{18}-1}{-y}$

simply find the y^3 term of the expansion of the numerator. When it gets divided by -y it becomes the y^2 term that you're after.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 12:19:43 am

Quote from: dejan91 on November 13, 2009, 12:16:26 am

So this is 'real' maths? Different. Definitely different. :P

haha this is the maths I've long awaited for! God I love it!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 12:56:12 am

Quote from: kamil9876 on November 13, 2009, 12:18:37 am

Quote from: TrueTears on November 13, 2009, 12:06:53 am
$\frac{(y-1)^{18}-1}{-y}$

simply find the y^3 term of the expansion of the numerator. When it gets divided by -y it becomes the y^2 term that you're after.

Oh right! Let me try that now :P

Numerator = $\left(\left(^{18}_0\right)y^{18} + \left(^{18}_1\right)y^{17}(-1)^1 + \left(^{18}_2\right)y^{16}(-1)^2 + ... + \left(^{18}_{15}\right)y^3(-1)^{15} + \left(^{18}_{16}\right)y^{2}(-1)^{16} + \left(^{18}_{17}\right)y(-1)^{17} + \left(^{18}_{18}\right)(-1)^{18}\right) - 1$

= $\left(^{18}_0\right)y^{18} + \left(^{18}_1\right)y^{17}(-1)^1 + \left(^{18}_2\right)y^{16}(-1)^2 + ... + \left(^{18}_{15}\right)y^3(-1)^{15} + \left(^{18}_{16}\right)y^{2}(-1)^{16} + \left(^{18}_{17}\right)y(-1)^{17}$

Then divide each term by $-y$ yields:

$-\left(^{18}_0\right)y^{17} - \left(^{18}_1\right)y^{16}(-1)^1 - \left(^{18}_2\right)y^{15}(-1)^2 - ... - \left(^{18}_{15}\right)y^2(-1)^{15} - \left(^{18}_{16}\right)y^{1}(-1)^{16} - \left(^{18}_{17}\right)y^0(-1)^{17}$

$\therefore a_2 = \left(^{18}_{15}\right) = 816$

amirite?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 12:58:34 am

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 12:59:40 am

Thanks kamil.

I'm in shock too ;)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 01:33:40 am

Just a few more, I'm new to these types of questions, so any hints/suggestions would be fine:

1. Suppose that $a,b,c,d$ are real numbers such that

$a^2 + b^2 = 1$

$c^2 + d^2 = 1$

$ac+bd=0$

Show that

$a^2 + c^2 = 1$

$b^2 + d^2 = 1$

$ab + cd = 0$

(Problem can get messy but there is an elegant and complete solution)

~~2. Find all integer solutions $(n,m)$ to $n^4+2n^3+2n^2+2n+1=m^2$~~

3. Find the smallest positive integer whose cube ends in 888.

~~4. Find $3x^2y^2$ if x, y are integers such that $y^2 + 3x^2y^2 = 30x^2 + 517$ .~~

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 02:26:43 am

Okay I've thought about Q 4 for about 1 n half hours or so and got to this step:

$3x^2y^2 - 30x^2 + y^2 = 517$

$3x^2(y^2 - 10) + y^2 = 517$

$3x^2(y^2-10) + y^2 - 10 + 10 = 517$

$(y^2-10)(3x^2+1) = 507$

Now I'm stuck... what to do now?

Title: Re: TT's Maths Thread
Post by: /0 on November 13, 2009, 02:29:47 am

Factors of 507?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 02:32:15 am

Quote from: /0 on November 13, 2009, 02:29:47 am

Factors of 507?

And then what?

Title: Re: TT's Maths Thread
Post by: humph on November 13, 2009, 02:36:46 am

How many ways can you factorise 507? For each pair of factorisations, solve for x and y.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 02:38:10 am

Ah I see

$(y^2-10)(3x^2+1) = (39)(13)$

Since x and y are integers $y^2 - 10 = 39$ and $3x^2 + 1 = 13$

Awesome, got it now, thanks humph and /0!

Title: Re: TT's Maths Thread
Post by: /0 on November 13, 2009, 03:28:42 am

For 1. a good formula is

$(a^2+b^2)(c^2+d^2) = (ac+bd)^2+(ad-bc)^2$

But I'm not sure how to apply it yet

For 2.
The rational root theorem says a factor is $(n+1)$ , then after complete factorisation I think grouping factor pairs might work

Title: Re: TT's Maths Thread
Post by: humph on November 13, 2009, 04:15:21 am

Here's a neat proof of 1. using matrix methods:
Consider $A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix}$ and its transpose $A^T = \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix}$ . Then
$A A^T = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix} = \begin{pmatrix}a^2 + b^2 & ac + bd \\ ac + bd & c^2 + d^2 \\ \end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix} = I$
where $I$ is the identity matrix. This implies that $A^T = A^{-1}$ (so $A$ is an orthogonal matrix), and so we must have that
$\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix} = I = A^{-1} A = A^T A = \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix} \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} = \begin{pmatrix}a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \\ \end{pmatrix}$
so by equating entries, we obtain the result.

(Long) Edit: /0 gave me a major hint by showing that $ad - bc = \pm 1$ , which is a necessary (but not sufficient) condition for the matrix $A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix}$ to be orthogonal. The proof above though shows that the conditions of the question are necessary and sufficient for $A$ to be orthogonal. That is,
$\begin{array}{lll} a^2 + b^2 = 1 & \hspace{5.7cm} a^2 + c^2 = 1 \\ c^2 + d^2 = 1 & \Longleftrightarrow A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \text{ is orthogonal} \Longleftrightarrow b^2 + d^2 = 1 \\ ab + cd = 0 & \hspace{5.7cm} ac + bd = 0 \\ \end{array}$

This basically follows from the fact that a $n \times n$ matrix with real entries $A$ ,
$A^T A = I \Longleftrightarrow \left \langle Au , Av \right \rangle = \left \langle u , v \right \rangle \text{ for all } u,v \in \mathbb{R}^n$ .
Here $\left \langle u , v \right \rangle = u^T v$ denotes the standard inner product of two (column) vectors $u = \begin{pmatrix}u_1 \\ \vdots \\ u_n \\ \end{pmatrix}, v = \begin{pmatrix}v_1 \\ \vdots \\ v_n \\ \end{pmatrix} \in \mathbb{R}^n$ . It is closely related to the dot product;
$\left \langle u , v \right \rangle = u^T v = \begin{pmatrix}u_1 & \cdots & u_n \\ \end{pmatrix} \begin{pmatrix}v_1 \\ \vdots \\ v_n \\ \end{pmatrix} = u_1 v_1 + \ldots + u_n v_n = (u_1, \ldots, u_n) \cdot (v_1, \ldots, v_n).$
The linearity of the inner product means that we only need to show that
$A^T A = I \Longleftrightarrow \left \langle Ae_i , Ae_j \right \rangle = \left \langle e_i , e_j \right \rangle \text{ for all } 1 \leq i,j \leq n,$
where $\{e_i : 1 \leq i \leq n\}$ is some basis for $\mathbb{R}^n$ . We can clearly choose $e_1 = \begin{pmatrix}1 \\ 0 \\ \vdots \\ 0 \\ \end{pmatrix}, \ldots, e_n = \begin{pmatrix}0 \\ \vdots \\ 0 \\ 1 \\ \end{pmatrix}$ , so that $\left \langle e_i , e_j \right \rangle = \delta_{ij}$ , where $\delta_{ij}$ is the Kronecker delta. We can then prove the result:

If $A^T A = I$ , then
$\left \langle Ae_i , Ae_j \right \rangle = (Ae_i)^T(Ae_j) = e_i^T A^T A e_j = e_i^T e_j = \left \langle e_i , e_j \right \rangle.$

Conversely, if $\left \langle Ae_i , Ae_j \right \rangle = \left \langle e_i , e_j \right \rangle$ , then
$\delta_{ij} = e_i^T (A^T A) e_j,$
which is the entry of the $i$ -th row and $j$ -th column of $A^T A$ , which implies that $A^T A = I$ .

So the original question can be restated as:
If $\delta_{ij} = e_i^T (A^T A) e_j$ , show that $\delta_{ij} = e_i^T (A A^T) e_j$ , and the result holds because both expressions are equivalent to $A$ being an orthogonal matrix. The result can clearly also be generalised to the $n \times n$ dimensional case, though of course it can't be written down so neatly.

There's also a similar problem for complex matrices; instead of considering orthogonal matrices, we consider unitary matrices. A matrix with complex entries is unitary if $A A^{*} = A^{*} A = I$ ; here $A^{*}$ denotes the conjugate transpose of $A$ , obtained by taking the transpose of $A$ and conjugating each entry.

Thus the complex version of question 1. is

$\begin{array}{lll} |a|^2 + |b|^2 = 1 & \hspace{5.2cm} |a|^2 + |c|^2 = 1 \\ |c|^2 + |d|^2 = 1 & \Longleftrightarrow A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \text{ is unitary} \Longleftrightarrow |b|^2 + |d|^2 = 1 \\ a\overline{b} + c\overline{d} = \overline{a}b + \overline{c}d = 0 & \hspace{5.2cm} a\overline{c} + b\overline{d} = \overline{a}c + \overline{b}d= 0 \\ \end{array}$

(Can you tell I'm procrastinating from studying algebraic topology? :P)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 01:11:55 pm

Quote from: humph on November 13, 2009, 04:15:21 am

Here's a neat proof of 1. using matrix methods:
Consider $A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix}$ and its transpose $A^T = \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix}$ . Then
$A A^T = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix} = \begin{pmatrix}a^2 + b^2 & ac + bd \\ ac + bd & c^2 + d^2 \\ \end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix} = I$
where $I$ is the identity matrix. This implies that $A^T = A^{-1}$ (so $A$ is an orthogonal matrix), and so we must have that
$\begin{pmatrix}1 & 0 \\ 0 & 1 \\ \end{pmatrix} = I = A^{-1} A = A^T A = \begin{pmatrix}a & c \\ b & d \\ \end{pmatrix} \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} = \begin{pmatrix}a^2 + c^2 & ab + cd \\ ab + cd & b^2 + d^2 \\ \end{pmatrix}$
so by equating entries, we obtain the result.

(Long) Edit: /0 gave me a major hint by showing that $ad - bc = \pm 1$ , which is a necessary (but not sufficient) condition for the matrix $A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix}$ to be orthogonal. The proof above though shows that the conditions of the question are necessary and sufficient for $A$ to be orthogonal. That is,
$\begin{array}{lll} a^2 + b^2 = 1 & \hspace{5.7cm} a^2 + c^2 = 1 \\ c^2 + d^2 = 1 & \Longleftrightarrow A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \text{ is orthogonal} \Longleftrightarrow b^2 + d^2 = 1 \\ ab + cd = 0 & \hspace{5.7cm} ac + bd = 0 \\ \end{array}$

This basically follows from the fact that a $n \times n$ matrix with real entries $A$ ,
$A^T A = I \Longleftrightarrow \left \langle Au , Av \right \rangle = \left \langle u , v \right \rangle \text{ for all } u,v \in \mathbb{R}^n$ .
Here $\left \langle u , v \right \rangle = u^T v$ denotes the standard inner product of two (column) vectors $u = \begin{pmatrix}u_1 \\ \vdots \\ u_n \\ \end{pmatrix}, v = \begin{pmatrix}v_1 \\ \vdots \\ v_n \\ \end{pmatrix} \in \mathbb{R}^n$ . It is closely related to the dot product;
$\left \langle u , v \right \rangle = u^T v = \begin{pmatrix}u_1 & \cdots & u_n \\ \end{pmatrix} \begin{pmatrix}v_1 \\ \vdots \\ v_n \\ \end{pmatrix} = u_1 v_1 + \ldots + u_n v_n = (u_1, \ldots, u_n) \cdot (v_1, \ldots, v_n).$
The linearity of the inner product means that we only need to show that
$A^T A = I \Longleftrightarrow \left \langle Ae_i , Ae_j \right \rangle = \left \langle e_i , e_j \right \rangle \text{ for all } 1 \leq i,j \leq n,$
where $\{e_i : 1 \leq i \leq n\}$ is some basis for $\mathbb{R}^n$ . We can clearly choose $e_1 = \begin{pmatrix}1 \\ 0 \\ \vdots \\ 0 \\ \end{pmatrix}, \ldots, e_n = \begin{pmatrix}0 \\ \vdots \\ 0 \\ 1 \\ \end{pmatrix}$ , so that $\left \langle e_i , e_j \right \rangle = \delta_{ij}$ , where $\delta_{ij}$ is the Kronecker delta. We can then prove the result:

If $A^T A = I$ , then
$\left \langle Ae_i , Ae_j \right \rangle = (Ae_i)^T(Ae_j) = e_i^T A^T A e_j = e_i^T e_j = \left \langle e_i , e_j \right \rangle.$

Conversely, if $\left \langle Ae_i , Ae_j \right \rangle = \left \langle e_i , e_j \right \rangle$ , then
$\delta_{ij} = e_i^T (A^T A) e_j,$
which is the entry of the $i$ -th row and $j$ -th column of $A^T A$ , which implies that $A^T A = I$ .

So the original question can be restated as:
If $\delta_{ij} = e_i^T (A^T A) e_j$ , show that $\delta_{ij} = e_i^T (A A^T) e_j$ , and the result holds because both expressions are equivalent to $A$ being an orthogonal matrix. The result can clearly also be generalised to the $n \times n$ dimensional case, though of course it can't be written down so neatly.

There's also a similar problem for complex matrices; instead of considering orthogonal matrices, we consider unitary matrices. A matrix with complex entries is unitary if $A A^{*} = A^{*} A = I$ ; here $A^{*}$ denotes the conjugate transpose of $A$ , obtained by taking the transpose of $A$ and conjugating each entry.

Thus the complex version of question 1. is

$\begin{array}{lll} |a|^2 + |b|^2 = 1 & \hspace{5.2cm} |a|^2 + |c|^2 = 1 \\ |c|^2 + |d|^2 = 1 & \Longleftrightarrow A = \begin{pmatrix}a & b \\ c & d \\ \end{pmatrix} \text{ is unitary} \Longleftrightarrow |b|^2 + |d|^2 = 1 \\ a\overline{b} + c\overline{d} = \overline{a}b + \overline{c}d = 0 & \hspace{5.2cm} a\overline{c} + b\overline{d} = \overline{a}c + \overline{b}d= 0 \\ \end{array}$

(Can you tell I'm procrastinating from studying algebraic topology? :P)

Thanks humph! The first method is awesome but don't really understand the 2nd one lolz.

Thanks again :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 01:21:35 pm

Quote from: /0 on November 13, 2009, 03:28:42 am

For 1. a good formula is

$(a^2+b^2)(c^2+d^2) = (ac+bd)^2+(ad-bc)^2$

But I'm not sure how to apply it yet

For 2.
The rational root theorem says a factor is $(n+1)$ , then after complete factorisation I think grouping factor pairs might work

Thanks I was figuring around with that formula for 1, getting close now.

Ahh rational root theorem, let me try fiddle around that now.

Thanks /0 :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 02:30:40 pm

Quote from: TrueTears on November 12, 2009, 09:11:30 pm

Quote from: TrueTears on November 12, 2009, 06:41:56 pm
Amazing Ahmad, thank you so much!

Also:

1. Prove that for $n \in N$

$\left \lfloor \sqrt{n} + \sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{4n+2} \right \rfloor$

Where do I start with this?

2. The function $f : Z \to Z$ satisfies $f(n) = n - 3$ if $n \ge 1000$ and $f(n) = f(f(n+5))$ if $n<1000$ . Find $f(84)$
Just random brainstorming for Q 1...

Consider first the perfect squares: $\sqrt{1}, \sqrt{4}, \sqrt{9}, \sqrt{16}, \sqrt{25}...$ $= 1, 2, 3, 4, 5...$

It can be seen that the largest difference between the perfect squares is 1.

Thus for $\sqrt{n}$ and $\sqrt{n+1}$ the difference between them will be smaller than 1.

This implies that $\left \lfloor \sqrt{n} \right \rfloor = \left \lfloor \sqrt{n+1} \right \rfloor$ IFF n+1 is NOT a perfect square.

$RHS = \left \lfloor \sqrt{4(n+\frac{1}{2})} \right \rfloor = \left \lfloor 2\sqrt{(n+\frac{1}{2})} \right \rfloor = \left \lfloor \sqrt{(n+\frac{1}{2})} + \sqrt{(n+\frac{1}{2})} \right \rfloor$

kk I'm leaving this question for now, coming back to it later... brain just can't break through it rofl

Taking a different approach today:

First ignore the floor function.

Consider $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$

After plugging in some numbers we see $\sqrt{n} + \sqrt{n+1} < \sqrt{4n+2}$

But let's prove it anyway:

Squaring both sides: $(\sqrt{n} + \sqrt{n+1})^2 < 4n+2$

$n + 2\sqrt{n(n+1)} + n + 1 < 4n+2$

$2\sqrt{n(n+1)} < 2n+1$

$4n(n+1) < (2n+1)^2$

$4n^2+4n < (4n^2 + 4n + 1)$

Which is true.

Now if we can prove that there exists no integer q such that $\sqrt{n} + \sqrt{n+1} < q < \sqrt{4n+2}$

then that means the integer parts of $\sqrt{n} + \sqrt{n+1}$ and $\sqrt{4n+2}$ are equal which implies that their floor function is also equal.

THANK YOU AHMAD FOR HELP! I GOT IT NOW!

Squaring both sides yields $(\sqrt{n} + \sqrt{n+1})^2 < q^2 < 4n+2$

$2n+1+2\sqrt{n^2+n} < q^2 < 4n+2$

$2\sqrt{n^2+n} < q^2-2n-1 < 2n+1$

$4n^2+4n < (q^2-2n-1)^2 < 4n^2+4n+1$

This is impossible since there are no integers which exists between 2 consecutive numbers! Thus no q exists!

$\therefore \forall n \in \mathbb{N}$ $\left \lfloor \sqrt{n} + \sqrt{n+1} \right \rfloor = \left \lfloor \sqrt{4n+2} \right \rfloor$

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 03:03:13 pm

nice solution :)
although the proving the inequality bit needs to be fixed up. If A implies something that is true, that doesn't neccesarily mean A is true.

I proved that inequality when I did my solution like this:

since the difference between consequtive squares increases, then that means that the difference between consecutive square roots decreases. So:
$ \sqrt{n+0.5}-\sqrt{n}>\sqrt{n+1}-\sqrt{n+0.5}$ (1)

since n,n+0.5,n+1 are in arithmetic progression this is true and the result follows:
$ 2\sqrt{n+0.5}>\sqrt{n}+\sqrt{n+1}$
$\sqrt{4n+2}>\sqrt{n}+\sqrt{n+1}$

yeah in my solution I didn't explicitly use that inequality but I used some consequences of (1)

nice work :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 03:15:02 pm

Thanks kamil ^.^

Quote

2. Find all integer solutions $(n,m)$ to $n^4+2n^3+2n^2+2n+1=m^2$

I think I got this question now.

So using rational root theorem n+1 is a factor.

Thus the polynomial can be factorised into $(n+1)^2(n^2+1) = m^2$

Now $n^2+1 = \left(\frac{m}{n+1}\right)^2$

Which means $n^2 + 1$ is a perfect square.

Let $q^2 = n^2 + 1$ where $n,q \in \mathbb{N}$

$q^2 - n^2 = 1$

$(q-n)(q+n) = 1$

since q and n are integers:

$q-n = 1$ and $q+n = 1$

or $q-n = -1$ and $q+n = -1$

Thus $n = 0$ and $q = 1$

Subbing this back in yields $1 = \left(\frac{m}{1}\right)^2$

Thus $m = \pm 1$ and $n = 0$

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 03:20:24 pm

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 06:24:31 pm

Geometric interpretation of Q 1: (Thanks for hints Ahmad xD)

Consider the 2 vectors $ai+bj$ and $ci+dj$ .

(http://img297.imageshack.us/img297/5408/asdft.jpg)

These 2 vectors are unit vectors since they both have a magnitude of 1.

$(ai+bj).(ci+dj) = ac+bd = 0$

Thus these 2 vectors are perpendicular to each other at the origin.

Now since they have the same magnitude, lets rotate the vector $ci+dj$ 90 deg anticlockwise.

Thus the coordinate $(c,d)$ goes to $(-d,c)$ .

This maps exactly onto $(a,b)$ which means $(a,b) = (-d,c)$

so $a = -d$ and $b = c$

Subbing this into each of the 3 equations given gives each of the other 3 results.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 08:42:29 pm

Another question, where do you even start with it?

1. How many of the first 1000 positive integers can be expressed in the form:

$\left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor$ ?

I've never been good with these "how many" questions, how do you approach these questions? Do you just simply list every possibility or...?

2. Determine the triples of integers $(x,y,z)$ satisfying the equation $x^3 + y^3 + z^3 = (x + y + z)^3$

How to factorise this elegantly? Do I have to expand the RHS...?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 08:48:06 pm

ahh combinatorics, one of my favourite problem solving topics :)

It ussually requires counting things in clever ways with patterns etc.

Unfortunately i don't have enough time atm for these questions, but I'll give you a hint, notice that the LCM of (2,4,6,8) is 24.

so when x varies from 0 to 3/24=1/8, all the terms are 0 except the last, then as you increase, to 4/24,5/24,6/24... etc. You may notice that the the value of the each individual term increases by 1 or stays constant, so theoretically if you keep going like this you will not miss any numbers , now try to look for some patterns.

e.g: the [8x] term has a period of 3 lots of 1/24, while the [6x] term has a period of 4 lots of 1/24. by period I mean how much the x has to increase to make the term increase by 1.

Title: Re: TT's Maths Thread
Post by: Over9000 on November 13, 2009, 09:40:37 pm

Sorry to hijack ur thread TT but does anyone know which books are good for beginners to number theory and stuff not too dissimilar to whats in this thread. Ive heard of the art of problem solving series of books, can anyone suggest any other books that might be helpful for a beginner like me, not too advanced.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 13, 2009, 09:46:56 pm

Art and Craft of Problem Solving

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 09:49:35 pm

Thanks for the hint kamil I think I got it now.

So let's just check $p \in [1, 24]$ and $q = 24$ where $x = \frac{p}{q}$

$\frac{1}{24}, \frac{2}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 0$

$\frac{3}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 1$

$\frac{4}{24}, \frac{5}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 2$

$\frac{6}{24}, \frac{7}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 4$

$\frac{8}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 5$

$\frac{9}{24}, \frac{10}{24}, \frac{11}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 6$

$\frac{12}{24}, \frac{13}{24}, \frac{14}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 10$

$\frac{15}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 11$

$\frac{16}{24},\frac{17}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 12$

$\frac{18}{24}, \frac{19}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 14$

$\frac{20}{24}\to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 15$

$\frac{21}{24}, \frac{22}{24}, \frac{23}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 16$

$\frac{24}{24} \to \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor = 20$

So up until the first 20 integers we recorded 12 possible results.

Thus for the first 1000 integers we require $12 \times 50 = 600$ results.

Title: Re: TT's Maths Thread
Post by: Over9000 on November 13, 2009, 10:18:15 pm

Quote from: Ahmad on November 13, 2009, 09:46:56 pm

Art and Craft of Problem Solving

Any others that could be helpful for beginners to this stuff?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 10:34:33 pm

yeah TT that's my initial idea, it's the right track, however I thought about how to simplify it and here is a method.

So expanding on that idea of an increment of 1/24, let x=a/24:

y=[a/12]+[a/6]+[a/4]+[a/3]

now we want to answer this question: when a goes from a=k to a=k+1, does y increase or stay constant?

for which values of k does this happen? if a goes to k+1 and k+1 is a multiple of 3 then the last term will increase, and some others may to. if k goes to k+1 and k+1 is a multiple of 4 then the second term will increase and others may or may not. So really the only numbers where we will not increase if if k goes to some number that is either a multiply of 3 nor a multiply of 4. All such numbers are, congruent to modulo 12:

3,4,6,8,9,12.

hence we will only notice a change 6 times when a increases by 12. hence only 12 times when a increases by 24. and TT's last line follows :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 10:49:15 pm

Quote from: Over9000 on November 13, 2009, 10:18:15 pm

Quote from: Ahmad on November 13, 2009, 09:46:56 pm
Art and Craft of Problem Solving
Any others that could be helpful for beginners to this stuff?

Once you become a UOM student, my young paladin, you will have access to a library rich with a plethora of such books that will further entertain and nourish these intellectual thoughts. If you want specific number theory problems there are some nice newby books there too that i started with. And once you lvl up to mathematical pr0ness you can even read books by famous mathematicians on a research lvl :)

Title: Re: TT's Maths Thread
Post by: Over9000 on November 13, 2009, 10:51:00 pm

Quote from: kamil9876 on November 13, 2009, 10:49:15 pm

Quote from: Over9000 on November 13, 2009, 10:18:15 pm
Quote from: Ahmad on November 13, 2009, 09:46:56 pm
Art and Craft of Problem Solving
Any others that could be helpful for beginners to this stuff?

Once you become a UOM student, my young paladin, you will have access to a library rich with a plethora of such books that will further entertain and nourish these intellectual thoughts. If you want specific number theory problems there are some nice newby books there too that i started with. And once you lvl up to mathematical pr0ness you can even read books by famous mathematicians on a research lvl :)

Thx master. I will use the force to guide me to enlightenment in maths.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 10:52:07 pm

Quote from: kamil9876 on November 13, 2009, 10:34:33 pm

yeah TT that's my initial idea, it's the right track, however I thought about how to simplify it and here is a method.

So expanding on that idea of an increment of 1/24, let x=a/24:

y=[a/12]+[a/6]+[a/4]+[a/3]

now we want to answer this question: when a goes from a=k to a=k+1, does y increase or stay constant?

for which values of k does this happen? if a goes to k+1 and k+1 is a multiple of 3 then the last term will increase, and some others may to. if k goes to k+1 and k+1 is a multiple of 4 then the second term will increase and others may or may not. So really the only numbers where we will not increase if if k goes to some number that is either a multiply of 3 nor a multiply of 4. All such numbers are, congruent to modulo 12:

3,4,6,8,9,12.

hence we will only notice a change 6 times when a increases by 12. hence only 12 times when a increases by 24. and TT's last line follows :)

Awesome method.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 13, 2009, 11:34:17 pm

Wow I think I got Q 2. (Please do check for me kamil ;P and btw kamil thanks for your "freshman's dream" :laugh: )

$x^3+y^3+z^3 - (x+y+z)^3 = 0$

Let $q = y+z$

$x^3+y^3+z^3 - (x+q)^3 = 0$

$x^3+y^3+z^3-[x^3+q^3+3xq(x+q)] = 0$

$x^3+y^3+z^3-x^3-q^3-3xq(x+q) = 0$

$y^3 + z^3 - (z+y)^3 - 3xq(x+q) = 0$

$y^3 + z^3 - [z^3+y^3+3zy(z+y)] - 3x(z+y)(x+z+y) = 0$

$y^3 + z^3 - z^3 - y^3 -3zy(z+y) - 3x(z+y)(x+y+z) = 0$

$(y+z)(-3zy-3x(x+y+z)) = 0$

$-3(y+z)(zy+x(x+y+z)) = 0$

$-3(y+z)(zy+x^2+xy+xz) = 0$

$-3(y+z)[(y(z+x)+x(x+z)] = 0$

$(y+z)(x+z)(y+x) = 0$

Thus $y = -z$ , $x = -z$ and $y = -x$

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 13, 2009, 11:41:52 pm

nice one. You're going to make a nice freshman :)

Title: Re: TT's Maths Thread
Post by: humph on November 14, 2009, 12:45:48 am

Quote from: TrueTears on November 13, 2009, 01:11:55 pm

Thanks humph! The first method is awesome but don't really understand the 2nd one lolz.

Thanks again :)

Give it a year, it's all stuff you cover in linear algebra in first year uni maths ;)

Quote from: kamil9876 on November 13, 2009, 10:49:15 pm

Quote from: Over9000 on November 13, 2009, 10:18:15 pm
Quote from: Ahmad on November 13, 2009, 09:46:56 pm
Art and Craft of Problem Solving
Any others that could be helpful for beginners to this stuff?

Once you become a UOM student, my young paladin, you will have access to a library rich with a plethora of such books that will further entertain and nourish these intellectual thoughts. If you want specific number theory problems there are some nice newby books there too that i started with. And once you lvl up to mathematical pr0ness you can even read books by famous mathematicians on a research lvl :)

UoM maths library is rubbish! They only let you borrow books for a month :( And the library's not even that big!

I use Tom Apostol's Introduction to Analytic Number Theory for my basic number theory needs, but it's probably a bit too advanced - it's written for a third year undergrad. I seem to remember William Stein's notes on Elementary Number Theory (for a course he taught at Harvard) being pretty decent, so you could always search for them.

Interestingly, I'm doing an Advanced Studies Course on the elementary proof of the Prime Number Theorem this semester - it's quite difficult though because the paper which has the proof that I'm covering is written in French :o

Quote from: TrueTears on November 13, 2009, 11:34:17 pm

Wow I think I got Q 2. (Please do check for me kamil ;P and btw kamil thanks for your "freshman's dream" :laugh: )

$x^3+y^3+z^3 - (x+y+z)^3 = 0$

Let $q = y+z$

$x^3+y^3+z^3 - (x+q)^3 = 0$

$x^3+y^3+z^3-[x^3+q^3+3xq(x+q)] = 0$

$x^3+y^3+z^3-x^3-q^3-3xq(x+q) = 0$

$y^3 + z^3 - (z+y)^3 - 3xq(x+q) = 0$

$y^3 + z^3 - [z^3+y^3+3zy(z+y)] - 3x(z+y)(x+z+y) = 0$

$y^3 + z^3 - z^3 - y^3 -3zy(z+y) - 3x(z+y)(x+y+z) = 0$

$(y+z)(-3zy-3x(x+y+z)) = 0$

$-3(y+z)(zy+x(x+y+z)) = 0$

$-3(y+z)(zy+x^2+xy+xz) = 0$

$-3(y+z)[(y(z+x)+x(x+z)] = 0$

$(y+z)(x+z)(y+x) = 0$

Thus $y = -z$ , $x = -z$ and $y = -x$

You could just hit that with its trinomial expansion, it would save you all that substitution.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 14, 2009, 01:09:35 am

uom math library is small yeah, but still fun. I often find that I have to go from one library to another to find all the books I want.

Still a massive improvement from high school so I'm still impressed :P plus it's a good way to spend time and make the most of uni i realised :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 01:41:32 am

Quote from: humph on November 14, 2009, 12:45:48 am

You could just hit that with its trinomial expansion, it would save you all that substitution.

Thanks for that! I'll certainly remember it now :)

Title: Re: TT's Maths Thread
Post by: humph on November 14, 2009, 02:16:57 am

Quote from: kamil9876 on November 14, 2009, 01:09:35 am

uom math library is small yeah, but still fun. I often find that I have to go from one library to another to find all the books I want.

Still a massive improvement from high school so I'm still impressed :P plus it's a good way to spend time and make the most of uni i realised :)

I just download heaps of maths ebooks ;) Think I have about 3gb of them on my computer. It's easier to search through them for help than to go to a library and look through books there.

Mind you, I much prefer reading books irl than reading pdfs or djvus on a computer.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 02:19:04 am

Quote from: humph on November 14, 2009, 02:16:57 am

Quote from: kamil9876 on November 14, 2009, 01:09:35 am
uom math library is small yeah, but still fun. I often find that I have to go from one library to another to find all the books I want.

Still a massive improvement from high school so I'm still impressed :P plus it's a good way to spend time and make the most of uni i realised :)
I just download heaps of maths ebooks ;) Think I have about 3gb of them on my computer. It's easier to search through them for help than to go to a library and look through books there.

Mind you, I much prefer reading books irl than reading pdfs or djvus on a computer.

Yeah reading books on computer hurts my eyes, I find reading on paper better and plus you can jot down stuff next to important theorems etc lol

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 03:18:57 am

Can someone explain this proof by contradiction? My book doesn't explain it that well, thanks.

Show that $b^2+b+1=a^2$ has no positive integer solution.

Solution: We wish to show that the equality can not be true. So assume that it is. If $b^2+b+1=a^2$ then $b<a$ and $a^2-b^2=b+1.$

Thus $(a-b)(a+b) = b+1$

Since $a>b \ge 1$ , we must have $a-b \ge 1$ and $a+b \ge 2+b$ .

Thus LHS becomes $1 \times (b+2)$ which is larger than $b+1$ .

Thus the equation is false.

1. What does it mean when it says "If $b^2+b+1=a^2$ then $b<a$ "? Why is $b<a$ ??

2. Where did the $a+b \ge 2+b$ come from?

Thanks.

Also how would you do this Q?

Prove by contradiction: If $a, b, c, d, e$ are real numbers such that the equation $ax^2+(c+d)x+(e+d) = 0$ has real roots greater than 1, show that the equation $ax^4+bx^3+cx^2+dx+e = 0$ has at least one real root.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 14, 2009, 12:01:16 pm

$b+1>0$ (ie positive integers), so if you add a positive integer to a positive integer, it increases, hence $a>b$ .

2.) add $2b$ to both sides of the $a-b\geq 1$ :
$ a+b \geq 1+2b$

$a+b \geq 1+b+b \geq 1+b+1$

$\implies a+b \geq 2+b$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 01:55:31 pm

Quote from: kamil9876 on November 14, 2009, 12:01:16 pm

$b+1>0$ (ie positive integers), so if you add a positive integer to a positive integer, it increases, hence $a>b$ .

2.) add $2b$ to both sides of the $a-b\geq 1$ :
$ a+b \geq 1+2b$

$a+b \geq 1+b+b \geq 1+b+1$

$\implies a+b \geq 2+b$

Thanks for the reply kamil. But I still don't quite get it.

So looking at $b^2 + b + 1 = a^2$

Let $q = b^2 + b$

so $q + 1 = a^2$

This means $a^2 > q$

How does this imply that $a$ itself is larger than $b$ ?

Why can't you also consider $(a+b) > 1$ ? Why did the book single out $(a-b)$ instead of the other factor?

Also I don't get your working, you added $2b$ to both sides then this step confused me, " $a+b \geq 1+b+b \geq 1+b+1$ " What did you do here?

Thanks.

DW Thanks Ahmad/dcc :P

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 09:21:01 pm

Find a formula for $\sum_{i=1}^n i^2$ ie, the sum of the first n squares.

(Hint: try telescoping)

Thanks.

Title: Re: TT's Maths Thread
Post by: dcc on November 14, 2009, 09:50:43 pm

Quote from: TrueTears on November 14, 2009, 09:21:01 pm

Find a formula for $\sum_{i=1}^n i^2$ ie, the sum of the first n squares.

(Hint: try telescoping

Consider:

$\begin{align*} \mbox{Let }A = \sum_{k=1}^n k^3 &= \sum_{k=1}^n (k + 1)^3 - (3k^2 + 3k + 1) &\mbox{ check this - it's the same thing}\\ &= \left(\sum_{k=2}^{n+1} k^3\right) - \left( \sum_{k=1}^n 3k^2 + 3k + 1\right) &\mbox{ rewriting index of summation}\\ &= \left(\left(\sum_{k=1}^{n} k^3\right) - 1 + \left(n + 1\right)^3\right) - \left( \sum_{k=1}^n 3k^2 + 3k + 1\right)\\ \implies A &= \left(A - 1 + \left(n + 1\right)^3\right) - \left( \sum_{k=1}^n 3k^2 + 3k + 1\right)\\ \implies \sum_{k=1}^n 3k^2 &= \left(n+1\right)^3 - 1 - \sum_{k=1}^n \left(3k + 1\right)\\ &= \left(n + 1\right)^3 - 1 - \left(3 \sum_{k=1}^n k\right) - n\\ &= \left(n + 1\right)^3 - 1 - \frac{3}{2}n\left(n + 1\right) - n & \mbox{ arithmetic series}\\ \implies \sum_{k=1}^n k^2 &= \frac{1}{3} \left( (n + 1)^3 - 1 - \frac{3}{2}n(n+ 1) - n\right) = \frac{1}{6}n(n+1)(2n+1) & \mbox{ trust me!}\\ \end{align*} $

What a shit way to do it. I'd just use induction, personally.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 09:57:48 pm

Thanks a lot dcc.

Can you show me induction method?

Title: Re: TT's Maths Thread
Post by: Ahmad on November 14, 2009, 10:05:05 pm

Let $1 \le x, y, z \le n$ be integers with z being the largest (but not strictly) of the 3, how many such tuples (x,y,z) are there which satisfy this condition?

Well if z = 1 then there's only 1 choice for x, notably x = 1 and similarly 1 choice for y, notably y = 1, giving 1^2 possibilities. If z = 2, then there are 2 choices for x and 2 choices for y, giving 2^2 possibilities. Continuing in this manner, in total there are $1^2 + 2^2 + \cdots + n^2$ such tuples.

But we can also count this another way by dividing this into cases.

x = y = z: C(n, 1) = n ways
x = y < z: We can pick any 2 integers from 1 to n and the larger we assign to z, the smaller we assign to x = y. C(n,2) possibilities here.
x < y = z or y < x = z: C(n, 2) ways each, similar reasoning to above
x < y < z or y < x < z: We can pick any 3 integers from 1 to n, and assign them to x, y, z based on their size. C(n, 3) possibilities each here.

Hence we've counted the same quantity in 2 ways, so $1^2 + 2^2 + \cdots + n^2 = \binom{n}{1} + 3 \binom{n}{2} + 2 \binom{n}{3} = \frac{n(n+\frac{1}{2})(n+1)}{3}$

Title: Re: TT's Maths Thread
Post by: dcc on November 14, 2009, 10:06:12 pm

By popular request, I'll also show the induction proof (which assumes for some reason that you know the answer!)

Let $P(n)$ be the statement that $\sum_{k=1}^n k^2 = \frac{1}{6}n\left(n+1\right)(2n+1)$ .

Base Case:

Consider $P(1)$ . $\sum_{k=1}^1 k^2 = 1$ , $\frac{1}{6}\cdot 1 \cdot 2 \cdot 3 = 1$ , hence $P(1)$ is true.

Inductive Step:

Assume $P(m)$ is true for some positive integer $m$ . Therefore we know that:

$\sum_{k=1}^m k^2 = \frac{1}{6}m\left(m+1\right)\left(2m+1\right)$

Adding $\left(m + 1\right)^2$ to each side of this expression, we find:

$\begin{align*} \sum_{k=1}^m k^2 + \left(m + 1\right)^2 &= \frac{1}{6}m\left(m + 1\right)\left(2m+1\right) + \left(m + 1\right)^2\\ \sum_{k=1}^{m+1} k^2 &= \left(m + 1\right) \left( \frac{1}{6}m\left(2m + 1\right) + \left(m + 1\right) \right)\\ &= \frac{1}{6} \left(m+1\right) \left(m(2m + 1) + 6(m + 1)\right)\\ &= \frac{1}{6} \left(m + 1\right) \left(2m^2 + 7m + 6\right)\\ &= \frac{1}{6} \left(m + 1\right) \left(m + 2\right)\left(2m + 3\right)\\ &= \frac{1}{6} \left(m + 1\right) \left(\left(m + 1\right) + 1\right)\left(2(m + 1) + 1\right)\\ \end{align*}$

This is precisely the expression we were hoping to attain to show $P(m+1)$ . Hence $P(m) \implies P(m + 1)$ .

Therefore by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 10:16:12 pm

Thanks so much! That really clears it all up.

Just started on induction and summations today, so I wasn't really sure how to approach these questions. That really helped!

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 14, 2009, 10:25:53 pm

lol i remember running into a method that's similair to ahmad's but more complicated when realising that a marble counting problem can be solved in two different ways. http://vcenotes.com/forum/index.php/topic,13271.msg171784.html#msg171784

Title: Re: TT's Maths Thread
Post by: TrueTears on November 14, 2009, 11:59:09 pm

An exercise dcc gave me: Find a formula for $\sum_{k=1}^n k^3$

There are probs plenty of better methods but this is the one I've learnt so far...

First remember the result $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Now to find $\sum_{k=1}^n k^3$ we need to find some polynomial containing $k^3$ .

So let $u_{k+1} - u_k = \mbox{some polynomial with} \ k^3$

Now let $u_k = k^4$

Thus $u_{k+1} - u_k = (k+1)^4 - k^4$

Now $\sum_{k=1}^n u_{k+1} - u_k = u_2 - u_1 + u_3 - u_2 + ... + u_{n+1} - u_n = u_{n+1} - u_1 = (n+1)^4 - 1$

But $\sum_{k=1}^n (k+1)^4 - k^4 = \sum_{k=1}^n (4k^3 + 6k^2 + 4k+1)$

$\implies \sum_{k=1}^n 4k^3 + 6k^2 + 4k+1 = (n+1)^4 - 1$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - 6\sum_{k=1}^n k^2 - \sum_{k=1}^n (4k+1)$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - n(n+1)(2n+1) - \frac{n}{2}(5+4n+1)$

$\therefore \sum_{k=1}^n k^3 = \frac{n^2}{4}\left(n^2+2n+1\right)$

I can see using this method to find something like $\sum_{k=1}^n k^{100}$ would be super tedious haha

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 02:55:32 am

Alright another Q, I think the book *might* be wrong with this question...

The Zeta function $Z(s)$ is defined by the infinite series $Z(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + ...$

When $s = 1$ , this becomes the harmonic series and diverges. Show that $Z(s)$ converges for all $s \ge 2$

My working:

If we can show that for $s = 2$ then $Z(s)$ converges then for $s > 2$ will converge as well.

The general sum for $s = 2$ is $\sum_{k=1}^n \frac{1}{k^2}$

So we need to prove that $\sum_{k=1}^n \frac{1}{k^2}$ converges.

Now using telescoping we find $\sum_{k=1}^n \frac{1}{k^2+k} = 1 - \frac{1}{n+1}$ (The proof is fairly easy)

However we need to find something similar to $\frac{1}{k^2-k}$ instead of $\frac{1}{k^2+k}$ since using comparison, if we compare $\sum_{k=1}^n \frac{1}{k^2}$ to a larger converging series, that means $\sum_{k=1}^n \frac{1}{k^2}$ is converging. This "larger series" we are looking for should be of the form $\sum_{k=a}^b \frac{1}{k^2-k}$ with some limits $a$ or $b$ .

Now what are these limits? Using change of index we find that $\sum_{k=1}^n \frac{1}{k^2+k} = \sum_{k=2}^{n+1} \frac{1}{k^2-k}$

This is where I think the first mistake is because book has $\sum_{k=1}^n \frac{1}{k^2+k} = \frac{1}{2} + \sum_{k=2}^{n+1} \frac{1}{k^2-k}$ . Where did $\frac{1}{2}$ come from? Mistake?

Now since $\frac{1}{k^2} < \frac{1}{k^2-k}$

Taking the summation of both sides from 1 to n yields:

$\sum_{k=1}^n \frac{1}{k^2} < \sum_{k=1}^n \frac{1}{k^2-k}$

But this will leave the RHS undefined so WLOG we require:

$\sum_{k=2}^n \frac{1}{k^2} < \sum_{k=2}^n \frac{1}{k^2-k}$

Now we need to find a formula for the RHS.

We know that $\sum_{k=2}^{n+1} \frac{1}{k^2-k} = \sum_{k=1}^n \frac{1}{k^2+k} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + ... + \frac{1}{n^2-n} + \frac{1}{n^2+n}$

And we also know $\sum_{k=2}^n \frac{1}{k^2-k} = \frac{1}{2} + \frac{1}{6} + \frac{1}{12} + ... + \frac{1}{n^2-n}$

$\implies \sum_{k=2}^n \frac{1}{k^2-k} = \sum_{k=1}^{n} \left(\frac{1}{k^2+k}\right) - \frac{1}{n^2+n}$

$\implies \sum_{k=2}^n \frac{1}{k^2-k} = 1 - \frac{1}{n+1} - \frac{1}{n^2+n}$

$\implies \sum_{k=2}^n \frac{1}{k^2-k} = 1 - \frac{1}{n}$

$\therefore \sum_{k=2}^n \frac{1}{k^2} < 1 - \frac{1}{n}$ Which means it is converging.

However book's last line is $\therefore \sum_{k=2}^n \frac{1}{k^2} < 1 - \frac{1}{n+1}$

Where did the $n+1$ on the denominator come from?

Thanks.

Last few questions for the night.

1. Find a formula for the sum $\sum_{k=1}^n \frac{k}{(k+1)!}$

2. Find the sum of $\sum_{k=2}^n \left(\frac{1}{\log_k(u)}\right)$

Many thanks!

Title: Re: TT's Maths Thread
Post by: dcc on November 15, 2009, 10:25:27 am

2. Evaluate $\sum_{k=2}^n \left(\frac{1}{\log_{k}(u)}\right)$

Let $S(u) = \sum_{k=2}^n \left(\frac{1}{\log_{k}(u)}\right) = \sum_{k=2}^n \left( \frac{1}{\frac{\log(u)}{\log(k)}} \right)$

Hence $S(u) = \frac{1}{\log(u)} \left( \sum_{k=2}^n \log(k) \right) = \frac{\log(n!)}{\log(u)} = \log_u(n!)$

By our logarithm laws.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 03:02:05 pm

Quote from: dcc on November 15, 2009, 10:25:27 am

2. Evaluate $\sum_{k=2}^n \left(\frac{1}{\log_{k}(u)}\right)$

Let $S(u) = \sum_{k=2}^n \left(\frac{1}{\log_{k}(u)}\right) = \sum_{k=2}^n \left( \frac{1}{\frac{\log(u)}{\log(k)}} \right)$

Hence $S(u) = \frac{1}{\log(u)} \left( \sum_{k=2}^n \log(k) \right) = \frac{\log(n!)}{\log(u)} = \log_u(n!)$

By our logarithm laws.

Thanks dcc, so elegant!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 03:40:02 pm

Quote

1. Find a formula for the sum $\sum_{k=1}^n \frac{k}{(k+1)!}$

I've tried telescoping this one a bit and I've got to:

Let $f(x) = \frac{k}{(k+1)!}$

Thus we need to find some expression $u_{k+1} - u_{k} = f(x)$

Let $u_k = \frac{1}{k!}$

$\implies u_{k+1} = \frac{1}{(k+1)!}$

$\therefore \frac{1}{(k+1)!} - \frac{1}{k!} = \frac{-k}{(k+1)!}$

Which is close but there is a negative on the $k$ .

So why not try $u_k - u_{k+1}$ ? This leads to:

$\frac{1}{k!} - \frac{1}{(k+1)!} = \frac{k}{(k+1)!}$ which is what we want.

Thus $\sum_{k=1}^n f(x) = \sum_{k=1}^n \frac{1}{k!} - \sum_{k=1}^n \frac{1}{(k+1)!}$

And then... I'm stuck, how do you go about evaluating those 2 sums?

Basically I think if I can get the first sum then the second one will follow, since it is just $\sum_{k=1}^n \frac{1}{(k+1)k!}$ , but how to get the first one lol

It looks awfully similar to Taylor's series for $\textrm{e}^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}$ ...

But it is just an approximation for $e^x$ , so can I use it?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2009, 03:53:09 pm

You did all the hard work, but got confused by some notation.

So you found that. $u_{k+1} - u_k=a_k$ where the sum we're after is the sum of $a_i$ .

therefore:

$f(k)=a_1+a_2+a_3+...a_k=u_{k+1}-u_1$

edit: basically $u_k=\frac{-1}{k!}$ (another way to handle that negative you initially got :) )

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 04:02:23 pm

Quote from: kamil9876 on November 15, 2009, 03:53:09 pm

You did all the hard work, but got confused by some notation.

So you found that. $u_{k+1} - u_k=a_k$ where the sum we're after is the sum of $a_i$ .

therefore:

$f(k)=a_1+a_2+a_3+...a_k=u_{k+1}-u_1$

edit: basically $u_k=\frac{-1}{k!}$ (another way to handle that negative you initially got :) )

hahaha thanks I think I got it.

kk so let $u_k = \frac{-1}{k!}$

$\implies u_{k+1} = \frac{-1}{(k+1)!}$

Thus $u_{k+1} - u_k = \frac{k}{(k+1)!}$ which is what we want!

$\therefore \sum_{k=1}^n \frac{1}{k!} - \frac{1}{(k+1)!} = \sum_{k=1}^n u_{k+1} - u_k = u_2 - u_1 + u_3 - u_2 + ... + u_{n+1} - u_n = u_{n+1} - u_1$

So $\sum_{k=1}^n \frac{k}{(k+1)!} = u_{n+1} - u_1 = 1 - \frac{1}{(n+1)!}$

I <3 telescoping

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 04:17:52 pm

I just realised something else

Using Taylor's series:

$\textrm{e}^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}$

Let $x = 1$

Thus $e - 1 = 1 + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!}$

Which follows on from my previous working: $\sum_{k=1}^n f(x) = \sum_{k=1}^n \frac{1}{k!} - \sum_{k=1}^n \frac{1}{(k+1)!}$

Since $\sum_{k=1}^n \frac{1}{k!} = 1 + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} = e - 1$

and $\sum_{k=1}^n \frac{1}{(k+1)!} = \sum_{k=1}^n \frac{1}{(k+1)k!} = \frac{1}{2} + \frac{1}{(3)(2!)} + \frac{1}{(4)(3!)} + ... + \frac{1}{(n+1)n!} = \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ... +\frac{1}{(n+1)!} = e - 2$

$\therefore$ $\sum_{k=1}^n f(x) = \sum_{k=1}^n \frac{1}{k!} - \sum_{k=1}^n \frac{1}{(k+1)!} = e - 1 - (e-2) = 1$

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2009, 04:27:56 pm

as n approaches infinity ;) Which you found from formula in previous post

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 05:11:01 pm

Thanks kamil :)

Another one: Find a formula for $\frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + ... + \frac{1}{n \times (n+1) \times (n+2)}$

So we require $\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) = \sum_{k=1}^n \left( \frac{1}{2(k+2)} + \frac{1}{2k} - \frac{1}{k+1} \right)$

I tried telescoping this one:

$\sum_{k=1}^n \left( \frac{1}{2k} - \frac{1}{k+1} \right)$

since it looks very similar to $\sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n-1}$ but doesn't seem to work...

Any hints/help?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2009, 05:19:05 pm

$\frac{1}{2(k+2)} + \frac{1}{2k} - \frac{1}{k+1}=\frac{1}{2(k+2)}-\frac{1}{2(k+1)}+\frac{1}{2k}-\frac{1}{2(k+1)}$

$=\frac{1}{u_{k+1}}-\frac{1}{u_k} + \frac{1}{2}(\frac{1}{k} - \frac{1}{k+1})$

So you're better off splitting your sum into these to, since the way you have splitted up you may end up evaluating $\infty - \infty$ . (ie the difference between two sequences who both limit to infinity)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 05:52:53 pm

hahaha that splitting was the crux of the question, now I got it!

Let $u_k = \frac{1}{k+1}$

$\implies u_{k+1} = \frac{1}{k+2}$

We require to find $\sum_{k=1}^n \left( \frac{1}{k(k+1)(k+2)} \right) = \frac{1}{2} \sum_{k=1}^n \left(\frac{1}{k+2} - \frac{1}{k+1}\right) + \frac{1}{2}\sum_{k=1}^n \left(\frac{1}{x} - \frac{1}{x+1}\right)$

Telescoping the first sum yields:

$\frac{1}{2} \sum_{k=1}^n u_{k+1} - u_k = \frac{1}{2}(u_{n+1} - u_1) = \frac{1}{2} \left(\frac{1}{n+2} - \frac{1}{2}\right)$

Telescoping the second sum yields:

$\frac{1}{2} \sum_{k=1}^n \left(\frac{1}{x} - \frac{1}{x+1}\right) = \frac{1}{2}\left(1-\frac{1}{n+1}\right)$

$\therefore \sum_{k=1}^n \left( \frac{1}{k(k+1)(k+2)} \right) = \frac{1}{2} \left(\frac{1}{n+2} - \frac{1}{2}\right) + \frac{1}{2}\left(1-\frac{1}{n+1}\right) = \frac{1}{2(n+2)} - \frac{1}{2(n+1)} + \frac{1}{4}$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 06:55:13 pm

Here's a fun one... any idea? I'm fiddling around for now haha

For each permutation $a_1,a_2,a_3,\cdots,a_{10}$ of the integers $1,2,3,\cdots,10$ , form the sum $|a_1-a_2|+|a_3-a_4|+|a_5-a_6|+|a_7-a_8|+|a_9-a_{10}|$ to find the average value of all these sums.

Something I just came up with:

Consider a permutation like this: (Just notice the first 'term' ie, $|1-a_2|$ for each row)

$|1-2|+|3-4|+|5-6| +|7-8| +|9-10|$

$|1-3| + |2-4| + |5-6|+|7-8| +|9-10|$

$|1-4|+ |2-3| + |5-6|+|7-8| +|9-10|$

$|1-5| + |2-4| + |3-6|+|7-8| +|9-10|$

$|1-6|+ |2-4| + |5-3|+|7-8| +|9-10|$

$|1-7| + |2-4| + |5-6|+|3-8| +|9-10|$

$|1-8|+ |2-4| + |5-6|+|7-3| +|9-10|$

$|1-9| + |2-4| + |5-6|+|7-8| +|3-10|$

$|1-10| + |2-4| + |5-6|+|7-8| +|9-3|$

Now ignore the other terms, and just consider the first term of each row, now instead of adding "horizontally" why not try adding vertically?

Thus we get: $|1-2| + |1-3|+ |1-4|+ |1-5|+ |1-6|+ |1-7|+ |1-8|+ |1-9| + |1-10|$

Now if we make another set of rows starting with $2$ as $a_1$ we yield the "vertical" sum of $|2-1| + |2-3|+ |2-4|+ |2-5|+ |2-6|+ |2-7|+ |2-8|+ |2-9| + |2-10|$

Now if we can compute this sum: $|1-2| + ... + |1-10| + |2-1| + ... + |2-10| + |3-1| + ... + |3-10| + |4-1| +...+ |4-10| + |5-1| +...+ |5-10| + ......... + |10-1| + |10-9|$

Then we found the total sum! Now my problem is how to compute it...

Actually that's not right... that doesn't give the sum of all the permutations... back to drawing board.

Gonna come back to this one, can't get it.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 09:15:25 pm

Evaluate the infinite product: $\prod_{n=2}^{\infty} \left(\frac{n^3-1}{n^3+1}\right)$

So I factored it $\prod_{n=2}^{\infty} \left(\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\right)$ then I don't know how to continue.

Title: Re: TT's Maths Thread
Post by: supdawg on November 15, 2009, 09:23:13 pm

|a-b|
18 ways to equal 1
16 ways to equal 2
14 ways to equal 3
...
2 ways to equal 9

E(|a-b|) = 11/3
E(Sum) = 55/3

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 09:59:03 pm

Okay I'm not sure if this method is rigorous, seems rather primitive without some formal definitions, but anyways:

Writing the first few terms of the product out you get:

$\left(\frac{(1)(7)}{(3)(3)}\right) \left(\frac{(2)(13)}{(4)(7)}\right) \left(\frac{(3)(21)}{(5)(13)}\right) \left(\frac{(4)(31)}{(6)(21)}\right) \left(\frac{(5)(43)}{(7)(31)}\right) \left(\frac{(6)(57)}{(8)(43)}\right) ...$

A pattern can be seen with the canceling:

Say we have the fractions in the form of $\frac{(a)(b)}{(c)(d)}$

Then the cancelling looks like this: (Ignore the first two fractions of the product series, consider from $\frac{(3)(21)}{(5)(13)}$ onwards...)

To cancel out the top right number, it will correspond with the bottom right number of the following fraction:

$\left(\frac{()(b)}{()()}\right)\left(\frac{()()}{()(b)}\right)$

$\therefore$ ' $b$ ' will always cancel.

To cancel the top left number, it will always correspond with the bottom left number of 2 fractions BEFORE it.

$\left(\frac{()()}{(a)()}\right)...\left(\frac{(a)()}{()()}\right)$ where $...$ represents 1 other fraction in between.

$\therefore$ ' $a$ ' will always cancel.

Therefore since the top 2 numbers of any fraction will cancel out the bottom 2 numbers of another fraction, then ALL fractions in the product series to infinity will cancel.

Now analyzing the first two fractions we see some terms which can not be 'canceled'

$\left(\frac{(1)(7)}{(3)(3)}\right) \left(\frac{(2)(13)}{(4)(7)}\right)$

The 1 can be omitted as it does nothing.

Using the "pattern" we formed earlier, both the top right numbers of both fractions will cancel. So we are left with:

$\left(\frac{1}{(3)(3)}\right) \left(\frac{(2)}{(4)}\right)$

Now the top left number, ie the 2 in the second fraction, can not cancel as there is only 1 fraction preceding it, since we realised we need TWO fractions preceding it to cancel it.

Now both the 3 and 4 in the respective fractions can be cancelled since they are the bottom left of the each fraction which means there must be another 2 fractions FOLLOWING it which can cancel them out. Thus 3 and 4 are gone.

Thus we are only left with $\left(\frac{(1)}{(3)}\right) \left(\frac{(2)}{(1)}\right)$

$\implies \prod_{n=2}^{\infty} \left(\frac{n^3-1}{n^3+1}\right) = \frac{2}{3}$

Is there any other methods with this question? The canceling is not bad but is there any more rigorous methods of doing it?

Thanks!

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2009, 10:06:00 pm

Quote from: TrueTears on November 15, 2009, 09:15:25 pm

Evaluate the infinite product: $\prod_{n=2}^{\infty} \left(\frac{n^3-1}{n^3+1}\right)$

So I factored it $\prod_{n=2}^{\infty} \left(\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\right)$ then I don't know how to continue.

Split the product into two. Now notice the simpler product:

$\prod_{n=2}^k \frac{n-1}{n+1}$

$= \frac{1}{3} * \frac{2}{4} * \frac{3}{5} * \frac{4}{6} * \frac{5}{7}.... \frac{k-1}{k+1}$

Now notice how 3 cancels, 4 cancels, 5 cancels etc. and in general, the denominator of the ith term cancels with the numerator of the (i+2)th term, this should cancel a lot of in between factors and leave only some factors on the ends which will give a nice closed expression of which we can take the limit of.

A similair thing can be done with the second factor since it can be rewritten as:

$\frac{(n(n+1)+1)}{(n(n-1)+1)}$

which also has this recursive nature due to n-1,n,n+1, being consecutive, so same strategy can be applied.

LOL TT beat me to it :P

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 10:09:20 pm

lol yeah thanks kamil, you way is actually much neater because by splitting them up, you can see the pattern much clearer :P

Now how to apply limits and get rid of the $\mbox{n}^{\mbox{th}}$ term?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2009, 10:18:30 pm

so you can do just like in my example find the product to n=k. Find this as a function of k which you can do since heaps of denominators cancel etc. Once you do you will basically get the quotient of two polynomials, so you just take their limit to infinity.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 10:29:54 pm

Quote from: kamil9876 on November 15, 2009, 10:18:30 pm

so you can do just like in my example find the product to n=k. Find this as a function of k which you can do since heaps of denominators cancel etc. Once you do you will basically get the quotient of two polynomials, so you just take their limit to infinity.

Ahh I see and that would be more tedious I guess, which is why this pattern method works a bit better ;)

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2009, 10:42:28 pm

It actually turns out quite nicely.

the fact that in the first product, the denominator of the ith term cancels with numerator of (i+2)th term gives:
$ \frac{2}{k(k+1)}$

the fact that in the second product the numerator of the ith term cancels with the denominator of the (i+1)th term leaves only the denominator of first term, and numerator of last term hence:

$\frac{k(k+1)+1}{3}$

Hence the product of these is:
$ \frac{2}{3} * \frac{k(k+1)+1}{k(k+1)}$

$=\frac{2}{3}(1+\frac{1}{k(k+1)})$

whose limit as $k$ approaches infinity is:

$\frac{2}{3}*1=\frac{2}{3}$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 10:56:42 pm

Thanks kamil.

What exactly did you do though lol, did you just n = k, k+1, k+2 etc?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2009, 11:01:00 pm

$\prod_{n=2}^k \frac{n-1}{n+1}$

$= \frac{1}{3} * \frac{2}{4} * \frac{3}{5} * \frac{4}{6} * \frac{5}{7}....*\frac{k-3}{k-1}*\frac{k-2}{k}* \frac{k-1}{k+1}$

$\frac{2}{ } * \frac{ }{k} * \frac{ }{k+1}$

$=\frac{2}{k(k+1)}$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 11:07:02 pm

ahhh thanks for that, that's just like telescoping but for products!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2009, 11:39:12 pm

Let $f(n)$ be the integer closest to $n^{\frac{1}{4}}$ . Find $\sum_{k=1}^{1995} \frac{1}{f(k)}$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 16, 2009, 01:24:15 am

sketch:

we need to know what the values of f(n) are first of all.

$f(n)=m$ iff:

$m-\frac{1}{2}<n^{0.25}<m+\frac{1}{2}$

$(m-\frac{1}{2})^4<n<(m+\frac{1}{2})^4$

How many values of $n$ satisfy this will depend on $m$ , notice that the difference $(m+\frac{1}{2})^4-(m-\frac{1}{2})^4$ is an indication of how many values of $n$ solve this. In fact it is the number of values of n that solve this, since the difference is itself an integer(since it's a difference of two numbers with the same decimal part). After doing the messy algebra, you get an expression for the number of times $m$ appears, now you can figure out which numbers and how many of them you are exactly summing, and then see if you can try to find the sum.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2009, 02:55:11 am

Quote from: kamil9876 on November 16, 2009, 01:24:15 am

sketch:

we need to know what the values of f(n) are first of all.

$f(n)=m$ iff:

$m-\frac{1}{2}<n^{0.25}<m+\frac{1}{2}$

$(m-\frac{1}{2})^4<n<(m+\frac{1}{2})^4$

How many values of $n$ satisfy this will depend on $m$ , notice that the difference $(m+\frac{1}{2})^4-(m-\frac{1}{2})^4$ is an indication of how many values of $n$ solve this. In fact it is the number of values of n that solve this, since the difference is itself an integer(since it's a difference of two numbers with the same decimal part). After doing the messy algebra, you get an expression for the number of times $m$ appears, now you can figure out which numbers and how many of them you are exactly summing, and then see if you can try to find the sum.

Ah okay I get how you came up with the inequality, but how did you get the difference?

Title: Re: TT's Maths Thread
Post by: QuantumJG on November 16, 2009, 09:47:54 am

This stuff is much more advanced than what you will see in first year uni maths. I think you may see it if you do accelerated maths. My friends who did accelerated maths in first year were trying to solve a problem similar to one of the problems you posted.

I think I may understand what this stuff is by the end of next year (during hopefully). MUMS always post problems like that in the maths building. If you are this interested in maths you may want to pursue something like a PhB in maths at ANU or some course with a lot of maths. With commerce the only area of study with a lot of maths is actuarial studies.

The only way you will see this maths is with a pure maths degree or if you decide to undertake studies to become a theoretical physicist. Yep you are not only a lover of maths, but you love pure maths.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 16, 2009, 12:04:01 pm

It's often best to picture numbers on a number line. So we want to know how many numbers are in between a and b:

a----|--(a+1)----|--(a+2)----|--(a+3)----|--(a+4)----|--(a+5)....

we see that there are k integers in between a and a+k. Hence letting b=a+k, there are b-a integers in between b and a.

Title: Re: TT's Maths Thread
Post by: Over9000 on November 16, 2009, 12:08:44 pm

Quote from: QuantumJG on November 16, 2009, 09:47:54 am

This stuff is much more advanced than what you will see in first year uni maths. I think you may see it if you do accelerated maths. My friends who did accelerated maths in first year were trying to solve a problem similar to one of the problems you posted.

I think I may understand what this stuff is by the end of next year (during hopefully). MUMS always post problems like that in the maths building. If you are this interested in maths you may want to pursue something like a PhB in maths at ANU or some course with a lot of maths. With commerce the only area of study with a lot of maths is actuarial studies.

The only way you will see this maths is with a pure maths degree or if you decide to undertake studies to become a theoretical physicist. Yep you are not only a lover of maths, but you love pure maths.

Some of it is olympiad level stuff

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2009, 12:57:23 pm

Quote from: QuantumJG on November 16, 2009, 09:47:54 am

This stuff is much more advanced than what you will see in first year uni maths. I think you may see it if you do accelerated maths. My friends who did accelerated maths in first year were trying to solve a problem similar to one of the problems you posted.

I think I may understand what this stuff is by the end of next year (during hopefully). MUMS always post problems like that in the maths building. If you are this interested in maths you may want to pursue something like a PhB in maths at ANU or some course with a lot of maths. With commerce the only area of study with a lot of maths is actuarial studies.

The only way you will see this maths is with a pure maths degree or if you decide to undertake studies to become a theoretical physicist. Yep you are not only a lover of maths, but you love pure maths.

Yeah this stuff is pretty cool, I'm not finding it too hard, it's actually really enjoyable =)

Quote from: kamil9876 on November 16, 2009, 12:04:01 pm

It's often best to picture numbers on a number line. So we want to know how many numbers are in between a and b:

a----|--(a+1)----|--(a+2)----|--(a+3)----|--(a+4)----|--(a+5)....

we see that there are k integers in between a and a+k. Hence letting b=a+k, there are b-a integers in between b and a.

Thanks I think I got the idea now, gonna try to solve now.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 16, 2009, 01:09:30 pm

Yeah it is, I prefer those more olympiad problems that use tricks, this stuff contains a lot of algebra and infinite series so it sort of is closer to uni maths, but the previous problems we did like the combinatorial ones or number theoretical ones like

Quote

1. How many of the first 1000 positive integers can be expressed in the form:
$ \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor$ ?

are more the olympiad stuff I like and you won't find it in first year uni.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2009, 01:13:34 pm

Quote from: kamil9876 on November 16, 2009, 01:09:30 pm

Yeah it is, I prefer those more olympiad problems that use tricks, this stuff contains a lot of algebra and infinite series so it sort of is closer to uni maths, but the previous problems we did like the combinatorial ones or number theoretical ones like

Quote
1. How many of the first 1000 positive integers can be expressed in the form:
$ \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor$ ?

are more the olympiad stuff I like and you won't find it in first year uni.

Yeah these are the problems that I find really fun.

Title: Re: TT's Maths Thread
Post by: dcc on November 16, 2009, 01:23:53 pm

In Accelerated Maths (1&2) @ uom, we studied linear algebra (vector spaces / inner product spaces) / real analysis / calculus, so there isn't much relation between the types of questions posted here and the stuff you will study at uni.

Uni mathematics is set up so that you are able to tackle other subjects later on (i.e. complex analysis follows from real analysis, group theory follows from linear algebra and so on), whereas Olympiad maths doesn't really 'lead' anywhere (in terms of learning more about different topics in the future).

At least, that's my take on it.

Title: Re: TT's Maths Thread
Post by: enwiabe on November 16, 2009, 02:03:47 pm

I think Olympiad Maths teaches the much more important skills of critical thinking and problem solving.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 16, 2009, 02:12:09 pm

True, I think it is best to be selective about which problems to care about since some may be more "natural" and some more "artificial"(contrived problems requiring to find some needle that has been cleverly placed in a haystack, ie not something you would yourself ask/invent before knowing the answer). e.g the latest problem is quite shit and not natural in my opinion so I wouldn't bother with it. But other problems can be quite natural and complement more serious stuff, they're alright at stimulating some creative thought every now and then. I don't neccesarily get into the whole competitive nature of IMO, but just pick and choose whatever I think is natural to think about, stress free, and try out problems not neccearily ones you would find on an IMO but still ones that provide good ideas that may build generic skills.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2009, 02:23:15 pm

Quote from: TrueTears on November 15, 2009, 11:39:12 pm

Let $f(n)$ be the integer closest to $n^{\frac{1}{4}}$ . Find $\sum_{k=1}^{1995} \frac{1}{f(k)}$ .

Okay I think I finally got it, god this method is pretty primitive but it is the only one I can think of. There must be some other method.

So first of all we know that $\left(m-\frac{1}{2}\right)^4<n<\left(m+\frac{1}{2}\right)^4$ for some positive integer m.

Thus the number of n that satisfies this inequality is determined by: $\left(m+\frac{1}{2}\right)^4 - \left(m-\frac{1}{2}\right)^4$

Now let's list all the possibilities we have:

$f(n) = 1$

$\frac 1{2^4} < n < \frac {3^4}{2^4} \iff n\in[1,5]$

$f(n) = 2$

$\frac {3^4}{2^4} < n < \frac {5^4}{2^4} \iff n\in[6,39]$

$f(n) = 3$

$\frac {5^4}{2^4} < n < \frac {7^4}{2^4} \iff n\in[40,150]$

$f(n) = 4$

$\frac {7^4}{2^4} < n < \frac {9^4}{2^4} \iff n\in[151,410]$

$f(n) = 5$

$\frac {9^4}{2^4} < n < \frac {11^4}{2^4} \iff n\in[411,915]$

$f(n) = 6$

$\frac {11^4}{2^4} < n < \frac {13^4}{2^4} \iff n\in[916,1785]$

$f(n) = 7$

$\frac {13^4}{2^4} < n < \frac {15^4}{2^4} \iff n\in[1786,3164]$

Therefore we require $\sum_{k = 1}^{1995}\frac 1{f(k)} = \sum_{k = 1}^{5}\frac 1{f(k)} + \sum_{k = 6}^{39}\frac 1{f(k)} + \sum_{k = 40}^{150}\frac 1{f(k)} + \sum_{k = 151}^{410}\frac 1{f(k)} + \sum_{k = 411}^{915}\frac 1{f(k)} + \sum_{k = 916}^{1785}\frac 1{f(k)} + \sum_{k = 1786}^{1995}\frac 1{f(k)}$

$\implies$ $\sum_{k = 1}^{1995}\frac 1{f(k)} = 5\frac 1{1} + 34\frac 1{2} + 111\frac 1{3} + 260\frac 1{4} + 505\frac 1{5} + 870\frac 1{6} + 210\frac 1{7}$

$\therefore \sum_{k = 1}^{1995}\frac 1{f(k)} = 400$

So yeah, what if the question was like $\sum_{k = 1}^{1000000000000000000000}\frac 1{f(k)}$

I don't wanna list forever :P Need a more general method!

Title: Re: TT's Maths Thread
Post by: dcc on November 16, 2009, 02:26:52 pm

Quote from: kamil9876 on November 16, 2009, 01:09:30 pm

1. How many of the first 1000 positive integers can be expressed in the form:
$ \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor$ ?

601 ?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2009, 02:32:31 pm

Quote from: dcc on November 16, 2009, 02:26:52 pm

Quote from: kamil9876 on November 16, 2009, 01:09:30 pm
1. How many of the first 1000 positive integers can be expressed in the form:
$ \left \lfloor 2x \right \rfloor + \left \lfloor 4x \right \rfloor +\left \lfloor 6x \right \rfloor +\left \lfloor 8x \right \rfloor$ ?

601 ?

Hmm I got 600 a few days ago: http://vcenotes.com/forum/index.php/topic,19896.msg201890.html#msg201890

Maybe you included 0 :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 16, 2009, 02:33:07 pm

expand that difference and you get:

$4m^3+m$

so this is the number of times $\frac{1}{m}$ appears in our sum. Therefore it contributes $4m^2+1$ to the sum.

So now we have to add $4m^2+1$ over m=1 to to m=6. We add to m=6 since 7^4>1995. Then we add on the remaining numbers like you did for f(n=7).

Title: Re: TT's Maths Thread
Post by: dcc on November 16, 2009, 02:36:50 pm

Ah yes 'positive integers'. 600 it is.

Title: Re: TT's Maths Thread
Post by: dcc on November 16, 2009, 02:40:46 pm

Proof by large numbers (Hi polya's conjecture):

Code: [Select]

import List

-- Checks the number's under n using accuracy level p
check :: Integer   -- Number under consideration 
      -> Integer   -- Accuracy
      -> Int       -- # of numbers found (<n)
check n p = length $ filter (<=n) $ nub $ map sfloor (genNumList n p)

-- Our special floor function
sfloor :: Double -> Integer
sfloor x = floor (2*x) + floor (4*x) + floor (6*x) + floor (8*x) :: Integer

-- Generates a special list of numbers (maximal element = n/8 => the highest we want to check (seriously))
genNumList :: Integer      -- Number under consideration
           -> Integer      -- Accuracy
           -> [Double] -- Output list of numbers
genNumList n p = [ (fromIntegral x) / (fromIntegral (8 * p)) | x <- [1..(n * p)]]

From this, we see:

Code: [Select]

check 1000 1   = 400
check 1000 2   = 501
check 1000 3   = 601
check 1000 20  = 601
check 1000 100 = 601
-- ad infinitum

This clearly shows that 601 (counting 0) is the solution. So the 'real' solution is 600. Takes about 5 seconds to run for accuracy level 3, about a minute for accuracy level 100. This is my 'exhaustive' proof by large numbers.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2009, 02:45:38 pm

Quote from: kamil9876 on November 16, 2009, 02:33:07 pm

expand that difference and you get:

$4m^3+m$

so this is the number of times $\frac{1}{m}$ appears in our sum. Therefore it contributes $4m^2+1$ to the sum.

So now we have to add $4m^2+1$ over m=1 to to m=6. We add to m=6 since 7^4>1995. Then we add on the remaining numbers like you did for f(n=7).

Ah okay, I'll try now, and yeah pretty shit question cause you gotta use a bit of trial and error no matter what to find out that 7^4 :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 16, 2009, 02:46:23 pm

~~Where's the flaw in this one? http://vcenotes.com/forum/index.php/topic,19896.msg201920.html#msg201920~~

edit: oh wait, you agree it's 600 :-[

Title: Re: TT's Maths Thread
Post by: dcc on November 16, 2009, 03:34:49 pm

I drew a picture of the floor problem in Hogo (a programming language I created using haskell), and it's kinda interesting:

(http://img32.imageshack.us/img32/1976/floors.jpg)

You can clearly see the periodic behaviour that kamil referred to (watch the difference in the levels of the bar => increases are not uniform).

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2009, 09:16:20 pm

Prove that the sum $\sqrt{1001^2 + 1} + \sqrt{1002^2 + 1} + ... + \sqrt{2000^2 + 1}$ is irrational.

I've planned my approach in 2 ways:

1. Must show that the sum is not an integer.
2. Must show it is a zero of a monic polynomial.

But how to go about it?

Title: Re: TT's Maths Thread
Post by: Ilovemathsmeth on November 16, 2009, 09:29:52 pm

TrueTears, you're SOOOO motivated! WOW!

I feel really ashamed, haven't started looking at Specialist, although I did buy the book and all my friend's working out and intend to start...sometime...

Title: Re: TT's Maths Thread
Post by: supdawg on November 16, 2009, 09:49:31 pm

If you can prove each term is irrational then sum must be irrational

So you must prove that $1000^2+1,...,2000^2+1 \neq k^2$

gcd(n, n+1) = 1, so $1000^2$ does not share any common divisors with $1000^2+1$ . But $1000^2+1 < 1001^2$ ... etc

So none of them are squares => sum is irrational

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 16, 2009, 11:02:27 pm

Quote

If you can prove each term is irrational then sum must be irrational

$(\sqrt{2}-\sqrt{3}+1)+(\sqrt{3}-\sqrt{2}+2)=3 \in \mathbb{Q}$

Title: Re: TT's Maths Thread
Post by: supdawg on November 16, 2009, 11:29:38 pm

doesn't apply to that sum

Title: Re: TT's Maths Thread
Post by: TrueTears on November 17, 2009, 01:40:21 pm

Quote from: TrueTears on November 16, 2009, 09:16:20 pm

Prove that the sum $\sqrt{1001^2 + 1} + \sqrt{1002^2 + 1} + ... + \sqrt{2000^2 + 1}$ is irrational.

I've planned my approach in 2 ways:

1. Must show that the sum is not an integer.
2. Must show it is a zero of a monic polynomial.

But how to go about it?

Okay so had another look at this.

So to prove the sum is not an integer.

$n < \sqrt{n^2+1} < n + \frac{1}{n}$

So, $1001 < \sqrt{1001^2+1} < 1001 + \frac{1}{1001}$

Let $\frac{1}{n} = q$

An overestimate for the sum would be $1001 + q_1 + 1002 + q_2 + 1003 + q_3 + ... + 2000 + q_{1000}$

An underestimate for the sum would be $1001 + 1002 + 1003 + ... + 2000$

But since $0 < q_1 + q_2 + q_3 + ... + q_{1000} < \frac{1}{1001}$

Thus the sum is not an integer.

Use induction to prove it is a zero of a monic polynomial.

Base case:

Let the sum be written as $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_n}$

Let $n = 1$ .

Thus $\sqrt{a_1} = y$

$y^2 - a_1 = 0$

Thus $a_1$ is a zero.

Now for the inductive hypothesis, assume $n = k$ is true.

Thus $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k}$ is true.

Now we wish to prove that $n = k+1$ is also true.

We will denote this as $x$ .

$x = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k} + \sqrt{a_{k+1}}$

$\implies x = y + \sqrt{a_{k+1}}$

$\implies y = x - \sqrt{a_{k+1}}$

Let this be a zero of a monic polynomial $P(x) = x^r + c_{r-1}x^{r-1} + c_{r-2}x^{r-2} + ... + c_0$

So using the inductive hypothesis, $y$ is a zero $P(x)$ , namely $P(y) = 0$

so $P(y) = P(x - \sqrt{a_{k+1}}) = (x - \sqrt{a_{k+1}})^r + c_{r-1}(x - \sqrt{a_{k+1}})^{r-1} + c_{r-2}(x - \sqrt{a_{k+1}})^{r-2} + ... + c_0 = 0$

Now what...? How do we expand that polynomial...?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 17, 2009, 03:17:01 pm

Quote from: supdawg on November 16, 2009, 11:29:38 pm

doesn't apply to that sum

yes, and that is the trick to the problem.

Title: Re: TT's Maths Thread
Post by: humph on November 18, 2009, 03:56:15 am

Quote from: dcc on November 16, 2009, 01:23:53 pm

In Accelerated Maths (1&2) @ uom, we studied linear algebra (vector spaces / inner product spaces) / real analysis / calculus, so there isn't much relation between the types of questions posted here and the stuff you will study at uni.

Uni mathematics is set up so that you are able to tackle other subjects later on (i.e. complex analysis follows from real analysis, group theory follows from linear algebra and so on), whereas Olympiad maths doesn't really 'lead' anywhere (in terms of learning more about different topics in the future).

At least, that's my take on it.

+1. There's no real structure or focus in Olympiad maths, which is one of the reasons why it's so difficult - there's no set of rules to apply or techniques to use for every question.
(Of course the point of Olympiad training is to teach rules and techniques that come in handy, and try to identify when to apply them, but it's this identifying bit that makes successful IMO participants a head above the rest.)

Quote from: enwiabe on November 16, 2009, 02:03:47 pm

I think Olympiad Maths teaches the much more important skills of critical thinking and problem solving.

True, but again it's not something that you can really teach, so much as have in the first place. The point of IMO is to nurture that natural talent.

Title: Re: TT's Maths Thread
Post by: Over9000 on November 18, 2009, 03:57:57 am

Someone solve the latest problem

Title: Re: TT's Maths Thread
Post by: TrueTears on November 19, 2009, 04:06:52 pm

Quote from: TrueTears on November 17, 2009, 01:40:21 pm

Quote from: TrueTears on November 16, 2009, 09:16:20 pm
Prove that the sum $\sqrt{1001^2 + 1} + \sqrt{1002^2 + 1} + ... + \sqrt{2000^2 + 1}$ is irrational.

I've planned my approach in 2 ways:

1. Must show that the sum is not an integer.
2. Must show it is a zero of a monic polynomial.

But how to go about it?
Okay so had another look at this.

So to prove the sum is not an integer.

$n < \sqrt{n^2+1} < n + \frac{1}{n}$

So, $1001 < \sqrt{1001^2+1} < 1001 + \frac{1}{1001}$

Let $\frac{1}{n} = q$

An overestimate for the sum would be $1001 + q_1 + 1002 + q_2 + 1003 + q_3 + ... + 2000 + q_{1000}$

An underestimate for the sum would be $1001 + 1002 + 1003 + ... + 2000$

But since $0 < q_1 + q_2 + q_3 + ... + q_{1000} < \frac{1}{1001}$

Thus the sum is not an integer.

Use induction to prove it is a zero of a monic polynomial.

Base case:

Let the sum be written as $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_n}$

Let $n = 1$ .

Thus $\sqrt{a_1} = y$

$y^2 - a_1 = 0$

Thus $a_1$ is a zero.

Now for the inductive hypothesis, assume $n = k$ is true.

Thus $y = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k}$ is true.

Now we wish to prove that $n = k+1$ is also true.

We will denote this as $x$ .

$x = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_3} + ... + \sqrt{a_k} + \sqrt{a_{k+1}}$

$\implies x = y + \sqrt{a_{k+1}}$

$\implies y = x - \sqrt{a_{k+1}}$

Let this be a zero of a monic polynomial $P(x) = x^r + c_{r-1}x^{r-1} + c_{r-2}x^{r-2} + ... + c_0$

So using the inductive hypothesis, $y$ is a zero $P(x)$ , namely $P(y) = 0$

so $P(y) = P(x - \sqrt{a_{k+1}}) = (x - \sqrt{a_{k+1}})^r + c_{r-1}(x - \sqrt{a_{k+1}})^{r-1} + c_{r-2}(x - \sqrt{a_{k+1}})^{r-2} + ... + c_0 = 0$

Now what...? How do we expand that polynomial...?

yay with kamil's hint I got it.

Using binomial expansion we see terms with $\sqrt{a_{k+1}}$ and those without it.

So $P(y) = x^r + Q(x) + \sqrt{a_{k+1}}R(x) = 0$

Where $Q(x)$ and $R(x)$ represents polynomials containing $x$ .

Thus $x^r + Q(x) = -\sqrt{a_{k+1}}R(x)$

$x^{2r} + 2x^rQ(x) + Q(x)^2 + a_{k+1}R(x)^2 = 0$

QED.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 19, 2009, 10:49:14 pm

Let $p(x) = x^6 + x^5 + ... + 1$ . Find the remainder when $p(x^7)$ is divided by $p(x)$ .

So $p(x) = \frac{x^7-1}{x-1}$

$p(x^7) = \frac{x^{49}-1}{x^7-1}$

Then what?

Title: Re: TT's Maths Thread
Post by: Ahmad on November 19, 2009, 11:29:26 pm

You can use polynomial modular arithmetic. We're trying to find $p(x^7) \pmod{p(x)}$ . Now $p(x) = 0 \pmod{p(x)}$ and upon multiplying by $x-1$ gives $x^7 = 1 \pmod{p(x)}$ . So that $p(x^7) = (x^7)^6 + \cdots + 1 = 1 + \cdots + 1 = 7 \pmod{p(x)}$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on November 20, 2009, 02:19:33 am

Thank you Ahmad!!!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 20, 2009, 05:30:48 am

Let $r$ and $s$ be the roots of $x^2 - (a+d)x+ (ad - bc) = 0$ . Prove that $r^3$ and $s^3$ are the roots of $y^2 - (a^3 + d^3 + 3abc + 3bcd)y + (ad - bc)^3 = 0.$

Title: Re: TT's Maths Thread
Post by: Ahmad on November 20, 2009, 10:43:23 am

$(x-r)(x-s) = x^2 - (a+d)x + (ad - bc) \implies a+d = r+s,\; rs = ad - bc$

$(y-r^3)(y-s^3) = y^2 - (r^3 + s^3)y + (rs)^3$

$(rs)^3 = (ad-bc)^3$

$r^3 + s^3 = (r+s)(r^2 - rs + s^2) = (r+s)( (r+s)^2 - 3rs ) = (a+d)^3 - 3(ad-bc)(a+d) = a^3 + d^3 + 3abc + 3bcd$

We can call a multivariable polynomial $f(x_1, \ldots, x_n)$ symmetric if $f(x_1, \ldots, x_n) = f(x_{\alpha(1)}, \ldots, x_{\alpha(n)})$ where $\alpha$ is a permutation of 1..n. Symmetric polynomials can always be written as a polynomial in the roots of a polynomial. (If this confuses you just ignore it completely).

Title: Re: TT's Maths Thread
Post by: Ahmad on November 20, 2009, 10:52:45 am

As an exercise solve the problem using matrices. (Hint: eigenvalues of a matrix A satisfy $x^2 - tr(A) x + det(A) = 0$ , cube of the eigenvalues satisfy $y^2 - tr(A^3) y + det(A^3) = 0$ )

Title: Re: TT's Maths Thread
Post by: Ahmad on November 20, 2009, 11:04:05 am

As a third way, similar to the first but requires less algebraic ingenuity (especially for similar, but more difficult problems involving higher powers):

To work out $r^3 + s^3$ it's not obvious what factorisation one should use.

However we know that r and s satisfy $x^2 = (a+d)x - (ad - bc)$ multiply this by x giving $x^3 = (a+d)x^2 - (ad - bc)x$ , now this must be satisfied by r and s, plugging in each r and s:

$r^3 = (a+d)r^2 - (ad-bc)r$
$s^3 = (a+d)s^2 - (ad-bc)s$

Adding these together yields $r^3 + s^3 = (a+d)(r^2 + s^2) - (ad - bc)(r+s)$

So we've reduced expressing $r^3 + s^3$ in terms of r+s and rs to expressing $r^2 + s^2$ . You can plug in each of r and s into $x^2 = (a+d)x - (ad - bc)$ to yield an expression for $r^2+s^3$ in terms of r+s, completing the problem.

As a fourth way, have a look at the Newton-Girard formulas which has an interesting proof using generating functions (which as I recall is shown on planetmath). I tend to stay clear of this path though, simply because I never bothered to remember the formulas, and don't like looking them up and for small powers there are reasonably fast other ways. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 20, 2009, 07:26:43 pm

Thanks for the help Ahmad!
(http://img30.imageshack.us/img30/7726/asdf1q.jpg)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 20, 2009, 07:53:02 pm

1. Let $f(x) = x^n + 5x^{n-1} + 3$ where $n > 1$ is an integer. Prove that $f(x)$ cannot be expressed as the product of two polynomials, each of which has all its coefficients integers and degree at least $1$ .

2. Let $P(x) = x^n + a_{n_1}x^{n - 1} + . . . + a_1x + a_0$ be a polynomial with integral coefficients. Suppose that there exist four distinct integers $a, b, c, d$ with $P(a) = P(b) = P(c) = P(d) = 5$ . Prove that there is no integer $k$ with $P(k) = 8$ .

My working:

Using division algorithm $\implies f(x) = Q(x)g(x) + R(x)$

Where $Q(x)$ denotes the quotient and $R(x)$ denotes the remainder.

Thus $P(x) = Q(x)(x-a) + 5$

We find out that:

$P(x) = 5 \pmod{x-a}$

$P(x) = 5 \pmod{x-b}$

$P(x) = 5 \pmod{x-c}$

$P(x) = 5 \pmod{x-d}$

Now what...?

Title: Re: TT's Maths Thread
Post by: zzdfa on November 20, 2009, 10:49:49 pm

hint for the 2nd one:
consider the polynomial P(x)-5. we know that this has zeros at a,b,c,d. so we can factorize it, P(x)-5=(x-a)(x-b)(x-c)(x-d)Q(x) where Q(x) is some polynomial. now show that P(x)-5 cannot equal 3.

and for the first one, write it as
(x^p + ..... + c) * (x^q+ ...... +d ) where the mononomials are arranged in descending powers

if p,q,c,d are all integers, what must c and d be?
what is the relationship between p and q? edit: actually i didn't really think that through, might not work out.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 21, 2009, 04:12:33 pm

Rough idea for q1:

I think we must invoke some property of the integers:

say say our polynomial is:

$(a_0+a_1x+a_2x^2....x^p)(b_0+b_1x.... x^q)$

Now we know that $a_0b_0=3$

so let's assume WLOG that $a_0=\pm 3$

we know that for n>2, the x^1 term must have coefficent of 0, and this coefficent is gotten by $a_0b_1x$
and $a_1b_0x$ .

Hence we have:

$a_0b_1+a_1b_0=0$
$\implies \pm 3b_1 \pm a_1=0$

$\implies |a_1|=3|b_1|$

so in our first polynomial, our first two terms must be multiples of 3, perhaps this argument can be continued to show that all the terms are multiples of 3. But i am not sure I haven't tried.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 22, 2009, 01:20:40 am

Quote from: zzdfa on November 20, 2009, 10:49:49 pm

hint for the 2nd one:
consider the polynomial P(x)-5. we know that this has zeros at a,b,c,d. so we can factorize it, P(x)-5=(x-a)(x-b)(x-c)(x-d)Q(x) where Q(x) is some polynomial. now show that P(x)-5 cannot equal 3.

and for the first one, write it as
(x^p + ..... + c) * (x^q+ ...... +d ) where the mononomials are arranged in descending powers

if p,q,c,d are all integers, what must c and d be?
what is the relationship between p and q? edit: actually i didn't really think that through, might not work out.

Thanks!

Right I think I got Q 2

Quote

2. Let $P(x) = x^n + a_{n_1}x^{n - 1} + . . . + a_1x + a_0$ be a polynomial with integral coefficients. Suppose that there exist four distinct integers $a, b, c, d$ with $P(a) = P(b) = P(c) = P(d) = 5$ . Prove that there is no integer $k$ with $P(k) = 8$ .

Using the division algorithm we see that $P(x) = Q(x)g(x) + R(x)$

Let $y(x) = Q(x)g(x)$

$P(x) - 5 = y(x)$ .

$y(x)$ has 4 distinct integer roots, $a, b, c, d$ .

Using the fundamental theorem of Algebra, we get:

$y(x) = (x-a)(x-b)(x-c)(x-d)h(x)$ where $h(x)$ is some other polynomial so it "bumps up" the degree of $y(x)$

Now using contradiction to prove that there is no integer $k$ with $P(k) = 8$

Assume $P(k) = 8$

$y(k) = (k-a)(k-b)(k-c)(k-d)h(k)$

$P(k) - 5 = (k-a)(k-b)(k-c)(k-d)h(k)$

$3 = (k-a)(k-b)(k-c)(k-d)h(k)$

Now since $a,b,c,d$ are distinct integer roots, at least $3$ of $(k-a)(k-b)(k-c)(k-d)$ MUST equal to $1$ or else the product will go above $3$ .

Ofcourse there could be other cases such as $(k-a)=-1$ and $(k-b)=-1$ and $(k-c)=1$ and $(k-d)=1$ , but they all lead to the same result.

So if at least 3 of $(k-a)(k-b)(k-c)(k-d)$ equals 1.

Then WLOG assume $k-a = 1$ and $k-b=1$

Thus $k-a = k-b$

$\implies a=b$ .

But the question says the $a,b,c,d$ are 4 distinct integers.

$\therefore$ by contradiction, there exists no k such that $P(k) = 8$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 22, 2009, 01:29:13 am

Quote from: kamil9876 on November 21, 2009, 04:12:33 pm

Rough idea for q1:

I think we must invoke some property of the integers:

say say our polynomial is:

$(a_0+a_1x+a_2x^2....x^p)(b_0+b_1x.... x^q)$

Now we know that $a_0b_0=3$

so let's assume WLOG that $a_0=\pm 3$

we know that for n>2, the x^1 term must have coefficent of 0, and this coefficent is gotten by $a_0b_1x$
and $a_1b_0x$ .

Hence we have:

$a_0b_1+a_1b_0=0$
$\implies \pm 3b_1 \pm a_1=0$

$\implies |a_1|=3|b_1|$

so in our first polynomial, our first two terms must be multiples of 3, perhaps this argument can be continued to show that all the terms are multiples of 3. But i am not sure I haven't tried.

Thanks kamil, that gives me some new ideas!

Some new questions:

1. Let $p(x)$ be a polynomial with integer coefficients satisfying $p(0) = p(1) = 1999$ . Show that $p(x)$ has no integer zeros.

2. Let $P(x)$ be a polynomial with real coefficients. Show that there exists a nonzero polynomial $Q(x)$ with real coefficients such that $P(x)Q(x)$ has terms that are all of a degree divisible by $10^9$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 22, 2009, 01:53:13 am

Quote from: kamil9876 on November 21, 2009, 04:12:33 pm

Rough idea for q1:

I think we must invoke some property of the integers:

say say our polynomial is:

$(a_0+a_1x+a_2x^2....x^p)(b_0+b_1x.... x^q)$

Now we know that $a_0b_0=3$

so let's assume WLOG that $a_0=\pm 3$

we know that for n>2, the x^1 term must have coefficent of 0, and this coefficent is gotten by $a_0b_1x$
and $a_1b_0x$ .

Hence we have:

$a_0b_1+a_1b_0=0$
$\implies \pm 3b_1 \pm a_1=0$

$\implies |a_1|=3|b_1|$

so in our first polynomial, our first two terms must be multiples of 3, perhaps this argument can be continued to show that all the terms are multiples of 3. But i am not sure I haven't tried.

expanding on this idea, I have found that the proof requires testing a few cases individually so it gets messy, each case is alright to deal with, just keeping track that you have taken everything into account is annoying :/.

Ok so you can prove that the polynomials must have degree greater than 1, since otherwise we would get a linear term and we know this is impossible since $f(\pm3),f(\pm1)\neq 0$

Ok so first case assume $p \leq q$ , in other words the first polynomial(with constant $\pm 3$ ) has lesser or equal degree.

now recall our proof that $a_0$ and $a_1$ are multiples of $3$ , we will know prove by strong induction that all the $a_i$ are multiples of $3$ .

the coefficient of the x^m term in the product (where m<p<n-1) is:
$ a_0b_m+a_1b_{m-1}+a_2b_{m-2}..... a_{m-1}b_1 + a_mb_0=0$

however by induction, $a_i$ is a multiple of $3$ for all $i<m$ , thus the equation above tells us that $a_mb_0=\pm a_m$ is a multiple of $3$ , hence a_m is a multiple of 3 completing the induction. This can be used for m=p to show that $a_p=1$ is a multiple of $3$ , a contradiction thus this proves the question for this case.

For the other case, p>q, it is more difficult to show that all the $a_i$ are multiples of $3$ and requires a slightly more complicated variation of the trick we used above.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 22, 2009, 04:34:47 am

Thanks kamil, however just looking back at your previous post, the question says "...the product of two polynomials, each of which has all its coefficients integers and degree at least 1." so that means all the powers and coefficients of the 2 polynomials are all greater or equal to 1, ie all positive.

So then how can $a_0b_1 + a_1b_0$ ever equal to 0?

Since $\left(a_0, b_1, a_1, b_0\right) \ge 1$

Title: Re: TT's Maths Thread
Post by: Over9000 on November 22, 2009, 05:41:15 am

Quote from: TrueTears on November 22, 2009, 04:34:47 am

Thanks kamil, however just looking back at your previous post, the question says "...the product of two polynomials, each of which has all its coefficients integers and degree at least 1." so that means all the powers and coefficients of the 2 polynomials are all greater or equal to 1, ie all positive.

So then how can $a_0b_1 + a_1b_0$ ever equal to 0?

Since $\left(a_0, b_1, a_1, b_0\right) \ge 1$

"\left(a_0, b_1, a_1, b_0\right) \ge 1"
I think they just mean the degree of the polynomials is greater than 1 not the coefficients as well.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 22, 2009, 02:56:02 pm

yeah they tried to pack everything into one sentences. "all coefficents are integers of the polynomials", "the polynomials themselves are of degree at least one". You need some constant term obviously since you can only get the +3 at the end from multiplying two constants.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 22, 2009, 07:13:48 pm

Oh right nevermind I misread the sentence sigh.

Also kamil when you said:

Quote

Ok so first case assume $p \leq q$ , in other words the first polynomial(with constant $\pm 3$ ) has bigger or equal degree.

Shouldn't it be $p \ge q?$

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 22, 2009, 07:26:13 pm

nah I should've said the second polynomial has bigger or equal degree, ie just focus on $p \leq q$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on November 23, 2009, 05:10:59 am

Thank you SOOOO much kamil, I finally get this question! Been stuck on it for several days :(

Just gonna type up this question formally. Including the 2nd case.

So assume that $f(x)$ can be factorised into 2 polynomials with integer coefficients.

$f(x) = p(x)q(x) = (x^p + a_{p-1}x^{p-1} + a_{p-2}x^{p-2} +...+a_2x^2 + a_1x+a_0)(x^q + b_{q-1}x^{q-1} + b_{q-2}x^{q-2} +...+b_2x^2 + b_1x+b_0)$

Now $a_0b_0 = 3$ .

Assume $a_0 = \pm 3 \implies b_0 = \pm 1$ or $a_0 = \pm 1$ and $b_0 = \pm 3$

Now the 2 factors can not be linear. Why?

When $p(x)$ is linear then $(x+a_0)q(x) = f(x)$

But $a_0 = \pm 3$ or $\pm 1$ but none of these equal $0$ when subbed into $f(x)$ . Thus there can be no linear factors.

Now for all $n>2$ the coefficient of the $x$ term in $f(x)$ must be $0$ .

Thus $a_1b_0 + a_0b_1 = 0$

Take the first case $a_0 = \pm 3$ and $b_0 = \pm 1$

$(\pm 1)a_1 + (\pm 3)b_1 = 0$

$\pm a_1 = \mp 3b_1$

This means $a_1$ and $a_0$ are multiples of $3$ .

Doing some experimenting:

When $n = 4$ , that means the coefficient of $x^2$ is also $0$ .

Thus $a_2b_0 + a_1b_1 + a_0b_2 = 0$

Thus $\pm a_2 = -(a_1b_1 + a_0b_2)$ (since $b_0 = \pm 1$ )

Since $a_1$ and $a_0$ are both multiples of $3$ we can factor out a $3$ leaving $a_2 = -3(\mbox{some positive integer})$

Which means $a_2$ is also a multiple of $3$ .

Now a few things can be noticed:

1. The subscript of a and b both add to up the power of the term we need to the find the coefficient of.

2. Continuing this we find that $a_3, a_4...$ are also multiples of $3$ .

This means if we can prove that all of the coefficients of $p(x)$ are multiples of $3$ then we can use contradiction to show that $p(x)$ can not exist as the coefficient of $x^p$ is $1$ which is not a multiple of $3$ .

Say we want to find the coefficient of $x^m$ term in $f(x)$ where $0 < m \le p$ . (I'm restricting the domain only up to $p$ because we just require to prove the coefficients of $p(x)$ are multiples of $3$ .)

Now consider the polynomial $p(x)$ where $m \le p < n-1$ (In other words, the lowest degree of $p$ is $m$ and it must be strictly smaller than $n-1$ or else $q(x)$ would be a linear factor)

Using the pattern we saw earlier in the experimenting to find an equation for the coefficient of $x^m$ yields:

$a_0b_m + a_1b_{m-1} + a_2b_{m-2} + ... + a_{m-1}b_1 + a_mb_0 = 0$

Now we can use strong induction to prove $a_m$ is a multiple of $3$ .

First Case: When $p \le q$

Base Case

when $m = 1$

$a_0b_1 + a_1b_0 = 0$

We have already proved earlier using this that $a_1$ is a multiple of $3$ .

Inductive Hypothesis

Assume all $a_0, a_1, a_2 ... a_{m-1}$ are multiples of $3$ .

Proof

Since $a_0, a_1, a_2 ... a_{m-1}$ are multiples of $3$ then the equation can be rewritten into:

$3(S) + a_mb_0 = 0$ (Factoring out a $3$ , while calling the integer sum inside the bracket $S$ )

But $b_0 = \pm 1$

$\therefore 3S \pm a_m = 0$

$3S = \mp a_m$

Therefore by induction, $a_m$ is also a multiple of $3$ .

Second Case: When $p>q$ .

Why consider this case? Let's pick a random value of $p$ , say $p = 6$ and $q = 3$ .

Now say we wanted to find the coefficient of $x^m$ where $m = 4$ .

Then the equation would be: $a_0b_4 + a_1b_3 + a_2b_2 + a_3b_1 + a_4b_0 = 0$

However $b_4$ does not exist as the highest power of $q$ is $3$ so there is no such thing as the coefficient of $x^4$ for $q(x)$ .

This causes a problem, so we need to change our general equation a bit.

This is where I am stuck. lolz

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 23, 2009, 02:29:17 pm

you can say that $b_4=0$ since the x^4 coefficent of a polynomial of degree 3 can be considered as 0.

therefore in general equating the x^m coefficent for $m\leq p<n-1$ gives:
$ a_0 b_m + a_1 b_{m-1} ... + a_m b_0 = 0$

which gives:

$a_0 b_m + .... a_{m-1} b_1=\pm a_m$

now since by the inductive hypothesis each $a_i$ is a multiply of 3 for $i<m$ , the sum on the left hand side will be a sum of multiple of 3's hence a multiple of 3. (whether or not certain terms are 'non-existant' doesn't change the fact that the LHS must be a multiple of 3, or more conviently the non-existant terms can be considered as terms that equal 0, which is a multiple of 3 hence the argument is complete).

And so basically the argument is exactly the same as before, giving our desired contradiction.

Title: Re: TT's Maths Thread
Post by: Over9000 on November 23, 2009, 05:16:50 pm

Quote from: TrueTears on November 23, 2009, 05:42:36 pm

Found a pretty neat way for Q1.

Quote
1. Let $p(x)$ be a polynomial with integer coefficients satisfying $p(0) = p(1) = 1999$ . Show that $p(x)$ has no integer zeros.

Assume $p(x)$ does have integer zeros.

Thus $p(x)=(x-s)q(x)$

where s is an integer root and $q(x)$ is a polynomial with integer coefficients.

$p(0) = (0-s)q(0)$

$1999 = (-s)q(0)$

Now notice that $1999$ is prime.

We require $q(0)$ to be an integer. This means $s = -1$ and $q(0) = 1999$ or $s = 1$ and $q(0) = -1999$

$p(1) = (1-s)q(1)$

$1999 = (1-s)q(1)$

Subbing in $s = -1 \implies q(1) = \frac{1999}{2}$

Subbing in $s = 1 \implies 0 = 1999$

However these results are impossible thus by contradiction there exists no integer zeros for $p(x)$ .

Thats quite an elegant way to go about that question TT.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 23, 2009, 05:24:29 pm

Also just realised that Perron's Criterion can be used for:

Quote

Let $f(x) = x^n + 5x^{n-1} + 3$ where $n > 1$ is an integer. Prove that $f(x)$ cannot be expressed as the product of two polynomials, each of which has all its coefficients integers and degree at least $1$ .

Perron's Criterion states that: for any polynomial $P(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} +... + a_{1}x + a_0$ with integer coefficients and $a_0 \neq 0$ , if $|a_{n-1}| > 1 + |a_{n-2}| + ... + |a_1| + |a_0|$ then $P(x)$ is irreducible over $\mathbb{Z}$ .

Applying that here. We see that $5 > 1+0+0+...+3 \implies 5>4$

Thus Perron's Criterion shows that $f(x)$ is irreducible over $\mathbb{Z}$ .

However is there an elementary way of proving Perron's Criterion?

I feel bad just applying it without any solid proof ^.^

Title: Re: TT's Maths Thread
Post by: TrueTears on November 23, 2009, 05:34:58 pm

Quote from: kamil9876 on November 23, 2009, 02:29:17 pm

you can say that $b_4=0$ since the x^4 coefficent of a polynomial of degree 3 can be considered as 0.

therefore in general equating the x^m coefficent for $m\leq p<n-1$ gives:
$ a_0 b_m + a_1 b_{m-1} ... + a_m b_0 = 0$

which gives:

$a_0 b_m + .... a_{m-1} b_1=\pm a_m$

now since by the inductive hypothesis each $a_i$ is a multiply of 3 for $i<m$ , the sum on the left hand side will be a sum of multiple of 3's hence a multiple of 3. (whether or not certain terms are 'non-existant' doesn't change the fact that the LHS must be a multiple of 3, or more conviently the non-existant terms can be considered as terms that equal 0, which is a multiple of 3 hence the argument is complete).

And so basically the argument is exactly the same as before, giving our desired contradiction.

Ohhh right I see, so it doesn't matter which is non-existant or not, all you really have to say is that they exist as 0. Thus it doesn't affect the factoring out the 3 on the LHS. Right?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 23, 2009, 05:42:36 pm

Found a pretty neat way for Q1.

Quote

1. Let $p(x)$ be a polynomial with integer coefficients satisfying $p(0) = p(1) = 1999$ . Show that $p(x)$ has no integer zeros.

Assume $p(x)$ does have integer zeros.

Thus $p(x)=(x-s)q(x)$

where s is an integer root and $q(x)$ is a polynomial with integer coefficients.

$p(0) = (0-s)q(0)$

$1999 = (-s)q(0)$

Now notice that $1999$ is prime.

We require $q(0)$ to be an integer. This means $s = -1$ and $q(0) = 1999$ or $s = 1$ and $q(0) = -1999$

$p(1) = (1-s)q(1)$

$1999 = (1-s)q(1)$

Subbing in $s = -1 \implies q(1) = \frac{1999}{2}$

Subbing in $s = 1 \implies 0 = 1999$

However these results are impossible thus by contradiction there exists no integer zeros for $p(x)$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on November 23, 2009, 11:51:53 pm

Can anyone tell me what I'm doing wrong with this question?

Let $a_1, a_2 ... a_n$ be a sequence of positive numbers. Show that $(a_1+a_2+...+a_n)\left(\frac{1}{a_1} + \frac{1}{a_2} + ... + \frac{1}{a_n}\right) \ge n^2$ .

So we require $\left[\sum_{i=1}^n (a_i)\right]\left[\sum_{k=1}^n \left(\frac{1}{a_k}\right)\right]$ .

First there will be $n^2$ terms in total after we expand the 2 brackets.

We can see that $\frac{a_1}{a_1} + \frac{a_2}{a_2} + ... + \frac{a_n}{a_n} = 1+1+...+1$

Which means there will be $\mbox{n}$ groups of $1$ .

What about the rest of the terms? Ie, the remaining $n^2-n$ terms.

A bit of experimenting quickly yields a pattern:

Consider expanding all the terms which do not produce the number $1$ .

$\frac{a_1}{a_2} + \frac{a_1}{a_3} +...+\frac{a_1}{a_n} + \frac{a_2}{a_1} + \frac{a_2}{a_3} +...+\frac{a_2}{a_n} + \frac{a_3}{a_1} + \frac{a_3}{a_2} + ... + \frac{a_3}{a_n} + ...+ \frac{a_n}{a_1} + \frac{a_n}{a_2} + \frac{a_n}{a_3}+...$

Rearranging leads to:

$\frac{a_1}{a_2} + \frac{a_2}{a_1} + \frac{a_1}{a_3} + \frac{a_3}{a_1} + \frac{a_2}{a_3} + \frac{a_3}{a_2} + \frac{a_1}{a_n} + \frac{a_n}{a_1} + \frac{a_2}{a_n} + \frac{a_n}{a_2} + \frac{a_3}{a_n} + \frac{a_n}{a_3}$

All of these terms are in the form of $\frac{a_i}{a_k} + \frac{a_k}{a_i}$ pairs where $k>i$

Now using AM-GM on one of these pair yields:

$\frac{\frac{a_i}{a_k} + \frac{a_k}{a_i}}{2} \ge \sqrt{\left(\frac{a_i}{a_k}\right)\left(\frac{a_k}{a_i}\right)}$

$\therefore \frac{a_i}{a_k} + \frac{a_k}{a_i} \ge 2$

So each pair has a minimum value of $2$ .

Thus the remaining $(n^2-n)$ terms has a minimum value of $2(n^2-n)$

Thus totally we have $(n)(1) + 2(n^2-n) = 2n^2-n$

However the proof is $n^2$ only... where did that extra $\mbox{n}$ and $2n^2$ come from?

Title: Re: TT's Maths Thread
Post by: humph on November 24, 2009, 12:00:06 am

Quote from: TrueTears on November 23, 2009, 11:51:53 pm

So each pair has a minimum value of $2$ .

Thus the remaining $(n^2-n)$ terms has a minimum value of $2(n^2-n)$

Maybe I'm missing something here, but if each pair has minimum value of $2$ , then "on average" each term is $1$ . Because there are $n^2-n$ terms, so $(n^2-n)/2$ pairs, each with minimum value $2$ , so we should have that the remaining $n^2-n$ terms has a minimum value of $2 \cdot (n^2-n)/2 = n^2 - n$ .

Quote from: TrueTears on November 23, 2009, 11:51:53 pm

Thus totally we have $(n)(1) + 2(n^2-n) = 2n^2-n$

So this should be $(n)(1) + (n^2-n) = n^2$ , as required.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 24, 2009, 12:02:50 am

LOL opps how the hell did I miss that! Silly me =.=

Thanks humph.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 24, 2009, 06:49:36 pm

Some inequality questions: (These will be revolving around AM-GM, Cauchy-Schwarz and Chebyshev's inequality)

~~1) $1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!} + ... < 3$~~

~~2) For which integer $\mbox{n}$ is $\frac{1}{n}$ closest to $\sqrt{1000000} - \sqrt{999999}$ ?~~

3) Show that $\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} + ... + \sqrt{a_n^2 + b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2 + (b_1+b_2+...+b_n)^2}$ for all real values of the variables. Furthermore, give a condition for equality to hold.

~~4) Prove that $0 \le yz+zx+xy-2xyz\le \frac{7}{27}$ where $x,y,z$ are non negative real numbers for which $x+y+z=1$~~

Many thanks guys!

For Q 1. Taylor series can be used.

Since $e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ... + \frac{1}{n!}$

And since $e \approx 2.7$

That means the sum must be smaller than $3$ .

But are there any other ways of doing this?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 24, 2009, 07:54:04 pm

q1 is nice, rather than using the known value for e, you can work out an upper bound for it by pretending you have no clue what it is:

first of all notice that n!>2^n for all n>2. proof:

we're basically comparing:

1*2*3*4*5*6*7.....*n
2*2*2*2*2*2*2.....*2

ie notice that factor for factor, the top row has greater numbers.

actually, make that $n!>2^{n-1}$ as it is more useful (ie just ignore the first colunm)
therefore:

$\sum_{n=0}^{\infty} \frac{1}{n!}=1+\sum_{n=1}^{\infty}\frac{1}{n!}<1+\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}=3$

i.e: it is a very useful strategy to compare things to the geometric sequence, or other known sequences or even things that look like right hand rectangle approximations to integrals.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 24, 2009, 10:03:38 pm

So how did you think of $n! > 2^{n-1}$ ?

That's like a needle in a haystack for me :P Finding a sum to infinity which yields 3...

Is that just experience or...?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 24, 2009, 11:15:38 pm

Thanks kamil, I was playing around $2^n$ and gets a "better" estimate :P

The LHS can be rewritten into:

$\sum_{n=0}^{\infty}\frac{1}{n!} < 3$

But we know $n! > 2^n \forall n \ge 4$

Reciprocate:

$\frac{1}{n!} < \frac{1}{2^n} \forall n \ge 4$

But we require sum from 0 to infinity.

$\sum_{n=0}^{\infty} \frac{1}{n!} < \sum_{n=0}^{\infty} \frac{1}{2^n}$

However this is false since the inequality holds for $n \ge 4$

Thus we require:

$\sum_{n=4}^{\infty} \frac{1}{n!} < \sum_{n=4}^{\infty} \frac{1}{2^n}$

The RHS is a geometric sum, denote this as $S$ .

$S = \frac{\frac{1}{2^4}\left[\left(\frac{1}{2}\right)^n -1\right]}{\left(\frac{1}{2}\right)-1}$

Now $\lim_{n \to \infty} S = \frac{-\frac{1}{16}}{-\frac{1}{2}} = \frac{1}{8}$

Since we are after the upper bound of $\sum_{n=0}^{\infty} \frac{1}{n!}$

$= \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\sum_{n=4}^{\infty}$

$\implies 1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{8}=2+\frac{19}{24} < 3$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 01:30:11 am

Quote

4) Prove that $0 \le yz+zx+xy-2xyz\le \frac{7}{27}$ where $x,y,z$ are non negative real numbers for which $x+y+z=1$

I've proved that $yz+zx+xy-2xyz \ge 0$ however I can't seem to get the RHS inequality. Here's my working so far:

Applying AM-GM on $x+y+z$

$\frac{x+y+z}{3} \ge (xyz)^{\frac{1}{3}}$

$(x+y+z) \ge 3(xyz)^{\frac{1}{3}}$

But $x+y+z = 1$

$\implies \frac{1}{3} \ge (xyz)^{\frac{1}{3}}$

$\therefore (xyz) \le \frac{1}{27}$

Now if we can prove $|yz + zx + xy| > |2xyz|$

Then that implies $yz+zx+xy - 2xyz \ge 0$ since $x,y,z \in \mathbb{R^+} \cup \{0\}$

Applying AM-GM on $yz+zx+xy$ yields:

$\frac{yz+zx+xy}{3} \ge \left[(yz)(zx)(xy)\right]^{\frac{1}{3}}$

$(yz+zx+xy) \ge 3\left(x^2y^2z^2\right)^{\frac{1}{3}}$

$(yz+zx+xy) \ge 3\left(xyz\right)^{\frac{2}{3}}$

Now let's try to prove that $3\left(xyz\right)^{\frac{2}{3}} > 2xyz$

$\frac{3}{2} > (xyz)^{\frac{1}{3}}$

But we showed that $xyz \le \frac{1}{27}$

Sub in the upper bound into $\frac{3}{2} > (xyz)^{\frac{1}{3}}$ yields $\frac{3}{2} > \left(\frac{1}{27}\right)^{\frac{1}{3}}$ $\implies \frac{3}{2} > \frac{1}{3}$

Which is clearly true.

Thus $(yz+zx+xy) \ge 3\left(xyz\right)^{\frac{2}{3}} > 2xyz$

Hence $yz + zx +xy - 2xyz \ge 0$

Now the trouble is how to prove $yz + zx +xy - 2xyz \le \frac{7}{27}$ ...

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 25, 2009, 01:34:11 am

4.)

I have developed a method in the past of making these substitutions:

let $x=t+\frac{1}{3}, y=s+\frac{1}{3}, z=u+\frac{1}{3}$

with $t+s+u=0$ obviously.

Now expand the expression and after some algebra you eventually get:

$\frac{-1}{3}((t+s)^2-ts)+\frac{7}{27}$ (*)

then using AM-GM:
$ \frac{t+s}{2} \geq \sqrt{ts}$

$(t+s)^2-4ts \geq 0$

and so $(t+s)^2-ts \geq 0$

which proves * is less than or equal to $\frac{7}{27}$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 01:47:00 am

LOL thanks kamil but that's soooooooooooooooooooooooooo needle in the haystack for me :( (the substitution part, I would not have thought of that in a million years =S)

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 25, 2009, 01:58:41 am

lol i think the questions are rigged for it:

http://vcenotes.com/forum/index.php/topic,2398.msg140527.html#msg140527

sort of developed the technique here, it ussually happens when the expression is symmetric in the variables and in symmetric curves/surfaces the highest point is commonly the centre, and so what I did here is try to find a nice centre ie: (1/3,1/3,1/3) and show that if you move to another point (1/3 +t,1/3 +s, 1/3 + u) you 'fall off' from the top.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 25, 2009, 02:18:26 am

let $u_i=(a_i, b_i)$

$||u_1||+||u_2||....+||u_n|| \geq ||u_1+u_2.....||$ (triangle inequality)

from which the result immediately follows.

I once did a proof of the triangle inequality without using the Cauchy Schwartz ienquality, and hence used it to prove Cauchy Schwartz. However many textbooks go the other way and so you can try this yourself.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 02:39:45 am

Quote from: kamil9876 on November 25, 2009, 01:58:41 am

lol i think the questions are rigged for it:

http://vcenotes.com/forum/index.php/topic,2398.msg140527.html#msg140527

sort of developed the technique here, it ussually happens when the expression is symmetric in the variables and in symmetric curves/surfaces the highest point is commonly the centre, and so what I did here is try to find a nice centre ie: (1/3,1/3,1/3) and show that if you move to another point (1/3 +t,1/3 +s, 1/3 + u) you 'fall off' from the top.

Right, do you think that substitution only works for symmetrical questions such as this one? Ie, you can switch the variables around yet you get the same thing.

So say you had a symmetrical question like this but it had 4 variables, let's say $x,y,z,q$ and given that $x+y+z+q = 1$

Would you then do $x = t+\frac{1}{4}$ $\mbox{etc}$ ...?

What about if the question was NOT symmetrical. Yet $x+y+z+q = 1$ still, would you do the same substitution?

Sorry there are no examples here to illustrate my point but I'm just speaking in 'general' terms.

Title: Re: TT's Maths Thread
Post by: humph on November 25, 2009, 03:54:09 am

Quote from: TrueTears on November 24, 2009, 06:49:36 pm

3) Show that $\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} + ... + \sqrt{a_n^2 + b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2 + (b_1+b_2+...+b_n)^2}$ for all real values of the variables. Furthermore, give a condition for equality to hold.

This is just the triangle inequality, as kamil said. I'm guessing you're using the Cauchy-Schwarz inequality in the form
$a_1 a_2 + b_1 b_2} \leq \sqrt{a_1^2 + b_1^2}}\sqrt{a_2^2 + b_2^2}}$
If we let $\mathbf{u}_1 = (a_1, b_1)$ and $\mathbf{u}_2 = (a_2, b_2)$ , then this is equivalent to
$\mathbf{u}_1 \cdot \mathbf{u}_2 \leq \|\mathbf{u}_1\| \|\mathbf{u}_2\|$
where $\mathbf{u}_1 \cdot \mathbf{u}_2$ is the dot product of $\mathbf{u}_1$ and $\mathbf{u}_2$ , and $\|\mathbf{u}_1\|, \|\mathbf{u}_2\|$ is the magnitude of $\mathbf{u}_1, \mathbf{u}_2$ respectively.
Now note that $\mathbf{u}_1 \cdot \mathbf{u}_1 = \|\mathbf{u}_1\|^2$ , and hence
$ \begin{split} \|\mathbf{u}_1 + \mathbf{u}_2\|^2 & = (\mathbf{u}_1 + \mathbf{u}_2) \cdot (\mathbf{u}_1 + \mathbf{u}_2) \\ & = \mathbf{u}_1 \cdot \mathbf{u}_1 + \mathbf{u}_1 \cdot \mathbf{u}_2 + \mathbf{u}_2 \cdot \mathbf{u}_1 + \mathbf{u}_2 \cdot \mathbf{u}_2 \\ & = \|\mathbf{u}_1\|^2 + 2 \mathbf{u}_1 \cdot \mathbf{u}_2 + \|\mathbf{u}_2\|^2 \end{split}$
as the dot product is linear and commutative. Then by the Cauchy-Schwarz inequality,
$\begin{split} \|\mathbf{u}_1 + \mathbf{u}_2\|^2 & \leq \|\mathbf{u}_1\|^2 + 2 \|\mathbf{u}_1\| \|\mathbf{u}_2\| + \|\mathbf{u}_2\|^2 \\ & = (\|\mathbf{u}_1\| + \|\mathbf{u}_2\|)^2 \end{split}$
and taking square roots yields the triangle inequality. By induction, this implies that
$\|\mathbf{u}_1 + \ldots + \mathbf{u}_n\| \leq \|\mathbf{u}_1\| + \ldots + \|\mathbf{u}_n\|$
where $\mathbf{u}_i = (a_i, b_i)$ . It remains to note that
$\|\mathbf{u}_1 + \ldots + \mathbf{u}_n\| = \sqrt{(a_1 + \ldots + a_n)^2 + (b_1 + \ldots b_n)^2}$
and that
$\|\mathbf{u}_1\| + \ldots + \|\mathbf{u}_n\| = \sqrt{a_1^2 + b_1^2} + \ldots + \sqrt{a_n^2 + b_n^2}$

Note that you can prove all this without writing it in the language of vectors, dot products, and magnitudes, but it's in some sense the most "natural" way. Also as kamil said, you don't need the Cauchy-Schwarz inequality to prove the triangle inequality, but it's normal to do so in that order, rather than the other way around.

The Cauchy-Schwarz inequality is a particular feature of a mathematical structure called an inner product space, which is a vector space with an inner product, a function that measures angles between vectors and magnitudes of vectors; the dot product on the vector space $\mathbb{R}^2$ is the standard example. Inner products in particular give rise to norms, which is the magnitude of a vector, and norms satisfy the triangle inequality. This stuff is very important mathematically; the theory of Banach spaces (complete normed vector spaces) and Hilbert spaces (complete inner product spaces) are the foundations of an area of mathematics called functional analysis, and have applications to many areas of mathematics and physics.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 25, 2009, 11:12:37 am

Quote from: kamil9876 on November 25, 2009, 01:34:11 am

$\frac{-1}{3}((t+s)^2-ts)+\frac{7}{27}$ (*)

I get $2st(s+t) - \frac{1}{3}((t+s)^2-ts)+\frac{7}{27}$ , I could be wrong but it'd be worth checking your working, I would like to see this work out! :)

Also for the original problem the left hand side inequality isn't too strong. Notice that not all of x, y and z can be greater than 1/2, which means that we can suppose one of them is $\le 1/2$ say z. Then $xy(1 - 2z) + yz + xz \ge 0$ as required. :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 25, 2009, 12:50:38 pm

yep ahmad you're right, i missed $2st(t+s)$

Title: Re: TT's Maths Thread
Post by: humph on November 25, 2009, 12:58:38 pm

Fucking inequalities. I just spent ages stuck trying to show that
$\int^{x}_{1}{\log\left(\frac{t}{\lfloor t \rfloor}\right) \: dt} = O(\log x)$

Title: Re: TT's Maths Thread
Post by: Ahmad on November 25, 2009, 01:19:10 pm

It's asymptotically $\sum_{i=1}^{\lfloor x\rfloor} \int_{i}^{i+1} \log \left(\frac{t}{i}\right) dt = \sum_{i=1}^{\lfloor x\rfloor} \left( (i+1) \log(1+\frac{1}{i}) - 1\right) \le \sum_{i=1}^{\lfloor x\rfloor} \frac{1}{i} = O(\log x)$

where I used $x \ge \log(1+x)$ which is true because $e^x \ge 1+x$ . Is that how you did it? :)

Title: Re: TT's Maths Thread
Post by: Ahmad on November 25, 2009, 01:21:06 pm

Kamil another thing which might make your approach tricky (or maybe not, haven't invested so much thought into it) is that s and t can be negative so AM-GM isn't applicable.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 25, 2009, 01:32:22 pm

I didn't realise the restriction that x,y,z is positive, so i havn't took that into account. I think that ruins it more.

Title: Re: TT's Maths Thread
Post by: /0 on November 25, 2009, 01:37:22 pm

Lagrange multipliers? Calculus is boss!

(*runs away to hide*)

Title: Re: TT's Maths Thread
Post by: humph on November 25, 2009, 01:40:22 pm

Quote from: Ahmad on November 25, 2009, 01:19:10 pm

It's asymptotically $\sum_{i=1}^{\lfloor x\rfloor} \int_{i}^{i+1} \log \left(\frac{t}{i}\right) dt = \sum_{i=1}^{\lfloor x\rfloor} \left( (i+1) \log(1+\frac{1}{i}) - 1\right) \le \sum_{i=1}^{\lfloor x\rfloor} \frac{1}{i} = O(\log x)$

where I used $x \ge \log(1+x)$ which is true because $e^x \ge 1+x$ . Is that how you did it? :)

Nah. I was trying to show that
$\sum_{n \leq x}{\log n} = x\log x - x + O(\log x).$
Here's what I did:
$\sum_{n \leq x}{\log n} = \sum_{n \leq x}{\int^{n+1}_{n}{\log n \: dt}} = \int^{\left\lfloor x\right\rfloor + 1}_{1}{\log \left\lfloor t\right\rfloor \: dt} = \int^{x}_{1}{\log t \: dt} + \int^{2}_{1}{\log \left(\frac{\left\lfloor t\right\rfloor}{t}\right) \: dt} + \int^{x}_{2}{\log \left(\frac{\left\lfloor t\right\rfloor}{t}\right) \: dt} + \int^{\left\lfloor x\right\rfloor + 1}_{x}{\log \left\lfloor t\right\rfloor \: dt}.$
Then $\int^{x}_{1}{\log t \: dt} = x \log x - x$ , and
$0 \leq \int^{\left\lfloor x\right\rfloor + 1}_{x}{\log \left\lfloor t\right\rfloor \: dt} = (\left\lfloor x\right\rfloor + 1 - x) \log \left\lfloor x\right\rfloor = (1 - \{x\}) \log(x - \{x\}) \leq \log x.$
Also
$0 \geq \int^{x}_{2}{\log \left(\frac{\left\lfloor t\right\rfloor}{t}\right) \: dt} = \int^{x}_{2}{\log\left(1 - \frac{\{t\}}{t}\right) \: dt},$
and then using $\log(1-z) = \sum^{\infty}_{n=1}{z^n/n} \leq \sum^{\infty}_{n=1}{z^n} = z/(1-z)$ for $|z| < 1$ and using the estimate $0 \leq \{t\} < 1$ , we get
$\int^{x}_{2}{\log \left(\frac{\left\lfloor t\right\rfloor}{t}\right) \: dt}\geq -\int^{x}_{2}{\frac{\{t\}}{t - \{t\}} \: dt} \geq -\int^{x}_{2}{\frac{1}{t-1} \: dt} = -\log (x-1) \geq -\log x.$
Finally,
$0 \geq \int^{2}_{1}{\log\left(\frac{\left\lfloor t\right\rfloor}{t}\right) \: dt} = -\int^{2}_{1}{\log t \: dt} = -2\log 2 - 2.$

It wouldn't be very linear of me to use the fact that $\sum_{n \leq x}{1/n} = O(\log x)$ , as I prove in the next section that
$\sum_{n \leq x}{\frac{1}{n}} = \log x + \gamma + O\left(\frac{1}{x}\right)$
though admittedly my proof is independent of the above result.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 25, 2009, 01:48:34 pm

Quote from: /0 on November 25, 2009, 01:37:22 pm

Lagrange multipliers? Calculus is boss!

(*runs away to hide*)

I actually tried Lagrange multipliers on this question, it works. And it's not as ugly as it usually is.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 03:43:37 pm

Heh another way to prove the RHS inequality without the substitution:

$xy+yz+xz-2xyz=\frac{1}{4}(1-2x)(1-2y)(1-2z)+\frac{1}{4}$ .

Applying AM-GM on $(1-2x)(1-2y)(1-2z)$ yields:

$(1-2x) + (1-2y) + (1-2z) \ge 3((1-2x)(1-2y)(1-2z))^{\frac{1}{3}}$

Since $xy+yz+xz - 2xyz$ is positive this means $(1-2x)(1-2y)(1-2z)$ must all be positive or at least 2 of them is negative and 1 is positive.

Thus AM-GM can be applied.

$\implies \frac{3-2(x+y+z)}{3} \ge ((1-2x)(1-2y)(1-2z))^{\frac{1}{3}}$

$\therefore \frac{1}{27} \ge (1-2x)(1-2y)(1-2z)$ (since $x+y+z = 1$ )

Subbing this upper bound back yields $\left(\frac{1}{4}\right)\left(\frac{1}{27}\right) + \frac{1}{4} = \frac{7}{27}$

Thus $yz+zx+xy-2xyz \le \frac{7}{27}$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 05:57:58 pm

Quote from: TrueTears on November 24, 2009, 06:49:36 pm

3) Show that $\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} + ... + \sqrt{a_n^2 + b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2 + (b_1+b_2+...+b_n)^2}$ for all real values of the variables. Furthermore, give a condition for equality to hold.

Well good ol' induction solves this pretty well. (Thanks Ahmad xD)

Base case:

Let $n = 2$

$\sqrt{a_1^2+b_1^2} + \sqrt{a_2^2+b_2^2} \ge \sqrt{(a_1+a_2)^2+(b_1+b_2)^2}$

$(a_1^2+b_1^2) + 2\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)} + (a_2^2+b_2^2) \ge (a_1+a_2)^2 + (b_1+b_2)^2$

$2\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)} \ge a_1^2+2a_1a_2+a_2^2 + b_1^2+2b_1b_2+b_2^2 - a_1^2-b_1^2-a_2^2-b_2^2$

$2\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)} \ge 2a_1a_2+2b_1b_2$

$(a_1^2+b_1^2)(a_2^2+b_2^2) \ge (a_1a_2+b_1b_2)^2$

Which is true because of Cauchy-Schwarz.

Inductive Hypothesis:

Assume the inequality is true for $n = k$

$\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} + ... + \sqrt{a_k^2 + b_k^2} \ge \sqrt{(a_1+a_2+...+a_k)^2 + (b_1+b_2+...+b_k)^2}$

is true.

Proof:

Need to prove it's true for $n = k+1$

$\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} + ... + \sqrt{a_{k+1}^2 + b_{k+1}^2} \ge \sqrt{(a_1+a_2+...+a_{k+1})^2 + (b_1+b_2+...+b_{k+1})^2}$

So let's add $\sqrt{a_{k+1}^2 + b_{k+1}^2}$ to our inductive hypothesis.

This yields:

$\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} + ... + \sqrt{a_k^2 + b_k^2} + \sqrt{a_{k+1}^2 + b_{k+1}^2} \ge \sqrt{(a_1+a_2+...+a_k)^2 + (b_1+b_2+...+b_k)^2} + \sqrt{a_{k+1}^2 + b_{k+1}^2}$

Now if we can show $\sqrt{(a_1+a_2+...+a_k)^2 + (b_1+b_2+...+b_k)^2} + \sqrt{a_{k+1}^2 + b_{k+1}^2} > \sqrt{(a_1+a_2+...+a_{k+1})^2 + (b_1+b_2+...+b_{k+1})^2}$

Then we have completed the proof.

Let $a_1+a_2+...+a_k = u$ and $b_1+b_2+...+b_k = q$

Thus the inequality becomes $\sqrt{u^2+q^2} + \sqrt{a_{k+1}^2+b_{k+1}^2} > \sqrt{(u+a_{k+1})^2+(q+b_{k+1})^2}$

$u^2+q^2 + 2\sqrt{(u^2+q^2)(a_{k+1}^2+b_{k+1}^2)} + a_{k+1}^2+b_{k+1}^2 > u^2+2ua_{k+1}+a_{k+1}^2+q^2+2qb_{k+1}+b_{k+1}^2$

$2\sqrt{(u^2+q^2)(a_{k+1}^2+b_{k+1}^2)} > 2(ua_{k+1}+qb_{k+1})$

$(u^2+q^2)(a_{k+1}^2+b_{k+1}^2) > (ua_{k+1}+qb_{k+1})^2$

Which is true by Cauchy-Schwarz.

Thus $\sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2} + ... + \sqrt{a_n^2 + b_n^2} \ge \sqrt{(a_1+a_2+...+a_n)^2 + (b_1+b_2+...+b_n)^2}$

Title: Re: TT's Maths Thread
Post by: /0 on November 25, 2009, 07:13:29 pm

Very nice TT,

For Q2,

$10^6-(10^6-1) = (\sqrt{10^6}-\sqrt{10^6-1})(\sqrt{10^6}+\sqrt{10^6-1})$

$\sqrt{10^6} - \sqrt{10^6-1} = \frac{1}{\sqrt{10^6}+\sqrt{10^6-1}}$

So our problem is reduced to finding the number closest to $\sqrt{10^6}+\sqrt{10^6-1}\approx 2\sqrt{10^6} = 2000$

I'm not very rigorous sorry lol...

Title: Re: TT's Maths Thread
Post by: Ahmad on November 25, 2009, 07:35:40 pm

I think it'd be great if we could have some combinatorics problems in here! Why? Simply put they require very little knowledge beyond knowing when to add, subtract, multiply and divide and they're a great way to improve at problem solving. (Only a suggestion, sorry if I'm ruining your thread TT :P)

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 25, 2009, 08:09:25 pm

Yeah. It's one of my favourite topics in problem solving, along with number theory (especially when mixed together with number theory). We had quite a few nice ones in the other threads as well as one somewhere earlier in the thread :)

I did this one recently and think it is really nice: "Prove that the number of ways of writing n as a sum of consecutive natural numbers is equal to the number of odd divisors(greater than 1) of n"

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 09:52:30 pm

Quote from: /0 on November 25, 2009, 07:13:29 pm

Very nice TT,

For Q2,

$10^6-(10^6-1) = (\sqrt{10^6}-\sqrt{10^6-1})(\sqrt{10^6}+\sqrt{10^6-1})$

$\sqrt{10^6} - \sqrt{10^6-1} = \frac{1}{\sqrt{10^6}+\sqrt{10^6-1}}$

So our problem is reduced to finding the number closest to $\sqrt{10^6}+\sqrt{10^6-1}\approx 2\sqrt{10^6} = 2000$

I'm not very rigorous sorry lol...

Thanks /0 xD

Quote from: Ahmad on November 25, 2009, 07:35:40 pm

I think it'd be great if we could have some combinatorics problems in here! Why? Simply put they require very little knowledge beyond knowing when to add, subtract, multiply and divide and they're a great way to improve at problem solving. (Only a suggestion, sorry if I'm ruining your thread TT :P)

Yeap, I'm moving onto combinatorics very soon, just gonna finish off the last exercise in algebra chapter and then combinatorics ftw!

Title: Re: TT's Maths Thread
Post by: zzdfa on November 25, 2009, 10:58:18 pm

Yeh Combinatorics is fun because there are lots of different ways to think about things. I like interpreting equations.

e.g. show that $\frac{\binom{2n}{n} \binom{2m}{m} }{\binom{n+m}{m}}$ is integer for all integer n,m.

and solution to kamil's problem:

suppose p*q=n , with q odd.

then $n=\underbrace{p+\cdots+p}_{\mathrm{q\,\,times}}$

do reverse gaussian pairing around the middle on the above, e.g:

2+2+2+2+2+2+2 -> -1+0+1+2+3+4+5

get rid of all the negatives in the equation by adding them to their positive counterparts : -1+0+1+2+3+4+5 -> 2+3+4+5

You can check that if $q\neq q'$ then we will get different sums:
denote the maximum number in the sum generated by q as m(q):
m(q)=n/q+(q-1)/2
m(q')=n/q'+(q'-1)/2

bit of algebra and we find that m(q)=m(q') iff q=q' or qq'=2n. but odd*odd=odd so the only choice is q=q'.

so we have that every odd divisor gives rise to a unique sum.

and i can't be bothered doing the other way (sum gives rise to unique odd divisor), it's pretty much the same anyway.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 11:00:03 pm

lol you guys make combinatorics sound so fun, I seriously can't wait now =.=

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 25, 2009, 11:26:34 pm

ah yeah very nice one. I did it by representing the integer as a rectangle, ie the product. Then defined a function for each rectangle that would change it into a desired shape. And of course, we had to prove this function is one to one and each rectangle and desired shape is accounted for and all those other details(a bijection). The picture helped me find the mapping, but yes I realised that to change it into something that I can communicate online, it would be best to make an argument such as yours, which turns out analogous in certain aspects.

Also, the reason why the factor 1 is excluded is because the mapping for such a rectangle will map a 1*n rectangle to itself, hence it will be a sum of "one consecutive number"

Title: Re: TT's Maths Thread
Post by: TrueTears on November 25, 2009, 11:27:25 pm

Let $x = \sqrt{10^6} + \sqrt{10^6-1}$

Thus $\frac{1}{x} = \sqrt{10^6} - \sqrt{10^6-1}$

We can approximate $1999 < x < 2000$ (Since $\sqrt{10^6-1} \approx 1000$ )

Reciprocate the inequality yields:

$\frac{1}{2000} < \frac{1}{x} < \frac{1}{1999}$

This means $n = 2000$ or $n = 1999$

Let's assume $n = 2000$

This means $\frac{1}{2000}$ is closer to $\frac{1}{x}$ than $\frac{1}{1999}$

Which implies: $\frac{1}{x} - \frac{1}{2000} < \frac{1}{1999} - \frac{1}{x}$

$\frac{2}{x} < \frac{3999}{(2000)(1999)}$

$\frac{1}{x} < \frac{3999}{(4000)(1999)}$

$\sqrt{10^6} - \sqrt{10^6-1} < \frac{3999}{(4000)(1999)}$

$10^6 - 2(10^3)\sqrt{10^6-1} + 10^6-1 < \left(\frac{3999}{(4000)(1999)}\right)^2$

$2\left[10^6 - 10^3\sqrt{10^6-1}\right] < \left(\frac{3999}{(4000)(1999)}\right)^2 + 1$

We can approximate $\left(\frac{3999}{(4000)(1999)}\right)^2 \approx \left(\frac{4000}{(4000)(2000)}\right)^2$

Which is almost negligible small. Almost $0$ .

Thus $10^6 - 10^3\sqrt{10^6-1} < \frac{1}{2}$

$10^6 - \frac{1}{2} < 10^3\sqrt{10^6-1}$

The RHS is almost $10^6$ (it is just a bit less)

While the LHS is smaller than $10^6$ by $\frac{1}{2}$

This means the inequality holds.

Therefore $\frac{1}{2000}$ is closer to $\frac{1}{x}$ than $\frac{1}{1999}$

So $n = 2000.$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2009, 12:06:17 am

~~1. Prove that $n! < \left(\frac{n+1}{2}\right)^n$ for $n = 2,3,4...$~~

~~2. Show that $\frac{1}{\sqrt{4n}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n}}$~~

3. Let $a_1, a_2 ... a_n$ be a sequence of positive numbers. Show that for all positive $x$

$(x+a_1)(x+a_2)...(x+a_n) \le \left(\frac{a_1+a_2+...+a_n}{n} + x\right)^n$

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 26, 2009, 01:19:29 am

consider the sequence:

1,2,3,4.....n

it's arithmetic mean is $\frac{\sum_{i=1}^{n}i}{n}=\frac{n+1}{2}$

It's geometric mean is: $(n!)^{\frac{1}{n}}$

GM<AM by AM-GM. From which the result follows.

I noticed this inequality has been used many times in this thread, I love this pic btw, saw it in a textbook(a better version tho :P )

(http://upload.wikimedia.org/wikipedia/commons/7/75/RMS-AM-GM-HM.gif)

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 26, 2009, 01:56:23 am

q3.)

consider the sequence of $(x+a_1), (x+a_2).... (x+a_n)$ do some AM-GM.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2009, 02:14:54 am

Quote from: kamil9876 on November 26, 2009, 01:19:29 am

consider the sequence:

1,2,3,4.....n

it's arithmetic mean is $\frac{\sum_{i=1}^{n}i}{n}=\frac{n+1}{2}$

It's geometric mean is: $(n!)^{\frac{1}{n}}$

GM<AM by AM-GM. From which the result follows.

I noticed this inequality has been used many times in this thread, I love this pic btw, saw it in a textbook(a better version tho :P )

(http://upload.wikimedia.org/wikipedia/commons/7/75/RMS-AM-GM-HM.gif)

haha that was so trivial lol

Yeah that pic is awesome, using similar triangles to derive AM-GM.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 26, 2009, 02:43:10 am

That picture is indeed cool!

Hint for 2: square everything then apply simple inequalities.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2009, 04:47:28 pm

Quote from: kamil9876 on November 26, 2009, 01:56:23 am

q3.)

consider the sequence of $(x+a_1), (x+a_2).... (x+a_n)$ do some AM-GM.

lol thanks I got it.

Using AM-GM on that sequence yields:

$\frac{(x+a_1)+(x+a_2)+(x+a_3)+...+(x+a_n)}{n} \ge \left[(x+a_1)(x+a_2)(x+a_3)...(x+a_n)\right]^{\frac{1}{n}}$

$\frac{xn + a_1+a_2+a_3+...+a_n}{n} \ge \left[(x+a_1)(x+a_2)(x+a_3)...(x+a_n)\right]^{\frac{1}{n}}$

$\left(x + \frac{a_1+a_2+a_3+...+a_n}{n}\right) \ge \left[(x+a_1)(x+a_2)(x+a_3)...(x+a_n)\right]^{\frac{1}{n}}$

$\left(x + \frac{a_1+a_2+a_3+...+a_n}{n}\right)^n \ge (x+a_1)(x+a_2)(x+a_3)...(x+a_n)$ as required.

Quote from: Ahmad on November 26, 2009, 02:43:10 am

That picture is indeed cool!

Hint for 2: square everything then apply simple inequalities.

Thanks Ahmad! I think I got it now :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2009, 10:05:45 pm

Induction can be used for Q 2 but it is very tedious and not very elegant...

Quote

Show that $\frac{1}{\sqrt{4n}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n}}$

For lower bound:

Base Case

Let $n = 2$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) \ge \frac{1}{\sqrt{8}}$

Inductive Hypothesis

Assume it is true for $n = k$

$\frac{1}{\sqrt{4k}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right)$ is true.

Inductive Step

Need to prove it is true for $n = k+1$

$\frac{1}{\sqrt{4(k+1)}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2(k+1)-1}{2(k+1)}\right)$

Multiply Inductive Hypothesis by $\left(\frac{2(k+1)-1}{2(k+1)}\right)$

$\left(\frac{2(k+1)-1}{2(k+1)}\right)\left(\frac{1}{\sqrt{4k}}\right) \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right)$

If we can show $\left(\frac{2(k+1)-1}{2(k+1)}\right)\left(\frac{1}{\sqrt{4k}}\right) > \frac{1}{\sqrt{4(k+1)}}$ Then our proof is complete

$\mbox{LHS} = \left(\frac{2k+1}{2k+2}\right)\left(\frac{1}{\sqrt{4k}}\right) = \left(\frac{1}{\sqrt{4(k+1)}}\right)\left[\frac{2(2k+1)}{\sqrt{k+1}}\right]\left(\frac{1}{\sqrt{4k}}\right)$

Thus $\mbox{LHS} > \mbox{RHS}$ (Since $k > 1$ )

$\therefore \frac{1}{\sqrt{4n}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right)$

For upper bound:

Base Case

Let $n = 2$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) < \frac{1}{2}$

Inductive Hypothesis:

Assume it is true for $n = k$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right) < \frac{1}{\sqrt{2k}}$ is true.

Inductive Step:

Need to prove it is true for $n = k+1$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2(k+1)-1}{2(k+1)}\right) < \frac{1}{\sqrt{2(k+1)}}$

Multiply Inductive Hypothesis by $\left(\frac{2(k+1)-1}{2(k+1)}\right)$

$\left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2k-1}{2k}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right) < \left(\frac{1}{\sqrt{2k}}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right)$

If we can show $\left(\frac{1}{\sqrt{2k}}\right)\left(\frac{2(k+1)-1}{2(k+1)}\right) < \frac{1}{\sqrt{2(k+1)}}$ then our proof is complete.

$\mbox{LHS} = \left(\frac{1}{\sqrt{2(k+1)}}\right)\left[\frac{\sqrt{2}(2k+1)}{2\sqrt{k+1}}\right]\left(\frac{1}{\sqrt{2k}}\right)$

Since $k > 1$ then $\mbox{LHS} < \mbox{RHS}$

$\therefore \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n}}$

Overall combining the lower and upper bounds we have proved: $\frac{1}{\sqrt{4n}} \le \left(\frac{1}{2}\right)\left(\frac{3}{4}\right) ...\left(\frac{2n-1}{2n}\right) < \frac{1}{\sqrt{2n}}$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2009, 10:25:22 pm

Alright last set of questions and then combinatorics time!

~~1. Let $a,b,c,d \ge 0$ prove that $\frac{1}{a}+\frac{1}{b} + \frac{4}{c}+\frac{16}{d} \ge \frac{64}{a+b+c+d}$~~

~~2. Let $x,y,z \ge 0$ with $xyz = 1$ . Find the minimum value of $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y}$~~

~~3. Let $a,b,c \ge 0$ . Prove that $\sqrt{3(a+b+c)} \ge \sqrt{a} + \sqrt{b} + \sqrt{c}$~~

Title: Re: TT's Maths Thread
Post by: /0 on November 26, 2009, 11:40:27 pm

2. Looks a lot like http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
But the constraint is different. Hmmm.

3.

Taking squares of both sides,

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

From am-gm

$\frac{a+b}{2} \geq \sqrt{ab}$

$\frac{b+c}{2} \geq \sqrt{bc}$

$\frac{a+c}{2} \geq \sqrt{ac}$

Adding:

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

Title: Re: TT's Maths Thread
Post by: Ahmad on November 26, 2009, 11:46:08 pm

It's also an application of CS,

$(1 + 1 + 1)(a + b + c) \ge (\sqrt{a} + \sqrt{b} + \sqrt{c})^2$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2009, 11:57:24 pm

Quote from: /0 on November 26, 2009, 11:40:27 pm

2. Looks a lot like http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
But the constraint is different. Hmmm.

3.

Taking squares of both sides,

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

From am-gm

$\frac{a+b}{2} \geq \sqrt{ab}$

$\frac{b+c}{2} \geq \sqrt{bc}$

$\frac{a+c}{2} \geq \sqrt{ac}$

Adding:

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

haha I actually just did that but I took the AM-GM of $a+b+c$ and compared it with $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

Title: Re: TT's Maths Thread
Post by: /0 on November 27, 2009, 12:03:41 am

Quote from: TrueTears on November 26, 2009, 11:57:24 pm

Quote from: /0 on November 26, 2009, 11:40:27 pm
2. Looks a lot like http://en.wikipedia.org/wiki/Nesbitt%27s_inequality
But the constraint is different. Hmmm.

3.

Taking squares of both sides,

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

From am-gm

$\frac{a+b}{2} \geq \sqrt{ab}$

$\frac{b+c}{2} \geq \sqrt{bc}$

$\frac{a+c}{2} \geq \sqrt{ac}$

Adding:

$a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$
haha I actually just did that but I took the AM-GM of $a+b+c$ and compared it with $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

$a+b+c \geq 3\sqrt[3]{abc}$

$\sqrt{ab}+\sqrt{bc}+\sqrt{ac} \geq 3\sqrt[3]{\sqrt{ab}\sqrt{bc}\sqrt{ac}} = 3\sqrt[3]{abc}$ ?

Anyway got a bit further with 2

$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$

If we use $x = \frac{1}{yz}$ on the denominator terms and simplify we get

$\frac{x}{xz+xy} +...$

So we need to minimize

$\frac{1}{y+z} + \frac{1}{x+z} + \frac{1}{x+y}$

~~Then we could say $t = x+y+z$ so we need to minimize $\frac{1}{t-x}+\frac{1}{t-y}+\frac{1}{t-z}$ .....~~ nevermind shit idea

hmm

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 12:13:34 am

I think I got Q 1

Some rearranging leads to:

$\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\right)\left(a+b+c+d\right) \ge (1+1+2+4)^2$

$\left[\left(\frac{1}{\sqrt{a}}\right)^2\left(\frac{1}{\sqrt{b}}\right)^2\left(\frac{2}{\sqrt{c}}\right)^2\left(\frac{4}{\sqrt{d}}\right)^2\right]\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2+\sqrt{d}^2\right) \ge (1+1+2+4)^2$

Which is true by Cauchy-Schwarz.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 27, 2009, 12:21:44 am

The only part of the constraint in 2 that's important is that it means none of x, y, z can be 0. So it really does reduce to Nesbitt's, which is well known and admits many proofs :)

Title: Re: TT's Maths Thread
Post by: /0 on November 27, 2009, 12:32:50 am

Quote from: Ahmad on November 27, 2009, 12:21:44 am

The only part of the constraint in 2 that's important is that it means none of x, y, z can be 0. So it really does reduce to Nesbitt's, which is well known and admits many proofs :)

Oh right indeed! *facepalm*
For some reason I thought you needed $x+y+z=1$ for nesbitt

Title: Re: TT's Maths Thread
Post by: Ahmad on November 27, 2009, 12:37:17 am

Nope, but you can assume that anyway since the inequality is homogeneous, so you might've seen people exploiting that. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 12:39:05 am

Quote from: Ahmad on November 27, 2009, 12:37:17 am

Nope, but you can assume that anyway since the inequality is homogeneous, so you might've seen people exploiting that. :)

What does it mean when you say the inequality is homogeneous? As in it's 'symmetrical'? =]

Title: Re: TT's Maths Thread
Post by: Ahmad on November 27, 2009, 01:27:57 am

Suppose the inequality is $f(x,y,z) \ge 0$ then it's homogeneous if $f(x,y,z) = f(kx,ky,kz)$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 01:29:07 am

Since I'm in a 'rearranging mood' (lolz) looking at $a+b+c \geq \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

becomes $a-\sqrt{ab}+b-\sqrt{bc}+c-\sqrt{ac} \ge 0$ Which is what we want to prove.

Let $x = a-\sqrt{ab}+b-\sqrt{bc}+c-\sqrt{ac}$

$\implies$ $x=a-\sqrt{ab}+b+b-\sqrt{bc}+c+c-\sqrt{ac}+a-b-c-a$

$x = a-2\sqrt{ab}+b+b-2\sqrt{bc}+c+c-2\sqrt{ac}+a-b-c-a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}$

$x = (\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2 - \left(a-\sqrt{ab}+b-\sqrt{bc}+c-\sqrt{ac}\right)$

$x = (\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2 - x$

$2x = (\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2$

$x = \frac{(\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2}{2}$

Since $(\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2 \ge 0$ $\implies \frac{(\sqrt{a}-\sqrt{b})^2 + (\sqrt{b}-\sqrt{c})^2 + (\sqrt{c}-\sqrt{a})^2}{2} \ge 0$

$\implies x \ge 0$

$\therefore a-\sqrt{ab}+b-\sqrt{bc}+c-\sqrt{ac} \ge 0$

No need for AM-GM :2funny:

Quote from: Ahmad on November 27, 2009, 01:27:57 am

Suppose the inequality is $f(x,y,z) \ge 0$ then it's homogeneous if $f(x,y,z) = f(kx,ky,kz)$ .

Ah right, thanks!

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 27, 2009, 01:37:23 am

quite kool TT. I noticed a lot of inequalities can be solved either by making those substituions I tried on an earlier Q, or by using the fact that $\sum_{i=0}^n x_i =0$ iff $x_i=0 \forall 1\leq i \leq n$ . Rather than AM-GM. In fact a proof of AM-GM for the simple case is done using the latter method.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 02:35:11 am

Alright very last inequality question. It's a USAMO question so I left it until the end ^.^

Prove that the zeros of $x^5+ax^4+bx^3+cx^2+dx+e = 0$ cannot all be real if $2a^2<5b$

Title: Re: TT's Maths Thread
Post by: Ahmad on November 27, 2009, 01:50:45 pm

Rather easy for a USAMO question!

Title: Re: TT's Maths Thread
Post by: /0 on November 27, 2009, 04:33:32 pm

Don't know, but substitute $x = y-\frac{a}{5}$ and the coefficient of the $y^3$ term is $\frac{5b-2a^2}{5}$
Or expanding roots might help

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 05:42:13 pm

I'm not sure if this is a rigorous proof but it includes a lot of wishful thinking.

Let the $5$ roots of $x^5+ax^4+bx^3+cx^2+dx+e$ be $q,w,r,t,u$ where $q,w,r,t,u \in \mathbb{R}$

$\therefore \mbox{By the fundamental theorem of algebra} \implies x^5+ax^4+bx^3+cx^2+dx+e = (x-q)(x-w)(x-r)(x-t)(x-u)$

Now equating coefficients for $x^4$ and $x^3$ coefficients yields:

$\frac{a_4}{a_5} = (-1)^{5-4}\left(\mbox{Sum of 5-4 different roots}\right)$

$\therefore a_4 = a = -(q+w+r+t+u)$

$\frac{a_3}{a_5} = (-1)^{5-3}\left(\mbox{Sum of 5-3 different roots}\right)$

$\therefore a_3 = b = (qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)$

Now we require $2a^2 - 5b < 0$

Let $y = 2a^2-5b$

$\implies y = 2(q+w+r+t+u)^2 - 5(qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)$

$\mbox{By the fundamental theorem of elementary symmetric polynomials} \implies q^2+w^2+r^2+t^2+u^2 = (q+w+r+t+u)^2 -2(qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)$

$\therefore (q+w+r+t+u)^2 = q^2+w^2+r^2+t^2+u^2 + 2(qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)$

$\implies y = 2(q^2+w^2+r^2+t^2+u^2) + 4(qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)- 5(qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)$

$\implies y = 2(q^2+w^2+r^2+t^2+u^2) -(qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)$

$\implies y = 2q^2+2w^2+2r^2+2t^2+2u^2 -qw-qr-qt-qu-wr-wt-wu-rt-ru-tu$

$\implies y = 2q^2-4qw+2w^2+2q^2-4qr+2r^2+2q^2-4qt+2t^2+2q^2-4qu+2u^2+2w^2-4wr+2r^2+2w^2-4wt+2t^2+2w^2-4wu+2u^2+2r^2-4rt+2t^2+2r^2-4ru+2u^2+2t^2-4tu+2u^2-6(q^2+w^2+r^2+t^2+u^2)+3(qw+qr+qt+qu+wr+wt+wu+rt+ru+tu)$

$\implies y = (\sqrt{2}q-\sqrt{2}w)^2+(\sqrt{2}q-\sqrt{2}r)^2+(\sqrt{2}q-\sqrt{2}t)^2+(\sqrt{2}q-\sqrt{2}u)^2+(\sqrt{2}w-\sqrt{2}r)^2+(\sqrt{2}w-\sqrt{2}t)^2+(\sqrt{2}w-\sqrt{2}u)^2+(\sqrt{2}r-\sqrt{2}t)^2+(\sqrt{2}r-\sqrt{2}u)^2+(\sqrt{2}t-\sqrt{2}u)^2-3y$

$\implies 4y = (\sqrt{2}q-\sqrt{2}w)^2+(\sqrt{2}q-\sqrt{2}r)^2+(\sqrt{2}q-\sqrt{2}t)^2+(\sqrt{2}q-\sqrt{2}u)^2+(\sqrt{2}w-\sqrt{2}r)^2+(\sqrt{2}w-\sqrt{2}t)^2+(\sqrt{2}w-\sqrt{2}u)^2+(\sqrt{2}r-\sqrt{2}t)^2+(\sqrt{2}r-\sqrt{2}u)^2+(\sqrt{2}t-\sqrt{2}u)^2$

$\implies y = \frac{(\sqrt{2}q-\sqrt{2}w)^2+(\sqrt{2}q-\sqrt{2}r)^2+(\sqrt{2}q-\sqrt{2}t)^2+(\sqrt{2}q-\sqrt{2}u)^2+(\sqrt{2}w-\sqrt{2}r)^2+(\sqrt{2}w-\sqrt{2}t)^2+(\sqrt{2}w-\sqrt{2}u)^2+(\sqrt{2}r-\sqrt{2}t)^2+(\sqrt{2}r-\sqrt{2}u)^2+(\sqrt{2}t-\sqrt{2}u)^2}{4}$

But $y = 2a^2-5b < 0$

Thus $\frac{(\sqrt{2}q-\sqrt{2}w)^2+(\sqrt{2}q-\sqrt{2}r)^2+(\sqrt{2}q-\sqrt{2}t)^2+(\sqrt{2}q-\sqrt{2}u)^2+(\sqrt{2}w-\sqrt{2}r)^2+(\sqrt{2}w-\sqrt{2}t)^2+(\sqrt{2}w-\sqrt{2}u)^2+(\sqrt{2}r-\sqrt{2}t)^2+(\sqrt{2}r-\sqrt{2}u)^2+(\sqrt{2}t-\sqrt{2}u)^2}{4} < 0$

However if $q,w,r,t,u \in \mathbb{R}$ it is not possible for $y < 0$

Contradiction!

Thus if $2a^2<5b$ the polynomial $x^5+ax^4+bx^3+cx^2+dx+e$ can not have all real roots.

Title: Re: TT's Maths Thread
Post by: /0 on November 27, 2009, 05:47:45 pm

lol very nice :D :D :D
Summation notation could make it more readable though lol

e.g. $\sum_{sym}^n$ for the nth symmetric sum or $\sum_{cyc} q^2$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 05:56:04 pm

Quote from: /0 on November 27, 2009, 05:47:45 pm

lol very nice :D :D :D
Summation notation could make it more readable though lol

e.g. $\sum_{sym}^n$ for the nth symmetric sum or $\sum_{cyc} q^2$

Oh nice, can you show me an example of using that summation notation? I haven't used it before haha

Title: Re: TT's Maths Thread
Post by: /0 on November 27, 2009, 06:12:28 pm

I'm just taking this from the art of problem solving wiki but

I think if you let $P(x) = x^5+ax^4...$

$\sum_{sym} P(x)= q+r+w+t+u$

$\sum_{sym}^2 P(x)= qr+qw+qt+qu+rw+rt+ru+wt+wu+tu$

$\sum_{sym}^3 P(x)= qrw+qrt+qru+qwt+qtu+...$
.
.
.

$\sum_{sym}^5 P(x)= qrwtu$

And cyclic sum cycles through all the variables, so like $\sum_{q,r,w,t,u} q^2 = q^2+r^2+w^2+t^2+u^2$ or $\sum_{cyc} q^2$ if the context is clear

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 06:15:12 pm

Quote from: /0 on November 27, 2009, 06:12:28 pm

I'm just taking this from the art of problem solving wiki but

I think if you let $P(x) = x^5+ax^4...$

$\sum_{sym} P(x)= q+r+w+t+u$

$\sum_{sym}^2 P(x)= qr+qw+qt+qu+rw+rt+ru+wt+wu+tu$

$\sum_{sym}^3 P(x)= qrw+qrt+qru+qwt+qtu+...$
.
.
.

$\sum_{sym}^5 P(x)= qrwtu$

And cyclic sum cycles through all the variables, so like $\sum_{q,r,w,t,u} q^2 = q^2+r^2+w^2+t^2+u^2$ or $\sum_{cyc} q^2$ if the context is clear

wow that's friggin awesome! Thanks!!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 09:01:57 pm

Just started combinatorics and I have a few general theory questions:

What is the number of different 9 letter 'words' we could make from the 9 letters "CHERNOBYL"

So this is just $\frac{9!}{(9-9)!}$

What is the number of different 3 letter 'words' we could make from the 9 letters "CHERNOBYL"

This would be $\frac{9!}{(9-3)!}$

What is the number of different 9 letter 'words' we could make from the 9 letters "RAMANUJAN"

This would be $\frac{9!}{3!2!}$ Since the A and N are indistinguishable so we must divide by the overcounting factor.

Now what if the question said: What is the number of different 3 letter 'words' we could make from the 9 letters "RAMANUJAN"

Do we still need to divide by the overcounting factor?

ie, the answer would be $\frac{\frac{9!}{(9-3)!}}{3!2!}$ or is that wrong?

Title: Re: TT's Maths Thread
Post by: /0 on November 27, 2009, 09:26:49 pm

I keep trying to prove it and I keep stuffing up, so I'll just say ~~yes~~ NO for now. That's edit: NOT right.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 09:30:51 pm

Yeah I think when you divide by the overcounting factor you are already taking into consideration the permutation 'formula'

So $\frac{9!}{3!2!}$ is really just $\frac{\frac{9!}{(9-9)!}}{3!2!}$ lolz

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 27, 2009, 10:42:55 pm

I don't think it's just a simple division.

Let me derive the C(n,k) formula from the P(n,k) just to show how to generally use an overcounting factor:

say we have n distinct elements, and we want to know how many different combinations of k objects we can choose. There are P(n,k) permutations.

now for clarity, suppose we have n=5, and k=3. and our objects a,b,c,d,e:

Here is a list of permuations in a rectangle:

abc,acb,bac,bca,cab,cba
abe,aeb,bae,bea,eab,eba
.
.
.

etc.

that is, imagine that we list all the permutations, such that all permutations that are the same combination, are in the same row. Obviously there are C(n,k) rows, and a total of P(n,k) elements. But how many in each row? well the answer is k! since there are k! ways of arranging a fixed combination of k elements. Therefore P(n,k)=C(n,k) * k!

====================================================================================

Back to TT's modification of the RAMANUJAN problem:

Recall that our proof above used a bijection between rows and combinations, we will try to do something similair here. suppose that the A's are A1,A2,A3 and the N's N1 and N2. So basically we want to count (N2)R(N1) and (N1)R(N2) as the same thing. Again, arrange the words in a rectangle:

(N1)R(N2),(N2)R(N1)
(N1)J(N2),(N2)J(N1)
(N1)M(N2),(N2)M(N1)

.
.
.
etc.

Now it is looking good so far, we have that 2! factor represented in the length of each row, however this pattern does not continue, as eventually once we start listing words such as AMA, our row length will be bigger!:

(A1)M(A2),(A2)M(A1),(A1)M(A3),(A3)M(A1),(A2)M(A3),(A3)M(A2)

Therefore, I think in order to solve this problem, you must break it up into different cases, since in some cases we get a bigger length in each row.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 27, 2009, 11:03:20 pm

Wow that proof up there for the relationship b/w C and P was ingenious especially with the reordering and regrouping!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 28, 2009, 02:28:32 am

Just some beginner combinatorics questions:

~~1. Prove $\left(^n_0\right)+\left(^n_1\right)+\left(^n_2\right)+...+\left(^n_n\right) = 2^n$~~

~~2. Which are there more of among the natural numbers between $1$ and $10^6$ : Numbers that can be represented as a sum of a perfect square and a (positive) perfect cube or numbers that can not be?~~

3. Two of the squares of a $7 \times 7$ checkerboard are painted yellow and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane of the board. How many inequivalent color schemes are possible?

Many thanks!

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 28, 2009, 03:02:11 am

1 is a classic.

A purely combinatorial and natural way:

the LHS is calculates how many different combinations of any size you can choose from a set of n objects. To prove this is $2^n$ you do as follows:

let the elements be a1,a2,a3....a_n. We can choose a1, or not; choose a2 or not; choose a3 or not etc. Because we have n choices to make, and for each choice we can pick two options, there are $2^n$ different ways. (you can represent this on a tree diagram to see)

Another way:

$(x+1)^n=^nC_0+^nC_1 x.... ^nC_n x^n$

now plug in x=1

I'll leave the next two more open ended:

2.) I'm leaning more towards not:

let $f: \mathbb{N}^3 \rightarrow \mathbb{N}, f((n,m,k))=n^2+m^2+k^3$

Try to look for the "leaps" in the range of f. ie: there may probably big gaps between two consecutive numbers in the range: just like for n^2 the sequence 1,4,9,16.... has leaps 3,5,7 etc. so there are less squares then non-squares. too late for me to think about details but I think this should be a good way :P

3.) Let the middle square have co-ordinates (0,0) and the one just to the right have co-ordinates (1,0) etc. (like a cartesian plane)

Every time you rotate by $90^o$ you generate a new scheme. So each set of equivalent schemes has up to 4 elements, no more. Some have up exactly 4, some less, it is your job to figure out how many etc. (e.g: if the two points are (x,y) and (-x,-y) then they by rotating once, you get a new equivalent scheme, but by rotating again you get the original scheme (since it has been rotated by 180 degrees in total)).
And yeah, similair story on this one cbf anymore.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 28, 2009, 04:02:22 am

The aim of this post isn't exactly for you to understand what I do, but more to get you excited about the idea of generating functions for counting problems, or just generating functions in general since they're one of my favourite objects. :)

A bit of notation first. If p(x) is a polynomial/taylor series, then [x^k] p(x) means the coefficient of x^k in p(x).

RAMANUJAN, that has 3As, 2Ns, 4 other distinct letters. From which it immediately follows that the answer is $3! [x^3] \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!}\right)\left(1 + \frac{x}{1!} + \frac{x}{2!}\right)\left(1 + \frac{x}{1!}\right)^4 = 151$

3. The index polynomial is $\frac{f_1^{49} + 2 f_1 f_4^{12} + f_1 f_2^{24}}{4}$ from which it follows that the answer is:

$[y^2] \frac{(1+y)^{49} + 2 (1+y) (1+y^4)^{12} + (1+y) (1+y^2)^{24}}{4} = 300$

Remark: The first problem is a simple application of this. The second is an application of Polya's Enumeration Theorem (Fundamental Theorem of Combinatorial Enumeration) which is extremely general, but can be solved even faster using a less general result of group theory called Burnside's lemma. Seems kind of abstract on those wiki pages but I assure you applying these results aren't.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 28, 2009, 03:23:50 pm

for q2 i thought it said two squares and a cube :uglystupid2:. But the original problem is so much simpler

Title: Re: TT's Maths Thread
Post by: TrueTears on November 28, 2009, 03:29:47 pm

Quote from: Ahmad on November 28, 2009, 04:02:22 am

The aim of this post isn't exactly for you to understand what I do, but more to get you excited about the idea of generating functions for counting problems, or just generating functions in general since they're one of my favourite objects. :)

A bit of notation first. If p(x) is a polynomial/taylor series, then [x^k] p(x) means the coefficient of x^k in p(x).

RAMANUJAN, that has 3As, 2Ns, 4 other distinct letters. From which it immediately follows that the answer is $3! [x^3] \left(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!}\right)\left(1 + \frac{x}{1!} + \frac{x}{2!}\right)\left(1 + \frac{x}{1!}\right)^4 = 151$

3. The index polynomial is $\frac{f_1^{49} + 2 f_1 f_4^{12} + f_1 f_2^{24}}{4}$ from which it follows that the answer is:

$[y^2] \frac{(1+y)^{49} + 2 (1+y) (1+y^4)^{12} + (1+y) (1+y^2)^{24}}{4} = 300$

Remark: The first problem is a simple application of this. The second is an application of Polya's Enumeration Theorem (Fundamental Theorem of Combinatorial Enumeration) which is extremely general, but can be solved even faster using a less general result of group theory called Burnside's lemma. Seems kind of abstract on those wiki pages but I assure you applying these results aren't.

Looks really interesting, especially how powerful generating functions are heh

Quote from: kamil9876 on November 28, 2009, 03:02:11 am

1 is a classic.

A purely combinatorial and natural way:

the LHS is calculates how many different combinations of any size you can choose from a set of n objects. To prove this is $2^n$ you do as follows:

let the elements be a1,a2,a3....a_n. We can choose a1, or not; choose a2 or not; choose a3 or not etc. Because we have n choices to make, and for each choice we can pick two options, there are $2^n$ different ways. (you can represent this on a tree diagram to see)

Another way:

$(x+1)^n=^nC_0+^nC_1 x.... ^nC_n x^n$

now plug in x=1

I'll leave the next two more open ended:

2.) I'm leaning more towards not:

let $f: \mathbb{N}^3 \rightarrow \mathbb{N}, f((n,m,k))=n^2+m^2+k^3$

Try to look for the "leaps" in the range of f. ie: there may probably big gaps between two consecutive numbers in the range: just like for n^2 the sequence 1,4,9,16.... has leaps 3,5,7 etc. so there are less squares then non-squares. too late for me to think about details but I think this should be a good way :P

3.) Let the middle square have co-ordinates (0,0) and the one just to the right have co-ordinates (1,0) etc. (like a cartesian plane)

Every time you rotate by $90^o$ you generate a new scheme. So each set of equivalent schemes has up to 4 elements, no more. Some have up exactly 4, some less, it is your job to figure out how many etc. (e.g: if the two points are (x,y) and (-x,-y) then they by rotating once, you get a new equivalent scheme, but by rotating again you get the original scheme (since it has been rotated by 180 degrees in total)).
And yeah, similair story on this one cbf anymore.

Thanks kamil I understand 1. now.

So for 2. I'm thinking there are $10^3$ squares between 1 and $10^6$ inclusive and $10^2$ cubes between 1 and $10^6$ inclusive, so how do you find how many can be represented as a sum of a square and a cube?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 28, 2009, 03:32:34 pm

Remember, you don't need to know how many, just need to know whether there are less than $\frac{10^6}{2}$ or greater than $\frac{10^6}{2}$ . So try find some upper or lower bound ;)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 28, 2009, 04:40:23 pm

Right thanks! I think I got it!

$\mbox{Let n be the number of perfect squares between}$ $1$ $\mbox{and}$ $10^6$

$\mbox{Let m be the number of perfect cubes between}$ $1$ $\mbox{and}$ $10^6$

$\implies n = 10^3$

$\implies m = 10^2$

$\mbox{Now let S be a number between}$ $1$ $\mbox{and}$ $10^6 \$ $\mbox{which can be expressed as the sum of a square and a cube, namely}$ $S = n^2+m^3$

$\mbox{Now we can also place a restriction on n and m to try to minimize the amount of squares and cubes which sums exceed}$ $10^6$

$\therefore n \in [1, 10^3]$ $\mbox{and}$ $m \in [1,10^2]$

$\mbox{Let this sum (with restrictions on n and m) be B}$ $\implies B = n^2+m^3$

$\mbox{Therefore}$ $|S| < |B|$

$|B| = |n \times m| = |n||m| = (10^3)(10^2) = 10^5$

$\therefore |S| < 10^5$

$\mbox{Although this is clearly an overestimate of the actual amount of numbers which satisfy S we can still use this upperbound}$

$\mbox{Now the numbers which can not be represented as a sum of a perfect square and a perfect cube would be}$ $10^6 - 10^5 = 10^5(10-1) = 9(10^5)$

$\mbox{Clearly}$ $10^5 < 9(10^5)$

$\mbox{So although we used an overestimate value it is still much smaller than the numbers which can not be represented as a sum of a perfect square and a perfect cube}$

$\mbox{Thus between}$ $1$ $\mbox{and}$ $10^6$ $\mbox{there are more numbers which can not be represented as a sum of a perfect square and a perfect cube.}$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 28, 2009, 08:26:03 pm

So for a $7 \times 7$ checkerboard there are $\binom{49}{2}$ ways of painting $2$ squares yellow. [In other words there are $\binom{49}{2}$ pairs of yellow squares]

Assume the $2$ squares which are painted are not symmetrical around the centre of the checkerboard then the starting position of the 'pair' of yellow squares can be rotated $4$ times $\implies 90 \to 180 \to 270 \to 360$ degrees. Thus this one pair produces four equivalent color schemes.

Now what if the starting pair of yellow squares is symmetrical around the centre of the checkerboard? Then you can first rotate it $90/270$ degrees to produce a new pair but if you spin it $180/360$ degrees you get the same as the starting pair. Therefore if the starting pair of yellow squares is symmetrical around the centre then it produces only $2$ equivalent color schemes.

So how many pairs are symmetrical around the centre? Well any single square has a corresponding symmetrical square (excluding the centre square).

$\therefore \frac{49-1}{2} = 24$ pairs of symmetrical yellow squares.

Thus there are $\binom{49}{2} - 24$ pairs of unsymmetrical yellow squares.

Now I'm a bit stuck I don't quite know how to link what I've got with finding "how many inequivalent color schemes are possible?"

Title: Re: TT's Maths Thread
Post by: Ahmad on November 28, 2009, 08:36:58 pm

You're really close. 24 symmetrical pairs -> what is the size of each equivalence class, and what does this tell you about the number of equivalence classes if you know there are 24 pairs. Same goes with the other scenario. Finally, what is the total number of equivalence classes? :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 28, 2009, 08:53:11 pm

Oh finally I get it!

Since there are $24$ symmetrical pairs but every $2$ pairs are equivalent it means there are $\frac{24}{2}$ different starting positions for each pair. Thus there are 12 inequivalent symmetrical pairs.

For the pairs which are not symmetrical each pair has $4$ equivalent pairs thus there are $\frac{\binom{49}{2} - 24}{4}$ different starting positions meaning there are $\frac{\binom{49}{2} - 24}{4}$ inequivalent unsymmetrical pairs.

In total we have $12 +$ $\frac{\binom{49}{2} - 24}{4}$ $= 300$ inequivalent pairs!

Title: Re: TT's Maths Thread
Post by: kenhung123 on November 28, 2009, 11:13:10 pm

Whats the significance of the equation in your siggy?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 28, 2009, 11:22:29 pm

It is a minimal counter-example to Euler's conjecture.

http://en.wikipedia.org/wiki/Euler%27s_sum_of_powers_conjecture

Title: Re: TT's Maths Thread
Post by: TrueTears on November 29, 2009, 01:33:35 am

As an exercise I've been trying to do a combinatorial proof of the multinomial theorem which is where the binomial theorem comes from. The 'proof' has some weird notation that I've 'made' up but hopefully someone can change it into something nice heh.

So we basically want to find a generalised 'formula' for $(x_1+x_2+...+x_q)^n$ (Where as the binomial theorem is just for $(x_1+x_2)^n$ , ie when $q = 2$ )

So let's start off by doing some experimentation. Let's look at $(x+y+z)^7$ . How do we expand this? Let's consider the first 2 brackets, namely:

$(x+y+z)(x+y+z)(x+y+z)^5 = (x \cdot x + xy + xz +yx + y \cdot y + yz + zx +zy+ z \cdot z)(x+y+z)^5$

Now leaving them unsimplified we can see that the 2 brackets expanded into 9 'unsimplified' terms. Which is expected since we have 3 choices from the first bracket and another 3 choices from the 2nd bracket.

This means that if we expanded all 7 brackets we would get a total of $3 \times 3 \times 3 ... \times 3 = 3^7$ unsimplified terms since there are 7 brackets and each bracket has 3 'choices'.

Now let's assume we have expanded all 7 brackets and we want to find the coefficient of the $x^4y^2z^1$ term.

We notice a few things: the powers all add up to 7 and we realise that this is just a direct application of the Mississippi 'formula'.

Just imagine we have a lot $x^4y^2z^1$ terms lying around to be collected as like terms after the expansion of the 7 brackets. However each would have a different permutation.

As we list some: $xxxxyyz, xyyxxxz ...$

Thus the total 'amount' of $x^4y^2z^1$ terms lying around would be $\frac{7!}{4!2!1!}$ .

Let's try another experiment, let's say we want to find how many of $x^2y^2z^3$ terms are lying around uncollected after the expansion of the 7 brackets.

We realise after undergoing the same process as before we get $\frac{7!}{2!2!3!}$ $x^2y^2z^3$ terms are lying around.

A pattern can be seen: The numerator is always 7! (As we expect since there are always 7 terms to permute).

The denominator's factorials are correspondent to the power of each term.

Therefore we can generalise this a bit and say the coefficient of any term in the expansion of $(x+y+z)^7$ is $\frac{7!}{p(x)!p(y)!p(z)!}$ where $p(x,y,z)$ denotes the power of $x,y,z$ respectively of the term.

Now that we have generalised the result for working out the coefficients of any term we need to generalise what $(x+y+z)^7$ will be expanded into.

Let's try work out how many terms $(x+y+z)^7$ when all like terms are collected.

However $(x+y+z)^7$ seems too tedious to work with, so let's try an easier example $(x+y+z)^3$

To work out how many different like-terms there are in total when $(x+y+z)^3$ is expanded let's consider what we discovered before with the exponents. We found that the exponent must add up to $\mbox{n}$ . So all exponents of the terms of $(x+y+z)^3$ when expanded must add up to 3.

How many different combinations can we get? Certainly there can be all the different permutations of $(0,0,3)$ , $(0,1,2)$ and $(1,1,1)$ .

Thus the total number of terms we should expect should be $\frac{3!}{2!1!} + 3! + 1 = 10$ Which when we expand $(x+y+z)^3$ we certainly do get 10 terms!

Now we can try to find a more general formula for the expansion of $(x+y+z)^3$

$\implies (x+y+z)^3 = \sum_{p_1,p_2,p_3}\left[\left(\frac{3!}{p_1!p_2!p_3!}\right)\left(x^{p_1}y^{p_2}z^{p_3}\right)\right]$

What exactly does this mean?

Basically it means that we take the summation of all permutations of non-negative integer indices $p_1$ through to $p_3$ such that $p_1+p_2+p_3 = 3$

$\therefore (x+y+z)^3 = \sum_{p_1,p_2,p_3}\left[\left(\frac{3!}{p_1!p_2!p_3!}\right)\left(x^{p_1}y^{p_2}z^{p_3}\right)\right] = \sum_{0,0,3}\left[\left(\frac{3!}{0!0!3!}\right)\left(x^{0}y^{0}z^{3}\right)\right] + \sum_{0,1,2}\left[\left(\frac{3!}{0!1!2!}\right)\left(x^{0}y^{1}z^{2}\right)\right] + \sum_{1,1,1}\left[\left(\frac{3!}{1!1!1!}\right)\left(x^{1}y^{1}z^{1}\right)\right]$

$= \left[\left(\frac{3!}{0!0!3!}\right)\left(x^{0}y^{0}z^{3}\right)\right] + \left[\left(\frac{3!}{0!0!3!}\right)\left(x^{3}y^{0}z^{0}\right)\right] + \left[\left(\frac{3!}{0!0!3!}\right)\left(x^{0}y^{3}z^{0}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{0}y^{1}z^{2}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{0}y^{2}z^{1}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{1}y^{0}z^{2}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{1}y^{2}z^{0}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{2}y^{1}z^{0}\right)\right]+ \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{2}y^{0}z^{1}\right)\right] + \left[\left(\frac{3!}{1!1!1!}\right)\left(x^{1}y^{1}z^{1}\right)\right]$

$= \left(\frac{3!}{0!0!3!}\right)\left[z^3+x^3+y^3\right] + \left(\frac{3!}{0!1!2!}\right)\left[yz^2+y^2z+xz^2+xy^2+x^2y+x^2z\right] + \left(\frac{3!}{1!1!1!}\right)\left[xyz\right]$

$= z^3+x^3+y^3 + 3(yz^2+y^2z+xz^2+xy^2+x^2y+x^2z) + 6xyz$

Now we are ready to play around with our general statement $(x_1+x_2+...+x_q)^n$

Using the same format: $(x_1+x_2+...+x_q)^n = \sum_{p_1,p_2...p_q}\left[\left(\frac{n!}{p_1!,p_2!...p_q!}\right)\left(x_1^{p_1}x_2^{p_2}...x_q^{p_q}\right)\right]$

Which basically means that we take the summation of all permutations of positive integer indices $p_1$ through to $p_q$ such that $p_1+p_2+...+p_q = 3$

Not a very rigorous proof, purely based on combinatorics essentially. So is this actually the multinomial theorem? And is that the right way to expand it? TBH the multinomial theorem seems rather 'useless' for expanding things but the multinomial coefficient I can see is very handy...

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 29, 2009, 01:38:39 am

yeah I wouldn't want to be calculating stuff with that anyway :P but still good, also good if you're interested in just a particular coefficient etc. plus good way to practice elementary combinatorics

Title: Re: TT's Maths Thread
Post by: zzdfa on November 29, 2009, 09:38:17 am

yep that's correct. you can use it to prove fermat's little theorem: $a^p \equiv a \mod p$ for prime p and any integer a,
by considering the multinomial expansion of $(\underbrace{1+\cdots+1}_{\mathrm{a \, times}})^p$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on November 29, 2009, 03:31:38 pm

Quote from: zzdfa on November 29, 2009, 09:38:17 am

yep that's correct. you can use it to prove fermat's little theorem: $a^p \equiv a \mod p$ for prime p and any integer a,
by considering the multinomial expansion of $(\underbrace{1+\cdots+1}_{\mathrm{a \, times}})^p$ .

Thanks zzdfa!

Haha Fermat's Little Theorem, how cute does that sound lolz

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 29, 2009, 05:04:04 pm

and how sentimental does "last" sound :'(

Title: Re: TT's Maths Thread
Post by: TrueTears on November 29, 2009, 11:38:47 pm

Just a few questions

1. Can someone show me a combinatorial proof of the hockey stick identity? $\binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+ \cdots + \binom{f}{r} = \binom{f+1}{r+1}$ . I've done it for a specific case say when $f = 5$ and $r = 2$ . But I don't know how to generalise it.

~~2. How many different ordered triples $(a,b,c)$ of non-negative integers are there such that $a+b+c=50$ ?~~

How would you do this question using partitions?

Thanks!

Title: Re: TT's Maths Thread
Post by: Ahmad on November 30, 2009, 01:13:46 am

Let me address finding the number of triples (a,b,c) of non-negative integers such that a + b + c = 5 instead of 50, as this will simplify my explanation.

This is a well known problem with a well known argument, which involves a little trick which seems somewhat new at least the first time you see it. The underlying principle is: if you're trying to count a set of objects which is hard to count, show that there is a one-to-one correspondence between this set of objects and a simpler set of objects which you can count, as both of the sets have the same number of elements you'll be done if you can count the latter set of objects.

Consider 7 objects in a row, o o o o o o o. Suppose I choose 2 of them in some way, and replace them by the symbol +. For example o + o + o o o. I can interpret this as "1 + 1 + 3" if i replace o by 1, and o o o by 3. This is convenient because 1 + 1 + 3 = 5, is a solution to our equation given by the triple (1,1,3). And a little thought shows that any way I select two objects to be replaced by + signs I get a solution to the equation, provided we interpret something like + + o o o o o as "0 + 0 + 5 = 5" which is the solution (0,0,5). Similarly, any solution to our equation gives rise to exactly one selection of two of the seven objects to be replaced by + signs. Hence we've shown that there is a one-to-one correspondence between solutions of the equation and selections of 2 objects from 7. Which means the number of such solutions is C(7, 2). And the same argument can be used to show the number of solutions to our original problem is C(50 + 2, 2) :)

This solution is slightly tricky to come up with. However there is a method to solve this problem routinely using the concept of generating functions, which I will not show.

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 30, 2009, 01:29:19 am

1.)

ok so we got a set of f+1 objects and we want to know how many different subsets of these there are that consist of r+1 objects. We can count these subsets like this: First consider all these subsets that contain the element 'a', now because we have chosen a, there are f objects left to choose from, and we must choose r of them. Hence there are (f,r) different ways of choosing this. Now we move on to the subsets that have b, but do not have a(since we have already counted these previously), We have chosen one object b, hence we are left with r objects left to choose from f-1 objects (since we cannot pick a or b for these remaining objects). This accounts for the the (f-1,r) term.
Now we move on to the subsets that contain c, but do not contain neither a or b. Having already chosen c, we have r objects left to choose from f-2 objects (since we are not choosing a,b or c for our remaining ones) this accounts for the (f-2,r) term.

Now keep applying this d,e,f... etc until you get f-k=r. In that case there would only be one set left, the set with objects z,y,w,x.... etc that you have not chosen.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 30, 2009, 01:42:15 am

Thanks Ahmad!

I noticed another less elegant way using partitions:

Let $a+b = q$

Then the equation $a+b+c=50$ becomes $q+c=50$

If we let $c = 0$ then $q = 50$ ... $[0]$

$c=1 \implies q = 49$ ... $[1]$

$c=2 \implies q = 48$ ... $[2]$

.
.
.

$c=50 \implies q = 0$ ... $[50]$

Let's look at when $c = 0$ and $q = 50$ specifically.

Once we fix in the value of $c$ we need to decide how many different pairs of $a$ and $b$ we can get to sum up to $50$ .

By listing some pairs of $(a,b)$ we get a feel of how many ordered pairs there are: $(0,50), (1,49) ... (50,0)$

$\therefore$ there are $51$ ways for $[0]$ .

Consider $[1]$ . Again listing pairs of $(a,b)$ we get: $(0,49), (1,48) ... (49,0)$

$\therefore$ there are $50$ ways for $[1]$

We see a pattern: the number of ways for each case is just $q+1$ .

Thus if we sum together all the partitions: $|[0]| + |[1]| + |[2]| + ... + |[50]| = 51+50+49+...+1 = \frac{(51)(52)}{2} = 1326$

Title: Re: TT's Maths Thread
Post by: TrueTears on November 30, 2009, 02:27:00 am

Quote from: kamil9876 on November 30, 2009, 01:29:19 am

1.)

ok so we got a set of f+1 objects and we want to know how many different subsets of these there are that consist of r+1 objects. We can count these subsets like this: First consider all these subsets that contain the element 'a', now because we have chosen a, there are f objects left to choose from, and we must choose r of them. Hence there are (f,r) different ways of choosing this. Now we move on to the subsets that have b, but do not have a(since we have already counted these previously), We have chosen one object b, hence we are left with r objects left to choose from f-1 objects (since we cannot pick a or b for these remaining objects). This accounts for the the (f-1,r) term.
Now we move on to the subsets that contain c, but do not contain neither a or b. Having already chosen c, we have r objects left to choose from f-2 objects (since we are not choosing a,b or c for our remaining ones) this accounts for the (f-2,r) term.

Now keep applying this d,e,f... etc until you get f-k=r. In that case there would only be one set left, the set with objects z,y,w,x.... etc that you have not chosen.

Oh right, thanks kamil, that was exactly the sort of combinatorial argument I was looking for :)

Title: Re: TT's Maths Thread
Post by: Ahmad on November 30, 2009, 03:46:24 am

I'm going to prove the result using the idea of generating functions. Hopefully this will be clear enough that you'll be able to understand it with very little background knowledge.

I've given some examples of generating functions, even though I've never told you exactly what they are. From the examples you might've picked up that they're some sort of polynomial or involve a polynomial type of expression. And since we're dealing with binomial coefficients it's natural to ask yourself where you've seen polynomials and binomial coefficients, with the obvious answer being in the binomial theorem! C(n, k) is the coefficient of x^k in (1+x)^n, which we write as C(n, k) = [x^k] (1 + x)^n.

Now that we have C(n, k) = [x^k] (1 + x)^n the problem becomes routine. We know that,

$\binom{r}{r}+\binom{r+1}{r}+\binom{r+2}{r}+ \cdots + \binom{f}{r}$

$= ([x^r] (1 + x)^r) + ([x^{r}] (1 + x)^{r+1}) + \cdots + ([x^{r}] (1 + x)^{f})$

$= [x^r] \left( (1+x)^r + \cdots + (1+x)^{f} \right)$

which we can easily sum, since it's a geometric sum

$= [x^r] x^{-1}\left( (1+x)^{f+1} - (1+x)^r \right)$

we can "absorb" the $x^{-1}$ by increasing the coefficient extraction index from r to r+1

$= [x^{r+1}] \left( (1+x)^{f+1} - (1+x)^r \right)$

$= [x^{r+1}](1+x)^{f+1} - [x^{r+1}](1+x)^r$

notice the second term here contributes nothing since $(1+x)^r$ has no $x^{r+1}$ term

$= \binom{f+1}{r+1}$

which was to be shown.

Why is this solution good?
- If you know anything about generating functions it's very simple to set up the first line of the argument
- Once you've set up the first line the rest is just mindless and simple algebra
- Adapts easily to lots of other identities (in fact there's a powerful algorithm involving generating functions which "simplifies" any combinatorial sum)

Why is this solution bad?
- It's lengthier than kamil's elegant proof
- It proves the correctness of the identity but it gives you no insight, it doesn't tell you "why" the result is true, kamil's solution does

Anyway I really ought to get to bed. Night :)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 30, 2009, 03:54:58 am

Wow even though I haven't learn generating functions yet I understood that splendid explanation! I can see why you like generating functions so much now :P

Title: Re: TT's Maths Thread
Post by: TrueTears on November 30, 2009, 05:05:00 pm

Another combinatorics question:

Do there exist at least 10,000 10 digit numbers divisible by 7, all of which can be obtained from one another by a reordering of the digits?

I'm just not quite sure how to link the divisible by 7 part with the reordering of the digits. I know how to find out how many different permutations exist for each 10 digit number but how to find one that is divisible by 7?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 30, 2009, 07:49:40 pm

yay thanks kamil for hint I got it ;)

Let the set which contains numbers that can be formed by reordering their digits be called $\mathbb{S}$ .

So for example, one type of $\mathbb{S}$ can contain $1,222,333,444$ or $1,333,222,444$ etc. This $\mathbb{S}$ contains a total of $\frac{10!}{3!3!3!}$ elements.

Another type of $\mathbb{S}$ can contain $1,111,122,222$ etc which would only have $\frac{10!}{5!5!}$ elements.

Therefore to count the number of multiples of $7$ between $1$ and $9,999,999,999$ would be denoted by $\left \lfloor \frac{9,999,999,999}{7} \right \rfloor$

Thus using the pigeon hole theorem, if we have $\left \lfloor \frac{9,999,999,999}{7} \right \rfloor$ many multiples of $7$ and $q$ groups of set $\mathbb{S}$ then at least one set $\mathbb{S}$ will contain $\left \lceil \frac{\left \lfloor \frac{9,999,999,999}{7} \right \rfloor}{q} \right \rceil$

Thus if we can find out how many $q$ there are such that $\left \lceil \frac{\left \lfloor \frac{9,999,999,999}{7} \right \rfloor}{q} \right \rceil \ge 10000$ then we have answered the question.

Consider using encoding to find out $q$ .

Let $a_i$ denote the amount of $i$ in a $10$ digit number such that $a_0+a_1+...+a_9 = 10$

So say for example, the number $1,222,333,444$ has $a_1 = 1$ , $a_2=a_3=a_4=3$ and the rest $a_0=a_5=a_6=...=a_9 = 0$ Now notice how $a_0+a_1+a_2+...+a_9 = 10$

Now we have created a bijection between $a_i$ and a certain set $\mathbb{S}$ which means if we can find out how many different solutions there are to $a_0+a_1+...+a_9 = 10$ then we have found out how many groups of $\mathbb{S}$ there are (which is equal to $q$ )

Now using a similar trick Ahmad used earlier:

Consider $o+o+o+o+o+o+o+o+o+o = 10$ $o$ 's.

That string above would represent the string $(a_0,a_1,...,a_9)$ such that $(a_0,a_1,...,a_9) = (1,1...1)$

Now if we move the $+$ sign somewhere say like this:

$++o+o+o+o+o+o+o+ooo$

That string above would represent the string $(a_0,a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9)$ $= (0,0,1,1,1,1,1,1,1,3)$

This means that by rearranging the $9$ $+$ signs we get a different set of solutions to $a_0+a_1+a_2+...+a_9 = 10$

So how many ways can we rearrange the $9$ $+$ signs?

Well there's $19$ different 'slots' to place $+$ and $o$ and we need to pick $9$ slots to place a $+$ which means the $o$ slots fall naturally into place.

Therefore there are $\binom{19}{9}$ ways of rearranging the $+$ signs.

And since we created a bijection between the different rearrangement of the strings and $q$

$\therefore q = \binom{19}{9}$

Subbing $q$ back in yields:

$\left \lceil \frac{\left \lfloor \frac{9,999,999,999}{7} \right \rfloor}{\binom{19}{9}} \right \rceil = 15465 > 10000$

So yes there do exist at least $10,000$ $10$ digit numbers divisible by $7$ , all of which can be obtained from one another by a reordering of the digits

Title: Re: TT's Maths Thread
Post by: TrueTears on November 30, 2009, 07:55:51 pm

Actually another way is just to use the balls in urn formula:

$a_0+a_1+a_2+...+a_9 = 10$

Since we have $10$ balls and we want to put them into $10$ urns labelled $a_0,a_1...a_9$

$\implies \binom{10+10-1}{10} = \binom{10+10-1}{10-1} = \binom{19}{9}$

But I still like the 'intuitive' way better :P

Title: Re: TT's Maths Thread
Post by: zzdfa on November 30, 2009, 08:14:04 pm

Quote from: TrueTears on November 30, 2009, 07:49:40 pm

Let the set which contains numbers that can be formed by reordering their digits be called $\mathbb{S}$ .

Just letting you know, you can call them equivalence classes, sounds pr0er: http://en.wikipedia.org/wiki/Equivalence_class

i.e. 'We call 2 numbers equivalent if they can be obtained by one another by reordering digits'
'we are counting the number of equivalence classes'

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 30, 2009, 09:08:08 pm

yep and more importantly it's a very precise way of describing it.

Title: Re: TT's Maths Thread
Post by: Over9000 on November 30, 2009, 11:41:12 pm

I've unanimously decided to take over this thread for my own personal benefit and facilitate my own self-discovery.
You got a problem with it? Then turn off your station!
Whata bout it?
Whata bout it?

Anyway, my first question is....

The reciprocal of 4 positive integers add up to $\frac{19}{20}$ . Three of these integrers are in ratios 1:2:3. What is the sum of the four integers.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 01, 2009, 12:21:16 am

I'm gonna do the same whole personal benefit thing too:

Let $A$ be a set consisting of n positive integers, prove that the set $\lbrace \frac{ab}{gcd(a,b)^2} | a,b \in A \rbrace$ contains at least n elements.

Title: Re: TT's Maths Thread
Post by: Ilovemathsmeth on December 01, 2009, 12:58:41 am

How do you write in bold - do you need to download some software?

Title: Re: TT's Maths Thread
Post by: kendraaaaa on December 01, 2009, 01:08:39 am

Quote from: Ilovemathsmeth on December 01, 2009, 12:58:41 am

How do you write in bold - do you need to download some software?

It's latex.

Title: Re: TT's Maths Thread
Post by: Ilovemathsmeth on December 01, 2009, 01:09:54 am

Is this the same as for writing in that amazing Maths font?

Title: Re: TT's Maths Thread
Post by: kendraaaaa on December 01, 2009, 01:14:37 am

This bold? or $this$ ?

Title: Re: TT's Maths Thread
Post by: Ilovemathsmeth on December 01, 2009, 01:15:53 am

a^2
\int_{-N}^{N} e^x\, dx

\int_{2}^{0} e^x\, dx

AHH HOW COME IT DIDN'T WORK??? =(

Title: Re: TT's Maths Thread
Post by: kendraaaaa on December 01, 2009, 01:18:16 am

put [ tex ]

and [ / tex]

without the spaces at the start and end

Title: Re: TT's Maths Thread
Post by: Ilovemathsmeth on December 01, 2009, 01:23:21 am

$a^2$
[tex]int_{-N}^{N} e^x, dx[tex]

[tex]int_{2}^{0} e^x, dx[tex]

Title: Re: TT's Maths Thread
Post by: Ilovemathsmeth on December 01, 2009, 01:23:50 am

Awesome, thanks heaps! =)

Is this what you do when you use the Maths font or have you downloaded LaTex?

I give up for the integral section.

Title: Re: TT's Maths Thread
Post by: kendraaaaa on December 01, 2009, 01:28:17 am

Nah you don't need to download anything. All the software is on the servers side.

http://latex.codecogs.com/editor.php

This will solve all your problems.

edit:pardon for the short off-topic tangent.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 01, 2009, 02:04:26 am

My problem: ~~I think I had a new Idea to use some geometry, i think it's gonna work/~~ Actually, induction on the number of prime factors involved is better.
over9000's problem: I got a messy diophantine equation :P and so cbf, my problem is better anyway oldfag

Title: Re: TT's Maths Thread
Post by: Over9000 on December 01, 2009, 03:46:46 am

Quote from: kamil9876 on December 01, 2009, 02:04:26 am

My problem: I think I had a new Idea to use some geometry, i think it's gonna work/
over9000's problem: I got a messy diophantine equation :P and so cbf, my problem is better anyway oldfag

fukn negger

Title: Re: TT's Maths Thread
Post by: Ilovemathsmeth on December 01, 2009, 09:55:28 am

Quote from: kendraaaaa on December 01, 2009, 01:28:17 am

Nah you don't need to download anything. All the software is on the servers side.

http://latex.codecogs.com/editor.php

This will solve all your problems.

edit:pardon for the short off-topic tangent.

Thank you so much!!! =D

Title: Re: TT's Maths Thread
Post by: Over9000 on December 01, 2009, 06:56:07 pm

solve my QUESTION !!!!!!!!!!!!!

Title: Re: TT's Maths Thread
Post by: dcc on December 01, 2009, 07:13:10 pm

$\boxed{\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{30} = \frac{19}{20}}$

The meaning of life, if you will.

Title: Re: TT's Maths Thread
Post by: /0 on December 01, 2009, 07:17:44 pm

$2+4+6+30 = 42$ !!!!!!!!!!!
:o

Title: Re: TT's Maths Thread
Post by: dcc on December 01, 2009, 07:21:01 pm

From the comment you left in my karma log, it is obvious that you've never read the Hitchhiker's Guide to the Galaxy.

Title: Re: TT's Maths Thread
Post by: /0 on December 01, 2009, 07:35:00 pm

lol I did a thesis on it in year 10 extension english

Title: Re: TT's Maths Thread
Post by: Over9000 on December 01, 2009, 11:22:49 pm

Quote from: dcc on December 01, 2009, 07:13:10 pm

$\boxed{\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{30} = \frac{19}{20}}$

The meaning of life, if you will.

HOW DID YOU DO IT???

did u guess?

Title: Re: TT's Maths Thread
Post by: kenhung123 on December 01, 2009, 11:29:55 pm

Gundam 00 is SOOOOOOOOHHHHHHHHH GOOOOOOOOOOODDDDDDDDDDDD I cleaned my room
My personal review of gundam 00: ITS FUKN GOOD
My calculator used to smell yummy, what happend to it?
Now it smells like shit and its dirty
So random!

Title: Re: TT's Maths Thread
Post by: Over9000 on December 01, 2009, 11:53:06 pm

Quote from: kenhung123 on December 01, 2009, 11:29:55 pm

Gundam 00 is SOOOOOOOOHHHHHHHHH GOOOOOOOOOOODDDDDDDDDDDD I cleaned my room
My personal review of gundam 00: ITS FUKN GOOD
My calculator used to smell yummy, what happend to it?
Now it smells like shit and its dirty
So random!

wtf, wtf r u on?? kuong?

Title: Re: TT's Maths Thread
Post by: Over9000 on December 03, 2009, 03:13:35 pm

Quote from: kamil9876 on December 01, 2009, 12:21:16 am

I'm gonna do the same whole personal benefit thing too:

Let $A$ be a set consisting of n positive integers, prove that the set $\lbrace \frac{ab}{gcd(a,b)^2} | a,b \in A \rbrace$ contains at least n elements.

QED.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 04, 2009, 09:52:42 pm

wdf gone for holidays for a week and my thread is filled with spam =.=

Title: Re: TT's Maths Thread
Post by: TrueTears on December 04, 2009, 10:04:17 pm

Quote from: TrueTears on November 30, 2009, 07:49:40 pm

yay thanks kamil for hint I got it ;)

Let the set which contains numbers that can be formed by reordering their digits be called $\mathbb{S}$ .

So for example, one type of $\mathbb{S}$ can contain $1,222,333,444$ or $1,333,222,444$ etc. This $\mathbb{S}$ contains a total of $\frac{10!}{3!3!3!}$ elements.

Another type of $\mathbb{S}$ can contain $1,111,122,222$ etc which would only have $\frac{10!}{5!5!}$ elements.

Therefore to count the number of multiples of $7$ between $1$ and $9,999,999,999$ would be denoted by $\left \lfloor \frac{9,999,999,999}{7} \right \rfloor$

Thus using the pigeon hole theorem, if we have $\left \lfloor \frac{9,999,999,999}{7} \right \rfloor$ many multiples of $7$ and $q$ groups of set $\mathbb{S}$ then at least one set $\mathbb{S}$ will contain $\left \lceil \frac{\left \lfloor \frac{9,999,999,999}{7} \right \rfloor}{q} \right \rceil$

Thus if we can find out how many $q$ there are such that $\left \lceil \frac{\left \lfloor \frac{9,999,999,999}{7} \right \rfloor}{q} \right \rceil \ge 10000$ then we have answered the question.

Consider using encoding to find out $q$ .

Let $a_i$ denote the amount of $i$ in a $10$ digit number such that $a_0+a_1+...+a_9 = 10$

So say for example, the number $1,222,333,444$ has $a_1 = 1$ , $a_2=a_3=a_4=3$ and the rest $a_0=a_5=a_6=...=a_9 = 0$ Now notice how $a_0+a_1+a_2+...+a_9 = 10$

Now we have created a bijection between $a_i$ and a certain set $\mathbb{S}$ which means if we can find out how many different solutions there are to $a_0+a_1+...+a_9 = 10$ then we have found out how many groups of $\mathbb{S}$ there are (which is equal to $q$ )

Now using a similar trick Ahmad used earlier:

Consider $o+o+o+o+o+o+o+o+o+o = 10$ $o$ 's.

That string above would represent the string $(a_0,a_1,...,a_9)$ such that $(a_0,a_1,...,a_9) = (1,1...1)$

Now if we move the $+$ sign somewhere say like this:

$++o+o+o+o+o+o+o+ooo$

That string above would represent the string $(a_0,a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9)$ $= (0,0,1,1,1,1,1,1,1,3)$

This means that by rearranging the $9$ $+$ signs we get a different set of solutions to $a_0+a_1+a_2+...+a_9 = 10$

So how many ways can we rearrange the $9$ $+$ signs?

Well there's $19$ different 'slots' to place $+$ and $o$ and we need to pick $9$ slots to place a $+$ which means the $o$ slots fall naturally into place.

Therefore there are $\binom{19}{9}$ ways of rearranging the $+$ signs.

And since we created a bijection between the different rearrangement of the strings and $q$

$\therefore q = \binom{19}{9}$

Subbing $q$ back in yields:

$\left \lceil \frac{\left \lfloor \frac{9,999,999,999}{7} \right \rfloor}{\binom{19}{9}} \right \rceil = 15465 > 10000$

So yes there do exist at least $10,000$ $10$ digit numbers divisible by $7$ , all of which can be obtained from one another by a reordering of the digits

Eh just relooking at my working is there a mistake somewhere? Cause I worked out the multiples of 7 from 1 to 9,999,999,999 but the question asked for multiples of 7 with 10 digits only...?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 04, 2009, 10:55:51 pm

Some new questions:

~~1. How many strictly increasing sequences of positive integers being with $1$ and end with $1000$ ?~~

2. For any set prove that the number of its subsets with an even number of elements is equal to the number of subsets with an odd number of elements. For example, the set $\{a,b,c\}$ has four subsets with an even number of elements (the null set has $0$ elements which is even), and four with an odd number of elements.

3. Use a combinatorial argument to show that $\forall \ \{n,m,k\} \in \mathbb{Z^+}$ with $\{n,m\} \ge k$

$\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$

4. We are given $\mbox{n}$ points arranged around a circle and the chords connecting each pair of points are drawn. If no three chords meet in a point, how many points of intersection are there? For example, when $n = 6$ , there are $15$ intersections.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 04, 2009, 11:18:14 pm

question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.

1.)
Clue:

the stricly increasing sequence 2,4,6,8...998 (with 1 and 1000 on the ends) can be represented as CNCNCNC...NCN

where C means chosen, N means not chosen.

2.)

You can turn this into an algebra problem, ie prove that:
$ \binom{n}{0}+\binom{n}{2}.....=\binom{n}{1}+\binom{n}{3}...$

which is equivalent to:

$\sum_0^n (-1)^{i+1}\binom{n}{i}=0$

4.)

Interesting Q. Try to look for some patterns. E.g: say you label your points like a clock: $A_1, A_2.... A_n$ . Now if you connect $A_1$ to $A_i$ then for that particular chord, you have (i-2) points on the right, and (n-i) on the left, now the only chords that will cross your chord will be the ones being connected to points on opposite sides, so now you have to work how many different such chords are there for each situation. Then do this for all pairs, and try to figure out what you have overcounted and how to fix it.

Title: Re: TT's Maths Thread
Post by: addikaye03 on December 04, 2009, 11:26:49 pm

http://vcenotes.com/forum/index.php/topic,20009.0.html

I asked some more Q here, if people would like to attempt them, they're interesting imo

Title: Re: TT's Maths Thread
Post by: TrueTears on December 04, 2009, 11:30:07 pm

Quote from: kamil9876 on December 04, 2009, 11:18:14 pm

question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.

Oh I see, but say if I worked out the actual "number" of 10 digit multiples of 7, it wouldn't really affect anything, just a smaller lower bound heh

Title: Re: TT's Maths Thread
Post by: TrueTears on December 04, 2009, 11:44:58 pm

Quote from: kamil9876 on December 04, 2009, 11:18:14 pm

1.)
Clue:

the stricly increasing sequence 2,4,6,8...998 (with 1 and 1000 on the ends) can be represented as CNCNCNC...NCN

where C means chosen, N means not chosen.

Oh yeah, I was trying to find out an encoding method but didn't get it. Thanks for that I got it now.

2 choices for each number.

Consider the sequence $1,2,3...999,1000$

That is encoded by $C,C,C...C,C$

Now say you have $1000$ slots for $1000$ numbers from 1 to $1000$ .

Since $1$ and $1000$ are fixed at both ends the number of increasing sequences is $2^{998}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 05, 2009, 12:28:34 am

Quote from: TrueTears on December 04, 2009, 11:30:07 pm

Quote from: kamil9876 on December 04, 2009, 11:18:14 pm
question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.
Oh I see, but say if I worked out the actual "number" of 10 digit multiples of 7, it wouldn't really affect anything, just a smaller lower bound heh

yeah but if it is too small, we cannot say at least 10000

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 12:33:04 am

Quote from: kamil9876 on December 05, 2009, 12:28:34 am

Quote from: TrueTears on December 04, 2009, 11:30:07 pm
Quote from: kamil9876 on December 04, 2009, 11:18:14 pm
question about multiples of 7: I think it assumes that say, 9 is really 000 000 000 9. I think if you included that smaller numerator then you wouldn't get a lower bound of 15456 but something smaller and hence you will not be able to conclude that it's greater than 10000.
Oh I see, but say if I worked out the actual "number" of 10 digit multiples of 7, it wouldn't really affect anything, just a smaller lower bound heh

yeah but if it is too small, we cannot say at least 10000

Exactly, that's why the answer would be "no".

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 05, 2009, 12:39:41 am

no, the conclusion woudl be "can't tell". (lower bound is too weak for pigeonholing)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 12:42:22 am

Quote from: kamil9876 on December 05, 2009, 12:39:41 am

no, the conclusion woudl be "can't tell". (lower bound is too weak for pigeonholing)

Oh I see then there is no way of doing the question then... unless you assume 0,000,000,009 is a number...?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 01:44:32 am

For Q 2 I thought about it this way.

Consider the set $\mbox{A} = \{a,b,c\}$

Let $\mbox{S}$ denote the subsets of set $\mbox{A}$ .

$\mbox{S} = \{\{\varnothing\}, \{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\}, \{b,c\}, \{a,b,c\}\}$

Partitioning $\mbox{S}$ into mutually disjoint sets that have $\mbox{i}$ elements where $\mbox{i} \in \{0,1,2,3\}$ . Let $Q_i$ denote this.

$\therefore \mbox{S} = \bigsqcup_{i=0}^3 = Q_0 \sqcup Q_1 \sqcup Q_2 \sqcup Q_3$

$\implies |\mbox{S}| = |Q_0| +|Q_1| + |Q_2| + |Q_3| = \binom{3}{0} + \binom{3}{1} + \binom{3}{2} + \binom{3}{3}$

Notice how the number of subsets with an even number of elements is $2^{3-1}$

Generalizing this result with the set $\mbox{A}$ with $\mbox{n}$ terms such that $\mbox{A} = \{a,b,c,d...n\}$

Then $|\mbox{A}| = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + ... + \binom{n}{n} = 2^n$

This means that the number of subsets with an even number of elements is $2^{n-1}$

Let $|A|$ and $|B|$ denote the cardinality of the number of subsets with an even and odd number of elements respectively.

Thus $2^{n-1} + |B| = 2^n$

$\therefore |B| = 2^{n-1}$

$\implies |A| = |B|$ which proves that for any set prove that the number of its subsets with an even number of elements is equal to the number of subsets with an odd number of elements.

I think this is a fairly weak proof as I didn't really show how $2^{n-1}$ was derived, I merely used a pattern found from the case $n = 3$

Can anyone actually show that for any set with $\mbox{n}$ elements, the number of subsets that contain an even number of elements is $2^{n-1}$ ? Thanks!

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 05, 2009, 01:50:39 am

Here is another proof for q2:

Suppose we want all the subsets of {a,b,c,d....}

now we will write a list of all the subsets of the set. On the top row, we will write down all the subsets that do not contain a, and on the bottom we will write down the same subset but with a added in:

null, {b}, {c}.... {b,c}, {b,d}.... {b,c,d...}
{a},{ab},{ac}....{a,b,c},{a,b,d}... {a,b,c,d..}

We can see, this mapping is a bijection. However the bottom row contains one more element (a is introduced) thus two subsets in the same COLUMN have opposite parity, proving the result.

This picture can also be used to prove that the number of subsets of a set is $2^n$ (ie the top row contains the subsets of the set with one less element, and introducing the bottom row doubles the number of subsets thus showing $S_n=2S_{n-1}$ where $S_n$ is the number of subsets of a set with n elements)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 02:17:41 am

Quote from: kamil9876 on December 05, 2009, 01:50:39 am

Here is another proof for q2:

Suppose we want all the subsets of {a,b,c,d....}

now we will write a list of all the subsets of the set. On the top row, we will write down all the subsets that do not contain a, and on the bottom we will write down the same subset but with a added in:

null, {b}, {c}.... {b,c}, {b,d}.... {b,c,d...}
{a},{ab},{ac}....{a,b,c},{a,b,d}... {a,b,c,d..}

We can see, this mapping is a bijection. However the bottom row contains one more element (a is introduced) thus two subsets in the same COLUMN have opposite parity, proving the result.

This picture can also be used to prove that the number of subsets of a set is $2^n$ (ie the top row contains the subsets of the set with one less element, and introducing the bottom row doubles the number of subsets thus showing $S_n=2S_{n-1}$ where $S_n$ is the number of subsets of a set with n elements)

Ahh that's very smart and answers why it is $2^{n-1}$ quite well. Thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 03:38:42 am

Quote

3. Use a combinatorial argument to show that $\forall \ \{n,m,k\} \in \mathbb{Z^+}$ with $\{n,m\} \ge k$

$\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$

I think I got the combinatorial proof but if anyone can show the algebraic proof it would be very much appreciated :)

My working is as follows:

Consider picking $\mbox{k}$ fruits from $\mbox{n}$ apples and $\mbox{m}$ oranges. (Assume that each apple and orange is distinguishable)

Suppose the following:

If we pick $0$ apples then we must pick $\mbox{k}$ oranges.

If we pick $1$ apple then we must pick $k-1$ oranges.

If we pick $2$ apples then we must pick $k-2$ oranges.

.
.
.

If we pick $\mbox{k}$ apples then we must pick $0$ oranges.

Summing up the different cases we get:

$\binom{n+m}{k} = \binom{n}{0}\binom{m}{k} + \binom{n}{1}\binom{m}{k-1} + \binom{n}{2}\binom{m}{k-2} + \cdots + \binom{n}{k}\binom{m}{0}$

But $\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n}{0}\binom{m}{k} + \binom{n}{1}\binom{m}{k-1} + \binom{n}{2}\binom{m}{k-2} + \cdots + \binom{n}{k}\binom{m}{0}$

$\therefore$ $\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$

So yeah if anyone can explain an algebraic method of doing it, that would be good :)

Title: Re: TT's Maths Thread
Post by: addikaye03 on December 05, 2009, 12:31:46 pm

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

Title: Re: TT's Maths Thread
Post by: Over9000 on December 05, 2009, 12:45:51 pm

Quote from: addikaye03 on December 05, 2009, 12:31:46 pm

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

Are u joking?
What part of TT'S Maths thread dont you understand

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 02:29:05 pm

Quote from: addikaye03 on December 05, 2009, 12:31:46 pm

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

Hey addikaye03 didn't you already advertise your questions in this thread just on the page before?

Quote from: addikaye03 on December 04, 2009, 11:26:49 pm

http://vcenotes.com/forum/index.php/topic,20009.0.html

I asked some more Q here, if people would like to attempt them, they're interesting imo

So I would appreciate it if you would stop spamming. Thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 02:52:56 pm

Quote from: TrueTears on December 04, 2009, 10:55:51 pm

4. We are given $\mbox{n}$ points arranged around a circle and the chords connecting each pair of points are drawn. If no three chords meet in a point, how many points of intersection are there? For example, when $n = 6$ , there are $15$ intersections.

(http://img708.imageshack.us/img708/1277/circle.jpg)
Looking at when $n = 3$ there are no intersections.

When $n = 4$ , there is only one intersection. At the intersection point 1, it is formed by the 4 points a,b,c,d.

When $n = 5$ , there are five intersections.

1 is formed by e,a,b,c.

2 is formed by d,a,b,c.

3 is formed by e,b,c,d.

4 is formed by e,a,c,d.

5 is formed by e,a,b,d.

We can see that every intersection is formed by 4 different points. For $n = 5$ , there are a total of $\binom{5}{4}$ combinations of a,b,c,d,e.

$\therefore$ for $\mbox{n}$ points, there are $\binom{n}{4}$ intersections.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 03:12:39 pm

A few more interesting Q's.

1. In an office, at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day and the boss delivers them in the order 1,2,3,4,5,6,7,8,9. While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based upon the information given, how many such after-lunch typing orders are possible? (There are no letters left to typed is one of the possibilities.)

2. Let $\mathbb{P}_n$ be the set of subsets of $\{1,2, \cdots, n\}$ . Let $c(n,m)$ be the number of functions $f : \mathbb{P}_n \to \{1,2, \cdots, m\}$ such that $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$ . Prove that $c(n,m) = \sum_{j=1}^m j^n$ .

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 03:28:24 pm

Hey, with the 2nd question, I think you made a mistake somewhere, you have c as a function of n and m but I don't see an m anywhere.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 03:34:27 pm

Hmm, I rechecked it just then, I typed the exact thing from art n craft :P

TBH I don't really understand the notation used in this question, what does $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$ mean?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 03:42:05 pm

I found the error, it should be f: Pn -> {1,2,....,m}

it means that the functions we are counting have to satisfy the property that:

if you have two sets A and B,

and you apply f to the intersection of them,

then the result is the same as if you

applied f to both of them seperately,
then picked the smallest output.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 03:52:04 pm

Quote from: zzdfa on December 05, 2009, 03:42:05 pm

I found the error, it should be f: Pn -> {1,2,....,m}

it means that the functions we are counting have to satisfy the property that:

if you have two sets A and B,

and you apply f to the intersection of them,

then the result is the same as if you

applied f to both of them seperately,
then picked the smallest output.

Oh so the question should read:

Let $\mathbb{P}_n$ be the set of subsets of $\{1,2, \cdots, m\}$ . Let $c(n,m)$ be the number of functions $f : \mathbb{P}_n \to \{1,2, \cdots, m\}$ such that $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$ . Prove that $c(n,m) = \sum_{j=1}^m j^n$ .

Is that right?

Title: Re: TT's Maths Thread
Post by: addikaye03 on December 05, 2009, 04:09:45 pm

Quote from: Over9000 on December 05, 2009, 12:45:51 pm

Quote from: addikaye03 on December 05, 2009, 12:31:46 pm
1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

Are u joking?
What part of TT'S Maths thread dont you understand

All 3 Questions can be found within 'extension exercises' in 'Terry Lee MATHEMATICS Extension 2 HSC', i don't see what the big deal is:

Q1. Complex Numbers (Included in HSC+VCE course)
Q2. Integration (Included in HSC+VCE course)
Q3. Mechanics (Included in HSC+VCE? course)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 04:12:09 pm

Quote from: addikaye03 on December 05, 2009, 12:31:46 pm

1) When two resistors of resistances z1 and z2 are connected in series, their equivalent resistance z is given by z=z1+z2, and when they are connected in parallel their equivalent resistance z is given by 1/z=1/z1+1/z2.

In an electric circuit that contains two resistors whose resistances are z1=R1+iwL and z2=R2-i/(wC) respectively, find the value of w so that their equivalent resistance is purely real when their resistors are

i) Connected in Series
ii)Connected in Parallel

2)Decompose the integrand into partial fractions with complex linear denominators, hence, find

int. dx/(x^8-1)

* i got the answer with plenty of tedious algebra + noting (x^4-1)(x^4+1) Hence roots of unity of (x^8-1), but want a quicker way*

3) A car of mass M kg, width w metres and centre of mass h metres above the ground travels round a level curve of radius r metres with a constant speed v m/s such that driver's right-hand side is near centre of the curve.

a) By drawing the front view of the car and two normal reactions of the road's surface on the right and the left wheels (neglect length of car), show that the right and left normal reactions respectively are:

M/2rw(rgw-2hv^2) and M/2rw(rgw+2hv^2)

b) hence, show that the car overturns when v^2>= rgw/2h

Quote from: addikaye03 on December 04, 2009, 11:26:49 pm

http://vcenotes.com/forum/index.php/topic,20009.0.html

I asked some more Q here, if people would like to attempt them, they're interesting imo

There is no big deal, I love to see different types of questions, but it's just that you have already posted the questions and you posted it again. There's no need to bump something in such a short period.

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 04:14:44 pm

nonono, it should be

Quote from: TrueTears on December 05, 2009, 03:52:04 pm

Quote from: zzdfa on December 05, 2009, 03:42:05 pm
I found the error, it should be f: Pn -> {1,2,....,m}

it means that the functions we are counting have to satisfy the property that:

if you have two sets A and B,

and you apply f to the intersection of them,

then the result is the same as if you

applied f to both of them seperately,
then picked the smallest output.
Oh so the question should read:

Let $\mathbb{P}_n$ be the set of subsets of $\{1,2, \cdots, \mathbf{n}]\}$ . Let $c(n,m)$ be the number of functions $f : \mathbb{P}_n \to \{1,2, \cdots, m\}$ such that $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$ . Prove that $c(n,m) = \sum_{j=1}^m j^n$ .

Is that right?

or else the answer wouldn't depend on n.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 04:16:56 pm

Quote from: zzdfa on December 05, 2009, 04:14:44 pm

nonono, it should be

Quote from: TrueTears on December 05, 2009, 03:52:04 pm
Quote from: zzdfa on December 05, 2009, 03:42:05 pm
I found the error, it should be f: Pn -> {1,2,....,m}

it means that the functions we are counting have to satisfy the property that:

if you have two sets A and B,

and you apply f to the intersection of them,

then the result is the same as if you

applied f to both of them seperately,
then picked the smallest output.
Oh so the question should read:

Let $\mathbb{P}_n$ be the set of subsets of $\{1,2, \cdots, \mathbf{n}]\}$ . Let $c(n,m)$ be the number of functions $f : \mathbb{P}_n \to \{1,2, \cdots, m\}$ such that $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$ . Prove that $c(n,m) = \sum_{j=1}^m j^n$ .

Is that right?

or else the answer wouldn't depend on n.

Oh, then what does $f : \mathbb{P}_n \to \{1,2, \cdots, m\}$ mean?

I get the $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$ bit now :P Thanks

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 04:19:25 pm

f: A -> B just means a function called f that takes something from the set A and outputs something from the set B.
In this case, the set A is P_n and the set B is {1,2,....m}.

so this particular function takes in sets like {1,3,5} and outputs a number between 1 and m.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 04:23:26 pm

Quote from: zzdfa on December 05, 2009, 04:19:25 pm

f: A -> B just means a function called f that takes something from the set A and outputs something from the set B.
In this case the elements of P_n are sets as well so it can be a little confusing.
so this particular function takes in sets like {1,3,5} and outputs a number between 1 and m.

Oh I see, thanks for that.

lol I don't even know where to start with this question. Completely stomped =(

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 04:31:34 pm

It looks like it can be done with induction.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 04:41:57 pm

So let's say $n = 3$ and $m = 4$

Then $\mathbb{P}_n = \{\{\varnothing\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$

$\implies$ $f : \mathbb{P}_n \to \{1,2,3,4\}$

So what's set A and set B? And what's $c(n,m)$ ?

So can set A and B be anything?

If so then let $A = \{1,2,3,4,5\}$ and $B = \{2,3,7,8\}$

Then $A \cap B = \{2,3\}$

Then what is $f(A \cap B)$ ?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 05:04:17 pm

edited my earlier post to clarify a bit.

so lets say that for some function f:P_3 -> {1,2,3,4}, f( {1,2} )= 2 and f( {2,3} )=3 and f( {2} )=4. Then this function is NOT counted because it doesn't satisfy the condition.

We want to know how many functions from P_3 to {1,2,3,4} DO satisfy the condition.

As an example, here are some easier questions:

how many quadratic functions f:R->R? infinite
how many quadratic functions f:R->R with zeroes at x=3 and x=5? infinite
how many quadratic functions f:R->R with zeroes at x=3 and x=5 and coefficient of x^2 is 1? one

how many functions are there f: {1,2,3} -> {1,2,3}? what about {1,2,....n} ->{1,2,.....,m}?

what about the number of functions from P_n to {1,2,...,m}?

...

...

what about the number of functions from P_n to {1,2,....,m} with the property that f(A int B) = min{fA,fB}?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 05:08:59 pm

A,B are arbitrary elements from P_n.

Quote from: TrueTears on December 05, 2009, 04:41:57 pm

So let's say $n = 3$ and $m = 4$

Then $\mathbb{P}_n = \{\{\varnothing\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$

$\implies$ $f : \mathbb{P}_n \to \{1,2,3,4\}$

So what's set A and set B? And what's $c(n,m)$ ?

So can set A and B be anything?

If so then let $A = \{1,2,3,4,5\}$ and $B = \{2,3,7,8\}$

Then $A \cap B = \{2,3\}$

Then what is $f(A \cap B)$ ?

pretend you had a function that sent {1,2,3,4,5} to 4 and sent {2,3,7,8} to 3. Then {2,3} has to be sent to 3 because min(3,4)=3

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 06:47:46 pm

Quote from: zzdfa on December 05, 2009, 05:08:59 pm

A,B are arbitrary elements from P_n.

Quote from: TrueTears on December 05, 2009, 04:41:57 pm
So let's say $n = 3$ and $m = 4$

Then $\mathbb{P}_n = \{\{\varnothing\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}\}$

$\implies$ $f : \mathbb{P}_n \to \{1,2,3,4\}$

So what's set A and set B? And what's $c(n,m)$ ?

So can set A and B be anything?

If so then let $A = \{1,2,3,4,5\}$ and $B = \{2,3,7,8\}$

Then $A \cap B = \{2,3\}$

Then what is $f(A \cap B)$ ?

pretend you had a function that sent {1,2,3,4,5} to 4 and sent {2,3,7,8} to 3. Then {2,3} has to be sent to 3 because min(3,4)=3

Ahh thanks, I think I have an idea now.

Title: Re: TT's Maths Thread
Post by: zzdfa on December 05, 2009, 10:23:12 pm

this problem is driving me crazy, fuck

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 10:28:35 pm

Quote

Let $\mathbb{P}_n$ be the set of subsets of $\{1,2, \cdots, \mathbf{n}\}$ . Let $c(n,m)$ be the number of functions $f : \mathbb{P}_n \to \{1,2, \cdots, m\}$ such that $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$ . Prove that $c(n,m) = \sum_{j=1}^m j^n$ .

lol I finally get this question, thanks zzdfa and Ahmad for your help.

Consider a specific case when $n = 3$ and when $m = 2$

$\mathbb{P}_n$ can arranged sets that contain the same number of elements from the highest to the lowest:

$\{1,2,3\}$

$\{1,2\} , \{1,3\}, \{2,3\}$

$\{1\}, \{2\}, \{3\}$

$\{\varnothing\}$

Now our aim is to find all functions $f : \mathbb{P}_n \to \{1,2\}$ such that $f(A \cap B) = \mbox{min}\{f(A), f(B)\}$

Let $A = \{1,2,3\}$ and $B = \{1,2\}$

$f(\{1,2,3\} \cap \{1,2\}) = \mbox{min}\{f(\{1,2,3\}), f(\{1,2\}))$

Now assume $f(\{1,2,3\}) < f(\{1,2\})$

Which means that $f(\{1,2,3\} \cap \{1,2\}) = f(\{1,2\}) = f(\{1,2,3\})$ . Contradiction!

This means that $f(\{1,2,3\}) \ge f(\{1,2\})$

Or more specifically the first row of $\mathbb{P}_n$ must be larger than the 2nd row $\implies$ $f(\{1,2,3\}) \ge \left[f(\{1,2\}), f(\{1,3\}), f(\{2,3\})\right]$

This follows on that $\left[\{1,2\} , \{1,3\}, \{2,3\}\right] \ge \left[\{1\}, \{2\}, \{3\}\right] \ge \left[\{\varnothing\}\right]$

Following on with our specific case of $n = 3$ and $m = 2$

If $\{1,2,3\}$ was mapped to $\{1\}$

$\left[\{1,2\} , \{1,3\}, \{2,3\}\right] = \left[\{1\}, \{2\}, \{3\}\right] = \left[\{\varnothing\}\right] = 1$

Thus for $m = 1$ there is only $1$ function $f$ which satisfies this.

If $\{1,2,3\}$ was mapped to $\{2\}$

Then $\{1,2\} = 1,2 , \{1,3\} = 1,2 , \{2,3\} = 1,2$ . This means the 2nd row has $(2)^3$ choices.

What about the 3rd row?

Consider $f(\{1,2\} \cap \{1,3\}) = \mbox{min}(f(\{1,2\}), f(\{1,3\}))$

Say $f(\{1,2\}) = 1 \ \mbox{and} \ f(\{1,3\}) = 2 \implies f(\{1\}) = \mbox{min}(1, 2) = 1$

This means that the value of the 3rd row is determined as soon as the values of the 2nd row is determined.

So if $\{1,2,3\}$ was mapped to $\{2\}$ there exist $2^3$ different function $f$ .

So overall for $f : \mathbb{P}_n \to \{1,2\}$ there exist $1+2^3$ different function $f$ .

Now for a more general result:

Consider the set $\{1,2...n\}$ .

This set can be arranged in the following way:

$\{1,2...n\}$

$\{1,2...n-1\}, \{2,3...n-1,n\} ... \{1,2,4...n-1,n\} ...$

$\{1,2...n-2\}, \{1,3...n-1\} ...$

.
.
.

$\{1\}, \{2\}, \{3\}$

$\{\varnothing\}$

Now consider all the different mappings of $\mathbb{P}_n$ to be $\{1,2...m\}$

Consider $\{1,2...n\}$ mapped to 1.

Then there's 1 function.

If $\{1,2...n\}$ was mapped to 2. Then there would be $2^{\binom{n}{n-1}} = 2^n$ different functions.

.
.
.

If $\{1,2...n\}$ was mapped to $m$ . Then there would be $m^{n}$ different functions.

In total we have $c(n,m) = 1+2^n+...+m^n = \sum_{j=1}^m j^n$

Title: Re: TT's Maths Thread
Post by: Ahmad on December 05, 2009, 10:28:43 pm

You've been nerd sniped haha

(http://imgs.xkcd.com/comics/nerd_sniping.png)

Title: Re: TT's Maths Thread
Post by: Ahmad on December 05, 2009, 10:34:33 pm

Oh that was directed at zzdfa. Also, they say the day you become a mathematician is the day after the night you spend awake scratching your head (or probably pulling your hair out, by then) over a problem ;D

I might as well point out that a university friend of mine suffers deeply from that nerd snipe issue. If you give him an interesting problem he zones out completely to the point where if you call his name right to his face he can't hear you. Once I gave him a geometry problem and we didn't see him for a week at uni. After a week he came back and said that he was busy solving the problem. :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 05, 2009, 10:35:30 pm

Quote from: Ahmad on December 05, 2009, 10:34:33 pm

Oh that was directed at zzdfa. Also, they say the day you become a mathematician is the day after the night you spend awake scratching your head (or probably pulling your hair out, by then) over a problem ;D

I've spent my entire afternoon until now on this problem lol and I still couldn't do it without your help =(

Title: Re: TT's Maths Thread
Post by: Ahmad on December 05, 2009, 10:38:12 pm

Frustrating over problems is one of the most important parts of the learning process, so don't despair. You've been getting better! :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 12:04:58 am

^ Yeah true, Ahmah. I've had a few of those moments though I ussually try to plan to postpone any possible nerd sniping of myself for the holidays or some unbusy time of the year.

By request, an algebraic argument for:

Quote

$\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$

consider the $x^k$ term of $(x+1)^{n+m}$

this is simply $\binom{n+m}{k} x^k$ Hence the RHS.

Now another way to work this out is:

$(x+1)^{n+m}=(x+1)^n (x+1)^m$

$=(\binom{n}{0} x^n + \binom{n}{1}x^{n-1}...)(\binom{m}{0} x^m + \binom{m}{1}x^{m-1}...)$

But the only way to get a term that contributes to $x^k$ we get this by multiplying a $\binom{n}{j}x^j$ term from first factor by a $\binom{m}{k-j} x^{k-j}$ term from the second factor.

Then when we sum up and collect like terms etc, we get the LHS $\blacksquare$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 02:00:43 am

Wait so you have:

$(x+1)^{n+m} = \binom{n+m}{0}x^{n+m} + \binom{n+m}{1}x^{n+m-1} + \cdots + \binom{n+m}{n+m-1}x + 1$

$(x+1)^n(x+1)^m = \left[\binom{n}{0}x^n + \binom{n}{1}x^{n-1}+ \cdots + \binom{n}{n-1}x + 1\right]\left[\binom{m}{0}x^m + \binom{m}{1}x^{m-1}+ \cdots + \binom{m}{m-1}x + 1\right]$

Then what? How do you show that $\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$ identity from what you have?

Is it like this?

Another way to write $(x+1)^{n+m} = \binom{n+m}{0}x^{n+m} + \binom{n+m}{1}x^{n+m-1} + \cdots + \binom{n+m}{n+m-1}x + 1 = \binom{n+m}{n+m}x^{n+m} + \binom{n+m}{n+m-1}x^{n+m-1} + \cdots + \binom{n+m}{1}x + 1$

Another way to write $(x+1)^n(x+1)^m = \left[\binom{n}{0}x^n + \binom{n}{1}x^{n-1}+ \cdots + \binom{n}{n-1}x + 1\right]\left[\binom{m}{0}x^m + \binom{m}{1}x^{m-1}+ \cdots + \binom{m}{m-1}x + 1\right] = \left[\binom{n}{n}x^n + \binom{n}{n-1}x^{n-1}+ \cdots + \binom{n}{1}x + 1\right]\left[\binom{m}{m}x^m + \binom{m}{m-1}x^{m-1}+ \cdots + \binom{m}{1}x + 1\right]$

$\therefore$ the coefficient of any term, namely, $x^k$ , where $k \le n+m$ , in the expansion of $(x+1)^{n+m}$ is $\binom{n+m}{k}$

Expanding $(x+1)^n(x+1)^m =\binom{n}{n}\binom{m}{m} x^{m+n} + \left(\binom{n}{n}\binom{m}{m-1}+\binom{n}{n-1}\binom{m}{m}\right) x^{m+n-1}+ \left(\binom{n}{n}\binom{m}{m-2}+ \binom{n}{n-1}\binom{m}{m-1} + \binom{n}{n-2}\binom{m}{m}\right)x^{n+m-2} + ... + 1$

Which means any coefficient of $x^k$ , where $k \le n+m$ , in the expansion of $(x+1)^n(x+1)^m$ can be written in the form of $x^k\left(\binom{n}{k}\binom{m}{0}+\binom{n}{k-1}\binom{m}{1}+ ... + \binom{n}{2}\binom{m}{k-2}+\binom{n}{1}\binom{m}{k-1} + \binom{n}{0}\binom{m}{k}\right)$

Equating $x^k$ coefficients yields: $\left(\binom{n}{k}\binom{m}{0}+\binom{n}{k-1}\binom{m}{1}+ ... + \binom{n}{2}\binom{m}{k-2}+\binom{n}{1}\binom{m}{k-1} + \binom{n}{0}\binom{m}{k}\right) = \binom{n+m}{k}$

$\mbox{LHS} = \sum_{j=0}^k\binom{n}{j}\binom{m}{k-j}$

$\therefore \sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 02:19:09 am

yeah lol i shoulve used $\binom{n}{n} x^n$ instead of $\binom{n}{0} x^n$ to make the notation neater :P and more clear

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 02:20:37 am

Quote from: kamil9876 on December 06, 2009, 02:19:09 am

yeah lol i shoulve used $\binom{n}{n} x^n$ instead of $\binom{n}{0} x^n$ to make the notation neater :P and more clear

Haha, but that $(x+1)^{n+m}$ was the crux step, was that just from experience you knew that would lead to solving the problem?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 02:57:09 am

Quote from: TrueTears on December 06, 2009, 02:20:37 am

Quote from: kamil9876 on December 06, 2009, 02:19:09 am
yeah lol i shoulve used $\binom{n}{n} x^n$ instead of $\binom{n}{0} x^n$ to make the notation neater :P and more clear
Haha, but that $(x+1)^{n+m}$ was the crux step, was that just from experience you knew that would lead to solving the problem?

i looked at RHS first because it looks simple. So I knew that i needed the power of the expansion to be n+m, and the x^k term. Also the fact that n and m are "broken up" in the LHS expression, i figured we need to break it up somehow.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 04:02:15 am

Quote from: TrueTears on December 05, 2009, 03:12:39 pm

1. In an office, at various times during the day, the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's in-box. When there is time, the secretary takes the top letter off the pile and types it. There are nine letters to be typed during the day and the boss delivers them in the order 1,2,3,4,5,6,7,8,9. While leaving for lunch, the secretary tells a colleague that letter 8 has already been typed, but says nothing else about the morning's typing. The colleague wonders which of the nine letters remain to be typed after lunch and in what order they will be typed. Based upon the information given, how many such after-lunch typing orders are possible? (There are no letters left to typed is one of the possibilities.)

I think I've got this question after some thought.

The question can be split up into 2 scenarios. The letter 9 arrives before lunch break or it arrives after lunch break.

Let the case of it arriving before lunch break be $[A]$ .

Let the case of it arriving after lunch break be $[B]$ .

$[A]$

If letter 9 arrived before lunch then that means letter 8 have been typed and we are not sure whether letters of the set $\{1,2,3,4,5,6,7,9\}$ have been typed or not typed.

Some experimenting quickly leads to what we must count.

Say letter 1 arrives and the secretary immediately types it up, then letter 2 comes and the secretary does not have time to type it up, then letter 3,4,5,6,7 comes and she does not have the time to type them up until letter 8 comes which the secretary has time to type up and then finally letter 9 arrives which she does not type up.

This means after lunch she must type up the letter set $\{9,7,6,5,4,3,2\}$ (The letter numbers are placed from the top letter in the box down to the bottom letter in the box) which is a subset of $\{1,2,3,4,5,6,7,9\}$

Imagine another scenario, letter 1 comes and she types it up, letter 2 comes and she does not type it up. Letter 3, 4, 5 comes and she types all three up. Letter 6,7 come and she doesn't type up. 8 comes and she obviously types it up. 9 comes and she does not type it up.

This means after lunch she must type up the letter set $\{9,7,6,2\}$ (The letter numbers are placed from the top letter in the box down to the bottom letter in the box) which again is a subset of $\{1,2,3,4,5,6,7,9\}$

Now notice how the question also asks for the order of the typing, but the order of $\{9,7,6,5,4,3,2\}$ and $\{9,7,6,2\}$ is fixed as the secretary has no choice what to type since she types the letter on the top till the one on the very bottom.

The aim is clear: we need to find all possible subsets of $\{1,2,3,4,5,6,7,9\}$ which is found by $2^8$ .

$\therefore |[A]| = 2^8$

$[B]$

Now we ask what if the letter 9 comes after lunch?

Consider the set of letters that can be typed or not typed up before lunch in this case. It would be $\{1,2,3,4,5,6,7\}$

Now to be consistent with our experimentation in $[A]$ lets use the same example.

Letter 1 comes and she types it up, letter 2 comes and she does not type it up. Letter 3, 4, 5 comes and she types all three up. Letter 6,7 come and she doesn't type up. 8 comes and she obviously types it up. However this time letter 9 does not come, since it is still before lunch.

Now after the secretary comes back from lunch break she will have to type up $\{7,6,2\}$ (The letter numbers are placed from the top letter in the box down to the bottom letter in the box).

However this time the boss can give her the letter 9 at any time. Say she just returns from lunch break and her boss gives her letter 9, then she would have to type up $\{9,7,6,2\}$ .

If she already types up 7 and then her boss gives her letter 9, her typing order would now be $\{7,9,6,2\}$

There are 2 more possibilities of when her boss can give her letter 9 namely: $\{7,6,9,2\}$ and $\{7,6,2,9\}$

All of those orderings are from the very top letter in the box down to the bottom letter in the box.

Now we notice something, if she just returns from lunch break and her boss gives her letter 9, then she would have to type up $\{9,7,6,2\}$ , but the ordered group $\{9,7,6,2\}$ is already counted in $[A]$ which means we are only left with 3 ordered groups namely: $\{7,9,6,2\}$ , $\{7,6,9,2\}$ and $\{7,6,2,9\}$

Again we notice something, if the subset of $\{1,2,3,4,5,6,7\}$ contains 3 elements then there are 3 typing orders possible for that one set.

How many 3 element subsets are there for $\{1,2,3,4,5,6,7\}$ ? Well there are $\binom{7}{3}$ subsets which means there are a total of $\binom{7}{3}(3)$ typing orders for a subset with 3 elements.

Doing some more experimentation shows that for a 2 element subsets there are 2 typing orders for that set. Which means there are $\binom{7}{2}(2)$ typing orders for a subset with 2 elements.

Thus we require this summation:

$\sum_{j=1}^7 \left((j)\binom{7}{j}\right) = \sum_{j=1}^7 \frac{(j)(7!)}{(7-j)!j!} = \sum_{j=1}^7 \frac{(j)(7!)}{(7-j)!(j)(j-1)!} = \sum_{j=1}^7 \frac{(7!)}{(7-j)!(j-1)!} = \sum_{j=1}^7 \frac{(7)((6)!)}{(7-j)!(j-1)!} = 7 \sum_{j=1}^7 \binom{6}{7-j} = 7 \left(\binom{6}{6} + \binom{6}{5} + ... + \binom{6}{0}\right) = 7(2^6)$

(Notice we start from when $j = 1$ , ie when the subset of $\{1,2,3,4,5,6,7\}$ contains only 1 element, since if it contains 0 elements it is already counted in $[A]$ )

$\therefore |[B]| = 7(2^6)$

In total we have $|[A]| + |[B]| = 2^8 + 7(2^6) = 704$ typing orders!

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 04:06:53 am

Another set of interesting questions:

~~1. In how many ways can four squares, not all in the same row or column be selected from an 8-by-8 chessboard to form a rectangle?~~

2. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Find the probability that no two birch trees are next to one another.

~~3. 25 people sit around a circular table. Three of them are chosen randomly. What is the probability that at least two of the three are sitting next to one another?~~

Title: Re: TT's Maths Thread
Post by: /0 on December 06, 2009, 05:28:06 am

1.

Let an $m\times n$ be a rectangle with height m rows and width n columns.

The number of a particular rectangle can be easily counted by translating the original rectangle throughout the board.

For a 2x2 rectangle there are 7 possible horizontal and 7 possible vertical positions, in total 49.

2x2 rectangles: 7*7
2x3 rectangles: 6*7
2x4 rectangles: 5*7
...
2x8 rectangles: 1*7

3x2 rectangles: 7*6
3x3 rectangles: 6*6
...
3x8 rectangles: 1*6

So in total there are:

$7(7+6+5+4+3+2+1)+6(7+6+5+4+3+2+1)+...+1(7+6+5+4+3+2+1) = (7+6+5+4+3+2+1)^2=784$

(There must be exactly 2 squares in a row and 2 squares in a column. If there were 3 in the same row/column, then a rectangle would not be formed unless the 4th were in the same row/column, which is not allowed)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 12:11:31 pm

alternative to 1:

If we select two points $(x,y)$ and $(x',y')$ such that $x\neq x'$ and $y\neq y'$ (ie two points that are not in same row nor in the same column) then we have defined the two opposite corners of a rectangle, which already defines the rectangle. The total number of ways of choosing such corners is: $\frac{1}{2}*64*(64-8-8+1)=32*49$ ways of doing this. However each rectangle has two different pair of opposite corners, hence we halve that number to get $16*49$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 12:21:35 pm

3.) Sketch:

Let's try to find how many ways there are when none of them sit next to one another. If we choose one person, then we cannot choose the two people sitting next to him, hence we have 22 people left to choose. Then when we choose the next person, there are two possibilities that may occur:

1.)

**xCx****xCx**

C means chosen, * means unchosen and avaibalbe, x means unavaibable. In this case we have 19 people left for the third person to choose.

2.)

***xCxCx***

In this case, we have 20 people left to choose for the third person.

Now just work out how many different ways there are to choose the people, then subtract from the total number of ways of choosing people in this manner (23*24*25), and then divide by this total.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 03:20:55 pm

Thanks kamil, /0 =]

Title: Re: TT's Maths Thread
Post by: Ahmad on December 06, 2009, 03:50:17 pm

Cool way to do 1: $\binom{8}{2}^2 = 784$ (why?)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 05:38:09 pm

Quote from: TrueTears on December 06, 2009, 04:06:53 am

2. A gardener plants three maple trees, four oak trees and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Find the probability that no two birch trees are next to one another.

Got this one :)

Let M represent maple trees, O represent oak trees and B represent bitch trees.

Consider without restrictions how many ways we can permute 3 M, 4 O and 5 B

$\therefore \frac{12!}{3!4!5!}$

Now we need to consider the restriction that no two birch trees are next to one another.

Consider this arrangement:

_O_O_O_O_M_M_M_
1 2 3 4 5 6 7 8

There are 8 distinct slots to place B $\implies \binom{8}{5}$ ways to place B.

Now there are $\frac{7!}{4!3!}$ ways to arrangement the 4 O and 3 M in between.

$\therefore \binom{8}{5}\left(\frac{7!}{4!3!}\right)$ different arrangements with restriction.

$\therefore$ the probability is $\frac{\binom{8}{5}\left(\frac{7!}{4!3!}\right)}{\frac{12!}{3!4!5!}} = \frac{7}{99}$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 05:49:09 pm

Quote from: Ahmad on December 06, 2009, 03:50:17 pm

Cool way to do 1: $\binom{8}{2}^2 = 784$ (why?)

Hey Ahmad can you explain how you got that?

I'm thinking it's because you must pick 2 rows and there are 8 rows in total.

Then you must pick 2 columns and there are 8 columns in total.

But I'm not sure how it works still...

Title: Re: TT's Maths Thread
Post by: zzdfa on December 06, 2009, 06:08:37 pm

yea i deleted my post because i realized that it wasn't clear*

you were right.

draw a diagram. color in 2 rows and color in 2 columns. yay! (the rows and columns chosen determine a unique rectangle, look at their intersections)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 06:10:17 pm

Quote from: zzdfa on December 06, 2009, 06:08:37 pm

yea i deleted my post because i realized that it wasn't right lol.

you were right.

draw a diagram. color in 2 rows and color in 2 columns. yay!

Oh lol I thought you just color in 2 unit squares rather than a whole row and column, now I get it. I misinterpreted my own statement lol

But wow $\binom{8}{2}^2$ is really clever way.

Quote from: zzdfa on December 06, 2009, 06:08:37 pm

(the rows and columns chosen determine a unique rectangle, look at their intersections)

Yeah I finally got it now haha thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 06:37:47 pm

Quote from: kamil9876 on December 06, 2009, 12:21:35 pm

3.) Sketch:

Let's try to find how many ways there are when none of them sit next to one another. If we choose one person, then we cannot choose the two people sitting next to him, hence we have 22 people left to choose. Then when we choose the next person, there are two possibilities that may occur:

1.)

**xCx****xCx**

C means chosen, * means unchosen and avaibalbe, x means unavaibable. In this case we have 19 people left for the third person to choose.

2.)

***xCxCx***

In this case, we have 20 people left to choose for the third person.

Now just work out how many different ways there are to choose the people, then subtract from the total number of ways of choosing people in this manner (23*24*25), and then divide by this total.

I didn't do your way so I'm not sure if this way is right, can someone check?

There's 2 parts to this question.

2 adjacent people sitting next to other and 3 adjacent people sitting next to each other.

So imagine a circle and there are 25 slots (labelled from $\{1,2,...,25\}$ ) spaced out evenly around the circle.

For 3 adjacent people sitting next to each other we could have $\{1,2,3\}, \{2,3,4\}, \{3,4,5\} ... \{24,25,1\}, \{25,1,2\}$ . Which means if you pick a seat for one person to be seated then the rest 2 are determined $\implies \binom{25}{1} = 25$ choices for this case.

For 2 adjacent people sitting next to other:

Imagine we pick someone to sit down at 1. Then we can place the 2 other people at $\{3,4\}, \{4,5\} ... \{23,24\}$ . This means if we picked 1 person to sit then there are $(23-3+1)=21$ choices for the other 2 people to sit down. There are a total of 25 seats so there are a total of $25 \times 21$ choices for this case.

There are a total of $\binom{25}{3}$ ways of picking 3 people randomly from 25 people.

$\therefore$ probability is $\frac{25+25 \times 21}{\binom{25}{3}} = \frac{11}{46}$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 07:55:13 pm

Last set of questions:

~~1. How many subsets of the set $\{1,2,3...30\}$ have the property that the sum of the elements of the subset is greater than $232$ ?~~

2. Eight people are in a room. One or more of them get an ice-cream cone. One or more of them get a chocolate-chip cookie. In how many different ways can this happen, given that at least one person gets both an ice-cream cone and a chocolate-chip cookie?

3. Let $\mathbb{S}$ be a set with $\mbox{n}$ elements. In how many different ways can one select two not necessarily distinct or disjoint subsets of $\mathbb{S}$ so that the union of the two subsets is $\mathbb{S}$ ? The order of selection does not matter. For example, the pair of subsets $\{a,c\}, \{b,c,d,e,f\}$ represents the same selection as the pair $\{b,c,d,e,f\}, \{a,c\}$

I don't know why I can't get question 1. This is what I have:

The total sum of the set is $\frac{(30)(30+1)}{2}=465$

Half of this is $232.5$ .

How do I find how many subsets with elements that add up to $232$ though...?

For Q 2 is this right?

Assume only 1 person gets both an ice-cream and a CC-cookie.

This means the rest of the 7 people can only pick ice-cream or CC-cookie, so they have 2 choices.

Thus we have $\binom{8}{1}2^7$ ways for this case.

If we have 2 people getting both an ice-cream and a CC-cookie.

This means the rest of the 6 people can only pick ice-cream or CC-cookie, so they have 2 choices.

Thus we have $\binom{8}{2}2^6$ ways for this case.

Summing together all choices we get: $\binom{8}{1}2^7 + \binom{8}{2}2^6 + ... + \binom{8}{8}2^0=6305$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 08:50:41 pm

1.) Say we pick a subset A with sum greater than 232, then we know that the sum of the elements in the complement of A must be less than 232. Therefore half the sets have some greater than 232, so $2^{30} /2=2^{29}$

2.) Yeah it's right but I have an alternative: Each person can have 3 different meals, hence $3^8$ different choices in total. However we have to subtract from this the total number of ways of no one choosing the option of 'both ice cream and cookie', there are 2^8 ways of doing this. Hence $3^8-2^8$ . I think you can generalise this sort of 'counting' in both ways technique to prove many identities, or even play around with $(1+x)^n$ (ie think differentiation etc.)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 08:54:10 pm

Quote from: kamil9876 on December 06, 2009, 08:50:41 pm

2.) Yeah it's right but I have an alternative: Each person can have 3 different meals, hence $3^8$ different choices in total. However we have to subtract from this the total number of ways of no one choosing the option of 'both ice cream and cookie', there are 2^8 ways of doing this. Hence $3^8-2^8$ . I think you can generalise this sort of 'counting' in both ways technique to prove many identities, or even play around with $(1+x)^n$ (ie think differentiation etc.)

Ah yeah $3^8$ , damn I was wondering what was the total different choices, can't believe I missed that!

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 08:56:12 pm

Quote from: kamil9876 on December 06, 2009, 08:50:41 pm

1.) Say we pick a subset A with sum greater than 232, then we know that the sum of the elements in the complement of A must be less than 232. Therefore half the sets have some greater than 232, so $2^{30} /2=2^{29}$

Right but can you prove it? It seems right but I just don't feel at ease without a proof of why this is so :P

And how do you know half of the subsets of $\{1,2,3...30\}$ have a sum greater than 232?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 09:05:29 pm

Let S(A) denote the sum of the elements in A. Let A' denote the complement of A.

S( A)+S(A')=465

S(A)=465-S(A')

S(A)>232 iff 232<465-S(A') which means S(A')<233.

Let B be the set of subsets that have sum greater than 232, then we know that A is in B iff A' is in B'. Thus there is a bijection between B and B', hence equal elements, so it splits the total number of sets in half.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 09:13:03 pm

Quote from: kamil9876 on December 06, 2009, 09:05:29 pm

Let S(A) denote the sum of the elements in A. Let A' denote the complement of A.

S( A)+S(A')=465

S(A)=465-S(A')

S(A)>232 iff 232<465-S(A') which means S(A')<233.

Let B be the set of subsets that have sum greater than 232, then we know that A is in B iff A' is in B'. Thus there is a bijection between B and B', hence equal elements, so it splits the total number of sets in half.

Can this result be generalised such that every single set of the form $\{1,2...n\}$ has an equal amount of subsets whose sum is larger than $\frac{\frac{(n)(n+1)}{2}}{2}$ and whose sum is smaller than $\frac{\frac{(n)(n+1)}{2}}{2}$ ?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 09:15:34 pm

Yeah I think so, though check for the cases where the total sum is even, or total sum is odd, I think that .5 in 232.5 might be of a bit of significance. but yeah that's the main idea.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 09:55:54 pm

Quote from: kamil9876 on December 06, 2009, 09:05:29 pm

Let S(A) denote the sum of the elements in A. Let A' denote the complement of A.

S( A)+S(A')=465

S(A)=465-S(A')

S(A)>232 iff 232<465-S(A') which means S(A')<233.

Let B be the set of subsets that have sum greater than 232, then we know that A is in B iff A' is in B'. Thus there is a bijection between B and B', hence equal elements, so it splits the total number of sets in half.

So you are saying this right:

Let A be the subset of the set $\{1,2,3...30\}$ which contains elements whose sum is larger than 232. Denote this sum by S(A)

The lowest S(A) can be is 233, then S(A') = 232

If S(A) = 234 then S(A') = 231

If S(A) = 235 then S(A') = 230

If S(A) = 236 then S(A') = 229

.
.
.

If S(A) = 464 then S(A') = 1

If S(A) = 465 then S(A') = 0 [Here A' would be the $\varnothing$ set]

So by pairing like this we get every subset. And those are that larger than 232 is exactly half the total amount of the subsets.

But this brings up another question:

Can you prove that the numbers in this set $\{0,1,2,3...232,233...464,465\}$ can be obtained from the sum of the elements of subsets of the set $\{1,2,...30\}$ ?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 10:44:44 pm

I'm not saying that we pair up the actual sums, but that we pair up each set A with it's complement A':

A A'
B B'
C C'
.
.
.
.

And say the first column contains sets whose sums are greater than 232, then the second column contains ones that are smaller, hence the splitting in half.

So even if there didn't exist a set A such that S(A)=300 that wouldn't affect our argument. But still, a good question.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 10:51:29 pm

Alright yeah thanks, I get the question 1 now :)

EDIT: 5000 posts!

Title: Re: TT's Maths Thread
Post by: zzdfa on December 06, 2009, 10:55:47 pm

and just to put your mind at ease about your little question, you can construct a set for any given x as follows:
find the greatest n such that sum 1 to n is less than x. add n+1. the sum of this set is, at most, n more than x. thus the element corresponding to the difference between the current sum and x can be removed from your set. done $\blacksquare$

edit add some clarifying examplezz

so for example {1,2,3,4,5,6,7,8} and we want to get 23.

go 1,2,3,4,5,6 sum =21

add the next number, 7 to make it go over 23

1,2,3,4,5,6,7 sum = 28

then remove the 5 to make the sum exactly 23:

1,2,3,4,6,7 sum=23

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 11:00:26 pm

ok I will try prove out of interest your question about whethr for all $0\leq b \leq 465$ there exists some set $A$ such that $S(A)=b$ .

I will try a proof by induction:

Base case: obviously b=1 has a solution A={1}

Inductive hypothesis: suppose b has a solution A, we will now try a solution B for b+1 (provided $b \neq 465$ )

If A was to have a "gap" like: {.... a,a+k...} where k>1 then we can merely delete a and introduce a+1 and we are done. Likewise, if say you have the set {...a} where a is the largest element and not 30, then just swap a with a+1.

Now if A did not have a gap or largest element less than 30 then it must not contain 1, (otherwise it is the WHOLE set and b=465 and there is no solution for b+1). Therefore in this case, just introduce 1 and you're done.

edit: beaten by zzdfa :P

Title: Re: TT's Maths Thread
Post by: Ahmad on December 06, 2009, 11:31:20 pm

I'll show an alternative induction, because it illustrates an important idea in induction. Which is that it's often simpler to prove a more general statement. Another useful thing to think about when trying a solution by induction is that it's often helpful to prove a stronger statement, since it gives you more power in the inductive step.

Let $P_n$ be the statement: for all natural numbers n, there exists a subset of $A_n = \{1, 2, ..., n\}$ with element sum any given number in $\{0, 1, ..., n(n+1)/2\}$ .

The base case is trivial, do it for n = 1 and n = 2. So suppose $P_k$ is true, so that from $A_k$ we can form the element sum of any given number in $\{0, 1, ..., k(k+1)/2\}$ . Now consider the subsets of $A_{k+1}$ which don't contain k+1, this is just the set of subsets of A_k, from which we can form $\{0, 1, ..., k(k+1)/2\}$ , by adding the element k+1 to each of these subsets we can form $\{0 + (k+1), 1 + (k+1), ..., \frac{k(k+1)}{2} + (k+1) = \frac{(k+1)(k+2)}{2}\}$ , so we can form all numbers from $0$ to $(k+1)(k+2)/2$ , as required. :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 06, 2009, 11:39:15 pm

Yeah when i went to take a leak I thought of about 10 other proofs by induction, and this one is still a new one :P

Title: Re: TT's Maths Thread
Post by: Ahmad on December 06, 2009, 11:42:35 pm

You think of maths while taking a leak. Nice. :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2009, 11:52:35 pm

Quote from: kamil9876 on December 06, 2009, 11:39:15 pm

Yeah when i went to take a leak I thought of about 10 other proofs by induction, and this one is still a new one :P

LOL

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 07, 2009, 12:03:44 am

Quote from: Ahmad on December 06, 2009, 11:42:35 pm

You think of maths while taking a leak. Nice. :P

sometimes I deliberately go without the need to, just purely as a source of inspiration.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 07, 2009, 12:27:39 am

Quote

3. Let $\mathbb{S}$ be a set with $\mbox{n}$ elements. In how many different ways can one select two not necessarily distinct or disjoint subsets of $\mathbb{S}$ so that the union of the two subsets is $\mathbb{S}$ ? The order of selection does not matter. For example, the pair of subsets $\{a,c\}, \{b,c,d,e,f\}$ represents the same selection as the pair $\{b,c,d,e,f\}, \{a,c\}$

Suppose we have $S=A \cup B$

Each element of S must have one of these three muttually exclusive properties:

1.) it is in $A$ only
2.) it is in $B$ only
3.) it is in $A \cap B$

(you may already tell that this is similair to your previous ice cream problem)

hence we have $3$ choices for each element of S. Hence $3^n$ different combinations of choices.

(note though that my solution regards $S=\lbrace 1,2,3...n-1 \rbrace \cup \lbrace n \rbrace$ and $S=\lbrace n \rbrace \cup \lbrace 1,2,3... n-1 \rbrace$ as two different things. )

Title: Re: TT's Maths Thread
Post by: /0 on December 07, 2009, 12:47:16 am

Quote from: Ahmad on December 06, 2009, 11:31:20 pm

Let $P_n$ be the statement: for all natural numbers n, there exists a subset of $A_n = \{1, 2, ..., n\}$ with element sum any given number in $\{0, 1, ..., n(n+1)/2\}$ .

Why do mathematicians insist on proving things like this? :/

Title: Re: TT's Maths Thread
Post by: TrueTears on December 07, 2009, 12:52:03 am

Quote from: /0 on December 07, 2009, 12:47:16 am

Quote from: Ahmad on December 06, 2009, 11:31:20 pm
Let $P_n$ be the statement: for all natural numbers n, there exists a subset of $A_n = \{1, 2, ..., n\}$ with element sum any given number in $\{0, 1, ..., n(n+1)/2\}$ .

Why do mathematicians insist on proving things like this? :/

Once you understand why, you will become a true mathematician.

jk jk :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 07, 2009, 03:40:47 am

Quote from: kamil9876 on December 07, 2009, 12:27:39 am

Quote
3. Let $\mathbb{S}$ be a set with $\mbox{n}$ elements. In how many different ways can one select two not necessarily distinct or disjoint subsets of $\mathbb{S}$ so that the union of the two subsets is $\mathbb{S}$ ? The order of selection does not matter. For example, the pair of subsets $\{a,c\}, \{b,c,d,e,f\}$ represents the same selection as the pair $\{b,c,d,e,f\}, \{a,c\}$

Suppose we have $S=A \cup B$

Each element of S must have one of these three muttually exclusive properties:

1.) it is in $A$ only
2.) it is in $B$ only
3.) it is in $A \cap B$

(you may already tell that this is similair to your previous ice cream problem)

hence we have $3$ choices for each element of S. Hence $3^n$ different combinations of choices.

(note though that my solution regards $S=\lbrace 1,2,3...n-1 \rbrace \cup \lbrace n \rbrace$ and $S=\lbrace n \rbrace \cup \lbrace 1,2,3... n-1 \rbrace$ as two different things. )

But doesn't the question say $S=\lbrace 1,2,3...n-1 \rbrace \cup \lbrace n \rbrace$ and $S=\lbrace n \rbrace \cup \lbrace 1,2,3... n-1 \rbrace$ are the same?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 07, 2009, 12:20:52 pm

probably. Fix it. gogogo

Title: Re: TT's Maths Thread
Post by: TrueTears on December 07, 2009, 04:18:29 pm

Quote from: kamil9876 on December 07, 2009, 12:20:52 pm

probably. Fix it. gogogo

Oh wait I got it.

Let $\mathbb{S} = \{a,b,c,...n\}$

So each element can either be in $A$ or in $B$ or in $A \cap B$ .

So we can have something like:

$A = \{a,b,c...n\}$ and $B = \{\varnothing\}$

$B = \{a,b,c...n\}$ and $A = \{\varnothing\}$

Which is an overcount since we don't care about order.

This means each pair is counted twice rather than once.

But there is one exception: If we picked all the elements in $\mathbb{S} = \{a,b,c,...n\}$ to be in $A \cap B$ then we have only counted that once.

$\therefore \frac{3^n-1}{2} + 1$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 07, 2009, 06:24:43 pm

Just another Q, I can do it algebraically but what about a combinatorial argument method?

Find a formula for $\binom{n}{0}^2+ \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2$

Since $\sum_{j=0}^k\binom{n}{j}\binom{m}{k-j} = \binom{n+m}{k} \ \forall \ \{n,m,k\} \in \mathbb{N^+} \ \mbox{and} \ \{n,m\} \ge k$

Let $k = n$ and $m = n$

$\sum_{j=0}^n\binom{n}{j}\binom{n}{n-j} = \binom{2n}{n}$

$\mbox{LHS} = \binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + ... + \binom{n}{n}\binom{n}{0} = \binom{n}{n}\binom{n}{n} + \binom{n}{1}\binom{n}{1} + ... + \binom{n}{n}\binom{n}{n} = \binom{n}{0}^2+ \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2$

$\therefore \binom{n}{0}^2+ \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2 = \binom{2n}{n}$

~~So yeah what is a combinatorial way?~~ Nvm I got it.

Notice how $\binom{n}{0}^2+ \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2$ can be rewritten into $\binom{n}{0}\binom{n}{n}+ \binom{n}{1}\binom{n}{n-1} + \binom{n}{2}\binom{n}{n-2} + ... + \binom{n}{n}\binom{n}{0}$

So just imagine $2n$ objects gets split into $2$ piles containing $\mbox{n}$ objects each.

If you pick $j$ objects from the first pile then you must pick $n-j$ objects in the second pile.

This means you are basically picking $\mbox{n}$ objects out of $2n$ objects.

$\therefore$ $\binom{n}{0}^2+ \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2 = \binom{n}{0}\binom{n}{n}+ \binom{n}{1}\binom{n}{n-1} + \binom{n}{2}\binom{n}{n-2} + ... + \binom{n}{n}\binom{n}{0} = \binom{2n}{n}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 07, 2009, 08:04:50 pm

Quote from: TrueTears on December 07, 2009, 12:52:03 am

Quote from: /0 on December 07, 2009, 12:47:16 am
Quote from: Ahmad on December 06, 2009, 11:31:20 pm
Let $P_n$ be the statement: for all natural numbers n, there exists a subset of $A_n = \{1, 2, ..., n\}$ with element sum any given number in $\{0, 1, ..., n(n+1)/2\}$ .

Why do mathematicians insist on proving things like this? :/
Once you understand why, you will become a true mathematician.

jk jk :P

It's an interesting phenomena: is claiming something to be trivial a sign of genius, or a sign of ignorance? :P

Title: Re: TT's Maths Thread
Post by: Ahmad on December 07, 2009, 10:44:04 pm

Here's one definition of trivial, but I forget who it's attributed to: a result is trivial when its proof is immediately and automatically conjured in one's mind. That means something that's obvious, such as that a torus and sphere are topologically different, might not be trivial (for most people). There's also the joke that trivial is synonymous with 'proved', so Fermat's last theorem is trivial. :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 08, 2009, 12:27:44 am

Just wondering if this is the right way to do this problem:

In how many ways can we place $r$ red balls and $w$ white balls in $\mbox{n}$ boxes so that each box contains at least one ball of each colour?

Obviously $\{r,w\} \ge \mbox{n}$

Consider a specific case:

$r = 4$ , $w = 6$ and $n = 3$

Let $r_1,r_2,r_3$ be the number of red balls in box $[1],[2],[3]$ respectively. We assume each box to be distinguishable.

We need to apply the restriction so that each box must contain at least one red ball and at least 1 white ball.

Let's consider just the "at least one red ball" part.

Obviously $r_1+r_2+r_3 = 4$

Let $o$ represent a ball. We can represent one scenario like this:

o_o_o_o where _ represents a place to place a + sign.

Eg, o+o_o+o will represent o+oo+o which means 1 ball in $[1]$ , 2 in $[2]$ , 1 in $[3]$ .

Now notice we can't have this scenario +o_o_o+o. This means there will be 0 ball in $[1]$ , 3 in $[2]$ and 1 in $[4]$ but each box must contain at least 1 red ball.

So there are 3 empty _ to place 2 + signs. $\implies \binom{3}{2}$ choices.

Doing the same for the white balls yields $\binom{5}{2}$ choices.

So in total for this specific case we have $\binom{3}{2}\binom{5}{2}$ ways.

Now we can generalise.

We have $r$ red balls, $w$ white balls and $\mbox{n}$ boxes.

Let $r_1,r_2,r_3...r_n$ be the number of red balls in box $[1],[2],[3]...[n]$

Obviously $r_1+r_2+r_3+...+r_n = r$

Now there would be $r-1$ slots to place a + sign $\implies \binom{r-1}{n-1}$ for red balls

For white balls there would be $\binom{w-1}{n-1}$ slots to place the + sign.

Total $\binom{r-1}{n-1}\binom{w-1}{n-1}$ ways.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 08, 2009, 04:10:57 am

$\binom{r}{0} - \binom{r}{1} + \binom{r}{2} + ... + (-1)^r\binom{r}{r} = 0$

I know this identity can be proved just by expanding $(1-1)^r = 0$

But I was wondering if anyone can show me how to prove it using induction and the use of the summation identity.

:)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 08, 2009, 01:36:44 pm

I like this way the best:
http://vcenotes.com/forum/index.php/topic,19896.msg207276.html#msg207276

It's also obvious for odd n.

You can do it by induction by using pascal's identity. ie suppose it is true for n, now we want to prove true for n+1:

$\sum_{j=0}^{j=n+1} (-1)^j\binom{n+1}{j}$

now all we need to do is sub in $\binom{n+1}{j}=\binom{n}{j-1}+\binom{n}{j}$ (for all j>0 of course). So now because we have n's at the top, we can invoke the inductive hypothesis (ie true for n) and split the sums up etc. carefully doing it all.

Title: Re: TT's Maths Thread
Post by: Ahmad on December 08, 2009, 01:44:58 pm

The identity says that if you have a set of r elements, then the number of subsets containing an odd number of elements is the same as the number of subsets containing an even number of elements. Can you find a combinatorial proof of this? :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 08, 2009, 02:17:35 pm

Is this the one you're talking about?

Quote from: kamil9876 on December 08, 2009, 01:36:44 pm

I like this way the best:
http://vcenotes.com/forum/index.php/topic,19896.msg207276.html#msg207276

Title: Re: TT's Maths Thread
Post by: Ahmad on December 08, 2009, 02:28:32 pm

Sure is :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 08, 2009, 05:37:51 pm

Quote

$\binom{r}{0} - \binom{r}{1} + \binom{r}{2} + ... + (-1)^r\binom{r}{r} = 0$

Base Case:

When $r = 2$

$\binom{2}{0} - \binom{2}{1} + \binom{2}{2} = 0$

Inductive Hypothesis:

Assume it is true for $r = k$

$\binom{k}{0} - \binom{k}{1} + \binom{k}{2} + ... + (-1)^{k}\binom{k}{k} = 0$

Proof:

Must show it is true for $r = k+1$

$\binom{k+1}{0} - \binom{k+1}{1} + \binom{k+1}{2} + ... + (-1)^{k+1}\binom{k+1}{k+1} = 0$

But $\binom{k}{0} - \binom{k}{1} + \binom{k}{2} + ... + (-1)^{k}\binom{k}{k} = 0$ is true using our inductive hypothesis.

Using the summation identity:

$\binom{n}{q} = \binom{n+1}{q+1}- \binom{n}{q+1}$

$\binom{k}{0} = \binom{k+1}{0+1}- \binom{k}{0+1}$

$\implies \binom{k+1}{0+1}- \binom{k}{0+1} - \left(\binom{k+1}{1+1}- \binom{k}{1+1}\right) + \left(\binom{k+1}{2+1}- \binom{k}{2+1}\right) + ... + \left[(-1)^{k}\left(\binom{k+1}{k+1}- \binom{k}{k+1}\right)\right] = 0$

$\implies \binom{k+1}{0+1}- \binom{k}{0+1} - \binom{k+1}{1+1}+ \binom{k}{1+1} + \binom{k+1}{2+1}- \binom{k}{2+1}+ ... + (-1)^{k}\left(\binom{k+1}{k+1}- \binom{k}{k+1}\right) = 0$

$\implies \binom{k+1}{1}- \binom{k}{1} - \binom{k+1}{2}+ \binom{k}{2} + \binom{k+1}{3}- \binom{k}{3}+ ... + (-1)^{k}\binom{k+1}{k+1}+(-1)^{k+1} \binom{k}{k+1} = 0$

$\implies \left(- \binom{k}{1}+ \binom{k}{2}- \binom{k}{3}+ ... +(-1)^k\binom{k}{k}\right) + \left(\binom{k+1}{1}- \binom{k+1}{2}+ \binom{k+1}{3}+...+ (-1)^{k}\binom{k+1}{k+1}\right)$

$\implies -1+\binom{k+1}{1}- \binom{k+1}{2}+ \binom{k+1}{3}+...+ (-1)^{k}\binom{k+1}{k+1} = 0$

$\implies 1-\binom{k+1}{1}+ \binom{k+1}{2}- \binom{k+1}{3}+...+ (-1)^{k+1}\binom{k+1}{k+1} = 0$

$\therefore \binom{k+1}{0}-\binom{k+1}{1}+ \binom{k+1}{2}- \binom{k+1}{3}+...+ (-1)^{k+1}\binom{k+1}{k+1} = 0$

$\mbox{Q.E.D}$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 08, 2009, 11:34:36 pm

Wow just finished reading a chapter on Complement PIE and I gotta say it's probs one of the funnest chapter I've read so far in combinatorics.

A few questions for thought :P

1. A random number generator randomly generates the integers $1,2,...,9$ with equal probability. Find the probability that after $\mbox{n}$ numbers are generated, the product is a multiple of $10$ .

2. Let $a_1,a_2,...a_n$ be an ordered sequence of n distinct objects. A derangement of this sequence is a permutation that leaves no object in its original place. For example, if the original sequence is $1,2,3,4$ then $2,4,3,1$ is not a derangement, but $2,1,4,3$ is. Let $D_n$ denote the number of derangements of an $\mbox{n}$ -element sequence. Show that $D_n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - ... + (-1)^n\frac{1}{n!}\right)$

3. Imagine you are going to give $\mbox{n}$ kids ice-cream cones, one cone per kid, and there are $k$ different flavours available. Assuming that no flavours get mixed, show that the number of ways we can give out the cones using all $k$ flavours is $k^n - \binom{k}{1}(k-1)^n + \binom{k}{2}(k-2)^n- \binom{k}{3}(k-3)^n+...+(-1)^k \binom{k}{k}0^n$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 08, 2009, 11:41:54 pm

1.) by fundumanetal theorem of arithmetic it is equivalent to: how many different ways so that we don't choose:

5 and an even number.

so find how many ways there are where we have no 5 or no even number.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 09, 2009, 03:20:17 am

Ah, yeah good strategy kamil.

So if the number is a multiple of 10 then it must have 5 amongst its prime factorization and the rest of the digits chosen must all be even.

We require the complement of this.

Let $A$ represent the set which contains no 5.

Let $B$ represent the set which contains even numbers.

Let $C$ represent the set which contains odd numbers.

The complement is $(A \cap B) \cup C$

$|(A \cap B) \cup C| = |(A \cap B)| + |C| - |\left((A \cap B) \cap C\right)| = 4^n + 5^n - 8^n$

The total number of ways is $9^n$

$\therefore Pr((A \cap B) \cup C) = \frac{4^n + 5^n - 8^n}{9^n}$

$\therefore$ probability required $= 1 - \frac{4^n + 5^n - 8^n}{9^n}$

But the answer is $1 - \frac{8^n+5^n-4^n}{9^n}$ . Where did I go wrong?

I'm confused to what the complement is now =.=

Title: Re: TT's Maths Thread
Post by: /0 on December 09, 2009, 03:53:21 am

So you want to find the number of ways to NOT pick 5 and and even number... hmm sounds like a good idea

You could do that by choosing $\boxed{1},\boxed{2},\boxed{3},\boxed{4},5,\boxed{6},\boxed{7},\boxed{8},\boxed{9}$ n times.
So there are $8^n$ ways of picking from those n times.

You could do that by choosing $\boxed{1},2,\boxed{3},4,\boxed{5},6,\boxed{7},8,\boxed{9}$ n times.
So there are $5^n$ of picking from those n times.

But there is an overlap between them, namely $\boxed{1},\boxed{3},\boxed{7},\boxed{9}$ .
There are $4^n$ ways of picking from those n times.

So there are $8^n+5^n-4^n$ ways of avoiding the deadly combination.

My reasoning's a bit shaky, but just going with my gut

Title: Re: TT's Maths Thread
Post by: Ahmad on December 09, 2009, 01:21:50 pm

In my opinion the best way to think about PIE is through indicator functions (sometimes called characteristic functions). In fact, once you've set up the very simple idea of indicator functions PIE is an immediate consequence, and there are lots of other useful consequences too.

So what's the idea of characteristic functions? Well suppose the set S contains all possible integer strings of length n output by the random generator. Then we define the function $H_2 : S \rightarrow \{0,1\}$ which intuitively represents "string does not contains an even number", and takes as input a possible integer string from the random number generator, and outputs the value 1 if "string does not contains an even number", and 0 otherwise. Notice that $\sum_{s \in S} H_2(s)$ counts the number of outcomes which don't contain an even number.

Similarly we can define $H_5 : S \rightarrow \{0,1\}$ which represents "string does not contain a 5". In particular I chose to define $H_2$ and $H_5$ this way because the number of outcomes which satisfies the property associated with each of these indicator function is easily counted.

The intriguing thing about these indicator functions is that they convert logical operations into standard algebra. For example, what if we wanted to find the characteristic function which represents "string contains an even number", well this is just NOT(H_2), which has the indicator function $1 - H_2$ . I'll leave you to verify the following,

$\neg H_2 \leftrightarrow (1 - H_2)$ where $x \leftrightarrow y$ means that x translates into characteristic function y. The funny symbol represents "not".

$H_2 \wedge H_5 \leftrightarrow H_2 \cdot H_5$ represents "and"

$H_2 \vee H_5 \leftrightarrow H_2 + H_5 - H_2\cdot H_5$ represents "or"

Now we want to find the number of strings with products a multiple of 10. How do we represent this in terms of $H_2$ and $H_5$ ? Well we want neither $H_2$ nor $H_5$ to occur, i.e. we want strings that contain an even number and a multiple of 5. Representing this in logical terms this is exactly $(\neg H_2) \wedge (\neg H_5)$ which using the above rules we can see has indicator function $(1-H_2)(1-H_5)$ . So to find the number of outcomes that have products that are multiples of 10 we simply sum over this indicator function, which is easily evaluated because we can expand out to get everything in terms of $H_2$ and $H_5$ which as I've previously mentioned have been chosen in particular because they're easily summed over!

$\sum_{s \in S} (1-H_2(s))(1-H_5(s)) = \sum_{s \in S} (1 - H_2(s) - H_5(s) + H_2(s)H_5(s))$

$= \sum_{s \in S} 1 - \sum_{s \in S} H_2(s) - \sum_{s \in S} H_5(s) + \sum_{s \in S} H_2(s)H_5(s)$

$= 9^n - 5^n - 8^n + 4^n$

Of course we simply divide by $9^n$ to get the desired probability. It might seem that this is overkill, but that's because I was showing the gory details, on paper I'd maybe write one or two lines, or just the last line. Also, the case I've presented this for is simple and you can just think it through and work it out like /0 did, but it's rather helpful for more complicated scenarios which it easily adapts to.

Notice that in this language PIE can be stated as follows: suppose $A_1, A_2, \ldots, A_n$ are indicator functions each associated with a property, and you want to count the number of outcomes not satisfying any of these properties, then,

$(\neg A_1) \wedge (\neg A_2) \wedge \cdots \wedge (\neg A_n) \rightarrow (1 - A_1)(1 - A_2)\cdots (1 - A_n) = 1 - (A_1 + A_2 + \cdots A_n) + (A_1 A_2 + \cdots) + \cdots + (-1)^n A_1\cdot A_2 \cdots A_n$

So it's essentially been reduced to algebra. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 09, 2009, 05:23:06 pm

Ahmad you have inspired me to use indicator functions for the next question lol I think it worked out pretty well.

Quote

2. Let $a_1,a_2,...a_n$ be an ordered sequence of n distinct objects. A derangement of this sequence is a permutation that leaves no object in its original place. For example, if the original sequence is $1,2,3,4$ then $2,4,3,1$ is not a derangement, but $2,1,4,3$ is. Let $D_n$ denote the number of derangements of an $\mbox{n}$ -element sequence. Show that $D_n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - ... + (-1)^n\frac{1}{n!}\right)$

Consider the specific case of just $4$ distinct objects.

Let $A$ represent the event where $a_1$ is in the same place.

Let $B$ represent the event where $a_2$ is in the same place.

Let $C$ represent the event where $a_3$ is in the same place.

Let $D$ represent the event where $a_4$ is in the same place.

What we require is the complement $\bar A \cap \bar B \cap \bar C \cap \bar D$

Let's define an indicator function namely $I_y(x)$ where $y \in \{A,B,C,D\}$ and $x \in \mathbb{U}$ where $\mathbb{U}$ is an universal set that contains all sets $\{A,B,C,D\}$

$I_y(x) = \begin{cases} 0, & \mbox{if} \ \mbox{x} \ \notin \ \mbox{y} \\ 1, & \mbox{if} \ \mbox{x} \ \in \ \mbox{y} \end{cases}$

We require:

$I_{\bar A}(x) \wedge I_{\bar B}(x) \wedge I_{\bar C}(x) \wedge I_{\bar D}(x)$

Let's define a function:

$f(x) = (I_{\bar A}(x))(I_{\bar B}(x))(I_{\bar C}(x))(I_{\bar D}(x))$

$\implies f(x) = (1-I_A(x))(1-I_B(x))(1-I_C(x))(1-I_D(x)) = I_{\bar A \cap \bar B \cap \bar C \cap \bar D}(x)$

Let $F$ represent the cardinality of $f(x)$

$\therefore \sum_{x \in \mathbb{U}} f(x) = F$

$\implies F = \sum_{x \in \mathbb{U}} \left((1-I_A(x))(1-I_B(x))(1-I_C(x))(1-I_D(x))\right)$

$F = \sum_{x \in \mathbb{U}} 1 - \sum_{x \in \mathbb{U}} \left((I_{A}(x))+(I_{B}(x))+(I_{C}(x))+(I_{D}(x))\right) + \sum_{x \in \mathbb{U}} \left[(I_{A}(x))(I_{B}(x))+(I_{A}(x))(I_{C}(x)) + (I_{A}(x))(I_{D}(x)) + (I_{B}(x))(I_{C}(x)) +(I_{B}(x))(I_{D}(x)) +(I_{C}(x))(I_{D}(x))\right]$

$- \sum_{x \in \mathbb{U}} \left[(I_{A}(x))(I_{B}(x))(I_{C}(x))+(I_{A}(x))(I_{B}(x))(I_{D}(x))+(I_{B}(x))(I_{C}(x))(I_{D}(x)) + (I_{A}(x))(I_{C}(x))(I_{D}(x))\right] + \sum_{x \in \mathbb{U}} \left[(I_{A}(x))(I_{B}(x))(I_{C}(x))(I_{D}(x))\right]$

$F = 4! - 3!(4) + \binom{4}{2}2! - \binom{4}{3} + \binom{4}{4}$

$F = 4!-4! + \frac{4!}{2!2!}(2!) - \frac{4!}{3!1!} + \frac{4!}{4!0!}$

$F = 4!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right)$

Which is exactly what we get if we sub in $n = 4$ .

Now we can generalise:

If there were $\mbox{n}$ distinct objects.

$a_1,a_2,a_3...a_n$

Let $A$ represent the event where $a_1$ is in the same place.

Let $B$ represent the event where $a_2$ is in the same place.

Let $C$ represent the event where $a_3$ is in the same place.

.
.
.

Let $N$ represent the event where $a_n$ is in the same place.

We require:

$I_{\bar A}(x) \wedge I_{\bar B}(x) \wedge I_{\bar C}(x) \wedge ... \wedge I_{\bar N}(x)$

$\implies F = \sum_{x \in \mathbb{U}} \left((1-I_A(x))(1-I_B(x))(1-I_C(x)) ... (1-I_N(x))\right)$

$F = \sum_{x \in \mathbb{U}} 1 - \sum_{x \in \mathbb{U}} \left((I_{A}(x))+(I_{B}(x))+(I_{C}(x))+...+(I_{N}(x))\right) + \sum_{x \in \mathbb{U}} \left[(I_{A}(x))(I_{B}(x))+(I_{A}(x))(I_{C}(x)) + ...+ (I_{A}(x))(I_{N}(x)) \cdots \right]$

$- \sum_{x \in \mathbb{U}} \left[(I_{A}(x))(I_{B}(x))(I_{C}(x))+...+(I_{A}(x))(I_{B}(x))(I_{D}(x))\right] + ... + (-1)^n \sum_{x \in \mathbb{U}} \left[(I_{A}(x))(I_{B}(x))(I_{C}(x))...(I_{D}(x))\right]$

$F = n! - (n-1)!(n) + \binom{n}{2}(n-2)! - \binom{n}{3}(n-3)! + ... + (-1)^n \binom{n}{n}$

$F = n!-n!+\frac{n!}{(n-2)!2!}(n-2)! - \frac{n!}{(n-3)!3!}(n-3)! + ... + (-1)^n\frac{n!}{n!0!}$

$F = n!\left(1-\frac{1}{1!}+\frac{1}{2!} - \frac{1}{3!} + ... + (-1)^n\frac{1}{n!}\right)$

$\mbox{Q.E.D}$

Title: Re: TT's Maths Thread
Post by: zzdfa on December 09, 2009, 05:33:59 pm

Note that the stuff in the brackets is the truncated expansion of $e^{-1}$ . So $\lim_{n\rightarrow \infty}D_n=\frac{n!}{e}$ . In fact $D_n =\left\lfloor \frac{n!}{e} \right\rfloor$ which i think is pretty neat :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 09, 2009, 05:34:54 pm

Quote from: zzdfa on December 09, 2009, 05:33:59 pm

Note that the stuff in the brackets is the truncated expansion of $e^{-1}$ . So $\lim_{n\rightarrow \infty}D_n=\frac{n!}{e}$ . In fact $D_n =\left\lfloor \frac{n!}{e} \right\rfloor$ which i think is pretty neat :)

lol I knew it looked like Taylor series or something related.

Thanks for that zzdfa :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 09, 2009, 05:41:58 pm

Also just another method, if we were to do it intuitively using complement PIE:

Let $A$ represent the event where $a_1$ is in the same place.

Let $B$ represent the event where $a_2$ is in the same place.

Let $C$ represent the event where $a_3$ is in the same place.

.
.
.

Let $N$ represent the event where $a_n$ is in the same place.

We require: $|A \cup B \cup C \cup ... \cup N| = \left(|A| + |B| + |C| + ... + |N|\right) - \left(|A \cap B| + |A \cap C| + ... + |A \cap N| \cdots\right) + ... + (-1)^{n-1}\left(|A \cap B \cap C \cap ... \cap N|\right)$

$|A \cup B \cup C \cup ... \cup N| = (n-1)!n - \binom{n}{2}(n-2)! + ... + (-1)^{n-1}\binom{n}{n} = n!- \left(\frac{n!}{(n-2)!2!}\right)(n-2)!+...+(-1)^{n-1}\left(\frac{n!}{n!}\right)$

$\implies |A \cup B \cup C \cup ... \cup N| = n! - \frac{n!}{2!} + ... + (-1)^{n-1}\left(\frac{n!}{n!}\right)$

Now we require the complement of this which is given by $n! - |A \cup B \cup C \cup ... \cup N| = n!-n! + \frac{n!}{2!} - ... + (-1)^{1}(-1)^{n-1}\left(\frac{n!}{n!}\right) = n!\left(1-\frac{1}{1!} + \frac{1}{2!} - ... + (-1)^n\frac{1}{n!}\right)$

$\mbox{Q.E.D}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 09, 2009, 09:16:50 pm

Quote from: TrueTears on December 09, 2009, 05:41:58 pm

$|A \cup B \cup C \cup ... \cup N| = (n-1)!n - \binom{n}{2}(n-2)! + ... + (-1)^{n-1}\binom{n}{n}$

Notice how this is analogous to the answer to question 3. This is because "event where $a_i$ is in the same place" is analogous to "flavour $a_i$ is not chosen by any kid"

replacing $(k-i)^n$ for $(n-i)!$ because giving freedom to the positions remaining $n-i$ objects is the analogous to giving freedom to the remaining $k-i$ flavours(ie we have n children and each can chose from k-i flavours)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 09, 2009, 09:37:41 pm

Quote

3. Imagine you are going to give $\mbox{n}$ kids ice-cream cones, one cone per kid, and there are $k$ different flavours available. Assuming that no flavours get mixed, show that the number of ways we can give out the cones using all $k$ flavours is $k^n - \binom{k}{1}(k-1)^n + \binom{k}{2}(k-2)^n- \binom{k}{3}(k-3)^n+...+(-1)^k \binom{k}{k}0^n$

Let the flavours be $1,2, \cdots, k$ .

Let $A_1$ represent the set where flavour $1$ does not get used.

Let $A_2$ represent the set where flavour $2$ does not get used.

.
.
.

Let $A_k$ represent the set where flavour $k$ does not get used.

We require $\overline{\bigcup_{i=1}^{k} A_i} = \bigcap_{i=1}^k \overline{A_i} = \overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_k}$

Let's define an indicator function namely $I_y(x)$ where $y \in \{A_1,A_2, \cdots, A_k\}$ and $x \in \mathbb{U}$ where $\mathbb{U}$ is an universal set that contains all sets $\{A_1,A_2, \cdots, A_k\}$

$I_y(x) = \begin{cases} 0, & \mbox{if} \ \mbox{x} \ \notin \ \mbox{y} \\ 1, & \mbox{if} \ \mbox{x} \ \in \ \mbox{y} \end{cases}$

Let $S_k = |\overline{A_1} \cap \overline{A_2} \cap \cdots \cap \overline{A_k}|$

Define $f(x) := I_{\bar A_1}(x)I_{\bar A_2}(x) \cdots I_{\bar A_k}(x) = (1-I_{A_1}(x))(1-I_{A_2}(x)) \cdots (1-I_{A_k}(x))$

$\therefore S_k = \sum_{x \in \mathbb{U}} (1-I_{A_1}(x))(1-I_{A_2}(x)) \cdots (1-I_{A_k}(x))$

$S_k = \sum_{x \in \mathbb{U}} 1 - \sum_{x \in \mathbb{U}} \left[I_{A_1}(x)+I_{A_2}(x)+ \cdots + I_{A_k}(x)\right] + \sum_{x \in \mathbb{U}}\left[I_{A_1}(x)I_{A_2}(x)+ \cdots + I_{A_1}(x)I_{A_k}(x) + \cdots \right] + \cdots + (-1)^n \sum_{x \in \mathbb{U}} \left[(I_{A_1}(x)I_{A_2}(x) \cdots I_{A_k}(x)\right]$

$S_k = k^n - \binom{k}{1}\left(k-1\right)^n + \binom{k}{2}(k-2)^n - ... + (-1)^n\binom{k}{k}0^n$

$\mbox{Q.E.D}$

Title: Re: TT's Maths Thread
Post by: hard on December 09, 2009, 09:57:38 pm

OMG KILL ME NOW U GUYS ARE PURE GENIUS

Title: Re: TT's Maths Thread
Post by: TrueTears on December 10, 2009, 03:17:41 am

IMO question:

A permutation $\{x_1, \ldots, x_{2n}\}$ of the set $\{1,2, \ldots, 2n\}$ where $\mbox{n}$ is a positive integer, is said to have property $T$ if $|x_i - x_{i + 1}| = n$ for at least one $i \in \{1,2, \ldots, 2n - 1\}$ . Show that, for each $\mbox{n}$ , there are more permutations with property $T$ than without.

Title: Re: TT's Maths Thread
Post by: Ahmad on December 10, 2009, 11:48:59 am

You can count exactly how many permutations satisfy the property:

$\sum_{k=1}^n \left( (-1)^{k+1} 2^k \binom{n}{k} \binom{2n-k}{k} k! (2n-2k)! \right)$

Title: Re: TT's Maths Thread
Post by: kyzoo on December 10, 2009, 11:43:22 pm

Do I have to deal with this in Uni Maths? Uni Maths looks so much harder than Specialist - Eigenvalues, three variable graphs, etc.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 10, 2009, 11:43:50 pm

Quote from: kyzoo on December 10, 2009, 11:43:22 pm

Do I have to deal with this in Uni Maths? Uni Maths looks so much harder than Specialist - Eigenvalues, three variable graphs, etc.

It's not uni maths, It's just olympiad style problem solving questions.

Title: Re: TT's Maths Thread
Post by: Cataclysmic on December 10, 2009, 11:46:07 pm

These problems are a piece of cake

lol i have no fkn idea what is going on >_>

Title: Re: TT's Maths Thread
Post by: TrueTears on December 10, 2009, 11:57:17 pm

So for the specific case when $n = 2$ .

We have the set $\{1,2,3,4\}$

To satisfy the condition, the $2$ numbers must be adjacent and we can have either $(1,3), (3,1), (2,4), (4,2)$ where $(x,y)$ represents an adjacent pair.

To find those that satisfy $T$ we need to find: $(1,3) \cup (3,1) \cup (2,4) \cup (4,2)$

Using PIE we can find those that doesn't satisfy $T$ $\overline{(1,3) \cup (3,1) \cup (2,4) \cup (4,2)} = \overline{(1,3)} \cap \overline{(3,1)} \cap \overline{(2,4)} \cap \overline{(4,2)}$

Let $Q = |\overline{(1,3)} \cap \overline{(3,1)} \cap \overline{(2,4)} \cap \overline{(4,2)}|$

Defining an indicator function $I_y(x)$ where $y \in \{(1,3), (3,1), (2,4), (4,2)\}$ with domain $x \in \mathbb{U}$ such that $\mathbb{U}$ contains all sets $(x,y)$

$I_y(x) = \begin{cases} 0, & \mbox{if} \ \mbox{x} \ \notin \ \mbox{y} \\ 1, & \mbox{if} \ \mbox{x} \ \in \ \mbox{y} \end{cases}$

$Q = \sum_{x \in \mathbb{U}} I_{\overline{(1,3)}}(x) I_{\overline{(3,1)}}(x)I_{\overline{(2,4)}}(x)I_{\overline{(4,2)}}(x)$

$Q = \sum_{x \in \mathbb{U}} (1-I_{(1,3)}(x))(1-I_{(3,1)}(x))(1-I_{(2,4)}(x))(1-I_{(4,2)}(x))$

$Q = \sum_{x \in \mathbb{U}} 1 - \sum_{x \in \mathbb{U}}I_{(1,3)}(x)+I_{(3,1)}(x)+I_{(2,4)}(x)+I_{(4,2)}(x)+\sum_{x \in \mathbb{U}} I_{(1,3)}(x)I_{(3,1)}(x) + I_{(1,3)}(x)I_{(2,4)}(x)+I_{(1,3)}(x)I_{(4,2)}(x) + I_{(3,1)}(x)I_{(2,4)}(x) +I_{(3,1)}(x)I_{(4,2)}(x)+I_{(2,4)}(x)I_{(4,2)}(x)$

$-\sum_{x \in \mathbb{U}}I_{(1,3)}(x)I_{(3,1)}(x)I_{(2,4)}(x)+I_{(1,3)}(x)I_{(2,4)}(x)I_{(4,2)}(x)+I_{(3,1)}(x)I_{(2,4)}(x)I_{(4,2)}(x)+I_{(1,3)}(x)I_{(3,1)}(x)I_{(4,2)}(x) + \sum_{x \in \mathbb{U}}I_{(1,3)}(x)I_{(3,1)}(x)I_{(2,4)}(x)I_{(4,2)}(x)$

Now to work out the cardinality of each consider the set $\{(1,3), (3,1)\}$ and $\{(2,4), (4,2)\}$

The first sum is obvious: $\sum_{x \in \mathbb{U}} 1 = 4!$

The second sum is also pretty obvious: $\sum_{x \in \mathbb{U}}I_{(1,3)}(x)+I_{(3,1)}(x)+I_{(2,4)}(x)+I_{(4,2)}(x) = 4!$

The third sum is not so obvious since we have terms that equal $0$ , eg, $I_{(1,3)}(x)I_{(3,1)}(x) = 0$ .

Thus we need to pick any $2$ pairs from the 1st set and any $2$ pairs from the 2nd set.

So there are $2 \times 2 = 4$ non-zero pairs. Each pair however has 2 ways to rearrange. So the third sum equals $2 \times 4 =8$

The fourth sum is $0$ since we need $3$ sets but we only have $2$ to pick from.

The fifth sum is also $0$ by the same argument as above.

Now we can generalise.

Consider the set $\{1,2,3, \cdots, 2n\}$

Let $(1,1+n), (1+n,1), (2,2+n), (2+n,2), \cdots, (n,2n),(2n,n)$ represent the adjacent pairs. There are a total of $2n$ pairs.

To find those that satisfy $T$ we need to find $(1,1+n) \cup (1+n,1) \cup (2,2+n) \cup (2+n,2) \cup \cdots \cup (n,2n) \cup (2n,n)$

Using PIE we can find the complement of $T$ : $\overline{(1,1+n) \cup (1+n,1) \cup (2,2+n) \cup (2+n,2) \cup \cdots \cup (n,2n) \cup (2n,n)} = \overline{(1,1+n)} \cap \overline{(1+n,1)} \cap \overline{(2,2+n)} \cap \overline{(2+n,2)} \cap \cdots \cap \overline{(n,2n)} \cap \overline{(2n,n)}$

Let $Q_n = |\overline{(1,1+n)} \cap \overline{(1+n,1)} \cap \overline{(2,2+n)} \cap \overline{(2+n,2)} \cap \cdots \cap \overline{(n,2n)} \cap \overline{(2n,n)}|$

Now we define an indicator function $I_y(x)$ where $y \in \{(1,1+n), (1+n,1), (2,2+n), (2+n,2), \cdots, (n,2n),(2n,n)\}$

$I_y(x) = \begin{cases} 0, & \mbox{if} \ \mbox{x} \ \notin \ \mbox{y} \\ 1, & \mbox{if} \ \mbox{x} \ \in \ \mbox{y} \end{cases}$

$Q_n = \sum_{x \in \mathbb{U}} (1-I_{(1,1+n)}(x))(1-I_{(1+n,1)}(x))(1-I_{(2,2+n)}(x))(1-I_{(2+n,2)}(x)) \cdots (1-I_{(n,2n)}(x))(1-I_{(2n,n)}(x))$

$Q_n = \sum_{x \in \mathbb{U}} 1 - \sum_{x \in \mathbb{U}} I_{(1,1+n)}(x)+I_{(1+n,1)}(x)+ \cdots + I_{(n,2n)}(x) + I_{(2n,n)}(x) + \sum_{x \in \mathbb{U}} I_{(1,1+n)}(x)I_{(1+n,1)}(x)+I_{(1,1+n)}(x)I_{(2,2+n)}(x)+ \cdots + I_{(1+n,n)}(x)I_{(2n,n)}(x) \cdots - \sum_{x \in \mathbb{U}}I_{(1,1+n)}(x)I_{(1+n,1)}(x)I_{(2,2+n)}(x) \cdots$

$+\sum_{x \in \mathbb{U}} (I_{(1,1+n)}(x))(I_{(1+n,1)}(x))(I_{(2,2+n)}(x))(I_{(2+n,2)}(x)) \cdots (I_{(n,2n)}(x))(I_{(2n,n)}(x))$

The first sum equals $(2n)!$

The second sum equals $\binom{2n}{1}(2n-1)!$

The third sum is a bit tricky since some pairs equal $0$ , thus consider all the different pairs placed into sets like this:

$\{(1,1+n), (1+n,1)\}, \{(2,2+n), (2+n,2)\}, \{(3,3+n), (3+n,n)\}, \cdots, \{(n,2n), (2n,n)\}$

We need $2$ pairs, since there are $\mbox{n}$ sets, we need to pick $2$ sets first $\implies \binom{n}{2}$ . But each set contains $2$ terms, thus we can have $2^2$ different pairings for each $2$ sets.

Therefore this sum equals $2^2\binom{n}{2}(2n-2)!$

The fourth sum is equal to $2^3\binom{n}{3}(2n-3)!$

The fifth sum is equal to $2^4\binom{n}{4}(2n-4)!$

.
.
.

The last sum is equal to $2^{2n}\binom{n}{2n}(2n-2n)!$

In total we have $Q_n = (2n)!-(2n)! + 2^2\binom{n}{2}(2n-2)! - 2^3\binom{n}{3}(2n-3)! + 2^4\binom{n}{4}(2n-4)! - \cdots + 2^{2n}\binom{n}{2n}(2n-2n)!$

$Q_n = \sum_{i=2}^{n}(-1)^{i}2^i\binom{n}{i}(2n-i)!$

So that means there are a total of $Q_n$ sets which does not satisfy $T$ .

Now we just have to prove that the number of sets that satisfy $T$ is larger than those that don't.

The number of sets that satisfies $T$ is equal to $(2n)!-Q_n$ .

So we need to prove $(2n)!-Q_n > Q_n$

$\implies (2n)! > 2Q_n$

First let $t_i$ represent $|(-1)^{i}2^i\binom{n}{i}(2n-i)!|$

$\therefore Q_n = t_2-t_3+t_4-t_5+ \cdots -t_{2n-1} + t_{2n}$

We see that $2t_2 = (-1)^{2}2^3\binom{n}{2}(2n-2)! = (2^2)\left(\frac{n!}{(n-2)!2!}\right)(2n-2)! = (2^3)\left(\frac{n(n-1)}{2}\right)(2n-2)! = 2n(2n-2)(2n-2)!$

But $(2n)! = (2n)(2n-1)(2n-2)!$

$(2n)(2n-1)(2n-2)! > 2n(2n-2)(2n-2)!$

$\implies (2n)! > 2t_2$

Next take $t_3$ and $t_4$ for example.

$t_3$ means at least 3 pairs satisfy $T$ and $t_4$ means at least $4$ pairs satisfy $T$ .

But at least 4 pairs is a subset of at least 3 pairs which means $t_4 \subset t_3$ $\implies$ $|t_3| \ge |t_4|$

Generalising this leads to $t_i \ge t_{i+1}$

So $2Q_n = 2t_2 + 2(t_4-t_3) + \cdots 2(t_{2n} - t_{2n-1})$

$\{(t_4-t_3), \cdots, (t_{2n}-t_{2n-1})\} \le 0$

$\implies 2t_2 + 2(t_4-t_3) + \cdots 2(t_{2n} - t_{2n-1}) \le 2t_2 < (2n)!$

$\therefore 2Q_n < (2n)!$

$\mbox{Q.E.D}$

fuck yeah good job guys :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 11, 2009, 12:41:34 am

let $t_i$ be the absolute value of the ith summand, our sum is therefore:

$Q_n=t_2-t_3+t_4...$

now one can show algebraically that $2t_2<(2n!)$ (1) (by comparing factors).

Moreover I showed algebraically, (but then realised its easier combinatorally) to show that $t_i\geq t_{i+1}$ . Therefore:

$2Q_n=2t_2+2(-t_3+t_4).... +2(-t_{2n-1}+t_{2n})$

and so the terms in brackets must all be non-positive and so we get:

$2Q_n\leq 2t_2<(2n)!$ (the last part is due too (1) )

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 12:55:39 am

Quote from: Ahmad on December 10, 2009, 11:48:59 am

$\sum_{k=1}^n \left( (-1)^{k+1} 2^k \binom{n}{k} \binom{2n-k}{k} k! (2n-2k)! \right)$

An obvious simplification is that $\binom{2n-k}{k} k! (2n-2k)! = (2n-k)!$ , which shows the equivalence of both of our results :)

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 01:09:31 am

A fair effort by each of us. Well done guys! ;D

Title: Re: TT's Maths Thread
Post by: TrueTears on December 11, 2009, 02:24:57 am

wowow what a question, very very fun!

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 11, 2009, 12:36:32 pm

actually, I just thought about it and I think the cominatorial argument for $t_i>t_{i+1}$ is wrong, but the algebraic is still true. Simply because $t_i$ does not exactly count how many with at least $i$ pairs, but it actually overcounts this, so we cannot be sure.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 11, 2009, 04:16:25 pm

Quote from: kamil9876 on December 11, 2009, 12:36:32 pm

actually, I just thought about it and I think the cominatorial argument for $t_i>t_{i+1}$ is wrong, but the algebraic is still true. Simply because $t_i$ does not exactly count how many with at least $i$ pairs, but it actually overcounts this, so we cannot be sure.

How do you do it algebraically?

I tried comparing factors but didn't really work =S

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 04:30:22 pm

It comes from the more general result http://www.artofproblemsolving.com/Wiki/index.php/Principle_of_Inclusion-Exclusion#Remark

Title: Re: TT's Maths Thread
Post by: TrueTears on December 11, 2009, 04:58:04 pm

Quote from: Ahmad on December 11, 2009, 04:30:22 pm

It comes from the more general result http://www.artofproblemsolving.com/Wiki/index.php/Principle_of_Inclusion-Exclusion#Remark

Thanks Ahmad :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 11, 2009, 06:33:54 pm

Use a combinatorial argument to prove that $n! = \sum_{r=0}^n\binom{n}{r}(n-r)!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \cdots + (-1)^{n-r}\frac{1}{n!}\right)$

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 09:10:10 pm

The right hand side summand counts the number of permutations with exactly r fixed points. Make sure you make use of your expression for derangements which you derived previously to interpret parts of the summand. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 11, 2009, 09:45:02 pm

Oh yeah, I got it, thanks Ahmad =)

This may not be combinatorics but I just saw it in AnC but they don't give a proof of where it came from. Can anyone enlighten me?

The Fibonacci recurrence formula $f_n = f_{n-1}+f_{n-2}$ for $n \ge 2$ and $f_0 = 0, f_1 = 1$

However a more "closed-form" formula for any number in the Fibonacci sequence is given by $f_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right]$

Is there an elementary way to prove that formula or is it beyond my capabilities?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 11, 2009, 09:47:23 pm

Quote from: TrueTears on December 11, 2009, 04:16:25 pm

Quote from: kamil9876 on December 11, 2009, 12:36:32 pm
actually, I just thought about it and I think the cominatorial argument for $t_i>t_{i+1}$ is wrong, but the algebraic is still true. Simply because $t_i$ does not exactly count how many with at least $i$ pairs, but it actually overcounts this, so we cannot be sure.
How do you do it algebraically?

I tried comparing factors but didn't really work =S

comparing factors does work. Ie you have to compare factors of $t_i$ and $t_{i+1}$ :
$ t_i= \frac{[(2n)(2n-2)(2n-4)....(2n-2i+2)] [(2n-i)(2n-i-1).... (1)]}{some denominator}$

$ t_{i+1}=\frac{[(2n)(2n-2)(2n-4)....(2n-2i)][(2n-i-1)...(1)]}{some bigger denominator}$

ie the expresion in the first set of square brackets comes from the numerator of $\binom{n}{i}$ with a 2 multipled into each factor (these $i$ lots of 2 come from $2^i$ ). The expression in the second pair of square brackets comes from $(2n-i)!$ And the denomonator comes from the denominator of $\binom{n}{i}$ which clearly increases as i increases.

Now all you need to do is compare numerators, and you can clearly spot the difference between them, it is that the bottom one has a 2n-2i factor in place of a 2n-i factor.

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 09:54:04 pm

It's called Binet's formula, and you can prove it by induction. But that's if you know the result already. If you don't you can discover it using generating functions, or linear algebra, or perhaps a bunch of other ways. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 11, 2009, 09:54:55 pm

Ahmah can you run me through deriving it by using generating functions? :P (I haven't studied it yet so I don't really know how to approach it myself lolz but I'm super interested)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 11, 2009, 10:30:41 pm

Here is a good one. Enjoy

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 10:35:46 pm

Sure thing, there's nothing to it, really.

Step 1. Define the generating function $F(x) = \sum_{n=0}^\infty f_n x^n$ where $f_n$ is the nth Fibonacci number. F is our clothesline upon which we hang up our sequence of numbers for display!

Step 2. Write down the recurrence equation, $f_{n+2} = f_{n+1} + f_{n}$ and multiply it by $x^n$ , then sum both sides for n from 0 to infinity! This step is used to turn our recurrence into an equation describing F.

$\sum_{n=0}^\infty f_{n+2}x^n = \sum_{n=0}^\infty f_{n+1}x^n + \sum_{n=0}^\infty f_{n} x^n$

The last term is F(x). It's not so obvious what to do with our second last term, but we ultimately want to write it in terms of F(x). Here's what we do:

$F(x) = f_0 + f_1 x + f_2 x^2 + \ldots$

$\frac{F(x) - f_0}{x} = f_1 + f_2 x + \ldots$

But the RHS of this equation is just the second last term, so the second last term is $\frac{F(x) - f_0}{x}$ . In a similar fashion, you find that $\sum_{n=0}^\infty f_{n+2}x^n = \frac{F(x) - f_0 - f_1 x}{x^2}$ .

So we've turned out recurrence equation into $\frac{F(x) - f_0 - f_1 x}{x^2} = \frac{F(x) - f_0}{x} + F(x)$ . We know that $f_0=0$ and $f_1 = 1$ , so we can use this to solve for F(x), giving $F(x) = \frac{x}{1-x-x^2}$ . I expect you to perform this calculation.

Step 3. The idea of this stage is to expand our function as a series, to do that we'll be making use of the fact that $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$ this is just the geometric series formula, so keep this in mind. We can use partial fractions to write $\frac{x}{1-x-x^2} = \frac{A}{x - r} + \frac{B}{x - r'}$ where r and r' are the roots of the denominator i.e. roots of $1-x-x^2$ , and I'll suppose r is the positive root. I'll leave you to work out what r, r', A and B are. Once we've done that we can use our geometric series formula in reverse:

$\frac{A}{x - r} = -\frac{A}{r} \cdot \frac{1}{1 - (x/r)} = -\frac{A}{r} (1 + (x/r) + (x/r)^2 + \ldots) = \sum_{n=0}^\infty -\frac{A}{r^{n+1}} x^n$

$\frac{B}{x - r'} = \sum_{n=0}^\infty -\frac{B}{r'^{n+1}} x^n$

So that $F(x) = \frac{A}{x - r} + \frac{B}{x - r'} = \sum_{n=0}^\infty \left(-\frac{A}{r^{n+1}} - \frac{B}{r'^{n+1}}\right) x^n = \sum_{n=0}^\infty f_n x^n$

Equating coefficients gives us, $f_n = -\frac{A}{r^{n+1}} - \frac{B}{r'^{n+1}}$ , which is Binet's formula.

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 10:52:08 pm

I didn't get to mention how one actually does this in practice, which is to use Mathematica or Maple, or some other CAS:

Mathematica:
Input: SeriesCoefficient[x/(1 - x - x^2), {x, 0, n}]
Output: (-(1/2 (1 - Sqrt[5]))^n + (1/2 (1 + Sqrt[5]))^n)/Sqrt[5]

So the key step really is finding the generating function, the rest is routine computer work, which I've done by hand before, but it takes a few minutes for me vs a few microseconds for the computer. 8-)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 11, 2009, 11:13:18 pm

Quote from: Ahmad on December 11, 2009, 10:35:46 pm

Sure thing, there's nothing to it, really.

Step 1. Define the generating function $F(x) = \sum_{n=0}^\infty f_n x^n$ where $f_n$ is the nth Fibonacci number. F is our clothesline upon which we hang up our sequence of numbers for display!

Step 2. Write down the recurrence equation, $f_{n+2} = f_{n+1} + f_{n}$ and multiply it by $x^n$ , then sum both sides for n from 0 to infinity! This step is used to turn our recurrence into an equation describing F.

$\sum_{n=0}^\infty f_{n+2}x^n = \sum_{n=0}^\infty f_{n+1}x^n + \sum_{n=0}^\infty f_{n} x^n$

The last term is F(x). It's not so obvious what to do with our second last term, but we ultimately want to write it in terms of F(x). Here's what we do:

$F(x) = f_0 + f_1 x + f_2 x^2 + \ldots$

$\frac{F(x) - f_0}{x} = f_1 + f_2 x + \ldots$

But the RHS of this equation is just the second last term, so the second last term is $\frac{F(x) - f_0}{x}$ . In a similar fashion, you find that $\sum_{n=0}^\infty f_{n+2}x^n = \frac{F(x) - f_0 - f_1 x}{x^2}$ .

So we've turned out recurrence equation into $\frac{F(x) - f_0 - f_1 x}{x^2} = \frac{F(x) - f_0}{x} + F(x)$ . We know that $f_0=0$ and $f_1 = 1$ , so we can use this to solve for F(x), giving $F(x) = \frac{x}{1-x-x^2}$ . I expect you to perform this calculation.

Step 3. The idea of this stage is to expand our function as a series, to do that we'll be making use of the fact that $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$ this is just the geometric series formula, so keep this in mind. We can use partial fractions to write $\frac{x}{1-x-x^2} = \frac{A}{x - r} + \frac{B}{x - r'}$ where r and r' are the roots of the denominator i.e. roots of $1-x-x^2$ , and I'll suppose r is the positive root. I'll leave you to work out what r, r', A and B are. Once we've done that we can use our geometric series formula in reverse:

$\frac{A}{x - r} = -\frac{A}{r} \cdot \frac{1}{1 - (x/r)} = -\frac{A}{r} (1 + (x/r) + (x/r)^2 + \ldots) = \sum_{n=0}^\infty -\frac{A}{r^{n+1}} x^n$

$\frac{B}{x - r'} = \sum_{n=0}^\infty -\frac{B}{r'^{n+1}} x^n$

So that $F(x) = \frac{A}{x - r} + \frac{B}{x - r'} = \sum_{n=0}^\infty \left(-\frac{A}{r^{n+1}} - \frac{B}{r'^{n+1}}\right) x^n = \sum_{n=0}^\infty f_n x^n$

Equating coefficients gives us, $f_n = -\frac{A}{r^{n+1}} - \frac{B}{r'^{n+1}}$ , which is Binet's formula.

Thanks Ahmath, I can't wait till I start generating functions.

Title: Re: TT's Maths Thread
Post by: /0 on December 11, 2009, 11:14:21 pm

Quote from: Ahmad on December 11, 2009, 10:52:08 pm

I didn't get to mention how one actually does this in practice, which is to use Mathematica or Maple, or some other CAS:

Mathematica:
Input: SeriesCoefficient[x/(1 - x - x^2), {x, 0, n}]
Output: (-(1/2 (1 - Sqrt[5]))^n + (1/2 (1 + Sqrt[5]))^n)/Sqrt[5]

So the key step really is finding the generating function, the rest is routine computer work, which I've done by hand before, but it takes a few minutes for me vs a few microseconds for the computer. 8-)

I've heard some universities provide require a computer math program. Do you need to buy it yourself or do they provide?

Title: Re: TT's Maths Thread
Post by: Ahmad on December 11, 2009, 11:30:01 pm

I haven't heard of UoM providing copies of Mathematica, but they have it installed on many of the computers on campus. So I'm not entirely sure, but I'd be curious to know if you found out. :)

Title: Re: TT's Maths Thread
Post by: Mao on December 12, 2009, 10:38:14 am

Quote from: /0 on December 11, 2009, 11:14:21 pm

Quote from: Ahmad on December 11, 2009, 10:52:08 pm
I didn't get to mention how one actually does this in practice, which is to use Mathematica or Maple, or some other CAS:

Mathematica:
Input: SeriesCoefficient[x/(1 - x - x^2), {x, 0, n}]
Output: (-(1/2 (1 - Sqrt[5]))^n + (1/2 (1 + Sqrt[5]))^n)/Sqrt[5]

So the key step really is finding the generating function, the rest is routine computer work, which I've done by hand before, but it takes a few minutes for me vs a few microseconds for the computer. 8-)

I've heard some universities provide require a computer math program. Do you need to buy it yourself or do they provide?

Generally universities usually only have a site license. But I hear the Internet does provide Mathematica and other packages. *wink*

Title: Re: TT's Maths Thread
Post by: TrueTears on December 12, 2009, 03:27:37 pm

Question: Consider 10 people sitting around a circular table. In how many ways can they change seats so that each person has a different neighbour to the right?

This is my attempt using PIE but are there any other ways? Because computing the last sum seems pretty tedious...

We want to find the complement such that each person has the same neighbour.

Start off with 10 people around the table label them $1,2,3, \cdots, 10$

We need to find $(1,2) \cup (2,3) \cup \cdots \cup (10,1)$ A total of $10$ pairs.

$|(1,2) \cup (2,3) \cup \cdots \cup (10,1)| = |(1,2)|+|(2,3)| + \cdots + |(10,1)| - \left[|(1,2)(2,3)|+ \cdots + |(1,2)(10,1)| \cdots\right] + \left[|(1,2)(2,3)(3,4)| + |(1,2)(2,3)(4,5)| + \cdots + |(1,2)(2,3)(10,1)| \cdots \right]$

$- \cdots + (-1)^{10}|(1,2)(2,3)\cdots(10,1)|$

$|(1,2) \cup (2,3) \cup \cdots \cup (10,1)| = \binom{10}{1}10(8!) - \binom{10}{2}10(7!) + \binom{10}{3}10(6!) - \cdots - \binom{10}{9}10(0!) + \binom{10}{10}$

The total number of ways of rearranging the 10 people is $10!$

Thus we require $10! - \left[\binom{10}{1}10(8!) - \binom{10}{2}10(7!) + \binom{10}{3}10(6!) - \cdots - \binom{10}{9}10(0!) + \binom{10}{10}\right]$

Title: Re: TT's Maths Thread
Post by: zzdfa on December 12, 2009, 07:45:13 pm

Quote from: TrueTears on December 11, 2009, 06:33:54 pm

Use a combinatorial argument to prove that $n! = \sum_{r=0}^n\binom{n}{r}(n-r)!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \cdots + (-1)^{n-r}\frac{1}{n!}\right)$

have you tried using this identity on that last sum?

Title: Re: TT's Maths Thread
Post by: kenhung123 on December 12, 2009, 09:24:25 pm

Hey TT, sorry for off topic. How many cheat sheets did you have?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 12, 2009, 09:30:46 pm

Quote from: kenhung123 on December 12, 2009, 09:24:25 pm

Hey TT, sorry for off topic. How many cheat sheets did you have?

For what? I don't really have any cheat sheets for Maths, but I had a lot of notes for Physics and Chem.

Title: Re: TT's Maths Thread
Post by: kenhung123 on December 12, 2009, 09:51:10 pm

Physics

Title: Re: TT's Maths Thread
Post by: TrueTears on December 12, 2009, 10:05:08 pm

Quote from: kenhung123 on December 12, 2009, 09:51:10 pm

Physics

Yeah had several cheat sheets.

Title: Re: TT's Maths Thread
Post by: kenhung123 on December 12, 2009, 10:08:59 pm

How many did you upload?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 12, 2009, 10:09:23 pm

Quote from: kenhung123 on December 12, 2009, 10:08:59 pm

How many did you upload?

Most I think.

Title: Re: TT's Maths Thread
Post by: kenhung123 on December 12, 2009, 10:34:39 pm

Do you have heinemann physics solutions?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 12, 2009, 10:35:26 pm

Quote from: kenhung123 on December 12, 2009, 10:34:39 pm

Do you have heinemann physics solutions?

I didn't use Heinemann physics hehe

Title: Re: TT's Maths Thread
Post by: kenhung123 on December 12, 2009, 10:46:00 pm

Quote from: TrueTears on December 12, 2009, 10:35:26 pm

Quote from: kenhung123 on December 12, 2009, 10:34:39 pm
Do you have heinemann physics solutions?
I didn't use Heinemann physics hehe

Oh thanks anyway.

P.S. I found your synchrotron cheat sheet??O.O

Title: Re: TT's Maths Thread
Post by: TrueTears on December 13, 2009, 03:32:19 am

Use generating functions to show that for each positive integer $\mbox{n}$ , the number of partitions of $\mbox{n}$ into unequal parts is equal to the number of partitions of $\mbox{n}$ into odd parts.

The generated function for the number of unequal parts is $(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5) \cdots$

The generated function for the number of odd parts is $(1+x+x^2+x^3+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^5+x^{10}+x^{15}+\cdots) \cdots$

So for example if $n = 6$ then there are $4$ unequal parts: $1+5, 1+2+3, 2+4, 6$

And the coefficient of $x^6$ in $(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5) \cdots$ is indeed $4$ .

Now the book tells me to ponder why $(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5) \cdots$ and $(1+x+x^2+x^3+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^5+x^{10}+x^{15}+\cdots) \cdots$ are the generated function for unequal/odd parts, however no matter how much I think about it, it doesn't become "obvious" to me, can someone explain the thought process of arriving at each of those functions?

Thanks :)

Title: Re: TT's Maths Thread
Post by: Ahmad on December 13, 2009, 01:25:35 pm

I think it comes from experience, from seeing lots of infinite products you learn to write and think in terms of them, just like one of the best ways to learn to write is by reading! I'll go through a few examples and hopefully that will help. :)

Let's think of what a finite product such as $(1+x)(1+x^2)(1+x^3)$ means. Well how do you typically expand out something like this? You select one term from each bracket and multiply them, then sum up all such products! For example if I select $x$ from the first bracket, $1$ from the second and $x^3$ from the third, then the product will be $x\cdot 1\cdot x^3 = x^{1+0+3}$ . To expand it out fully you have to do this 2 x 2 x 2 = 8 times, for each possible selection of one term from each bracket, then add all the possible products.

A slightly trickier example: find $[x^n] (1 + x + x^2 + \cdots)(1 + x + x^2 + \cdots)$ . So what possible selections of a term from each bracket are there, so that the product of the two selections is $x^n$ . Well a general term of the first bracket is $x^p$ and a general term of the second is $x^q$ where p and q are non-negative (remember p=0 represents selecting 1 from the first bracket), and so the general product is $x^{p+q}$ which we want to equal $x^n$ and so the coefficient of $x^n$ is the number of solutions to $p+q = n$ .

What about $[x^n] (1 + x + x^2 + \cdots)(1 + x^2 + x^4 + \cdots)$ ? Well a general term of the first bracket is $x^p$ and a general term of the second is $x^{2q}$ , so a general term of the product is $x^{p+2q}$ , and the coefficient of $x^n$ is the number of non-negative solutions to $p+2q = n$ .

Concerning our original problem which is integer partitions, how do we write a partition of 6? Well we can write any such partition as $6 = a_1 \cdot 1 + a_2 \cdot 2 + a_3 \cdot 3 + \cdots$ , where $a_i$ is non-negative. For example 6 = 1+1+4, here we'd have $a_1 = 2,\, a_4 = 1$ and all other $a_i = 0$ . Can you see how this might relate to our previous example? Here we want to write an integer n as $n = a_1 \cdot 1 + a_2 \cdot 2 + a_3 \cdot 3 + \cdots$ , previously we had $p+2q = n$ . So a natural choice here is to consider the infinite product $(1 + x + x^2 + \cdots)(1 + x^2 + x^4 + \cdots)(1 + x^3 + x^6 + \cdots)$ .

Some exercises you might like to think about:
0. Can you appropriately interpret how to expand out $(1+x)^n = \underbrace{(1+x)(1+x)\cdots(1+x)}_{\mbox{n times}}$ to give a proof of the binomial theorem?

1. Let $f(n)$ denote the number of non-negative solutions to $a_1 + a_2 + \cdots + a_k = n$ . Find the generating function for the counting sequence $f(n)$ .

2. What is $(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots$ when expanded out? (Hint: think binary).

3. Write $\left(1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \cdots\right)\left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots\right)\left(1 + \frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} + \cdots\right)\cdots$
in expanded form (hint: unique factorization of the positive integers), where the 2, 3, 5, etc are the prime numbers. Alter this slightly to show that $\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{\mbox{p prime}} \frac{1}{1 - p^{-s}}$ which is Euler's product for the Riemann Zeta function.

:)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 13, 2009, 08:23:54 pm

Thank you Ahmad, your explanations are always so easy to understand :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 13, 2009, 08:25:39 pm

use exercise 3 and the fundamental theorem of arithmetic to show that there are infinitely many primes :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 13, 2009, 10:51:11 pm

Quote

1. Let $f(n)$ denote the number of non-negative solutions to $a_1 + a_2 + \cdots + a_k = n$ . Find the generating function for the counting sequence $f(n)$ .

$a_1$ can take any value between $[0,\infty)$ , so can $a_2, a_3, \cdots, a_k$

so define $a_1(x) = a_2(x) = a_3(x) = \cdots = a_k(x) = 1+x^1+x^2+x^3+x^4+\cdots$

Thus $f(n) = a_1(x)a_2(x)a_3(x) \cdots a_k(x) = (1+x^1+x^2+x^3+x^4+\cdots)^k = \left(\frac{1}{1-x}\right)^k = \frac{1}{(1-x)^k}$

Quote

2. What is $(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots$ when expanded out? (Hint: think binary).

Expanding out the first few factors we get: $x^7+x^6+x^5+x^4+x^3+x^2+x^1+1$

My conjecture is that every non-negative integer can be expressed as unique powers with base 2 and that there is only one way to achieve this.

Now how to prove it?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 13, 2009, 11:10:04 pm

if it were not true, computers would be nonexistent.
computers exist, hence it is true. qed.

(strong induction works fine too)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 13, 2009, 11:53:24 pm

lol thanks Ahmad for hint :P

$\frac{(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})}{1-x} = \frac{(1-x^2)(1+x^2)(1+x^4)(1+x^8)\cdots(1+x^{2^n})}{1-x} = \frac{(1-x^{2^n})(1+x^{2^n})}{1-x} = \frac{1-x^{2^{n+1}}}{1-x}$

$= x^{2^{n+1}-1} + \frac{1}{1-x} = x^{2^{n+1}-1} + \left(1+x^1+x^2+x^3+ \cdots\right)$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 14, 2009, 12:34:13 am

an extension:

prove that for all $b>1$ :

$(1+x+x^2...x^{b-1})(1+x^b+x^{2b}...+x^{(b-1)b})(1+x^{b^2}+x^{2b^2}....+x^{(b-1)b^2})(1+x^{b^3}+x^{2b^3}....+x^{(b-1)b^3})....=1+x+x^2+x^3...$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 14, 2009, 04:11:31 am

Quote from: zzdfa on December 13, 2009, 11:10:04 pm

if it were not true, computers would be nonexistent.
computers exist, hence it is true. qed.

(strong induction works fine too)

Yeap so using strong induction.

Let $\mbox{n}$ be the integer which will be represented as the sums of unique powers of $2$ .

$n = M + x$ where $M$ and $x$ (' $x$ ' can be a sum of distinct powers of $2$ ) are both distinct powers of $2$ . And clearly $M \le n$

For example, $37 = 2^5+2^2+2^0$

Here $n = 37, M = 2^5$ and $x = 2^2+2^0$

Our inductive hypothesis is that for all integers less than $\mbox{n}$ , can also be expressed like $M+x$ .

However if $x$ contains a $M$ , then we can not satisfy the condition of having unique powers.

I conjecture that $x < M$

Consider if $x > M$ .

Thus let $x = M+y$ (where $y$ is some integer).

$\therefore n = M + M + y \implies n = 2M+y$

$\therefore n > 2M$ Contradiction! (Since $2M$ is just a larger power of $2$ so the equation becomes $n = 2M + x$ and clearly $2M \le n$ )

Thus $x<M$

Now by strong induction, all integers less than $\mbox{n}$ can be expressed in the form $n = M+x$ which completes the proof.

Title: Re: TT's Maths Thread
Post by: kenhung123 on December 14, 2009, 10:05:59 am

How'd you do TT?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 14, 2009, 12:53:00 pm

Now you have to prove uniqueness. and congratz on the score :D

Title: Re: TT's Maths Thread
Post by: TrueTears on December 14, 2009, 11:48:00 pm

Quote from: kamil9876 on December 14, 2009, 12:53:00 pm

Now you have to prove uniqueness. and congratz on the score :D

Finally time to do some maths!

What do you mean prove uniqueness?

Title: Re: TT's Maths Thread
Post by: humph on December 14, 2009, 11:52:17 pm

You've proved existence (that every integer can be written in that form). You also need to prove uniqueness (that there is only one such form for each integer).

P.S. How did you go in each of your subjects? :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 15, 2009, 12:00:45 am

meaning that a number cannot have two different representations. ie let's write it down in descending powers:

$2^{a_1}+2^{a_2}....2^{a_m}=2^{b_1}+2^{b_2}... 2^{b_k}$

now we want to show that if $a_i=b_i$ for all $i$ DOES not hold, then we arrive at a contradiction:

Now suppose the expressions are not the same: we can cancel out any terms common to both sides, thus all terms involved are different and only one side contains the maximum term, call this $2^M$ . Say WLOG that LHS contains this term. Thus we have

$LHS\geq 2^M$ (1)

Now let the RHS have a maximum term, call it $2^K$ thus the largest possible value for the RHS is $1+2^1+2^2...+2^K=2^{K+1}-1$

Therefore $RHS\leq 2^{K+1}-1$

$\implies RHS<2^{K+1}$ , but $2^{K+1}\leq 2^M$ since $K$ is strictly less than $M$ . Therefore:
$ RHS<2^M$ (2)

But (1) and (2) contradict $RHS=LHS$ . Thus we cannot have different expressions for the same number.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 15, 2009, 02:00:53 am

Thanks!!!!

I'm really enjoying generating functions, it makes such good use of algebra to tell a story.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 15, 2009, 03:31:26 pm

Hmm the Catalan Recurrence, how does one prove it using generating functions to get a closed form of the recurrence formula: $t_n = \sum_{u+v=n+1} t_ut_v$

Title: Re: TT's Maths Thread
Post by: Ahmad on December 15, 2009, 04:15:31 pm

We have $t_0 = 1$ and $t_n = \sum_{i=1}^n t_{i-1} t_{n-i}$ for $n > 0$ . Define $T(x) = \sum_{n=0}^\infty t_n x^n$ . As per usual to find the generating function we multiply the recurrence equation by $x^n$ and sum over all values of n for which the recurrence is valid. The rest is just about having the fortitude to carry out the computation, which with enough experience doesn't require much thinking at all. However, I'm aware it can be difficult to do (and understand) in the beginning, so feel free to ask about any of the steps below if you're confused and I'll go through how it works.

$\sum_{n\ge 1} t_n x^n = \sum_{n\ge 1} \sum_{i=1}^n t_{i-1} t_{n-i} x^n$

The left hand side is just $T(x) - t_0 x^0 = T(x) - 1$ . At this point I want to interchange the order of summation (to break the inner sum's dependence on the outer sum's dummy variable).

$= \sum_{i=1}^\infty \sum_{n=i}^\infty t_{i-1} t_{n-i} x^n$

$= \sum_{i=1}^\infty t_{i-1} \sum_{n=i}^\infty t_{n-i} x^n$

Now I'm going to perform an index shift in the inner sum, by starting the sum at n=0, and replacing n by n+i in the summand.

$= \sum_{i=1}^\infty t_{i-1} \sum_{n=0}^\infty t_{n} x^{n+i}$

$=\left( \sum_{i=1}^\infty t_{i-1} x^i \right) \left(\sum_{n=0}^\infty t_{n} x^{n} \right)$

And we've broken the sum into independent parts, which is what I tried to hint at as my reason for interchanging summation order. The second bracket is just $T(x)$ , so we'll work on the first bracket. To do this we perform an index shift starting the sum at i=0 and changing i to i+1 in the summand.

$=\left( \sum_{i=0}^\infty t_{i} x^{i+1} \right) \left(\sum_{n=0}^\infty t_{n} x^{n} \right)$

$=x \cdot \left( \sum_{i=0}^\infty t_{i} x^{i} \right) \left(\sum_{n=0}^\infty t_{n} x^{n} \right)$

$= x (T(x))^2$

And we've done the hard part. We have $T(x) - 1 = x (T(x))^2$ . Now you can solve this functional equation for $T(x)$ . You'll get a quadratic and hence two solutions, think about which one you should select. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 15, 2009, 07:45:59 pm

Quote from: Ahmad on December 15, 2009, 04:15:31 pm

We have $t_0 = 1$ and $t_n = \sum_{i=1}^n t_{i-1} t_{n-i}$ for $n > 0$ . Define $T(x) = \sum_{n=0}^\infty t_n x^n$ . As per usual to find the generating function we multiply the recurrence equation by $x^n$ and sum over all values of n for which the recurrence is valid. The rest is just about having the fortitude to carry out the computation, which with enough experience doesn't require much thinking at all. However, I'm aware it can be difficult to do (and understand) in the beginning, so feel free to ask about any of the steps below if you're confused and I'll go through how it works.

$\sum_{n\ge 1} t_n x^n = \sum_{n\ge 1} \sum_{i=1}^n t_{i-1} t_{n-i} x^n$

The left hand side is just $T(x) - t_0 x^0 = T(x) - 1$ . At this point I want to interchange the order of summation (to break the inner sum's dependence on the outer sum's dummy variable).

$= \sum_{i=1}^\infty \sum_{n=i}^\infty t_{i-1} t_{n-i} x^n$

$= \sum_{i=1}^\infty t_{i-1} \sum_{n=i}^\infty t_{n-i} x^n$

Now I'm going to perform an index shift in the inner sum, by starting the sum at n=0, and replacing n by n+i in the summand.

$= \sum_{i=1}^\infty t_{i-1} \sum_{n=0}^\infty t_{n} x^{n+i}$

$=\left( \sum_{i=1}^\infty t_{i-1} x^i \right) \left(\sum_{n=0}^\infty t_{n} x^{n} \right)$

And we've broken the sum into independent parts, which is what I tried to hint at as my reason for interchanging summation order. The second bracket is just $T(x)$ , so we'll work on the first bracket. To do this we perform an index shift starting the sum at i=0 and changing i to i+1 in the summand.

$=\left( \sum_{i=0}^\infty t_{i} x^{i+1} \right) \left(\sum_{n=0}^\infty t_{n} x^{n} \right)$

$=x \cdot \left( \sum_{i=0}^\infty t_{i} x^{i} \right) \left(\sum_{n=0}^\infty t_{n} x^{n} \right)$

$= x (T(x))^2$

And we've done the hard part. We have $T(x) - 1 = x (T(x))^2$ . Now you can solve this functional equation for $T(x)$ . You'll get a quadratic and hence two solutions, think about which one you should select. :)

Oh right, very clever, thanks Madah! I can do the rest now.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 15, 2009, 11:49:53 pm

Say we have $(1+y)^{\lambda}$ then the generalised binomial theorem says we can expand this by using:

$(1+y)^{\lambda} = 1 + \sum_{n \ge 1}\left(\frac{\lambda(\lambda-1)\cdots(\lambda-n+1)}{n!}y^{n}\right)$

Now my question is, how do you prove this generalised binomial theorem? Combinatorially? Algebraically?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 16, 2009, 12:05:31 am

you could try proving for rational lambda first:
so let lambda=p/q and sub it in, take both sides to the power of q, and you're left with proving that $(1+x)^p =\mathrm{ stuff}$ which looks doable (but messy).

then once youve done that, you can probably use some sort of limiting argument to show that
if the identity holds for rational lambda then it holds for irrational lambda as well.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 16, 2009, 12:35:24 am

Hmm, I didn't actually prove it or anything I think I just rewrote it in another form and it happens to be the same.

Say we have $(x+y)^{\lambda}$ where $\lambda \in \mathbb{R}$

Then we have $(x+y)^{\lambda} = \binom{\lambda}{0}x^{\lambda}y^0 + \binom{\lambda}{1}x^{\lambda-1}y^1 + \cdots + \binom{\lambda}{\lambda}x^0y^{\lambda} = \sum_{i=1}^{\lambda}\binom{\lambda}{i}x^{\lambda-i}y^i$

But $\binom{\lambda}{i} = \frac{(\lambda)(\lambda-1)(\lambda-2) \cdots (2)(1)}{i!(\lambda-i)!} = \frac{(\lambda)(\lambda-1)(\lambda-2) \cdots (\lambda-i)!}{i!(\lambda-i)!} =\frac{(\lambda)(\lambda-1)(\lambda-2) \cdots (\lambda-i+1)}{i!}$

$\therefore (x+y)^{\lambda} = \sum_{i=0}^{\lambda}\binom{\lambda}{i}x^{\lambda-i}y^i = \sum_{i=0}^{\lambda}\left[\left(\frac{(\lambda)(\lambda-1)(\lambda-2) \cdots (\lambda-i+1)}{i!}\right)\left(x^{\lambda-i}y^i\right)\right] = x^{\lambda} + \lambda x^{\lambda-1}y + \left(\frac{\lambda(\lambda-1)}{2}\right)x^{\lambda-2}y^2 + \cdots$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 16, 2009, 12:45:05 am

yeah that's the idea Newton had behind it, to extend the natural exponent case. He even experimented by deriving from this uncertain idea the expression for $(1+x)^{\frac{1}{2}}$ and then squaring and noticing that he did indeed get $(x+1)$

I think it can be done with taylor series. Since the kth derivative of $(1+x)^a$ is $a(a-1)(a-2)..(a-k+1)(x+1)^{a-k}$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 16, 2009, 12:46:25 am

Right so it exists for all real powers, what about complex?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 16, 2009, 12:48:05 am

I don't know. I havn't even learnt how to multiply real numbers yet.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 16, 2009, 02:56:29 am

Finally finished the recurrence chapter in combinatorics, took a long read cause read generating functions during it hahaha. Anyways here are some cool questions for thought (some probs require generating functions).

1. For a set $S$ of integers, define $S+1$ to be $\{x+1:x \in S\}$ . How many subsets of $S$ of $\{1,2,3, \cdots. n\}$ satisfy $S \cup (S+1) = \{1,2,3, \cdots, n\}$ ? What about a generalisation?

~~2. Find the number of subsets of $\{1,2,3, \cdots. n\}$ that contain no two consecutive elements of $\{1,2,3, \cdots. n\}$ .~~

~~3. Fix positive $a,b,c$ . Define a sequence of real numbers by $x_0 = a, x_1 = b \ \mbox{and} \ x_{n+1}=cx_nx_{n-1}$ . Find a formula for $x_n$ .~~

4. For each $n \ge 1$ , we call a sequence of $\mbox{n}$ (s and n)s 'legal' if the parentheses match up. For example, if $n =4$ , the sequence (()()()) is legal, but ()())(() is not. Let $l_n$ denote the number of legal arrangements for the $2n$ parentheses. Find a recurrence formula for $l_n$ .
[WHAT THE FUCK IS THIS QUESTION EVEN ON ABOUT???? WHAT DOES IT EVEN MEAN AND WHAT IS IT ASKING FOR?????]

Title: Re: TT's Maths Thread
Post by: zzdfa on December 16, 2009, 09:20:30 am

'how many ways can you arrange n pairs of brackets so that they make sense'

1 pair:
)( does not make sense
() does make sense

2 pair:
)()( does not make sense
(()) makes sense
()() makes sense

find a recurrence formula for l_n

and re. the generalized binomial theorem:

Quote from: TrueTears on December 16, 2009, 12:35:24 am

But $\binom{\lambda}{i} = \frac{(\lambda)(\lambda-1)(\lambda-2) \cdots (2)(1)}{i!(\lambda-i)!} = \frac{(\lambda)(\lambda-1)(\lambda-2) \cdots (\lambda-i)!}{i!(\lambda-i)!} =\frac{(\lambda)(\lambda-1)(\lambda-2) \cdots (\lambda-i+1)}{i!}$

that line is only true if lamda is a positive integer.

Quote from: kamil9876 on December 16, 2009, 12:48:05 am

I don't know. I havn't even learnt how to multiply real numbers yet.

lol wut?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 16, 2009, 03:32:17 pm

2.)
Let $x_n$ be the number of ways of doing it for a set containing first n naturals.

We can partition these subsets into the following: Those that do contain n, those that do not contain n. If we have n in our set, then we can pick any subset from {1,2,3,4...n-2} such that there are no consecutives there, there are $x_{n-2}$ ways of doing this. If however our set does not have n, then we can choose any subset from {1,2,3...n-1} that does not have any consecutives, there are $x_{n-1}$ ways of doing this. Now we have covered every possibility in a mutually exlusive way, thus we add up:

$ x_n=x_{n-1}+x_{n-2}$

and $x_1$ and $x_2$ can be counted manually.

I'll let you think about the rest since some of them use a similair strategy.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 16, 2009, 06:21:20 pm

Quote from: kamil9876 on December 16, 2009, 03:32:17 pm

2.)
Let $x_n$ be the number of ways of doing it for a set containing first n naturals.

We can partition these subsets into the following: Those that do contain n, those that do not contain n. If we have n in our set, then we can pick any subset from {1,2,3,4...n-2} such that there are no consecutives there, there are $x_{n-2}$ ways of doing this. If however our set does not have n, then we can choose any subset from {1,2,3...n-1} that does not have any consecutives, there are $x_{n-1}$ ways of doing this. Now we have covered every possibility in a mutually exlusive way, thus we add up:

$ x_n=x_{n-1}+x_{n-2}$

and $x_1$ and $x_2$ can be counted manually.

I'll let you think about the rest since some of them use a similair strategy.

Haha clever, I'm noticing that for a lot of recurrence problems, it is simply just partitions with a bit of fiddling around?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 16, 2009, 06:27:58 pm

yep. Usually you want to partition it in such a way that the each individual part is a case that can be dealt with by knowledge of the previous term in the sequence. A common example of this is using your knowledge about n-1 elements by removing an object form a set of n elements. Or simply using knowledge about n elements, and then adding an object to see what happens(probably the more common sense way). Notice that this reminds you of induction, and in fact induction goes hand in hand with this sometimes.

Another example of this method:

how many ways are there of partitioning the set {1,2,3,....2009} into $3$ (mutually exclusive) subsets such that no two elements in the same set are consecutive?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 16, 2009, 06:44:19 pm

Right, so basically for Q 2 we just have the Fibonacci sequence and the answer would just be the closed form of it haha. Ofcourse the domain of n would be restricted.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 16, 2009, 09:56:08 pm

1.)

don't know if there is a simpler solution but...

S satisfies the property iff all of the following are satisfied:

1.) $1 \in S$
2.) $max(S)=n-1$
3.) $whenever m-1 \notin S, m \in S$ (for m<n-1 of course)

Therefore 1 and n-1 are definitely in S. The elements 2,3,4...n-2 can be chosen in any way, as long as condition 3) is satisfied.

To gain more insight, we can represent our possible choices in a tree diagram:

(http://vcenotes.com/forum/index.php?action=dlattach;topic=19896.0;attach=4262;image)

Now each path represents some subset. If the path includes m, it means you have chosen m, if it includes m', it means m' is not in the set. e.g the middle path represents {1,2,4}. The construction is such that property 3) is satisfied. This particular tree shows the answer for n=6,5,4,3,2,1. Ie to find the answer for n=6, just count how many there are in the last column: 5. the answer for n=5 is how many there are in the second last column: 3. etc. and the tree can be extended infinitely. Therefore we wish to find how many there are in the (n-2)th column.

Firstly some notation: let $x_k$ denote the number of elements in the kth column that are dashed ' (in other words, the number of $k'$ in our tree). Let $y_k$ denote the number of non-dashed elements in the kth column (in other words the number of $k$ in our tree).

We have the following recurrence relations:

$ y_{k+1}=x_k+y_k$ (because there are $y_k+x_k$ elements in the kth row, and each branches of to an element (k+1) (without a '))

$x_{k+1}=y_k$ (since only a non-dashed element can branch of to a dashed one)

now define $f(k)=x_k+y_k$ . clearly we wish to know the sequence f(1),f(2),f(3)... etc. and the recurrences above can be used to find a recurrence for f(k). You can guess that this recurrence is fibonacci-like just from computing a few values using the tree method (i did this manually, trust me it's not that long to work out the first few couple, and you see they are fibonacci).

Now you can just manipulate the recurrences above algebraically to prove this, or you could even use linear algebra I can imagine. Maybe even generating functions.

$f(k+2)=x_{k+2}+y_{k+2}$
$=y_{k+1}+x_{k+1}+y_{k+1}$
$=f(k+1)+x_k+y_k=f(k+1)+f(k)$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 16, 2009, 11:02:20 pm

ok I am emabarrased to say this but thinking over my argument and disliking it's complexity and length, I realised there is a simpler one :P. Using properties 1,2,3: we see that we can either choose n-2 or not choose n-2, hence just partition the subsets into these two sets and you're done! :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 16, 2009, 11:31:38 pm

Haha that is awesome kamil, I hardly think of tree diagrams damn.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 17, 2009, 03:38:27 am

Quote

4. For each $n \ge 1$ , we call a sequence of $\mbox{n}$ (s and n)s 'legal' if the parentheses match up. For example, if $n =4$ , the sequence (()()()) is legal, but ()())(() is not. Let $l_n$ denote the number of legal arrangements for the $2n$ parentheses. Find a recurrence formula for $l_n$ .

Consider the specific case when $n = 10$ .

Now imagine we have $20$ slots _ _ _ _ _ $\cdots$ _

Each slot is distinct and lets name them $1,2,3, \cdots 20$ .

A partition can be done like this: Let $A_k$ be the set containing all possible ways of having a left bracket ' $($ ' with a correspondent right bracket ' $)$ ' on the slot $k$ .

For example, $($ $)$ _ _ _ _ $\cdots$ _ would be $A_2$

However $($ _ _ $)$ _ _ _ $\cdots$ _ would be $A_4$ .

Noticing that $k \in \{\mbox{even numbers}\}$

Thus $l_{10} = |A_2| + |A_4| + |A_6| + \cdots + |A_{20}|$

Now let's take a look at say $|A_6|$

$($ _ _ _ _ $)$ _ _ _ $\cdots$ _
1 2 3 4 5 6 7 8 9 20

Now clearly $|A_6| = (l_2)(l_{\frac{20-7+1}{2}}) = (l_2)(l_7)$ since the slots in between must also contain legal brackets.

Now what about $|A_2|$ ?

Using the same principle we should get $|A_2| = (l_0)(l_9)$

So clearly $l_{10} = (l_0)(l_9) + (l_1)(l_8) + (l_2)(l_7) + \cdots + (l_9)(l_0)$

This looks very similar to the Catalan Recurrence.

Now we can generalise.

Imagine now there are $\mbox{n}$ pairs which means now there are $2n$ slots.

$l_{n} = |A_2| + |A_4| + |A_6| + \cdots + |A_{2n}|$

$|A_{2}| = (l_0)(l_{\frac{2n-2}{2}}) = (l_0)(l_{n-1})$

$|A_4| = (l_1)(l_{\frac{2n-4}{2}}) = (l_1)(l_{n-2})$

$\therefore l_n = (l_0)(l_{n-1}) + (l_1)(l_{n-2}) + (l_2)(l_{n-3}) + \cdots + (l_{n-1})(l_{0})$

Now we have the recurrence formula for the question.

What happens if we evaluate $l_1, l_2$ etc?

Let us define $l_0 = 1$

$l_1 = 1$

$l_2 = 2$

$l_3 = (l_0)(l_2) + (l_1)(l_1) + (l_2)(l_0) = 2 + 1 + 2 =5$

$l_4 = (l_0)(l_3) + (l_1)(l_2) + (l_2)(l_1) + (l_3)(l_0) = 5 + 2 + 2+ 5 = 14$

If we continue the sequence we get $1,1,2,5,14,42,132, \cdots$ which is the Catalan sequence!

$\therefore l_n = \frac{1}{n+1}\binom{2n}{n}$ for $n \ge 0$

Title: Re: TT's Maths Thread
Post by: /0 on December 17, 2009, 04:54:36 am

Very nice TT!

Quote from: TrueTears on December 16, 2009, 02:56:29 am

3. Fix positive $a,b,c$ . Define a sequence of real numbers by $x_0 = a, x_1 = b \ \mbox{and} \ x_{n+1}=cx_nx_{n-1}$ . Find a formula for $x_n$ .

I've thought about a really messy possibility

$\begin{array}{rcl}x_0 &=& a \\\\ x_1 &=& b \\\\ x_2 &=& cab \\\\ x_3 &=& c^2ab^2 \\\\ x_4 &=& c^4a^2b^3 \\\\ x_5 &=& c^7a^3b^5 \\\\ x_6 &=& c^{12}a^5b^8 \end{array}$

If the closed form is like $x_n = c^{c_n}a^{a_n}b^{b_n}$ , we have

$a_{n} = a_{n-1}+a_{n-2}$

$b_{n}=b_{n-1}+b_{n-2}$

So $a_n$ and $b_n$ can be found with binet's formula

For $c_n$ we have

$c_n = 1+c_{n-1}+c_{n-2}$

If we write the following equations:

$\begin{array}{rcl} c_n &=& 1+\rlap{/}{c_{n-1}}+\rlap{/}{c_{n-2}} \\\\ \rlap{/}{c_{n-1}} &=&1 +\rlap{/}{c_{n-2}}+\rlap{/}{c_{n-3}} \\\\ \rlap{/}{2\rlap{/}{c_{n-2}}} &=& 2+\rlap{/}{2\rlap{/}{c_{n-3}}}+\rlap{/}{2\rlap{/}{c_{n-4}}} \\\\ \rlap{/}{3\rlap{/}c_{n-3}}} &=& 3+\rlap{/}{3\rlap{/}{c_{n-4}}}+\rlap{/}{3\rlap{/}{c_{n-5}}} \\\\ \rlap{/}{5\rlap{/}{c_{n-4}}}&=&5+\rlap{/}{5\rlap{/}{c_{n-5}}}+\rlap{/}{5\rlap{/}{c_{n-6}}}\\ &. \\ &. \\\\ \rlap{/}{F}(n\rlap{/}{-}2)\rlap{/}{c_{3}}&=& \rlap{/}{F}(n\rlap{/}{-}2)+\rlap{/}{F}(n\rlap{/}{-}2)\rlap{/}{c}_2+F(n-2)c_1 \\\\ \rlap{/}{F}(n\rlap{/}{-}1)\rlap{/}{c_{2}}&=& F(n-1)(1+c_1+c_0)\end{array}$

Where $F(i)$ is the (i)th Fibonacci number, and we add all the equations up, then we should get

$c_n = \sum_{i=1}^{n-2}F(i) +F(n-2)c_1+F(n-1)(1+c_1+c_0)=\sum_{i=1}^{n-1}F(i)$ ( $c_1=c_0=0$ )

Now all there is to do is find the closed form for the fibonacci sum....
Not very elegant rofl

Title: Re: TT's Maths Thread
Post by: humph on December 17, 2009, 05:22:59 am

Quote from: /0 on December 17, 2009, 04:54:36 am

Where $F(i)$ is the (i)th Fibonacci number, and we add all the equations up, then we should get

$c_n = \sum_{i=1}^{n-2}F(i) +F(n-1)(1+c_1+c_0)=\sum_{i=1}^{n-1}F(i)$

Now all there is to do is find the closed form for the fibonacci sum....
Not very elegant rofl

By this, the answer comes out quite nicely, and you can write it in a closed form.

Title: Re: TT's Maths Thread
Post by: Ahmad on December 17, 2009, 12:42:42 pm

Quote from: /0 on December 17, 2009, 04:54:36 am

$c_n = 1+c_{n-1}+c_{n-2}$

Once you get to here a simplifying trick is to set $d_n = 1 + c_n$ , then the recursion becomes $d_n = d_{n-1} + d_{n-2}$ with $d_1 = d_0 = 1$ , which means $d_n$ is a shifted Fibonacci sequence, $d_n = F_{n+1}$ so that $c_n = F_{n+1} - 1$ .

Well done though :)

Title: Re: TT's Maths Thread
Post by: /0 on December 17, 2009, 01:03:45 pm

Quote from: Ahmad on December 17, 2009, 12:42:42 pm

Quote from: /0 on December 17, 2009, 04:54:36 am
$c_n = 1+c_{n-1}+c_{n-2}$

Once you get to here a simplifying trick is to set $d_n = 1 + c_n$ , then the recursion becomes $d_n = d_{n-1} + d_{n-2}$ with $d_1 = d_0 = 1$ , which means $d_n$ is a shifted Fibonacci sequence, $d_n = F_{n+1}$ so that $c_n = F_{n+1} - 1$ .

Well done though :)

Ah, that's clever ;p

Title: Re: TT's Maths Thread
Post by: TrueTears on December 17, 2009, 02:35:37 pm

haha awesome solution /0 ;)

So basically:

$a_n = F_{n-1} \ \forall \ n \ge 1$ and define $a_0 = 1$

$b_n = F_{n} \ \forall \ n \ge 0$

$c_n = F_{n+1}-1 \ \forall \ n \ge 0$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 17, 2009, 05:35:01 pm

Challenge question: (Putnam)

Define a selfish set to be a set that has its own cardinality as an element. Find, with proof, the number of subsets of $\{1,2, \cdots, n\}$ that are minimal selfish sets, that is, selfish sets none of whose proper subsets is selfish.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 17, 2009, 06:42:16 pm

i remember doing that one :P it's easier than the previous q's u've been doing IMO

Title: Re: TT's Maths Thread
Post by: TrueTears on December 17, 2009, 06:43:47 pm

Quote from: kamil9876 on December 17, 2009, 06:42:16 pm

i remember doing that one :P it's easier than the previous q's u've been doing IMO

lolz yeah I'm having a go now, don't post solution yet :P or put it in white if you are going to post solution, I wanna have some fun too :P

Title: Re: TT's Maths Thread
Post by: dcc on December 18, 2009, 12:14:11 am

I completely screwed up my first attempt at this, but I realised the trick as I was driving home from a jam :P

Let $C_n$ be the number of minimal selfish subsets of $A_n = \{1,2,\dots,n\}$ . Consider some $k \in \{1,2, \dots,n\}$ . We wish to find all the minimal selfish subsets of $A_n$ of cardinality $k$ .

We first note that our minimal selfish subsets cannot contain any element less then $k$ , as the existence of such an element allows us to easily construct a selfish subset, which shows that our original subset is not minimal.

Membership in our set is restricted to $k$ and elements between $k$ and up to (and including) $\textbf{n}$ . Therefore we have $n - k$ elements to choose from to make our minimal selfish subsets from. In addition, since our set has cardinality $k$ , then for each potential subset we must choose $k - 1$ numbers to create a full set.

Therefore the number of ways of creating a minimal subset of $A_n$ of cardinality $k$ is $\binom{n-k}{k-1}$ . Summing this over all valid lengths, we find that $C_n = \sum_{k=1}^{n-1} \binom{n-k}{k-1}$ , which is of course a generator of the general Fibonacci sequence.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 18, 2009, 03:54:02 am

lol I've been at this question for ages, I'm so tempted to look at your solution dcc but I feel that I have almost got it hahaa

Title: Re: TT's Maths Thread
Post by: TrueTears on December 18, 2009, 10:44:59 pm

Ok here is what I have come up with after last night's thought.

Consider listing out the minimal selfish subsets of $\{1,2,3, \cdots, n\}$

Let $M$ represent a minimal selfish subset.

$n=1 \implies \{1\} \to 1 M$

$n=2 \implies \{1\} \to 1 M$

$n=3 \implies \{1\} \ \mbox{and} \ \{2,3\} \to 2 M$

$n=4 \implies \{1\} \ \mbox{and} \ \{2,3\} \ \mbox{and} \ \{2,4\} \to 3 M$

$n=5 \implies \{1\} \ \mbox{and} \ \{2,3\} \ \mbox{and} \ \{2,4\} \ \mbox{and} \ \{2,5\} \ \mbox{and} \ \{3,4,5\} \to 5 M$

A few things can be noticed.

$M$ is a minimal selfish set if and only if $|M|$ is the smallest element of the set.

Why is this? Consider if there was an element smaller than $|M|$ in the set $M$ . Call this element $m$ .

Thus there exists subsets of $M$ which contains the element $m$ which could also be a selfish set.

Therefore, if $|M|$ is the smallest element of the set $M$ then $M$ is a minimal selfish set.

Consider the specific case when $n = 5$ , there are exactly $2+3=5$ minimal selfish sets.

Partition the minimal selfish subsets of when $n =5$ of the set $\{1,2,3,4,5\}$ to those that have $5$ and those that don't.

Those that do not have $5$ are simply the $n=4$ case which are $\{1\} \ \mbox{and} \ \{2,3\} \ \mbox{and} \ \{2,4\}$

Those that do are $\{2,5\} \ \mbox{and} \ \{3,4,5\}$ . Now note that if we have $M+1$ for the $n=3$ case then we have $\{2\} \ \mbox{and} \ \{3,4\}$ . However these are now no longer minimal selfish conditions as $|M|$ is not the smallest element in the set. So if we add $5$ into the subsets $M+1$ we will change them into minimal selfish subsets that contains $5$ .

Summing together the partitions we get $\{1\} \ \mbox{and} \ \{2,3\} \ \mbox{and} \ \{2,4\} \ \mbox{and} \ \{2,5\} \ \mbox{and} \ \{3,4,5\}$

Now let's try generalise this.

Looking at the sequence of $M$ 's we have $1,1,2,3,5, \cdots$ we see it resembles the Fibonacci sequence, so I am going to conjecture that it does follow the Fibonacci sequence and try to prove it.

Consider for all $n \ge 2$ , let there be $q$ minimal selfish subsets for the set $\{1,2,3, \cdots, n\}$ . Let there be $w$ minimal selfish subsets for the set $\{1,2,3, \cdots, n-1\}$

I conjecture that there will be exactly $q+w$ minimal selfish subsets for the set $\{1,2,3, \cdots, n+1\}$ .

Let's split our approach in two ways, those subsets of $\{1,2,3, \cdots, n+1\}$ that contain $n+1$ and those that do not.

Consider the cases where the subsets does not contain $n+1$ .

Let $M$ be a minimal selfish subset of $\{1,2,3, \cdots, n\}$ . Then $M$ must also be a minimal selfish subset of $\{1,2,3, \cdots, n+1\}$ since $\{1,2,3, \cdots, n\} \subset \{1,2,3, \cdots, n+1\}$

$\therefore$ There are $q$ minimal selfish subsets of $\{1,2,3, \cdots, n+1\}$ without the element $n+1$

Now let us consider the case where the subsets of $\{1,2,3, \cdots, n+1\}$ that contain $n+1$ .

Let $A$ be a minimal selfish subset of $\{1,2,3, \cdots, n-1\}$ .

Let us define $P = \{n+1\} \cup \{A+1\}$

Now notice # $P =$ # $(A+1)$ (Although we changed the number of elements in each set, the total number of sets do not change)

$\therefore$ $P$ is a minimal selfish subset since $|P| = |A+1|$ is the smallest element in $P$ .

This produces exactly $w$ minimal subsets of $\{1,2,3, \cdots, n+1\}$ which contain the element $n+1$ .

But why does this work? The proof is as follows:

Consider the minimal selfish set P of the set $\{1,2,3, \cdots, n+1\}$ that contains the element $n+1$ .

Now $1 \notin P$ since $\min P = |P| = 2$

$A = x-1 \ \mbox{where} \ x \in P \ \mbox{and that} \ x \neq n+1$

$A$ is a minimal selfish subsets since # $A =$ # $(P-1)$ .

$\therefore$ $P = \{n+1\} \cup \{A+1\}$

$\therefore q+w = \mbox{total number of minimal selfish subsets of the set} \ \{1,2,3, \cdots, n-1\}$

If we define $f_n, f_{n-1}$ to be the number of the minimal selfish subsets of $\{1,2,3, \cdots, n\}$ and $\{1,2,3, \cdots, n-1\}$ respectively.

Then we have $f_{n+1} = f_n + f_{n-1}$

$\mbox{Q.E.D}$

Title: Re: TT's Maths Thread
Post by: zzdfa on December 18, 2009, 11:13:21 pm

hey , about 12 lines from the bottom up, let us define p={n+1} U {A+1}

whats A+1?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 18, 2009, 11:16:39 pm

Sorry I knew that notation was shit, what I mean is: add 1 to every element in A

Also #A means the number of sets that satisfies A.

Title: Re: TT's Maths Thread
Post by: zzdfa on December 18, 2009, 11:24:10 pm

yea all makes sense now, nice :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 18, 2009, 11:25:12 pm

Quote from: zzdfa on December 18, 2009, 11:24:10 pm

yea all makes sense now, nice :)

lol, I just don't think it's rigorous enough, I mean I basically generalised a specific case, everything is just based off it, it just doesn't seem right to me >_<

Title: Re: TT's Maths Thread
Post by: zzdfa on December 18, 2009, 11:33:57 pm

:s
you proved that there is a bijection between

the set of all MSS containing n+1 <-> the set of all MSS of n-1

and

the set of all MSS not containing n+1 <-> the set of all MSS of n

therefore

f(n+1)=f(n)+f(n-1)

its completely rigorous. dont worry ;p

edit: then again, rigorous is subjective. so if you dont feel its rigorous you find out why and then plug the gap until your 100% convinced.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 18, 2009, 11:41:23 pm

Oh right I see, lol I didn't even think of bijection while doing this question.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 19, 2009, 12:18:05 am

with TT's permision, I'll post the following problems i found quite fun in the past month or so and are sort of related to combinatorics. I invite everyone to solve. I tried to put it into an ascending order of difficulty (first few are give aways, last one probably not so).

1.)Fifty numbers are chosen from the set {1, . . . , 99}, no two of which sum to 99 or 100. Prove that the
chosen numbers must be 50, 51, 52, . . . , 99.

2*.) $\mathbb{N}$ is partitioned into k (disjoint) arithmetic progressions, all of infinite length, with common differences $d_1, d_2.. d_k$ . Prove that $\sum_{i=1}^k \frac{1}{d_i}=1$

3.) You want to color the integers from 1 to 100 so that no number is divisible by a different number of
the same color. What is the smallest possible number of colors you must have?

4.) Prove that any positive integer can be represented as a sum of Fibonacci numbers, no two of which are consecutive.

5.) Given 81 positive integers all of whose prime factors are in the set {2, 3, 5}, prove that there are 4
numbers whose product is the fourth power of an integer.

6.) A 2002 x 2004 chessboard is given with a 0 or 1 written in each square so that each row and column contain an odd number of squares containing a 1. Prove that there is an ~~even~~ odd number of white unit squares containing 1.
(source had "even", but I believe it should be odd)

7.)Let n be a positive integer, and let X be a set of n+2 integers, each of absolute value at most n. Show
that there exist three distinct numbers a, b, c in X such that a + b = c.

8.) The set of positive integers is divided into finitely many (disjoint) subsets. Prove that one of them, say
$A_i$ , has the following property: There exist a positive integer m such that for any k one can find numbers $a_{1}, \cdots, a_{k} \in A_{i}$ such that $0 < a_{j+1}-a_{j} \leq m \$ ; $(1\le j \le k-1).$

*An extension to probelm 2 is that the two greatest common differences must be equal, though I don't think this result is elementary enough for this thread.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 19, 2009, 12:29:27 am

Yeah in conjunction to kamil's set of questions, these are the final set of combinatorics questions:

1. Let $u(n)$ denote the number of ways in which a set of $\mbox{n}$ elements can be partitioned into non-empty subsets. For example, $u(3)=5$ , corresponding to:

$\{a,b,c\}$

$\{a,b\}, \{c\}$

$\{a,c\},\{b\}$

$\{a\},\{b,c\}$

$\{a\},\{b\},\{c\}$

Find a recurrence relation for $u(n)$ .

2. For $n,k$ positive integers with $n \ge k$ , define $\Big\{^n_k \Big\}$ to be the number of different partitions of a set with $\mbox{n}$ elements into $k$ non-empty subsets. For example, $\Big\{^4_3 \Big\}=6$ , because there are six 3-part partitions of the set $\{a,b,c,d\}$ :

$\{a\},\{b\},\{c,d\}$

$\{a\},\{c\},\{b,d\}$

$\{a\},\{d\},\{b,c\}$

$\{b\},\{c\},\{a,d\}$

$\{b\},\{d\},\{a,c\}$

$\{c\},\{d\},\{a,b\}$

Show that for all $n>0$

a) $\Big\{^n_1 \Big\}=1$

b) $\Big\{^n_2 \Big\} = 2^{n-1}-1$

c) $\Big\{^n_{n-1} \Big\} = \binom{n}{2}$

3. Find a combinatorial argument to explain the recurrence:

$\Big\{^{n+1}_k \Big\} = \Big\{^n_{k-1} \Big\} + k \Big\{^n_k \Big\}$

Title: Re: TT's Maths Thread
Post by: zzdfa on December 19, 2009, 01:21:24 am

my proof of 2):

we have k disjoint arithmetic sequences. Let $A_i$ be the set of naturals in the i th sequence.

Define $\mu_i(n) := \mbox{The number of elements less than or equal to } n \mbox{ that are in } A_i$

it is easily shown that

$\lim_{n\rightarrow \infty} \frac{\mu_i(n)}{n} = \frac{1}{d_i}$

and

$\sum^{k}_{i=1}\frac{\mu_i(n)}{n}=1$ for all n (because every n is in exactly one $A_i$ ) (1)

so we can make $\left|\frac{\mu_i(n)}{n} - \frac{1}{d_i}\right|$ arbitrarily small.

thus we can also make $\sum^{k}_{i=1} \left|\frac{\mu_i(n)}{n} - \frac{1}{d_i}\right| \geq \left| \sum^{k}_{i=1}\frac{\mu_i(n)}{n} - \frac{1}{d_i} \right|=\left|1- \sum^{k}_{i=1}\frac{1}{d_i} \right|$ arbitrarily small.

first inequality is the triangle inequality, 2nd one is due to the (1).

therefore

$\sum^{k}_{i=1}\frac{1}{d_i}=1 \right|$
corollary: replace N by Z and the theorem still holds, this can be seen by considering a 2 sided sequence in Z as 2 one sided sequences in N and
applying the above. i think. 0 probably screws something up.

lolz number 4 was on my mth1112 exam ;p

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 19, 2009, 01:36:15 am

yeah i chose $\mathbb{N}$ because I wasn't sure what 'infinite arithmetic progression' means exactly, i meant to put $\mathbb{Z}$ but then wondered what would happen if we had this: {-1,-2,-3....} and {0,1,2,3....}. (althought if u count the former as negative it does work :P ). However I think a good definition would be that an arithmetic progression with a_i and common difference d_i is $\lbrace a_i+ md_i|m \in \mathbb{Z}\rbrace$ with positive d. Then the theorem is definitely true and zero doesn't matter since we're only performing addition (and if we have such a partition, we can always "translate it" (shift) and we still get another partition ie: change reference frame :P )

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 19, 2009, 01:39:19 am

Quote

lolz number 4 was on my mth1112 exam ;p

what a cool subject! :P

also: my soln is same as yours only that I don't use limits but just choose a suitable n that is divisible by all the common differences (like for instance the product of the differences). But the same idea in that you saw that it's a statement about "density". I've seen others solve this one with generating functions but I prefer our type of soln since it seems more natural.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 19, 2009, 02:00:24 am

Quote from: dcc on December 18, 2009, 12:14:11 am

I completely screwed up my first attempt at this, but I realised the trick as I was driving home from a jam :P

Let $C_n$ be the number of minimal selfish subsets of $A_n = \{1,2,\dots,n\}$ . Consider some $k \in \{1,2, \dots,n\}$ . We wish to find all the minimal selfish subsets of $A_n$ of cardinality $k$ .

We first note that our minimal selfish subsets cannot contain any element less then $k$ , as the existence of such an element allows us to easily construct a selfish subset, which shows that our original subset is not minimal.

Membership in our set is restricted to $k$ and elements between $k$ and up to (and including) $\textbf{n}$ . Therefore we have $n - k$ elements to choose from to make our minimal selfish subsets from. In addition, since our set has cardinality $k$ , then for each potential subset we must choose $k - 1$ numbers to create a full set.

Therefore the number of ways of creating a minimal subset of $A_n$ of cardinality $k$ is $\binom{n-k}{k-1}$ . Summing this over all valid lengths, we find that $C_n = \sum_{k=1}^{n-1} \binom{n-k}{k-1}$ , which is of course a generator of the general Fibonacci sequence.

lol just had a read of your solution dcc, it makes mine look so retarded, I understand all except for the last bit.

How do you show $C_n = \sum_{k=1}^{n-1} \binom{n-k}{k-1}$ , is a generator of the general Fibonacci sequence?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 19, 2009, 02:01:46 am

i remember doing the same soln as dcc when i first saw this q, i found out it's fibonacci by using pascal's identity.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 19, 2009, 02:23:21 am

So we must prove $C_{n-2} + C_{n-1} =C_n$

$C_n = \sum_{k=1}^{n}\binom{n-k}{k-1}$

$C_{n-1} = \sum_{k=1}^{n-1}\binom{n-1-k}{k-1}$

$C_{n-2} = \sum_{k=1}^{n-2}\binom{n-2-k}{k-1}$

$\implies \sum_{k=1}^{n}\binom{n-k}{k-1} = \sum_{k=1}^{n-1}\binom{n-1-k}{k-1} + \sum_{k=1}^{n-2}\binom{n-2-k}{k-1}$

$\mbox{RHS} = \left[\binom{n-2}{0} + \binom{n-3}{1} + \binom{n-4}{2} + \binom{n-5}{3} + \cdots + \binom{0}{n-2}\right] + \left[\binom{n-3}{0} + \binom{n-4}{1} +\binom{n-5}{2} + \cdots + \binom{0}{n-3}\right]$

$\mbox{RHS} = \binom{n-2}{0} + \binom{n-3}{0} + \binom{n-3}{1} + \binom{n-4}{1} + \binom{n-4}{2} + \binom{n-5}{2} + \binom{n-5}{3} + \cdots$

$\mbox{RHS} = \binom{n-2}{0} + \binom{n-2}{1} + \binom{n-3}{2} + \binom{n-4}{3} + \cdots$

$\mbox{RHS} = \sum_{k=1}^{n}\binom{n-k}{k-1} = \mbox{LHS}$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 19, 2009, 03:16:50 am

Quote from: kamil9876 on December 19, 2009, 12:18:05 am

4.) Prove that any positive integer can be represented as a sum of Fibonacci numbers, no two of which are consecutive.

lol kamil I got your 4).

$0,1,1,2,3,5,8,13,21, \cdots$

Let's use strong induction.

Base case:

Call our positive integer $i$ .

Let $i = 4$ .

$4 = 1+3 = f_1+f_4$ . Where $f_n$ is a fibonacci number and $n \ge 0$ .

Inductive hypothesis:

Assume that every positive integer less than or equal to $f_n$ satisfies the condition that they can be represented as a sum of Fibonacci numbers, no two of which are consecutive.

Proof:

Consider the interval between $f_n$ and $f_{n+1}$ (inclusive).

Every number in this interval can be written as $f_n + q$ where $q \in \mathbb{Z^+}$

Clearly $q \le f_{n+1} - f_n$

Which means $q \le f_{n-1}$

But using our inductive hypothesis, every positive integer less than or equal to $f_n$ satisfies the condition.

Which means $q$ satisfies the condition.

This implies all the $f_n + q$ satisfies the condition and so does $f_{n+1}$

$\therefore \ \forall \ i \in \mathbb{Z^+}$ satisfies the condition.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 19, 2009, 11:01:32 pm

Quote from: TrueTears on December 19, 2009, 12:29:27 am

1. Let $u(n)$ denote the number of ways in which a set of $\mbox{n}$ elements can be partitioned into non-empty subsets. For example, $u(3)=5$ , corresponding to:

$\{a,b,c\}$

$\{a,b\}, \{c\}$

$\{a,c\},\{b\}$

$\{a\},\{b,c\}$

$\{a\},\{b\},\{c\}$

Find a recurrence relation for $u(n)$ .

Consider the specific case for $u(5)$ .

Let us consider this $5$ element set: $\{1,2,3,4,5\}$

Obviously the element $5$ must be in every partition.

Let us denote the set that the element $5$ is in 'size $k$ ' where $k$ tells us how many elements are in that partitioned set.

For example, the partitioned set $\{1,5\},\{2,3,4\}$ has size $2$ .

Consider size 1, we have:

$\{5\}, \{1,2,3,4\}$

But how many ways can we partition $\{1,2,3,4\}$ ? Well there are $u(4)$ ways to partition it.

Thus for size 1 there are $u(4)$ ways.

Consider size 2, we have:

$\{5, \mbox{some other element}\}, \{\mbox{Three element set}\}$

For the 'some other element' we have $\binom{4}{1}$ choices and then for the 'Three element set' we have $u(3)$ ways to partition it.

In total we have $\binom{4}{1}(u(3))$ ways

Consider size 3.

We would have $\binom{4}{2}(u(2))$ ways

Size 4: $\binom{4}{3}(u(1))$ ways

Size 5: $\binom{4}{4}(u(0))$ ways

In total: $u(5) = \binom{4}{0}(u(4)) + \binom{4}{1}(u(3)) + \binom{4}{2}(u(2)) + \binom{4}{3}(u(1)) + \binom{4}{4}(u(0)) = \sum_{k=0}^4\binom{4}{k}u(4-k)$

Now let's generalise:

Say we have a $\mbox{n}$ element set.

$\therefore u(n) = \sum_{k=0}^n\binom{n-1}{k}u(n-1-k) \ \mbox{where} \ u(0) = 1$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 19, 2009, 11:05:47 pm

2.)
a.)

Quote

a) $\Big\{^n_1 \Big\}=1$

if we have n elements a_1,a_2... then the only way to partition them into one subset is to leave the set as it is.
b.)

Quote

b) $\Big\{^n_2 \Big\} = 2^{n-1}-1$

we have 2 sets, A and B and each of the n elements must belong to one set only. Hence we have to make n choices, each either A or B, there are 2^n ways of doing this. However one such allocation is "every element in A", and "every element in B", and we cannot have that since that would mean B or A is empty, respectively. Hence $2^n-2$ ways of doing this. Furthermore the sets A and B are indistinguishible (if we "swap" the sets we still count it as the same partition). Hence we divide by 2, giving the result.

c.)

Quote

$\Big\{^n_{n-1} \Big\} = \binom{n}{2}$

The example actually gives you a clue for this (it's the case for n=4). By splitting n objects into n-1 non-empty subset, it's neccesary that all of them have cardinality 1 except one of them, which has cardinality 2. Therefore it's just a matter of choosing 2 object to go into this special set, and then the rest of the elements are the single-element sets. hence $\binom{n}{2}$

3.)

Quote

$\Big\{^{n+1}_k \Big\} = \Big\{^n_{k-1} \Big\} + k \Big\{^n_k \Big\}$

I actually derived this formula as an attempt to solve q1, but didn't seem too helpful.

Suppose we put all the partitions that contribute to {n,k} into a set, A, and those that contribute to {n,k-1} in another set, B.

Now suppose we introduce a new element, x. Then the set of partitions that contribute to {n+1,k} can be made in two ways: Choose a partition from A, place x into some subset in this partition. Since there are {n,k} things in A, and k subsets to choose from each, there are k{n,k} ways of doing this. Another way to make a partition that would contribute to {n+1,k} is to select a partition from B, and simple introduce the set {x} into it. This gives {n,k-1} new partitions. Hence in total it's the sum k{n,k}+{n,k-1}

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 12:28:28 am

Awesome, thanks kamil :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 03:09:47 am

For positive integers $r$ , define $x^{\underline{r}}= x(x-1) \cdots (x-r+1)$

Show that $x^n = \Big\{^n_1 \Big\}x^{\underline{1}} + \Big\{^n_2 \Big\}x^{\underline{2}} + \cdots + \Big\{^n_n \Big\}x^{\underline{n}}$

I swear that looks so similar to $n2^{n-1} = 1\binom{n}{1}+2\binom{n}{2}+ \cdots + n\binom{n}{n}$ haha

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 20, 2009, 11:56:01 am

an observation: $x^r=P(x,r)$ so perhaps show that you are counting P(x,n) in small parts.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 01:05:34 pm

Sorry I edited the question, had a typo :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 01:45:37 pm

Quote from: TrueTears on December 19, 2009, 11:01:32 pm

Quote from: TrueTears on December 19, 2009, 12:29:27 am
1. Let $u(n)$ denote the number of ways in which a set of $\mbox{n}$ elements can be partitioned into non-empty subsets. For example, $u(3)=5$ , corresponding to:

$\{a,b,c\}$

$\{a,b\}, \{c\}$

$\{a,c\},\{b\}$

$\{a\},\{b,c\}$

$\{a\},\{b\},\{c\}$

Find a recurrence relation for $u(n)$ .
Consider the specific case for $u(5)$ .

Let us consider this $5$ element set: $\{1,2,3,4,5\}$

Obviously the element $5$ must be in every partition.

Let us denote the set that the element $5$ is in 'size $k$ ' where $k$ tells us how many elements are in that partitioned set.

For example, the partitioned set $\{1,5\},\{2,3,4\}$ has size $2$ .

Consider size 1, we have:

$\{5\}, \{1,2,3,4\}$

But how many ways can we partition $\{1,2,3,4\}$ ? Well there are $u(4)$ ways to partition it.

Thus for size 1 there are $u(4)$ ways.

Consider size 2, we have:

$\{5, \mbox{some other element}\}, \{\mbox{Three element set}\}$

For the 'some other element' we have $\binom{4}{1}$ choices and then for the 'Three element set' we have $u(3)$ ways to partition it.

In total we have $\binom{4}{1}(u(3))$ ways

Consider size 3.

We would have $\binom{4}{2}(u(2))$ ways

Size 4: $\binom{4}{3}(u(1))$ ways

Size 5: $\binom{4}{4}(u(0))$ ways

In total: $u(5) = \binom{4}{0}(u(4)) + \binom{4}{1}(u(3)) + \binom{4}{2}(u(2)) + \binom{4}{3}(u(1)) + \binom{4}{4}(u(0)) = \sum_{k=0}^4\binom{4}{k}u(4-k)$

Now let's generalise:

Say we have a $\mbox{n}$ element set.

$\therefore u(n) = \sum_{k=0}^n\binom{n-1}{k}u(n-1-k) \ \mbox{where} \ u(0) = 1$

Haha, just realised that this is the bell number sequence. http://en.wikipedia.org/wiki/Bell_number

Title: Re: TT's Maths Thread
Post by: Ahmad on December 20, 2009, 01:51:38 pm

Algebra.

It's true for n=1. Suppose that,

$x^n = \sum_{k=1}^n \genfrac{\{}{\}}{0pt}{}{n}{k} x^{\underline k}$

then we have,

$x^{n+1} = x\cdot x^n = \sum_{k=1}^n \genfrac{\{}{\}}{0pt}{}{n}{k} (x - k + k) x^{\underline k}$

$= \sum_{k=1}^n \genfrac{\{}{\}}{0pt}{}{n}{k} (x - k) x^{\underline k} + \sum_{k=1}^n k \genfrac{\{}{\}}{0pt}{}{n}{k} x^{\underline k}$

$= \sum_{k=1}^n \genfrac{\{}{\}}{0pt}{}{n}{k} x^{\underline {k+1}} + \sum_{k=1}^n k \genfrac{\{}{\}}{0pt}{}{n}{k} x^{\underline k}$

$= \sum_{k=2}^{n+1} \genfrac{\{}{\}}{0pt}{}{n}{k-1} x^{\underline {k}} + \sum_{k=1}^n k \genfrac{\{}{\}}{0pt}{}{n}{k} x^{\underline k}$

we can start the left sum off from k = 1 since the summand for k = 1 is zero. Same goes for the right sum going until k = n+1 since that summand is zero too.

$= \sum_{k=1}^{n+1} \left(\genfrac{\{}{\}}{0pt}{}{n}{k-1} + k \genfrac{\{}{\}}{0pt}{}{n}{k}\right) x^{\underline k}$

we can use the recursion on the previous page of this thread giving

$= \sum_{k=1}^{n+1} \genfrac{\{}{\}}{0pt}{}{n+1}{k} x^{\underline k}$

which completes the induction

Title: Re: TT's Maths Thread
Post by: Ahmad on December 20, 2009, 01:56:48 pm

Combinatorics.
Outline: I'm going to prove that for positive integer values of $x \ge n$ both sides of the identity agree. Since both sides are polynomials in x if they agree on the positive integers then they must be identical (why?).

Suppose $x\ge n$ is a positive integer. I claim both sides count the number of functions $f : \{1,2,\ldots,n\} \rightarrow \{1,2,\cdots,x\}$ . We can map 1 to any of x values, 2 to any of x values and so on, hence there are $x^n$ such functions, which is the left hand side. We can also condition on the number of elements of the image/range. If the range contains k elements, then we can partition the domain into k sets, and assign to each set a particular distinct value of the range. There are $\genfrac{\{}{\}}{0pt}{}{n}{k}$ ways to partition the domain into k sets, and there are $x(x-1)\cdots (x-k+1) = x^{\underline k}$ ways to assign a distinct element of the codomain to each of these k sets, that is easy to see: label the sets of the partition $A_1,\ldots,A_k$ , then $A_1$ can be assigned any of x elements, $A_2$ can be assigned any of x-1 elements and so on, and this constructs a function which has a range containing exactly k elements. Btw, when I say for example $A_1$ is assigned to 3, that means we're considering the function which maps each element of $A_1$ to 3. Summing over the possible values of k gives the right hand side and hence proves the result. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 02:48:37 pm

haha awesome, thanks Ahmad, the combinatorial way is so much easier :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 03:19:09 pm

Just started number theory 8-)

Firstly can anyone show me why $GCD(a,b) \times LCM(a,b) = ab$ ?

You can assume the FTA to be true while showing this (I haven't seen the proof for FTA yet :P)

Title: Re: TT's Maths Thread
Post by: Ahmad on December 20, 2009, 03:27:29 pm

Quote from: TrueTears on December 20, 2009, 02:48:37 pm

haha awesome, thanks Ahmad, the combinatorial way is so much easier :P

I personally found the algebraic way easier because it didn't require much thinking on my part. I knew what to do immediately, and the rest was just having enough technique to go through it. The combinatorial way is nicer though :)

Title: Re: TT's Maths Thread
Post by: Ahmad on December 20, 2009, 03:38:52 pm

I'll do an example and hopefully this will make it clear.

12 = 2^2 x 3
42 = 2 x 3 x 7

To find the gcd of the two of these numbers you take the smallest power of each prime divisor. For example, take the prime p = 2, then 12 has a factor of 2^2 and 42 has a factor of 2^1, so you take 2 to the power of min(2,1) = 1. So gcd(12,42) will have a factor of 2^min(2,1) = 2^1. Similarly, it will have a factor of 3^1 and 7^0. Hence gcd(12,42) = 2^1 x 3^1 x 7^0 = 6. On the other hand to work out the lcm you take the maximum power of each prime divisor, so we'd take 2^max(2,1) = 2^2, 3^max(1,1) = 3^1 and 7^max(0,1) = 7^1. Then the relationship given is just expressing the fact that, $\gcd(12,42) \cdot \mbox{lcm}(12,42) = ( 2^{\mbox{min}(2,1)} \cdot 3^{\mbox{min}(1,1)} \cdot 7^{\mbox{min}(0,1)} ) \cdot ( 2^{\mbox{max}(2,1)} \cdot 3^{\mbox{max}(1,1)} \cdot 7^{\mbox{max}(0,1)} ) = 2^{2+1} 3^{1+1} 7^{0+1}$ . This is true exactly because min(a,b) + max(a,b) = a+b, for example looking at powers of 2, $2^{\mbox{min}(2,1) + \mbox{max}(2,1)} = 2^{2+1}$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 04:53:32 pm

Quote from: Ahmad on December 20, 2009, 03:38:52 pm

I'll do an example and hopefully this will make it clear.

12 = 2^2 x 3
42 = 2 x 3 x 7

To find the gcd of the two of these numbers you take the smallest power of each prime divisor. For example, take the prime p = 2, then 12 has a factor of 2^2 and 42 has a factor of 2^1, so you take 2 to the power of min(2,1) = 1. So gcd(12,42) will have a factor of 2^min(2,1) = 2^1. Similarly, it will have a factor of 3^1 and 7^0. Hence gcd(12,42) = 2^1 x 3^1 x 7^0 = 6. On the other hand to work out the lcm you take the maximum power of each prime divisor, so we'd take 2^max(2,1) = 2^2, 3^max(1,1) = 3^1 and 7^max(0,1) = 7^1. Then the relationship given is just expressing the fact that, $\gcd(12,42) \cdot \mbox{lcm}(12,42) = ( 2^{\mbox{min}(2,1)} \cdot 3^{\mbox{min}(1,1)} \cdot 7^{\mbox{min}(0,1)} ) \cdot ( 2^{\mbox{max}(2,1)} \cdot 3^{\mbox{max}(1,1)} \cdot 7^{\mbox{max}(0,1)} ) = 2^{2+1} 3^{1+1} 7^{0+1}$ . This is true exactly because min(a,b) + max(a,b) = a+b, for example looking at powers of 2, $2^{\mbox{min}(2,1) + \mbox{max}(2,1)} = 2^{2+1}$ .

Ahh thanks the min(a,b)+max(a,b)=a+b was the essential part :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 06:41:54 pm

Show that if $\{x,y\} \in \mathbb{Z}$ and $\{a,b\} \in \mathbb{N}$ and $ax+by = 1$ then $(a,b) = 1$

Is this proof valid?

Assume $(a,b) = q$ where $q \in \mathbb{Z^+}$ and $q \neq 1$

Then let us denote the PPF of $a$ as $(a')(q)$ and the PPF of $b$ as $(b')(q)$ where $q$ is the greatest common factor for $a$ and $b$ and $a',b'$ is a product of primes in $a$ and $b$ 's PPF respectively.

Then $ax+by=1 \implies (a')(q)(x) + (b')(q)(y) = 1$

$q(a'x+b'y)=1$

$(a'x+b'y) = \frac{1}{q}$

But $(a'x+b'y) \in \mathbb{Z}$ if and only if $q = 1$ (In fact $(a'x+b'y) = 1$ as well)

Thus $(a,b) = 1$

Title: Re: TT's Maths Thread
Post by: zzdfa on December 20, 2009, 09:14:23 pm

yes. a more general way to do it is to note that

if d|a and d|b

then

d|a*x
d|b*y

where x,y are integers

and hence

d|(a*x+b*y)

to apply this to the above example, just let d=(a,b).

we get

(a,b)|(a*x+b*y)

stated in words, any common divisor of a,b must divide any linear combination of a,b.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 20, 2009, 10:36:14 pm

Thanks zzdfa :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 21, 2009, 08:21:38 pm

Hey can someone show me how to prove the FTA?

Title: Re: TT's Maths Thread
Post by: humph on December 21, 2009, 08:44:31 pm

The proof(s) on Wikipedia are very "natural": see http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
I also seem to remember Hardy and Wright's book "An Introduction to the Theory of Numbers" having quite a few proofs (and it's a great book too, albeit quite outdated).

Title: Re: TT's Maths Thread
Post by: TrueTears on December 21, 2009, 08:45:29 pm

Cool, thanks humph checking it out now :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 21, 2009, 08:50:12 pm

the most crucial part of the step is to prove Euclid's lemma: if p|ab then p|a or p|b. (p is prime).

Most textbook approach is with bezout's identity, but Gauss had a different proof in his Disquisitiones Arithmeticae which was posted in the first page of this thread

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 01:24:47 am

I like the following proof the best ;)

Assume that there are numbers which can not be expressed as a product of primes. Let the smallest possible number of this kind be $m$ .

$m$ can not be $1$ since $1$ is neither composite nor prime. $m$ can not be prime since the PPF of a prime number is just itself. Thus $m$ must be a composite number.

Let the composition of $m = ab$ where $\{a,b\} < m$

Since $m$ was the smallest number that can not be expressed as a product of primes, this means $a$ and $b$ can be expressed as a product of primes and consequently we get $m = ab$ where $a$ and $b$ can be both expressed as primes. Contradiction!

Thus $m$ can also be expressed as a product of primes.

Lemma 1: If $p$ is a prime and $p|a_1a_2a_3 \cdots a_n$ then $p|a_i$ for some $i$ .

Lemma 2: If $p$ and $q$ are primes and $e$ is a natural number and $p|q^e$ then $p=q$ .

Let's assume that for some number $k$ that there are $2$ (at least) ways of expressing its PPF.

$\therefore k = p_1^{f_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = q_1^{e_1}q_2^{e_2}q_3^{e_3} \cdots q_m^{e_m}$

Clearly for all $\mbox{n}$ , $p_n|k$

$\implies p_n| q_1^{e_1}q_2^{e_2}q_3^{e_3} \cdots q_m^{e_m}$

By Lemma 1 $p_n|q_m^{e_m}$ for any $m$ .

By Lemma 2 $p_n=q_m$

This means that for all $\mbox{n}$ and all $m$ there are values of $p$ which equals to those of $q$ . For example, $p_3$ could equal to $q_5$ , or $p_6=q_3$ etc. This also means we have created a bijection between $p$ and $q$ such that $n = m$ .

Therefore if the number $k$ has $2$ PPF's then the prime number 'base' will be exactly the same, the only different would be in the powers, namely $f_n$ and $e_m$ .

Now since each $p_n$ has a corespondent equivalent $q_m$ we can rewrite $k$ as:

$p_1^{f_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = p_1^{e_1}p_2^{e_2}p_3^{e_3} \cdots p_n^{e_n}$

$p_1^{f_1-e_1}p_2^{f_2}p_3^{f_3} \cdots p_n^{f_n} = p_2^{e_2}p_3^{e_3} \cdots p_n^{e_n}$

$\mbox{LHS}|p_1$ however $\mbox{RHS}$ can not be divided by $p_1$ unless for some $2 \le j \le n$ such that $p_j = p_1$

But since $p_1 \neq p_2 \neq p_3 \neq \cdots \neq p_n$ we have a contradiction!

Therefore FTA is proven to be true.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 01:28:47 am

Proving the uniqueness and understanding the process took me the whole night, it drove me crazy z.z.z.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 22, 2009, 01:40:49 am

yeah, just remember that a few things were assumed without loss of generality. example: you assumed that $f_1>e_1$ since we knew that some two corresponding powers must be non-equal. But yeah it's quite obvious to see how important Euclid's lemma was and I think its proof is the most non-trivial part.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 01:41:59 am

Yeap, very powerful :)

Title: Re: TT's Maths Thread
Post by: humph on December 22, 2009, 01:58:15 am

The Fundamental Theorem of Arithmetic is pretty damn important on its own. For example, it is equivalent to the fact that
$\sum^{\infty}_{n=1}{\frac{1}{n^s}} = \prod_{p}{\frac{1}{1 - p^{-s}}},$
where the product is taken over all primes $p$ , and $s$ is some complex variable with real part greater than 1. This is the Euler product for the Riemann zeta function, famous for its central role in the unsolved Riemann hypothesis.
More interestingly, the fact that the meromorphic extension of the Riemann zeta function to the open half-plane $\Re(s) > 0$ has no zeroes along the line $\Re(s) = 1$ is equivalent to the Prime Number Theorem
$\pi(x) \sim \frac{x}{\log x},$
where $\pi(x)$ is the function that counts how many primes are lesser than or equal to $x$ ; this is perhaps the central result of the field of multiplicative number theory.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 03:30:09 am

Anyways my first set of fun number theory questions~!

~~1. Prove that the fraction $\frac{n^3+2n}{n^4+3n^2+1}$ is in lowest terms for every positive integer $\mbox{n}$~~

~~2. Let $\{a,b,c\} \in \mathbb{N}$ , show that $\frac{[a,b,c]^2}{[a,b][b,c][c,a]} = \frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$~~

~~3. Prove that consecutive Fibonacci numbers are always relatively prime.~~

4. Show that $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ can never be an integer. (I'm thinking of showing $\lim_{n \to \infty}\sum_{k=2}^{n}\frac{1}{k}$ diverges to a value... or something along those lines, probs wrong heh)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 12:28:40 pm

Actually 4 is just the harmonic series, how do you show that it diverges to a non-integer value?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 22, 2009, 01:04:29 pm

I remember posting 4 as a challenge once lol

Suppose the sum of the first n terms is an integer m, multiply both sides by n!:

$m*n!=2*3*4..n + 1*3*4..n + 1*2*4...n + .... (n-1)!$

Now consider the biggest prime number p, in the sequence 1,2,3...n.

m*n! is divisible by p. All the terms on the RHS are divisible by p except for the pth term. This is because the kth term is a product of all numbers between 1 and n except for k. Therefore if $k\neq p$ the kth term is divisible by p. If $k=p$ then the kth term is missing p. But no other factor of this term is divisible by p since they are made of prime factors smaller than p. Denote the kth term as $t_k$ :

$t_p=m*n! - t_1 - t_2... - t_{p-1} - t_{p+1}...-t_n$

So the LHS is not divisible by p, but the RHS is since it is a linear combination of multiples of p, a contradiction.

edit: actually, I think there is a flaw with this proof since it assumes that for every integer n, there exists a prime p such that p<n<2p, and i do not know if this is true.

However an alternate proof is this: multiply both sides by the biggest power of 2(call this $2^k$ ), then multiply it by the biggest power of 3, then multiply it by the biggest power of 5 etc. eventually we get some equation like in the previous attempt at a proof, that is one that involves only integers. Now the $t_{2^k}$ term is not an even number, but the rest must be and so we get a contradiction since we get:

even numer=sum of even numbers + odd number.

1.) $n^3+2n=n(n^2+2)$ , $n^4+3n^2+1=(n^2+2)(n^2+1)-1$

any prime that divides the numerator must divide either n or $n^2+2$ .

Suppose p divides n. Then:

$\frac{n^4+3n^2+1}{p}=\frac{n^4+3n^2}{p} + \frac{1}{p} \notin \mathbb{N}$ hence p does not divide the denominator.

Suppose p divides $n^2+2$ :
$ \frac{n^4+3n^2+1}{p}=\frac{(n^2+2)(n^2+1)}{p} - \frac{1}{p} \notin \mathbb{N}$ hence p does not divide the numerator.

Therefore no prime can divide both the numerator and denominator. Hence they are relatively prime and so in lowest terms.

3.) suppose some integer $d>1$ , divides $f_{n+1}$ and $f_{n+2}$ then $f_n=f_{n+2}-f_{n+1}$ hence $d|f_n$ . Now $f_{n-1}=f_{n+1}-f_n$ hence $d|f_{n-1}$ . Now $f_{n-2}=f_{n}-f_{n-1}$ hence $d|f_{n-2}$ ... etc and we can imagine continuing this process until we get to $d|f_3$ and $d|f_4$ which is false since no integer greater than 1 divides both of these(2 and 3).

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 03:11:46 pm

Thanks kamil for Q 4. I actually thought of another method, but I'm kind stuck at the end.

$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$ is just the Zeta function $\zeta (1)$

Consider the sequence $1 + \frac{1}{4} + \frac{1}{4} + \cdots + \frac{1}{4}$

Clearly $\frac{1}{3} + \frac{1}{4} \ge \frac{1}{4} + \frac{1}{4}$

$\implies \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \ge \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4}$

So clearly the sum of $\zeta (1) \ge 1 + \frac{1}{4} + \frac{1}{4} + \cdots + \frac{1}{4}$

However $1 + \frac{1}{4} + \frac{1}{4} + \cdots + \frac{1}{4}$ clearly diverges, which means $\zeta (1)$ diverges as well.

But how can I relate this to the question?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 04:04:36 pm

Quote from: kamil9876 on December 22, 2009, 01:04:29 pm

I remember posting 4 as a challenge once lol

Suppose the sum of the first n terms is an integer m, multiply both sides by n!:

$m*n!=2*3*4..n + 1*3*4..n + 1*2*4...n + .... (n-1)!$

Now consider the biggest prime number p, in the sequence 1,2,3...n.

m*n! is divisible by p. All the terms on the RHS are divisible by p except for the pth term. This is because the kth term is a product of all numbers between 1 and n except for k. Therefore if $k\neq p$ the kth term is divisible by p. If $k=p$ then the kth term is missing p. But no other factor of this term is divisible by p since they are made of prime factors smaller than p. Denote the kth term as $t_k$ :

$t_p=m*n! - t_1 - t_2... - t_{p-1} - t_{p+1}...-t_n$

So the LHS is not divisible by p, but the RHS is since it is a linear combination of multiples of p, a contradiction.

edit: actually, I think there is a flaw with this proof since it assumes that for every integer n, there exists a prime p such that p<n<2p, and i do not know if this is true.

Actually if you use Bertrand's postulate for any natural $\mbox{n}$ at least one prime exists $n < p < 2n$ .

Now set $n = \frac{n}{2}$

$\frac{n}{2} < p < n$

Now assume we pick the largest $p$ within this interval.

$\implies n < 2p < 2n$

$\therefore p < n < 2p$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 22, 2009, 04:14:03 pm

lulz yes, but it's nice to have a simple and elementary proof. Though it just goes to show how intuitive bertrand's postulate is :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 04:19:10 pm

Quote from: kamil9876 on December 22, 2009, 04:14:03 pm

lulz yes, but it's nice to have a simple and elementary proof. Though it just goes to show how intuitive bertrand's postulate is :P

Lol yeah when you did it, you didn't even think about proving it haha

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 09:27:42 pm

Quote from: kamil9876 on December 22, 2009, 01:04:29 pm

3.) suppose some integer $d>1$ , divides $f_{n+1}$ and $f_{n+2}$ then $f_n=f_{n+2}-f_{n+1}$ hence $d|f_n$ . Now $f_{n-1}=f_{n+1}-f_n$ hence $d|f_{n-1}$ . Now $f_{n-2}=f_{n}-f_{n-1}$ hence $d|f_{n-2}$ ... etc and we can imagine continuing this process until we get to $d|f_3$ and $d|f_4$ which is false since no integer greater than 1 divides both of these(2 and 3).

lol kamil, I love this proof of yours so simple!!

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 22, 2009, 09:30:07 pm

actually, it can be stated in an even shorter way :P. Suppose n is the smallest integer such that $f_n$ and $f_{n+1}$ has some d>1 that divides them both. Then d divides $f_{n-1}$ and so the result is also true for n-1. but n-1<n and this contradicts n being the smallest! LOL

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 09:31:45 pm

Quote from: kamil9876 on December 22, 2009, 09:30:07 pm

actually, it can be stated in an even shorter way :P. Suppose n is the smallest integer such that $f_n$ and $f_{n+1}$ has some d>1 that divides them both. Then d divides $f_{n-1}$ and so the result is also true for n-1. but n-1<n and this contradicts n being the smallest! LOL

hahahaha, that's such a cool question.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 11:55:14 pm

Quote

2. Let $\{a,b,c\} \in \mathbb{N}$ , show that $\frac{[a,b,c]^2}{[a,b][b,c][c,a]} = \frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$

Let $a = p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}$

$b = p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}$

$c = p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}$

Where $p_i$ is a prime for all $i \in [1,n]$ and $\{e_i,f_i,g_i\} \in \mathbb{N}$

$[a,b] = p_1^{\text{max}(e_1,f_1)}p_2^{\text{max}(e_2,f_2)} \cdots p_n^{\text{max}(e_n,f_n)}$

$[b,c] = p_1^{\text{max}(f_1,g_1)}p_2^{\text{max}(f_2,g_2)} \cdots p_n^{\text{max}(f_n,g_n)}$

$[c,a] = p_1^{\text{max}(g_1,e_1)}p_2^{\text{max}(g_2,e_2)} \cdots p_n^{\text{max}(g_n,e_n)}$

$[a,b,c] = p_1^{\text{max}(e_1,f_1,g_1)}p_2^{\text{max}(e_2,f_2,g_2)} \cdots p_n^{\text{max}(e_n,f_n,g_n)}$

WLOG assume $e_n \ge f_n \ge g_n \ \forall \ n$

$[a,b][b,c] = \left(p_1^{\text{max}(e_1,f_1)}p_2^{\text{max}(e_2,f_2)} \cdots p_n^{\text{max}(e_n,f_n)}\right)\left(p_1^{\text{max}(f_1,g_1)}p_2^{\text{max}(f_2,g_2)} \cdots p_n^{\text{max}(f_n,g_n)}\right) = \left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)$

$\implies \left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)[c,a] = \left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{\text{max}(g_1,e_1)}p_2^{\text{max}(g_2,e_2)} \cdots p_n^{\text{max}(g_n,e_n)}\right)$

$\implies \mbox{RHS} = \left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right) = \left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)^2\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)$

$[a,b,c]^2 = \left(p_1^{\text{max}(e_1,f_1,g_1)}p_2^{\text{max}(e_2,f_2,g_2)} \cdots p_n^{\text{max}(e_n,f_n,g_n)}\right)^2 = \left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)^2$

$\therefore \mbox{LHS} = \frac{\left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)^2}{\left(p_1^{e_1}p_2^{e_2} \cdots p_n^{e_n}\right)^2\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)} = \frac{1}{\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)}$

$(a,b) = p_1^{\text{min}(e_1,f_1)}p_2^{\text{min}(e_2,f_2)} \cdots p_n^{\text{min}(e_n,f_n)}$

$(b,c) = p_1^{\text{min}(f_1,g_1)}p_2^{\text{min}(f_2,g_2)} \cdots p_n^{\text{min}(f_n,g_n)}$

$(c,a) = p_1^{\text{min}(g_1,e_1)}p_2^{\text{min}(g_2,e_2)} \cdots p_n^{\text{min}(g_n,e_n)}$

$(a,b,c) = p_1^{\text{min}(e_1,f_1,g_1)}p_2^{\text{min}(e_2,f_2,g_2)} \cdots p_n^{\text{min}(e_n,f_n,g_n)}$

$(a,b)(b,c) = \left(p_1^{\text{min}(e_1,f_1)}p_2^{\text{min}(e_2,f_2)} \cdots p_n^{\text{min}(e_n,f_n)}\right)\left(p_1^{\text{min}(f_1,g_1)}p_2^{\text{min}(f_2,g_2)} \cdots p_n^{\text{min}(f_n,g_n)}\right) = \left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)$

$\implies \left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)(c,a) = \left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)\left(p_1^{\text{min}(g_1,e_1)}p_2^{\text{min}(g_2,e_2)} \cdots p_n^{\text{min}(g_n,e_n)}\right) = \left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)$

$\implies \left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)^2$

$(a,b,c)^2 = \left(p_1^{\text{min}(e_1,f_1,g_1)}p_2^{\text{min}(e_2,f_2,g_2)} \cdots p_n^{\text{min}(e_n,f_n,g_n)}\right)^2 = \left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)^2$

$\therefore \mbox{RHS} = \frac{\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)^2}{\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)\left(p_1^{g_1}p_2^{g_2} \cdots p_n^{g_n}\right)^2} = \frac{1}{\left(p_1^{f_1}p_2^{f_2} \cdots p_n^{f_n}\right)} = \mbox{LHS}$

$\mbox{Q.E.D}$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 22, 2009, 11:57:53 pm

Actually is that right? Since my whole argument was based on the assumption $e_n \ge f_n \ge g_n \ \forall \ n$ , I'm not sure if I can do that...

Title: Re: TT's Maths Thread
Post by: zzdfa on December 23, 2009, 12:00:50 am

my first thought: did you try using the fact that (a,b)=a*b/[a,b] before going through all that :P

a sometimes useful way of thinking:

gcd can be interpreted as the intersection of the prime factors and lcm can be interpreted as the union of the prime factors.

example:
pfactors of 96 are {2,2,2,2,2,3}
pfactors of 60 are {2,2,3,5}

intersection is {2,2,3}

so gcd is 2*2*3=12.

similar thing with lcm. the union of the pfactors of 96 and 60 is {2,2,2,2,2,3,5}, so lcm is 480.

using the above interpretation, [a,b]=a*b/(a,b) is just principle of inclusion-exclusion.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 23, 2009, 12:01:57 am

Quote from: zzdfa on December 23, 2009, 12:00:50 am

my first thought: did you try using the fact that (a,b)=a*b/[a,b] before going through all that :P

lol I didn't even know that result :(, how do you prove it?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 23, 2009, 12:10:02 am

Quote from: TrueTears on December 22, 2009, 11:57:53 pm

Actually is that right? Since my whole argument was based on the assumption $e_n \ge f_n \ge g_n \ \forall \ n$ , I'm not sure if I can do that...

no that assumption was fine. are you sure you know what WLOG means?

\\

Title: Re: TT's Maths Thread
Post by: TrueTears on December 23, 2009, 12:10:58 am

Quote from: zzdfa on December 23, 2009, 12:00:50 am

my first thought: did you try using the fact that (a,b)=a*b/[a,b] before going through all that :P

a sometimes useful way of thinking:

gcd can be interpreted as the intersection of the prime factors and lcm can be interpreted as the union of the prime factors.

pfactors of 96 are {2,2,2,2,2,3}
pfactors of 60 are {2,2,3,5}

intersection is {2,3}

so gcd is 2*3=6.

similar thing with lcm.

using the above interpretation, [a,b]=a*b/(a,b) is just principle of inclusion-exclusion.

Wait isn't the intersection $\{2,2,3\}$ ?

Quote from: zzdfa on December 23, 2009, 12:10:02 am

Quote from: TrueTears on December 22, 2009, 11:57:53 pm
Actually is that right? Since my whole argument was based on the assumption $e_n \ge f_n \ge g_n \ \forall \ n$ , I'm not sure if I can do that...
no that assumption was fine. are you sure you know what WLOG means?

\\

Yeah I do, without loss of generality but I when I reread what I did I thought it was wrong.

Title: Re: TT's Maths Thread
Post by: zzdfa on December 23, 2009, 12:12:03 am

ahh yep youre right^^

lol i meant 'do you know what without loss of generality means' not 'do you know what WLOG stands for' :p

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 23, 2009, 12:14:53 am

strictly speaking, you can assume that WLOG $e_i \ge f_i \ge g_i$ for some i, but not for all. However this proof can be executed just by focusing on one prime number, say p, rather than all of them. Notice also that the equation is symmetric in the variables proving it for one prime power is enough since the primes only 'interact' with themselves. (~~Another way to justify using only one prime power is to notice that gcd(x,y)*gcd(z,w)=gcd(xz,zw) if x and z are relatively prime; w and y are relatively prime, and likewise for lcm~~).

by request, my soln (probably the same):

let $a_1=p^A, b_1=p^B, c_1=p^C$ and assume WLOG that $A\leq B \leq C$

lcm(a,b,c)=max(A,B,C)=C
lcm(a,b)=B
lcm(a,c)=C
lcm(b,c)=C

gcd(a,b,c)=min(A,B,C)=A
gcd(a,b)=A
gcd(a,c)=A
gcd(b,c)=B

plugging these values in we see the equation is true as both sides become 1/B.

Now if we do this on all the individual 'pure prime powers' the equation is true, since p was arbitrary and we can always do the WLOG thing since the equation is symmetric, and so we can relabel as we like and it won't change the truth of the equation. Now you can multiply all these equations together to get one big equation, which changes it into the equation for a,b,c using the fact:

"gcd(x,y)*gcd(z,w)=gcd(xz,zw) if x and z are relatively prime; w and y are relatively prime, and likewise for lcm"

Notice that B is the middle number (again showing some nice symmetry :D ). So yeah, actually just do your big general approach, works best, and then split it into a product into many little products that involve only the single prime, and then just prove that for each of these products, you get the 'middle number' both for lcm and gcd. and yeah result easily follows from here. (this is ussually a good approach, to focus on single primes and then take the product of all the singles).

ie for clarity:

$RHS=\prod_{i=1}^n \frac{p_i^{2min(e_i, f_i, g_i)}}{p_i^{min(e_i,f_i)}p_i^{min(g_i,f_i)}p_i^{min(e_i,g_i)}}$
$=\prod_{i=1}^n \frac{1}{p_i^B_i}$

where $B_i$ is the 'middle number' for $e_i, f_i, g_i$ .

And same thing for LHS.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 23, 2009, 12:18:29 am

Quote from: zzdfa on December 23, 2009, 12:12:03 am

ahh yep youre right^^

lol i meant 'do you know what without loss of generality means' not 'do you know what WLOG stands for' :p

lol haha, it means, you are not losing any information or falsifying anything when you make that assumption right?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 23, 2009, 12:22:08 am

Quote from: kamil9876 on December 23, 2009, 12:14:53 am

strictly speaking, you can assume that WLOG $e_i \ge f_i \ge g_i$ for some i, but not for all. However this proof can be executed just by focusing on one prime number, say p, rather than all of them. Notice also that the equation is symmetric in the variables proving it for one prime power is enough since the primes only 'interact' with themselves. (Another way to justify using only one prime power is to notice that gcd(x,y)*gcd(z,w)=gcd(xz,zw) if x and z are relatively prime; w and y are relatively prime, and likewise for lcm).

doesnt he only assume it for e_n f_n and g_n though? so the assumption is valid right?/

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 23, 2009, 12:36:52 am

that's what he preached, but not what he practiced.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 23, 2009, 01:19:49 am

Quote from: kamil9876 on December 23, 2009, 12:14:53 am

Notice that B is the middle number (again showing some nice symmetry :D ). So yeah, actually just do your big general approach, works best, and then split it into a product into many little products that involve only the single prime, and then just prove that for each of these products, you get the 'middle number' both for lcm and gcd. and yeah result easily follows from here. (this is ussually a good approach, to focus on single primes and then take the product of all the singles).

I don't really get this last part, what do you mean smaller products? what product? Can you show me?

Thanks :)

Actually dw I get it now :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 23, 2009, 02:13:41 am

A few more questions :)

~~1. Show that $1000!$ ends with $249$ zeros, generalise as well.~~

~~2. Prove that there are infinitely many primes of the form $4k+3$ where $k \in \mathbb{Z^+} \cup \{0\}$ , ie, this sequence $\{3,7,11,19, \cdots\}$ .~~

3. Prove that among any $12$ consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer $\mbox{n}$ are all positive integers other than $1$ and $\mbox{n}$ which divide $\mbox{n}$ . For example, the proper divisors of $14$ are $2$ and $7$ .)

~~4. Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.~~

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 23, 2009, 12:00:19 pm

nice ones :D

Here are some clues

1.) basically this is equivalent to proving that there are exactly 249 5's in the product $(1)(2)(3)(2^2)(5)(3*2)(7)(2^3)(3^2)(2*5)..... (2*5)^3$ . All multiples of 5 contribue at least one 5. multiples of 25 contribue two 5's, multiples of 125 contribue three 5's, multiples of 625 contribue four 5's....

2.) This problem can be done very efficiently with modular arithmetic, but without it note that any odd number of the form 4k+3 has at least one prime factor of the form 4k+3; for if it didn't then it would have it's prime factors all of the form 4k+1 and (4a+1)(4b+1)(4c+1).... would be of the form 4k+1 if you expand out.

Clue number 2: Euclid's proof on infinitude of primes.

3.) try a stronger result: all multiples of 12 have this property. If you can prove this then it implies this result since amongst any 12 consecutive numbers we have one multiple of 12.

4.) This is actually more of an algebra problem, my solution is pretty algebraic. suppose they were in an arithmetic progression of length d:

$p_1^{1/3}-p_3^{1/3}=kd$ for some k

$p_1^{1/3}-p_2^{1/3}=md$ for some m

therefore:
$ p_1^{1/3}-p_3^{1/3}=\frac{k}{m}(p_1^{1/3}-p_2^{1/3})$
$ p_1 - (p_3 p_1^2)^{1/3}=\frac{k}{m}(p_1-(p_2 p_1^2)^{1/3}) $

now you can rearrange this into an equation of the form:

$a (p_3 p_1^2)^{1/3} + b (p_2 p_1^2)^{1/3} = c$

where a,b,c are rational. and the cube roots are obviously irrational and distinct. Prove that this cannot be the case (algebra exercise).

Title: Re: TT's Maths Thread
Post by: brightsky on December 23, 2009, 12:18:12 pm

wow...good job. :D

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 12:38:32 am

Quote

1. Show that $1000!$ ends with $249$ zeros, generalise as well.

Let's count the number of $2$ 's in the sequence $1,2,3,4, \cdots, 999, 1000$ .

$\sum_{k=1}^9 \left \lfloor \frac{1000}{2^k} \right \rfloor = \left \lfloor \frac{1000}{2} \right \rfloor + \left \lfloor \frac{1000}{2^2} \right \rfloor + \cdots + \left \lfloor \frac{1000}{2^9} \right \rfloor$

Let's count the number of $5$ 's in the sequence $1,2,3,4, \cdots, 999, 1000$ .

$\sum_{k=1}^4 \left \lfloor \frac{1000}{5^k} \right \rfloor = \left \lfloor \frac{1000}{5} \right \rfloor + \left \lfloor \frac{1000}{5^2} \right \rfloor + \cdots + \left \lfloor \frac{1000}{5^4} \right \rfloor$

By comparing summands we see that there exists much more $2$ 's in the total amount of PPF's of every number in the sequence $1,2,3,4, \cdots, 999, 1000$

Now that means there exists $\sum_{k=1}^4 \left \lfloor \frac{1000}{5^k} \right \rfloor = 249$ zeros!

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 12:56:50 am

Quote

2. Prove that there are infinitely many primes of the form 4k+3

Quote

2.) This problem can be done very efficiently with modular arithmetic, but without it note that any odd number of the form 4k+3 has at least one prime factor of the form 4k+3; for if it didn't then it would have it's prime factors all of the form 4k+1 and (4a+1)(4b+1)(4c+1).... would be of the form 4k+1 if you expand it out

ok here is the completion:

assume there are only finitely many primes of this form. Take the product of all of these, call this M. M will either be of the form 4k+1 or 4k+3:

case1: M is the form 4k+1

then M+2 is of the form 4k+3 and hence by the lemma in the second quote above: it must be divisible by some prime of the form 4k+3. Let 4a+3 by such a prime, then $\frac{M+2}{4a+3}=\frac{M}{4a+3} + \frac{2}{4a+3}$ is not an integer since M is divisible by 4a+3. But this is true for all 4a+3 hence there must be some bigger one.

case2: M is of the form 4k+3

then M+4 is of the form 4k+3, and we can do a similair proof as in Case1.

Some interesting history:

Fermat proposed this sort of problem (perhaps even the same 1). to prove for 4k+1 it is much more difficult. Euler conjectured that there are infinitely many primes of the form 1+ak for all $a \in \mathbb{N}$ . Dirichlet later proved something in fact even stronger, that there are infinitely many primes in any arithmetic sequence where the gcd of the common difference and initial value is 1. see more here

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 01:07:57 am

Quote

2. Prove that there are infinitely many primes of the form $4k+3$ where $k \in \mathbb{Z^+} \cup \{0\}$ , ie, this sequence $\{3,7,11,19, \cdots\}$ .

Hehe, pretty similar to Euclid's proof :P

Assume there are finitely many primes $4a+3, 4b+3, 4c+3, \cdots, 4n+3$ where $4n+3$ is the largest prime in this sequence.

Now consider a number $Q = \left((4a+3)(4b+3)(4c+3) \cdots (4n+3)\right) + 1$

$Q$ is either prime or composite.

$Q$ can not be prime because it would contradict the fact that $4n+3$ is the largest prime.

However we know that $Q$ must have at least $1$ prime divisor, $p$ , by the Fundamental Theorem of Arithmetic.

However this prime can not be any number in the sequence since that will result in a remainder of $1$ .

Thus this contradicts our assumption that there are finitely many primes in this sequence, therefore there must be infinite many primes in this sequence.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 01:09:40 am

but how do u know p is of the form 4k+3? could be of the form 4k+1...

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 04:36:22 pm

Quote

note that any odd number of the form 4k+3 has at least one prime factor of the form 4k+3

How do you prove that?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 05:01:19 pm

suppose there was a number of the form 4k+3 that had no prime factor of them form 4k+3. Hence it has prime factors all of the form 4k+1. and so it can be written as:

(4a+1)(4b+1)....

which when expanded out has a +1 at the end, and all otehr terms multiples of 4. Hence it is of the form 4k+1, a contradiction.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 05:14:07 pm

But if you're talking about the PPF of 4k+3 where's the exponents?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 05:49:24 pm

you can write then in if u want but still u can write 3^2 * 7 as 3*3*7 so i guess u could say it is (4*0 + 3)(4*0 + 3)(4*1 + 3). ie a,b,c... etc do not have to be distinct. But if u really want just add in the exponents if u want and u still get same result.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 06:22:25 pm

Ah ok, I get it now.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 09:04:11 pm

Quote from: TrueTears on December 23, 2009, 02:13:41 am

3. Prove that among any $12$ consecutive positive integers there is at least one which is smaller than the sum of its proper divisors. (The proper divisors of a positive integer $\mbox{n}$ are all positive integers other than $1$ and $\mbox{n}$ which divide $\mbox{n}$ . For example, the proper divisors of $14$ are $2$ and $7$ .)

Let's list the first $12$ consecutive positive integers.

$1,2,3,4,5,6,7,8,9,10,11,12$

The proper divisors of each are as follows:

$1: \text{none}$

$2: \text{none}$

$3: \text{none}$

$4: 2$

$5: \text{none}$

$6: 2,3$

$7: \text{none}$

$8: 2,4$

$9: 3$

$10: 2,5$

$11: \text{none}$

$12: 2,3,4,6$

So only $12$ satisfies the condition as $2+3+4+6=15 > 12$

So if we can find every multiple of $12$ then it must also satisfy this condition.

Every $12$ consecutive positive integers must contain a multiple of $12$ .

Let this multiple of $12$ be $a$ where $a = 12k \ \forall \ k \in \mathbb{N}$

Then the proper divisors would contain at least $2k,3k,4k,6k$

Thus $2k+3k+4k+6k=15k > 12k$

I have a few questions to this question though, why does only multiples of 12 work? (I just tried by experimentation and 12 worked, but why does it work?)

Also the $15k$ is not the real sum of the all the proper divisors, it just includes some, eg when $k = 2$ we have $30>24$

But we are missing the proper divisors $2$ and $3$ . So the real sum of the divisors should be $35$ rather than $30$ , but $30$ is already larger than $24$ so the condition is satisfied...

But don't I need to somehow include the sum of ALL possible divisors in my generalization or is it alright to just prove that if some divisors' sum exceeds $24k$ then I've proven what I needed?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 09:26:21 pm

yeah. let by that logic, a generalisation would be sd(ab)> a * sd(b). where sd is sum of divisors.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 09:53:14 pm

Quote from: kamil9876 on December 24, 2009, 09:26:21 pm

yeah. let by that logic, a generalisation would be sd(ab)> a * sd(b). where sd is sum of divisors.

Ah okay, so the 15k acts like a lower bound? Thus it doesn't matter how much more we add to it since it is larger than 12k anyway :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 10:44:53 pm

Quote

4. Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.

The arithmetic progression is in the form $a_n = a_1 + (n-1)d$

Let the three primes be $x,y,z$ .

WLOG assume $x>y>z$

Now assume $x,y,z$ are three terms of an arithmetic progression and $z^{\frac{1}{3}}$ is the first term of the progression.

$y^{\frac{1}{3}} = z^{\frac{1}{3}} + kd \cdots [1]$

$x^{\frac{1}{3}} = z^{\frac{1}{3}} + md \cdots [2]$

Where $\{k,m\} \in \mathbb{N}$ and $d$ is the common difference.

$[1] \times m \implies my^{\frac{1}{3}} = mz^{\frac{1}{3}} + mkd$

$[2] \times k \implies kx^{\frac{1}{3}} = kz^{\frac{1}{3}} + mkd$

$[1] \times m - [2] \times k \implies my^{\frac{1}{3}} - kx^{\frac{1}{3}} = (m-k)z^{\frac{1}{3}}$

$\left(my^{\frac{1}{3}} - kx^{\frac{1}{3}}\right)^3 = \left((m-k)z^{\frac{1}{3}}\right)^3$

$m^3y - k^3x - 3my^{\frac{1}{3}}kx^{\frac{1}{3}}\left(my^{\frac{1}{3}} - kx^{\frac{1}{3}}\right) = (m-k)^3z$

$m^3y - k^3x - 3my^{\frac{1}{3}}kx^{\frac{1}{3}}\left((m-k)z^{\frac{1}{3}}\right) = (m-k)^3z$

$3my^{\frac{1}{3}}kx^{\frac{1}{3}}\left((m-k)z^{\frac{1}{3}}\right) = -(m-k)^3z + m^3y - k^3x$

$y^{\frac{1}{3}}x^{\frac{1}{3}}z^{\frac{1}{3}} = \frac{-(m-k)^3z + m^3y - k^3x}{3mk(m-k)}$

$(yxz)^{\frac{1}{3}} = \frac{-(m-k)^3z + m^3y - k^3x}{3mk(m-k)}$

Since $x,y,z$ are all distinct primes the cube root of the product of any $3$ primes is irrational, however the $\mbox{RHS}$ is clearly rational.

Thus contradiction!

But... how do you prove the product of 3 primes is never a cube?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 11:01:38 pm

u might wanna do the m=k case seperately, so that your RHS really is rational and not undefined. (ie work from your third last line, since you didn't do any division before that point, and the result follows easily from it)

Quote

But... how do you prove the product of 3 primes is never a cube?

$n^3=(\prod p_i^{e_i})^3=\prod p_i^{3e_i}$

Thus all cubes have their prime powers as multiples of 3 (in fact, k is a cube iff all prime powers are multiples of 3).

But $xyz=(x^1)(y^1)(z^1)$ hence not multiples of 3.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 11:03:02 pm

Quote from: kamil9876 on December 24, 2009, 11:01:38 pm

u might wanna do the m=k case seperately, so that your RHS really is rational and not undefined. (ie work from your third last line, since you didn't do any division before that point, and the result follows easily from it)

Quote
But... how do you prove the product of 3 primes is never a cube?

$n^3=(\prod p_i^{e_i})^3=\prod p_i^{3e_i}$

Thus all cubes have their prime powers as multiples of 3 (in fact, k is a cube iff all prime powers are multiples of 3).

But $xyz=(x^1)(y^1)(z^1)$ hence not multiples of 3.

How can $m=k$ ? They are 3 distinct primes.

Ahh easy proof, thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 11:11:31 pm

Okay some challenge ones to finish off :D

1. Prove that any integer greater than or equal to 7 can be written as a sum of 2 relatively prime integers, both greater than 1. For example, 22 and 15 are relatively prime, and thus $37=22+15$ represents the number 37 in the desired way.

2. Since $24=3+5+7+9$ , the number 24 can be written as the sum of at least two consecutive odd positive integers. Can 2005 be written as the sum of at least 2 consecutive odd positive integers? What about 2006?

~~3. The sequence $a_1,a_2, \cdots$ of natural numbers satisfies $\text{GCD}(a_i,a_j) = \text{GCD}(i,j)$ for all $i \neq j$ . Prove that $a_i=i \ \forall \ i$~~

4. Let $r \in \mathbb{N}$ . Show that $\binom{p^r}{k}$ is a multiple of $p$ for $0 < k < p^r$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 11:33:47 pm

1.)

say we want to do this for n.

case 1: n is odd

then n-2 and 2 are appropriate summands, clearly relatively prime since n-2 is odd.

case 2i: n=2k with k even

k+1 and k-1 are appropriate summands since no d>1 divides them both. Proof: suppose k+1=ad, k-1=bd. Then:

k+1-(k-1)=ad-bd
2=d(a-b)

hence the only possibility is d=2, d is a divisor of k+1 (an odd number), so it cannot be 2.

case2ii: n=2k, with k odd:

then k-2 and k+2 are our desired summands. Suppose that for some d>1, ad=k+2, bd=k-2

k+2-(k-2)=d(a-b)
4=d(a-b)

hence we're left with the possiblity that d is 2 or 4. But it cannot be these since d is a factor of an odd number, so it has to be odd.

These cases exhuast all possibilities for n.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 24, 2009, 11:37:07 pm

a good way to experiment/play with number 3 is to try to prove specific cases to see what techinques will work. e.g: try to prove $a_1=1$ , then try to prove $a_2=2$ etc. and try to refine your method for some more general cases.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 24, 2009, 11:37:52 pm

Quote from: kamil9876 on December 24, 2009, 11:37:07 pm

a good way to experiment/play with number 3 is to try to prove specific cases to see what techinques will work. e.g: try to prove $a_1=1$ , then try to prove $a_2=2$ etc. and try to refine your method for some more general cases.

Yeah, that's actually what I am attempting atm :P

Title: Re: TT's Maths Thread
Post by: TrueTears on December 25, 2009, 09:49:15 pm

Quote

3. The sequence $a_1,a_2, \cdots$ of natural numbers satisfies $\text{GCD}(a_i,a_j) = \text{GCD}(i,j)$ for all $i \neq j$ . Prove that $a_i=i \ \forall \ i$

Finally I can put an end to this question, it's been driving me nuts!

First assume $a_i \neq i$

Therefore for some $i$ , let's call it $m$ , $a_m \neq m$

$\implies (a_m, a_{2m}) = (m,2m) = m$

$\implies \frac{a_m}{m} = k$ for some $k$

$\therefore a_m = km$

Now $(a_{km}, a_{2km}) = (km,2km) = km$

$\therefore km | a_{km} \implies a_m | a_{km}$

$\implies (a_m, a_{km}) = a_m = km$

But $(a_m, a_{km}) = (m, km) = m$ Contradiction!

Thus $a_m = m$ so $a_i = i \ \blacksquare$

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 25, 2009, 09:51:45 pm

nice one :D even simpler than mine :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 25, 2009, 10:00:00 pm

omg i feel like a rock has been lifted off me, this question was driving me insane seriously.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 26, 2009, 02:38:31 am

Quote from: TrueTears on December 24, 2009, 11:11:31 pm

2. Since $24=3+5+7+9$ , the number 24 can be written as the sum of at least two consecutive odd positive integers. Can 2005 be written as the sum of at least 2 consecutive odd positive integers? What about 2006?

Let $S$ be the sum of a sequence of consecutive odd numbers.

$S = (2k+1)+(2k+3)+(2k+5)+ \cdots + (2k+2n-1)$ where $k \ge 0$ and $n \ge 2$

$S = \frac{n}{2}\left(2k+1+(2k+2n-1)\right) = \frac{n}{2}\left(4k+2n)\right) = n(2k+n)$

The PPF of $2005 = (401)(5)$

$\implies (401)(5) = n(2k+n)$

$\implies n = 5$

$\therefore 2k+5 = 401 \implies k = 198$

Subbing that back into our original expression for $S$ yields:

$S = (2(198)+1)+(2(198)+3)+(2(198)+5)+(2(198)+7)+(2(198)+9) = 2005$

$2006$ won't work because either $\text{n}$ and $(2k+n)$ are both odd or both even, but $2006 = 2 \times 1003$ , ie $\text{n}$ is even and $(2k+n)$ is odd. Contradiction!

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 27, 2009, 12:49:31 am

Quote from: TrueTears on December 24, 2009, 11:11:31 pm

4. Let $r \in \mathbb{N}$ . Show that $\binom{p^r}{k}$ is a multiple of $p$ for $0 < k < p^r$ .

the thing is:

$\frac{p^r(p^{r}-1)....(p^{r}-k+1)}{1*2*3....*k}$

Define the sequences

$A=(p^r, p^r-1..., p^r-k+1)$

$B=(1,2,3...k)$

now both sequences are strings of consecutive numbers.

Lemma1:

in a sequence of any $k$ consecutive numbers, the the number of multiples of $a$ is either $\lfloor\frac{k}{a}\rfloor$ or $\lceil \frac{k}{a} \rceil$

Proof: exercise

Lemma2: if the sequence from lemma1 has a multiples of $a$ as its first element, then the number of multiples of $a$ is exactly $\lceil \frac{k}{a} \rceil$ . If the sequence is {1,2,3....k} then the number of multiples of a is exactly $\lfloor \frac{k}{a} \rfloor$

Proof: more exercise

Lemma3: The two lemmas above imply that the number of multiples of $p^i$ in A is greater than or equal to the number of multiples of $p^i$ in B.

Now from each multiple of p, factor out one p and u get:

$\frac{p^{a_1}}{p^{b_1}} * (... )$

where $a_1 \geq b_1$ by lemma3.

The thing inside the bracket has a new numerator and a new denominator. So we have two new sequences $A_1$ and $B_1$ where these sequences are the previous ones with the power of p lowered by 1 in each multiple of p. The number of multiples of $p$ in $A_1$ is precisely equal to the number of multiples of $p^2$ in $A$ , and likewise for $B_1$ and $B$ . Therefore if we do this process again to get:

$\frac{p^{a_1}}{p^{b_1}}\frac{p^{a_2}}{p^{b_2}} * (... )$

with $a_2\geq b_2$ by lemma 3. If we keep doing this process we eventually get to:

$\frac{p^{a_1}}{p^{b_1}}\frac{p^{a_2}}{p^{b_2}}...\frac{p^{a_r}}{p^{b_r}} * (... )$

where in this case $a_r=1, b_r=0$ since there is only one multiple of $p^r$ in $A$ , and no multiples of $p^r$ in $B$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on December 27, 2009, 02:31:21 pm

Quote from: TrueTears on December 24, 2009, 11:11:31 pm

4. Let $r \in \mathbb{N}$ . Show that $\binom{p^r}{k}$ is a multiple of $p$ for $0 < k < p^r$ .

$\binom{p^r}{k} = \frac{(p^r-k+1)(p^r-k+2) \cdots (p^r-2)(p^r-1)p^r}{(k)(k-1)(k-2) \cdots (3)(2)(1)}$

If $k-1$ has factors of $p$ then $k-1 = qp^s$ where $q$ is some constant and $s<r$ .

Looking at the factor $(p^r-(k-1))$ in the numerator, if we paired up $(p^r-(k-1))$ and $(k-1)$ we would have $\frac{(p^r-(k-1))}{k-1}$ .

$\implies \frac{p^r}{k-1} - \frac{k-1}{k-1} = \frac{p^r}{qp^s}-1 = \frac{p^{r-s}}{q}-1$

If $k-2$ had factors of $p$ then $k-2 = cp^w$ where $c$ is some constant and $w<r$

$\implies \frac{p^r-(k-2)}{k-2} = \frac{p^{r-w}}{c} - 1$

There could be more factors of $p$ in the denominator but WLOG let's say these $2$ are the only ones.

Then we would have $\binom{p^r}{k} = \frac{\left(\frac{p^{r-s}}{q}-1\right)\left(\frac{p^{r-w}}{c} - 1\right) \cdots (p^r-2)(p^r-1)(p^r)}{(k) \cdots (3)(2)(1)}$

$\implies \frac{\left(\frac{p^{r-s}p^{r-w}}{qc} - \frac{p^{r-s}}{q} - \frac{p^{r-w}}{c} + 1\right) \cdots (p^r-2)(p^r-1)(p^r)}{(k) \cdots (3)(2)(1)}$

$\implies \frac{p^{r-s}\left(\frac{p^{r-w}}{qc} - \frac{1}{q} - \frac{p^{r-w}}{cp^{r-s}} + \frac{1}{p^{r-s}}\right) \cdots (p^r-2)(p^r-1)(p^r)}{(k) \cdots (3)(2)(1)}$

Therefore it is a multiple of $p$ .

If there were more "pairs" then we could always factor out a $p^{i}$ where $i < r$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 27, 2009, 11:11:20 pm

1. Prove that $\text{GCD}(a_1, .. . ,a_k) = \text{GCD}(\text{GCD}(a_1, a_2), a_3, . .. ,a_k)$

How to prove this using Euclid's Algorithm?

~~2. Show that $c|a$ and $c|b$ iff $c|(a,b)$~~

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 27, 2009, 11:21:49 pm

let's prove 2 first.

suppose c|a and c|b.

ax+by=(a,b) for some x,y.

a=a'c, b=b'c for some b',a'.

(a,b)=c(a'x+b'y) hence c divides (a,b)

Now to prove the converse:

$(a,b)|a \implies a=(a,b)a'$ for some $a'$

Since c|(a,b), (a,b)=cc' for some c':

Hence a=cc'a'=c(cc'a')

thus a is a multiple of c.

To prove b is a multiple of c, just repeat the same argument with b in place of a.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 27, 2009, 11:38:35 pm

let's prove 1 now:

for LHS, we seek the greatest number that divides all numbers in the set $\lbrace a_1,a_2 \rbrace \cup The rest$

for RHS, we seek the greatest number that divides $\lbrace gcd(a_1,a_2) \rbrace \cup The rest$

But from 2 we know that the greatest number that divides both elements in $\lbrace a_1,a_2 \rbrace$ is also the greatest number that divides $gcd(a_1,a_2)$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on December 28, 2009, 02:48:30 am

Quote from: kamil9876 on December 27, 2009, 11:21:49 pm

let's prove 2 first.

suppose c|a and c|b.

ax+by=(a,b) for some x,y.

a=a'c, b=b'c for some b',a'.

(a,b)=c(a'x+b'y) hence c divides (a,b)

Now to prove the converse:

$(a,b)|a \implies a=(a,b)a'$ for some $a'$

Since c|(a,b), (a,b)=cc' for some c':

Hence a=cc'a'=c(cc'a')

thus a is a multiple of c.

To prove b is a multiple of c, just repeat the same argument with b in place of a.

3rd last line, is that a typo? should be c(c'a')?

Quote from: kamil9876 on December 27, 2009, 11:38:35 pm

let's prove 1 now:

for LHS, we seek the greatest number that divides all numbers in the set $\lbrace a_1,a_2 \rbrace \cup The rest$

for RHS, we seek the greatest number that divides $\lbrace gcd(a_1,a_2) \rbrace \cup The rest$

But from 2 we know that the greatest number that divides both elements in $\lbrace a_1,a_2 \rbrace$ is also the greatest number that divides $gcd(a_1,a_2)$ .

Thanks for that, I'm happy with your explanation but how would you include Euclid's algorithm in it?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 28, 2009, 05:59:44 pm

Quote

3rd last line, is that a typo? should be c(c'a')?

yes

Quote

Thanks for that, I'm happy with your explanation but how would you include Euclid's algorithm in it?

this proof provides a method for finding the gcd of more than two numbers. For example.
$ gcd(a_1,a_2,a_3)=gcd(gcd(a_1,a_2),a_3)$

and so we can use our technique for finding gcd of two numbers and then do this simplification, and then use it again for the RHS (since we have two numbers now).

If we had n numbers, just keep using this repetetively and notice how the ammount of numbers in gcd(...) decreases by one in each step, until eventually you will ahve only two numbers there.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 28, 2009, 06:56:54 pm

Oh thanks kamil, so Euclid's algorithm is just simply repeated uses of another algorithm? In the case before to find the GCD it is the repeated use of the division algorithm, while here it is the repeated use of finding the GCD?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 28, 2009, 10:52:13 pm

yeah

Title: Re: TT's Maths Thread
Post by: TrueTears on December 28, 2009, 11:00:27 pm

Just started on some Linear Diophantine equations :)

Let $a, b$ and $c$ be integers, with $a$ and $b$ not both $0$ , and let $d = \text{GCD}(a, b)$ . Then the equation $ax+by=c$ has an integer solution $x, y$ if and only if $c$ is a multiple of $d$ , in which case there are infinitely many solutions. These are the pairs $x = x_0+\frac{bn}{d}$ , $y = y_0-\frac{an}{d}$ where n $\in \mathbb{Z}$ and $x_0, y_0$ is any particular solution.

How do you prove that $x = x_0+\frac{bn}{d}$ , $y = y_0-\frac{an}{d}$ are the general solutions?

Thanks

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 28, 2009, 11:27:28 pm

There is a graphical motivation fo the following proof:

$ax_0+by_0=c$

now any other solution can be written in the form $x=x_0+h, y=y_0+g$ (ie: our "run" and "rise" respectively).

$ax_0+by_0=c=a(x_0+h)+b(y_0+g)$

$\implies ah=-bg$

Thus we are after all common multiples of $a$ and $b$ . The lowest common multiple of a and b is $\frac{ab}{d}$ and all common multiples of $a$ and $b$ are multiples of of the lcm: $\frac{nab}{d}$ (prove this as an exercise)

Thus $ah=\frac{nab}{d}$

$h=\frac{nb}{d}$

and now do a similair job to find $g$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on December 28, 2009, 11:41:50 pm

Quote from: kamil9876 on December 28, 2009, 11:27:28 pm

There is a graphical motivation fo the following proof:

$ax_0+by_0=c$

now any other solution can be written in the form $x=x_0+h, y=y_0+g$ (ie: our "run" and "rise" respectively).

$ax_0+by_0=c=a(x_0+h)+b(y_0+g)$

$\implies ah=-bg$

Thus we are after all common multiples of $a$ and $b$ . The lowest common multiple of a and b is $\frac{ab}{d}$ and all common multiples of $a$ and $b$ are multiples of of the lcm: $\frac{nab}{d}$ (prove this as an exercise)

Thus $ah=\frac{nab}{d}$

$h=\frac{nb}{d}$

and now do a similair job to find $g$ .

Wait how can $ax_0+by_0=c=a(x_0+h)+b(y_0+g)$ since if we change $x_0$ and $y_0$ then ain't we changing $c$ to some other multiple of $d$ ?

As in the numerical value of $c$ for $ax_0+by_0=c$ is different to that of $c=a(x_0+h)+b(y_0+g)$

So why can you equate them?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 28, 2009, 11:50:54 pm

Quote

now any other solution can be written in the form x=x_0+h, y=y_0+g (ie: our "run" and "rise" respectively).

a straight forward consequence of "another solution"[to the SAME equation]

Title: Re: TT's Maths Thread
Post by: TrueTears on December 28, 2009, 11:59:21 pm

Quote from: kamil9876 on December 28, 2009, 11:50:54 pm

Quote
now any other solution can be written in the form x=x_0+h, y=y_0+g (ie: our "run" and "rise" respectively).

a straight forward consequence of "another solution"[to the SAME equation]

Ahhh yeah, thanks I get the proof now :)

A good question:

If $a_1, \cdots, a_k$ and $c$ are integers when does the Diophantine equation $a_1x_1+ \cdots +a_kx_k = c$ have integer solutions $x_1, \cdots, x_k$ ?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 29, 2009, 12:29:46 am

Let g=gcd(a1,...,an)
g divides a1 ... an, thus it divides any integer linear combo of a1,...,an.
applying the same logic you use to prove that ax+by=c has integer solution iff gcd(a1,...,an)|c:
the equation has a solution iff c is a multiple of gcd(a1,...,an).

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 12:58:45 am

Ahhh yeah, thanks zzdfa :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 01:34:06 am

Eisenstein's criterion states that if $f(x) = a_nx^n+ \cdots + a_1x+a_0$ where $a_i \in \mathbb{Z}$ , if $p$ is a prime and $p|\{a_0,a_1, \cdots, a_{n-1}\}$ but not $a_n$ and it $p^2$ does not divide $a_0$ , then $f(x)$ is irreducible.

I've actually used this criterion several times in the past while doing algebra problem solving questions, but I've never bothered to try to prove it, now that I'm doing number theory, how would you prove it?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 29, 2009, 01:59:34 am

suppose it is redducible, then:

$f(x)=(b_qx^q....+b_0)(c_rx^r...+c_0)$

$b_qc_r=a_n$ is not divisible by $p$ , hence $b_q$ and $c_r$ are both not divisible by $p$ . $b_0c_0=a_0$ is divisible by $p$ , but not by $p^2$ , thus one of these terms is not divisible by $p$ , say WLOG $c_0$ .

$ a_1=c_0b_1+c_1b_0$

but $a_1$ is divisibly by $p$ , $c_1b_0$ is (since $c_1$ is). $c_0b_1$ must therefore be divisible by $p$ , but since $c_0$ isn't then $b_1$ must be.

hence so far we now know that $b_1$ and $b_0$ are divisible by $p$ .

$a_2=c_0b_2+c_1b_1+c_2b_0$

now we use a similair argument to show that b_2 is divisible by p. Hence we keep doing this until we get b_q is divisibly by p, which contradicts what we know. Eventually this proof is tantamount to the technique used in this old problem we solved earlier (problem 1)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 02:19:51 pm

Quote from: kamil9876 on December 29, 2009, 01:59:34 am

suppose it is redducible, then:

$f(x)=(b_qx^q....+b_0)(c_rx^r...+c_0)$

$b_qc_r=a_n$ is not divisible by $p$ , hence $b_q$ and $c_r$ are both not divisible by $p$ . $b_0c_0=a_0$ is divisible by $p$ , but not by $p^2$ , thus one of these terms is not divisible by $p$ , say WLOG $c_0$ .

$ a_1=c_0b_1+c_1b_0$

but $a_1$ is divisibly by $p$ , $c_1b_0$ is (since $c_1$ is). $c_0b_1$ must therefore be divisible by $p$ , but since $c_0$ isn't then $b_1$ must be.

hence so far we now know that $b_1$ and $b_0$ are divisible by $p$ .

$a_2=c_0b_2+c_1b_1+c_2b_0$

now we use a similair argument to show that b_2 is divisible by p. Hence we keep doing this until we get b_q is divisibly by p, which contradicts what we know. Eventually this proof is tantamount to the technique used in this old problem we solved earlier (problem 1)

What if say $a_0=2, b_0 = 4$ and $c_0 = 6$

Ain't $b_0$ and $c_0$ both divisible by $2$ ?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 02:56:59 pm

Yeap I get the rest kamil, very nice proof and same notation as the international maths olympiad question we did earlier in the month :P

So yeah it's just that one query...

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 29, 2009, 03:08:32 pm

$c_0=a_0b_0=2*4\neq 6$

in general if $a_0=qp, b_0=rp$

$a_0b_0=qr (p^2)$ hence $p^2|c_0$ in this case.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 03:19:53 pm

Quote from: kamil9876 on December 29, 2009, 03:08:32 pm

$c_0=a_0b_0=2*4\neq 6$

in general if $a_0=qp, b_0=rp$

$a_0b_0=qr (p^2)$ hence $p^2|c_0$ in this case.

OH I see, lol I forgot the $p^2$ can not divide $a_0$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 04:53:17 pm

Consider the p-th cyclotomic polynomial $\phi_p(x) = 1+x+x^2+ \cdots +x^{p-1}$ where $p$ is a prime. Show that $\phi_p(x)$ is irreducible.

Why doesn't Eisenstein's criterion work?

$p^2$ does not divide $1$ .

$p$ does not divide the coefficient of $x^{p-1}$

But there are no $p$ that can divide the coefficients of all the $x$ terms since they are all $1$ ...

Doesn't this mean its reducible?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 29, 2009, 05:25:40 pm

In general: "A imples B" is not equivalent to "(not A) implies (not B)"

Specifically: So even though the criterion cannot be applied, that does not mean that the conclusion that the criterion would arrive at if applicable (that the polynomial is irreducible) is false.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 05:27:26 pm

Quote from: kamil9876 on December 29, 2009, 05:25:40 pm

In general: "A imples B" is not equivalent to "(not A) implies (not B)"

Specifically: So even though the criterion cannot be applied, that does not mean that the conclusion that the criterion would arrive at if applicable (that the polynomial is irreducible) is false.

Ah okay, so then how would you prove it is irreducible if Eisenstein's criterion can't be applied?

Title: Re: TT's Maths Thread
Post by: humph on December 29, 2009, 06:42:27 pm

Try making the substitution $x \mapsto x+1$ and then apply Eisenstein. Finally, use the fact that if $f(x+1)$ is irreducible, then so is $f(x)$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 10:01:17 pm

Quote from: humph on December 29, 2009, 06:42:27 pm

Try making the substitution $x \mapsto x+1$ and then apply Eisenstein. Finally, use the fact that if $f(x+1)$ is irreducible, then so is $f(x)$ .

But if you sub in x+1 doesn't that change the question? It's a totally different function?

Title: Re: TT's Maths Thread
Post by: zzdfa on December 29, 2009, 10:38:37 pm

Its a totally different function but if you can prove that function is irreducible then it implies that the original function is irreducible too (why?).

Title: Re: TT's Maths Thread
Post by: TrueTears on December 29, 2009, 11:20:26 pm

Ahh okay I think I understand the logic now.

Thanks guys.

$\phi_p(x) = 1+x+x^2+ \cdots +x^{p-1} \implies \phi_p(x) = \frac{x^p-1}{x-1}$

$\phi_p(x+1)$ is reducible iff $\phi_p(x)$ is reducible.

So assume $\phi_p(x)$ is reducible.

$\phi_p(x+1) = \frac{(x-1)^p-1}{x}$

$\frac{(x-1)^p-1}{x} = \frac{\binom{p}{0}x^p + \binom{p}{1}x^{p-1} + \binom{p}{2}x^{p-2} + \cdots + \binom{p}{p-1}x + \binom{p}{p}-1}{x}$

$\implies \phi_p(x+1) = \binom{p}{0}x^{p-1} + \binom{p}{1}x^{p-2} + \binom{p}{2}x^{p-3} + \cdots + \binom{p}{p-1}$

Now applying Eisenstein's criterion: $p$ does divide all coefficients except for the coefficient of $x^{p-1}$

$p^2$ certainly does not divide $\binom{p}{p-1}$ since $\binom{p}{p-1}=p$

Therefore $\phi_p(x+1)$ is irreducible. Contradiction!

So $\phi_p(x)$ is irreducible.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 30, 2009, 11:14:21 pm

Prove that if a positive integer $m$ is not a perfect square, then $\sqrt{m}$ is irrational.

Cool simple question, wanted to share it around xD

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 30, 2009, 11:27:37 pm

if it's not a square, then it cannot be an integer. and so it must be of the form:

$\frac{\prod p^{e^i}}{\prod q_i^{f^i}}$ . Where the numerator and denominator are relatively prime and the denominator is not 1 (not the empty product). Can the nth power of such a number be an integer? This proves it not just for square root, but for nth root.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 01:12:42 am

Yeap that's right.

So we assume $\sqrt{m}$ is rational thus $\sqrt{m} = \frac{a}{b} \implies m = \frac{a^2}{b^2}$ for some positive integer $a$ and $b$ . Assume that the trivial case $a = b$ is not included, thus $a \neq b$ .

Now for $m$ to be a perfect square, $\frac{a}{b}$ must be an integer, thus $a>b$ .

$a = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}p_{k+1}^{e_{k+1}}p_{k+2}^{e_{k+2}}p_{k+3}^{e_{k+3}} \cdots p_{k+n}^{e_{k+n}}$

$b = p_1^{f_1}p_2^{f_2} \cdots p_k^{f_k}$

$\implies \frac{a}{b} = \frac{p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}p_{k+1}^{e_{k+1}}p_{k+2}^{e_{k+2}}p_{k+3}^{e_{k+3}} \cdots p_{k+n}^{e_{k+n}}}{p_1^{f_1}p_2^{f_2} \cdots p_k^{f_k}} = p_1^{e_1-f_1}p_2^{e_2-f_2}p_3^{e_3-f_3} \cdots p_k^{e_k-f_k}p_{k+1}^{e_{k+1}}p_{k+2}^{e_{k+2}}p_{k+3}^{e_{k+3}} \cdots p_{k+n}^{e_{k+n}}$

$\implies \frac{a^2}{b^2} = p_1^{2(e_1-f_1)}p_2^{2(e_2-f_2)}p_3^{2(e_3-f_3)} \cdots p_k^{2(e_k-f_k)}p_{k+1}^{2e_{k+1}}p_{k+2}^{2e_{k+2}}p_{k+3}^{2e_{k+3}} \cdots p_{k+n}^{2e_{k+n}}$

Since $e_i-f_i \ge 0$ that means the exponents in the PPF of $\frac{a^2}{b^2}$ are all multiples of $2$ .

Thus $m$ is a perfect square.

This means $m$ is a perfect square iff $\sqrt{m}$ is an integer.

Therefore if $\sqrt{m}$ is irrational, $m$ is not a perfect square.

Actually kamil raised a good point.

What if $\sqrt{m}$ is rational?

Let's assume $\sqrt{m}$ is rational which means $\sqrt{m} = \frac{a}{b}$

Thus $(a,b) = 1$

So $m = \frac{a^2}{b^2}$ but $\left(a^2,b^2\right) = 1 \implies m \in \mathbb{Q}$

But $m$ must be an integer, so contradiction.

The only option left now is that if $\sqrt{m}$ is irrational then $m$ to be not a perfect square.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 31, 2009, 01:39:02 am

something feels sticky

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 02:15:50 am

Quote from: kamil9876 on December 31, 2009, 01:39:02 am

something feels sticky

lolololol

Prove that if $m$ is a square then $m$ leaves a remainder $0$ or $1$ when divided by $4$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 31, 2009, 02:27:57 am

if m is even. let $m=2k: m^2=4k^2$ which is a multiple of 4, hence remainder 0.

if m is odd. let $m=2k+1: m^2=4(k^2+k)+1$ which leaves remainder 1 when divided by 4.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 02:33:58 am

something feels sticky

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 07:11:32 pm

Calculate the least non-negative residue of $3^8$ $\mod (13)$ .

Title: Re: TT's Maths Thread
Post by: /0 on December 31, 2009, 07:15:53 pm

$3^3 \equiv 27 \equiv 1 \pmod {13}$

$\Rightarrow (3^3)^2 \equiv 1^2 \equiv 1 \pmod {13}$

$\therefore 3^8 \equiv (3^3)^2\times 3^2 \equiv 1 \times 3^2 \equiv 9 \pmod {13}$ xD

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 08:02:59 pm

Quote from: /0 on December 31, 2009, 07:15:53 pm

$3^3 \equiv 27 \equiv 1 \pmod {13}$

$\Rightarrow (3^3)^2 \equiv 1^2 \equiv 1 \pmod {13}$

$\therefore 3^8 \equiv (3^3)^2\times 3^2 \equiv 1 \times 3^2 \equiv 9 \pmod {13}$ xD

Cool. Thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 08:04:48 pm

Find the final decimal digit of $1! + 2! + 3! + ... + 10!$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 31, 2009, 10:12:38 pm

$\equiv 1!+2!+3!+4! mod 10$

since all the other terms are multiples of 10. This is because they contain a 5 and a 2 in the product, hence a 10.

$1!\equiv 1$

$2!\equiv 2$

$3!\equiv 6$

$4!\equiv 24=4+20\equiv 4$

therefore the sum $\equiv 1 + 2 + 6 + 4 \equiv 13 \equiv 3$

hence 3 is the last digit.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 10:32:50 pm

Quote from: kamil9876 on December 31, 2009, 10:12:38 pm

$\equiv 1!+2!+3!+4! mod 10$

since all the other terms are multiples of 10. This is because they contain a 5 and a 2 in the product, hence a 10.

$1!\equiv 1$

$2!\equiv 2$

$3!\equiv 6$

$4!\equiv 24=4+20\equiv 4$

therefore the sum $\equiv 1 + 2 + 6 + 4 \equiv 13 \equiv 3$

hence 3 is the last digit.

But you could also have $1!+2!+3!+4! \equiv a+b+c+d \pmod {10}$

So $1! \equiv 1 \pmod {10}$

$2! \equiv -8 \pmod {10}$

$3! \equiv -4 \pmod {10}$

$4! \equiv 24 \pmod {10}$

So $1!+2!+3!+4! \equiv 1-8-4+24 \equiv 13 \pmod {10}$

In other words how do you know what values to pick for $a,b,c,d$ ?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 31, 2009, 10:53:24 pm

still get mod 3 in the end

Title: Re: TT's Maths Thread
Post by: TrueTears on December 31, 2009, 10:54:30 pm

Quote from: kamil9876 on December 31, 2009, 10:53:24 pm

still get mod 3 in the end

But is there a way I can minimise the work at the end so I get 3 at the end?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 31, 2009, 10:59:27 pm

just delete all the digits from each summand, except the last one.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 01, 2010, 04:28:45 pm

~~1. Prove that the polynomial $f(x) = x^5 - x^2 + x - 3$ has no integer roots.~~

2. Prove that there is no non-constant polynomial $f(x)$ , with integer coefficients, such that $f(x)$ is prime for all integers $x$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on January 01, 2010, 05:13:09 pm

Quote from: TrueTears on January 01, 2010, 04:28:45 pm

2. Prove that there is no non-constant polynomial $f(x)$ , with integer coefficients, such that $f(x)$ is prime for all integers $x$ .

Assume $f(x)$ is a non-constant polynomial with integer coefficients and is prime for all integers $x$ .

This means for any integer $a \implies f(a) = p$ for any prime $p$ .

$f(a) \equiv 0 \pmod {p}$ since $f(a) = p$

But $a+pk \equiv a \pmod {p}$ for some constant $k$ .

$f(a+pk) \equiv f(a) \equiv 0 \pmod {p}$

Which means $f(a+pk) = p$ since $p$ is prime and the only thing $p$ can divide is itself.

This means $f(x) - p = 0$ has infinitely many solutions since if you plug in any number in the form of $a+pk$ then $f(x)-p=0$ is satisfied.

However $f(x)$ can only have at most $d$ roots where $d$ is the leading degree.

Contradiction!

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 08:03:25 pm

An interesting question:-

"There are only about 20 people in Victoria that can prove this." - Dr. Jianming He

Prove that 1 + 1 = 2.

Are the 20 people among the VN community?

Title: Re: TT's Maths Thread
Post by: TrueTears on January 01, 2010, 08:10:40 pm

Quote from: brightsky on January 01, 2010, 08:03:25 pm

An interesting question:-

"There are only about 20 people in Victoria that can prove this." - Dr. Jianming He

Prove that 1 + 1 = 2.

Are the 20 people among the VN community?

Where in history do we get to start?

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 08:22:42 pm

"The proof starts from the Peano Postulates, which define the natural numbers N."

Title: Re: TT's Maths Thread
Post by: TrueTears on January 01, 2010, 08:26:58 pm

Quote from: brightsky on January 01, 2010, 08:22:42 pm

"The proof starts from the Peano Postulates, which define the natural numbers N."

One could go further back in history before Peano.

Title: Re: TT's Maths Thread
Post by: /0 on January 01, 2010, 08:30:38 pm

Quote from: brightsky on January 01, 2010, 08:03:25 pm

An interesting question:-

"There are only about 20 people in Victoria that can prove this." - Dr. Jianming He

Prove that 1 + 1 = 2.

Are the 20 people among the VN community?

That's ridiculous. Show the proof to people and they will know how to prove it. Show it to 100 people and 100 people will know how to prove it.

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 08:32:22 pm

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

Here's the proof but I still don't understand it.

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 08:33:14 pm

Quote from: TrueTears on January 01, 2010, 08:26:58 pm

Quote from: brightsky on January 01, 2010, 08:22:42 pm
"The proof starts from the Peano Postulates, which define the natural numbers N."
One could go further back in history before Peano.

Lolol, definitely agreed. Edit: Since mathematical concepts were around?

Title: Re: TT's Maths Thread
Post by: /0 on January 01, 2010, 08:35:04 pm

It's all definitions and postulates, there's hardly a proof in it

Title: Re: TT's Maths Thread
Post by: TrueTears on January 01, 2010, 08:36:44 pm

Quote from: /0 on January 01, 2010, 08:35:04 pm

It's all definitions and postulates, there's hardly a proof in it

Exactly.

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 01, 2010, 08:47:51 pm

I'm quite interested in proving the fundamental properties of numbers, and constructing the reals from the rationals, dedekind cuts etc. I recently read Edmund Landau's "Foundations of Analysis", a book that is completely devoted to this and he proves everything starting from the Peano Postulates. However "proving" "1+1=2" is useless. He never mentions the number 2, but denotes it always as 1+1. For example: "Theorem: 1+1 is irrational" or when refering to the midpoint of a and b he writes it as " $\frac{a+b}{1+1}$ ". The only times he ever mentions the number 2 is in the preface:

"The multiplication table is not to be found in this book, not even the theorem 2*2=4; but i would recommend, as an exercise in connection with chapter 1, that you make the following definitions:

2=1+1
4=(((1+1)+1)+1)

and then prove the theorem"

Also found in the preface:

"Forgive me for 'theeing' and 'thouing' you. One reason for my doing so is that this book is written partly for my daugheters who have been studying chemistry at the University several semesters alreadyand think that they have learnt the differential and integral calculus in College; and yet they still don't know why:

x*y=y*x"

Having said that, this is what mathematicains are really interested in when it comes to Peano axioms: proving these general laws of arithmetic like xy=yx. They simply treat "2" as a denotation for "1+1" or "the succesor of 1", not as some deep theorem.

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 08:54:01 pm

Quote from: kamil9876 on January 01, 2010, 08:47:51 pm

One reason for my doing so is that this book is written partly for my daugheters who have been studying chemistry at the University several semesters alreadyand think that they have learnt the differential and integral calculus in College; and yet they still don't know why:

x*y=y*x"

That's probably true for 99.99999% of people. Great post! +1

Title: Re: TT's Maths Thread
Post by: zzdfa on January 01, 2010, 09:02:59 pm

Quote from: brightsky on January 01, 2010, 08:03:25 pm

An interesting question:-

"There are only about 20 people in Victoria that can prove this." - Dr. Jianming He

Prove that 1 + 1 = 2.

Are the 20 people among the VN community?

who is dr jianming he??

Title: Re: TT's Maths Thread
Post by: TrueTears on January 01, 2010, 09:03:38 pm

Quote from: zzdfa on January 01, 2010, 09:02:59 pm

Quote from: brightsky on January 01, 2010, 08:03:25 pm
An interesting question:-

"There are only about 20 people in Victoria that can prove this." - Dr. Jianming He

Prove that 1 + 1 = 2.

Are the 20 people among the VN community?

who is dr jianming he??

Overrated tutor I believe.

Title: Re: TT's Maths Thread
Post by: zzdfa on January 01, 2010, 09:13:53 pm

ohh, dr he

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 09:15:33 pm

Quote from: TrueTears on January 01, 2010, 09:03:38 pm

Quote from: zzdfa on January 01, 2010, 09:02:59 pm
Quote from: brightsky on January 01, 2010, 08:03:25 pm
An interesting question:-

"There are only about 20 people in Victoria that can prove this." - Dr. Jianming He

Prove that 1 + 1 = 2.

Are the 20 people among the VN community?

who is dr jianming he??
Overrated tutor I believe.

Have you been there before? Lol, yeah, I think he is a bit overrated, but he is really really good at doing his job. I think his methods and ways of teaching help explain a lot of things that normal teachers can't accurately and aptly explain to a student.

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 09:20:01 pm

Lolol. Differs from person to person I guess. 可谓仁者见仁，智者见智! ;D

Title: Re: TT's Maths Thread
Post by: TrueTears on January 01, 2010, 09:26:54 pm

Quote from: brightsky on January 01, 2010, 09:20:01 pm

Lolol. Differs from person to person I guess. 可谓仁者见仁，智者见智! ;D

hahaha I see you have been reading my Chinese essays!

Title: Re: TT's Maths Thread
Post by: brightsky on January 01, 2010, 09:28:34 pm

Hehe, sure have! Got to say, they are wayy too pro!! :p

Title: Re: TT's Maths Thread
Post by: EvangelionZeta on January 02, 2010, 03:42:20 pm

Quote from: TrueTears on January 01, 2010, 09:17:25 pm

Nope haven't been to him, but from what I've heard from friends, he's pretty shit.

Really? I've heard he teaches concepts very well (haven't been to his class personally, although I sat his Methods practice exams. :p). Also his questions tend to be challenging enough to make actual VCAA exams appear very easy.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 02, 2010, 05:18:36 pm

Quote from: EvangelionZeta on January 02, 2010, 03:42:20 pm

Quote from: TrueTears on January 01, 2010, 09:17:25 pm
Nope haven't been to him, but from what I've heard from friends, he's pretty shit.

Really? I've heard he teaches concepts very well (haven't been to his class personally, although I sat his Methods practice exams. :p). Also his questions tend to be challenging enough to make actual VCAA exams appear very easy.

Alright just wondering, if we have $2$ integers $a$ and $b$ such that $(a,b) = 1$ then there exists some integers $x$ and $y$ such that $ax+by = 1.$

Then does that mean if we have $\text{n}$ integers $a_1,a_2, \cdots a_n$ such that $(a_1,a_2, \cdots a_n) = 1$ then does there exists some integers $x_1,x_2,x_3 \cdots x_n$ such that $a_1x_1+a_2x_2+ \cdots + a_nx_n = 1$ ?

If there is, then how do we solve the linear diophantine equation $a_1x_1+a_2x_2+ \cdots + a_nx_n = 1$ for $x_1$ ? Ie, what would the general solution be? (Is this question beyond my abilities? Is there even a general solution?)

Thanks :)

Title: Re: TT's Maths Thread
Post by: zzdfa on January 02, 2010, 06:45:13 pm

yea it's pretty simple,

try proving it for n=3.
let g=(a1,a2)

use the fact that
(a1,a2,a3)= (g,a3)

and

g=a1y1+a2y2 for some y1,y2

the general case should follow easily, by induction.

edit: actually, if you're using induction, it's easier to prove the general case instead:

a1x1+...+anxn=m has a solution iff (a1,...,an)|m

the inductive proof also shows you how to find the solution.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 03, 2010, 01:34:35 am

Finally I can do some modular arithmetic questions.

I must say this probs has been the most challenging topic for me so far, but now I love it so much <3

Some warmup questions out there for thought (And yes there's like 3 putnam and 2 IMO's questions coming soon :P, this ought to be fun)

~~1. Show that if $a^2+b^2=c^2$ then $3|ab$~~

~~2. If $x^3+y^3=z^3$ show that one of the three must be a multiple of $7$~~

~~3. Find all prime $x$ such that $x^2+2$ is a prime~~

~~4. Prove there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes~~

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 03, 2010, 04:33:57 pm

We have to show that either a or b is a multiple of 3. Assume that neither a nor b was a multiple of 3. This means:

$a\equiv \pm 1\mod 3, b\equiv \pm 1\mod 3$

$a^2+b^2 \equiv 2 \mod 3$

But no square is congruent to 2 mod 3. Since $0^2 \equiv 0, (\pm 1)^2=1 \mod 3$

same trick for all other q's.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 03, 2010, 05:12:58 pm

How do you know if a or b wasn't a multiple of 3 then it'd be congruent to 1 mod 3?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 03, 2010, 06:02:10 pm

0,1,2 or equivalently 0,1,-1 are all the residues mod 3. Each integer has exactly one of these, if not a multiple of 3, then not 0 and hence 1 or -1.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 03, 2010, 06:02:50 pm

Quote from: kamil9876 on January 03, 2010, 06:02:10 pm

0,1,2 or equivalently 0,1,-1 are all the residues mod 3. Each integer has exactly one of these, if not a multiple of 3, then not 0 and hence 1 or -1.

lol how stupid of me, how did I not think of that ffs

Title: Re: TT's Maths Thread
Post by: TrueTears on January 03, 2010, 06:29:23 pm

Quote from: kamil9876 on January 03, 2010, 04:33:57 pm

We have to show that either a or b is a multiple of 3. Assume that neither a nor b was a multiple of 3. This means:

$a\equiv \pm 1\mod 3, b\equiv \pm 1\mod 3$

$a^2+b^2 \equiv 2 \mod 3$

But no square is congruent to 2 mod 3. Since $0^2 \equiv 0, (\pm 1)^2=1 \mod 3$

But $c^2 \equiv 2 \pmod {3}$

$c \equiv \sqrt{2} \pmod {3}$

So there is a value of c that satisfies, since the question never stated a b and c are integers?

Nvm

Title: Re: TT's Maths Thread
Post by: TrueTears on January 03, 2010, 09:09:42 pm

Quote

2. If $x^3+y^3=z^3$ show that one of the three must be a multiple of $7$

The questions wants us to show that $7|xyz$

Let's assume the contrapositive, that $7$ does not divide $xyz$

$x \equiv \pm 1, \pm 2, \pm 3 \pmod {7}$

$y \equiv \pm 1, \pm 2, \pm 3 \pmod {7}$

$z \equiv \pm 1, \pm 2, \pm 3 \pmod {7}$

$x^3+y^3 \equiv z^3 \pmod {7}$

Now assume some number that is not congruent to $0 \pmod {7}$ satisfies the equation.

Now $x^3 \equiv \left(\pm 1\right)^2, \left(\pm 2\right)^2, \left(\pm 3\right)^2 \pmod {7}$

$\implies x^3,y^3,z^3 \equiv 1, 8, 27 \equiv \pm 1 \pmod {7}$

$\therefore \pm 1 \pm 1 \equiv \pm 1 \pmod {7}$

But clearly this is false.

So there are no numbers that is not congruent to $0 \pmod {7}$ which satisfies the equation. So at least one of them must be divisible by $7$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 03, 2010, 09:20:45 pm

Mods doing mod ;D

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 04, 2010, 01:42:46 am

hint for 3:

3^2+2=11

so x=3 is a solution.

5^2+2=27=3(9) so 5 is not a soution.

7^2+2=51=3(14) so 7 is not a solution.

11^2+2=123=3(41) so 11 is not a solution.

13^2+2=171=3(57) so 13 is not a solution.

Now conjecture and prove.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 04:22:27 pm

Quote

3. Find all prime $x$ such that $x^2+2$ is a prime

Thanks kamil for hint.

I conjecture that the final answer can not be a multiple of $3$ and that any prime number larger than $3$ when subbed into the expression produces an integer that is a multiple of $3$ .

Assume that the final answer is a multiple of $3$ .

$\implies x^2+2 \equiv 0 \pmod {3}$

$\implies x^2 \equiv -2 \pmod {3}$

$\implies x^2 \equiv 1 \pmod {3}$

$\implies x \equiv 1, -1 \pmod {3}$

$\implies x \equiv 1 \ \text{or} \ x \equiv 2 \pmod {3}$

But all prime numbers (except for $3$ ) can be written in the form of either $6k+1$ or $6k-1$ .

$\implies 6k+1 \equiv 1 \pmod {3} \ \text{and} \ 6k-1 \equiv 2 \pmod {3}$

$\implies x \equiv 6k+1 \ \text{or} \ x \equiv 6k-1 \pmod {3}$

Thus any primes in the form of $6k+1$ or $6k-1$ when subbed into the expression $x^2+2$ produces an integer that is a multiple of $3$ which does not satisfy the condition.

However the only prime which can not be expressed in the form of $6k+1$ or $6k-1$ is $3$ .

$\therefore x = 3$

Or how about this way...

Assume that $x^2+2$ is prime iff $x$ is prime.

Since from the experimenting we conjecture that any prime larger than $3$ produces an integer that is a multiple of $3$ . Let us consider $\mathbb{Z}_3$ .

Since $x^2+2$ is prime, it can only be divided by itself or $1$ .

$x^2+2 \not\equiv 0 \pmod {3}$

$x^2 \not\equiv 1 \pmod {3}$

$x \not\equiv 1 \ \text{or} \ x \not\equiv -1 \pmod {3}$

We require $x \equiv 0 \pmod {3}$ in order to maintain $x^2+2$ as a prime.

$\therefore x = 3k \equiv 0 \pmod {3}$ for $x^2+2$ to be prime.

However the only prime number in the form of $3k$ is $3$ .

Thus $x = 3$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 07:02:51 pm

Quote

4. Prove there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes

Okay, I'm not sure if this is the right way to go about this question, could someone please check it's validity.

I conjecture that all perfect cubes are congruent to either $0, \pm 1 \pmod {9}$

Let $x$ be an integer.

Then in $\mathbb{Z}_9$ we have:

$x \equiv 0, \pm 1, \pm 2, \pm 3, \pm 4 \pmod {9}$

$\implies x^3 \equiv 0, \pm 1, \pm 8, \pm 27, \pm 64 \pmod {9}$

$\implies x^3 \equiv 0, \pm 1 \pmod {9}$

Now let $a^3, b^3, c^3$ be any $3$ perfect cubes.

Thus $a^3+b^3+c^3 \equiv 0, \pm 1, \pm 2, \pm 3 \pmod {9}$

However the least absolute residue $\pm 4$ is unaccounted for. Since there are infinitely many numbers congruent to $\pm 4 \pmod {9}$ then there are there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes.

Title: Re: TT's Maths Thread
Post by: zzdfa on January 04, 2010, 07:12:41 pm

looks right to me!

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 07:15:11 pm

Awesome, the only thing that took me a long time to get was the $\pmod {9}$

I knew we had to deal with some kinda of $\mathbb{Z}_n$ , so I tried $\mathbb{Z}_2, \mathbb{Z}_3$ none worked.

So is there a faster way to find out how the cubes were related to $\mathbb{Z}_9$ rather than experimentation?

Also since this question was in the modular arithmetic chapter I knew I had to playing around with mods, if it wasn't there was no way I could have thought of $\mathbb{Z}_9$ .

So yeah, what is a fast way to see the link between $\mathbb{Z}_9$ and cubes?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 04, 2010, 08:46:09 pm

You have three terms involved, hence you need something with an absolute residue class of 4. This means you have to look at at least mod 8.

by the 18th century "power tables" were used for computations. Perhaps not as famous trig or log tables but you can find a few modular arithmetic tables at the back of disquisitiones arithmeticae

In general. If we consider for some relatively prime integers $a$ and $m$ the residues $\mod m$ of $a,a^2,a^3...$ We find that two of them must be equal by the pigeonhole principle:

$a^k=a^n \mod m$ (with k>n)

$a^{k-n}=1 \mod m$ (because $a^n$ and $m$ are relatively prime)

hence $a^x=1 \mod m$ always has a solution when (a,m)=1

9=3^2 and so it is expected that there are many numbers relatively prime to it and any number not relatively prime to it will by =0 mod 9 (because the numebr will have 3 as factor, hence it cube will have 27 and hence 9 as a factor)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 08:48:34 pm

Thanks kamil, that was very clever :P

Alright, this next set is all IMO and Putnam questions :P (This is gonna be fun...)

1. When $4444^{4444}$ is written in decimal notation, the sum of its digits is $A$ . Let $B$ be the sum of the digits of $A$ . Find the sum of the digits of $B$ . ( $A$ and $B$ are written in decimal notation.)

2. The number $d_1d_2 \cdots d_9$ has nine (not necessarily distinct) decimal digits. The number $e_1e_2 \cdots e_9$ is such that each of the nine $9$ -digit numbers formed by replacing just one of the digits $d_i$ is $d_1d_2 \cdots d_9$ by the corresponding digit $e_i \ (1 \le i \le 9)$ is divisible by $7$ . The number $f_1f_2 \cdots f_9$ is related to $e_1e_2 \cdots e_9$ in the same way: that is each of the nine numbers formed by replacing one of the $e_i$ by the corresponding $f_i$ is divisible by $7$ . Show that for each $i$ , $d_i - f_i$ is divisible by $7$ . (For example, if $d_1d_2 \cdots d_9 = 199501996$ then $e_6$ may be $2$ or $9$ , since $199502996$ and $199509996$ are multiples of $7$ ).

3. Suppose $a,b,c,d$ are integers with $0 \le a \le b \le 99, 0 \le c \le d \le 99$ . For any integer $i$ , let $n_i = 101i+100(2^i)$ . Show that if $n_a + n_b$ is congruent to $n_c+n_d \pmod {10100}$ then $a=c$ and $b=d$ .

Title: Re: TT's Maths Thread
Post by: brightsky on January 04, 2010, 10:23:41 pm

Lemme have a crack at question 1.

First, let $X = 4444^{4444}$ .

$\because 10^4 > 4444$

$\therefore (10^4)^{4444} > 4444^{4444}$

$\implies 10^{17776} > 4444^{4444}$

...meaning that X must have less than or equal to 17777 digits in it. (TT Edit)

A sum of the digits of any number $n$ must be greater or equal to $9n$ . That is to say that, A (the sum of the digits of X) must have no more than 17776*9 = 159984 digits.

It is to note here that the number with the largest sum of digits below 159984 is 99999, which has a sum of 45. From this, we can see that B (the sum of the digits of A) must have no more than 5*9 = 45 digits.

Similarly, it is to note here that the number with the largest sum of digits below 45 is 39, which has a digit sum of 12. Hence, we can say that C (where C is the sum of the digits of B) must have less than or equal to 12 digits.

Remember now the divisibility test for 9. An integer is divisible by 9 if and only if the sum of its digits (in decimal notation), is divisible by 9. Quick proof for this is seen here, $10 \equiv 1 (\mod 9), 10^2 \equiv 1 (\mod 9), ..., 10^n \equiv 1 (\mod 9)$

That is to say that:

$N = a_n10^n + ... + a_110+ a_0 \equiv a_0 + a_1 + a_2 + ... + a_n \equiv 1 (\mod 9)$ , for any natural number N.

Hence, we can see that:

$X \equiv A \equiv B \equiv C (\mod 9)$

As 4444 = 9 x 493 + 7

$\therefore 4444 \equiv 7 \mod (9)$ [1]

Now lets do some tests with the number 7.

$7 \equiv -2 (\mod 9)$

$7^2 \equiv 4 (\mod 9)$

$7^3 \equiv 1 (\mod 9)$ [2]

From [1],

$4444^{4444} \equiv \7^{4444} (\mod 9)$

$\equiv 7^{3*1481+1} (\mod 9)$

$\equiv 7^{3*1481} . 7 (\mod 9)$

From [2],

$\equiv 7 (\mod 9)$

That means that, $C \equiv 7 (\mod 9)$ . Because C > 12, hence C must be 7.

Edit: I'm still learning how to use latex properly. Why is there a huge gap in the modulos? This is just a simple solution, there must be more elegant alternatives.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 10:39:54 pm

Quote from: brightsky on January 01, 2010, 08:32:22 pm

The proof starts from the Peano Postulates, which define the natural
numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

Here's the proof but I still don't understand it.

http://mathforum.org/library/drmath/view/51551.html

Title: Re: TT's Maths Thread
Post by: brightsky on January 04, 2010, 10:44:05 pm

Yep. Found the proof there. But I still don't understand it..

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 10:45:03 pm

Can you help me solve $10x \equiv 3 \pmod {12}$ for $x$ ?

Title: Re: TT's Maths Thread
Post by: brightsky on January 04, 2010, 11:07:25 pm

Think you can solve this by diophantine equations...(Can I?)

Assuming that x is a integer:

$10x - 3 = 12 N$ , where N is a positive integer.

$x = \frac {12N + 3}{10}$

$x = \frac {10N +10 + 2N - 7}{10}$

$x = N + 1 + \frac {2N - 7}{10}$

$\because$ N + 1 and x are both integers

$\therefore \frac {2N - 7}{10}$ must also be an integer

Let $\frac {2N - 7}{10} = k_1$ where $k_1$ is an integer.

$2N - 7 = 10k_1$

$2N = 10k_1 - 7$

$N = \frac {10k_1 - 7}{2}$

$N = \frac {10k_1 - 6 - 1}{2}$

$N = 5k_1 - 3 - \frac {1}{2}$

$\because$ N and 5k_1 - 3 are integers

$\therefore \frac {1}{2}$ is an integer.

Contradiction. Hence no solutions.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 11:09:45 pm

Kewl.

I think this proof follows on nicely from the one you posted. Just an extension.

Prove that for a base 10 number it is congruent mod 11 to the number's 'units digit' - 'tens digit' + 'hundreds digit' - 'thousands digit' .....

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 04, 2010, 11:13:48 pm

there is a much simpler way:

$10x-3=12k$

10x is even, 3 is odd, therefore 10x-3 is odd. But 12k is even. contradiction.

Title: Re: TT's Maths Thread
Post by: /0 on January 04, 2010, 11:39:47 pm

"Introduction to Number Theory" - Zuckerman, theorem 2.13 states that:
"Let $g$ denote $(a,m)$ . Then $ax \equiv b \pmod m$ has no solutions if $g \not|\ b$ "
Clearly that applies to $10x \equiv 3 \pmod {12}$ here too

Also, I think for divisibility by 11... if you have a number, $x = a_0+a_1 \cdot 10 + a_2 \cdot 10^2...$

Since $10 \equiv -1\pmod {11}$

$10^2 \equiv 1 \pmod {11}$ , $10^3 \equiv -1 \pmod{11}$ ...

So

$x \equiv a_0 +a_1(-1)+a_2(1)+a_3(-1)... \pmod {11}$

Title: Re: TT's Maths Thread
Post by: brightsky on January 04, 2010, 11:39:59 pm

We are looking at the number $N = x_1 + x_2 (10^1)+ x_3 (10^2)+ ... + x_n (10^{n-1})$

Consider these two patterns:

Pattern 1:

$99 \equiv 0 (\mod11$

$9999 \equiv 0 (\mod11)$

$999999 \equiv 0 (\mod11)$

...

Pattern 2:

$11 \equiv 0 (\mod11)$

$1001 \equiv 0 (\mod11)$

$100001 \equiv 0 (\mod11)$

...

Now back to our original equation, we see that:

$10^1 = 11 - 1$

$10^2 = 99 + 1$

$10^3 = 1001 - 1$

$10^4 = 9999 + 1$

...

Hence, rewriting the equation:

$x_1 + x_2 (11 - 1) + x_3 (99+1) + x_4 (1001 -1) + x_5 (9999+1) + ...$

$x_1 + x_2 (11) - x_2 + x_3 (99) + x_3 + x_4(1001) - x_4 + x_5 (9999) + x_5 + ...$

Because the numbers in brackets are all multiples of 11, hence, the number is congruent to mod 11 if the remaining numbers are multiples of 11.

Hence,

$N \equiv x_1 - x^2 + x_3 - x_4 + x_5 ...x_n (\mod 11)$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 04, 2010, 11:42:31 pm

Quote from: /0 on January 04, 2010, 11:39:47 pm

"Introduction to Number Theory" - Zuckerman, theorem 2.13 states that:
"Let $g$ denote $(a,m)$ . Then $ax \equiv b \pmod m$ has no solutions if $g \not|\ b$ "
Clearly that applies to $10x \equiv 3 \pmod {12}$ here too

Also, I think for divisibility by 11... if you have a number, $x = a_0+a_1 \cdot 10 + a_2 \cdot 10^2...$

Since $10 \equiv -1\pmod {11}$

$10^2 \equiv 1 \pmod {11}$ , $10^3 \equiv -1 \pmod{11}$ ...

So

$x \equiv a_0 +a_1(-1)+a_2(1)+a_3(-1)... \pmod {11}$

Nice /0, exactly the one I had in mind too XD

Title: Re: TT's Maths Thread
Post by: TrueTears on January 05, 2010, 12:00:28 am

Quote from: brightsky on January 04, 2010, 10:23:41 pm

Lemme have a crack at question 1.

First, let $X = 4444^{4444}$ .

$\because 10^4 > 4444$

$\therefore (10^4)^{4444} > 4444^{4444}$

$\implies 10^{17776} > 4444^{4444}$

...meaning that X must have less than or equal to 17776 digits in it.

Not quite, should have 17776+1 digits in it.

Title: Re: TT's Maths Thread
Post by: brightsky on January 05, 2010, 12:03:11 am

Quote from: TrueTears on January 05, 2010, 12:00:28 am

Quote from: brightsky on January 04, 2010, 10:23:41 pm
Lemme have a crack at question 1.

First, let $X = 4444^{4444}$ .

$\because 10^4 > 4444$

$\therefore (10^4)^{4444} > 4444^{4444}$

$\implies 10^{17776} > 4444^{4444}$

...meaning that X must have less than or equal to 17776 digits in it.
Not quite, should have 17776+1 digits in it.

How so?

Title: Re: TT's Maths Thread
Post by: TrueTears on January 05, 2010, 12:03:56 am

How many digits does $10^1$ have?

How many digits does $10^2$ have?

.
.
.

How many digits does $10^{17776}$ have?

Title: Re: TT's Maths Thread
Post by: brightsky on January 05, 2010, 12:04:57 am

Ahh I see. :p

Title: Re: TT's Maths Thread
Post by: TrueTears on January 05, 2010, 01:21:43 pm

Quote

3. Suppose $a,b,c,d$ are integers with $0 \le a \le b \le 99, 0 \le c \le d \le 99$ . For any integer $i$ , let $n_i = 101i+1002^i$ . Show that if $n_a + n_b$ is congruent to $n_c+n_d \pmod {10100}$ then $a=c$ and $b=d$ .

If $a \equiv b \pmod {xy}$ then that implies $a \equiv b \pmod {x}$ and $a \equiv b \pmod {y}$ . I know it works from experimentation but how to prove this?

Does it have anything to do with if $a \equiv b \pmod {n}$ iff $a \equiv b \pmod {p_i^{e_i}}$ for each $i$ in n's PPF?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 05, 2010, 02:08:38 pm

$xy|(a-b) \implies x|(a-b)$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 05, 2010, 05:37:14 pm

Quote from: kamil9876 on January 05, 2010, 02:08:38 pm

$xy|(a-b) \implies x|(a-b)$

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 05, 2010, 08:17:26 pm

symmetry, mate.

ie: xy=yx. or simply, y can play the role of x. x isn't any more special than y. in general if n|(a-b) and k|n then k|(a-b).

let n=xy. k can be either y or x.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 05, 2010, 09:05:02 pm

Quote from: kamil9876 on January 05, 2010, 08:17:26 pm

symmetry, mate.

ie: xy=yx. or simply, y can play the role of x. x isn't any more special than y. in general if n|(a-b) and k|n then k|(a-b).

let n=xy. k can be either y or x.

Ah right so we can split up the congruence into:

$n_a+n_b \equiv n_c+n_d \pmod {101}$ or $n_a+n_b \equiv n_c+n_d \pmod {100}$

Title: Re: TT's Maths Thread
Post by: Over9000 on January 05, 2010, 10:01:21 pm

Quote from: brightsky on January 04, 2010, 10:23:41 pm

Lemme have a crack at question 1.

First, let $X = 4444^{4444}$ .

$\because 10^4 > 4444$

$\therefore (10^4)^{4444} > 4444^{4444}$

$\implies 10^{17776} > 4444^{4444}$

...meaning that X must have less than or equal to 17776 digits in it.

A sum of the digits of any number $n$ must be greater or equal to $9n$ . That is to say that, A (the sum of the digits of X) must have no more than 17776*9 = 159984 digits.

It is to note here that the number with the largest sum of digits below 159984 is 99999, which has a sum of 45. From this, we can see that B (the sum of the digits of A) must have no more than 5*9 = 45 digits.

Similarly, it is to note here that the number with the largest sum of digits below 45 is 39, which has a digit sum of 12. Hence, we can say that C (where C is the sum of the digits of B) must have less than or equal to 12 digits.

Remember now the divisibility test for 9. An integer is divisible by 9 if and only if the sum of its digits (in decimal notation), is divisible by 9. Quick proof for this is seen here, $10 \equiv 1 (\mod 9), 10^2 \equiv 1 (\mod 9), ..., 10^n \equiv 1 (\mod 9)$

That is to say that:

$N = a_n10^n + ... + a_110+ a_0 \equiv a_0 + a_1 + a_2 + ... + a_n \equiv 1 (\mod 9)$ , for any natural number N.

Hence, we can see that:

$X \equiv A \equiv B \equiv C (\mod 9)$

As 4444 = 9 x 493 + 7

$\therefore 4444 \equiv 7 \mod (9)$ [1]

Now lets do some tests with the number 7.

$7 \equiv -2 (\mod 9)$

$7^2 \equiv 4 (\mod 9)$

$7^3 \equiv 1 (\mod 9)$ [2]

From [1],

$4444^{4444} \equiv \7^{4444} (\mod 9)$

$\equiv 7^{3*1481+1} (\mod 9)$

$\equiv 7^{3*1481} . 7 (\mod 9)$

From [2],

$\equiv 7 (\mod 9)$

That means that, $C \equiv 7 (\mod 9)$ . Because C > 12, hence C must be 7.

Edit: I'm still learning how to use latex properly. Why is there a huge gap in the modulos? This is just a simple solution, there must be more elegant alternatives.

http://docs.google.com/viewer?a=v&q=cache:u88H1v0EWqcJ:school.maths.uwa.edu.au/~gregg/Olympiad/1995/congrsolns.pdf+It+is+to+note+here+that+the+number+with+the+largest+sum+of+digits+below+159984+is+99999,+which+has+a+sum+of+45.+From+this,+we+can+see+that+B+%28the+sum+of+the+digits+of+A%29+must+have+no+more+than+5*

page 5

i like gundamz

Title: Re: TT's Maths Thread
Post by: TrueTears on January 06, 2010, 03:47:08 am

Quote

3. Suppose $a,b,c,d$ are integers with $0 \le a \le b \le 99, 0 \le c \le d \le 99$ . For any integer $i$ , let $n_i = 101i+(100)(2^i)$ . Show that if $n_a + n_b$ is congruent to $n_c+n_d \pmod {10100}$ then $a=c$ and $b=d$

If $n_a + n_b \equiv n_c+n_d \pmod {10100}$

Then $n_a+n_b \equiv n_c+n_d \pmod {101}$ and $n_a+n_b \equiv n_c+n_d \pmod {100}$

Thus $101a+100(2^a) +101b+100(2^b) \equiv 101c+100(2^c) +101d+100(2^d) \pmod {101}$

$101(a+b-c-d) +100(2^a)+100(2^b) \equiv 100(2^c) + 100(2^d) \pmod {101}$

$101(a+b-c-d) \equiv 0 \pmod {101}$

$\implies 2^a + 2^b \equiv 2^c + 2^d \pmod {101} \cdots [1]$

$101a+100(2^a) +101b+100(2^b) \equiv 101c+100(2^c) +101d+100(2^d) \pmod {100}$

$101(a+b-c-d) +100(2^a)+100(2^b) \equiv 100(2^c) + 100(2^d) \pmod {100}$

But $100(2^a) \equiv 100(2^b) \equiv 100(2^c) \equiv 100(2^d) \equiv 0 \pmod {100}$

$\implies 101(a+b-c-d) \equiv 0 \pmod {100}$

$\implies a+b \equiv c+d \pmod {100}$

$\implies 2^{a+b} \equiv 2^{c+d} \pmod {101} \cdots [2]$ $\text{iff a+b = c+d}$ since $q \equiv q \pmod {n}$ for any $\text{n and some integer q}$ .

$[1] \times 2^a \implies 2^{2a} + 2^{a+b} \equiv 2^{c+a} + 2^{d+a} \pmod {101}$

$\implies 2^{2a} + 2^{c+d} - 2^{c+a} - 2^{d+a} \equiv 0 \pmod {101}$

$\implies 2^{2a} + 2^{c}2^{d} - 2^{c}2^{a} - 2^{d}2^{a} \equiv 0 \pmod {101}$

$\implies 2^{2a} - 2^{a}\left(2^c+2^d\right) +2^c2^d \equiv 0 \pmod {101}$

$\text{Let} \ x = 2^a \implies x^2 -x\left(2^c+2^d\right) +2^c2^d \equiv 0 \pmod {101}$

$\implies \left(x-2^c\right)\left(x-2^d\right) \equiv 0 \pmod {101}$

$\implies \left(2^a-2^c\right) \equiv 0 \pmod {101} \ \text{or} \ \left(2^a-2^d\right) \equiv 0 \pmod {101}$

$\implies 2^a \equiv 2^c \pmod {101} \ \text{or} \ 2^a \equiv 2^d \pmod {101} \ \text{iff a = c or a = d}$

$\text{If a = c} \ \implies 2^b \equiv 2^d \pmod {101} \ \text{iff b = d}$

$\text{Thus a = c and b=d}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 06, 2010, 12:30:14 pm

I was thinking of another solution earlier:

I will steal two of your formulas as they are quite easy to derive:

Quote

$2^a + 2^b \equiv 2^c + 2^d \pmod {101} \cdots [1] $

$a+b \equiv c+d \pmod {100} \cdots [2]$

The solution relies on the fact that $2^x \equiv 1 \mod 101$ iff $x\equiv 0 \mod 100$ . The verification of this requires some computation, (in fact by fermat's little theorem the smallest solution must be a divisor of 100, therefore it suffices to show that no other divisors, $d$ ,give $2^d\equiv 1$ .

Ie: we have established that the order of 2 mod 101, is 100.

Now we have:
$ 2^a-2^c\equiv 2^d-2^b \mod 101$

$2^c(2^{a-c}-1)\equiv 2^b (2^{d-b}-1) \mod 101$ [3]

Now here is the crux move. Consider the following two statemnts:

a.) $a\equiv c \mod 100$
b.) $b\equiv d \mod 100$

Then either BOTH are true, or BOTH are false. We cannot have one without the other. This is a consequence of [2].
If both are true we are done(since this would imply $a=c$ and $b=d$ ), therefore suppose both are false. (hypothesis i)

then the factors $(2^{a-c}-1)$ and $(2^{d-b}-1)$ are both not congruent to 0 modulo 101, since $a-c \neq 0$ and $d-b \neq 0$ by hypothesis i.

Moreover $a-c=d-b \mod 100 \implies (2^{a-c}-1)\equiv (2^{d-b}-1) \mod 101$ but both factors are not 0 modulo 101 hence they are relatively prime to 101 because 101 is prime. Hence we can cancel them off from both sides in [3] to get:

$2^c \equiv 2^b \mod 101 \implies c \equiv b \mod 100$

which implies $d \equiv a mod 100$ . Hence $c=b$ and $d=a$ in this case as required.

hence the theorem is true assuming $2^x=1 mod 101$ has x=100 as the smallest positive solution, which can be verified by computation.

This doesn't seem like such a good putnam problem, i could solve it with some boring algebra (which i let you do for me) followed by some pretty obvious theorems in first year uni modular arithmetic/group theory, not to mention the ugly computation that awaits in verifying that 100 is indeed the order of 2 modulo 101.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 06, 2010, 12:32:17 pm

Awesome kamil, do you know if my proof is valid or not?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 06, 2010, 12:37:40 pm

Yeah it is. Though I know it also assumes in many instancse, like my proof, that the order of 2 modulo 101 is 100.

Title: Re: TT's Maths Thread
Post by: zzdfa on January 06, 2010, 02:47:22 pm

nice. so the important part was ensuring the numbers were <100 so that $x \equiv y \mod 100 \iff x=y$

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 06, 2010, 02:50:04 pm

yeah exactly, that's the first thing that comes to mind since it reminded me of that trick in the proof of:

a,2a,3a,4a...(p-1)a are all distinct modulo p. (notice that the numbers we have is the modulus less one, just like here).

Title: Re: TT's Maths Thread
Post by: TrueTears on January 06, 2010, 06:33:50 pm

Quote from: kamil9876 on January 06, 2010, 12:37:40 pm

Yeah it is. Though I know it also assumes in many instancse, like my proof, that the order of 2 modulo 101 is 100.

Yeah heaps of iff's....

And yeah, freaking awesome solution kamil.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 08, 2010, 05:10:18 pm

Quote from: TrueTears on January 04, 2010, 08:48:34 pm

2. The number $d_1d_2 \cdots d_9$ has nine (not necessarily distinct) decimal digits. The number $e_1e_2 \cdots e_9$ is such that each of the nine $9$ -digit numbers formed by replacing just one of the digits $d_i$ is $d_1d_2 \cdots d_9$ by the corresponding digit $e_i \ (1 \le i \le 9)$ is divisible by $7$ . The number $f_1f_2 \cdots f_9$ is related to $e_1e_2 \cdots e_9$ in the same way: that is each of the nine numbers formed by replacing one of the $e_i$ by the corresponding $f_i$ is divisible by $7$ . Show that for each $i$ , $d_i - f_i$ is divisible by $7$ . (For example, if $d_1d_2 \cdots d_9 = 199501996$ then $e_6$ may be $2$ or $9$ , since $199502996$ and $199509996$ are multiples of $7$ ).

omfg fukn finally got this question.

Let $D = d_1d_2 \cdots d_9, \ E = e_1e_2 \cdots e_9, \ F = f_1f_2 \cdots f_9$

Now let's take a look at $D=199501996$ .

If we replace $d_6$ by $2$ or $9$ then we have created a number that is divisible by $7$ .

Which means $E = e_1e_2e_3e_4e_5(2)e_7e_8e_9$ or $E = e_1e_2e_3e_4e_5(9)e_7e_8e_9$

This means the other digits of $E$ can be anything but $e_6$ must be either $2$ or $9$ .

Let $D', E'$ be the numbers which are divisible by $7$ .

Using the same example and assume that we are replacing $d_6$ with $2$

$\implies D' = D + 1000 = D + (e_6-d_6)\left(10^{3}\right)$

$\implies D' = D + (e_i-d_i)\left(10^{9-i}\right)$

$\implies D + (e_i-d_i)\left(10^{9-i}\right) \equiv 0 \pmod {7} \cdots [1]$

Likewise we have $E' = E + (f_i-e_i)\left(10^{9-i}\right)$

$\implies E + (f_i-e_i)\left(10^{9-i}\right) \equiv 0 \pmod {7} \cdots [2]$

$[1]+[2] \implies D + (e_i-d_i)\left(10^{9-i}\right)+E + (f_i-e_i)\left(10^{9-i}\right) \equiv 0 \pmod {7}$

$\implies D+E + \left(10^{9-i}\right)(e_i) - \left(10^{9-i}\right)(d_i) + \left(10^{9-i}\right)(f_i)-\left(10^{9-i}\right)(e_i) \equiv 0 \pmod {7}$

$\implies D+E+\left(10^{9-i}\right)\left[f_i-d_i\right] \equiv 0 \pmod {7}$

But $D+E+\left(10^{9-i}\right)\left[f_i-d_i\right] \equiv \left(10^{9-i}\right)\left[f_i-d_i\right] \pmod {7}$

Thus $\left(10^{9-i}\right)\left[f_i-d_i\right] \equiv 0 \pmod {7}$

But $10^{9-i} \not\equiv 0 \pmod {7}$

$\therefore \left[f_i-d_i\right] \equiv 0 \pmod {7}$

$\text{Q.E.D}$

YAY FINISHED ART N CRAFT NOW I CAN FINALLY MOVE ON TO STEWARTS!!!!!

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 01:30:47 am

$\lim_{x \to 7} \left(\frac{\sqrt{x+2}-3}{x-7}\right)$

I've rationalised the numerator and I got $\left(\frac{\sqrt{x+2}-3}{x-7}\right) = \frac{x-7}{3x+\sqrt{x+2}(x-7)-21}$

Now what?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 09, 2010, 01:47:49 am

let x=u+7

$\lim_{x \to 7} \left(\frac{\sqrt{x+2}-3}{x-7}\right)$

$= \lim_{u \to 0} \left(\frac{\sqrt{u+9}-3}{u}\right)$

Now you can rationalise to get something better :)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 01:55:36 am

I see you did some wishful thinking :)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 02:33:21 am

Another one:

$\lim_{x \to 4^-}\left(\frac{\sqrt{x^2+9}-5}{x+4}\right)$

When we are trying to find left/right limits, can we still apply the limit laws?

Ie, in this question, can we use the direct substitution method and just sub in $x=4$ ?

Title: Re: TT's Maths Thread
Post by: humph on January 09, 2010, 03:11:31 am

I'm guessing it's supposed to be $-4$ instead of $4^{-}$ , as otherwise the question's trivial (although it's pretty trivial anyway because it's the same as the previous question).

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 04:14:56 pm

Quote from: humph on January 09, 2010, 03:11:31 am

I'm guessing it's supposed to be $-4$ instead of $4^{-}$ , as otherwise the question's trivial (although it's pretty trivial anyway because it's the same as the previous question).

Thanks humph :)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 04:38:53 pm

Use squeeze theorem to show that $\lim_{x \to 0} \left( \sqrt{x^3+x^2} \sin\left(\frac{\pi}{x}\right)\right)$

How do we know what inequality to pick?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 09, 2010, 04:58:15 pm

$-1 \leq sin\frac{\pi}{x} \leq 1$

Pretty common thing to do.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 05:10:27 pm

Quote from: kamil9876 on January 09, 2010, 04:58:15 pm

$-1 \leq sin\frac{\pi}{x} \leq 1$

Pretty common thing to do.

Oh yeah, so basically all squeeze theorems involving cos and sin are the same lol

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 06:00:28 pm

$\lim_{x \to 0.5^-}\left(\frac{2x-1}{|2x^3-x^2|}\right)$

If $2x^3-x^2 \le 0 \implies x \le \frac{1}{2}$

So $\lim_{x \to 0.5^-}\left(-(2x^3-x^2)\right) = 0$

So we get $\frac{\lim_{x \to 0.5^-} \left(2x-1\right)}{\lim_{x \to 0.5^-} \left(-(2x^3-x^2)\right)}$ which is indeterminant.

So how do I do this?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 09, 2010, 06:03:29 pm

factorise the denominator.
$x^2(2x-1)$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 06:07:37 pm

Thanks but then we have $\lim_{x \to 0.5^-} \frac{1}{-x^2}$

We treat left/right limits as just the actual limit so $\lim_{x \to 0.5} \frac{1}{-x^2} = \lim_{x \to 0.5^-} \frac{1}{-x^2}$

But how do we know $\lim_{x \to 0.5} \frac{1}{-x^2}$ exists (without sketching a graph ofcourse!)

Do I need to know the definition of a limit which is taught later on Stewarts?

So it's sufficient now to just do it graphically?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 09, 2010, 06:09:58 pm

u could do an $\epsilon-\delta$ proof ;D

The idea behind that is that if we compute values $\frac{-1}{x^2}$ for x-values below -0.5, we can make $|\frac{-1}{x^2}-4|$ less than any desired positive value. Ie we can minimize the 'approximation error' as much as we like.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 06:22:37 pm

lol kk how about

$\lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{|x|}\right)$

If $x \ge 0 \implies |x| = x$

$\therefore \lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{x}\right) = 0$

But I sketched the graph of $\frac{1}{x} - \frac{1}{|x|}$ , there is no graph on the right side of 0...

nvm I got it, it was cause the graph was y = 0 lolz

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 09:12:35 pm

$\lim_{x \to 0} \lfloor \cos(x) \rfloor \ \forall \ -\pi \le x \le \pi$

Book says it's undefined.

Shouldn't it be 0?

Since: $\lim_{x \to 0^-} \lfloor \cos(x) \rfloor =\lim_{x \to 0^+} \lfloor \cos(x) \rfloor = 0$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 09:31:41 pm

Quote from: Damo17 on January 09, 2010, 09:31:24 pm

Quote from: TrueTears on January 09, 2010, 09:12:35 pm
$\lim_{x \to 0} \lfloor \cos(x) \rfloor \ \forall \ -\pi \le x \le \pi$

Book says it's undefined.

Shouldn't it be 0?

Since: $\lim_{x \to 0^-} \lfloor \cos(x) \rfloor =\lim_{x \to 0^+} \lfloor \cos(x) \rfloor = 0$

Does the question have absolute signs?

Nah just the floor function.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 09:37:28 pm

Quote from: kamil9876 on January 09, 2010, 09:36:06 pm

book is right only limit from the positive side is 0. Limit from the negative side is -1 since for all x such that $-\pi\leq x< 0$ the $-1\leq cosx<0$ so -1 is the greatest integer less than it. (rememebr, the value at x=0 is irrelevant to the limit).

No it is still 0 from the left.

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 09, 2010, 09:44:12 pm

ahh eyah shit i was thinking of sinx since it would be correct for that one :P. Yeah you are correct they're wrong.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 09:45:57 pm

Quote from: kamil9876 on January 09, 2010, 09:44:12 pm

ahh eyah shit i was thinking of sinx since it would be correct for that one :P. Yeah you are correct they're wrong.

Thanks kamilz

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 10:11:20 pm

$\lim_{x \to 2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\right)$

I've tried rationalising the numerator and denominator... none works, can't use L'hopital's theorem... what to do?

Title: Re: TT's Maths Thread
Post by: brightsky on January 09, 2010, 10:22:36 pm

You can use L'Hospital's rule. I pretty sure.

Let $f(x) = \sqrt{6-x} - 2$ and $g(x) = \sqrt{3-x}-1$ .

Using L'Hospital's rule,

$\frac {f'(x)}{g'(x)} = \sqrt{\frac{x-3}{x-6}}$

$\implies \lim_{x\to2} \sqrt{\frac{x-3}{x-6}}$

Substitute $x = 2$ ,

$\implies \sqrt{\frac{2-3}{2-6}} \implies \sqrt{\frac{-1}{-4}} \implies \sqrt {\frac{1}{4}} \implies \frac {1}{2}$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 10:24:53 pm

Quote from: brightsky on January 09, 2010, 10:22:36 pm

You can use L'Hospital's rule. I pretty sure.

Yeah I know you can use it, but I specifically avoid it whenever I can because it's not fun and simply ruins the question.

I'm after a more elegant solution.

Title: Re: TT's Maths Thread
Post by: brightsky on January 09, 2010, 10:31:53 pm

Oh ok, my bad..hmmm...

Title: Re: TT's Maths Thread
Post by: TrueTears on January 09, 2010, 11:56:20 pm

Quote from: TrueTears on January 09, 2010, 10:11:20 pm

$\lim_{x \to 2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\right)$

Actually I think I got it.

Let $x = u+2$

$\sqrt{6-x} = \sqrt{4-u}$

$\left(4-u\right)^{\frac{1}{2}} = \sqrt{4}\left(1-\frac{u}{4}\right)^{\frac{1}{2}} = \sqrt{4} \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}(1)^{\frac{1}{2}-k}\left(\frac{-u}{4}\right)^k = 2 \left[ \binom{\frac{1}{2}}{0}\left(\frac{-u}{4}\right)^0 + \binom{\frac{1}{2}}{1}\left(\frac{-u}{4}\right)^1 + \binom{\frac{1}{2}}{2}\left(\frac{-u}{4}\right)^2 + \cdots \right] = 2\left[ 1 - \frac{u}{8} - \frac{u^2}{128} + \cdots \right] = 2 - \frac{u}{4} - \frac{u^2}{64} + \cdots$

$\sqrt{3-x} = \sqrt{1-u}$

$\left(1-u\right)^{\frac{1}{2}} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}(1)^{r-k}(-u)^k = \binom{\frac{1}{2}}{0}(-u)^0 + \binom{\frac{1}{2}}{1}(-u)^1 + \binom{\frac{1}{2}}{2}(-u)^2 + \cdots = 1 - \frac{u}{2} - \frac{u^2}{8} + \cdots$

$\implies \lim_{x \to 2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\right) = \lim_{u \to 0}\left(\frac{2 - \frac{u}{4} - \frac{u^2}{64} + \cdots-2}{1 - \frac{u}{2} - \frac{u^2}{8} + \cdots -1}\right) =\lim_{u \to 0}\left(\frac{\left(- \frac{u}{4} - \frac{u^2}{64} + \cdots\right)\left(\frac{1}{u}\right)}{\left(- \frac{u}{2} - \frac{u^2}{8} + \cdots\right)\left(\frac{1}{u}\right)}\right) = \lim_{u \to 0} \left(\frac{-\frac{1}{4} - \frac{u}{64} + \cdots}{-\frac{1}{2} - \frac{u}{8} + \cdots}\right) = \frac{1}{2}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 09, 2010, 11:59:30 pm

Ironically, binomial theorem is more sophisticated/advanced/harder to prove than L'hopital's rule.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 12:00:23 am

fuck l'hopital's rule, that cheap shit

Title: Re: TT's Maths Thread
Post by: /0 on January 10, 2010, 12:00:50 am

LOL
I see calculus isn't liked very much around here :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 10, 2010, 12:01:30 am

Yeah but Bernoulli's discovery of it is much less cheap than the application of elementary algebra.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 12:02:01 am

yeah ok, but still fuck l'hopital's rule.

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 10, 2010, 12:02:32 am

Quote from: /0 on January 10, 2010, 12:00:50 am

LOL
I see calculus isn't liked very much around here :P

Actually Generalized Binomial theorem is pretty calculusy, again more irony.

Title: Re: TT's Maths Thread
Post by: /0 on January 10, 2010, 12:07:13 am

Quote from: kamil9876 on January 10, 2010, 12:02:32 am

Quote from: /0 on January 10, 2010, 12:00:50 am
LOL
I see calculus isn't liked very much around here :P

Actually Generalized Binomial theorem is pretty calculusy, again more irony.

oh my bad

still, nice non-standard solution TT

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 01:54:46 am

Use an $\epsilon - \delta$ proof to show that $\lim_{x \to 3} \left(4x-5\right) = 7$

Just started on $\epsilon - \delta$ , was wondering if someone can show me how to go through this proof.

I get the basic jist but I don't really know how to set it out etc and what exactly do we have to prove...

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 10, 2010, 02:26:58 am

so we want to show that for any given $\epsilon>0$ we can find a $\delta>0$ such that:

$|4x-12|<\epsilon \forall x such that |x-3|<\delta$

We can naturally find this by showing that we seek to find the x values that satisfy:

$|x-3|<\frac{\epsilon}{4}$

But this already suggests we should set $\delta=\frac{\epsilon}{4}$ .

since if $|x-3|<\delta=\frac{\epsilon}{4}$ (1)

Then multiplying everything by 4 yields:

$ |4x-12|<\epsilon$ (2) as required. Thus for all x that satisfy (1), they also satisfy (2)

Title: Re: TT's Maths Thread
Post by: /0 on January 10, 2010, 02:28:57 am

We need to prove that IF $0<|x-3|<\delta$ , THEN $|(4x-5)-7| < \epsilon$

First we can analyse the second expression and work backwards

$|(4x-5)-7| < \epsilon$

Then $|4x-12| < \epsilon$

$4|x-3| < \epsilon$

$|x-3| < \frac{\epsilon}{4}$

Now if we set $\delta = \frac{\epsilon}{4}$ , we can sub into the original proposition:

IF $0<|x-3|< \frac{\epsilon}{4}$ THEN $|(4x-5)-7|< \epsilon$

Since $0<|x-3|<\frac{\epsilon}{4}$ and $|(4x-5)-7|<\epsilon$ are exactly the same statement, the IF-THEN construction is satisfied.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 02:51:36 am

Thanks for the reply guys but I still don't get quite what we need to prove exactly.

From Stewarts:

The precise definition of a limit: $\lim_{x \to a}f(x) = L$ is

"iff for every $\epsilon>0$ there is a number $\delta >0$ such that if $0<|x-a|<\delta$ then $\left | f(x) - L \right | < \epsilon$ "

So what EXACTLY out of that statement do we have to prove?

Actually I think I kinda get it...

So the first thing you have to prove is that there does EXIST a $\delta >0$ for every $\epsilon > 0$ . In this case we worked backwards to find that $\delta = \frac{\epsilon}{4}$

Then we must prove that if $0<|x-a|<\delta$ then $\left | f(x) - L \right | < \epsilon$

Which we showed by subbing $\delta = \frac{\epsilon}{4}$ back into the original inequality $0<|x-a|<\delta$ which does yield the result of $\left | f(x) - L \right | < \epsilon$

So we have Q.E.D

Is that interpretation right?

But the only dilemma is, how do we prove "for every $\epsilon > 0$ "?

We just proved for one fixed $\epsilon$ and proved for every number LESS than it, but there are numbers bigger than $\epsilon$ , so the question is how do we prove for ALL $\epsilon > 0$ ?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 10, 2010, 12:28:45 pm

The trick is that $\epsilon$ was arbitrary. In other words we have answered this:

for $\epsilon=1$ , set $\delta=0.25$ and the condition is satisfied.

for $\epsilon=0.1$ set $\delta =0.025$ and the condition is satisfied.

for $\epsilon=0.01$ set $\delta =0.0025$ and the condition is satisfied.

.
.
.

ie we have answered an uncountably infinite number of questions, with just one sentence.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 03:05:53 pm

Quote from: kamil9876 on January 10, 2010, 12:28:45 pm

The trick is that $\epsilon$ was arbitrary. In other words we have answered this:

for $\epsilon=1$ , set $\delta=0.25$ and the condition is satisfied.

for $\epsilon=0.1$ set $\delta =0.025$ and the condition is satisfied.

for $\epsilon=0.01$ set $\delta =0.0025$ and the condition is satisfied.

.
.
.

ie we have answered an uncountably infinite number of questions, with just one sentence.

So basically whatever $\epsilon$ we sub in first acts like an "upper bound", and thus everything smaller than this upper bound satisfies the condition.

However there are infinitely many upper bounds we can sub in, thus all $\epsilon >$ 0 is satisfied.

Is that right?

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 06:07:44 pm

1. How to use $\epsilon - \delta$ proof to show $\lim_{x \to a} c = c$ ?

2. How to use $\epsilon - \delta$ proof to show $\lim_{x \to 0} |x| = 0$ ?

Title: Re: TT's Maths Thread
Post by: /0 on January 10, 2010, 08:45:30 pm

I'll give these a go, hope they're right

1.

$\lim_{x\to a}c=c$ if the following statement is true:

"If $|x-a|<\delta$ then $|c-c|<\epsilon$ "

$\Leftrightarrow$ "If $|x-a|<\delta$ then $0<\epsilon$ "

Since $\epsilon > 0$ is always true, the limit is c.

2.

"If $|x-0| <\delta$ then $||x|-0| < \epsilon$ "

$\Leftrightarrow$ "If $|x|<\delta$ then $|x|<\epsilon$ "

So we can choose $\delta =\epsilon$

Then the proof goes:

If $|x-0|<\delta$

Pick $\delta = \epsilon$

$\Rightarrow |x-0| <\epsilon$

$\therefore ||x|-0|<\epsilon$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 09:20:50 pm

Thanks I think I'm getting the hang of these now, can check if the following is right? 8-)

Prove $\lim_{x \to 2} x^3 = 8$

We need to prove that for every $\epsilon>0$ there exists a $\delta > 0$ such that if $0<|x-2|< \delta$ then $|x^3-8| < \epsilon$

First we need to find that there does exist a $\delta > 0$ for every $\epsilon > 0$

$|x^3-8| = |(x-2)(x^2+2x+4)| < \epsilon$

$|x-2||x^2+2x+4| < \epsilon$

If we can find some constant $c$ such that $|x^2+2x+4|<c$

Then if we can find that $c$ to satisfy $|x-2||x^2+2x+4|<c|x-2|<\epsilon$

We can have $|x-2| < \frac{\epsilon}{c}$

Now assume $x$ is fairly close to $2$ so we can have $|x-2|<1$

Thus $1<x<3 \implies (1+1)^2+3 < (x+1)^2+3 < (3+1)^2+3 \implies 7 < x^2+2x+4 < 19$

Thus $|x^2+2x+4| < 19$

Which means a suitable $c$ is $19$

So we have $|x-2|<1$ and $|x-2| < \frac{\epsilon}{19}$

Thus $\delta = \text{min}\left(1, \frac{\epsilon}{19}\right)$

So for every $\epsilon > 0$ if $0<|x-2|< \text{min}\left(1, \frac{\epsilon}{19}\right)$ then $|x^3-8| < \epsilon$

Case 1: If $\delta = 1 \implies \frac{\epsilon}{19} > 1 \implies \epsilon > 19$

We have $|x-2| < 1 \implies |x-2||x^2+2x+4| < |x^2+2x+4| < \epsilon$

Case 2: If $\delta = \frac{\epsilon}{19} \implies \frac{\epsilon}{19} < 1 \implies \epsilon < 19$

We have $|x-2|<\frac{\epsilon}{19} \implies |x-2||x^2+2x+4| < \frac{\epsilon}{19}(19) = \epsilon$

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 10, 2010, 11:13:34 pm

case1 is unnecesary (why?)

I once gave birth to a nicer proof for the $x^n$ case that I like more :P It's basically proving continuity of $x^n$ over $\mathbb{R}$ and also over $\mathbb{Q}$ !!

Title: Re: TT's Maths Thread
Post by: TrueTears on January 10, 2010, 11:15:05 pm

Prove that $\lim_{x \to a} \sqrt{x} = \sqrt{a} \ \forall \ a > 0$

This one is a bit tricky because I can't seem to get the $\left | \sqrt{x}-\sqrt{a}\right | < \epsilon$ term to contain a $(x-a)$ term :idiot2:

kamil, shed some ingenuity pls :P

Title: Re: TT's Maths Thread
Post by: /0 on January 10, 2010, 11:31:04 pm

hmm maybe

$(x-a) = (\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})$

$\Rightarrow |x-a|<\epsilon|\sqrt{x}+\sqrt{a}|$

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 10, 2010, 11:31:22 pm

~~if we can prove that $|\sqrt{x}-\sqrt{a}|<|x-a|$ for an arbitrarily small $|x-a|$ then we are done.~~

It is based on the
Hint: let $x=u^2, a=b^2$

edit: actually i think it only works if $a\geq 1$

Though i think letting $x=u^2, a=b^2$ is a good idea, similair idea to /0's.

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 11, 2010, 12:36:30 am

yeah here is a completion of /0's start:

$|x-a|<\epsilon |\sqrt{x}+\sqrt{a}|<\epsilon |\sqrt{4a}+\sqrt{a}|=3\epsilon\sqrt{a}$ provided $x<4a$

So we need: $|x-a|<3\epsilon \sqrt{a}$ and $x<4a$ to be satisfied simultaenously.

Hence we need: $-3\epsilon \sqrt{a}+a<x<3\epsilon \sqrt{a}+a$ (1) and $x<4a$ (2) to be satisfied simultaenously.

(1) and (2) are satisfied simultaenously if:

$-3\epsilon \sqrt{a}+a<x < min\lbrace 3\epsilon \sqrt{a}+a, 4a \rbrace$

$-3\epsilon \sqrt{a} <x-a < min\lbrace 3\epsilon \sqrt{a}+a, 4a \rbrace -a$

which is satisfied if $|x-a|<min\lbrace 3\epsilon \sqrt{a}+a, 4a \rbrace -a$

so we found the $\delta$

edit: some small mistake. Hope there is a neater and general way for all continous and invertible functions.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 01:58:42 am

haha awesome, yeah the crux was finding the minimum, the rest is trivial xD

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 05:33:58 pm

How do I show $H(x) = \cos\left(e^{\sqrt{x}}\right)$ is continuous in its domain and find it's domain.

So I split it up into 3 functions.

$f(x) = \cos(x)$

$g(x) = e^x$

$j(x) = \sqrt{x}$

So $H(x) = f \circ g \circ j$

I know that if the functions that make up the composite function is continuous then so is the composite, but how do I apply that here and prove it and find the domain that it's continuous over?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 11, 2010, 05:41:00 pm

$\mathbb{R}^+$ should be the domain.

This is the maximal domain for which $\sqrt{x}$ is continous(exclude 0 since in that case the limit doesn't exist, (no limit from negative side)). It has range $\mathbb{R}^+$

therefore $e^{\sqrt{x}}$ is continous since $e^u$ is continous if the domain of $u$ is $\mathbb{R^+}$ . But the range of of this function is $(e^0,\infty)=(1,\infty)$ But $cos(w)$ is continous if the domain of $w$ is $(1,\infty)$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 07:01:23 pm

Thanks.

What about this one:

Locate the discontinuities in the following function:

$H(x) = \frac{1}{1+e^{\frac{1}{x}}}$

Let $f(x) = \frac{1}{1+x}, g(x) = e^x, j(x) = \frac{1}{x}$

$H(x) = f \circ g \circ j$

First we need ensure $g \circ j$ is defined.

$\text{ran j} \subset \text{dom g}$

$\mathbb{R}\setminus\{0\} \subset \mathbb{R}$

So $\text{dom} \ g \circ j = \text{dom} \ j = \mathbb{R} \setminus \{0\}$

Now we need to ensure $f \circ \left[g \circ j\right]$ is defined.

So $\text{ran} \ g \circ j \subset \text{dom} \ f$

But how to find $\text{ran} \ g \circ j$ ?

Title: Re: TT's Maths Thread
Post by: /0 on January 11, 2010, 08:13:23 pm

$e^\frac{1}{x}$ where $x \in R \setminus \lbrace 0\rbrace$

For all defined x, $\frac{d}{dx}e^\frac{1}{x} = -\frac{e^\frac{1}{x}}{x^2}$ , so it's decreasing in its continuous sections.

Since there is an asymptote at x = 0,

We can check $\lim_{x \to 0^+}$ and $\lim_{x \to -\infty}$ for the maximums
and $\lim_{x \to 0^-}$ and $\lim_{x \to \infty}$ for the minimums.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 08:51:31 pm

Quote from: /0 on January 11, 2010, 08:13:23 pm

$e^\frac{1}{x}$ where $x \in R \setminus \lbrace 0\rbrace$

For all defined x, $\frac{d}{dx}e^\frac{1}{x} = -\frac{e^\frac{1}{x}}{x^2}$ , so it's decreasing in its continuous sections.

Since there is an asymptote at x = 0,

We can check $\lim_{x \to 0^+}$ and $\lim_{x \to -\infty}$ for the maximums
and $\lim_{x \to 0^-}$ and $\lim_{x \to \infty}$ for the minimums.

Ah thanks, could you also work it out this way? (It's the way Stewarts suggests)

$\frac{1}{x}$ can take a range of values from $\left(-\infty, \infty\right) \setminus \{0\}$

Now let $u = \frac{1}{x}$

We have $y = e^u$

So when you sub in $u = -\infty$ we get a very small value but it will approach 0.

We can't have $y=e^0$

When you sub in $u = \infty$ we have $y = e^u$ is very large, so it will also approach $\infty$ .

Thus the range of $y = e^{\frac{1}{x}}$ is $\left(0, \infty\right) \setminus \{1\}$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 09:24:57 pm

$a$ and $b$ are positive numbers, prove that the equation $\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2} = 0$ has at least $1$ root in the interval $(-1,1)$ .

Since both $\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}$ are continuous then their sums are also continuous.

Let $f(x) = \frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}$

Thus using the intermediate value theorem if we can find a $c \in (-1,1)$ such that $f(c) = 0$ then there must be at least 1 root.

So if $f(-1)<0<f(1)$ then there must be a $c$ .

But how can I evaluate $f(1)$ ? $\frac{b}{x^3+x-2}$ is undefined at $x = 1$ .

Title: Re: TT's Maths Thread
Post by: /0 on January 11, 2010, 09:46:46 pm

You don't need to input -1 and 1, you can try -0.9 and 0.9

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 09:47:16 pm

Quote from: /0 on January 11, 2010, 09:46:46 pm

You don't need to input -1 and 1, you can try -0.9 and 0.9

lol, but those numbers are so ugly :P

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 10:40:02 pm

Show using the squeeze theorem that $\lim_{x \to \infty}\left[\frac{1}{\sqrt{x^2+1}+x}\right] = 0$

How would you do this using the squeeze theorem? =S

I can do it without using the theorem...

$\lim_{x \to \infty}\left[\frac{1}{\sqrt{x^2+1}+x}\right] = \lim_{x \to \infty}\left[\frac{\frac{1}{x}}{\frac{\sqrt{x^2+1}}{x} + 1}\right] = \lim_{x \to \infty}\left[\frac{\frac{1}{x}}{\frac{\sqrt{x^2+1}}{\sqrt{x^2}} + 1}\right] = \lim_{x \to \infty}\left[\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x^2}}+ 1}\right]$

$\implies \left[\frac{\lim_{x \to \infty} \frac{1}{x}}{\sqrt{\lim_{x \to \infty}\left(1+\frac{1}{x^2}\right)}+ \lim_{x \to \infty}1}\right] = \frac{0}{1+1} = 0$

Title: Re: TT's Maths Thread
Post by: /0 on January 11, 2010, 10:50:37 pm

hmm this might work

$0 \leq \frac{1}{\sqrt{x^2+1}+x} \leq \frac{1}{x}$ for all $x>0$

$\lim_{x \to \infty} 0 \leq \lim_{x \to \infty}\frac{1}{\sqrt{x^2+1}+x} \leq \lim_{x \to \infty} \frac{1}{x}$

$0\leq \lim_{x \to \infty} \frac{1}{\sqrt{x^2+1}+x} \leq 0$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 11, 2010, 10:53:33 pm

Quote from: /0 on January 11, 2010, 10:50:37 pm

hmm this might work

$0 \leq \frac{1}{\sqrt{x^2+1}+x} \leq \frac{1}{x}$ for all $x>0$

$\lim_{x \to \infty} 0 \leq \lim_{x \to \infty}\frac{1}{\sqrt{x^2+1}+x} \leq \lim_{x \to \infty} \frac{1}{x}$

$0\leq \lim_{x \to \infty} \frac{1}{\sqrt{x^2+1}+x} \leq 0$

pr0pr0pr0pr0pr0!!!!!

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 04:43:26 am

1. Is there a way to evaluate $\lim_{x \to \infty}\left(e^{-2x}\cos(x)\right)$ without using the squeeze theorem?

2. Also how to evaluate $\lim_{x \to \left(\frac{\pi}{2}\right)^+}e^{\tan(x)}$ ? (How to set it out formally?)

Do you just say since $\tan(x)$ approaches $-\infty$ , then $e^{u}$ where $u = \tan(x)$ will approach $0$ ?

3. Also how do I show $x = -2$ and $1$ are the vertical asymptotes of $y = \frac{2x^2+x-1}{x^2+x-2}$ ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have $y = \frac{\sqrt{2x^2+1}}{3x-5}$

Clearly $x = \frac{5}{3}$ is a vertical asymptote.

But why?

We can show that $\lim_{x \to \frac{5}{3}^+} \frac{\sqrt{2x^2+1}}{3x-5} = \infty$

This is because if we take a value right next to the right of $\frac{5}{3}$ then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $\infty$ .

$\lim_{x \to \frac{5}{3}^-} \frac{\sqrt{2x^2+1}}{3x-5} = -\infty$

This is because if we take a value right next to the left of $\frac{5}{3}$ then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $-\infty$ .

Now how do we apply the same "technique" to the question above?

Thanks :)

Title: Re: TT's Maths Thread
Post by: humph on January 12, 2010, 05:57:33 am

Quote from: TrueTears on January 12, 2010, 04:43:26 am

1. Is there a way to evaluate $\lim_{x \to \infty}\left(e^{-2x}\cos(x)\right)$ without using the squeeze theorem?

Probably, though I don't see why you would, seeing as the squeeze theorem is the natural way to evaluate these kind of limits. When you want to show something tends to zero, it's often easiest to just compare it to some other functions that also go to zero, and that the comparison forces the original function to have the same limit - that's basically the squeeze theorem (or sandwich theorem, or sandwich rule, or whatever you call it).

Quote from: TrueTears on January 12, 2010, 04:43:26 am

2. Also how to evaluate $\lim_{x \to \left(\frac{\pi}{2}\right)^+}e^{\tan(x)}$ ? (How to set it out formally?)

Do you just say since $\tan(x)$ approaches $-\infty$ , then $e^{u}$ where $u = \tan(x)$ will approach $0$ ?

I suppose, though that might not be rigorous enough for some people (in which case you could do an $\varepsilon-\delta$ argument, but surely that'd be overkill). One other trick to remember is to change variables/write a function as a composition of functions.

Quote from: TrueTears on January 12, 2010, 04:43:26 am

3. Also how do I show $x = -2$ and $1$ are the vertical asymptotes of $y = \frac{2x^2+x-1}{x^2+x-2}$ ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have $y = \frac{\sqrt{2x^2+1}}{3x-5}$

Clearly $x = \frac{5}{3}$ is a vertical asymptote.

But why?

We can show that $\lim_{x \to \frac{5}{3}^+} \frac{\sqrt{2x^2+1}}{3x-5} = \infty$

This is because if we take a value right next to the right of $\frac{5}{3}$ then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $\infty$ .

$\lim_{x \to \frac{5}{3}^-} \frac{\sqrt{2x^2+1}}{3x-5} = -\infty$

This is because if we take a value right next to the left of $\frac{5}{3}$ then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $-\infty$ .

Now how do we apply the same "technique" to the question above?

Thanks :)

Well $5/3$ is a vertical asymptote because the denominator tends to zero as $x$ tends to $5/3$ , while the numerator remains bounded away from zero as $x$ tends to $5/3$ . You can again make some intelligent change of variables to show that it comes down to the fact that $f(x) = g(x)/x$ has a vertical asymptote at $x=0$ if $|g(x)| > c > 0$ for all $x$ in some interval containing $x=0$ .

Title: Re: TT's Maths Thread
Post by: moekamo on January 12, 2010, 06:05:07 am

Quote from: TrueTears on January 12, 2010, 04:43:26 am

3. Also how do I show $x = -2$ and $1$ are the vertical asymptotes of $y = \frac{2x^2+x-1}{x^2+x-2}$ ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have $y = \frac{\sqrt{2x^2+1}}{3x-5}$

Clearly $x = \frac{5}{3}$ is a vertical asymptote.

But why?

We can show that $\lim_{x \to \frac{5}{3}^+} \frac{\sqrt{2x^2+1}}{3x-5} = \infty$

This is because if we take a value right next to the right of $\frac{5}{3}$ then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $\infty$ .

$\lim_{x \to \frac{5}{3}^-} \frac{\sqrt{2x^2+1}}{3x-5} = -\infty$

This is because if we take a value right next to the left of $\frac{5}{3}$ then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $-\infty$ .

Now how do we apply the same "technique" to the question above?

Thanks :)

cant you just divide so $y = \frac{2x^2+x-1}{x^2+x-2}= 2-\frac{5}{3(x+2)}+\frac{2}{3(x-1)}$ , then use the same method?

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 11:52:31 am

Quote from: humph on January 12, 2010, 05:57:33 am

Quote from: TrueTears on January 12, 2010, 04:43:26 am
1. Is there a way to evaluate $\lim_{x \to \infty}\left(e^{-2x}\cos(x)\right)$ without using the squeeze theorem?
Probably, though I don't see why you would, seeing as the squeeze theorem is the natural way to evaluate these kind of limits. When you want to show something tends to zero, it's often easiest to just compare it to some other functions that also go to zero, and that the comparison forces the original function to have the same limit - that's basically the squeeze theorem (or sandwich theorem, or sandwich rule, or whatever you call it).

Quote from: TrueTears on January 12, 2010, 04:43:26 am
2. Also how to evaluate $\lim_{x \to \left(\frac{\pi}{2}\right)^+}e^{\tan(x)}$ ? (How to set it out formally?)

Do you just say since $\tan(x)$ approaches $-\infty$ , then $e^{u}$ where $u = \tan(x)$ will approach $0$ ?
I suppose, though that might not be rigorous enough for some people (in which case you could do an $\varepsilon-\delta$ argument, but surely that'd be overkill). One other trick to remember is to change variables/write a function as a composition of functions.

Quote from: TrueTears on January 12, 2010, 04:43:26 am
3. Also how do I show $x = -2$ and $1$ are the vertical asymptotes of $y = \frac{2x^2+x-1}{x^2+x-2}$ ?

This is how Stewarts teaches me it(for another question) but how do you apply the same technique to the question above?

So say we have $y = \frac{\sqrt{2x^2+1}}{3x-5}$

Clearly $x = \frac{5}{3}$ is a vertical asymptote.

But why?

We can show that $\lim_{x \to \frac{5}{3}^+} \frac{\sqrt{2x^2+1}}{3x-5} = \infty$

This is because if we take a value right next to the right of $\frac{5}{3}$ then the denominator will be very small but still positive, however the top will remain positive, thus we have a positive number divide a very small positive number so the result is a very large positive number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $\infty$ .

$\lim_{x \to \frac{5}{3}^-} \frac{\sqrt{2x^2+1}}{3x-5} = -\infty$

This is because if we take a value right next to the left of $\frac{5}{3}$ then the denominator will be very small but it will be negative, thus we have a positive number divide a very small negative number so the result is a very large negative number. Since we can take infinitely many small numbers right next to $\frac{5}{3}$ , the limit is $-\infty$ .

Now how do we apply the same "technique" to the question above?

Thanks :)
Well $5/3$ is a vertical asymptote because the denominator tends to zero as $x$ tends to $5/3$ , while the numerator remains bounded away from zero as $x$ tends to $5/3$ . You can again make some intelligent change of variables to show that it comes down to the fact that $f(x) = g(x)/x$ has a vertical asymptote at $x=0$ if $|g(x)| > c > 0$ for all $x$ in some interval containing $x=0$ .

Okay cool thanks humph, so for Q 3, how would you do the same for the other question?

Title: Re: TT's Maths Thread
Post by: the.watchman on January 12, 2010, 11:57:43 am

Well,

If you solve $x^2+x-2=0$ first, you get solutions $x=-2$ or $1$

$\lim_{x \to -2}\frac{2x^2+x-1}{x^2+x-2} = \infty$ as the denominator approaches 0, making y undefined.
From right, it approaches $-\infty$ (numerator = 5, denominator is negative approaching 0).
From left, it approaches $\infty$ (numerator = 5, denominator is positive approaching 0).

Same with $x=1$ , I guess.
From right, it approaches $\infty$ (numerator = 1, denominator is positive approaching 0).
From left, it approaches $-\infty$ (numerator = 1, denominator is negative approaching 0).

I'm not exactly that good at maths, so I could be really wrong :)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 02:23:59 pm

Quote from: the.watchman on January 12, 2010, 11:57:43 am

Well,

If you solve $x^2+x-2=0$ first, you get solutions $x=-2$ or $1$

$\lim_{x \to -2}\frac{2x^2+x-1}{x^2+x-2} = \infty$ as the denominator approaches 0, making y undefined.
From right, it approaches $-\infty$ (numerator = 5, denominator is negative approaching 0).
From left, it approaches $\infty$ (numerator = 5, denominator is positive approaching 0).

Same with $x=1$ , I guess.
From right, it approaches $\infty$ (numerator = 1, denominator is positive approaching 0).
From left, it approaches $-\infty$ (numerator = 1, denominator is negative approaching 0).

I'm not exactly that good at maths, so I could be really wrong :)

Yeah thanks, that is the way my book goes about 'explaining' it, but I was looking for a more rigorous 'proof' heh

Maybe I'll try a $\epsilon-\delta$ proof, but it seems too hard for this one =S

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 02:39:56 pm

Quote from: TrueTears on January 12, 2010, 04:43:26 am

2. Also how to evaluate $\lim_{x \to \left(\frac{\pi}{2}\right)^+}e^{\tan(x)}$ ? (How to set it out formally?)

Actually how would one use an $\epsilon-\delta$ proof on this?

We have to prove that for every $\epsilon >0$ there exists a number $\delta >0$ such that if $\left(\frac{\pi}{2}\right)<x<\left(\frac{\pi}{2}\right)+\delta$ then $|e^{\tan(x)}-0|<\epsilon$

Right?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 12, 2010, 03:02:56 pm

Quote

Maybe I'll try a $\epsilon-\delta$ proof, but it seems too hard for this one =S

Contrary to first impressions, $\epsilon-\delta$ proofs are actually simply and neat if done for a more general result. When doing them on specific results like specific numbers and specific functions, certain textbooks can make the proofs messy. The same woudl happen for this one. But if you rather focus on proving limit laws first AND then apply them to this question, it is not so messy.

By request:

Quote

$\lim_{x \to \left(\frac{\pi}{2}\right)^+}e^{\tan(x)}$

Suppose we want to find the $\delta$ as mentioned in TT's post.

We will take the following properties for granted:

Now we have that for every given $\epsilon>0$ , there exists an $N$ such that $u<N \implies e^u<\epsilon$ (this is the meaning of $lim_{u \to -\infty} e^u=0$ )

We also have that for every given $N$ (such as say that one given in the previous sentence) there exists a $\delta$ such that $\frac{\pi}{2}<x<\frac{\pi}{2}+\delta \implies tan(x)<N$ . (this is the meaning of $lim_{x \to \frac{\pi}{2}^+} tan(x )=-\infty$ .

Now applying this to the proof:

So in summary, we have that for every $\epsilon>0$ there exist $\delta$ and $N$ that give $\frac{\pi}{2}<x<\frac{\pi}{2}+\delta \implies tan(x)<N \implies e^{\tan (x)}<\epsilon$ (ie $u=tan(x)$ )Which completes the proof.

Note: this is an existence proof, not a 'find an explicit expression for $\delta$ in terms of $\epsilon$ proof' that you are used to.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 03:21:16 pm

Quote from: kamil9876 on January 12, 2010, 03:02:56 pm

Quote
Maybe I'll try a $\epsilon-\delta$ proof, but it seems too hard for this one =S

Contrary to first impressions, $\epsilon-\delta$ proofs are actually simply and neat if done for a more general result. When doing them on specific results like specific numbers and specific functions, certain textbooks can make the proofs messy. The same woudl happen for this one. But if you rather focus on proving limit laws first AND then apply them to this question, it is not so messy.

By request:
Quote
$\lim_{x \to \left(\frac{\pi}{2}\right)^+}e^{\tan(x)}$

Suppose we want to find the $\delta$ as mentioned in TT's post.

We will take the following properties for granted:

Now we have that for every given $\epsilon>0$ , there exists an $N$ such that $u<N \implies e^u<\epsilon$ (this is the meaning of $lim_{u \to -\infty} e^u=0$ )

We also have that for every given $N$ (such as say that one given in the previous sentence) there exists a $\delta$ such that $\frac{\pi}{2}<x<\frac{\pi}{2}+\delta \implies tan(x)<N$ . (this is the meaning of $lim_{x \to \frac{\pi}{2}^+} tan(x )=-\infty$ .

Now applying this to the proof:

So in summary, we have that for every $\epsilon>0$ there exist $\delta$ and $N$ that give $\frac{\pi}{2}<x<\frac{\pi}{2}+\delta \implies tan(x)<N \implies e^{\tan (x)}<\epsilon$ (ie $u=tan(x)$ )Which completes the proof.

Note: this is an existence proof, not a 'find an explicit expression for $\delta$ in terms of $\epsilon$ proof' that you are used to.

omg that is so smart, thanks kamil.

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 12, 2010, 04:11:03 pm

By taking those properties as for granted, you can see that it can be extended to the general case of any continous functions that satisfy blah blah blah...

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 04:52:23 pm

Quote from: kamil9876 on January 12, 2010, 04:11:03 pm

By taking those properties as for granted, you can see that it can be extended to the general case of any continous functions that satisfy blah blah blah...

Yeah, so it's just the conditions for $\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x))$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 04:57:38 pm

Also say we are using the precise definition to prove $\lim_{x \to \infty} x^3 = \infty$

This means that for every $M>0$ then th ere is a corresponding $N$ such that if $x>N$ then $f(x)>M$

Is it just simply $x^3>M \implies x > M^{\frac{1}{3}}$

Thus if we have $N = M^{\frac{1}{3}}$ then the condition is satisfied?

Title: Re: TT's Maths Thread
Post by: Damo17 on January 12, 2010, 05:16:07 pm

Quote from: TrueTears on January 12, 2010, 04:57:38 pm

Also say we are using the precise definition to prove $\lim_{x \to \infty} x^3 = \infty$

This means that for every $M>0$ then th ere is a corresponding $N$ such that if $x>N$ then $f(x)>M$

Is it just simply $x^3>M \implies x > M^{\frac{1}{3}}$

Thus if we have $N = M^{\frac{1}{3}}$ then the condition is satisfied?

Yes, that is correct.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 07:52:35 pm

Find $a$ and $b$ such that $\lim_{x \to 0} \frac{\sqrt{ax+b}-2}{x} = 1$

Need to introduce something new here, but what lol

Thanks :P

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 08:21:43 pm

Quote from: brightsky on January 12, 2010, 08:00:23 pm

Is there even any solutions, TT? :)

yup

needs a bit of wishful thinking for this Q

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 08:25:35 pm

Quote from: brightsky on January 12, 2010, 08:00:23 pm

Is there even any solutions, TT? :)

My reasoning, may be epic fail ><

If $lim_{x\to0} \frac {sqrt{ax + b} -2}{x} = 1$

Then $\sqrt {ax + b} - 2 = x$

$\implies \sqrt { ax + b} = 2 + x$

$\implies ax + b = \pm (2 + x)$

When $ax + b = 2 + x$ ...............[1],

$ax - x = 2 - b$

$x (a - 1) = 2 - b$

$x = \frac {2 - b}{a - 1}$

As $x \to 0$ , substitute x = 0.

$\frac {2-b}{a - 1} = 0$ ..........[2]

$2 - b = 0 \implies b = 2$

Substitute $b = 2$ into [1]:

$ax + 2 = 2 + x$

$ax = x$

$a = 1$

Substituting a = 1, b = 2 into [2]:

Equation is undefined.

Doing the same for $ax + b = - (2 + x)$

We find that: $x = \frac{-b-2}{a+1}$

...but $a = -1$ and $b= -2$ are the only solutions.

Substituting them into when $x \to 0$ , we find that the equation is again, undefined.

Hence no solutions.

I am 90% sure my reasoning is flawed...:)

Don't think you can substitute $\sqrt{ax + b} - 2 = x$

Needs some other substitution.

Since the division law doesn't apply here, we need to somehow find a substitution that gets rid of the $x$ on the bottom...

Title: Re: TT's Maths Thread
Post by: Damo17 on January 12, 2010, 08:35:43 pm

Quote from: TrueTears on January 12, 2010, 07:52:35 pm

Find $a$ and $b$ such that $\lim_{x \to 0} \frac{\sqrt{ax+b}-2}{x} = 1$

Need to introduce something new here, but what lol

Thanks :P

Rationalising the Numerator gives:
$\lim_{x \to 0} \frac{ax+b-4}{x(\sqrt{ax+b}+2)}=1$

Thus this is true only if the numerator tends to 0, so $b=4$ .

So we sub in $b=4$ and get:

$\frac{a}{4}=1$

thus $a=4$ and $b=4$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 08:38:49 pm

Quote from: brightsky on January 12, 2010, 08:28:32 pm

Quote
Since the division law doesn't apply here...

It doesn't? Can you elaborate on this, I'm interested. :)

Ok, goes back to hopeless pondering..

Because you have x on the denominator.

Quote from: Damo17 on January 12, 2010, 08:35:43 pm

Quote from: TrueTears on January 12, 2010, 07:52:35 pm
Find $a$ and $b$ such that $\lim_{x \to 0} \frac{\sqrt{ax+b}-2}{x} = 1$

Need to introduce something new here, but what lol

Thanks :P

Rationalising the Numerator gives:
$\lim_{x \to 0} \frac{ax+b-4}{x(\sqrt{ax+b}+2)}=1$

Thus this is true only if the numerator tends to 0, so $b=4$ .

So we sub in $b=4$ and get:

$\frac{a}{4}=1$

thus $a=4$ and $b=4$ .

But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use $\lim_{x \to 0} \frac{ax+b-4}{x(\sqrt{ax+b}+2)} = \frac{\lim_{x \to 0}ax+b-4}{\lim_{x \to 0}x(\sqrt{ax+b}+2)}$

Title: Re: TT's Maths Thread
Post by: /0 on January 12, 2010, 08:54:50 pm

this might work...

$\lim_{x \to 0}\frac{\sqrt{ax+b}-2}{x}=1$

$u = ax+b$ , $u \to b$ as $x \to 0$

$\lim_{u \to b} \frac{\sqrt{u}-2}{\frac{u-b}{a}} = a\lim_{u \to b}\frac{\sqrt{u}-2}{u-b}$

$=a\lim_{u \to b} \frac{\sqrt{u}-2}{(\sqrt{u}-\sqrt{b})(\sqrt{u}+\sqrt{b})}$

Let's say... $b = 4$

$=a\lim_{u \to 4} \frac{1}{\sqrt{u}+\sqrt{4}}$

$=a\left(\frac{1}{4}\right)$

$a=4$

(taking a cue from one of kamil's earlier solutions ;) )

Title: Re: TT's Maths Thread
Post by: Damo17 on January 12, 2010, 08:59:01 pm

Quote from: TrueTears on January 12, 2010, 08:38:49 pm

But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use $\lim_{x \to 0} \frac{ax+b-4}{x(\sqrt{ax+b}+2)} = \frac{\lim_{x \to 0}ax+b-4}{\lim_{x \to 0}x(\sqrt{ax+b}+2)}$

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form $\lim_{x \to 0} \frac{f(x)}{g(x)}$ where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach $\frac{0}{0}$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 09:11:56 pm

Quote from: Damo17 on January 12, 2010, 08:59:01 pm

Quote from: TrueTears on January 12, 2010, 08:38:49 pm
But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use $\lim_{x \to 0} \frac{ax+b-4}{x(\sqrt{ax+b}+2)} = \frac{\lim_{x \to 0}ax+b-4}{\lim_{x \to 0}x(\sqrt{ax+b}+2)}$

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form $\lim_{x \to 0} \frac{f(x)}{g(x)}$ where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach $\frac{0}{0}$ .

Cool, but division by 0 in this is still not well-defined?

Also I try not to use L'hopital's theorem whenever possible :P It is too cheap ^.^ That's why I tried to go for more wishful methods haha

Title: Re: TT's Maths Thread
Post by: Damo17 on January 12, 2010, 09:18:03 pm

Quote from: TrueTears on January 12, 2010, 09:11:56 pm

Quote from: Damo17 on January 12, 2010, 08:59:01 pm
Quote from: TrueTears on January 12, 2010, 08:38:49 pm
But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use $\lim_{x \to 0} \frac{ax+b-4}{x(\sqrt{ax+b}+2)} = \frac{\lim_{x \to 0}ax+b-4}{\lim_{x \to 0}x(\sqrt{ax+b}+2)}$

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form $\lim_{x \to 0} \frac{f(x)}{g(x)}$ where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach $\frac{0}{0}$ .

Cool, but division by 0 in this is still not well-defined?

Also I try not to use L'hopital's theorem whenever possible :P It is too cheap ^.^ That's why I tried to go for more wishful methods haha

It is not well-defined.

I do understand your hatred of L'Hopital's rule (I also dislike it) but unfortunately that is the only way to do this question as far as I can see.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 09:21:43 pm

Quote from: Damo17 on January 12, 2010, 09:18:03 pm

Quote from: TrueTears on January 12, 2010, 09:11:56 pm
Quote from: Damo17 on January 12, 2010, 08:59:01 pm
Quote from: TrueTears on January 12, 2010, 08:38:49 pm
But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use $\lim_{x \to 0} \frac{ax+b-4}{x(\sqrt{ax+b}+2)} = \frac{\lim_{x \to 0}ax+b-4}{\lim_{x \to 0}x(\sqrt{ax+b}+2)}$

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form $\lim_{x \to 0} \frac{f(x)}{g(x)}$ where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach $\frac{0}{0}$ .

Cool, but division by 0 in this is still not well-defined?

Also I try not to use L'hopital's theorem whenever possible :P It is too cheap ^.^ That's why I tried to go for more wishful methods haha

It is not well-defined.

I do understand your hatred of L'Hopital's rule (I also dislike it) but unfortunately that is the only way to do this question as far as I can see.

Ah okay, awesome thanks, it was a good method nonetheless, but yeah I was looking for more of a substitution method like /0 :P

Title: Re: TT's Maths Thread
Post by: /0 on January 12, 2010, 10:30:50 pm

$\begin{matrix} 151 & 130 & 222 & 122 & 146 \\ 373 & 290 & 310 & 239 & 286 \\ 393 & 385 & 586 & 309 & 328 \\ 274 & 175 & 312 & 141 & 229 \\ 293 & 225 & 292 & 207 & 215 \end{matrix}$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 10:40:11 pm

$\left[\begin{matrix} 3 & 1 & 4 & 1 \\ 5 & 9 & 2 & 6 \\ 5 & 3 & 5 & 8 \\ 9 & 7 & 9 & 3\end{matrix}\right] \times \left[\begin{matrix} 20 & 8 & 1 & 0 \\ 6 & 21 & 11 & 14 \\ 0 & 16 & 21 & 19 \\ 19 & 1 & 25 & 0 \end{matrix}\right]=\left[\begin{matrix} 85 & 110 & 123 & 90 \\ 268 & 267 & 296 & 164 \\ 270 & 191 & 343 & 137 \\ 279 & 366 & 350 & 269\end{matrix}\right]$ .

Title: Re: TT's Maths Thread
Post by: /0 on January 12, 2010, 10:42:15 pm

d'oh, now everyone knows

:P

Title: Re: TT's Maths Thread
Post by: TrueTears on January 12, 2010, 10:44:12 pm

Quote from: /0 on January 12, 2010, 10:42:15 pm

d'oh, now everyone knows

:P

THA.....

Title: Re: TT's Maths Thread
Post by: /0 on January 12, 2010, 10:45:54 pm

Quote from: TrueTears on January 12, 2010, 10:44:12 pm

Quote from: /0 on January 12, 2010, 10:42:15 pm
d'oh, now everyone knows

:P
THA.....

FKN.....

Title: Re: TT's Maths Thread
Post by: TrueTears on January 13, 2010, 02:50:58 am

Hmm $\lim_{x \to 0} \frac{\sin(4x)}{\sin(6x)}$

Title: Re: TT's Maths Thread
Post by: /0 on January 13, 2010, 02:56:57 am

Perhaps think about it a bit longer first, but here's an idea

=\frac{4}{6}lim_{x \to 0} \frac{\sin 4x}{4x} \times \frac{6x}{\sin 6x}

Title: Re: TT's Maths Thread
Post by: TrueTears on January 13, 2010, 03:01:22 am

oooo nice man

so...

$\lim_{x \to 0} \frac{6x}{\sin 6x} = \lim_{x \to 0} \frac{1}{\frac{\sin(6x)}{6x}} = \lim_{u \to 0} \frac{1}{\frac{\sin(u)}{u}} = 1$ ?

Title: Re: TT's Maths Thread
Post by: /0 on January 13, 2010, 03:08:56 am

lol yep

:o

Title: Re: TT's Maths Thread
Post by: humph on January 13, 2010, 03:16:06 am

tpa gki pussg??

Title: Re: TT's Maths Thread
Post by: TrueTears on January 13, 2010, 09:27:07 pm

$\lim_{\theta \to 0} \left( \frac{\sin\theta}{\theta + \tan\theta}\right)$

That $\tan\theta$ is so annoying, how to get around this...?

nvm I got it :tickedoff:

Title: Re: TT's Maths Thread
Post by: TrueTears on January 13, 2010, 11:53:16 pm

When using logarithmic differentiation when do we use modulus?

Eg, $y = (2x+1)^5(x^4-3)^6$

$\log_e(y) = \log_e\left((2x+1)^5\right)+\log_e\left((x^4-3)^6\right)$

Now do we need modulus around the $(2x+1)^5$ since it is negative for some $x$ ? As in:

$\log_e(y) = \log_e\left | (2x+1)^5\right |+\log_e\left((x^4-3)^6\right)$

What about when we simplify and get:

$\log_e(y) = \log_e\left((2x+1)^5\right)+6\log_e\left((x^4-3)\right)$

Originally $(x^4-3)^6$ is positive for all $x$ but now once you take out the $6$ , $x^4-3$ is negative for some $x$ , so do we need to do:

$\log_e(y) = \log_e\left((2x+1)^5\right)+6\log_e\left | (x^4-3)\right |$

So yeah just confused when we need to put modulus...

Title: Re: TT's Maths Thread
Post by: /0 on January 14, 2010, 12:12:35 am

If you take the logs of both sides that's like $f \circ g (x)$ where $f(x) = \log_ex$ , so the domain of $y=(2x+1)^5(x^4-3)^6$ will be restricted to where $y \geq 0$ .

However, you can apply the modulus first:

$|y| = |(2x+1)^5(x^4-3)^6|$

$\log_e|y| = \log_e|(2x+1)^5|+\log_e|(x^4-3)^6|$

The derivative of $\log_e|x|$ is the same as the derivative of $\log_e(x)$ or $\log_e(-x)$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 14, 2010, 12:15:56 am

Quote from: /0 on January 14, 2010, 12:12:35 am

If you take the logs of both sides that's like $f \circ g (x)$ where $f(x) = \log_ex$ , so the domain of $y=(2x+1)^5(x^4-3)^6$ will be restricted to where $y \geq 0$ .

However, you can apply the modulus first:

$|y| = |(2x+1)^5(x^4-3)^6|$

$\log_e|y| = \log_e|(2x+1)^5|+\log_e|(x^4-3)^6|$

The derivative of $\log_e|x|$ is the same as the derivative of $\log_e(x)$ or $\log_e(-x)$

Thanks for that :)

y can't equal 0 right?

Since log(0) is undefined?

So shudn't it be y>0?

Also putting in modulus doesnt affect the derivative anyways so we can just omit it most of the time yeah?

Title: Re: TT's Maths Thread
Post by: /0 on January 14, 2010, 12:25:38 am

Yeah

Title: Re: TT's Maths Thread
Post by: TrueTears on January 14, 2010, 03:58:59 pm

Quote from: /0 on January 14, 2010, 12:25:38 am

Yeah

Oh okay, then what the hell is the point of worrying about if f(x)<0 anyway lulz, it's not like it's gonna affect the derivative haha

Title: Re: TT's Maths Thread
Post by: TrueTears on January 14, 2010, 04:44:39 pm

For the extreme value theorem, what if the function is a horizontal line in the interval $[a,b]$ .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

Title: Re: TT's Maths Thread
Post by: Damo17 on January 14, 2010, 07:08:59 pm

Quote from: TrueTears on January 14, 2010, 04:44:39 pm

For the extreme value theorem, what if the function is a horizontal line in the interval $[a,b]$ .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

The function is bounded and continuous, so the 'extreme value theorem' applies.
So: absolute minimum=constant=absolute maximum.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 14, 2010, 07:10:59 pm

Quote from: Damo17 on January 14, 2010, 07:08:59 pm

Quote from: TrueTears on January 14, 2010, 04:44:39 pm
For the extreme value theorem, what if the function is a horizontal line in the interval $[a,b]$ .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

The function is bounded and continuous, so the 'extreme value theorem' applies.
So: absolute minimum=constant=absolute maximum.

mmm yeah thanks heaps :), but isn't that a bit weird since there is no 'extreme' value :P (if you get what I mean)

Since here absolute min = absolute max, so there shouldn't be any extreme value, it just doesn't make any intuitive sense hahaa

Title: Re: TT's Maths Thread
Post by: Damo17 on January 14, 2010, 07:47:16 pm

Quote from: TrueTears on January 14, 2010, 07:10:59 pm

Quote from: Damo17 on January 14, 2010, 07:08:59 pm
Quote from: TrueTears on January 14, 2010, 04:44:39 pm
For the extreme value theorem, what if the function is a horizontal line in the interval $[a,b]$ .

Does it still attain an absolute minimum/maximum value? (Since all the values are equal...)

The function is bounded and continuous, so the 'extreme value theorem' applies.
So: absolute minimum=constant=absolute maximum.
mmm yeah thanks heaps :), but isn't that a bit weird since there is no 'extreme' value :P (if you get what I mean)

Since here absolute min = absolute max, so there shouldn't be any extreme value, it just doesn't make any intuitive sense hahaa

It does seem weird to me, but it passes the requirements for the 'Extreme Value Theorem (as far as I can see), so that is the conclusion I draw from that. But others that know more than me will give a better answer. :)

Title: Re: TT's Maths Thread
Post by: humph on January 14, 2010, 08:35:11 pm

Extreme value just means that there exists some $x_0$ such that $f(x) \geq x_0$ for all $x \in [a,b]$ (or something along those lines). There's no claim of uniqueness of an extreme value (hence the possible equality in the lesser-than-or-equal-to sign).
Often in complex analysis, you'll be given some complex-differentiable function and told behaviour of where it attains its maxima and minima, and you can pretty much determine the function from that (or at least very useful global behaviour). But that's because complex-differentiable functions behave incredibly nicely in general (whereas real-differentiable functions can be extremely poorly behaved).

Title: Re: TT's Maths Thread
Post by: TrueTears on January 15, 2010, 01:49:11 am

Thanks ^^

Prove $\left | \sin(a) - \sin(b) \right| \le \left | a-b \right|$

I'm thinking of using MVT... but how though?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 15, 2010, 01:51:12 am

you can do it geometrically :)

The red line is |sina-sinb| while |a-b| is that arc length right next to it (obviously longer)

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 15, 2010, 02:01:14 am

But if you must... bore yourself away with calculus:

$(sin(x))'=cosx$

Therefore there exists a $c$ such that:

$ sin(a)-sin(b)=cos(c)(a-b)$

$cos(c)=\frac{sin(a)-sin(b)}{(a-b)}$

but $-1\leq cos(c)\leq 1$

so:

$-1\leq \frac{sin(a)-sin(b)}{a-b}\leq 1$

result now easily follows.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 15, 2010, 02:04:35 am

Nice, well I don't have a choice do I, Stewarts is a calculus book afterall :P

Title: Re: TT's Maths Thread
Post by: TrueTears on January 15, 2010, 11:38:27 pm

Show that $\sqrt{1+x} < 1+ \frac{1}{2}x \ \forall \ x>0$ using MVT.

I know this question seems pretty trivial (it is :P) but how would one do it using MVT?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 15, 2010, 11:53:06 pm

$(\sqrt{1+x})'=\frac{1}{2\sqrt{x+1}}$

Therefore there exists a $c$ such that $0<c<x$ and:

$x* \frac{1}{2\sqrt{1+c}}=\sqrt{1+x}-1$

$\implies \frac{1}{2}>\frac{1}{2\sqrt{1+c}}=(\sqrt{1+x}-1)(\frac{1}{x})$

$\therefore \frac{1}{2}>(\sqrt{1+x}-1)(\frac{1}{x})$

From which the result follows.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 15, 2010, 11:59:41 pm

Thanks that was pr0

Title: Re: TT's Maths Thread
Post by: TrueTears on January 16, 2010, 08:13:56 pm

$\lim_{x \to \infty} x^{\left[\frac{\log_e(2)}{1+\log_e(x)}\right]}$

How to find the limit to this indeterminate case?

Title: Re: TT's Maths Thread
Post by: Damo17 on January 16, 2010, 08:42:55 pm

Quote from: TrueTears on January 16, 2010, 08:13:56 pm

$\lim_{x \to \infty} x^{\left[\frac{\log_e(2)}{1+\log_e(x)}\right]}$

How to find the limit to this indeterminate case?

$y=x^{\frac{\ln(2)}{1+\ln(x)}}$

$ln(y)=\frac{ln(2)ln(x)}{1+ln(x)}$

$\therefore \lim_{x \to \infty} \frac{ln(2)ln(x)}{1+ln(x)} =\lim_{x \to \infty} \frac{ln(2) \frac{1}{x}}{\frac{1}{x}}=\lim_{x \to \infty} ln(2)=ln(2)$

$\therefore \lim_{x \to \infty} x^{\frac{\ln(2)}{1+\ln(x)}}=\lim_{x \to \infty} e^{ln(y)}=e^{ln(2)}=2$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 16, 2010, 08:45:38 pm

Thanks, that is the way Stewarts does it and is pretty formulaic but yeah again I was wondering if there is a method besides L'hospital to bring out the aesthetic nature of this question.

Title: Re: TT's Maths Thread
Post by: humph on January 17, 2010, 03:47:44 am

You don't need l'hopital, just take logs, divide top and bottom by $\log x$ , and use the fact that $\log x \to \infty$ as $x \to \infty$ implies that $1/\log x \to 0$ as $x \to \infty$ .
That is, we use the properties of the exponential and logarithm functions:
$\begin{split}\lim_{x \to \infty} x^{\log 2/(1 + \log x)} & = \lim_{x \to \infty} \exp \left(\log x^{\log 2/(1 + \log x)}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2 \log x}{1 + \log x}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp \left(\lim_{x \to \infty} \frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp(\log 2) \\ & = 2\end{split}$
(you can take the limit inside the exp because the exponential function is continuous)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 17, 2010, 04:02:12 am

Quote from: humph on January 17, 2010, 03:47:44 am

You don't need l'hopital, just take logs, divide top and bottom by $\log x$ , and use the fact that $\log x \to \infty$ as $x \to \infty$ implies that $1/\log x \to 0$ as $x \to \infty$ .
That is, we use the properties of the exponential and logarithm functions:
$\begin{split}\lim_{x \to \infty} x^{\log 2/(1 + \log x)} & = \lim_{x \to \infty} \exp \left(\log x^{\log 2/(1 + \log x)}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2 \log x}{1 + \log x}\right) \\ & = \lim_{x \to \infty} \exp \left(\frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp \left(\lim_{x \to \infty} \frac{\log 2}{\frac{1}{\log x} + 1}\right) \\ & = \exp(\log 2) \\ & = 2\end{split}$
(you can take the limit inside the exp because the exponential function is continuous)

Exactly what I was looking for!!!!

Thanks humph!!

Title: Re: TT's Maths Thread
Post by: TrueTears on January 17, 2010, 05:37:22 pm

$\int_0^3 \left(x^3-6x\right)dx$ using Riemann Sum.

My attempt, not sure if it's right. I'm not too worried about the answer but rather the setting out.

$\int_0^3 \left(x^3-6x\right)dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x$

$\Delta x = \frac{3-0}{n} \implies x_i = i\left(\frac{3}{n}\right)$

$\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x = \lim_{n \to \infty} \sum_{i=1}^n f\left(i\left(\frac{3}{n}\right)\right)\left(\frac{3}{n}\right)$

$\implies \lim_{n \to \infty} \sum_{i=1}^n \left[\left(\frac{3i}{n}\right)^3-6\left(\frac{3i}{n}\right)\right]\left(\frac{3}{n}\right)$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \sum_{i=1}^n \left[\left(\frac{27i^3}{n^3}\right)-\left(\frac{18i}{n}\right)\right]$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \left[\sum_{i=1}^n \frac{27}{n^3}i^3 - \sum_{i=1}^n \frac{18}{n}i\right]$

First remember the result $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Now to find $\sum_{k=1}^n k^3$ we need to find some polynomial containing $k^3$ .

To telescope: let $u_{k+1} - u_k = \mbox{some polynomial with} \ k^3$

Now let $u_k = k^4$

Thus $u_{k+1} - u_k = (k+1)^4 - k^4$

Now $\sum_{k=1}^n u_{k+1} - u_k = u_2 - u_1 + u_3 - u_2 + ... + u_{n+1} - u_n = u_{n+1} - u_1 = (n+1)^4 - 1$

But $\sum_{k=1}^n (k+1)^4 - k^4 = \sum_{k=1}^n (4k^3 + 6k^2 + 4k+1)$

$\implies \sum_{k=1}^n 4k^3 + 6k^2 + 4k+1 = (n+1)^4 - 1$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - 6\sum_{k=1}^n k^2 - \sum_{k=1}^n (4k+1)$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - n(n+1)(2n+1) - \frac{n}{2}(5+4n+1)$

$\therefore \sum_{k=1}^n k^3 = \frac{n^2}{4}\left(n^2+2n+1\right)$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \left[\frac{27}{n^3}\left(\frac{n^2}{4}\left(n^2+2n+1\right)\right) - \frac{18}{n}\left(\frac{n(n+1)}{2}\right)\right]$

$\implies \lim_{n \to \infty} \left(\frac{3}{n}\right) \left[\frac{27}{n}\left(\frac{1}{4}\left(n+1\right)^2\right) - 18\left(\frac{(n+1)}{2}\right)\right]$

$\implies \lim_{n \to \infty} \frac{3(n+1)}{n}\left(\frac{27}{4}-9+\frac{27}{4n}\right)$

$\implies \lim_{n \to \infty} \left(3+\frac{3}{n}\right)\left(\frac{27}{4n}-\frac{9}{4}\right) = -\frac{27}{4}$

Title: Re: TT's Maths Thread
Post by: brightsky on January 17, 2010, 06:07:49 pm

$\int_0^3 (x^3 - 6x) dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*) \triangle x$

$\triangle x = \frac{3 - 0}{n} = \frac{3}{n}$

$x_i^*= 0 + (\triangle x) i = \frac{3i}{n}$

$f(x_i^*) = (\frac{3i}{n})^3 - 6(\frac{3i}{n}) = \frac{27i^3}{n^3} - \frac{18i}{n}$

So $\lim_{n\to\infty} \sum_{i=1}^n \left(\frac{27i^3}{n^3} - \frac{18i}{n} \right) \left(\frac{3}{n}\right)$

$\lim_{n\to\infty} \left(\frac{3}{n}\right) \left[\sum_{i=1}^n \frac{27i^3}{n^3} - \sum_{i=1}^n \frac{18i}{n} \right]$

$\lim_{n\to\infty} \left(\frac{3}{n}\right) \left[\frac{27}{n^3} \sum_{i=1}^n i^3 - \frac{18}{n} \sum_{i=1}^n i \right]$

$\lim_{n\to\infty} \left(\frac{3}{n}\right) \left[\frac{27}{n^3} \left(\frac{1}{4}n^2(n+1)^2\right) - \frac{18}{n} \left(\frac{1}{2} n(n+1)\right)\right]$

$\lim_{n\to\infty} \left(\frac{81}{n^4}\left(\frac{1}{4}n^2(n+1)^2\right) - \frac{54}{n^2}\left(\frac{1}{2}n(n+1)\right)\right)$

$\frac{81}{4} - \frac{108}{4}$

$-\frac{27}{4}$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 17, 2010, 06:12:12 pm

I just found out both Chopin and Riemann died at the age of 39 and of the same cause of death: tuberculosis

fukn tuberculosis killed my fave composer and a great mathematician.

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 17, 2010, 07:10:50 pm

I remember once proving for the general case of $\int x^n dx$ . Motivation for this came when i was once playing with $\sum_{i=0}^k i^n$ and realised it's an n+1 degree polynomial of k with where $k^{n+1}$ coefficient is $\frac{1}{n+1}$

Title: Re: TT's Maths Thread
Post by: brightsky on January 17, 2010, 07:14:04 pm

Quote from: TrueTears on January 17, 2010, 06:12:12 pm

I just found out both Chopin and Riemann died at the age of 39 and of the same cause of death: tuberculosis

fukn tuberculosis killed my fave composer and a great mathematician.

Riemann's your favourite mathematician? Hehe...Euler FTW!! But in the realms of music, Chopin wins hands down.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 17, 2010, 07:17:43 pm

Nah he's not. I don't think I have a fave mathematician yet...

But Chopin <3 fave composer and most talented composer in my eyes.

Title: Re: TT's Maths Thread
Post by: brightsky on January 17, 2010, 07:54:05 pm

Quote from: TrueTears on January 17, 2010, 05:37:22 pm

First remember the result $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

Now to find $\sum_{k=1}^n k^3$ we need to find some polynomial containing $k^3$ .

To telescope: let $u_{k+1} - u_k = \mbox{some polynomial with} \ k^3$

Now let $u_k = k^4$

Thus $u_{k+1} - u_k = (k+1)^4 - k^4$

Now $\sum_{k=1}^n u_{k+1} - u_k = u_2 - u_1 + u_3 - u_2 + ... + u_{n+1} - u_n = u_{n+1} - u_1 = (n+1)^4 - 1$

But $\sum_{k=1}^n (k+1)^4 - k^4 = \sum_{k=1}^n (4k^3 + 6k^2 + 4k+1)$

$\implies \sum_{k=1}^n 4k^3 + 6k^2 + 4k+1 = (n+1)^4 - 1$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - 6\sum_{k=1}^n k^2 - \sum_{k=1}^n (4k+1)$

$\implies 4\sum_{k=1}^n k^3 = (n+1)^4 - 1 - n(n+1)(2n+1) - \frac{n}{2}(5+4n+1)$

$\therefore \sum_{k=1}^n k^3 = \frac{n^2}{4}\left(n^2+2n+1\right)$

Or you can directly use Faulhaber's formula for power sums. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 17, 2010, 07:57:11 pm

mmm bernoulli would work well for sums of powers but telescoping would work better in general and requires more wishful thinking xD

zeitz style ftw

Title: Re: TT's Maths Thread
Post by: brightsky on January 17, 2010, 08:01:29 pm

Is there a way of using telescoping to find the general $\sum_{k=1}^n k^n$ ?

Title: Re: TT's Maths Thread
Post by: TrueTears on January 17, 2010, 08:08:49 pm

http://vcenotes.com/forum/index.php/topic,19896.msg225557.html#msg225557

I'm haven't thought about that but I'm pretty sure a generalised version would involve telescoping at some point.

kamil has probably played around with it, ask him :)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 19, 2010, 03:10:41 am

1. The base of a solid S is a circular disk with radius r. Parallel cross-section perpendicular to the base are isosceles triangles with height h and unequal side in the base.

Set up an integral for the volume of S.

What the heck does the shape even look like?

2. (http://img514.imageshack.us/img514/854/torus.jpg)

where do i place the axis.... too much art in this one...

3. A barrel with height h and maximum radius R is constructed by rotating about the x axis the parabola $y = R - cx^2$ , $\frac{-h}{2} \le x \le \frac{h}{2}$ where $c > 0$ . Show that the radius of each of the barrel is $r = R- \frac{ch^2}{4}$ .

Eh what is this question asking? " radius of each of the barrel" ??? there only is 1 barrel? wdf

Title: Re: TT's Maths Thread
Post by: /0 on January 19, 2010, 03:37:57 am

My paint skills are no match for multivariable calculus... but imagine having a stack of sold, thin isosceles triangles lying around, each with the same height but different base lengths. If you place them standing up on a glass table so that one is in front of the other, and then look from underneath the table, you should see a circle. The circle is formed by their bases.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 19, 2010, 03:42:00 am

Quote from: /0 on January 19, 2010, 03:37:57 am

My paint skills are no match for multivariable calculus... but imagine having a stack of sold, thin isosceles triangles lying around, each with the same height but different base lengths. If you place them standing up on a glass table so that one is in front of the other, and then look from underneath the table, you should see a circle. The circle is formed by their bases.

Wait... but what do you do at the 'vertex' of the circle, you can't form an isosceles triangle...

I still have no idea what the shape looks like, all of this volumes and cylindrical shit is making me so confused.

Asif this is maths, it's like a fucking art class and i aint no piccaso

Title: Re: TT's Maths Thread
Post by: /0 on January 19, 2010, 03:47:33 am

Quote from: TrueTears on January 19, 2010, 03:42:00 am

Quote from: /0 on January 19, 2010, 03:37:57 am
My paint skills are no match for multivariable calculus... but imagine having a stack of sold, thin isosceles triangles lying around, each with the same height but different base lengths. If you place them standing up on a glass table so that one is in front of the other, and then look from underneath the table, you should see a circle. The circle is formed by their bases.
Wait... but what do you do at the 'vertex' of the circle, you can't form an isosceles triangle...

I still have no idea what the shape looks like, all of this volumes and cylindrical shit is making me so confused.

Asif this is maths, it's like a fucking art class and i aint no piccaso

lol picasso sucks at art

Anyway at the 'vertices' of the circle, you'd get a degenerate isosceles triangle (one with no area).

For 2) the easiest way is using a volume of revolution, but it also works in cylindrical coordinates, if you've reached that chapter yet.

For 3) the 'solid' is made up of an infinite number of barrels of different radiuses. The largest barrel has radius R.

fuck this im going to sleeep

Title: Re: TT's Maths Thread
Post by: TrueTears on January 19, 2010, 03:53:21 am

wdf what kinda shape is that i cant even fucking visualise it, i keep thinking of a half hemisphere but the height must be the same WDF

maybe i need to get a degree in painting so i can paint some shapes to do these shitty questions

this is way too physics-esque, all of this cylindrical shell and volumes shit. where's the friggin maths

Title: Re: TT's Maths Thread
Post by: moekamo on January 19, 2010, 04:43:42 am

Quote from: TrueTears on January 19, 2010, 03:10:41 am

2. (http://img514.imageshack.us/img514/854/torus.jpg)

where do i place the axis.... too much art in this one...

for this one, flip the donut on the side and look side. This is the same as the pic of the circle attached rotated around the x axis

Make the circle equation: $x^2+(y-R)^2=r^2$ you'll also have to split it into top and bottom half of circles when you rearrange for $y^2=r^2-x^2\pm 2R\sqrt{r^2-x^2}+R^2$

Does the volume end up as $2\pi^2r^2R$ ?

Title: Re: TT's Maths Thread
Post by: TrueTears on January 19, 2010, 04:41:42 pm

Yeap it is, awesome explanation moekamo :)

I actually did it another way earlier today, used cylindrical shells, I thought that was easier to visualise than trying to set up an axis =S

Title: Re: TT's Maths Thread
Post by: Ahmad on January 19, 2010, 04:53:08 pm

What happens when R = 0?

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 19, 2010, 05:02:12 pm

Quote from: TrueTears on January 17, 2010, 08:08:49 pm

http://vcenotes.com/forum/index.php/topic,19896.msg225557.html#msg225557

I'm haven't thought about that but I'm pretty sure a generalised version would involve telescoping at some point.

kamil has probably played around with it, ask him :)

Well i don't know about a quick general way, but it is an alright exercise to prove that it is always a polynomial of degree (n+1). And as an extra prove that the $x^{n+1}$ coefficient is $\frac{1}{n+1}$ (this allows u to find that integral).

By proving that, you have validated the method of just subbing in values into the polynomial and working out the linear equation system to find the coefficients.

Anyway a method I used to play around with this is:

let $f(k)=\sum_{x=1}^k x^n$

$f(k+1)-f(k)=(k+1)^n$ This leads to the guess of a polynomial.

edit: oh yes, how could i forget. This can also be done combinatorily!

Title: Re: TT's Maths Thread
Post by: TrueTears on January 19, 2010, 06:23:04 pm

Quote from: kamil9876 on January 19, 2010, 05:02:12 pm

Quote from: TrueTears on January 17, 2010, 08:08:49 pm
Anyway a method I used to play around with this is:

let $f(k)=\sum{x=1}^k x^n$

$f(k+1)-f(k)=(k+1)^n$ This leads to the guess of a polynomial.

edit: oh yes, how could i forget. This can also be done combinatorily!

That's just telescoping :P

Title: Re: TT's Maths Thread
Post by: TrueTears on January 20, 2010, 01:05:57 am

(http://img31.imageshack.us/img31/3508/wateryd.jpg)

thx

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 20, 2010, 03:30:53 pm

(http://vcenotes.com/forum/index.php?action=dlattach;topic=19896.0;attach=4652;image)

We want the volume of the shape that is the intersection of the sphere and cone, ie how much of the sphere enters the cone is the ammount of water spilt out.

This volume is generated by rotating the area of the intersection of circle and area around y axis:

$\pi \int_{c-r}^h f(y)^2 dy$

so really the only difficult part is finding the limits in terms of the variables given. We have that $c=\frac{r}{sin\theta}$ by geometry so the rest should be fine, just integrating.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 20, 2010, 09:39:11 pm

Oh that's smart kamilz.

What about this one, just a bit stuck =X

Prove the reduction formula:

$\int \sec^n(x) dx = \frac{\tan(x)\sec^{n-2}(x)}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}(x)dx \ \text{and} \ n \neq 1$

Title: Re: TT's Maths Thread
Post by: brightsky on January 20, 2010, 10:10:22 pm

"Split" the integral into:

$\int\sec^n (x) dx = \int \sec^2 x \sec^{n-2} x dx$

$\because \int \sec^2x dx = \tan(x) + c$

$\therefore \int \sec^n(x) dx = \int \sec^{n-2} x \frac{d}{dx} (\tan(x)) dx$

$= \tan(x)\sec^{n-2}(x) - \int \tan (x) \frac{d}{dx}(\sec^{n-2}(x) dx$

$= \tan (x) \sec^{n-2}(x) - \int \tan(x) [(n-2)\sec^{n-3} x (\sec(x) \tan(x))] dx$

$\because \tan^2x = \sec^2(x) - 1$

$\therefore \implies \tan(x) \sec^{n-2}(x) - (n-2) \int \sec^{n-2} x (\sec^2(x) -1) dx$

$= \tan(x) \sec^{n-2}(x) - (n-2) \int \sec^n(x) dx + (n-2) \int \sec ^{n-2}x dx$

Add $(n-2) \int \sec^n(x) dx$ to both sides so the LHS becomes $(n-2+1) \int \sec^n(x) dx = (n-1)\int \sec^n(x) dx$ .

Divide both sides by $(n-1)$ and you're done.

Title: Re: TT's Maths Thread
Post by: humph on January 20, 2010, 10:24:03 pm

$\sec^2 x = 1 + \tan^2 x$ , then make the substitution $u = \sec x$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on January 21, 2010, 12:04:12 am

For trigonometric substitutions say we got :

$\sqrt{x^2-a^2}$ why do we make the substitution $x = a\sec(\theta) \ \forall \ 0 \le \theta < \frac{\pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$

I know it "works" but what is the reason behind it?

I'm not after a proof, I'm just after a justification or an informal demonstration.

Title: Re: TT's Maths Thread
Post by: brightsky on January 21, 2010, 12:21:41 am

By the Pythagoras Theorem,

$\sqrt{x^2-a^2}$ suggests a triangle with a hypotenuse of $x$ and base length of $a$ .

This implies $\sec \theta = \frac{x}{a} \implies x = a \sec \theta$ .

$\sec \theta$ is positive in quadrants 4 and 1.

Hence why $0 \le \theta < \frac{\pi}{2}$ or $\pi \le \theta < \frac{3\pi}{2}$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 21, 2010, 09:58:19 am

Yeah, it's pretty trivial to get sec, however that is not what I am after. You could have also done $y = \sqrt{x^2-a^2}$ so $y^2 +a^2 = x^2$ so $a^2 = x^2-y^2$ so $1 = \frac{x^2}{a^2} - \frac{y^2}{a^2}$ which is a hyperbola so the parametric equation $x = a\sec(\theta)$ would work. However that still doesn't explain why you make the substitution when you are integrating. Why do we do this kind of inverse substitution? I understand the restriction to make it 1-to-1 function or else the inverse would not be defined when changing the variable $\theta$ back into $x$ however the crux step is the first substitution, why does it work? (Not necessarily where it comes from)

What I'm looking for is analogous to say integration by parts which is the "reverse" of the product rule.

Oh wait nvm I got it, it's because by picking $x = a\sec(\theta)$ we can simplify the radical $x^2-a^2$ . It actually works quite similar to a parametric equation xD

Title: Re: TT's Maths Thread
Post by: TrueTears on January 22, 2010, 10:13:03 pm

Hmm I'm stuck at the end for this question...

$\int \tan^{-1}\left(\sqrt{x}\right) dx$

Let $u = \tan^{-1}\left(\sqrt{x}\right)$ and $dv = dx$

$\implies \int \tan^{-1}\left(\sqrt{x}\right) dx = x \tan^{-1}\left(\sqrt{x}\right) - \int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$

So... how does one evaluate $\int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$ ?

Title: Re: TT's Maths Thread
Post by: Damo17 on January 22, 2010, 11:31:33 pm

Quote from: TrueTears on January 22, 2010, 10:13:03 pm

Hmm I'm stuck at the end for this question...

$\int \tan^{-1}\left(\sqrt{x}\right) dx$

Let $u = \tan^{-1}\left(\sqrt{x}\right)$ and $dv = dx$

$\implies \int \tan^{-1}\left(\sqrt{x}\right) dx = x \tan^{-1}\left(\sqrt{x}\right) - \int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$

So... how does one evaluate $\int \frac{x}{2\sqrt{x}\left(x+1\right)} dx$ ?

Well.. I recognised that $- \frac{x}{2\sqrt{x}(x+1)}= \frac{1}{2\sqrt{x}(x+1)} \ - \ \frac{1}{2\sqrt{x}}$ .

So take the integral of that and put it all together to get:

$\int \tan^{-1}\left(\sqrt{x}\right) dx= x \tan^{-1}\left(\sqrt{x}\right) + \tan^{-1}\left(\sqrt{x}\right)- \sqrt{x} \ + \ C =(x+1)\tan^{-1}\left(\sqrt{x}\right)- \sqrt{x} \ + \ C$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 22, 2010, 11:39:36 pm

Thanks Damo! That was awesome!

Yeah u substitution could have also worked by letting $u = \sqrt{x}$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 25, 2010, 08:17:54 pm

$\int \frac{1}{1-e^{2x}}dx$

hmmmz

Title: Re: TT's Maths Thread
Post by: brightsky on January 25, 2010, 08:29:23 pm

Quote from: TrueTears on January 25, 2010, 08:17:54 pm

$\int \frac{1}{1-e^{2x}}dx$

hmmmz

Substitute $u = 2x, du = 2dx$ :

$\frac{1}{2} \int \frac{1}{1-e^u} du$

Substitute $v = e^u, dv = e^u du$

$\frac{1}{2} \int \frac{1}{v} \cdot \frac{1}{1-v} dv$

$\frac{1}{2} \int \left(\frac{1}{v} - \frac{1}{v-1}\right) dv$

Substitute $w = s - 1, dw = 1 dv$

$\frac{1}{2} \int \left(\frac{1}{v} - \frac{1}{w} \right) dw$

Integrate then substitute back in.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 25, 2010, 08:38:44 pm

Yeap, good method, I just did $u = e^x$ and then partial fractions.

Title: Re: TT's Maths Thread
Post by: Ahmad on January 25, 2010, 09:16:51 pm

Alternative approach re-using this idea.

Let $I = \int \frac{1}{1 - e^{2x}} dx = \int \frac{e^{-x}}{e^{-x} - e^{x}} dx$

and $J = \int \frac{e^x}{e^{-x} - e^{x}} dx$

$I + J = -\log (e^{-x}-e^x)$

$I - J = x$

$I = \frac{1}{2} (x - \log (e^{-x}-e^x))$

(upto constants)

:)

Title: Re: TT's Maths Thread
Post by: brightsky on January 25, 2010, 09:24:40 pm

Quote from: Ahmad on January 25, 2010, 09:16:51 pm

Alternative approach re-using this idea.

Let $I = \int \frac{1}{1 - e^{2x}} dx = \int \frac{e^{-x}}{e^{-x} - e^{x}} dx$

and $J = \int \frac{e^x}{e^{-x} - e^{x}} dx$

$I + J = -\log (e^{-x}-e^x)$

$I - J = x$

$I = \frac{1}{2} (x - \log (e^{-x}-e^x))$

(upto constants)

:)

Oohh that is smart...

Title: Re: TT's Maths Thread
Post by: TrueTears on January 26, 2010, 12:23:22 am

$\int \frac{1+\sin(x)}{1-\sin(x)} dx$

I've played around for a while, tried to multiply by $\frac{1 \pm \sin(x)}{1 \pm \sin(x)}$ ...

Title: Re: TT's Maths Thread
Post by: humph on January 26, 2010, 07:10:33 am

Haven't had a stab at it, but think about double/half-angle formulae, or try making the classic $t=tan(x/2)$ substitution (which works with most integrals of rational functions of trigonometric functions).

Title: Re: TT's Maths Thread
Post by: TrueTears on January 26, 2010, 09:19:36 am

Oh yeah I got it, thanks :)

Title: Re: TT's Maths Thread
Post by: Ahmad on January 26, 2010, 01:12:06 pm

My first intuition was to either multiply and divide by $1-\sin(x)$ or $1+\sin(x)$ , with the idea being to exploit $\sin^2(x) + \cos^2(x) = 1$ . I wrote down both expressions then discarded the first because it wasn't clear how to proceed, the second gives:

$\int \frac{(1+\sin(x))^2}{\cos^2(x)} dx$

$\int sec^2(x) + 2\underbrace{\frac{\sin(x)}{\cos^2(x)}}_{\frac{d}{dx} \frac{1}{\cos(x)}} + \underbrace{\tan^2{x}}_{\sec^2(x) - 1} dx$

$= 2 \tan(x) + 2 \sec x - x + C$ . :)

In the general case of rational functions of trigonometric functions as humph mentioned the idea of the t substitution is the important one (basically converts it into a rational function), downside being it can get messy fast.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 26, 2010, 01:19:24 pm

$\int \frac{\sqrt{1+x^2}}{x^2} dx$

I did:

Let $x = \tan(\theta)$ where $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$

$\sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sqrt{\sec^2(\theta)} = |\sec(\theta)| = \sec(\theta)$

$dx = \sec^2(\theta) d\theta$

$\int \frac{\sec(\theta)}{\tan^2(\theta)}\sec^2(\theta) d\theta = \int \frac{\sec^3(\theta)}{\tan^2(\theta)} d\theta$

Hmm now what?

Title: Re: TT's Maths Thread
Post by: brightsky on January 26, 2010, 01:33:44 pm

$\int \frac{\sec(\theta)}{\tan^2\(\theta)} \sec^2(\theta) d\theta = \int \frac{\sec^2(\theta)}{\tan^2\(\theta)} \sec(\theta) d\theta= \int \csc^2(\theta) \sec(\theta) d\theta = \int (\cot^2(\theta) + 1) \sec(\theta) d\theta$ etc....

Title: Re: TT's Maths Thread
Post by: TrueTears on January 26, 2010, 10:21:09 pm

$\int \frac{\sqrt{x}}{1+x^3} dx$

Let $u = \sqrt{x} \implies 2udu = dx$

$u^2 = x \implies u^6+1 = x^3+1$

$\therefore 2 \int \frac{u^2}{u^6+1} du$

I don't think this is the right way... to partial fraction that would take ages...

Title: Re: TT's Maths Thread
Post by: Damo17 on January 26, 2010, 10:48:28 pm

Quote from: TrueTears on January 26, 2010, 10:21:09 pm

$\int \frac{\sqrt{x}}{1+x^3} dx$

Let $u = \sqrt{x} \implies 2udu = dx$

$u^2 = x \implies u^6+1 = x^3+1$

$\therefore 2 \int \frac{u^2}{u^6+1} du$

I don't think this is the right way... to partial fraction that would take ages...

Let $u= \sqrt{x^3}$ , a far easier way. :)

Title: Re: TT's Maths Thread
Post by: Over9000 on January 27, 2010, 03:06:31 pm

THA FUKN PUSSAYY

$\int_0^3 \frac{dx}{x-1}$

Title: Re: TT's Maths Thread
Post by: TrueTears on January 27, 2010, 10:45:28 pm

Find the arc length function for the curve $y = x^2 - \frac{1}{8}\log_e(x)$ taking $(1,1)$ as the starting point.

So $s(x) = \int_1^x \sqrt{1+\left[f'(t)\right]^2} dt$

Let $f(t) = t^2 - \frac{1}{8}\log_e(t)$

$1+\left[f'(t)\right]^2 = \left(2t + \frac{1}{8t}\right)^2$

$\implies \sqrt{1+\left[f'(t)\right]^2} = \sqrt{\left(2t + \frac{1}{8t}\right)^2} = \left | 2t + \frac{1}{8t} \right |$

So then we need $s(x) = \int_1^x \left | 2t + \frac{1}{8t} \right | dt$

But my book only has: $s(x) = \int_1^x 2t + \frac{1}{8t} dt$ why did they not include the modulus signs?

Title: Re: TT's Maths Thread
Post by: /0 on January 27, 2010, 11:05:42 pm

Because in $[1,x]$ , $2t+\frac{1}{8t}$ is always positive.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 27, 2010, 11:06:45 pm

But don't you need to state it though?

Title: Re: TT's Maths Thread
Post by: humph on January 28, 2010, 08:49:04 am

Why would you bother? It's a matter of simplification.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 30, 2010, 12:05:36 am

Solve for $t$ $\implies$ $2\pi = t - \sin(t)$

Okay, clearly from inspection $t = 2\pi$ is a solution, but how do we do this algebraically? How do we rule out other solutions?

[This question stems from trying to use parametric equations to work out the area under an arch of a cycloid, more precisely... trying to work out the new terminals of the integral...]

Title: Re: TT's Maths Thread
Post by: kamil9876 on January 30, 2010, 12:08:52 am

cycloids are historically cool :)

anyway:

$f(t)=t-sin(t)$

$f'(t)=1-cos(t)> 0 almost everywhere$

therefore $f(t)$ is an increasing function and so the equation $f(t)=2\pi$ can only have at most one solution.

edit: terrible calculus mistake (still didn't change the ans tho :P)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 30, 2010, 12:11:46 am

Ahhh, so can the answer $2\pi$ be only done from inspection?

Title: Re: TT's Maths Thread
Post by: TrueTears on January 30, 2010, 05:01:16 pm

Sketch the following curve with parametric equations $x = 6\sin(t)$ and $y = t^2+t$ without converting to Cartesian form.

So here is what I did...

$\frac{dy}{dx} = \frac{2t+1}{6\cos(t)}$

Now I tried to work out the places where $\frac{dy}{dx} = 0$

So $2t+1 = 0 \implies t = -\frac{1}{2}$ .

So this occurs at $\left(-6\sin\left(\frac{1}{2}\right), - \frac{1}{4}\right)$

Next I tried to find out the vertical tangents of the curve.

So $6\cos(t) = 0 \implies t = \frac{(2n+1)\pi}{2}$

But this doesn't give me any idea of what the curve looks like...

Title: Re: TT's Maths Thread
Post by: TrueTears on January 30, 2010, 08:06:29 pm

Sketch the curve defined by $x = 2\cos(t)-\cos(2t)$ and $y = 2\sin(t) - \sin(2t)$

$\frac{dy}{dx} = \frac{\cos(t)-\cos(2t)}{\sin(2t)-\sin(t)}$

So I worked out when the graph has horizontal tangents.

This occurs when $\cos(t)-\cos(2t) = 0$

I solved this for $t \in [0, 2\pi]$ since that constitutes 1 period of the graph.

These points are: $\left(1,0\right), \left(-\frac{1}{2}, \frac{3\sqrt{3}}{2}\right), \left(-\frac{1}{2},-\frac{3\sqrt{3}}{2}\right)$

Then I worked out when the graph has vertical tangents.

This occurs when $\sin(2t)-\sin(t) = 0$

Again I solved this for $t \in [0, 2\pi]$ since that constitutes 1 period of the graph.

These points occur at: $\left(1,0\right), \left(\frac{3}{2},\frac{\sqrt{3}}{2}\right), \left(-3,0\right), \left(\frac{3}{2},-\frac{\sqrt{3}}{2}\right)$

So the resultant graph should look like:

(http://img199.imageshack.us/img199/3978/parametric.jpg)

My question is: at the point $(1,0)$ , I worked out there is a horizontal tangent but there is ALSO a vertical one? How is that possible? It doesn't look like it has a vertical tangent on the graph?

Also would the integral to work out the surface area of the solid when rotated around the $x$ axis be:

$\int_0^{\pi}2\pi\left(2\sin(t)-\sin(2t)\right)\sqrt{[x'(t)]^2+[y'(t)]^2}dt$

Title: Re: TT's Maths Thread
Post by: Damo17 on January 30, 2010, 09:04:21 pm

At (1,0), t=0. So we need to find the slope at this point. Use l' Hospital's rule as t approaches 0 and we get 0. So there is a horizontal tangent.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 30, 2010, 09:33:46 pm

Yeah at t = 0 the slope doesn't exist so there is no horizontal tangent.

Title: Re: TT's Maths Thread
Post by: Damo17 on January 30, 2010, 09:43:43 pm

Quote from: TrueTears on January 30, 2010, 09:33:46 pm

Yeah at t = 0 the slope doesn't exist so there is no horizontal tangent.

There is a horizontal tangent, using dy/dx gives 0/0 for l' Hospital's, so use the 2nd derivative and you get the limit equaling 0.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 30, 2010, 09:44:37 pm

There is a horizontal tangent approaching x = 1, but there is no horizontal tangent at x=1 :)

Title: Re: TT's Maths Thread
Post by: TrueTears on January 31, 2010, 03:51:44 pm

Sketch $r = \sqrt{\theta}$ for $0 \le \theta \le 16\pi$

Title: Re: TT's Maths Thread
Post by: Damo17 on January 31, 2010, 05:25:20 pm

Quote from: TrueTears on January 30, 2010, 09:44:37 pm

There is a horizontal tangent approaching x = 1, but there is no horizontal tangent at x=1 :)

Yes, I should have thought about x '(t) when t=0, as x '(0)=0 a horizontal tangent does not exist.

Quote from: TrueTears on January 31, 2010, 03:51:44 pm

Sketch $r = \sqrt{\theta}$ for $0 \le \theta \le 16\pi$

You should recognise that this is an Euler Spiral, and because of the domain it will have 8 revolutions.
Just 'plug in' main points such as $\theta = 0, \frac{\pi}{2}, \pi ,\frac{3\pi}{2}, etc$ and connect the points to form a spiral. As $\theta$ increases each revolution will become closer to the last. Long process because of domain.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 31, 2010, 09:08:03 pm

yeah, I got the graph, pretty cool.

Title: Re: TT's Maths Thread
Post by: TrueTears on January 31, 2010, 10:58:53 pm

Find the points on the curve where the tangent line is horizontal and vertical.

$r^2 = \sin(2\theta)$

$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$

$x = r\cos(\theta)$ and $y = r\sin(\theta)$

$\frac{dy}{d\theta} = \frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)$

$\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)$

But $2r\frac{dr}{d\theta} = 2\cos(2\theta)$

so $\frac{dr}{d\theta} = \frac{\cos(2\theta)}{r}$

$\frac{dy}{d\theta} = \frac{\cos(2\theta)}{r}\sin(\theta) + r\cos(\theta)$

$\frac{dx}{d\theta} = \frac{\cos(2\theta)}{r}\cos(\theta)-r\sin(\theta)$

$\frac{dy}{dx} = \frac{\frac{\cos(2\theta)}{r}\sin(\theta) + r\cos(\theta)}{\frac{\cos(2\theta)}{r}\cos(\theta)-r\sin(\theta)}$

$\implies \frac{dy}{dx} = \frac{\cos(2\theta)\sin(\theta)+r^2\cos(\theta)}{\cos(2\theta)\cos(\theta)-r^2\sin(\theta)}$

$\implies \frac{dy}{dx} = \frac{\cos(2\theta)\sin(\theta)+\sin(2\theta)\cos(\theta)}{\cos(2\theta)\cos(\theta)-\sin(2\theta)\sin(\theta)}$

$\implies \frac{dy}{dx} = \frac{\sin(3\theta)}{\cos(3\theta)}$

So for horizontal tangents: $\sin(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$

For vertical tangents: $\cos(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$

A few questions: First how do you know which domain to solve for $\theta$ ? Because initially I thought if you sketched the cartesian equation of $r^2 = \sin(2\theta)$ the period is $\pi$ . But solving $\theta$ over $[0, \pi]$ doesn't give all the points for which there is a horizontal tangent?

Also when $\theta = 0$ there is a horizontal tangent at the origin, but when $\theta = \frac{\pi}{2}$ there is a vertical tangent at the origin. How is this possible?

Title: Re: TT's Maths Thread
Post by: Damo17 on January 31, 2010, 11:36:13 pm

Quote from: TrueTears on January 31, 2010, 10:58:53 pm

So for horizontal tangents: $\sin(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$

For vertical tangents: $\cos(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$

Values of $\theta$ do not exist in the 2nd and 4th quadrants for this graph. That should clear up many questions. :)

Title: Re: TT's Maths Thread
Post by: Damo17 on February 01, 2010, 12:04:39 am

Quote from: TrueTears on January 31, 2010, 10:58:53 pm

Also when $\theta = 0$ there is a horizontal tangent at the origin, but when $\theta = \frac{\pi}{2}$ there is a vertical tangent at the origin. How is this possible?

It is possible that at a point, both a Horizontal and Vertical tangent exist.
In the case of $r^2=sin(2 \theta)$ , the origin (0,0) is passed twice during the course of 1 period. In one of these cases a Horizontal tangent exists and on the other pass through (0,0) a Vertical tangent exists.

Well.. that is the way I see it. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 01, 2010, 12:12:21 am

O yeah, cause what happened was I zoomed in on my calc, it didn't look like a vertical tangent all O.O but then I realised the calc was on shit settings lol.

And also how do you know what values of $\theta$ to solve for? Is it typically between 0 and $2\pi$ since that sweeps every angle for the polar graph?

Title: Re: TT's Maths Thread
Post by: TrueTears on February 01, 2010, 12:25:37 am

Quote from: Damo17 on January 31, 2010, 11:36:13 pm

Quote from: TrueTears on January 31, 2010, 10:58:53 pm
So for horizontal tangents: $\sin(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$

For vertical tangents: $\cos(3\theta) = 0$ for $\theta \in [0, 2\pi]$

$\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$

Values of $\theta$ do not exist in the 2nd and 4th quadrants for this graph. That should clear up many questions. :)

Yeah, I should have been more specific and eliminated those such as $\frac{5\pi}{6}$ etc, but still I'm a bit hazy on why you solve across the the domain 0 to $2\pi$ .

Title: Re: TT's Maths Thread
Post by: Damo17 on February 01, 2010, 12:28:25 am

Quote from: TrueTears on February 01, 2010, 12:12:21 am

O yeah, cause what happened was I zoomed in on my calc, it didn't look like a vertical tangent all O.O but then I realised the calc was on shit settings lol.

And also how do you know what values of $\theta$ to solve for? Is it typically between 0 and $2\pi$ since that sweeps every angle for the polar graph?

Quote from: TrueTears on February 01, 2010, 12:25:37 am

Yeah, I should have been more specific and eliminated those such as $\frac{5\pi}{6}$ etc, but still I'm a bit hazy on why you solve across the the domain 0 to $2\pi$ .

Yeah, usually $[0, 2 \pi]$ but remember any restrictions. e.g in your question $sin(2 \theta) \ge 0$ , so there is a restriction on $\theta$ to the 1st and 3rd quadrants.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 01, 2010, 12:29:12 am

Yeah, so what's the reason for solving across that domain?

Title: Re: TT's Maths Thread
Post by: TrueTears on February 01, 2010, 09:32:38 pm

Thanks ^^

Weird question, I don't know why my way doesn't work...

Find the area bounded by the 2 curves $r^2 = \sin(2\theta)$ and $r^2 = \cos(2\theta)$

Now after some working, the 2 curves look like the following: (p means $\pi$ )

(http://img534.imageshack.us/img534/363/polargraph.jpg)

Red is the curve $r^2 = \cos(2\theta)$

Purple is the curve $r^2 = \sin(2\theta)$

Now we need to find the area defined by $4 \times \int_0^{\frac{\pi}{8}} \frac{1}{2}r^2d\theta = 4 \times \int_0^{\frac{\pi}{8}} \frac{1}{2}\sin(2\theta)d\theta$

But answers has $4 \times \int_0^{\frac{\pi}{8}} \frac{1}{2}\sin^2(2\theta)d\theta$

How is it $\sin^2(2\theta)$ ...?

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 01, 2010, 11:12:19 pm

boo

Title: Re: TT's Maths Thread
Post by: TrueTears on February 02, 2010, 05:02:25 pm

I have a few questions regarding conic sections.

Let F be a fixed point called the focus and L be a fixed line called the directrix in a plane. Let e be a fixed positive number called the eccentricity. The set of all points P in the plane such that $\frac{|PF|}{|PL|} = e$ is a conic section. The conic is an ellipse if $e<1$ , a parabola if $e=1$ or a hyperbola if $e>1$ .

I understand how a parabola works, since there is only 1 focus and 1 directrix and also the line joining the point P to the directrix is perpendicular to the directrix.

But what if you have an ellipse? They have 2 foci so does that mean we only pick 1 focus to deal with? Also does the line joining the point P to the directrix have to be perpendicular to the directrix?

Title: Re: TT's Maths Thread
Post by: Christiano on February 02, 2010, 10:45:58 pm

I need help on rearranging this formula to make q the subject

$2w^2q=2\sqrt{3}-3q$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 02, 2010, 10:48:55 pm

You can bring the $3q$ to the LHS so you get

$2w^2q+3q = 2\sqrt{3}$

Then factorise the $q$

$q(2w^2+3) = 2\sqrt{3}$

Now divide both sides by $(2w^2+3)$

$q = \frac{2\sqrt{3}}{(2w^2+3)}$

Title: Re: TT's Maths Thread
Post by: Christiano on February 02, 2010, 10:53:27 pm

And .. to make i the subject

$f=\sqrt{\frac{h^3i^2-3g}{2}}$

Title: Re: TT's Maths Thread
Post by: Christiano on February 02, 2010, 10:55:24 pm

Quote from: TrueTears on February 02, 2010, 10:48:55 pm

You can bring the $3q$ to the LHS so you get

$2w^2q+3q = 2\sqrt{3}$

Then factorise the $q$

$q(2w^2+3) = 2\sqrt{3}$

Now divide both sides by $(2w^2+3)$

$q = \frac{2\sqrt{3}}{(2w^2+3)}$

+1 .. How do you bring karma up ? hehe

Title: Re: TT's Maths Thread
Post by: Edmund on February 02, 2010, 11:00:19 pm

$f=\sqrt{\frac{h^3i^2-3g}{2}}$

Square both sides:

$f^2=\frac{h^3i^2-3g}{2}$

Multiply both sides by 2:

$2f^2=h^3i^2-3g$

$2f^2+3g=h^3i^2$

Divide both sides by h³

$\frac{2f^2+3g}{h^3}=i^2$

Square root both sides

$i=\sqrt{\frac{2f^2+3g}{h^3}}$

Quote from: Christiano on February 02, 2010, 10:55:24 pm

+1 .. How do you bring karma up ? hehe

Once you reach 50 posts :P

Title: Re: TT's Maths Thread
Post by: Christiano on February 02, 2010, 11:20:14 pm

Thanks a lot!

Title: Re: TT's Maths Thread
Post by: Christiano on February 03, 2010, 12:06:57 am

More simple equations I'm stuck on.

$3-(x-2)^2$

$64y^4-16y^2+1$

$(3p-5)^2-(p+2)^2$

$a^3-3a^2+a-3$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 03, 2010, 02:57:49 pm

$\lim_{x \to \infty} x^x$ (For $x >1$ )

Can we use the comparison tool here?

$x^x > x \ \forall \ x>1$

$\lim_{x \to \infty} x^x > \lim_{x \to \infty} x = \infty$

So $\lim_{x \to \infty} x^x = \infty$

Or can we just treat $x^x$ as a continuous function for $x>1$

So $\lim_{x \to \infty} x^x = x^{\lim_{x \to \infty} x} = x^{\infty} = \infty$

But $\infty$ is not a number so can you still substitute it in the exponent of $x$ ?

Title: Re: TT's Maths Thread
Post by: enwiabe on February 03, 2010, 03:39:43 pm

You need to have x -> N where N is a very large number, I believe, and then tend N to infinity.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 03, 2010, 03:41:30 pm

Quote from: enwiabe on February 03, 2010, 03:39:43 pm

You need to have x -> N where N is a very large number, I belive, and then tend N to infinity.

Ah yeah, that makes more sense, thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 04, 2010, 12:57:23 am

1. Find the limit of the sequence $\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}} \cdots \}$

2. Show that $\sqrt{2} = 1+\frac{1}{2+\frac{1}{2+ \cdots}}$

Many thanks!!!

Title: Re: TT's Maths Thread
Post by: Mao on February 04, 2010, 01:01:56 am

1. It can be shown that the sequence is $a_n = 2^{(2^n-1)/(2^n)}$ , hence as $n\to \infty$ , $a_n \to 2$

2. Show that $\sqrt{2} = 1+\frac{1}{2+\frac{1}{2+ \cdots}}$

Let $x=1+\frac{1}{2+\frac{1}{2+ \cdots}}$

$\implies \frac{1}{x-1} = 2 + \frac{1}{2+ \frac{1}{2+ \cdots}}$

$\implies \left(\frac{1}{x-1} - 2\right)^{-1} = 2 + \frac{1}{2+ \frac{1}{2+ \cdots}}$

$\implies \frac{1}{x-1} = \left(\frac{1}{x-1} - 2\right)^{-1}$

$\implies x-1 +2 = \frac{1}{x-1}$

$\implies x^2 - 1 = 1$

$\implies x = \sqrt{2}$ (discarding negative solution)

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 04, 2010, 01:34:00 am

some nice paradoxes:

1.)

$x=2\times 2\times 2\times 2...$

$\implies x=2x$

$\implies x=0$

2.)

$x=1+2+4+8....$

$\implies x=1+2x$

$\implies x=\frac{-1}{2}$

Moral: check convergence.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 04, 2010, 10:27:57 pm

Thanks Mao very clever!

How do I prove the sequence $\frac{n!}{2^n}$ where $n \in \mathbb{Z}^+$ is diverging?

Clearly by listing a few terms it is diverging, but how to prove it rigorously?

Nvm thanks Ahmad :D

$n! > 3^n$ for $n \ge 7$

Clearly this holds for $n = 7$

Assume it's true for $n = k$

$k! > 3^k$

Now $k!(k+1) > 3^{k}(k+1) > 3^k \cdot 3$

Thus proved by induction.

Now $\frac{n!}{2^n} > \left(\frac{3}{2}\right)^n$

Since $\left(\frac{3}{2}\right)^n$ is diverging for $n \in \mathbb{Z^+}$

$\frac{n!}{2^n}$ is also diverging.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 04, 2010, 11:03:53 pm

even easier:

let

$f(n)=\frac{n!}{2^n}$

$\frac{f(n+1)}{f(n)}=\frac{n+1}{2}\geq 2 \forall n\geq 4$

any sequence $f(4), f(5)...$ with this property must diverge. (as long as $f(4)\neq 0$ ) why?

Title: Re: TT's Maths Thread
Post by: TrueTears on February 05, 2010, 11:52:07 pm

Paradox:

$0 = 0 + 0 + 0 + 0 \cdots = (1-1)+(1-1)+(1-1)+(1-1)+ \cdots = 1-1+1-1+1-1+1-1+\cdots = 1+(-1+1)+(-1+1)+(-1+1)+(-1+1)+\cdots = 1+0+0+0+0+ \cdots = 1$

hahaha why is this wrong?

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 06, 2010, 12:00:25 am

Similair reason as to why the ones in my previous post are. Convergence is the key word.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 06, 2010, 12:01:07 am

Quote from: kamil9876 on February 06, 2010, 12:00:25 am

Similair reason as to why the ones in my previous post are. Convergence is the key word.

Don't you mean divergent? :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 06, 2010, 12:03:18 am

I used the noun "convergence" (hinting at the topic that relates to this) rather than used the adjective "convergent" to describe this particular case.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 06, 2010, 12:21:00 am

Quote from: kamil9876 on February 06, 2010, 12:03:18 am

I used the noun "convergence" (hinting at the topic that relates to this) rather than used the adjective "convergent" to describe this particular case.

But the series is divergent :P

Title: Re: TT's Maths Thread
Post by: TrueTears on February 06, 2010, 01:57:36 am

Find the values of $x$ such that $\sum_{n=0}^{\infty} \frac{\cos^n(x)}{2^n}$ is convergent.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 06, 2010, 02:07:00 am

$n^{th} summand=(\frac{cosx}{2})^n$

for convergence we require $-1<\frac{cos(x)}{2}<1$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 06, 2010, 09:23:06 pm

Ah yeah, nice thanks!

Title: Re: TT's Maths Thread
Post by: TrueTears on February 06, 2010, 10:31:24 pm

kamil check this out:

Find the sum of $\sum_{n = 1}^{\infty} \frac{1}{n^2}$

This is definitely convergent by the Monotonic squeeze theorem. So...

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 06, 2010, 10:32:32 pm

Google "Basel Problem".

Title: Re: TT's Maths Thread
Post by: TrueTears on February 06, 2010, 10:36:39 pm

haven't learnt taylor expansion of sin yet :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 06, 2010, 11:15:00 pm

aww

Title: Re: TT's Maths Thread
Post by: TrueTears on February 07, 2010, 07:26:49 pm

Hmm.. bit stuck on how to prove this limit comparison test :(

Suppose $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are series with positive terms and $\sum_{n=1}^{\infty} b_n$ is divergent, prove that if $\lim_{n \to \infty} \frac{a_n}{b_n} = \infty$ then $\sum_{n=1}^{\infty} a_n$ is also divergent.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 07, 2010, 07:36:07 pm

$lim_{n \to \infty} \frac{a_n}{b_n}=\infty$

implies that there exists some natural number $N$ such that: $\forall n\geq N, \frac{a_n}{b_n}>1$

$\implies a_n>b_n>0 \forall N\geq N$

Can you do the rest from here? The following lemma finishes it off nicely: if $\sum_{n=1}^{\infty} x_n=\infty$ then we can start from a latter term, say the $N^{th}$ term, and still get a diverging series ie: $\sum_{n=N}^{\infty}=\infty$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 07, 2010, 08:54:05 pm

Oh yeah thanks kamilz, you basically finished it off lolz

Title: Re: TT's Maths Thread
Post by: TrueTears on February 07, 2010, 11:10:58 pm

$\sum_{n=1}^{\infty} \frac{2+(-1)^n}{n\sqrt{n}}$

Is the series converging or diverging?

So what test do I use here... how to compare? I can see it's something along the lines of $\frac{1}{n^{\frac{3}{2}}}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 07, 2010, 11:18:58 pm

yeah integral test can easily be used here.

Don't let the $(-1)^n$ put you off as you can simply use:

$\sum_{n=1}^{\infty} \frac{3}{n^{3/2}}$

as an upper bound for the series.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 07, 2010, 11:40:27 pm

haha you got me, that (-1)^n i didn't know what to do with it lolz

and yeah don't really need integral test do we? it's a p series with p>1 so it's converging?

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 07, 2010, 11:48:01 pm

yeah true, p series. You can prove it with integral test but I guess if you are already familiair with it just mention p series.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 07, 2010, 11:50:20 pm

sure, thanks heaps kamilz

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 12:57:50 am

(Sorry for the trivial-like questions... it's the first time I've been exposed to these tests xD)

$\sum_{n=1}^{\infty} \frac{n-1}{n4^n}$

Is this the right way to show the series is converging.

$\sum_{n=1}^{\infty} \frac{n-1}{n} \frac{1}{4^n} = \sum_{n=1}^{\infty} \left(\frac{1}{4^n} - \frac{1}{n4^n}\right)$

$\sum_{n=1}^{\infty} \frac{1}{4^n}$ is converging since it's a geometric series with $|r|<1$

$\sum_{n=1}^{\infty} \frac{1}{n4^n} \le \sum_{n=1}^{\infty} \frac{1}{4^n}$ which the latter is converging thus $\sum_{n=1}^{\infty} \frac{1}{n4^n}$ is also converging.

Thus $\sum_{n=1}^{\infty} \frac{n-1}{n4^n}$ is converging...

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 08, 2010, 01:01:00 am

yep, that's correct (sum of converging series is a converging series).

Another way:

$\frac{n-1}{n4^n}=\frac{n-1}{n} 4^{-n}<4^{-n}$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 01:02:02 am

Quote from: kamil9876 on February 08, 2010, 01:01:00 am

yep, that's correct (sum of converging series is a converging series).

Another way:

$\frac{n-1}{n4^n}=\frac{n-1}{n} 4^n<4^n$

thanks bro, but how did you get the "another way", it doesn't seem obvious to me :(

edit: oh i see your edited version, lol i was like wdf, how can you bring the 4^n outside hahaha

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 02:25:08 am

last one for the night:

$\sum_{n=1}^{\infty} \frac{1+n+n^2}{\sqrt{1+n^2+n^6}}$

Is this series converging or diverging?

Title: Re: TT's Maths Thread
Post by: Mao on February 08, 2010, 04:28:57 am

A second degree over a third degree --> probably divergent.

$\sum_{n=1}^{\infty}\frac{1 + n + n^2}{\sqrt{1 + n^2 + n^6}} = \sum_{n=1}^{\infty}\frac{n}{\sqrt{1 + n^2 + n^6}} + \sum_{n=1}^{\infty}\frac{1+n^2}{\sqrt{1 + n^2 + n^6}}$

Convergence $\sum_{n=1}^{\infty}\frac{n}{\sqrt{1 + n^2 + n^6}}$ is trivial.

For $n\in \mathbb{Z}^+$ :
$\frac{1+n^2}{\sqrt{1 + n^2 + n^6}} \ge \frac{1 + n^2}{\sqrt{1 + n^2 + n^6 + 2n^2 + 3n^4}}=\frac{1}{\sqrt{1+n^2}}$

$\frac{1}{\sqrt{1+n^2}} \ge \frac{1}{\sqrt{1 + n^2 + 2n}} = \frac{1}{n+1}$

$\implies \sum_{n=1}^{\infty}\frac{1+n^2}{\sqrt{1 + n^2 + n^6}} > \sum_{n=1}^{\infty} \frac{1}{n+1}$ , which is divergent.

Hence the series is divergent.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 08, 2010, 12:29:04 pm

Quote

A second degree over a third degree --> probably divergent.

That's a good intuition to have for solving this. Consequently it suggests that the term of highest degree is most significant, ie:

$\frac{1+n+n^2}{\sqrt{1+n^2 + n^6}}\geq \frac{n^2}{\sqrt{1+n^2 + n^6}}=\frac{1}{\sqrt{n^{-4} + n^{-2} + n^2}}$

$\geq \frac{1}{\sqrt{4n^2}}=\frac{1}{2n}$ Hence divergence.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 03:50:49 pm

Thanks u guys :)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 04:49:52 pm

$\sum_{n=1}^{\infty} \sin\left(\frac{1}{n}\right)$

Hmm what to compare this to? LOl

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 08, 2010, 04:52:57 pm

hint:

$lim_{x\to0} \frac{sinx}{x}=1$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 04:53:54 pm

haha I see... limit comparison test!

Title: Re: TT's Maths Thread
Post by: Christiano on February 08, 2010, 10:14:40 pm

Transpose from intercept form to the gradient form:

5x-2y=4

Totally forgot how to. :'( O blessed math angels, come to my rescue.

Title: Re: TT's Maths Thread
Post by: the.watchman on February 08, 2010, 10:15:54 pm

$-2y = -5x+4$ , $y=\frac{5}{2}x-2$

don't hijack TT's thread :P

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 10:16:23 pm

$5x-2y=4$

$-2y = -5x+4$

$y = \frac{5}{2}x - 2$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 10:16:41 pm

Quote from: the.watchman on February 08, 2010, 10:15:54 pm

$-2y = -5x+4$ , $y=\frac{5}{2}x-2$

don't hijack TT's thread :P

na it's ok, he can post here.

Title: Re: TT's Maths Thread
Post by: superflya on February 08, 2010, 10:16:56 pm

Quote from: Christiano on February 08, 2010, 10:14:40 pm

Transpose from intercept form to the gradient form:

5x-2y=4

Totally forgot how to. :'( O blessed math angels, come to my rescue.

TT shall freak when he sees this :P

Edit: or maybe not.

Title: Re: TT's Maths Thread
Post by: the.watchman on February 08, 2010, 10:20:37 pm

Quote from: superflya on February 08, 2010, 10:16:56 pm

Quote from: Christiano on February 08, 2010, 10:14:40 pm
Transpose from intercept form to the gradient form:

5x-2y=4

Totally forgot how to. :'( O blessed math angels, come to my rescue.

TT shall freak when he sees this :P

Edit: or maybe not.

LOL, don't go there superflya :P

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 10:22:26 pm

superflya u pr0 kent

Title: Re: TT's Maths Thread
Post by: the.watchman on February 08, 2010, 10:31:14 pm

Quote from: TrueTears on February 08, 2010, 10:16:41 pm

Quote from: the.watchman on February 08, 2010, 10:15:54 pm
$-2y = -5x+4$ , $y=\frac{5}{2}x-2$

don't hijack TT's thread :P
na it's ok, he can post here.

Soz :P

Title: Re: TT's Maths Thread
Post by: superflya on February 08, 2010, 10:44:07 pm

LOL

Title: Re: TT's Maths Thread
Post by: Christiano on February 08, 2010, 10:58:54 pm

OH MAN sorry, I thought this was a 'general math question' thread. 'Scuse me genius'es

Thanks anyhow ;D

Title: Re: TT's Maths Thread
Post by: TrueTears on February 08, 2010, 11:00:58 pm

na man it's cool, if you have questions ur more than welcomed to ask.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 09, 2010, 12:01:09 am

yo kamil

$\lim_{n \to \infty} \frac{n^n}{n!}$

lolz

Title: Re: TT's Maths Thread
Post by: Cthulhu on February 09, 2010, 12:04:26 am

Hey TT. I have this reaaaaaaaaally hard maths problem and I cannot figure it out and I have a test tomorrow!!!!!
If Ben has two apples and Erin has 6 oranges and 12 apples and gives Ben 3 apples and 2 oranges what is the airspeed velocity of an unladen swallow?

Title: Re: TT's Maths Thread
Post by: Mao on February 09, 2010, 12:11:00 am

Quote from: Cthulhu on February 09, 2010, 12:04:26 am

Hey TT. I have this reaaaaaaaaally hard maths problem and I cannot figure it out and I have a test tomorrow!!!!!
If Ben has two apples and Erin has 6 oranges and 12 apples and gives Ben 3 apples and 2 oranges what is the airspeed velocity of an unladen swallow?

$\frac{\mbox{It's}}{9000}$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 09, 2010, 12:11:14 am

HAHAHAHAHAHA

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 09, 2010, 12:13:50 am

$\frac{n^n}{n!}=\frac{n}{1} \times \frac{n}{2}... \times \frac{n}{n} \geq \frac{n}{1}=n$

and obviously $lim_{n \to \infty} n=\infty$ therefore the limit in question is also infinity.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 09, 2010, 12:14:51 am

/facepalm

thx kamil

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 09, 2010, 12:19:16 am

Quote from: Cthulhu on February 09, 2010, 12:04:26 am

Hey TT. I have this reaaaaaaaaally hard maths problem and I cannot figure it out and I have a test tomorrow!!!!!
If Ben has two apples and Erin has 6 oranges and 12 apples and gives Ben 3 apples and 2 oranges what is the airspeed velocity of an unladen swallow?

(http://i.techrepublic.com.com/blogs/unladen_swallow.jpg)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 09, 2010, 03:35:33 am

If $\sum a_n$ is a conditionally convergent series and $r \in \mathbb{R}$ then there is a rearrangement of $\sum a_n$ that has a sum equal to $r$ .

Argh, I haven't done a proof in ages, but this one really caught my attention xD

Title: Re: TT's Maths Thread
Post by: Mao on February 09, 2010, 08:54:55 am

If $\sum a_n$ is conditionally convergent, $\sum |a_n|$ is divergent, and $\lim_{n\to \infty} |a_n| = 0$ .

So basically you have two divergent series, [for simplicity sake, I'll use alternating series as an example] $\sum a_{2n-1}$ and $\sum a_{2n}$ . Since both are divergent, but made up of infinitesimals, you can make up any real number with some arrangements of $\sum a_n$ (by selecting a certain number of elements from one/both series). That's the basic gist of it.

As for how to do this mathematically, that is for pure math majors to worry about. :)

Title: Re: TT's Maths Thread
Post by: humph on February 09, 2010, 12:11:28 pm

Quote from: Mao on February 09, 2010, 08:54:55 am

As for how to do this mathematically, that is for pure math majors to worry about. :)

Oh hai thar :)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 09, 2010, 02:35:51 pm

thx

also why doesn't this work...

Is the following series absolutely convergent?

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n(10)^n}{n!}$

$\sum_{n=0}^{\infty} \left | \frac{(-1)^n(10)^n}{n!} \right | = \sum_{n=0}^{\infty} \frac{(10)^n}{n!}$

Since $n! > 2^n$ for $n \ge 4$

$\frac{1}{n!} < \frac{1}{2^n}$ for $n \ge 4$

So $\frac{10^n}{n!} < \frac{10^n}{2^n}$ for $n \ge 4$

Using the comparison test, $\sum_{n=0}^{\infty} \frac{10^n}{2^n}$ is not convergent... so what do I do?

Title: Re: TT's Maths Thread
Post by: humph on February 09, 2010, 02:55:19 pm

Quote from: TrueTears on February 09, 2010, 02:35:51 pm

thx

also why doesn't this work...

Is the following series absolutely convergent?

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n(10)^n}{n!}$

$\sum_{n=0}^{\infty} \left | \frac{(-1)^n(10)^n}{n!} \right | = \sum_{n=0}^{\infty} \frac{(10)^n}{n!}$

Since $n! > 2^n$ for $n \ge 4$

$\frac{1}{n!} < \frac{1}{2^n}$ for $n \ge 4$

So $\frac{10^n}{n!} < \frac{10^n}{2^n}$ for $n \ge 4$

Using the comparison test, $\sum_{n=0}^{\infty} \frac{10^n}{2^n}$ is not convergent... so what do I do?

You want to show that $n! > 10^n$ for sufficiently large $n$ - work out when this inequality is first true, then use induction (this is when a calculator or computer comes in handy).

Title: Re: TT's Maths Thread
Post by: TrueTears on February 09, 2010, 03:36:43 pm

thanks humph.

Also using the test for divergence on any alternating series, wouldn't the limit would never exist.

eg, $\sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1}$

$\lim_{n \to \infty} (-1)^n \frac{n}{n+1}$ doesn't exist so the series is diverging, but it is actually converging if we use the alternating series test?

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 09, 2010, 03:55:16 pm

Quote from: humph on February 09, 2010, 02:55:19 pm

Quote from: TrueTears on February 09, 2010, 02:35:51 pm
thx

also why doesn't this work...

Is the following series absolutely convergent?

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$

$\sum_{n=0}^{\infty} \frac{(-10)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n(10)^n}{n!}$

$\sum_{n=0}^{\infty} \left | \frac{(-1)^n(10)^n}{n!} \right | = \sum_{n=0}^{\infty} \frac{(10)^n}{n!}$

Since $n! > 2^n$ for $n \ge 4$

$\frac{1}{n!} < \frac{1}{2^n}$ for $n \ge 4$

So $\frac{10^n}{n!} < \frac{10^n}{2^n}$ for $n \ge 4$

Using the comparison test, $\sum_{n=0}^{\infty} \frac{10^n}{2^n}$ is not convergent... so what do I do?

You want to show that $n! > 10^n$ for sufficiently large $n$ - work out when this inequality is first true, then use induction (this is when a calculator or computer comes in handy).

That's still too weak though.

for sufficiently large n the following is true:

$\frac{10^n}{n!}=\frac{10}{1} \times \frac{10}{2} ..... \times \frac{10}{10} \times \frac{10}{11} \times \frac{10}{12}.... \frac{10}{n} \leq \frac{10^{10}}{10!} \times (\frac{10}{11})^{n-10}$

So you can compare it to a geometric series with first term $\frac{10^{10}}{10!}$ and common ratio $\frac{10}{11}$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 09, 2010, 04:05:52 pm

So you're saying we can't use the test for divergence on alternating series?

how come stewarts did it on page 712?

oh u deleted ur post lol

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 09, 2010, 10:18:57 pm

Quote from: TrueTears on February 09, 2010, 03:35:33 am

If $\sum a_n$ is a conditionally convergent series and $r \in \mathbb{R}$ then there is a rearrangement of $\sum a_n$ that has a sum equal to $r$ .

Argh, I haven't done a proof in ages, but this one really caught my attention xD

Let $(u_i)$ be the subsequence of $(a_i)$ that consists of positive terms and let $(l_i)$ be the subsequence of $(a_i)$ that consists of negative terms. Specifically, let:

$u_i=\Big\{^{a_i, a_i>0}_{0, otherwise}$

and likewise for $l_i$

Lemma: $\sum u_i$ and $\sum l_i$ are both diverging.

The following construction produces the rearangement for the given $r$ :

Let $(x_i)$ be this rearangment. Imagine that it is initially empty, we will now create it by adding in the numbers in order:

Keep adding the first few numbers from $(u_i)$ until the sum exceeds $r$ . Then keep adding the first few numbers from $l_i$ until the cumulutative sum drops below $r$ . Then keep repeating this for the next lot of numbers in $u_i$ and $l_i$ ad infinitum.

Now I will leave it for you to prove that such a construction can be done (you will need the lemma) and that the limit of (x_i) is indeed $r$ .

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 10, 2010, 12:34:47 pm

Extension question: Prove that if $\sum_{i=1}^{\infty} a_{\phi(i)}$ is a rearangement of a convergent series $\sum_{i=1}^{\infty} a_i$ such that for some fixed integer $p$ , $|\phi(i)-i|\leq p$ (ie each term is "shifted" at most $p$ places) then the two sums are equal.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 03:15:55 pm

$\sum_{n=1}^{\infty} \tan\left(\frac{1}{n}\right)$

diverging or converging?

First I thought integral test would work since $\tan\left(\frac{1}{n}\right)$ is decreasing for $n \ge 1$ and positive. But the integral is not elementary.

Using the divergence test we get the limit as $0$ which is inconclusive.

Next, I tried to use the ratio test.

$\lim_{n \to \infty} \left | \frac{\tan\left(\frac{1}{n+1}\right)}{\tan\left(\frac{1}{n}\right)} \right |$

Since both functions are positive for $n \ge 1$ , the modulus signs are redundant.

$\lim_{n \to \infty} \frac{\tan\left(\frac{1}{n+1}\right)}{\tan\left(\frac{1}{n}\right)} = \lim_{n \to \infty} \frac{n^2}{(n+1)^2} \cdot \lim_{n \to \infty} \frac{\cos\left(\frac{1}{n}\right)^2}{\cos\left(\frac{1}{n+1}\right)^2} = 1$

Which again is inconclusive.

So what do I do...?

EDIT: 6000th post yay!

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 10, 2010, 04:06:05 pm

$lim_{n\to \infty} \frac{tan(\frac{1}{n})}{\frac{1}{n}}=lim_{x\to 0} \frac{tan(x)}{x}=\lim_{x\to 0}\frac{sin(x)}{x} * \lim_{x \to 0} \frac{1}{cos(x)}=1$

Hence either both the harmonic series and the series in question converge, or both diverges. Meaning the series in question diverges.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 04:17:52 pm

thanx

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 04:58:41 pm

$\sum_{n=1}^{\infty} \frac{(n!)^n}{n^{4n}}$

Using the root test we get:

$\lim_{n \to \infty} \left | \left(\frac{(n!)^n}{n^{4n}}\right)^\frac{1}{n} \right | = \lim_{n \to \infty} \frac{n!}{n^4}$

So... how to evaluate that haha

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 10, 2010, 05:19:54 pm

for $n>4$

$\frac{n!}{n^4}=\frac{n}{n} \times \frac{n-1}{n} \times \frac{n-2}{n} \times \frac{n-3}{n} \times (n-4)!$

$=(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n}) \times (n-4)!>(1-\frac{1}{4})(1-\frac{2}{4})(1-\frac{3}{4})(n-4)!$

Hence the limit is infinity.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 05:48:03 pm

Your inequality shouldn't it be

$(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n}) \times (n-4)! \ge (1-\frac{1}{4})(1-\frac{2}{4})(1-\frac{3}{4})(n-4)!$

coz they can be equal if n = 4, it's not undefined for n = 4...? why are you saying it is only greater?

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 10, 2010, 05:54:34 pm

sure, you can do that if you want. Getting the maximal domain is not important, my one still does the job.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 06:10:15 pm

yeah, true, thanks :)

Title: Re: TT's Maths Thread
Post by: Mao on February 10, 2010, 10:58:19 pm

Just a note: have you looked at the ratio test?

Taking $\sum_{n=0}^{\infty} \frac{(-10)^n}{n!}$ for example

$|r_n| = \frac{10^n}{n!}$

$|r_{n+1}| = \frac{10^{n+1}}{(n+1)!}$

$\lim_{n\to \infty} \frac{|r_{n+1}|}{|r_{n}|} = \frac{10}{n+1} = 0$

:. Absolute convergence

It can make life a lot easier: http://en.wikipedia.org/wiki/Ratio_test

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 10:58:34 pm

$\sum_{n=1}^{\infty} \left( \frac{n}{n+1} \right)^{n^2}$

So applying the root test we get:

$\lim_{n \to \infty} \left | \left( \frac{n}{n+1} \right)^n \right | = \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^n$

So let $y = \left( \frac{n}{n+1} \right)^n$

Thus $\log_e(y) = \log_e\left(\left( \frac{n}{n+1} \right)^n\right) = n \log_e\left( \frac{n}{n+1} \right)$

$\implies \lim_{n \to \infty} \log_e(y) = \lim_{n \to \infty} n \log_e\left( \frac{n}{n+1} \right)$

I've used hospital on $\lim_{n \to \infty} n \log_e\left( \frac{n}{n+1} \right)$ but it's too complicated...

Nvm fkn stupid arithmetic error grr

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 11:17:16 pm

Are we allowed to do this? The question was just yelling out for a substitution, but I've never done one before like this lolz

$\sum_{n=2}^{\infty} \frac{1}{\log_e(n)^{\log_e(n)}}$

Let $u = \log_e(n)$

Thus when $n = 2 \implies u = \log_e(2)$

$\sum_{u = \log_e(2)}^{\infty} \frac{1}{u^u}$

Using the root test:

$\lim_{u \to \infty} \left | \frac{1}{u} \right | = 0$

Thus the series is converging.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 10, 2010, 11:34:53 pm

Quote

$\sum_{u = \log_e(2)}^{\infty} \frac{1}{u^u} $

This expression isn't correct. $\log_e(2)$ isn't an integer so it doesn't look conventional, if not undefined.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 10, 2010, 11:53:14 pm

I see, so what do I do?

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 11, 2010, 12:37:34 am

Still the right idea.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 11, 2010, 05:47:26 pm

$\int \frac{x-\tan^{-1}(x)}{x^3} dx$

Express the integral as a power series.

This is what I have done so far:

$\int \frac{x-\tan^{-1}(x)}{x^3} dx = \int \frac{1}{x^2} dx - \int \frac{\tan^{-1}(x)}{x^3} dx$

Let $f(x) = \tan^{-1}(x)$

$f'(x) = \frac{1}{1+x^2} = \frac{1}{1-[-(x)^2]} = \sum_{n=0}^{\infty} \left[-(x)^2\right]^n = \sum_{n=0}^{\infty} (-1)^n(x)^{2n}$

Thus $f(x) = \sum_{n=0}^{\infty} \int (-1)^n(x)^{2n} dx + C$

$\implies \tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$

$\implies \frac{1}{x^3}\tan^{-1}(x) = \frac{1}{x^3}\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n-2}}{2n+1}$

$\implies \int \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n-2}}{2n+1} dx = \sum_{n=0}^{\infty} \int (-1)^n \frac{x^{2n-2}}{2n+1} dx = C + \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n-1}}{(2n+1)(2n-1)} =C + \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n-1}}{(4n^2-1)}$

But my problem is, how do I find a power series for $\int \frac{1}{x^2} dx$ ?

Thanks :)

Nvm I got it.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 12, 2010, 03:35:48 pm

Find the Taylor expansion of $f(x) = \sin(x)$ centered at $\frac{\pi}{2}$

Here is what I have done:

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^n$

$f^0(x) = \sin(x) \implies f^0\left(\frac{\pi}{2}\right) = 1$

$f^1(x) = \cos(x) \implies f^1\left(\frac{\pi}{2}\right) = 0$

$f^2(x) = -\sin(x) \implies f^2\left(\frac{\pi}{2}\right) = -1$

$f^3(x) = -\cos(x) \implies f^3\left(\frac{\pi}{2}\right) = 0$

$f^4(x) = \sin(x) \implies f^4\left(\frac{\pi}{2}\right) = 1$

Then the cycle repeats.

Thus $\sin(x) = 1 - \frac{1}{2}\left(x-\frac{\pi}{2}\right)^2 + \frac{1}{24}\left(x-\frac{\pi}{2}\right)^4 + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\left(x-\frac{\pi}{2}\right)^{2n}$

My question is: when I changed the power series into summation notation, it was more or less playing around (like just finding a pattern and changing it into summation notation). What if the power series was incredibly hard? Is there a systematic way of changing it into summation notation? For example, page 741 of Stewarts, the summation notation is crazy.

Thanks

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 12, 2010, 04:16:58 pm

boo

Title: Re: TT's Maths Thread
Post by: TrueTears on February 12, 2010, 04:20:41 pm

Is this the right way to prove $\sin(x)$ is equal to it's Taylor series?

$\left | f^{n+1}(x) \right | \le M$

Since $f^{n+1}(x) = \pm \sin(x), \pm \cos(x) \implies \left | f^{n+1}(x) \right | \le 1$

$0 \le \left | R_n(x) \right | \le \frac{\left | x- \frac{\pi}{2} \right |^{n+1}}{(n+1)!}$ for $\left | x-a \right | \le d$ for some $d \in \mathbb{R}$

Since $\lim_{n \to \infty} \frac{\left | x- \frac{\pi}{2} \right |^{n+1}}{(n+1)!} = 0$

Then by the squeeze theorem: $\lim_{n \to \infty} \left | R_n(x) \right | = 0 \implies \lim_{n \to \infty} R_n(x) = 0$

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 12, 2010, 04:49:05 pm

Just to demonstrate understanding of the theorem:

Quote from: TrueTears on February 12, 2010, 04:20:41 pm

$0 \le \left | R_n(x) \right | \le \frac{\left | x- \frac{\pi}{2} \right |^{n+1}}{(n+1)!} \le \frac{\left d^{n+1}}{(n+1)!}$ whenever $d$ is such that $\left | x-a \right | \le d $

And then you can sandwhich/ninja shit etc. the rest with the fixed $d$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 12, 2010, 05:08:49 pm

Thanks kamilz =]

Also this question, I'm stuck at the last part.

Express the function $\frac{1}{(2+x)^3}$ as a power series.

$\frac{1}{8\left(1+\frac{x}{2}\right)^3} = \frac{1}{8}\left(1+\frac{x}{2}\right)^{-3} = \frac{1}{8} \sum_{n=0}^{\infty} \binom{-3}{n}\left(\frac{x}{2}\right)^n$

$\sum_{n=0}^{\infty} \binom{-3}{n}x^n = 1 + \frac{-3}{1!}\left(\frac{x}{2}\right) + \frac{(-3)(-4)}{2!}\left(\frac{x}{2}\right)^2 + \frac{(-3)(-4)(-5)}{3!}\left(\frac{x}{2}\right)^3 + \cdots$

$\implies 1-3\left(\frac{x}{2}\right)+6\left(\frac{x}{2}\right)^2-10\left(\frac{x}{2}\right)^3 + \cdots$

$\implies \frac{1}{(2+x)^3} = \frac{1}{8}\left[1-3\left(\frac{x}{2}\right)+6\left(\frac{x}{2}\right)^2-10\left(\frac{x}{2}\right)^3 + \cdots\right]$

However the answer has $\frac{1}{(2+x)^3} = \sum_{n=0}^{\infty}(-1)^n \frac{(n+1)(n+2)}{2^{n+4}}x^n$

How do you convert my power series into the summation one? More importantly, how do you do it systematically...

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 12, 2010, 05:31:17 pm

Haiz, too many unnecesary steps. Let me start from:

Quote from: TrueTears on February 12, 2010, 05:08:49 pm

$\sum_{n=0}^{\infty} \binom{-3}{n}x^n$

$=\sum_{n=0}^{\infty} \frac{(-3)(-4)....(-3-n+1)}{n!} x^n$

$=\sum_{n=0}^{\infty} (-1)^{(n)} \frac{(3)(4)...(2+n)}{n!} x^n$

$=\sum_{n=0}^{\infty} (-1)^{(n)} (3)(4)...(2+n) \times \frac{1}{n!} x^n$

$=\sum_{n=0}^{\infty} (-1)^{(n)} \times \frac{(2+n)!}{2} \times \frac{1}{n!} x^n$

$=\sum_{n=0}^{\infty} (-1)^{(n)} \times \frac{(n)!(n+1)(n+2)}{2} \times \frac{1}{n!} x^n$

now watch the $n!$ cancel. In general, try to play with the term inside the summation as much as possible.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 12, 2010, 05:47:06 pm

haha thanks man, that makes more sense now

Title: Re: TT's Maths Thread
Post by: TrueTears on February 12, 2010, 06:11:39 pm

Eh, I'm stomped on this one lol

Find the sum of $\sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{4^{2n+1}(2n+1)!}$

Title: Re: TT's Maths Thread
Post by: Ahmad on February 12, 2010, 06:32:55 pm

It looks like a familiar power series evaluated at $x = \frac{\pi}{4}$ !

Title: Re: TT's Maths Thread
Post by: Damo17 on February 12, 2010, 11:14:35 pm

Quote from: TrueTears on February 12, 2010, 06:11:39 pm

Eh, I'm stomped on this one lol

Find the sum of $\sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n+1}}{4^{2n+1}(2n+1)!}$

You should remember all of the important Maclaurin Series seen on page 743 of Stewart's, they really do make things easier.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 12, 2010, 11:22:27 pm

$=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} * (\frac{\pi}{4})^{2n+1}$

so it really does tell you that it's a series evaluated at $x=\frac{\pi}{4}$ And the $2n$ tells you that every second derivative is 0. More precisely every odd derivative is 0. This already hints at something trigonometric.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 15, 2010, 06:23:51 pm

Let P be a point on the line L that passes through the points Q and R. Show that the distance d from the point P to the line L is $d = \frac{ \left | \mathbf{a} \times \mathbf{b} \right |}{\left |\mathbf{a} \right |}$ where $\mathbf{a} = \overrightarrow{QR}$ and $\mathbf{b} = \overrightarrow{QP}$

Thanks :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 15, 2010, 07:07:54 pm

let $<CQP=\theta$

$d=|QP|\sin\theta=|b|sin\theta$

$\implies |a|d=|a||b|\sin\theta$

$d=\frac{|a||b|\sin\theta}{|a|}$

and the result follows.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 15, 2010, 07:36:37 pm

Oh yeah that makes sense, thanks.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 15, 2010, 09:06:45 pm

Find the distance between the skewed lines:

Line 1. $x= 1+t, y = 1+6t, z = 2t$ where $t \in \mathbb{R}$

Line 2. $x=1+2s, y = 5+15s, z=-2+6s$ where $s \in \mathbb{R}$

Here's my working, can someone check if it's correct, cause there are no answers for this question.

The line which is parallel to line 1 is $v_1 = \mathbf{i} + 6\mathbf{j} + 2\mathbf{k}$

The line which is parallel to line 2 is $v_2 = 2\mathbf{i} +15\mathbf{j} +6\mathbf{k}$

Now let line 1 and line 2 can be considered to be on two parallel planes $P_1$ and $P_2$ .

Thus the normal vector for $P_1$ and $P_2$ must be the same.

This normal vector is given by:

$v_1 \times v_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 6 & 2 \\ 2 & 15 & 6 \end{vmatrix}$

$= \mathbf{i} \begin{vmatrix} 6 & 2 \\ 15 & 6 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2 \\ 2 & 6 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 6 \\ 2 & 15 \end{vmatrix}$

$= 6\mathbf{i} - 2\mathbf{j} +3 \mathbf{k}$

When $t = 0$ we have the point $(1,1,2)$ lying on the plane $P_1$ .

Thus the plane is given by: $6(x-1) -2(y-1) +3(z-2) = 0 \implies 6x-2y+3z-10=0$

Now when $s = 0$ we have the point $(1,5,-2)$ lying on line 2.

The distance that this point is from $P_1$ is given by:

$D = \frac{\left | 6(1) - 2(5) +3 (-2) - 10 \right |}{\sqrt{6^2+(-2)^2+3^2}} = \frac{20}{7}$

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 15, 2010, 10:44:31 pm

Quote

When t = 0 we have the point (1,1,2) lying on the plane P_1.

(1,1,0)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 17, 2010, 11:24:23 am

Reparametrize the curve $\mathbf{r}(t) = \left( \frac{2}{t^2+1}-1\right)\mathbf{i} + \frac{2t}{t^2+1}\mathbf{j}$ with respect to arc length measured from the point $(1,0)$ in the direction of increasing $t$ .

This is my working, can someone check if it's right because there are no answers =(

Define $\mathbf{r}(u) = \left( \frac{2}{u^2+1}-1\right)\mathbf{i} + \frac{2u}{u^2+1}\mathbf{j}$

$\implies \mathbf{r'}(u) = \frac{-4u}{(u^2+1)^2}\mathbf{i} - \frac{2(u^2-1)}{(u^2+1)^2}\mathbf{j}$

$\implies \left | \mathbf{r'}(u) \right | = \frac{2}{u^2+1}$

$(1,0)$ corresponds to when $t = 0$

$\implies s(t) = \int_0^t \frac{2}{u^2+1} du = 2\tan^{-1}(t)$

$\implies s = 2\tan^{-1}(t)$

$\implies \tan\left(\frac{s}{2}\right)=t$

$\implies \mathbf{r}(s) = \left( \frac{2}{\tan^2\left(\frac{s}{2}\right)+1}-1\right)\mathbf{i} + \frac{2\tan\left(\frac{s}{2}\right)}{\tan^2\left(\frac{s}{2}\right)+1}\mathbf{j}$

$\implies \mathbf{r}(s) =\left(2\sec^2\left(\frac{s}{2}\right)-1\right)\mathbf{i} + \sin(s) \mathbf{j}$

Title: Re: TT's Maths Thread
Post by: Damo17 on February 17, 2010, 12:03:40 pm

Quote from: TrueTears on February 17, 2010, 11:24:23 am

$\implies \mathbf{r}(s) = \left( \frac{2}{\tan^2\left(\frac{s}{2}\right)+1}-1\right)\mathbf{i} + \frac{2\tan\left(\frac{s}{2}\right)}{\tan^2\left(\frac{s}{2}\right)+1}\mathbf{j}$

$\implies \mathbf{r}(s) =\left(2\sec^2\left(\frac{s}{2}\right)-1\right)\mathbf{i} + \sin(s) \mathbf{j}$

$ \left( \frac{2}{\tan^2\left(\frac{s}{2}\right)+1}-1\right)\mathbf{i}=\left( \frac{1-\tan^2\left(\frac{s}{2}\right)}{\sec^2\left(\frac{s}{2}\right)}\right)\mathbf{i}=\left( {\cos^2\left(\frac{s}{2}\right)\ - sin^2\left(\frac{s}{2}\right)}\right)\mathbf{i}=cos(s)\mathbf{i}$

Everything else is correct. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 17, 2010, 12:09:07 pm

!!! Thanks Damo!!

Title: Re: TT's Maths Thread
Post by: TrueTears on February 18, 2010, 01:43:16 am

Use polar coordinates to find the limit: $\lim_{(x,y) \to (0,0)} \frac{e^{-x^2-y^2}-1}{x^2+y^2}$

oh man i have no idea on this one wdf

Title: Re: TT's Maths Thread
Post by: Mao on February 18, 2010, 01:46:35 am

Any point in the x-y plane can be represented as a modulus and an angle. Modulus is r, distance from origin, angle is theta, angle from positive x axis in the ccw direction.

Hence, $x = r \cos \theta,\; y = r \sin \theta$

$L = \lim_{r \to 0} \frac{e^{-r^2} - 1}{r^2}$

Title: Re: TT's Maths Thread
Post by: TrueTears on February 18, 2010, 01:48:11 am

OH

MY

FKN

GOD

HOW DID I NOT REMEMBER THAT WDF, MUST BE LACK OF SLEEP

thanks mao btw

Title: Re: TT's Maths Thread
Post by: TrueTears on February 18, 2010, 01:45:05 pm

Say we have the function $f(x_1,x_2,x_3 \cdots x_n) = c_1x_1 + c_2x_2 + \cdots c_nx_n$

Then this can be represented by a single vector variable $\mathbf{x} = <x_1,x_2,x_3 \cdots x_n>$ and $\mathbf{c} = <c_1,c_2,c_3 \cdots c_n>$ so that we get $f(\mathbf{x}) = \mathbf{x}.\mathbf{c}$

I understand where the dot product comes from but the $f(\mathbf{x})$ doesn't make sense, isn't that saying we have $f(\mathbf{x}) = f(x_1\mathbf{i},x_2\mathbf{j},x_3\mathbf{k} \cdots x_n\mathbf{n})$ rather than $f(x_1,x_2,x_3 \cdots x_n)$ ?

In other words inside the function we now have a vector instead of a set of coordinates?

Title: Re: TT's Maths Thread
Post by: Mao on February 19, 2010, 12:58:35 am

Yes, in this case it's the same thing.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 19, 2010, 08:07:02 pm

Thanks Mao!

Can someone show me how to use an $\epsilon - \delta$ proof to show $\lim_{(x,y,z) \to (3,0,1)} e^{-xy} \sin\left(\frac{\pi z}{2}\right) = 1$

I've done some stuff, but it doesn't really lead anywhere... so I won't be bothered posting it.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 19, 2010, 10:34:01 pm

Quote from: TrueTears on February 19, 2010, 08:07:02 pm

Thanks Mao!

Can someone show me how to use an $\epsilon - \delta$ proof to show $\lim_{(x,y,z) \to (3,0,1)} e^{-xy} \sin\left(\frac{\pi z}{2}\right) = 1$

I've done some stuff, but it doesn't really lead anywhere... so I won't be bothered posting it.

Bit of a too magical solution, (the algebraic trickery comes from my proof about limit laws which is more natural).

let $f=e^{-xy}$ , $g=\sin\left(\frac{\pi z}{2}\right)$

$|fg-1|=|(g-1)+(f-1)+(f-1)(g-1)|\leq |g-1|+|f-1| + |f-1||g-1|$

Now the trick is to show that each of the terms can be made as small as one wishes(ie: each less than $\frac{\epsilon}{3}$ ).

Title: Re: TT's Maths Thread
Post by: TrueTears on February 19, 2010, 10:35:21 pm

How did you get all that inequality and stuff? It doesn't seem obvious to me at all, can you leave in all your working?

thx.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 19, 2010, 11:03:17 pm

Well I tried to imitate a proof I once did of the following:

$lim_{x \to a} f(x)g(x)=\lim_{x\to a} f(a) \lim_{x\to a} g(x)$

let $L_f$ and $L_g$ be the limits of $f(x)$ and $g(x)$ respectively.

Now intuitively we can represent $f$ as $L_f + \Delta f$ (ie $\Delta f$ is some error term) and similairly $g=L_g + \Delta g$

now the errors can be made as small as possible, and because we want to prove that theorem we might as well multiply:

$fg=L_fL_g + L_f \Delta g + L_g \Delta f + \Delta f \Delta g$

$\implies |fg-L_fL_g|=|L_f \Delta g + L_g \Delta f + \Delta f \Delta g|$

Now plug back in $\Delta_f=f-L_f$ and $\Delta_g=g-L_g$ And you basically get an expression without the "error terms". This expression is what inspires:

Quote

$|fg-1|=|(g-1)+(f-1)+(f-1)(g-1)|$

Coz you think the $L_f$ and $L_g$ "ought to be" $1$ (Nonetheless, the expression can be verified simply by expanding RHS right? So it is technically independant of this post, but still good to understand where the idea comes from).

As for the inequality, well that is just the triangle inequality; an indespensible tool in epsilon delta proofs.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 21, 2010, 11:23:00 pm

For the differentiability of a function of 2 variables, say we have $z=f(x,y)$ , then if $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ exist near $(a,b)$ and are continuous at $(a,b)$ then $f$ is differentiable at $(a,b)$ .

For continuous I know you just have to prove $\lim_{(x,y) \to (a,b)} = f(a,b)$

But how do you prove $z=f(x,y)$ , then if $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ exist near $(a,b)$ ?

(Btw as a side note, that condition sounds so informal, "exist NEAR" ??? wdf)

Title: Re: TT's Maths Thread
Post by: TrueTears on February 22, 2010, 03:17:49 am

The definition of a closed set in R^2 is one that contains all its boundary points, ie if the set was D, then a boundary point of D is a point (a,b) such that every disk with centre (a,b) contains points in D and also points not in D.

Is this a correct interpretation of the definition?

(http://img691.imageshack.us/img691/5170/partiald.jpg)

Title: Re: TT's Maths Thread
Post by: humph on February 22, 2010, 09:48:30 am

Yup.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 23, 2010, 01:24:02 am

Thanks humph, also a closed set can be infinite in extent right? For example the closed set between the lines x = -3 and x = 3.

Title: Re: TT's Maths Thread
Post by: humph on February 23, 2010, 01:54:17 am

Yup again. In fact, the entire plane is an open and closed set. The only other subset of the plane with this property is the empty set.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 23, 2010, 02:09:55 am

Ahh thanks humph!

Title: Re: TT's Maths Thread
Post by: Mao on February 23, 2010, 05:14:18 am

Quote from: TrueTears on February 21, 2010, 11:23:00 pm

For the differentiability of a function of 2 variables, say we have $z=f(x,y)$ , then if $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ exist near $(a,b)$ and are continuous at $(a,b)$ then $f$ is differentiable at $(a,b)$ .

For continuous I know you just have to prove $\lim_{(x,y) \to (a,b)} = f(a,b)$

But how do you prove $z=f(x,y)$ , then if $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ exist near $(a,b)$ ?

(Btw as a side note, that condition sounds so informal, "exist NEAR" ??? wdf)

$\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ exist near $(a,b)$ means the partial derivatives are continuous at (a,b). By the definition of limits, this continuity ensures any path taken to (a,b) will yield the same gradient (which eliminates any possibilities that a partial derivative in another direction is undefined, because the partial derivative in any direction won't be continuous in that case). Hence the graph must be 'smooth' at (a,b) and is differentiable.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 25, 2010, 02:36:08 pm

$\int \int_D f(x,y) dA$ can be interpreted as the volume of a solid that lies above a bounded region $D$ (on the $xy$ -plane) and under the surface $z = f(x,y)$ if $f(x,y) \ge 0$

So what is the geometric interpretation of a triple integral. Say:

$\int \int \int_E f(x,y,z) dV = \int \int_D \left[ \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) dz \right] dA$

Where $E$ is a bounded region in three dimensional space and $D$ is the projection of $E$ onto the $xy$ -plane. $z=u_1(x,y)$ and $z=u_2(x,y)$ are the upper boundary and lower boundary surfaces of the solid $E$ , respectively.

Title: Re: TT's Maths Thread
Post by: kamil9876 on February 25, 2010, 03:02:01 pm

Don't be afraid of dimension>3. Physics probably has a lot of examples. Such as let f(x,y,z) be the density of a solid at the point (x,y,z). Triple integral can be used to find it's mass probably.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 25, 2010, 03:05:31 pm

Yeah I ain't afraid haha, I just wanted to know some meaning attached to it.

Title: Re: TT's Maths Thread
Post by: /0 on February 25, 2010, 03:07:34 pm

The triple integral gives you the 4-dimensional hypervolume of a shape, but that's not easy to visualise. There is another way of thinking about it which avoids pure geometry and I think is easier to visualise.

If we go back to double integrals, $z = f(x,y)$ gives you the height of the solid about the x-y plane. But $P=f(x,y)$ could also describe other things, like the probability density of finding a particle at a point on the x-y plane, in which case $\iint_D f(x,y)dA$ would be the probability of finding the particle in a region D. (Just a 2D analogue to 1D probability density functions)

Likewise, in three dimensions, $P=f(x,y,z)$ could be the probability density of finding a particle at a point in space, and $\iiint_E f(x,y,z)dV$ the probability of finding it in a region E.

$f(x,y,z)$ could also describe other things, like temperature, concentration, field strength etc.
When combined with the stoke's and the divergence theorem, the triple integral becomes even more powerful.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 25, 2010, 05:31:04 pm

Ahh yeah, and when you set f(x,y,z) = 1 in triple integrals you do get a volume interpretation!

Title: Re: TT's Maths Thread
Post by: TrueTears on February 25, 2010, 11:18:55 pm

$\int \int \int_E f(x,y,z) dV = \int_c^d \int_{\alpha}^{\beta} \int_a^b f(p\sin\phi\cos\theta, p\sin\phi\sin\theta, p\cos\phi)p^2\sin\phi dp d\theta d\phi$ where $E$ is a spherical wedge given by $E = \{(p,\theta,\phi) | a \le p \le b, \alpha \le \theta \le \beta, c \le \phi \le d \}$

Now I understand what that means, however if we extend this to a more general spherical region such as:

$E = \{(p,\theta,\phi) | a \le p \le b, \alpha \le \theta \le \beta, g_1(\theta,\phi) \le \phi \le g_2(\theta,\phi) \}$

What the hell are $g_1(\theta,\phi)$ and $g_2(\theta,\phi)$ ?

If $g_1(\theta,\phi)$ and $g_2(\theta,\phi)$ are a function of spherical coordinates, then where is the $p$ ?

Title: Re: TT's Maths Thread
Post by: /0 on February 25, 2010, 11:53:52 pm

Hmm it should be $g(p,\theta)$ right?

Title: Re: TT's Maths Thread
Post by: TrueTears on February 26, 2010, 12:01:38 am

Quote from: /0 on February 25, 2010, 11:53:52 pm

Hmm it should be $g(p,\theta)$ right?

nope just $\theta$ and $\phi$

page 1008 if u wanna double check

Title: Re: TT's Maths Thread
Post by: /0 on February 26, 2010, 02:27:27 am

But if you integrate $\int_{g_1(\theta,\phi)}^{g_2(\theta,\phi)} f \,d\phi$ there will be a $\phi$ leftover after integration

Title: Re: TT's Maths Thread
Post by: TrueTears on February 26, 2010, 02:34:12 am

hmmmmm weird...

Title: Re: TT's Maths Thread
Post by: Cthulhu on February 26, 2010, 03:20:59 am

I was reading through this earlier and I couldn't figure it out. The only thing that I could think of was some sort of limit that is beyond my late night math thinkedness.

Title: Re: TT's Maths Thread
Post by: Mao on February 26, 2010, 11:45:58 am

Quote from: TrueTears on February 25, 2010, 02:36:08 pm

So what is the geometric interpretation of a triple integral. Say:

$\int \int \int_E f(x,y,z) dV = \int \int_D \left[ \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) dz \right] dA$

Check out Stoke's theorem. It can give a surface area.

Title: Re: TT's Maths Thread
Post by: TrueTears on February 26, 2010, 05:55:01 pm

Quote from: Mao on February 26, 2010, 11:45:58 am

Quote from: TrueTears on February 25, 2010, 02:36:08 pm
So what is the geometric interpretation of a triple integral. Say:

$\int \int \int_E f(x,y,z) dV = \int \int_D \left[ \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) dz \right] dA$

Check out Stoke's theorem. It can give a surface area.

Thanks! I'll check it out.

Quote from: Cthulhu on February 26, 2010, 03:20:59 am

I was reading through this earlier and I couldn't figure it out. The only thing that I could think of was some sort of limit that is beyond my late night math thinkedness.

haha, yeah I think I got it now, p isn't mentioned because it is already restricted.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 07, 2010, 02:15:16 am

Ahhh I've decided to visit my good ol maths after a few month hiatus...

Anyways was playing around while I was bored of commerce study tonight and I seem to have forgotten how to prove the triple integral formula...

$\int \int \int_R f(x,y,z) dV = \int \int \int_S f(x(u,v,w),y(u,v,w),z(u,v,w)) \left | \frac{\partial(x,y,z)}{\partial(u,v,w)}\right | dudvdw$

where $\left | \frac{\partial(x,y,z)}{\partial(u,v,w)}\right |$ is the Jacobian of a transformation T

Any enlightenment would be good xD

Title: Re: TT's Maths Thread
Post by: TrueTears on April 17, 2010, 03:01:55 am

Ahh enough economics reading, time for some late night maths, very nostalgic haha, was flipping through AnC and came across this pretty one, I've started by applying AM-GM, but some more ideas would be wonderful xD

Let $n \geq 2$ , find the least constant C such that for all $\{x_1, \dots, x_n\} \in \mathbb{R}^+ \cup \{0\}$ , $\sum_{1\leq i<j \leq n} x_ix_j (x_i^2 + x_j^2) \leq C \left(\sum_{i=1}^n x_i \right)^4$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on May 01, 2010, 09:40:46 pm

brainless stream of consciousness... after a long period of accounting reading xD

Suppose that at most two of the $x_i$ , say $x_a$ and $x_b$ , are nonzero. Then the left-hand side of the inequality becomes $x_a x_b (x_a^2 + x_b^2)$ and the right-hand side becomes $\frac{(x_a + x_b)^4}{8}$ .

By AM-GM,

$\begin{align*}x_{a}x_{b}(x_{a}^{2}+x_{b}^{2}) = \frac{1}{2}\left[\sqrt{(2x_{a}x_{b})(x_{a}^{2}+x_{b}^{2})}\right]^{2}&\le\frac{1}{2}\left(\frac{x_{a}^{2}+2x_{a}x_{b}+x_{b}^{2}}{2}\right)^{2}\\ &= \frac{(x_{a}+x_{b})^{2}}{8}\end{align*}$

Suppose that our statement holds when at most k of the $x_i$ are equal to zero. Suppose now that k+1 of the $x_i$ are equal to zero, for $k\ge 2$ . WLOG, let these be $x_1 \ge \dotsb \ge x_k \ge x_{k+1}$ .

$y_j = \begin{cases} x_j, & j \notin \{k, k+1\} \\x_k + x_{k+1}, & j=k \\0, & j=k+1 \end{cases}$

We wish to show that by replacing the $x_i$ with the $y_i$ , we increase the left-hand side of the desired inequality without changing the right-hand side, then we can use the inductive hypothesis...

soo ehh, how to complete the induction, i feel like im missing something...

kamil, /0 any ideas?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 04, 2010, 08:34:40 pm

Probability paradox

You are participating in a game show where you are given the choice of three doors: Behind one of the doors is a Mazerati, behind the other two doors are goats. You pick a door and the game host, who knows what is behind the doors, opens another door which has a goat. He then says to you, "Do you want to switch to the other?" Is it better to switch at this point or to stick with your original choice or maybe it does not matter?

Eh shouldn't it not matter what door you pick?

But that's not the right answer... how do you do this question?

Title: Re: TT's Maths Thread
Post by: brightsky on May 04, 2010, 08:51:31 pm

Yeah it's always better to change. For simplicity's sake, let's call the doors with goats inside them Door A and B, and the one with the Mazerati in it, door C.

Case 1: You choose Door C.

Then you would be shown either Door A or Door B which has a goat inside it. Obviously, if you switch, you do not get the Mazerati, and if you stick with your original choice, then you do get it.

Case 2: You choose Door B.

Then you are shown either Door C or Door A, Door C which has the Mazerati in it. In this case, if you stick with your original choice, you don't get the Mazerati, and if you switch, you do.

Case 3: You choose Door A.

Then you are shown either Door C or Door B, Door C which has the Mazerati in it. In this case, again, if you stick with your original choice, you don't get the Mazerati, and if you switch, you do.

These are the ONLY 3 possibilities that can happen. As you can see, there is a 2 out of 3 chance you will pick Door B or Door C, and hence will need to switch to get the Mazerati, and only 1 out of 3 chance you will pick the right door.

Hence why there is a bigger possibility for you to get the Mazerati if you switch.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 04, 2010, 08:57:54 pm

i still dont get it =S

if u picked door C then if you switch u have a 2/3 of getting a goat, isn't that worse off? u have a higher percentage of losing.

if u picked door B then u have a 1/3 chance of getting the car or 1/3 chance of picking door A to get the goat, so isn't it equal chance?

Title: Re: TT's Maths Thread
Post by: brightsky on May 04, 2010, 09:05:29 pm

Quote from: TrueTears on May 04, 2010, 08:57:54 pm

i still dont get it =S

if u picked door C then if you switch u have a 2/3 of getting a goat, isn't that worse off? u have a higher percentage of losing.

if u picked door B then u have a 1/3 chance of getting the car or 1/3 chance of picking door A to get the goat, so isn't it equal chance?

Look at it from a broader spectrum. Start from step 1. You choose a door. Now there is a higher possibility that you will choose the door that doesn't contain the Mazerati. Because there is a higher possibility of this happen, this in turn means that there is a higher possibility that if you switch, you will get the car.

Quote

if u picked door C then if you switch u have a 2/3 of getting a goat, isn't that worse off? u have a higher percentage of losing.

But you need to compare the chances with Door A and B as well. It is MORE of a chance that you will not pick door C. Hence switching would mean that you are more likely to switch to the door that has the car. Yes, if you pick door C and switch, you will lose, but that is only 1/3 of the chances.

Quote

if u picked door B then u have a 1/3 chance of getting the car or 1/3 chance of picking door A to get the goat, so isn't it equal chance?

If you pick door B and you switch, you'll get the car. If you pick door A and switch, you'll get the car. If you pick Door C and switch, you don't.

If you pick door B and don't switch, you lose. Same deal for A. If you pick Door C and don't switch, you win.

Calculate the chances from here.

Title: Re: TT's Maths Thread
Post by: the.watchman on May 04, 2010, 09:07:03 pm

Ah but you know that one of the other doors is a goat, right?
So in each case it is a question of 50:50, but in the overall scheme, the three cases yield a 2/3 chance IF you switch :)

I love this question, don't you TT and brightsky? :D

Title: Re: TT's Maths Thread
Post by: TrueTears on May 04, 2010, 09:09:59 pm

no i dont like probability.

anyways i dont get it...

"If you pick door B and you switch, you'll get the car."

u dont tho, u can pick the goat still

So if you pick door C first, you have a probability of 1 of picking the goat if u switch, so u lose, so why would u wanna switch here?

But the answer is switching is always better, but clearly here, if u switch ur gonna lose instead of win

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 04, 2010, 09:22:21 pm

What if the host himself does not know where the car is, and may accidentally remove it. what should you do?

Title: Re: TT's Maths Thread
Post by: the.watchman on May 04, 2010, 09:24:07 pm

After an overall view of all cases, it is obvious that switching is better in general
Oh, and with the pick door B case, you are shown that one of the doors (A in this case) has a goat behind it, so if you switch you will pick C which has the car etc.

Quote from: kamil9876 on May 04, 2010, 09:22:21 pm

What if the host himself does not know where the car is, and may accidentally remove it. what should you do?

Lol, not gonna happen... :D

Title: Re: TT's Maths Thread
Post by: TrueTears on May 04, 2010, 09:25:44 pm

ahhh i don't care anymore, dont get this Q, dont care either now

Title: Re: TT's Maths Thread
Post by: brightsky on May 04, 2010, 09:34:49 pm

Quote

"If you pick door B and you switch, you'll get the car."

u dont tho, u can pick the goat still

So if you pick door C first, you have a probability of 1 of picking the goat if u switch, so u lose, so why would u wanna switch here?

The other door is already open. So you know what's in one door already, the only options are to stick with door B or switch to whatever door isn't open.

Again, picking door C is the only case where when you switch you don't get the car. In the remaining two cases, you do get the car if you switch.

And yes I do hate probability...and statistics too. xD

Title: Re: TT's Maths Thread
Post by: TrueTears on May 04, 2010, 09:36:30 pm

OMFG

I READ THE FUKN QUESTION WRONG, LOL I THOUGHT THE HOST DOESNT OPEN ANY DOORS AFTER U PICK UR DOOR, I THOUGHT HE'D JUST ASK U IF U WANNA SWITCH TO EITHER OF THE 2 OTHER DOORS

OMFG

FKN.........

shit shit no wonder, yeah i get it now totally lolololol epic fail ><

Title: Re: TT's Maths Thread
Post by: the.watchman on May 04, 2010, 09:41:17 pm

Quote from: TrueTears on May 04, 2010, 09:36:30 pm

OMFG

I READ THE FUKN QUESTION WRONG, LOL I THOUGHT THE HOST DOESNT OPEN ANY DOORS AFTER U PICK UR DOOR, I THOUGHT HE'D JUST ASK U IF U WANNA SWITCH TO EITHER OF THE 2 OTHER DOORS

OMFG

FKN.........

shit shit no wonder, yeah i get it now totally lolololol epic fail ><

Lol, it is trippy though... :)

Title: Re: TT's Maths Thread
Post by: Yitzi_K on May 04, 2010, 10:20:34 pm

It's counter intuitive, but it is better to switch. There's an iPhone app which runs simulations of this scenario, and if you switch, the likelihood of getting the car is 66.67%

Title: Re: TT's Maths Thread
Post by: /0 on May 04, 2010, 11:50:03 pm

I think this is called the Monty Hall problem

Title: Re: TT's Maths Thread
Post by: TrueTears on May 04, 2010, 11:53:21 pm

yeah it is, kamil just told me on msn, also showed me a neat tree diagram which explains it instantly

Title: Re: TT's Maths Thread
Post by: Ahmad on May 10, 2010, 11:11:26 pm

Probability is awesome.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 27, 2010, 04:32:49 am

Find the Taylor series for $\sqrt{x}$ centered around $x=1$ .

I've gotten pretty close but I don't know how to put it back into summation format. This is my working:

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(1)}{n!}(x-1)^n$

$f^1(1) = \frac{1}{2}$

$f^2(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$

$f^3(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)$

$f^4(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)$

So we have:

$\sqrt{x} = 1 + \frac{\frac{1}{2}}{1!}(x-1) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x-1)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(x-1)^3 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{4!}(x-1)^4 + \cdots$

So I know the summation form must have $1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\frac{1}{2^n}\text{something}}{n!}(x-1)^n$ but I can't find the pattern for $(1), (1)(-1), (1)(-1)(-3), (1)(1)(-3)(-5)$

Help please, thanks!

Title: Re: TT's Maths Thread
Post by: googoo on June 27, 2010, 08:46:13 am

To me it is so simple. There is a greater chance (2/3) of picking the goat at the first pick. If you are given a second chance to switch, why not.

Title: Re: TT's Maths Thread
Post by: Damo17 on June 27, 2010, 09:53:55 am

Quote from: TrueTears on June 27, 2010, 04:32:49 am

Find the Taylor series for $\sqrt{x}$ centered around $x=1$ .

I've gotten pretty close but I don't know how to put it back into summation format. This is my working:

$f(x) = \sum_{n=0}^{\infty} \frac{f^n(1)}{n!}(x-1)^n$

$f^1(1) = \frac{1}{2}$

$f^2(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$

$f^3(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)$

$f^4(1) = \left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)$

So we have:

$\sqrt{x} = 1 + \frac{\frac{1}{2}}{1!}(x-1) + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}(x-1)^2+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(x-1)^3 + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{4!}(x-1)^4 + \cdots$

So I know the summation form must have $1+\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\frac{1}{2^n}\text{something}}{n!}(x-1)^n$ but I can't find the pattern for $(1), (1)(-1), (1)(-1)(-3), (1)(1)(-3)(-5)$

Help please, thanks!

Notice that the numerators starting from the term $-\frac{1}{4}$ are like the factorials we normally see with an odd positive integer multiplied by all the odd positive integers down to 1.

From Wikipedia:

Quote

A function related to the factorial is the product of all odd values up to some odd positive integer n. It is often called double factorial (even though it only involves about half the factors of the ordinary factorial, and its value is therefore closer to the square root of the factorial), and denoted by n!!.

So we get as part of the summation:

$\frac{(-1)^{n-1}(2n-3)!!}{2^n \cdot n!}(x-1)^n$

But this is only for $n \ge 2$ . So the full summation form is:

$\sqrt{x}= 1 + \frac{1}{2}(x-1) + \sum_{n=2}^{\infty}\frac{(-1)^{n-1} \cdot (2n-3)!!}{2^n \cdot n!}(x-1)^n$

Title: Re: TT's Maths Thread
Post by: Ahmad on June 27, 2010, 12:37:33 pm

Alternatively use the generalised binomial theorem with $\sqrt{x} =(1+(x-1))^{1/2}$

Title: Re: TT's Maths Thread
Post by: pooshwaltzer on June 27, 2010, 02:55:52 pm

Here's a doozer for you...FAIR GAME

A: Select 5 numbers from 1 to 45. Numbers selected are unique.
B: Select 1 number from 1 to 45, indpendently from the above.

This forms 1 attempt. A win is getting 5/5 in A and 1/1 in B right. In a FAIR game what should be the game prize if each attempt is $2?

If prize = $50million, what should be the cost per attempt pricing in a FAIR gaming system?

Title: Re: TT's Maths Thread
Post by: TrueTears on June 27, 2010, 06:18:15 pm

ok so it's a matter of notation?

can't I have something like this:

$1+\frac{1}{2}(x-1) + \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^n\frac{(1)(3)\cdots(2n-1)}{2^n}}{(n+1)!}(x-1)^{n+1}$

...or can i? Lol, it just doesn't seem very rigorous, im sure people would know what it means, but it's not a really formal notation though...

Title: Re: TT's Maths Thread
Post by: /0 on June 27, 2010, 07:16:32 pm

Yeah I don't think anybody would have a problem with that, even though it is slightly messy. You could always replace it with $\Pi$ , or use Damo's double factorial.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 06, 2010, 02:29:25 pm

$\text{Prove} \ (a,b) = \text{min}(ax+by) \ \forall \ \{a,b\} \in \mathbb{N} \ \text{and} \ \{x,y\} \in \mathbb{Z}$

I seem to have forgotten how to prove this fundamental statement lol... been a while since I last touched number theory.

Haha nvm... I got it...

Let $\text{min}(ax+by) = j$ and let $g = (a,b)$ , since $g|a$ and $g|b \implies g|ax+by \implies g|j \implies j \ge g$

If we can force $g=j$ then we have $\text{min}(ax+by) = (a,b)$ .

Suppose we show $j$ is a common divisor of $a$ and $b$ , and since $g|j$ then we will force $j = g$ .

Now consider $j < \{a,b\}$ and $j$ does not divide $a$ but $j|b$ . So $j$ is not a common divisor of $a$ and $b$ .

Now in this case:

$a = qj+r$ where $q \ge 1$ and $0 \le r < j$ . Now we have r $= a - qj \implies r = a + (-b)$ since $j|b$ .

But $r<j$ and $r$ is also a linear combination of $a$ and $b$ but we just said $j$ was the smallest linear combination of $a$ and $b$ . So by contradiction $j$ must divide both $a$ and $b$ .

Now we can rest as we have shown $j = g$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on July 07, 2010, 08:27:19 pm

Let $p(x)$ be an irreducible function then $f(p(x))$ is also irreducible iff $f(x)$ is irreducible.

How do you prove this?

For me, only Eisenstein's criterion comes into mind...

Title: Re: TT's Maths Thread
Post by: /0 on July 07, 2010, 08:41:31 pm

I assume you mean irreducible over the reals?

A counterexample:

$f(x) = x(x-1)$ - reducible

$p(x) = x^2+1$ - irreducible

$f(p(x)) = (x^2+1)x^2$ - irreducible

Title: Re: TT's Maths Thread
Post by: TrueTears on July 07, 2010, 08:47:52 pm

Hm true, I actually can't remember what I meant with that statement, I have it written down in my exercise book somewhere but I don't understand what I meant LOL.

Maybe it was:

Let p(x) be an irreducible function and if f(x) is irreducible then f(p(x)) is also irreducible .

Title: Re: TT's Maths Thread
Post by: TrueTears on July 08, 2010, 01:46:14 pm

For the linear congruence equation $ax \equiv b \pmod{n}$ , the general solution is given by $x = x_0 + \frac{nt}{d}$ where $x_0$ is a particular solution and $d = (a,n)$ and $t \in \mathbb{Z}$ .

My book says for this general solution it forms 'd congruence classes mod n'. What does that mean?

I interpreted it as this:

$\frac{n}{d}$ is always a constant since $n = kd$ so $\frac{n}{d} = k$ for some integer constant $k$ .

So $x = tk+x_0$ which can be written as $x \equiv x_0 \pmod{k}$ which says ' $\frac{n}{d}$ congruence classes mod $\frac{n}{d}$ '

There are $\frac{n}{d}$ congruence classes because $0 \le x_0 < k$ so there are $\{0, 1, 2, 3, \cdots k-1\}$ possible remainders, so there are $k$ congruence classes since there are $k$ possible remainders.

How did they get 'd congruence classes mod n'?

Title: Re: TT's Maths Thread
Post by: humph on July 08, 2010, 03:39:45 pm

Quote from: /0 on July 07, 2010, 08:41:31 pm

I assume you mean irreducible over the reals?

A counterexample:

$f(x) = x(x-1)$ - reducible

$p(x) = x^2+1$ - irreducible

$f(p(x)) = (x^2+1)x^2$ - irreducible

Uhhh... what? That last polynomial is obviously reducible as you've factorised it in the reals.

Title: Re: TT's Maths Thread
Post by: /0 on July 08, 2010, 05:39:28 pm

Oh... i thought you would need to completely factor it into $(x-r_1)(x-r_2)(x-r_3)$ ...
mah bad

Title: Re: TT's Maths Thread
Post by: TrueTears on July 08, 2010, 08:59:39 pm

wait so does that my original statement holds? if so, how to prove it?

thanx

Title: Re: TT's Maths Thread
Post by: Ahmad on July 08, 2010, 09:25:30 pm

$f(x) = x - 1$ irreducible
$p(x) = x^2 + 1$ irreducible
$f(p(x)) = x \cdot x$ reducible

The other direction is true though

Title: Re: TT's Maths Thread
Post by: humph on July 08, 2010, 09:28:27 pm

Well it's pretty simple if you think about it. What does it mean for a polynomial to be irreducible? It's actually easier to deal with reducible polynomials than irreducible ones, so you're probably best off proving the contrapositive statement (but make sure you phrase the contrapositive statement correctly!).

Title: Re: TT's Maths Thread
Post by: TrueTears on July 09, 2010, 06:25:33 pm

ahhh yeah, thanks I got it :)

Title: Re: TT's Maths Thread
Post by: TrueTears on July 09, 2010, 08:13:12 pm

Another question which I need some clarification over.

How many classes of solutions are there to $x^2-1 \equiv 0 \pmod{168}$

So I did this:

1. Let a be the number of the classes of solutions to $x^2-1 \equiv 0 \pmod{2^3}$

2. Let b be the number of the classes of solutions to $x^2-1 \equiv 0 \pmod{3}$

3. Let c be the number of the classes of solutions to $x^2-1 \equiv 0 \pmod{7}$

So the classes of solutions to $x^2-1 \equiv 0 \pmod{168}$ is equal to $abc$

For case 1. We have $0, \pm 1, \pm 2, \pm 3, 4$ as congruence classes and only $\pm 1$ and $\pm 3$ works so that's 4 classes of solutions.

For case 2. We have $0, \pm 1$ as congruence classes and only $\pm 1$ works so that's 2 classes of solutions.

For case 3. We have $0, \pm 1, \pm 2, \pm 3$ as congruence classes and only $\pm 1$ works so that's 2 classes of solutions.

So all together we have $2^4 = 16$ classes of solutions.

However what I'm wondering is, isn't this way a bit primitive because if I worked out the PPF of some number other than 168 and ended up with say $2^{100}$ as one of the prime powers, then I would have to work out the number of classes of solutions to $x^2-1 \equiv 0 \pmod{2^{100}}$ Which means I have to list out $0, \pm 1, \pm 2 \cdots$ then test each of them to see if they work, wouldn't that take ages? Is there a faster way other than plugging a solution from each class of solutions into the equation and seeing if it works?

Thanks.

Title: Re: TT's Maths Thread
Post by: zzdfa on July 12, 2010, 04:54:58 pm

There is a result that tells you how many solutions $x^2=n \mod p^a$ has for prime p and integer exponent a.

If p=2, then the number of solutions to $x^2=n \mod p^a$ is :
1 if a=1 and n=1 mod 2
2 if a=2 and n=1 mod 4
4 if a>=3, and n=1 mod 8
0 otherwise

If p an odd prime then $x^2=n \mod p^a$ has either 0 or 2 solutions. (It follows that the only solutions of $x^2=1 \mod p^a$ are $x= \pm 1$

As you can see, this result reduces the above problem to looking things up in a chart, and it isn't too hard to prove. You can do it via induction on the exponent 'a'. If you want hints or more info, check out "Elementary Number Theory; A Problem Oriented Approcah" by Joe Roberts, pg192. It's a nice book, the whole book is just one big problem set designed to guide you through elementary number theory.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 12, 2010, 05:31:18 pm

omg thanks man, ill definitely check that book out, again thanks heaps!!

btw we haven't spoken for ages, where have u been bro? :P

Title: Re: TT's Maths Thread
Post by: TrueTears on July 16, 2010, 12:29:22 am

Show that $(A \cap B) \cup C = A \cap (B \cup C)$ is false for all sets A, B and C using direct proof.

I'm not sure how to finish off my proof... can someone please help :)

If we were to prove that the statement is true then we would need to show that if $x \in (A \cap B) \cup C$ then $x \in A \cap (B \cup C)$ and if $x \in A \cap (B \cup C)$ then $x \in (A \cap B) \cup C$

To formalise this we denote the following:

Let P(x) denote the proposition function " $x \in (A \cap B) \cup C$ " and Q(x) denote the proposition function " $x \in A \cap (B \cup C)$ "

Thus we need to show that $\forall x \left(P(x) \to Q(x)\right)$ and $\forall x \left(Q(x) \to P(x)\right)$

Let us first focus on $\forall x \left(P(x) \to Q(x)\right)$

Because we are using a direct proof let us assume the hypothesis P(x) is true. If we can show that Q(x) is false then the universally quantified statement is false and we have complete our proof.

Since P(x) is assumed to be true then $x \in (A \cap B) \cup C$ can be interpreted as $x \in (A \cap B)$ or $x \in C$ .

$x \in (A \cap B) \cup C$ is only true if:

1. $x \in (A \cap B)$ is true and $x \in C$ is true

2. $x \in (A \cap B)$ is true and $x \in C$ is false

3. $x \in (A \cap B)$ is false and $x \in C$ is true

Now Q(x) is interpreted as $x \in A$ and $x \in B \cup C$

Now what...?

Title: Re: TT's Maths Thread
Post by: TrueTears on July 16, 2010, 01:27:42 am

Can someone check if my working is correct for the following proof?

Prove $X \cup (Y-X) = X \cup Y$ for all sets $X$ and $Y$ .

Let $P(x)$ be the proposition function ' $x \in X \cup (Y-X)$ ' and $Q(x)$ be the proposition function ' $x \in X \cup Y$ '

Let the domain of discourse of both these proposition functions be $\mathbb{U}$ , the universal set.

Now to prove the statement to be true we must show the following 2 universally quantified statements to be true:

1. $\forall x \in \mathbb{U} \left(P(x) \to Q(x)\right)$

2. $\forall x \in \mathbb{U} \left(Q(x) \to P(x)\right)$

To show case 1 to be true, first we know that if $P(x)$ is false then the case is vacuously true, so we will ignore this trivial case.

We will assume $P(x)$ to be true and if we can show that $Q(x)$ is also true then case 1 is true.

Now $P(x): \left(x \in X\right) \lor \left(x \in (Y-X)\right)$ and $Q(x): \left(x \in X\right) \lor \left(x \in Y\right)$

Since $P(x)$ is true then either

a. $x \in X$ is true and $x \in (Y-X)$ is false

b. $x \in X$ is false and $x \in (Y-X)$ is true

[Note: these 2 propositions can not be both true since they are mutually exclusive]

If we use case a. Since $x \in X$ is true $\implies \left(x \in X\right) \lor \left(x \in Y\right)$ is true $\implies Q(x)$ is true.

Now using case b. Since $x \in (Y-X)$ is true $\implies x \in Y$ is true $\implies \left(x \in X\right) \lor \left(x \in Y\right)$ is true $\implies Q(x)$ is true.

Thus we have shown case 1. to be true.

To show case 2 to be true, first we know that if $Q(x)$ is false then the case is vacuously true, so again we will ignore this trivial case.

We will assume Q(x) to be true and Q(x) is true if either of the following 3 conditions are satisfied.

i. $x \in X$ is true and $x \in Y$ is false

ii. $x \in X$ is false and $x \in Y$ is true

iii. $x \in X$ is true and $x \in Y$ is true

Now using case i. Since $x \in X$ is true $\implies \left(x \in X\right) \lor \left(x \in (Y-X)\right)$ is true $\implies P(x)$ is true.

Using case ii. Since $x \in Y$ is true $\implies x \in (Y-X)$ is true $\implies \left(x \in X\right) \lor \left(x \in (Y-X)\right)$ is true $\implies P(x)$ is true.

Now using case iii. Since $x \in X$ is true and $x \in Y$ is true $\implies \left(x \in X\right) \lor \left(x \in (Y-X)\right)$ is true $\implies P(x)$ is true.

Case 2 is now proven to be true.

We have completed our proof to show that the original statement is true.

Title: Re: TT's Maths Thread
Post by: zzdfa on July 16, 2010, 11:55:56 am

why so formal are you studying logic?

oh wait you are :\

_why?

Title: Re: TT's Maths Thread
Post by: TrueTears on July 16, 2010, 02:59:49 pm

lololololol

yea logic, they formalise everything pretty damn hardcore so im just going along with it haha

logic is alright, its not as boring as i expected it to be

Title: Re: TT's Maths Thread
Post by: TrueTears on July 17, 2010, 04:38:58 pm

Just wondering do De Morgan's generalised laws work for double universal quantifiers?

Say we want to prove that a function is not one-to-one, in other words we need to prove that the negation of $\forall x_1 \forall x_2 \left[\left(f(x_1)=f(x_2)\right) \to \left(x_1=x_2\right)\right]$ is true.

$\neg \forall x_1 \forall x_2 \left[\left(f(x_1)=f(x_2)\right) \to \left(x_1=x_2\right)\right]$

$\equiv \exists x_1 \neg \forall x_2 \left[\left(f(x_1)=f(x_2)\right) \to \left(x_1=x_2\right)\right]$

$\equiv \exists x_1 \exists x_2 \neg \left[\left(f(x_1)=f(x_2)\right) \to \left(x_1=x_2\right)\right]$

$\equiv \exists x_1 \exists x_2 \left[\left(f(x_1)=f(x_2)\right) \land \neg \left(x_1=x_2\right)\right]$

$\equiv \exists x_1 \exists x_2 \left[\left(f(x_1)=f(x_2)\right) \land \left(x_1 \neq x_2\right)\right]$

Is that correct?

Title: Re: TT's Maths Thread
Post by: Mao on July 17, 2010, 04:51:48 pm

On a completely unrelated notes, TT, just WHAT are you going to do when you run out of maths to study?

Title: Re: TT's Maths Thread
Post by: TrueTears on July 17, 2010, 05:03:27 pm

lol, study moar

Title: Re: TT's Maths Thread
Post by: zzdfa on July 17, 2010, 06:54:12 pm

yeah i think you just proved it, didn't you? [that demorgans laws work for nested universal quantifiers]. just use the first 3 lines and replace the proposition (is that what it's called?) $\left[\left(f(x_1)=f(x_2)\right) \rightarrow \left(x_1=x_2\right)\right]$ with an arbitrary proposition $P(x_1,x_2)$

Title: Re: TT's Maths Thread
Post by: TrueTears on July 17, 2010, 07:54:06 pm

Quote from: zzdfa on July 17, 2010, 06:54:12 pm

yeah i think you just proved it, didn't you? [that demorgans laws work for nested universal quantifiers]. just use the first 3 lines and replace the proposition (is that what it's called?) $\left[\left(f(x_1)=f(x_2)\right) \rightarrow \left(x_1=x_2\right)\right]$ with an arbitrary proposition $P(x_1,x_2)$

ah yup, just wanted some confirmation, thanks

Title: Re: TT's Maths Thread
Post by: TrueTears on July 19, 2010, 02:40:55 am

Can someone show me how to do this?

Show that if a relation R on a set X is symmetric and transitive but not reflexive, then the collection of sets [a], $a \in X$ (equivalence classes of X) does not partition X.

Title: Re: TT's Maths Thread
Post by: /0 on July 19, 2010, 08:22:31 am

(hope this is right lol)

If $a \not\sim a$ , then $a \not\in [a]$

Suppose there is an arbitrary element $b\in [a]$ , so that $b \sim a$ .

But if $b \sim a$ , then $[a]=[b]$ :
Consider an arbitrary $x \in [b]$ . Then $b \sim x$ . Since $b \sim a$ , then by transitivity, $a \sim x$ . Hence, $[a]=[b]$ .

So if there exists an element $b \in [a]$ , then it must also be true that $b \in [b]$ . But since $b \not\sim b$ , $b \not \in [b]$ , so there exists no element $b \in [a]$ .

Hence, $[a]$ is empty and does not partition the set.

Title: Re: TT's Maths Thread
Post by: Ahmad on July 19, 2010, 11:21:46 am

What's wrong with this logic? :)

I claim that a symmetric and transitive relation is also reflexive.

Let a be in X, and let b be in X such that a ~ b. Now being symmetric means b ~ a also, and transitivity implies a ~ a as required.

Title: Re: TT's Maths Thread
Post by: /0 on July 19, 2010, 11:32:22 am

Ah true... I felt something was fishy
It's like how the redundant condition $d(x,x)=0$ in metric spaces can be derived from the triangle inequality

Title: Re: TT's Maths Thread
Post by: Ahmad on July 19, 2010, 12:00:24 pm

It's not true, it's tricky and there is something wrong with it. :)

Title: Re: TT's Maths Thread
Post by: zzdfa on July 19, 2010, 08:25:34 pm

Quote from: Ahmad on July 19, 2010, 11:21:46 am

What's wrong with this logic? :)

I claim that a symmetric and transitive relation is also reflexive.

Let a be in X, and let b be in X such that a ~ b. Now being symmetric means b ~ a also, and transitivity implies a ~ a as required.

sneaky :)

this talk about equivalence relations reminds me of a nice alternate definition of an equivalence relation:

a relation ~ is called an equivalence relation if and only if there exists a function f such that

$a\sim b \iff f(a)=f(b)$

(verify that it is indeed equivalent to the usual definition)

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 02:36:36 am

Oh right, thanks guys, I proved it now.

Another question: Show that if we select 151 distinct computer science courses numbered between 1 and 300 inclusive then at least 2 are consecutively numbered.

I know pigeonhole theorem is involved but how do I apply it?

Title: Re: TT's Maths Thread
Post by: brightsky on July 20, 2010, 05:33:22 pm

Pigeons: 151 pigeons defined by the selected distinct computer science courses.
Holes: 150 holes are defined by {1, 2} , {3,4}, {5,6} ... {299,300}

Since 151 = 150 *1 + 1, then by the pigeon hole principle, at least one hole will have 2 pigeons in it, no matter how which numbers you select.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 05:34:47 pm

AWESOME, THANK YOU VERY MUCH!!

Actually wait, if you have {1,2}, {3,4} which means each computer science course can have 2 numbers assigned to it, doesn't that mean the 151th computer science course will have no numbers assigned to it?

Eg, let the computer science courses be c1, c2, c3 ... c151

c1 = {1,2}

c2 = {3,4}

.
.
.

c150 = {299,300}

c151 = ?

Title: Re: TT's Maths Thread
Post by: pooshwaltzer on July 20, 2010, 06:17:00 pm

There seems to be a number of mutual exclusivities at play here...No course overlapping apparently...and it's a permutations rather than combinatorics sequence since ascension of order is a requirement.

Come to think of it...

{1, 2} , {3, 4}, {5, 6} ... {299, 300} = 150 distinct non-overlapping pairs ONLY

{1, 2}, {2, 3} , {3,4}, {4, 5} ... {299, 300} = 299 distinct overlapping & non-overlapping pairs

299 - 150 = 149 distinct overlapping pairs ONLY

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 06:25:59 pm

ok, so how do i do the q?

i am still a bit confused...

Title: Re: TT's Maths Thread
Post by: brightsky on July 20, 2010, 06:26:33 pm

Hmm..not entirely sure what you mean, but imagine the 151 selected computer science courses as pigeons. There are 300 choices, and for the sake of highlighting there must be a consecutive pair, we group the 300 choices as {1, 2}, {3, 4} ... In an extreme case, we would choose the computer course 1, 3, 5, 7, 9, ... , 299 as the first 150 of the selected courses, but the 151st one must be either 2, 4, 6, 7, 8, etc etc, so there must be at least 2 that are consecutively numbered.

Example of a selection:

1, 2 , 3, 5, 7, 9, ..., 299 - then the consecutively numbered pairs would be 1, 2 and 2, 3.

The {1, 2}, {3, 4}...{299, 300} are not computer courses, rather pigeon-holes that we have constructed to make the problem easier to solve/understand.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 06:56:30 pm

wait so does that mean that all the numbers from 1 to 300 inclusive doesn't have to be used?

Coz I interpreted when the question said

Quote

we select 151 distinct computer science courses numbered between 1 and 300 inclusive

That all 300 numbers (1,2,3...300) must be used so some computer courses will have 2 or more numbers assigned to it.

Title: Re: TT's Maths Thread
Post by: brightsky on July 20, 2010, 07:01:51 pm

Nup, we are selected 151 courses FROM the 300 courses given. The courses are named 1, 2, 3, 4, 5, ..., 300, so if we randomly pick 151 courses from that, we will get consecutive pairs.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 07:06:29 pm

Yeah, so we have 300 course numbers (pigeons) and we have to assign them to 151 computer courses (pigeonholes) so we will get at least 2 pigeons in 1 pigeonhole, in other words, 2 course numbers for 1 computer course.

Sorry I'm a bit slow at pigeonhole theorem, was never good at it.

Title: Re: TT's Maths Thread
Post by: /0 on July 20, 2010, 07:24:32 pm

It might be clearer to try by contradiction:

Suppose there exists a list of 151 integers in the interval [1,300] in which all consecutive integers differ by at least 2.

(The "at least 2" can be changed to just "2": We aim to show that not all integers can differ by 2, and as a consequence they obviously can't differ by 3, since that would make the list occupy an even larger interval.)

But the "smallest" such list of numbers would be {1,3,5,7,9,...,301}, or {-1,1,3,5,...,299} and this cannot be contained in the interval [1,300].

So out of the 151 integers in [1,300] there must exist at least 1 pair of integers in which differ by less than 2. (taking the negation of the 'false' hypothesis at the start)

Title: Re: TT's Maths Thread
Post by: brightsky on July 20, 2010, 07:33:49 pm

Quote from: TrueTears on July 20, 2010, 07:06:29 pm

Yeah, so we have 300 course numbers (pigeons) and we have to assign them to 151 computer courses (pigeonholes) so we will get at least 2 pigeons in 1 pigeonhole, in other words, 2 course numbers for 1 computer course.

Sorry I'm a bit slow at pigeonhole theorem, was never good at it.

The pigeons are the 151 selected numbers. We are assigning the pigeons to pigeonholes constructed from the 300 courses to choose from.

This can be proved by way of pure logic. We have 151 courses to choose out of the 300 available courses. Of course there will be at least 2 pairs that are next to each other. Pigeon hole principle essentially looks at the worst case scenario, so if we were, for the first 150 selections, to avoid the consecutive pairings, then we have two choices, either to select 2, 4, 6, 8, 10, 12, ..., 300 or 1, 3, 5, 7, 9, ..., 299. In either case, we have only considered the first 150 selections. The 151st has to be one of the numbers remaining, and no matter what 151st number we pick, we will then get two numbers next to it, for example 1, 3, 5, 7, 9, 11, 12 , 13, 15, ... 299, etc. The 151st is FORCED into a position that mandates at least 2 consecutive pairings.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 07:42:37 pm

I see, that makes perfect sense, so there must be a fallacy in my logic in the my statement above, can you show me how?

Thanks heaps btw!

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 07:43:29 pm

A form of the pigeonhole theorem is stated as "If f is a function from a finite set X to a finite set Y with $|X| > |Y|$ then $f(x_1) = f(x_2)$ for some $x_1, x_2 \in X$ , $x_1 \neq x_2$

Now a question says: An inventory consists of a list of 89 items (Listed from 1 to 89), each marked "available" or "unavailable". There are 45 available items, show that there are at least two available items in the list exactly 9 items apart (For example items at position 13 and 22 satisfy the condition).

Now I just don't know how to apply the pigeonhole theorem to this question, what is set X and Y in this case? And what is the function?

Title: Re: TT's Maths Thread
Post by: brightsky on July 20, 2010, 08:07:42 pm

Hmmm...is it supposed to be 46 available items...

Consider the worst-case scenario. We have the list of items:

1, 2, 3, 4, 5, ..., 89

If 1 - 9 were available, then for the sake of a worst-case scenario, let 19 - 27 be available, then 37 - 45, then 55 - 63, then 73 - 81. (We have 45 available at present, with no pair of numbers exactly 9 apart). Notice that if we pick any number from the remaining numbers, we will get a pair that is exactly 9 apart. So if we have 46 items, then there will be at least two items exactly 9 items part.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 08:11:27 pm

Hmm says 45

The solutions do it like this:

Let X be the set of the positions of the available items and those items who are 9 items away from it.

Let a_i denote the position of the ith available item.

X = {a_1, a_2, ... , a_45, a_1+9, a_2+9, ... , a_45+9}

|X| = 90

Now let Y be the set of the 89 items.

Y = {1, 2, ... , 89}

|Y| = 89

Since |X| > |Y| Using the second form of the pigeonhole theorem (which is the form I stated above) we have a_i = a_j+9 for some i,j.

Thus a_i-a_j = 9 so 2 available items are 9 items away.

The solutions make perfect sense, and so does your way, so again where is the fallacy? I can't see it ><

Title: Re: TT's Maths Thread
Post by: pooshwaltzer on July 20, 2010, 08:34:03 pm

TT,

Your "fallacy" may be in the assumption of fixed permutation. Who's to say order of precedence was necessary?

E.G. For items A and B where both denote available; there are 2 outcomes

A,1,2,3,4,5,6,7,8,9,B

B,1,2,3,4,5,6,7,8,9,A

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 08:43:48 pm

Hmm I see, thanks for that, but I think I made the mistake of assuming the course numbers (from 1 to 300) are ONLY for computer science, rather they can be for other courses too, maybe commerce etc. So since the course numbers are not only for computer science, you can not place the course numbers (pigeons) into the 151 computer courses (pigeonholes) because the course numbers may not all be 'pigeons', eg, only the course numbers which are for computer science are pigeons, the rest are 'chickens' and you can stuff chickens into pigeonholes lol

does that even make sense...?

Title: Re: TT's Maths Thread
Post by: brightsky on July 20, 2010, 10:25:34 pm

Yeah I think that's one way of interpreting it. :p

So the question would be something like: there are 300 courses, numbered in order from 1 to 300, of which 151 are computer science courses. Show that there must be at least two pairs of computer science courses that are next to each other in the aforementioned order.

Generally speaking, the pigeons are the "selected" items you are dealing with, or rather, the things you are focussing on. If you have a question like: There are 21 points on a circle of radius 10, prove that there must be at least one pair of points that are blah blah blah. Then the pigeons would be the 21 points. If you have something like: Jimmy selected 60 numbered balls at random from a bag of 100 number balls, prove that blah blah blah, then the pigeons would be the 60 balls.

The pigeonholes on the other hand are constructed from the other information that you are given, and you need to construct your holes so that it "works".

Lol, hope this makes sense. :D

Title: Re: TT's Maths Thread
Post by: TrueTears on July 20, 2010, 10:29:03 pm

haha yeah i like your explanation, thanks i think i have a much better idea how to tackle these problems now they have always been one of my weak points, like what do i assign the pigeons/pigeonholes to lol.

thanks again :)

Title: Re: TT's Maths Thread
Post by: brightsky on July 20, 2010, 10:34:11 pm

No probem. :)

Title: Re: TT's Maths Thread
Post by: TrueTears on July 21, 2010, 04:35:38 am

Find a recursive relation for the number of distinct ordered pairs (a,b) of non-negative integers satisfying 2a+5b=100.

Okay, so I took an entirely different approach to this question using generating functions unlike my book.

Let $q_n$ be the number of nonnegative ordered pairs which solve $2a+5b = n$ .

We define $A(x) = 1+x^2+x^4+x^6+ \cdots$ and $B(x) = 1+x^5+x^{10}+x^{15} + \cdots$

Now $A(x)B(x) = \left(1+x^2+x^4+x^6+ \cdots\right)\left(1+x^5+x^{10}+x^{15} + \cdots\right) = 1+x^2+x^4+x^5+x^6 + \cdots$

Clearly, A(x)B(x) is the generating function sequence for the sequence $q_1, q_2, q_3, \cdots$

$A(x)B(x) = \left(\frac{1}{1-x^2}\right)\left(\frac{1}{1-x^5}\right) = q_0+q_1x+q_2x^2+q_3x^3+q_4x^4 + \cdots$

$\implies 1 = (1-x^2)(1-x^5)\left(q_0+q_1x+q_2x^2+q_3x^3+q_4x^4 + \cdots\right)$

We can see from inspection that $q_1=q_3 = 0$ and $q_2=q_4=q_5=q_6 = 1$

But up to this step, I have no idea how to get the recursive relation for $q_n$

Title: Re: TT's Maths Thread
Post by: Ahmad on July 21, 2010, 12:05:39 pm

Due to the specially chosen numbers it really is easier to just do it the 'standard' way.

But you asked for a recursion:

$1 = (1 - x^2 - x^5 + x^7) Q(x)$

$q_{n} - q_{n-2} - q_{n-5} + q_{n-7} = 0, n \ge 7$

Combinatorial interpretation: To make up n cents using 2 and 5 cents you can either choose 2 cents and count the number of ways of making n-2 cents, or 5 cents and count the number of ways of making n-5 cents, but then you'll have double counted if you chose 2 cents then 5 cents, or 5 cents then 2 cents, so you subtract the number of ways to make n-7 cents.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 21, 2010, 04:33:51 pm

Quote from: Ahmad on July 21, 2010, 12:05:39 pm

Due to the specially chosen numbers it really is easier to just do it the 'standard' way.

But you asked for a recursion:

$1 = (1 - x^2 - x^5 + x^7) Q(x)$

$q_{n} - q_{n-2} - q_{n-5} + q_{n-7} = 0, n \ge 7$

Combinatorial interpretation: To make up n cents using 2 and 5 cents you can either choose 2 cents and count the number of ways of making n-2 cents, or 5 cents and count the number of ways of making n-5 cents, but then you'll have double counted if you chose 2 cents then 5 cents, or 5 cents then 2 cents, so you subtract the number of ways to make n-7 cents.

Hey Ahmad, thanks for the help, I'm still unsure how you went from $1 = (1 - x^2 - x^5 + x^7) Q(x)$ to the recursive relation, can you explain?

Nvm, thanks I got it :)

Title: Re: TT's Maths Thread
Post by: TrueTears on July 22, 2010, 05:17:53 am

A domino is a rectangle divided into two squares with each square numbered one of 0,1,2,3,4,5,6. Two squares on a single domino can have the same number. Show that distinct dominoes can be arranged in a circle so that touching dominoes have adjacent squares with identical numbers.

This is how I interpreted this question. But the solution says something else and I don't know how they got it.

This is my graph

(http://img39.imageshack.us/img39/2699/graphtheoryquestion.jpg)

Now we can turn this into a graph, G, by letting the dominoes be the edges and each of the numbers be vertices.

Clearly, G is a connected graph and each vertex's degree is 2. Thus G has a Euler's cycle, therefore the dominoes can be arranged in a circle so that touching dominoes have adjacent squares with identical numbers.

Now the solutions says this: (My questions are in brackets)

"We model the situation as a graph G with seven vertices labeled 0,1,2,3,4,5,6. The edges represent the dominoes. There is one edge between each distinct pair of vertices (Why do you connect every single vertex to each other?) and there is one loop at each vertex (Why do you have a loop at each vertex?). Notice that G is connected. Now the dominoes can be arranged in a circle so that touching dominoes have adjacent squares with identical numbers iff G contains an Euler cycle (Why does G have to contain an Euler cycle so that adjacent squares will have identical numbers?). Since the degree of each vertex is 8, then G has an Euler cycle. Therefore, the dominoes can be arranged in a circle so that touching dominoes have adjacent squares with identical numbers."

Many thanks.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 22, 2010, 06:28:12 pm

I'm confused about the definition of a Euler cycle, my book says "a cycle in a graph G that includes all of the edges and all of the vertices of G is called an Euler cycle"

However on wikipedia it says "An Eulerian cycle, in an undirected graph is a cycle that uses each edge exactly once."

Which one is it?

If it is the latter, then is it possible to travel all the edges once and touch every vertex (not necessarily once) given the graph G has a Euler cycle?

Title: Re: TT's Maths Thread
Post by: pooshwaltzer on July 22, 2010, 06:47:08 pm

Quote from: TrueTears on July 22, 2010, 06:28:12 pm

Which one is it?

Definition: A path through a graph which starts and ends at the same vertex and includes every edge exactly once.

Read the sentences again TT, they are actually both correct, ie. stating the same thing but from different priors.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 22, 2010, 06:52:43 pm

But using each edge once doesn't mean you have to touch every vertex.

So saying "A cycle in a graph G that includes all of the edges and all of the vertices in G" is very different from saying "A cycle that uses each edge exactly once"

I illustrate in my graph attached.

It contains a cycle that uses each edge exactly once, namely (1,5,6,4,1,3,6,1), but doesn't touch the vertex 2.

Title: Re: TT's Maths Thread
Post by: pooshwaltzer on July 22, 2010, 07:09:36 pm

http://www.cs.sunysb.edu/~skiena/combinatorica/animations/euler.html

Title: Re: TT's Maths Thread
Post by: TrueTears on July 22, 2010, 07:13:19 pm

then how about my example?

Title: Re: TT's Maths Thread
Post by: pooshwaltzer on July 22, 2010, 07:25:00 pm

Quote from: TrueTears on July 22, 2010, 07:13:19 pm

then how about my example?

All I can say is, you need to (1) use ALL vertices at least ONCE and (2) include all the edges during tracement. Your example is not an EULER cycle by Konisberg's definition.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 22, 2010, 08:51:35 pm

But if you follow the wikipedia definition: "An Eulerian cycle, in an undirected graph is a cycle that uses each edge exactly once." my graph satisfies that criteria.

a) graph is undirectied. tick

b) it is a cycle. tick

c) Uses each edge exactly once. tick

But anyway I think I get it now.

If a graph has a Euler cycle (following my books definition that the graph is connected and every vertex has an even degree) then a cycle which includes all the edges exactly once implies that each vertex will be touched at least once.

By being connected ensures there is no isolated vertex and to prove that each vertex has to be touched at least once, since the graph is connected, by traversing each edge once you will arrive at a vertex and by traversing ALL edges ensures that you will have touched every single vertex.

Can anyone confirm this?

Title: Re: TT's Maths Thread
Post by: Ahmad on July 23, 2010, 02:43:35 pm

I wouldn't worry about it, the connected case is the interesting one.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 24, 2010, 09:44:12 pm

"If a connected, planar graph is drawn on a plane then the plane can be divided into continuous regions called faces. A face is characterised by the cycle that forms its boundary."

This is what my book says, and then it goes on to illustrate an example by saying this graph:

[IMG]http://img245.imageshack.us/img245/3914/planargraph.jpg[/img]

... has 4 faces, namely A, B, C, D (D is the face bounded by the cycle (1,2,3,4,6,1)

however shouldn't there be at least 6 faces? ie a face bounded by (1,2,5,4,6,1) and (1,2,3,4,5,1)...?

Title: Re: TT's Maths Thread
Post by: TrueTears on August 15, 2010, 06:48:30 pm

Find $\lim_{(x,y) \to (0,0)} \frac{3x^2y}{x^2+y^2}$ if it exists.

After testing different paths, the limit appears to be 0, so let's try an $\epsilon - \delta$ proof. I got a bit rusty at this, I just don't know how to finish it off lol.

We need to prove if for every $\epsilon > 0$ there exists a $\delta$ such that if $0 < \sqrt{x^2+y^2} < \delta$ then $\left|\frac{3x^2y}{x^2+y^2} - 0\right|< \epsilon$

We know that $x^2 \le x^2 + y^2 \implies 1 \le \frac{x^2+y^2}{x^2} \implies 1 \ge \frac{x^2}{x^2+y^2} \implies 3|y| \ge \frac{3|y|x^2}{x^2+y^2} \implies 3\sqrt{y^2} \ge \frac{3|y|x^2}{x^2+y^2}$

But $3\sqrt{x^2+y^2} \ge 3\sqrt{y^2} \implies \frac{3|y|x^2}{x^2+y^2} \le 3\sqrt{x^2+y^2}$

Thus we can take $\epsilon = 3\sqrt{x^2+y^2}$ but then how do we relate this back to $\delta$ again? I forgot lols

Title: Re: TT's Maths Thread
Post by: kamil9876 on August 15, 2010, 07:16:07 pm

That's really simple, remember that the distance from $(x,y)$ to $(0,0)$ is $\sqrt{x^2+y^2}$ . So you need to make sure that $|(x,y)-(0,0)|<\delta \implies |f(x,y)-0|<\epsilon$ . So if i believe the rest of that garbage u posted (ill assume u are right), it is enough to set $\delta=\frac{\epsilon}{3}$ .

Because look:

Quote from: TrueTears on August 15, 2010, 06:48:30 pm

$\frac{3|y|x^2}{x^2+y^2} \le 3\sqrt{x^2+y^2}$

I'll assume this is true, i cbf checking if u derived it properly. Now the interesting thing that this is saying is that:

$0\leq |f(x,y)| \leq 3d$

Where $d$ is the distance between $(x,y)$ and $(0,0)$ .

So indeed if $d<\frac{\epsilon}{3}$ then $0 \leq |f(x,y)| \leq \epsilon$ as desired.

Btw: it is more economical to do it like this:

$0 \leq |f(x,y)| \leq |\frac{3x^2y}{x^2}|$ and apply the sandwhich theorem (or if u have real fetish for e-d, you can easily translate it to that if u UNDERSTAND it properly, i'm sick of how ppl treat e-d as some alien symbolic manipulation)

Title: Re: TT's Maths Thread
Post by: TrueTears on August 15, 2010, 08:04:05 pm

lololol ah yeah, i get it now, thanks heaps kamil

Title: Re: TT's Maths Thread
Post by: TrueTears on August 26, 2010, 07:00:32 pm

The total bill for a house is $260.65 over a 93 day period.

6 people live in this house.

F has lived 93 days (from the first day till the 93rd day which is 20/08)

L, Z, J all 3 of them has lived 30 days they all moved in from 21/07 to 20/08

W has lived 32 days moved in from 19/07 to 20/08

WG has lived 10 days 19/07 to 29/07

How much should they pay respectively? Obviously the person living longer should pay a bigger proportion.

Title: Re: TT's Maths Thread
Post by: Mao on August 26, 2010, 11:15:30 pm

Total people-day: F: 93 person-day, L+Z+J: 90 person-day, W+WG: 42 person-day, total: 225 person-day

Total bill is 260.65, thus each person-day is 260.65/224 = s

F pays 93*s, L pays 30*s and so on. [which is how I work my bills.. if that helps?]

Title: Re: TT's Maths Thread
Post by: pooshwaltzer on August 26, 2010, 11:32:08 pm

Ever had a landlord charge on the basis of a different cost driver? That is, one not based on "person-days"?

I've encountered,

* Time in apartment (down to the minute) based on facility entry cards
* floor area of let room (sq ft) excluding common areas
* Use historical cost info to work out average bill expenditure per month and charge as overhead plus inflation premium
* Divide bill by number of tenants as LL failed grade 3 math.
* Just charge a weekly flat rate irrespective of actual use; LL never attended primary.

Title: Re: TT's Maths Thread
Post by: TrueTears on September 13, 2010, 08:09:40 pm

For the change of variables in a double integral involving the Jacobian ie:

$\int \int_R f(x,y) dA = \int \int_S f(x(u,v), y(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v)}\right| dudv$

Do we strictly have to evaluate $du$ BEFORE THE $dv$ ? Or can we also do it like this:

$\int \int_S f(x(u,v), y(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v)}\right| dvdu$

Ofcourse after taken into consideration changing the terminals.

I ask this question because sometimes evaluating with respect to $u$ first yields an non elementary integral so reversing the order of the $du$ and $dv$ and changing the respective terminals works much better, but I'm not sure if the general Fubini's Theorem here applies to transformed double integrals.

Thanks.

Title: Re: TT's Maths Thread
Post by: Mao on September 13, 2010, 10:34:10 pm

That really depends on what S is.

I'm fairly sure you can prove that

$\int_{u_0}^{u_1}\int_{v_0}^{v_1}g(u,v) \; dv\; du = \int_{v_0}^{v_1}\int_{u_0}^{u_1}g(u,v) \; du\; dv$

But of course, if the boundary conditions of a pair of u or v are functions of the other variable, then they'd have to be the inner integral.

Title: Re: TT's Maths Thread
Post by: TrueTears on September 14, 2010, 12:21:51 am

Ah okay thanks Mao, yeah I'm actually not that interested on a proof here, but I was wondering if Fubini's Theorem still applies after a Jacobian transformation.

Title: Re: TT's Maths Thread
Post by: Mao on September 14, 2010, 01:18:29 am

I'm fairly sure it will. The Jacobian is just a function of u and v, so it shouldn't affect anything.

Title: Re: TT's Maths Thread
Post by: TrueTears on September 14, 2010, 01:21:12 am

Thanks Mao :smitten:

Title: Re: TT's Maths Thread
Post by: TrueTears on September 24, 2010, 02:08:28 am

Um is the answer wrong for this question? I can't seem to get the right answer... it is off by a factor of 2 grrr anyways:

Evaluate $\int \int \int_E x^2 dV$ where $E$ is bounded by the xz-plane and the hemispheres $y = \sqrt{9-x^2-z^2}$ and $y = \sqrt{16-x^2-z^2}$ .

So my solution is this:

Using spherical coordinates, we have $3 \le \rho \le 4, 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2}$

Thus $\int \int \int_E x^2 dV = \int_{0}^{\frac{\pi}{2}} \int_0^{\pi} \int_3^4 (\rho \sin(\phi)\cos(\theta))^2\rho^2\sin(\phi) d\rho d\theta d\phi = \frac{781\pi}{15}$

But the answer is $\frac{1562\pi}{15}$ hmmm how did they get this?

Title: Re: TT's Maths Thread
Post by: Cthulhu on September 24, 2010, 02:24:03 am

Quote from: TrueTears on September 24, 2010, 02:08:28 am

Um is the answer wrong for this question? I can't seem to get the right answer... it is off by a factor of 2 grrr anyways:

Evaluate $\int \int \int_E x^2 dV$ where $E$ is bounded by the xz-plane and the hemispheres $y = \sqrt{9-x^2-z^2}$ and $y = \sqrt{16-x^2-z^2}$ .

So my solution is this:

Using spherical coordinates, we have $3 \le \rho \le 4, 0 \le \theta \le \pi, 0 \le \phi \le \frac{\pi}{2}$

Thus $\int \int \int_E x^2 dV = \int_{0}^{\frac{\pi}{2}} \int_0^{\pi} \int_3^4 (\rho \sin(\phi)\cos(\theta))^2\rho^2\sin(\phi) d\rho d\theta d\phi = \frac{781\pi}{15}$

But the answer is $\frac{1562\pi}{15}$ hmmm how did they get this?

I remember doing this(pretty sure it was) question last semester... I got it wrong too. My tutor showed me what I did wrong and its in a notebook somewhere but I can't remember... sorry.

Title: Re: TT's Maths Thread
Post by: TrueTears on September 24, 2010, 02:25:45 am

~~haha, damnnnn i swear i'm checking every minuscule detail and i can't find anything wrong!~~

NOOOOOOOOO omg what a stupid mistake... sketched it wrong lmao, $\phi$ should range from 0 to $\pi$ , because y is the axis of symmetry! NOT Z! SIGH I'M TOO USED TO SEEING z = f(x,y) instead of y = f(z,x)!

Title: Re: TT's Maths Thread
Post by: TrueTears on October 16, 2010, 11:20:25 pm

Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

Title: Re: TT's Maths Thread
Post by: kamil9876 on October 17, 2010, 12:31:20 am

Let $p_n$ be the answer. Obviously $n\leq99$ otherwise this problem is no fun.

In order for him to go broke, he must eventually be at $n-1$ at some point. At that point his probability of going broke is $p_{n-1}$ . So now using this you can easily show that $p_n=q_np_{n-1}$ where $q_n$ is the probability of him eventually being at n-1. Now you must find $q_n$ . I guess the best way of doing this is again be a recursion, since $q_{99}$ is easy to find. And finding $q_n$ in terms of $q_{n+1}$ is also easy (ie it is like a "downwards" recursion).

Title: Re: TT's Maths Thread
Post by: Mao on October 17, 2010, 12:39:46 am

Quote from: TrueTears on October 16, 2010, 11:20:25 pm

Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

Not helping, but I remember doing this when I was trying to get a strategy for winning roulette. Not sure about doing it analytically, but I ran a simulation 1,000,000 times and approximated the long-term probability :P

Title: Re: TT's Maths Thread
Post by: Cthulhu on October 17, 2010, 12:41:06 am

Quote from: Mao on October 17, 2010, 12:39:46 am

Quote from: TrueTears on October 16, 2010, 11:20:25 pm
Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

Not helping, but I remember doing this when I was trying to get a strategy for winning roulette. Not sure about doing it analytically, but I ran a simulation 1,000,000 times and approximated the long-term probability :P

Link to arxiv or published paper or it didn't happen.

Title: Re: TT's Maths Thread
Post by: TrueTears on October 20, 2010, 04:27:09 am

The Fibonnaci numbers are as F(k+1) = F(k)+F(k-1) with initial values F(0) = F(1) = 1

Show that $F(k) \ge 2^{\frac{k}{2}}$ if k is even

How do I use induction to show this? If we use strong induction on k and assume $F(k) \ge 2^{\frac{k}{2}}$ is true for 0,2,4,...,k then we must show $F(k+2) \ge 2^{\frac{k+2}{2}} = 2^{\frac{k}{2}} \cdot 2$

But F(k+2) = F(k)+F(k+1)

We have information from our inductive hypothesis about F(k) but know nothing about F(k+1)... this is where I'm stuck.

Title: Re: TT's Maths Thread
Post by: /0 on October 20, 2010, 04:57:53 am

Let $k=2n$

The statement is equal to:

$F(2n) \geq 2^{n}$ for $n \geq 1$ .

There is another fibonacci identity that may be useful: $F(2n)=F(1)+F(3)+...+F(2n-1)+1$ (source: wikipedia lol)

$F(2(n+1))=F(1)+F(3)+...+F(2n+1)+1 \geq 2^n+F(2n+1)=2^n+F(2n)+F(2n-1)\geq 2^{n+1}$

Title: Re: TT's Maths Thread
Post by: TrueTears on October 20, 2010, 05:06:56 am

Ahh I see thanks for that, is there another way of doing it without knowing the identity... cause this is from an exam :P

If we were to show $F(2n) \ge 2^n$ for $n \ge 1$

then we'd need to show F(2(n+1)) = F(2n+2) = F(2n+1)+F(2n) to be true, but we don't have any information on F(2n+1)...

Title: Re: TT's Maths Thread
Post by: /0 on October 20, 2010, 05:12:22 am

Hmmm... oh right this might work...

$f(2(n+1))=f(2n+1)+f(2n)$

I wanna get rid of that odd f(2n+1) so I keep expanding that:

$f(2n+1)=f(2n)+f(2n-1)=f(2n)+f(2n-2)+f(2n-3)+...$

Actually I can just stop at the first step: $f(2n+1)=f(2n)+f(2n-1)$ . Then I have: $f(2(n+1))=2f(2n)+f(2n-1)\geq 2*2^n+f(2n-1)\geq 2^{n+1}$

no need for obscure identities xD

Title: Re: TT's Maths Thread
Post by: TrueTears on October 20, 2010, 05:15:16 am

oh em gee that is smart!!!!

thanks bro

Title: Re: TT's Maths Thread
Post by: kamil9876 on October 21, 2010, 09:52:27 pm

$f(2(n+1))=f(2n+1)+f(2n)>f(2n)+f(2n)=2^n + 2^n=2^{n+1}$ is an easy induction step.

Title: Re: TT's Maths Thread
Post by: Mao on October 21, 2010, 11:49:52 pm

Quote from: Cthulhu on October 17, 2010, 12:41:06 am

Quote from: Mao on October 17, 2010, 12:39:46 am
Quote from: TrueTears on October 16, 2010, 11:20:25 pm
Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

Not helping, but I remember doing this when I was trying to get a strategy for winning roulette. Not sure about doing it analytically, but I ran a simulation 1,000,000 times and approximated the long-term probability :P
Link to arxiv or published paper or it didn't happen.

sif publish that. did you watch 21? people get beaten up for that shit :P

Title: Re: TT's Maths Thread
Post by: TrueTears on October 24, 2010, 08:37:25 pm

Are the following true in general?

If a vector field F can be written into 2 parts, F_1 being conservative and F_2 is non conversative then:

$Curl \ \mathbb{F} = Curl \ \mathbb{F}_1 + Curl \ \mathbb{F}_2$

Title: Re: TT's Maths Thread
Post by: TrueTears on October 24, 2010, 08:43:15 pm

Also with the same initial conditions on F, are these true?

$\int_C \mathbf{F}.d\mathbf{r} = \int_C \mathbf{F}_1.d\mathbf{r} + \int_C \mathbf{F}_2.d\mathbf{r}$

Title: Re: TT's Maths Thread
Post by: /0 on October 24, 2010, 08:44:54 pm

You mean if $\mathbf{F}=\mathbf{F}_1+\mathbf{F}_2$ ? Then yep that's true (just from distributivity of the curl operator), and also $\nabla \times \mathbf{F_1}=0$ , so $\nabla \times \mathbf{F}=\nabla \times \mathbf{F_2}$

And also as far as I know ur next post should be true too (distributivity of the integral operator)

Title: Re: TT's Maths Thread
Post by: TrueTears on October 24, 2010, 08:46:57 pm

Thanks man, lol yeah i forgot the latex for mathbf

Title: Re: TT's Maths Thread
Post by: TrueTears on October 24, 2010, 09:00:13 pm

wth is up with this Q?

isn't the answer 0?

Title: Re: TT's Maths Thread
Post by: /0 on October 25, 2010, 02:04:42 am

Hmm I get $\frac{\pi}{4}$

At $z=0$ , $\mathbf{F}=\frac{1}{3}y^3\mathbf{j}+y^2\mathbf{k}$

The surface normal pointing upwards is just $\mathbf{k}$ , so $\mathbf{F}\cdot \mathbf{k}=y^2$

So essentially you're integrating $y^2$ over the disc $x^2+y^2=1$ , probably a good idea to switch to polar coordinates.

Title: Re: TT's Maths Thread
Post by: TrueTears on October 27, 2010, 03:51:28 pm

Quote from: /0 on October 25, 2010, 02:04:42 am

Hmm I get $\frac{\pi}{4}$

At $z=0$ , $\mathbf{F}=\frac{1}{3}y^3\mathbf{j}+y^2\mathbf{k}$

The surface normal pointing upwards is just $\mathbf{k}$ , so $\mathbf{F}\cdot \mathbf{k}=y^2$

So essentially you're integrating $y^2$ over the disc $x^2+y^2=1$ , probably a good idea to switch to polar coordinates.

Oh that's right, opps I thought it was the surface integral of a scalar function haha, silly me, thanks /0!!!

(Trust you to be good with this kinda maths :P)

Title: Re: TT's Maths Thread
Post by: TrueTears on November 01, 2010, 05:17:49 pm

Any help with this?

Thanks :D

Actually nevermind, I got it, but I was still wondering how do you approach these questions generally? Do you need to visualise what solid region you are integrating and then set the spherical coordinates?

Title: Re: TT's Maths Thread
Post by: Mao on November 01, 2010, 05:42:51 pm

pretty much, yeah

Or you can treat it like a jacobian transformation, do a few simultaneous eqns to find the boundaries. This method is longer, but more general.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 01, 2010, 08:08:15 pm

Thanks Mao!

Another one...

[IMG]http://img408.imageshack.us/img408/4751/mth2010examhelp.jpg[/img]

I got part a) b) c) however I am just unsure on part d)

For a) I got $r(\theta, \phi) = \sin(\phi)\cos(\theta)i + \sin(\phi)\sin(\theta)j+\cos(\phi)k$

b) $r(x,y) = xi+yj+zk \ \text{but} \ z^2+y^2+x^2 = 1 \ \text{so} \ z = \sqrt{1-(x^2+y^2)}$

So $r(x,y) = xi+yj+\sqrt{1-(x^2+y^2)}k$

c) Not required for d)

d) So area of surface $= \int \int_D |r_x \times r_y| dA$

$|r_x \times r_y| = \frac{1}{\sqrt{1-x^2-y^2}}$ (I've double checked this, it's correct :))

So $\int \int_D \frac{1}{\sqrt{1-x^2-y^2}} dA$

But what are the bounds for the double integral?

Are we allowed to convert to polar coordinates? Cause I'm interpreting the question as set up the integral purely in cartesian axis, if it's in polar coordinates then it's quite easy and becomes trivial.

Thanks!

Title: Re: TT's Maths Thread
Post by: /0 on November 01, 2010, 08:54:50 pm

$\frac{\pi}{6} \leq \phi \leq \frac{\pi}{3}$

$\Rightarrow \frac{1}{2} \leq z \leq \frac{\sqrt{3}}{2}$

$\Rightarrow \frac{1}{4} \leq x^2+y^2 \leq \frac{3}{4}$

I think... this is the annulus you might have to integrate over. If this is right it might require the sum of two integrals to get the whole area

e.g. $\int_{-1}^1 \int_{-\sqrt{(\frac{3}{4})^2-y^2}}^{\sqrt{(\frac{3}{4})^2-y^2}}....... dxdy-\int_{-1}^1 \int_{-\sqrt{(\frac{1}{4})^2-y^2}}^{\sqrt{(\frac{1}{4})^2-y^2}}....... dxdy$

But yeah polar coordinates would be nice

Title: Re: TT's Maths Thread
Post by: TrueTears on November 01, 2010, 08:59:10 pm

yup I know that, it's between 2 circles right? So without converting to polar coordinates it's incredibly hard to write the integrals, you need like 4....

however with polar its so easy?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 01, 2010, 10:17:24 pm

yeah the question was probably designed to test your parametrizing skills more than anything.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 01, 2010, 10:27:39 pm

thought so... thanks heaps guys!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 03, 2010, 04:13:13 pm

1. A sphere with radius 2 ( $x^2+y^2+z^2=4$ ) has its cylindrical core of radius 1 removed ( $x^2+y^2=1$ ), what is the volume of the resultant solid?

a) Use cylindrical coordinates to find the volume of the resultant solid.
b) Use spherical coordinates to find the volume of the resultant solid.

For a) I just used cylindrical coordinates to find the volume of the cylinder, then used $V = \frac{4}{3}\pi r^2$ which is the volume of the sphere then minus the volume of the cylinder from it.

For b) I used spherical coordinates to find the volume of the sphere, then used $V = \pi r^2 h$ which is the volume of the cylinder, then used the volume of sphere minus the volume of cylinder.

Now what I am wondering is... is that the right way to interpret the question? Or does the question mean use cylindrical coordinates ONLY to work out the volume? If so... how do I do that? Cause I thought as long as I used cylindrical coordinates to find the volume that's fine.

Same goes for b)

2. $\mathbf{F} = (x+y)i+(y+z)j+(z+x)k$ . Verify the Divergence Theorem by showing $\int \int \int_E \Delta . \mathbf{F} dV = \int \int_S \mathbf{F}.\mathbf{n} dS$ where S is the union of the surfaces $z=0$ and $z = 4-x^2-y^2$ and E is the solid encompassed by those 2 surfaces.

Now I can evaluate the LHS which in spherical coordinates is given by: $\int_0^{\frac{\pi}{2}} \int_0^{2\pi} \int_0^4 3p^2\sin(\phi)dp d\theta d\phi = 128\pi$ (btw is this triple integral right? As in did I set it up right, not the answer :P)

Now how do I evaluate the RHS for the surface integral over the surface $z = 4-x^2-y^2$ (without a calculator)... it's almost impossible to do by hand after you compute $\mathbf{F}(\mathbf{r}(u,v)).(\mathbf{r}_u \times \mathbf{r}_v)$ where u and v are x and y respectively.

3. Evaluate $\int_1^2 \int_{\frac{1}{y}}^{\sqrt{y}} e^{\sqrt{xy}}dxdy$ by making the transformation $x = \frac{u}{v}$ and $y = uv$

I swear you can NOT do this question without a CAS.

After sketching on the u-v plane (with u as the vertical axis) we get the region to be the region between $u = v^3, u = \frac{2}{v}$ and $u = 1$ (I've checked this many times)

Now the Jacobian is $\frac{2u}{v}$ and after making the transformations we can evaluate the integral in the u-v plane with: $\int_1^{2^{\frac{3}{4}}} \int_{u^{\frac{1}{3}}}^{\frac{2}{u}} 2ue^u dvdu$

Now it's impossible to integrate that without a calculator.

Then we can try reverse the order of the integral so we get a dudv instead of dvdu, it's also unable to be integrated by hand (try it yourself). So is there ANYWAY to do this question by hand?

Title: Re: TT's Maths Thread
Post by: /0 on November 03, 2010, 06:30:18 pm

For 1, how did you use $V=\pi r^2h$ for the cylinder, since it doesn't have constant height? Did you use it with another integral? If you were to use spherical coords to find the volume of the cylinder it would go like:

$V=\underbrace{\pi (1)^2(2\sqrt{3})}_{\mbox{volume of `cylindrical prism' bit}}+2\underbrace{\int_0^1\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{6}} r^2\sin\theta\,d\theta\,d\phi\,dr}_{\mbox{volume of caps}}$ , where $\phi$ is the angle with the x-axis and $\theta$ the angle with the z-axis.

With 2, I'm not sure if you can use spherical coordinates, since it's a parabola... if you use cylindrical then you have

$\int_{0}^2\int_0^{4-s^2}\int_{0}^{2\pi} 3s\,d\phi\,dz\,ds$ ( $s=x^2+y^2$ )

Or in cartesian,

$\int_{-2}^2\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}\int_0^{4-x^2-y^2} 3\,dz\,dx\,dy$

The surface is:

$\mathbf{r}=x\mathbf{i}+y\mathbf{j}+(4-x^2-y^2)\mathbf{k}$

$\mathbf{r}_x=\mathbf{i}-2x\mathbf{k}$

$\mathbf{r}_y=\mathbf{j}-2y\mathbf{k}$

$\mathbf{r}_x\times \mathbf{r}_y=2x\mathbf{i}+2y\mathbf{j}+\mathbf{k}$

$\mathbf{F}\cdot(\mathbf{r}_x\times\mathbf{r}_y)=2x(x+y)+2y(y+z)+z+x$

So the integral is

$\int_{-2}^2\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} 2x(x+y)+2y(y+4-x^2-y^2)+4-x^2-y^2+x\,dx\,dy$

This is horrible... lol

But switching to polar coordinates may help again:

$\int_0^2\int_0^{2\pi} (2s\cos{\phi}(s\cos\phi+s\sin\phi)+2s\sin{\phi}(s\sin\phi+4-s^2)+4-s^2+s\cos\phi)s\,d\phi\,ds=\int_0^2\int_0^{2\pi} (s^2+2s^2\cos\phi\sin\phi+8s\sin\phi-2s^3\sin\phi+4+s\cos\phi)s\,d\phi\,ds$

Now $\int_0^{2\pi}\sin\phi\,d\phi=\int_0^{2\pi}\cos\phi\,d\phi=\int_0^{2\pi}\cos{\phi}\sin\phi\,d\phi=0$ , so you're left with the slightly more manageable

$\int_0^2\int_0^{2\pi} s^3+4s\,d\phi\,ds=24\pi$

Still it's absurd they'd ask you to do that in a non-calc exam... I had to go back and fix my working on this question so many times, and I had a calculator -.- lol

I'm still stuck with the 3rd one though... like u said i can't find an antiderivative for it

Title: Re: TT's Maths Thread
Post by: TrueTears on November 03, 2010, 06:48:28 pm

Thanks heaps dude, hmm yeah I thought you couldn't use spherical since it was a parabola... I had the cylindrical coordinates too lol I wrote both...

then I spent like 20 mins on the RHS which like you said is insane without a calculator ><

And yeah I'm pretty sure the third one is a mistake in the exam, if you changed the terminal from rt[y] to y then it works out nicely with the rt[y] i don't think you can find an antiderivative.

also there were so many mistakes in the exam, had to tell the invigilators to fix em sigh... so pissed

The cylinder does have a constant height doesn't it?

I interpreted the cylindrical core as in a cylinder as the "core" of the sphere, which intersects the sphere on the planes $z = -\sqrt{3}$ and $\sqrt{3}$

so the height is just $2\sqrt{3}$

so eg, for b) $\int \int \int_E 1 dV$ in spherical coordinates, where E is the solid of a sphere with centre (0,0,0) and radius 2, then minus the $V=\pi r^2h$

Title: Re: TT's Maths Thread
Post by: /0 on November 03, 2010, 06:59:20 pm

Hmmm I guess there are different interpretations of the questions... sounds like the exam wasn't well written :/

My waves and optics exam on monday will probably be similar, sigh... -.-

Title: Re: TT's Maths Thread
Post by: TrueTears on November 03, 2010, 07:00:00 pm

Quote from: /0 on November 03, 2010, 06:59:20 pm

Hmmm I guess there are different interpretations of the questions... sounds like the exam wasn't well written :/

My waves and optics exam on monday will probably be similar, sigh... -.-

how did you interpret it? can u kinda sketch it? LOL but do u get my interpretation?

Title: Re: TT's Maths Thread
Post by: /0 on November 03, 2010, 07:05:02 pm

Hmmm I interpreted it like this:

(http://img825.imageshack.us/img825/3032/cyl.png)

But I get your interpretation as well, since it isn't really clear what they mean by 'core'

Title: Re: TT's Maths Thread
Post by: TrueTears on November 03, 2010, 07:35:46 pm

Ahh yeah I thought about it like that too in the exam, but I still stuck with my current interpretation, sigh, they really need a picture to show what they really want, or else there are too many interpretations... the reason why I rejected how you interpreted in the exam was because see the blue shaded part? the "cap" doesnt make the core a cylinder anymore technically right?

Title: Re: TT's Maths Thread
Post by: /0 on November 03, 2010, 08:48:17 pm

Quote from: TrueTears on November 03, 2010, 07:35:46 pm

the "cap" doesnt make the core a cylinder anymore technically right?

good point

Title: Re: TT's Maths Thread
Post by: TrueTears on November 03, 2010, 09:17:48 pm

yeah stupid exam, sigh... lol oh wells

gl for your exams man, how many have you done and how many left??

Thanks for all your help again man, really appreciate it :smitten:

Title: Re: TT's Maths Thread
Post by: /0 on November 03, 2010, 09:52:21 pm

lol gl for the rest of your exams too
for me the pain is yet to come, I've got exams on monday 8th, then 16th, 17th, and 18th...

I've been really lazy so it's gonna be cram central :D

Title: Re: TT's Maths Thread
Post by: TrueTears on November 03, 2010, 11:57:14 pm

haha thanks :)

oh I see, I finish next week on the 9th can't wait :D Only got commerce exams left, all my maths exams are done! (I loved my pure maths exam btw, my first ever pure maths exam <3)

then it's happy-fun-maths-time as soon as exams end!!

are you coming back to Melbourne for holidays after exams end? we gotta catch up :(

Title: Re: TT's Maths Thread
Post by: jimmy999 on November 10, 2010, 10:43:27 pm

Quote from: TrueTears on November 03, 2010, 04:13:13 pm

$\int_1^{2^{\frac{3}{4}}} \int_{u^{\frac{1}{3}}}^{\frac{2}{u}} 2ue^u dvdu$

Now it's impossible to integrate that without a calculator.

Then we can try reverse the order of the integral so we get a dudv instead of dvdu, it's also unable to be integrated by hand (try it yourself). So is there ANYWAY to do this question by hand?

Chuck this integral into Wolfram Alpha without the limits of the u variable. You'll find it that it gives you the gamma function instead of an exact antiderivative. When doing this question, I got as far as I could, then said that part of the integral couldn't be solved using elementary functions. In all fairness, they have to disregard the final bit of that question when marking the exam as it's not doable without a calculator.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 11, 2010, 01:44:45 pm

Yeah, I'm really pissed at the exam because I spent like 10-15 mins on that question thinking wdf, it said "EVALUATE the integral" meaning you should be able to get an exact answer and you can't do it without a CAS. Wasted so much time.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2010, 01:06:19 pm

Show that if A and B are invertible matrices of the same size, then AB is invertible and $(AB)^{-1} = B^{-1}A^{-1}$

How do I prove this?

I was thinking that if we can show $(B^{-1}A^{-1})(AB) = (AB)(B^{-1}A^{-1}) = I$ then we are done... but how?

Title: Re: TT's Maths Thread
Post by: m@tty on November 15, 2010, 01:12:21 pm

Quote from: TrueTears on November 15, 2010, 01:06:19 pm

Show that if A and B are invertible matrices of the same size, then AB is invertible and $(AB)^{-1} = B^{-1}A^{-1}$

How do I prove this?

I was thinking that if we can show $(B^{-1}A^{-1})(AB) = (AB)(B^{-1}A^{-1}) = I$ then we are done... but how?

$(AB)^{-1}(AB)=B^{-1}A^{-1}AB=B^{-1}IB=B^{-1}B=I$

It's invertible because it is square.

And as seen above $(AB)^{-1}$ must equal $B^{-1}A^{-1}$ because otherwise you won't get down to I.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 15, 2010, 01:23:05 pm

thought so but here you're assuming $(AB)^{-1} = B^{-1}A^{-1}$ doesn't that mean you also have to prove uniqueness too? Or are you assuming uniqueness too :P

Title: Re: TT's Maths Thread
Post by: m@tty on November 15, 2010, 01:26:18 pm

lol I don't know. This is only what I remember from the start of the year in uni maths.

Title: Re: TT's Maths Thread
Post by: /0 on November 15, 2010, 03:27:42 pm

Proof by example should work just fine. Alternatively,

$(AB)^{-1}(AB)=I$

Right multiplying by $B^{-1}$ :

$(AB)^{-1}A=B^{-1}$

Right multiplying by $A^{-1}$

$(AB)^{-1}=B^{-1}A^{-1}$

(Repeat for right inverse)

By construction, $AB$ is invertible. Alternatively, it is invertible because $\mbox{det}(AB)=\mbox{det}(A)\mbox{det}(B)\neq 0$

Uniqueness can be proven by assuming there are two inverses: $X$ , $Y$

If $X(AB)=I$ and $Y(AB)=I$ then $X=B^{-1}A^{-1}$ and $Y=B^{-1}A^{-1}$ so $X=Y$ .

(Repeat for right inverse)

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 15, 2010, 08:00:16 pm

Well your guess is good. Remember the defn of an inverse of $A$ is a matrix $X$ such that $XA=AX=I$ . Using the associativity of matrix multiplication, you can show that the $X$ you guessed works and hence is an inverse => the inverse by uniquness. ie it isn't good to write $(AB)^{-1}$ if you don't know if it exists, maybe that's what TT was worried about.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 16, 2010, 01:36:11 pm

Ahh yup cool thanks for filling in the last loophole, thanks guys :P

Title: Re: TT's Maths Thread
Post by: itolduso on November 16, 2010, 02:29:19 pm

Quote from: TrueTears on November 15, 2010, 01:06:19 pm

Show that if A and B are invertible matrices of the same size, then AB is invertible and $(AB)^{-1} = B^{-1}A^{-1}$

How do I prove this?

I was thinking that if we can show $(B^{-1}A^{-1})(AB) = (AB)(B^{-1}A^{-1}) = I$ then we are done... but how?

ABX=Y, BX=A^-1(Y), X=B^-1(A^-1(Y))=(B^-1A^-1)Y
.: (AB)^-1 exists and (AB)^-1=B^-1A^-1

Title: Re: TT's Maths Thread
Post by: TrueTears on November 17, 2010, 08:13:27 pm

I don't get this example, why does $b_3-b_2-b_1 = 0$ ? If it's reduced to row-echelon form, can't $b_3-b_2-b_1 = 1$ also satisfy the condition of the augmented matrix being in row echelon form?

(http://img221.imageshack.us/img221/4160/linearalgebra.jpg)

Title: Re: TT's Maths Thread
Post by: /0 on November 17, 2010, 08:33:33 pm

But if $b_3-b_2-b_1=1$ won't the last row be 0+0+0=1?

Title: Re: TT's Maths Thread
Post by: TrueTears on November 17, 2010, 09:50:47 pm

o i see haha, true that

Title: Re: TT's Maths Thread
Post by: TrueTears on November 19, 2010, 12:09:05 am

Prove that for the 3 x 3 matrix, $\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$ , $a_{11}C_{31}+a_{12}C_{32}+a_{13}C_{33} = 0$ where $C_{ij}$ denotes the cofactor entry of $a_{ij}$

Any ideas?

I've think I've heard of a more general result, where if one multiplies the entries in any row (or column) of any matrix by the corresponding cofactors from a different row (or column), the sum of these products is always zero. I don't know the proof of this result and hopefully will know in the near future, but I suspect this question is a more specific case?

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 19, 2010, 12:31:57 am

Looks like just an algebraic manipulation exercise. If the generalisation is indeed true, it looks like something that can be proven by induction (as is the case with a lot of things when it comes to determinants).

Title: Re: TT's Maths Thread
Post by: TrueTears on November 19, 2010, 12:37:49 am

mind showing me ? :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 19, 2010, 01:05:05 am

actually, I just thought of a really elegant solution! but i will post it 2moro as i am going to sleep now. The idea is that that sum calculates the determinant of a matrix with two equal rows (and we know such matrices have 0 determinants). In our case the matrix is that where you replace the last row with the first row. (though I think there may be some sign change or something going on, need to sort out those details.

Title: Re: TT's Maths Thread
Post by: Ahmad on November 19, 2010, 01:46:52 am

Kamil's right if you copy and paste the first row of the matrix to replace the third row then it's just computing the determinant using cofactor expansion on the 3rd row.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 20, 2010, 03:01:00 pm

Quote from: kamil9876 on November 19, 2010, 01:05:05 am

actually, I just thought of a really elegant solution! but i will post it 2moro as i am going to sleep now. The idea is that that sum calculates the determinant of a matrix with two equal rows (and we know such matrices have 0 determinants). In our case the matrix is that where you replace the last row with the first row. (though I think there may be some sign change or something going on, need to sort out those details.

Quote from: Ahmad on November 19, 2010, 01:46:52 am

Kamil's right if you copy and paste the first row of the matrix to replace the third row then it's just computing the determinant using cofactor expansion on the 3rd row.

Oh yes that's right, very smart, thanks guys!

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2010, 12:03:31 am

(http://img826.imageshack.us/img826/4160/linearalgebra.jpg)

Anyways so during my maths reading today I came across this, just have a few questions:

Firstly, I don't get how $p(x)=a_3(x-x_0)^3 + a_2(x-x_0)^2+a_1(x-x_0)+a_0$ is equivalent to (2)? Isn't it horizontally translated...? And furthermore, how is (3) equivalent to (2)...?

Thanks guys.

Title: Re: TT's Maths Thread
Post by: /0 on November 26, 2010, 12:13:03 am

By equivalent they mean they are all cubic polynomials with constant cofficients. If you expand (3) out you will get something similar to the first equation, and if you 'absorb' the constants together it will look just like the first equation, e.g. $-b_1x_0+b_2=a_0$ , where $a_0$ is from the first equation.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 26, 2010, 12:17:59 am

oh so THAT'S what they mean, sigh, thanks man lol

Title: Re: TT's Maths Thread
Post by: onur369 on November 26, 2010, 10:44:29 am

Anyone can find me the eBook for: Essentials Further Mathematics? Please, PM me if you can.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 29, 2010, 02:33:41 am

So I was just reading and I came across this in my book, this isn't a question from the book or whatever it's just that something that crossed my mind, are these the ONLY subspaces for R^2 and R^3? If so how do I prove uniqueness, I know how to prove they are subspaces but how can i show they are the ONLY ones? (if what I hypothesized was true) if not, then do there exist infinite subspaces for every R^n? Or are subspaces finite for all R^n? haha all of these are just random questions from the top of my head, maybe i'll find the answers to them further in the book but yeah if anyone would like to discuss these with me that'd be good.

[IMG]http://img80.imageshack.us/img80/7439/asdfasdfi.jpg[/img]

Title: Re: TT's Maths Thread
Post by: kamil9876 on November 29, 2010, 01:44:21 pm

They are indeed the only "kinds" of subspaces. It is easily shown if you know what a basis is.

I guess if you do not know this, you can prove it like this.

Suppose your subspace has some non-zero vector v, then your subspace must contain the line joining v and 0 by scalar multiplication. If your subspace has some other point w not on this line, show that your subspace must contain the plane that contains the line through 0 to w, and the line through 0 to v. If you subspace contains yet another vector z, now show that it must contain the whole R^3.

Title: Re: TT's Maths Thread
Post by: TrueTears on November 29, 2010, 03:17:47 pm

Quote from: kamil9876 on November 29, 2010, 01:44:21 pm

They are indeed the only "kinds" of subspaces. It is easily shown if you know what a basis is.

I guess if you do not know this, you can prove it like this.

Suppose your subspace has some non-zero vector v, then your subspace must contain the line joining v and 0 by scalar multiplication. If your subspace has some other point w not on this line, show that your subspace must contain the plane that contains the line through 0 to w, and the line through 0 to v. If you subspace contains yet another vector z, now show that it must contain the whole R^3.

ahh yeah i haven't read about basis yet in my book but i kinda know what they are from watching mit lectures, but yeah your explanation makes sense.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 08, 2010, 11:42:05 pm

Just with this question, I understand how they get the maximum value for the rank of A but what's up with the linearly dependent statements?

[IMG]http://img72.imageshack.us/img72/1843/linearalgebrarank.jpg[/img]

Eg, why must the 7 row vectors be linearly dependent if A is a 7x4 matrix? Can anyone show the argument? Maybe I am missing something obvious but I don't see how the rank of a matrix has to do with linear independence/dependence?

Thanks

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 09, 2010, 01:02:09 am

Are you aware of "column rank=row rank". What is your definition of rank btw? Mostly the definition of (column) rank is the maximal independent set of columns. Similairly for row rank (interesting theorem: column rank=row rank)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 09, 2010, 11:53:33 am

yeah column rank = row rank i know that, the definition of rank i use is that it is the dimension of the rowspace and the column space.

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 09, 2010, 12:52:17 pm

Oh ok yeah that's the same as the definition I was referring to. So for the 7 by 4 matrix we look at the columns and see that:

dim column space $\leq$ 4

hence: dim $rowspace \leq 4$

So the 7 rows are living inside a space that has no more than 4 dimensions. Therefore they are linearly dependent.

That is their argument but it's pretty stupid because you can easily see that the 7 rows are dependent by just looking at the fact that they live in $\mathbb{R}^4$ which is 4 dimensional. But I guess it just serves the purpose of illustrating some things about rank of matrix.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 18, 2010, 12:59:27 am

Okay so I just wanna confirm this, the Gram-Schmidt process, as my book has like 2 pages missing from it and only has the intro to this process. Just wanna confirm if this is correct:

Let V be any nonzero finite-dimensional inner product space, and suppose that $\{u_1, u_2, \cdots, u_n\}$ is any basis for V. It suffices to show that V has an orthogonal basis. Let $\{v_1, v_2, \cdots, v_n\}$ be the orthogonal basis for V.

Step 1: Let $v_1 = u_1$

Step 2: We can find a vector $v_2$ orthogonal to $v_1$ by computing the component of $u_2$ that is orthogonal to the space $w_1$ spanned by $v_1$ . This is given by $v_2 = u_2 - \text{proj}_{w_1}u_2 = u_2 - \frac{<u_2, v_1>}{||v_1||^2}v_1$

Step 3: We can find a vector $v_3$ orthogonal to $v_1$ and $v_2$ by computing the component of $u_3$ that is orthogonal to the space $w_2$ spanned by $v_1$ and $v_2$ . This is given by $v_3 = u_3 - \text{proj}_{w_2}u_3$

We keep going until we have found $v_n$ .

Now is my step 3 correct? It's missing in my book lol

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 18, 2010, 01:36:36 am

yes.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 19, 2010, 02:02:04 am

thx kamil

also another query, my book says that the normal system $A^TA\mathbf{x}=A^T\mathbf{b}$ of $A\mathbf{x}=\mathbf{b}$ may have infinitely many solutions in which all of its solutions are least squares solutions of $A\mathbf{x}=\mathbf{b}$ .

But then my book goes on to prove a theorem that states:

If A is a $m \times n$ matrix with linearly independent column vectors then for every $m \times 1$ matrix $\mathbf{b}$ the linear system $A\mathbf{x}=\mathbf{b}$ has an unique least squares solution given by $\mathbf{x} = (A^TA)^{-1} A^T\mathbf{b}$ .

Doesn't this theorem mean that every normal system will only ever have ONE unique least square solution? Then why did my book say that the normal system could have infinitely many solutions...? Am i missing something obvious coz it's quite late lol

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 19, 2010, 02:01:00 pm

I forgot what least squares are... havn't seen it since first year and it was quite boring. However the system may have infinitely many solutions if the column vectors are $not$ linearly independent. Your theorem applies to ones with linearly independent columns.

Title: Re: TT's Maths Thread
Post by: dcc on December 23, 2010, 10:24:49 pm

Quote from: kamil9876 on December 19, 2010, 02:01:00 pm

I forgot what least squares are... havn't seen it since first year and it was quite boring. However the system may have infinitely many solutions if the column vectors are $not$ linearly independent. Your theorem applies to ones with linearly independent columns.

It's one of those things they made us cover to ensure they had enough content for the exam. :(

Title: Re: TT's Maths Thread
Post by: TrueTears on December 23, 2010, 10:31:39 pm

thanks was so late i misread that initial condition :(

Title: Re: TT's Maths Thread
Post by: TrueTears on March 03, 2011, 03:05:16 am

Prove that if $n>1$ then $\sum_{d|n} \mu(d) = 0$ where $\mu(d)$ is the Mobius function.

Thanks :)

Title: Re: TT's Maths Thread
Post by: humph on March 03, 2011, 12:49:01 pm

Long (but more easily generalised) proof:

Definition
An arithmetic function is a complex-valued function defined on the natural numbers, that is, a function $f : \mathbb{N} \to \mathbb{C}$ .

Example
(1) The identity function $1_{n=1}$ defined by
$1_{n=1}(n) = \left\lfloor \frac{1}{n} \right\rfloor = \begin{cases} 1 & \text{if $n = 1$,} \\ 0 & \text{if $n > 1$.} \end{cases}$
(2) The unity function $1_{\mathbb{N}}$ defined by
$1_{\mathbb{N}}(n) = 1 \quad \text{for all $n \in \mathbb{N}$.} $
(3) The Mobius function} $\mu$ defined by
$\mu(n) = \begin{cases} 1 & \text{if $n = 1$,} \\ (-1)^k & \text{if $n = p_1^{a_1} \cdots p_k^{a_k}$ with $a_1 = \ldots = a_k = 1$,} \\ 0 & \text{otherwise.} \end{cases}$

Definition
An arithmetic function is multiplicative if, for $m,n \in \mathbb{N}$ , $f(mn) = f(m)f(n)$ whenever $(m,n) = 1$ (that is, whenever $m$ and $n$ are coprime).

Lemma
A multiplicative functions is completely determined by its values on prime powers.

Proof
This is a simple application of the fundamental theorem of arithmetic.

Definition
The Dirichlet convolution of two arithmetic functions $f,g : \mathbb{N} \to \mathbb{C}$ is the function $f * g : \mathbb{N} \to \mathbb{C}$ defined by
$f * g(n) = \sum_{d \mid n}{f(d) g\left(\frac{n}{d}\right)}.$

Proposition
If $f$ and $g$ are multiplicative, then their Dirichlet convolution $f * g$ is also multiplicative.

Proof
If $(m,n) = 1$ , then every divisor $d$ of $mn$ can be written uniquely in the form $d = ab$ , where $a \mid m$ and $b \mid n$ , and so
$\begin{align*} f * g(mn) & = \sum_{d \mid mn}{f(d) g\left(\frac{mn}{d}\right)} \\ & = \sum_{\substack{a \mid m \\ b \mid n}}{f(ab) g\left(\frac{mn}{ab}\right)} \\ & = \sum_{a \mid m}{f(a)g\left(\frac{m}{a}\right)} \sum_{b \mid n}{f(b) g\left(\frac{n}{b}\right)} \\ & = f*g(m) f*g(n). \end{align*}$

Combining everything, we can prove the following theorem.

Theorem
We have the identity
$\sum_{d \mid n}{\mu(d)} = \begin{cases} 1 & \text{if $n = 1$,} \\ 0 & \text{if $n > 1$.} \end{cases}$

Proof
We must show that $\mu * 1_{\mathbb{N}} = 1_{n=1}$ . As $\mu * 1_{\mathbb{N}}$ is multiplicative, being the Dirichlet convolution of two multiplicative functions, it suffices to show this merely for prime powers. So for a prime power $p^a$ , we have
$\mu * 1_{\mathbb{N}}(p^a) = \sum_{d \mid p^a}{\mu(d) 1_{\mathbb{N}}\left(\frac{n}{d}\right)} = \sum^{a}_{i=0}{\mu(p^i)} = \begin{cases} 1 & \text{if $a = 0$,} \\ 0 & \text{otherwise,} \\ \end{cases} $
as $\mu(1) = 1$ , $\mu(p) = -1$ , and $\mu(p^a) = 0$ for $a \geq 2$ .

Obviously this proof can be trimmed down a fair bit, but it's very useful for showing other identities among arithmetic functions.

Title: Re: TT's Maths Thread
Post by: kamil9876 on March 03, 2011, 02:49:20 pm

Notice that this is really just equivalent to "every finite non-empty set has as many even subsets as odd subsets" and there are many ways of showing this. Already discussed here http://vce.atarnotes.com/forum/index.php/topic,19896.msg207276.html#msg207276

Title: Re: TT's Maths Thread
Post by: TrueTears on March 07, 2011, 10:19:18 pm

Ahh yeah, thanks for that guys, really good explanations!!

I just have another question:

[IMG]http://img138.imageshack.us/img138/9163/92562440.jpg[/img]

Title: Re: TT's Maths Thread
Post by: humph on March 07, 2011, 11:16:33 pm

I'm guessing you just missed the obvious, but (c) is pretty trivial. It's not true if $n = 1$ just by plugging in. So suppose that $n > 1$ . We are asked to classify the integers $n > 1$ such that $\sigma_k(n) = n^k + 1$ . Note that
$\sigma_k(n) = \sum_{d \mid n}{d^k} = 1^k + n^k + \sum_{\substack{d \mid n \\ 1 < d < n}}{d^k}.$

So for the equality to hold, we must have that
$\sum_{\substack{d \mid n \\ 1 < d < n}}{d^k} = 0.$

That is, there must be no divisors of $n$ other than itself and $1$ . This is precisely the definition of a prime number.

Part (b) is pretty easy too: the major hint is Fermat's little theorem.

Title: Re: TT's Maths Thread
Post by: TrueTears on March 09, 2011, 04:13:53 pm

oh yes, stupid me haha.

thanks humph!

Title: Re: TT's Maths Thread
Post by: TrueTears on March 15, 2011, 02:26:53 am

[IMG]http://img821.imageshack.us/img821/3013/68209502.jpg[/img]

I don't get this question, $N^2 \neq 0$ , does the question mean $N^k = 0$ for $k \ge 3$

I can show that $N^k = 0, N^k = N^3N^{k-3} = 0 \times N = 0$

But how can I show the latter? ie, that a nilpotent matrix can be invertible (or non invertible?)

Thanks

Title: Re: TT's Maths Thread
Post by: Mao on March 15, 2011, 03:36:08 am

Aha, a 2021 question I see? I do remember them having a special love for determinants in that course.

If $N^k=0$ , this then implies $(det(N))^k = 0 \implies det(N) = 0$

Thus N is not invertible.

Title: Re: TT's Maths Thread
Post by: TrueTears on March 15, 2011, 04:59:20 am

aha i see, shudda seen that trick, thanks mao

Title: Re: TT's Maths Thread
Post by: TrueTears on March 15, 2011, 05:03:20 am

also got another one, maybe its just 5 am in the morning but i cant see what b) is asking?

[IMG]http://img546.imageshack.us/img546/5854/83956954.jpg[/img]

a) is quite simple and I can do c) as well, but what exactly is b) asking for? deduce what about it...?

Cheers

Title: Re: TT's Maths Thread
Post by: /0 on March 15, 2011, 09:26:19 am

$I+BA$ is also invertible. Right-Multiplying (a) by $B^{-1}$ , then left-multiplying by $B$ ,

$B(I+AB)^{-1}B^{-1}(I+BA)=1$

Hence, $B(I+AB)^{-1}B^{-1}$ is the left inverse of $I+BA$

You can similarly manipulate (a) to show that $I+BA$ has a right inverse.

Title: Re: TT's Maths Thread
Post by: TrueTears on March 30, 2011, 07:15:39 pm

[IMG]http://img836.imageshack.us/img836/8905/detlz.jpg[/img]

If det(A) = -7 find det(B)

I was thinking of applying elementary row operations to A to change it to B and then use the fact that det(E_1E_2...E_rA) = det(E_1)det(E_2)...det(A)

however it seems like guess work to find the E_j's...

Title: Re: TT's Maths Thread
Post by: kamil9876 on March 30, 2011, 07:31:48 pm

hint: $det(A)=det(A^T)$

Title: Re: TT's Maths Thread
Post by: TrueTears on March 30, 2011, 11:01:32 pm

~~yeah i thought about that as well, but they're not transpose of each other?~~

ohh wait i see, i got it!

cheers kamil!

Title: Re: TT's Maths Thread
Post by: TrueTears on April 05, 2011, 01:58:40 am

[IMG]http://img225.imageshack.us/img225/9051/22703967.jpg[/img]

I actually have no idea how to go about this question, in fact I weren't even taught what the Gamma function is/does in my classes nor have I ever seen the Jacobi theta-function in my life! So I am really clueless on this question. Any help would be appreciated!

Cheers!

Title: Re: TT's Maths Thread
Post by: Mao on April 05, 2011, 02:45:42 am

Read from here onwards: http://en.wikipedia.org/wiki/Riemann_zeta_function#Mellin_transform
It probably won't help, but it may help put things into context.

What subject is this? and why are you learning crazy transforms? xD

Title: Re: TT's Maths Thread
Post by: TrueTears on April 05, 2011, 02:57:54 am

ohh i see, ill take a good read of that cheers man.

its MTH2121, number theory and algebra, its an alright unit, but this was a suprising q since we haven't covered any of this in lectures nor is it in the prescribed texts lol

Title: Re: TT's Maths Thread
Post by: Mao on April 05, 2011, 03:29:11 am

Quote from: TrueTears on April 05, 2011, 02:57:54 am

ohh i see, ill take a good read of that cheers man.

its MTH2121, number theory and algebra, its an alright unit, but this was a suprising q since we haven't covered any of this in lectures nor is it in the prescribed texts lol

Have a play with it. I managed to simplify the RHS to $\int \left( \sum_{n=1}^{\infty} \exp(-\pi n^2 t) \right) t^{s-1}\, dt$ before I gave up. I'm sure it will be relatable to the course material.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 05, 2011, 06:00:45 am

hmm i still have no clue lol

Title: Re: TT's Maths Thread
Post by: /0 on April 05, 2011, 10:57:23 am

If you interchange summation and integral, then set $u=\pi n^2t$ , you can use u-substitution to reduce Mao's expression to $\pi^{-s}\Gamma (s)\sum_{n=-\infty}^\infty \frac{1}{n^{2s}}$ . The sum over n=0 is a bit of a pain. Still thinking about how to do the rest though.

Interesting question for an algebra course... woulda thought it to be more complex analysis

Title: Re: TT's Maths Thread
Post by: humph on April 05, 2011, 11:57:11 am

This is an extremely standard question in analytic number theory, but I can't imagine why it'd be in your course. Are you sure it's not an April Fool's joke??
Anyway, a good reference for this is Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory (which you may find here: http://gen.lib.rus.ec/get?md5=207be3298f0abee321df6ba0f0887561), chapter 10.

But /0's method should work, just if you're careful you should be able to get rid of the $n = 0$ term (as this is a singularity).

Title: Re: TT's Maths Thread
Post by: Mao on April 05, 2011, 03:57:12 pm

Quote from: /0 on April 05, 2011, 10:57:23 am

If you interchange summation and integral, then set $u=\pi n^2t$ , you can use u-substitution to reduce Mao's expression to $\pi^{-s}\Gamma (s)\sum_{n=-\infty}^\infty \frac{1}{n^{2s}}$ . The sum over n=0 is a bit of a pain. Still thinking about how to do the rest though.

Interesting question for an algebra course... woulda thought it to be more complex analysis

Aha, nice. That solves it. typing it up now.

Title: Re: TT's Maths Thread
Post by: Mao on April 05, 2011, 04:11:19 pm

$\begin{align*} RHS & = \frac{1}{2s-1}+\frac{1}{2} \int_{0}^{1} \left( \theta (\imath t) - t^{-1/2} \right) t^{s-1} \, dt -\frac{1}{2s} + \frac{1}{2} \int_{1}^{\infty} \left( \theta (\imath t) - 1 \right) t^{s-1} \, dt \\ & = \frac{1}{2s-1} - \frac{1}{2s} + \frac{1}{2} \int_{0}^{\infty} \theta(\imath t) t^{s-1}\, dt - \frac{1}{2} \int_{0}^{1} t^{s-3/2}\, dt - \frac{1}{2} \int_{1}^{\infty} t^{s-1} \end{align*}$

the Jacobi Theta function is simplified as
$\theta(\imath t) = \sum_{-\infty}^{\infty} \exp(-\pi n^2 t)$
due to the symmetry of $n^2$ , we can rewrite this as
$\theta(\imath t) = 1 + 2\sum_{n=1}^{\infty} \exp(-\pi n^2 t)$
the integral containing the theta function thus becomes
$\frac{1}{2} \int_{0}^{\infty} \theta(\imath t) t^{s-1}\, dt = \frac{1}{2}\int_{0}^{\infty} t^{s-1}\, dt + \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt$

Substituting this gives
$\begin{align*} RHS & = \frac{1}{2s-1} - \frac{1}{2s} + \frac{1}{2} \int_{0}^{1} t^{s-3/2} - t^{s-1} \, dt + \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt \\ & = \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt\qquad \mbox{substituting }u=\pi n^2 t \\ & = \sum_{n=1}^{\infty} \int_{0}^{\infty} \left( \frac{u}{\pi n^2} \right)^{s-1} \exp(-u) \, \frac{du}{\pi n^2} \\ & = \sum_{n=1}^{\infty} \frac{1}{(\pi n^2)^s} \int_{0}^{\infty} u^{s-1} \exp(-u) \, du\qquad \mbox{the integral is the gamma function by definition, thus} \\ & = \pi^{-s} \sum_{n=1}^{\infty} \frac{1}{n^{2s}} \Gamma (s) \\ & = \pi^{-s} \Gamma(s) \times \zeta (2s) \\ & = LHS \end{align*}$

the proof is thus complete

Title: Re: TT's Maths Thread
Post by: TrueTears on April 05, 2011, 07:21:53 pm

omg that's awesome, thanks guys ^^^^^

@humph+/0, i have no idea, i don't think this is an april fools joke lol it's on our set questions sheet that gets marked for this week haha but yeah like i said, i thought this wouldn't be in the unit that im doing lol

oh and yeah that's probs a small typo mao, but it should be $+\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt$ XD

Title: Re: TT's Maths Thread
Post by: humph on April 05, 2011, 09:48:06 pm

Mao's way is right, but you can't split up the integrals at the beginning like he did because I'm pretty sure then some of the integrals don't converge (the $\int^{\infty}_{0}{t^{s - 1} \, dt}$ term). The grouping is to ensure that everything works nicely. Also the reason why you can switch integration and summation is due to absolute convergence (this is very important).

Title: Re: TT's Maths Thread
Post by: TrueTears on April 05, 2011, 10:31:45 pm

ohh yeah, wait a sec, now im a bit confused, in mao's working :

$RHS = \frac{1}{2s-1} - \frac{1}{2s} - \frac{1}{2} \int_{0}^{1} t^{s-1} + t^{s-3/2}\, dt + \sum_{n=1}^{\infty} \int_{0}^{\infty}\exp(-\pi n^2 t) t^{s-1}\, dt$

How did he get the " $-\frac{1}{2}\int_0^1t^{s-1}+t^{s-\frac{3}{2}}dt$ " term ? it's probably pretty obvious...

I keep getting $+\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt$

since after substitution we have $-\frac{1}{2}\int_1^{\infty}t^{s-1}dt + \frac{1}{2}\int_0^{\infty}t^{s-1}dt = \frac{1}{2}\int_0^{1}t^{s-1} dt$

Then grouping with $-\frac{1}{2}\int_0^1t^{s-\frac{3}{2}}dt$ yields:

$\frac{1}{2}\int_0^{1}t^{s-1} dt -\frac{1}{2}\int_0^1t^{s-\frac{3}{2}}dt = +\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt$

Title: Re: TT's Maths Thread
Post by: /0 on April 05, 2011, 10:42:55 pm

Hmm it probably should be

$-\frac{1}{2}\int_0^1 -t^{s-1}+t^{s-\frac{3}{2}}\,dt$

Also,

$\int_0^1t^{s-1}\,dt = \int_0^\infty t^{s-1}\,dt - \int_1^\infty t^{s-1}\,dt$

But as humph mentioned... the divergence of the RHS integrals might still be a problem though. Perhaps there's a way of grouping the RHS so that they don't diverge.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 05, 2011, 11:48:55 pm

yeah, thanks for the confirmation /0, and yeah, so both $\int_0^\infty t^{s-1}\,dt$ and $-\int_1^\infty t^{s-1}\,dt$ diverge... but when you put them together its fine, so is it necessary to change the grouping?

Title: Re: TT's Maths Thread
Post by: humph on April 06, 2011, 12:31:36 am

Let's first restrict ourselves to the open half-plane $\Re(s) > 1/2$ . Then for any $n \geq 1$ ,
$n^{-2s} \Gamma(s) \pi^{-s} = \int^{\infty}_{0}{n^{-2s} \pi^{-s} e^{-x} x^{s - 1} \, dx} = \int^{\infty}_{0}{e^{-\pi n^2 t} t^{ s - 1} \, dt}$
by making the substitution $x = \pi n^2 t$ . Summing over all $n \geq 1$ and interchanging the order of summation and integration, which is justified by absolute convergence (which relies on the crucial fact that $\Re(s) > 1/2$ ), we find that
$ \begin{align} \zeta(2s) \Gamma(s) \pi^{-s} & = \int^{\infty}_{0}{\sum^{\infty}_{n = 1}{e^{-\pi n^2 t}} t^{ s - 1} \, dt} \\ & = \frac{1}{2} \int^{1}_{0}{(\theta(it) - t^{-1/2}) t^{s - 1} \, dt} + \frac{1}{2} \int^{1}_{0}{(t^{-1/2} - 1) \, dt} + \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt} \\ & = \frac{1}{2} \int^{1}_{0}{(\theta(it) - t^{-1/2}) t^{s - 1} \, dt} + \frac{1}{2s - 1} - \frac{1}{2s} + \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt}, \end{align} $
where again the fact that $\Re(s) > 1/2$ was crucial in allowing us to evaluate the middle integral. The key point now is although everything is currently only well-defined on $\Re(s) > 1/2$ , the terms on the right-hand side actually define a function that is well-defined (in fact, holomorphic) on $\{s \in \mathbb{C} : s \neq 0, 1/2, 1\}$ (so that the right-hand side, and hence the left-hand side, extends to a meromorphic function on $\mathbb{C}$ with singularities at $s = 0, 1/2, 1$ ). The middle two terms clearly satisfy this. For the first term, we actually require the functional equation
$ \theta(it) = \frac{1}{\sqrt{t}} \theta\left(\frac{i}{t}\right) $
for $t > 0$ ; this is a standard application of the Poisson summation formula in Fourier analysis. Using this, we see that
$ \frac{1}{2} \int^{1}_{0}{(\theta(it) - t^{-1/2}) t^{s - 1} \, dt} = - \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{-s - 1/2} \, dt}. $
So we must show that
$ - \frac{1}{2} \int^{\infty}_{1}{(\theta(it) - 1) t^{-s - 1/2} \, dt}, \quad \int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt} $
both converge for all $s \in \mathbb{C}$ . Let's look at the second integral. We already know this converges for $\Re(s) > 1/2$ . If $\Re(s) \leq 1/2$ , then it's easy to show that
$ \left|\int^{\infty}_{1}{(\theta(it) - 1) t^{s - 1} \, dt}\right| \leq \sum^{\infty}_{n = 1}{\int^{\infty}_{1}{e^{-\pi n^2 t} \, dt} = \frac{1}{\pi} \sum^{\infty}_{n = 1}{\frac{1}{n^2}} < \infty, $
and hence this integral converges. A similar argument proves this for the first integral.

Title: Re: TT's Maths Thread
Post by: Mao on April 06, 2011, 01:12:31 am

You are quite right, I did make a typo (because I was expecting the integrals to cancel out and got lazy).

I also didn't care too much about convergence. I don't make a very good mathematician. :P

Title: Re: TT's Maths Thread
Post by: TrueTears on April 18, 2011, 01:58:30 am

hey guys, just some questions regarding this problem, I've done most of the essential working but just some details i want to clarification over:

"Let $*$ be a binary relation on $\mathbb{R} \backslash \{0\}$ defined by $a * b = a|b|$ . Is $<\mathbb{R}, *>$ a group? Explain"

My first question is that it says at the start that $*$ is a binary operation ON $\mathbb{R} \backslash \{0\}$ but then the binary structure $<\mathbb{R}, *>$ means that $*$ is now a binary operation on $\mathbb{R}$ instead of $\mathbb{R} \backslash \{0\}$

so thus in showing whether $<\mathbb{R}, *>$ is a group or not, do we take the binary operation $*$ on $\mathbb{R}$ or $\mathbb{R} \backslash \{0\}$ as this is actually an essential part of showing whether $<\mathbb{R}, *>$ is a group or not

eg, if we use $\mathbb{R}$ then when we are trying to show that for all $x \in \mathbb{R}$ there is an identity element $e$ such that $e * x = x * e = x$

thus $e*x = e|x$ |

and assume $e|x| = x$ then we have $e = \frac{x}{|x|}$

but if $x = 0$ , then $e$ is undefined and thus $x = 0$ does not have an identity element in $\mathbb{R}$

however if we use $\mathbb{R} \backslash \{0\}$ we can clearly see that $e = \frac{x}{|x|}$ is the identity element since $x$ can not be 0 and thus $e$ is defined

cheers

Also another question:

[IMG]http://img850.imageshack.us/img850/1990/matrixgroup.jpg[/img]

First, what does it mean by "all operations performed modulo 2"? My understanding is that say we multiply 2 matrices that are 2x2 and we get something like

$\begin{bmatrix} 3 & 2 \\ 4 & 5 \end{bmatrix}$

then actually the matrix should be:

$\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$

in other words every entry in the matrix has to be modulo 2?

Also what does "let S be the set of 2x2 matrices A with entries in $\{0, 1\}$ such..." mean?

does that mean all the entries in these matrices are either 0 or 1?

and lastly, how are we meant to find the elements of the group? do we actually have to list all 2^4 possible matrices with entries either 0 or 1 and pick the ones where $det(A) \not\equiv 0 \pmod{2}$ ? or am i interpreting it wrong?

Also how are we meant to find the identity and inverse element?

because by the definition of the identity element, it is an element e in S such that for all x in S, $e * x = x * e = x$

how can we find the matrix e in S that satisfies the above for every x in S?

thanks :)

Title: Re: TT's Maths Thread
Post by: /0 on April 18, 2011, 02:46:07 am

With the 1st question, I assume they really mean $\langle \mathbb{R}\setminus \lbrace 0\rbrace,*\rangle$ ... I might be wrong, but sometimes they get lazy with notation.
[Actually don't take my word on this one, since i'm really not sure]

With the second question, all operations performed modulo 2 means every entry in the resulting matrix must be modulo two. In the example you gave the matrix would actually be the identity.

And yep, if all the matrices' entries are in $\lbrace 0,1\rbrace$ then they are only 0 or 1, and furthermore it is possible to list out all the elements of $S$ by simply constructing all invertible 2x2 matrices with 0s and 1s. From a quick glance there should only be ~~2 elements~~ (lol or not). (but check)

Title: Re: TT's Maths Thread
Post by: TrueTears on April 18, 2011, 02:53:53 am

Cheers /0 !

I found 6 elements in S though lol

$\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$

$\begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$

$\begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix}$

$\begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}$

$\begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}$

$\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$

i listed out the entire 2^4 possibilities and calculated each det, those above satisfies the condition for S [namely, the set of all 2x2 matrices A with entires in $\{0, 1\}$ such that $det(A) \not\equiv 0 \pmod{2}$ ]

assuming those 6 elements are right... how do we find the inverse and inverse for the group?

actually is my reasoning right for finding the inverse?

we require an element e in S such that for all x in S, $e * x = x * e = x$

thus say we let $e = \begin{bmatrix} a & b \\ c & d\end{bmatrix}$ and pick a specific x from S, ie, pick $x = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$

thus we find that $e = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ satisfies $e * x = x * e = x$

Now since we ALREADY know that $<S, *>$ is a group and that the identity element in a group is unique, then $e = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ must be the identity element for the group!

Now this only works because we already know that $<S, *>$ is a group beforehand right? since if we didn't know $<S,*>$ was a group then just solving $e * x = x * e = x$ for a specific $x$ doesn't mean that $e$ could be the identity element for the rest of the elements, in other words, there might not exist an identity element!

Title: Re: TT's Maths Thread
Post by: TrueTears on April 18, 2011, 05:06:25 am

[IMG]http://img140.imageshack.us/img140/3783/rankofa.jpg[/img]

Also I don't get this question, I've reduced A down to its row echelon form of:

$\begin{bmatrix} 1 & 1 & t \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{bmatrix}$ and $t \neq \{1,-2\}$

We see that the basis for the row space would be the vectors (1,1,t), (0,1, -1) and (0,0,1) so the rank(A) = 3

but the question says how does the rank of A vary with t? well even if t changes (as long as $t \neq \{1,-2\}$ ) then the rank of A is independent of t?

I don't get what the question is trying to ask... any clarifications?

Title: Re: TT's Maths Thread
Post by: evaever on April 18, 2011, 08:44:20 am

rank of A is not a continuous function of t

Title: Re: TT's Maths Thread
Post by: humph on April 18, 2011, 02:22:46 pm

For your first question, I'd imagine that it's a typo: it's definitely supposed to be $\mathbb{R} \setminus \{0\}$ throughout, not $\mathbb{R}$ .

Also note that your calculation shows that should the identity $e$ exist, it must satisfy $e = x/|x|$ for every $x \in \mathbb{R} \setminus \{0\}$ , but this is impossible, as $x/|x| = 1$ if $x > 0$ but $x/|x| = -1$ if $x < 0$ .

For your second question, I think /0 has explained this pretty well. So basically you're studying 2x2 matrices whose entries are either 1 or 0, whose determinant is either 1 or -1. You can shortcut here a bit by remembering that $\det(A) = \det(A^T)$ (which tells you that you only need to study 2^3 possibilities), and similarly that $\det(\mathbf{a}_1 \; \mathbf{a}_2) = - \det(\mathbf{a}_2 \; \mathbf{a}_1)$ for a matrix made of two column vectors $\det(\mathbf{a}_1 \; \mathbf{a}_2) = - \det(\mathbf{a}_2 \; \mathbf{a}_1)$ .
Once you've found them all, it's pretty obvious which one is the identity (same matrix as usual) - this is because it's ALWAYS the identity for any group of matrices (for obvious reasons). A little matrix multiplication tells you which matrices are inverses of each other. There are shortcuts to avoid all this calculation, but on the whole it shouldn't be too much work anyway.

For your third question, evaever has the right idea. Think of the rank as a function of $t$ ; that is, consider the function $\mathrm{rank}(A)(t)$ . As evaever says, it's a piecewise defined function that isn't continuous at two points.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 18, 2011, 04:12:51 pm

thanks guys :)

and also @humph, I've found $S = \{ \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix}, \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\}$

and by my argument in the previous post (is that right?) I found $e = \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ to be the identity element.

now if we let $\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}=a , \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}=b, \begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix} =c, \begin{bmatrix} 0 & 1 \\ 1 & 1\end{bmatrix}=d, \begin{bmatrix} 1 & 0 \\ 1 & 1\end{bmatrix} = f, \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}=g$

now the table of operations looks like:

[IMG]http://img19.imageshack.us/img19/3672/tableoperations.jpg[/img]

however isn't the inverse element in a group always unique? ie, there is only one just like the identity element? As my book says:

[IMG]http://img863.imageshack.us/img863/9718/uniqueness.jpg[/img]

however from the table we can clearly see that the inverse element for a is a since $a*a = a$

also $c*d = d*c = a$ so c's inverse element is d

but now we have 2 inverse elements, one being a and the other being d, contradicting the fact that the group can only have one unique inverse for all of its elements?

Title: Re: TT's Maths Thread
Post by: humph on April 18, 2011, 05:46:13 pm

No, for EACH $a$ , there is a unique inverse $a'$ . Your statement is really quite silly, if you think about it: you're saying that there is a unique, fixed $a'$ such that for every $a$ , $a'$ is the inverse of $a$ . What about the most natural definition of a group, $\mathbb{R}$ (or $\mathbb{Q}$ , or $\mathbb{Z}$ ) under addition? Each element $a$ has a unique inverse $-a$ , but of course it's impossible for there to be one single "global" inverse for every point.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 18, 2011, 05:48:18 pm

oh wait i think i misinterpreted it, every inverse is unique for that specific element but there doesn't have to be 1 inverse for the whole group right?

so for each $y \in S$ , the inverse of y, given by y' has to satisfy $y * y' = y' * y = e$

so $y*y' = yy' = e$

treating y and y' as matrices

we can solve for y': $y^{-1}yy'=y^{-1}e$

so $y' = y^{-1}$

this also satisfies $y'*y = e$ since $y^{-1}y = e$

thus for every element y in S, the inverse of y is given by $y^{-1}$ , this is unique for EACH element not for ALL elements

is that right?

lol opps yeah i realised my stupid mistake right after you posted humph haha thanks again :)

Title: Re: TT's Maths Thread
Post by: Mao on April 18, 2011, 11:27:37 pm

Quote from: TrueTears on April 18, 2011, 05:06:25 am

[IMG]http://img140.imageshack.us/img140/3783/rankofa.jpg[/img]

Also I don't get this question, I've reduced A down to its row echelon form of:

$\begin{bmatrix} 1 & 1 & t \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{bmatrix}$ and $t \neq \{1,-2\}$

We see that the basis for the row space would be the vectors (1,1,t), (0,1, -1) and (0,0,1) so the rank(A) = 3

but the question says how does the rank of A vary with t? well even if t changes (as long as $t \neq \{1,-2\}$ ) then the rank of A is independent of t?

I don't get what the question is trying to ask... any clarifications?

During row operations, it seems you would need to divide by functions of t. The zeroes of these functions are values of t you should investigate.

You could also try to compute the determinant to see what values would play up.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 19, 2011, 01:17:44 am

yup thanks mao, i got it now :)

just have another Q

[IMG]http://img263.imageshack.us/img263/7585/solutionsforagroup.jpg[/img]

(e is the identity element)

I played around with examples and for $a^m=e$ to be true for each $a \in G$ then we need to have $a * (a*a*...*a) = e$ where $(a*a*...*a) = a'$ [the inverse of a]

however i am still a bit unsure on how to go about proving what's required... any help would be greatly appreciated!

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 19, 2011, 10:40:02 pm

So you want to show that one of $a^1,a^2...a^n$ is the identity, what if none of them were? Then what does the pigeonhole principle say?

Title: Re: TT's Maths Thread
Post by: TrueTears on April 20, 2011, 05:22:06 am

cheers kamil ;)

Title: Re: TT's Maths Thread
Post by: TrueTears on April 22, 2011, 03:04:47 am

[IMG]http://img714.imageshack.us/img714/5115/proofgroups.jpg[/img]

Just not too convinced on this proof...

For Case I, it is proving the isomorphism for if the order of G is infinite, however why does the order of G being infinite have anything to do with it specifying that $\forall m \in \mathbb{Z}^+$ , $a^m \neq e$ ? Can't $a^m = e$ also be a possibility? In other words, there could be 2 exponents which when a is raised to yield the same element in G. So shouldn't Case I have 2 parts, 1. Showing that G is isomorphic to $<\mathbb{Z}, +>$ if all the elements in G can be represent with distinct exponents of a. and 2. Show that it's isomorphic if some of the elements in G can be represented with 2 or more exponents of a.

So why has Case I left out a part?

Wait... or is Case I stating that if G is a cyclic group with generator a and the order of G is infinite, then $a^m \neq e \ \forall m \in \mathbb{Z}^+$ ?

And Case II is stating that if G is a cyclic group with generator a but the order is now FINITE, then there must exist some positive integer m such that $a^m = e$ . Actually this was just proven in a question a few posts ago lol.

So how do you show that if G is a cyclic group with generator a and the order of G is infinite, then $a^m \neq e \ \forall m \in \mathbb{Z}^+$ ?

I am not too sure on their proof: "Suppose that $a^h = a^k$ ...." because can't this method also be applied and show that even if the order of G is finite then $a^m \neq e \ \forall m \in \mathbb{Z}^+$ ? Which would then be a contradiction to what I proved a few posts earlier?

For example: Consider $h, k \le m$ WLOG assume h>k, suppose that $a^h = a^k \implies a^{h-k} = e \implies h-k = 0 \implies h = k$

So the conclusion is $a^m \neq e \ \forall m \in \mathbb{Z}^+$ , but we clearly know this is a false statement...

Title: Re: TT's Maths Thread
Post by: schnappy on April 22, 2011, 05:40:08 am

I can see the real world applications of pure maths presenting themselves in this thread.

Title: Re: TT's Maths Thread
Post by: /0 on April 22, 2011, 09:29:23 am

Hmmm well if $a$ generates G and there exists m such that $a^m=e$ , then just intuitively, $G$ can only contain $m$ elements, so it can't be infinite cyclic.

$\begin{array}{cccc} a^1 & a^2 & ... & a^m \\\\ a^{m+1} & a^{m+2} & ... & a^{2m} \\\\ a^{2m+1} & a^{2m+2} & ... & a^{3m} \\\\ ... & ... & ... & ... \end{array}$

Each column represents one element.

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 22, 2011, 01:33:52 pm

^ of course you mean $m\neq 0$ btw

Quote

$a^{h-k}=e \implies h-k=0$

this is where it fails for a finite cyclic group. Obvious example is if we consider the group with two element ${e,a}$ in which $a^2=e$

Title: Re: TT's Maths Thread
Post by: TrueTears on April 22, 2011, 05:36:59 pm

ok thanks guys :)

so that means

1. if G is a cyclic group with generator a and the order of G is infinite, then $a^m \neq e \ \forall m \in \mathbb{Z}^+$ ?

and

2. if G is a cyclic group with generator a and the order of G is finite, then $a^m =e$ for some $m \in \mathbb{Z}^+$ ?

But why is m only restricted to $\mathbb{Z}^+$ ? I would have thought for 1. it would be $a^m \neq e$ for $m \in \mathbb{Z} \setminus \{0\}$ since a is a generator... and same with 2. m should be in $m \in \mathbb{Z} \setminus \{0\}$ as well since a is a generator.

ohhh wait i see, i think the restriction on m is enough if m is in $\mathbb{Z}^+$ since if G is of finite order and a is a generator then the group G is "symmetric", what i mean is:

$a^{-1} = a$ since $a^{-1}a= aa^{-1}=a^0$

$a^{-2} = a^2$

$a^{-3} = a^3$

.
.
.

$a^{-m} = a^m$

?

but using the above logic doesnt make sense, consider the cyclic finite group $<\mathbb{Z}_4, +>$

1 is a generator.

clearly $1^{-1} \neq 1$

since $1^{-1} = -1$

so... im confused lol

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 22, 2011, 06:29:43 pm

so as to your question "why is the restriction to $\mathbb{Z}^+$ ?" well yeah you are right we can instead just restrict it to $\mathbb{Z}-\{0\}$ but so what? we didn't need that to go through the proof.

And I have no idea what you mean by $a=a^{-1}$ . It doesn't need to hold for a generator $a$ . I think what you might have wanted to say is that if $a$ is a generator then so is $a^{-1}$ , that is true.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 22, 2011, 06:49:55 pm

ahh yes i finally get it, thanks !

Title: Re: TT's Maths Thread
Post by: TrueTears on April 22, 2011, 08:14:55 pm

[IMG]http://img21.imageshack.us/img21/5066/614ig.jpg[/img]

Just a little unsure on 2 parts in this proof.

First how does $(a^s)^m = e$ imply n divides ms?

Also in the final paragraph, why does <d> contain all the positive m less than n such that (m,n) = d?

Cheers

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 22, 2011, 09:19:56 pm

More generally, if $a^k=e$ then $n|k$ . To see this write $k=nq+r$ with $0\leq r<n$ . Notice that this means $a^r=e$ which means that $r=0$ since $0\leq r<n$ .

Of course the more intuitive way of seeing this (and it's good to understand it like this once you get used to it) is look at /0's table (with $m$ the MINIMAL positive $m$ such that $a^m=e$ ) Notice that the elements in the SAME column are the same, hence the ones equal to $e$ are those in the rightmost columns, Viola.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 23, 2011, 05:35:12 am

cheers kamil i get it :)

Also just another question:

Cayley's theorem states that every group, G, is isomorphic to a subgroup of the group $S_G$ of all permutations of G.

I know how to prove this etc however I am just wondering, how do we actually FIND this subgroup? The theorem only tells us the existence of a subgroup but does not tell us how to find it. Is there any slick ways to do so?

Cheers

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 23, 2011, 10:09:32 pm

There is a way of finding it if you know the Cayley table. For example look at the table here http://en.wikipedia.org/wiki/Klein_four-group

Notice that each row is a permutation of the elements of the group. Let us encode e,a,b,ab as 1,2,3,4 respectively. Using this encoding the first row reads as:

1 2 3 4

the second is:

2 1 4 3

the third is:

3 4 1 2

the fourth is:

4 3 2 1

So that the subgroup of $S_4$ that is isomorphic to the group consists of the permutations:

1 2 3 4
1 2 3 4

1 2 3 4
2 1 4 3

1 2 3 4
3 4 1 2

1 2 3 4
4 3 2 1

In fact you can use this very idea to write out a proof of Cayley's Theorem.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 24, 2011, 05:05:10 pm

ahhh i see, yeah actually my book went into details of how to find it in latter chapters haha guess i was too eager to find out...

just another question ive thought up while reading about direct products, would say, changing the order of the factors in ANY direct product yield a group isomorphic to the original one?

or more formally, let $G_1, G_2, \cdots, G_n$ be groups, then the direct product of the groups, $G_1 \times G_2 \times \cdots \times G_n$ is also a group. Call this group $g_1$ .

Now if we reshuffle the order of the factors and say we have $G_3 \times G_1 \times G_2 \times \cdots \times G_n$ , clearly this would be also be a group, call this group $g_2$ .

Then does $g_2 \simeq g_1$ ?

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 24, 2011, 05:24:00 pm

A general hint for finding isomorphisms: try the obvious.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 24, 2011, 05:29:11 pm

yeah i know how to prove isomorphisms... defining the function $\phi$ and etc but i dono how to define such a function in this case

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 24, 2011, 06:24:55 pm

I know so try to define that function, think up of a definition. You get g_2 from g_1 by permuting the factors, so what's an obvious way to get an element from g_1 and changing it to an element of g_2?

Title: Re: TT's Maths Thread
Post by: TrueTears on April 24, 2011, 06:41:20 pm

ohhh right i get it, it's pretty easy to think about it intuitively.

Since if we change the order of the factors in the direct product, then the 'names' of each element in the direct product is just simply changed by a permutation of the components in each of the n-tuples.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 26, 2011, 05:45:56 am

[IMG]http://img849.imageshack.us/img849/8864/factorgroups.jpg[/img]

Eh is there a mistake in the theorem?

If we let $H = \mathbb{Z}_4$ and $K = \mathbb{Z}_6$

then according to 15.8 Theorem we have $\bar{K} = \{(k, e) | k \in K\} = \{(0, 0), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0)\}$

So that means $G / \bar{K} = (\mathbb{Z}_4 \times \mathbb{Z}_6) / \bar{K} \simeq \mathbb{Z}_4$

However look at Example 15.7 It SHOULD be $(\mathbb{Z}_4 \times \mathbb{Z}_6) / ( \{(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5)\}) \simeq \mathbb{Z}_4$

In other words shouldn't the theorem state $\bar{H} = \{(e, h) | h \in H\}$ not the other way around?

Also another question:

We know that every cyclic group is abelian.

Also all subgroups of abelian groups are normal.

Does that imply if a group G is cyclic then it is abelian, now since all groups are a subgroup of itself, then G (being an abelian group) must also be a normal group?

[IMG]http://img189.imageshack.us/img189/67/factorcomputation.jpg[/img]

I completely lost on what this example is trying to explain.

First of all, 'computing' a factor group just means finding which direct product of groups is the factor group is isomorphic to.

Now we know that $|\mathbb{Z}_4 \times \mathbb{Z}_6|=24$ and $|H| = 3$ so $|(\mathbb{Z}_4 \times \mathbb{Z}_6) / <(0,2)>| = 8$

There since this factor group is finitely generated and it is also abelian then it is isomorphic to a direct product of either of 3 forms below:

$\mathbb{Z}_8$

$\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

$\mathbb{Z}_4 \times \mathbb{Z}_2$

Now this is where I am lost. What does the example mean by "the first factor $\mathbb{Z}_4$ of $\mathbb{Z}_4 \times \mathbb{Z}_6$ is left alone" ? Why do you leave it "alone"?

Then "The $\mathbb{Z}_6$ factor ... is collapsed by a subgroup of order 3" What the hell does it mean that the $\mathbb{Z}_6$ factor is "collapsed by a subgroup"?

And how does "collapsing" yield a "factor group in the second factor of order 2 that must be isomorphic to $\mathbb{Z}_2$ " ?

Basically can someone please translate what the hell this example is trying to say into english? Or shall i say, less ambiguous words such as "collapsing"?

that would be great since ive been trying to do this question myself, i've listed out the 8 elements in the factor group and reached the same conclusion as the question itself did, however listing out 8 elements is certainly very troublesome and the example seemed to skip listing the out the whole factor group and find what it's isomorphic to. however I just cant seem to understand the logic behind the example at all!

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 26, 2011, 12:07:34 pm

As to the theorem: Yeah you're right on how it should be. In the author's defence his statement is all true except for the last sentence (he should've defined $\bar{K}$ differently, but with the way he has defined $\bar{H}$ his second last sentence is true).

Quote

Also another question:

We know that every cyclic group is abelian.

Also all subgroups of abelian groups are normal.

Does that imply if a group G is cyclic then it is abelian, now since all groups are a subgroup of itself, then G (being an abelian group) must also be a normal group?

G is always a normal subgroup of G (whole group is always a normal subgroup of itself), even if it is not abelian. The only interesting thing is whether the smaller subgroups are normal. BTW: there is no such thing as a "Normal Group" only "Normal Subgroup in G", indeed there are examples of groups that are normal subgroups of some group G, but not normal subgroups in another group G'.

Hrmm I find it very weird that you already know the classification theorem for finitely generated abelian groups (it's not really for beginners and you probably won't be able to prove it until you learn much more algebra) and it is not very useful for you here. Here is a longer way of doing it but perhaps it will give you insight:

Basically we want to find the H-Cosets of G.

$ H=\{ (0,0), (0,2), (0,4) \}$

So if $(a,b)$ is an arbitrary element of G then the Coset that $(a,b)$ is in is:

$ \{(a,b), (a,b+2), (a,b+4)\}$

So this tells you something nice: that $(x,y)$ and $(x',y')$ are in the same Coset IFF both $x \equiv x' \mod 4$ and $y \equiv y' \mod 2$ .
Now how does this help us? Well you can now see that the Coset that (x,y) is in depends only on the value of x modulo 4, and the value of y modulo 2. Hence the isomorphism.

Hope that builds some intuition.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 26, 2011, 05:40:40 pm

Thanks for the help kamil but i still dont quite get the last question.

Don't you mean x = x' ? why mod 4 the 'a' doesn't change in any coset.

Also $y \equiv y' \pmod{2}$ <- I get that but how does that relate to the isomorphism? can you break down their explanation for me?

Cheers

Actually nvm, after thinking about it over and over it I think I get it.

For the first value in (x,y) we can have anything from the set \{0, 1, 2, 3\} since any single ordered pair (x,y) with any one of the values for x from that set will be in a different coset. As for the second value y in (x,y) we can have anything from the set \{0, 1\}. Thus the isomorphism to Z4 x Z2

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 26, 2011, 06:58:51 pm

You're right, in fact i had "x=x'" but then i changed it coz I think it might be confusing.

Title: Re: TT's Maths Thread
Post by: TrueTears on April 26, 2011, 07:20:14 pm

cheers man, your explanations were great, understand it much better now!

Title: Re: TT's Maths Thread
Post by: TrueTears on April 26, 2011, 07:59:34 pm

Hmmm, I've got another Q...

Compute the factor group $(\mathbb{Z}_4 \times \mathbb{Z}_6) / <(2,3)>$

Now using your logic kamil, if we have some $(a,b) \in (\mathbb{Z}_4 \times \mathbb{Z}_6)$

then the H cosets of G is $(a,b)H = \{(a, b), (a+2, b+3)\}$

Which means (x,y) and (x', y') are in the same coset iff both $x \equiv x' \pmod{2}$ AND $y \equiv y' \pmod{3}$

So $x \in \{0, 1\}$ and $y \in \{0, 1, 2\}$

Thus $(\mathbb{Z}_4 \times \mathbb{Z}_6) / <(2,3)> \simeq \mathbb{Z}_2 \times \mathbb{Z}_3$

But clearly that's wrong, since $|\mathbb{Z}_2 \times \mathbb{Z}_3| = 6$ when in fact we know that $|(\mathbb{Z}_4 \times \mathbb{Z}_6) / <(2,3)>| =12$ !!!!

What went wrong :S

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 26, 2011, 08:14:21 pm

Quote

Which means (x,y) and (x', y') are in the same coset iff both $x \equiv x' \pmod{2}$ AND $y \equiv y' \pmod{3}$

This is what went wrong, it is true that if (x,y) and (x',y') are in the same Coset then $x \equiv x' \pmod{2}$ AND $y \equiv y' \pmod{3}$ , but the converse is not true (example look at $H+(0,0)=\{(0,0), (2,3)\}$ so clearly you see that (0,0) and (0,3) are not in the same Coset yet they satisfy the congruences). It is not as nice as the previous example.

~~There is a way of doing it but it is probably beyond the scope of where you are at~~. The other way of doing is to do what you noticed, that the order is 12 and then use the fundamental theorem of finite abelian groups and try to see which one it is isomorphic too.

Though a good way of illustrating the reasoning I used in the previous example is with this slight variation, what is:

$(\mathbb{Z}_4 \times \mathbb{Z}_6) / <(2,0),(0,3)>$ isomorphic to?

Title: Re: TT's Maths Thread
Post by: TrueTears on April 26, 2011, 10:14:02 pm

ohhhh i get it, true.

now in regards to your question, $(\mathbb{Z}_4 \times \mathbb{Z}_6) / <(2,0),(0,3)>$ isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3$ haha, however i am just wondering how do u know when the congruence 'technique' fails and when it doesnt?

Is it because $<(2,3)> = \{(0,0), (2,3)\}$ and we see that (x,y) and (x',y') is in the same coset IFF the x values differ by STRICTLY 2 and y values differ by STRICTLY 3. So if we wanted (0,0) to be the "representative" of (0,3)H, (2,0)H and (2,3)H, it can only "represent" (2,3)H in the sense that they are the same coset. However for (0,3)H, (2,0)H, these 2 are clearly different from (0,0)H, they are NOT the same coset.

Now $<(2,0),(0,3)> = \{(0,0), (0,3), (2,0), (2,3)\}$ so any (x,y) that satisfies the congruences $x \equiv x' \pmod{2}$ AND $y \equiv y' \pmod{3}$ must be in the same coset and if they are in the same coset they must satisfy the congruence.

Title: Re: TT's Maths Thread
Post by: kamil9876 on April 26, 2011, 11:20:28 pm

Ok good seems that you are understanding, but try this now:

What is $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(0,4)>$ isomorphic to?

How about $\mathbb{Z}_4 \times \mathbb{Z}_6 / <(0,5)>$ ?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 03, 2011, 03:24:40 am

[IMG]http://img35.imageshack.us/img35/957/47409814.jpg[/img]

Just wondering, the result of the dimension of the nullspace or rowspace of a matrix does not change regardless of what field it is over correct?

[IMG]http://img97.imageshack.us/img97/6063/linearmap.jpg[/img]

Also how do I do this q?

cheers

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 03, 2011, 06:33:56 pm

Quote

Just wondering, the result of the dimension of the nullspace or rowspace of a matrix does not change regardless of what field it is over correct?

Technically that doesn't make sense, you should look at the $1$ in $\mathbb{R}$ as a completely different animal to the $1$ in $\mathbb{Z}_2$ it's just (almost) a symbolic accident that they look the same. e.g your question may not make sense if for example there was a $2$ in your matrix or say you were comparing matrices over two completely different fields (example: Field of rational functions on $\mathbb{C}$ compared with just $\mathbb{Z}_2$ )

So in general your question doesn't make sense, you shouldn't mix fields (unless you've got some morphism or comparing a field to a subfield or something else that makes sense).

However perhaps it is interesting to restrict this question to the special case of 0,1 matrices (perhaps something to think about when you have spare time)

9. This looks like a very standard first year exercise, just use the definitions.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 03, 2011, 08:56:24 pm

no, actually i think i phrased my question badly, what i mean is, is the underlying "theorems/rules" regarding vector spaces all true regardless or watever field.

eg, say we are reducing a matrix to row echelon form over the field Z_3, does that mean we are still trying to make all the pivots as 1, etc etc??

and yeah i know the definition, but how to apply it? ><

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 03, 2011, 09:47:21 pm

ahh ok, Yes in fact you can go over each proof and notice that the only property of real numbers that are used in most of them is the the simple algebra like division, multiplication etc. and hence the same proofs work for fields in general. (some though are special like the fact that every linear transformation has an eigenvalue only works in fields like the complex numbers, though you probably won't be worried by this)

well, take an arbitrary element a+bx, now take an arbitrary scalar $\lambda$ , what is $T(\lambda (a+bx))$ , what is $\lambda T(a+bx)$ ? are they the same? if so then scalar multiplication is satisfied, now we need to do the same for addition...

Title: Re: TT's Maths Thread
Post by: TrueTears on May 03, 2011, 09:59:15 pm

thanks kamil!

yes that's what i dont get, what IS $T(\lambda (a+bx))$ ???

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 03, 2011, 10:21:50 pm

what happened on Sunday?

Hint: distribute the $\lambda$

Title: Re: TT's Maths Thread
Post by: TrueTears on May 03, 2011, 10:52:20 pm

Quote from: kamil9876 on May 03, 2011, 10:21:50 pm

what happened on Sunday?

Hint: distribute the $\lambda$

i replied to u on msn, lol my friend didnt reply me, didnt end up going on sunday just worked for parents. maybe sometime this week... ill text u

ok ill try distributing it

Title: Re: TT's Maths Thread
Post by: TrueTears on May 04, 2011, 06:30:11 am

o i got it, thanks kamil

[IMG]http://img684.imageshack.us/img684/7002/expanding.jpg[/img]

How would I do this?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 08, 2011, 03:46:03 am

[IMG]http://img607.imageshack.us/img607/2628/assignment5.jpg[/img]

Just wondering for the first one, it should be false right? Because for H to be a subgroup of G it has to satisfy 3 conditions.

1. H has to be closed under *
2. The identity element e of G is in H
3. for all a in H, it is true that the inverse of a is in H.

So H only satisfies 1. But I can't seem to find a counter example...

Also for the 2nd part of the question, I also can't seem to find a counterexample again...

Furthermore, just a question that arose when I thought about the 2nd part of the Q, if a group G has finite elements then does each subgroup of G also have finite elements? Or can a group G that has a finite order have a subgroup that has an infinite order?

Cheers.

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 08, 2011, 12:32:22 pm

Quote

Furthermore, just a question that arose when I thought about the 2nd part of the Q, if a group G has finite elements then does each subgroup of G also have finite elements? Or can a group G that has a finite order have a subgroup that has an infinite order?

What is a subgroup? it is a subset first of all, so of course this is true.

The other quetions are interesting. The first one is actually TRUE :D

Because look, if H is non-empty then it must contain some element $g$ , and since it is close under the operation it must contain: $g*g*g*g...*g$ , But $g^n=e$ for some $n$ (as $H$ is finite) Hence it contains the identity. Likewise if $g\neq e$ then $g^n=e$ for some $n>1$ and so $g^{n-1}=g^{-1}$ .

The second I leave to you. In fact I remember solving some other problems like this of the form "finite set with certain conditions => it is a group" and there are stronger results.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 08, 2011, 06:59:15 pm

Just a few questions about your proof.

Firstly, if H is closed under the operation of G, then does that mean for any element $g \in H$ , then $g^n \in H$ for all $n \in \mathbb{Z}$ , or is the restriction on $n$ , $n \in \mathbb{Z}^+$ ?

When you tried to show that the identity element of G is also in H, there's a problem.

Since H is a of finite order, then there exists some element $g \in H$ such that $g^m = e$ for some $m \in \mathbb{Z}^+$

That means H contains a identity element. However we still haven't shown that this identity element is the same identity element of G?

Since we don't know if there exists a $m \in \mathbb{Z}^+$ such that $g^m = e$ in G because we don't know if G is of finite order or not. We know that the subset H is finite but this does not imply G is finite. G can be an infinite set.

Also just to generalise, if we have a group G (does not necessarily have to be cyclic) and consider an element g in G. If $g^m = e$ for some $m \in \mathbb{Z}^+$ then G is of finite order. And the converse is also true, ie, if G is of finite order, then there exists some g in G such that $g^m = e$ for some $m \in \mathbb{Z}^+$ .

Is that true?

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 08, 2011, 10:52:30 pm

Yes sorry I read it slightly wrong(i thought it said $G$ is finite), see the new post I changed

Quote

(as G is finite)

Quote

(as H is finite)

and now the proof is correct.

Quote

Firstly, if H is closed under the operation of G, then does that mean for any element g \in H, then g^n \in H for all n \in \mathbb{Z}, or is the restriction on n, n \in \mathbb{Z}^+?

Yes correct we can only infer it for strictly positive integers $n$ , but that is no issue in my proof.

Basically we know that for any element $g$ there are ONLY two options:

(1) $ g,g^2,g^3....$ are all distinct

(2) $g^n=e$ for some positive integer $n$

(I think you even asked about this theorem some posts ago)

Now obviously (1) cannot occur as $H$ is finite. Hence (2) occurs, that is my proof.

Quote

That means H contains a identity element. However we still haven't shown that this identity element is the same identity element of G?

No, you see $H$ is just some random subset of $G$ and my argument above is looking at $g^1,g^2..$ as elements of $G$ with the operations from $G$ . (and then noticing they are also in $H$ )

Quote

Also just to generalise, if we have a group G (does not necessarily have to be cyclic) and consider an element g in G. If g^m = e for some m \in \mathbb{Z}^+ then G is of finite order. And the converse is also true, ie, if G is of finite order, then there exists some g in G such that g^m = e for some m \in \mathbb{Z}^+.

Is that true?

Actually your converse is true (It is clearly true by the pigeonhole principle, and you can take $g$ to be anything you don't just have to say "there exists", it holds "FOR ALL g" ie an even stronger statement)

The original statement is false though. Any easy counterexample is this:

$(\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z}) \times...$ (infinite product). It is an infinite group but every element is of order 2.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 09, 2011, 12:13:07 am

Ah okay, i get it now, but also just want to make clear of a few things:

If G is a cyclic group and has generator g, then:

1. $\forall m \in \mathbb{Z}^+$ , if $g^m \neq e$ then G is of infinite order. The converse is also true.

2. If $g^m = e$ for some $m \in \mathbb{Z}^+$ then G is of finite order. The converse is also true.

Now i rmb talking to you about these 2 facts and we proved both were true. Now I wanna ask, what if we don't know whether G is cyclic or not and we pick a element g from G such that g might not be a generator of G.

Then 1. and 2. are not true anymore right? But we can adjust it to:

1. If G is of infinite order, then $g^m \neq e \ \forall m \in \mathbb{Z}^+$

2. If G is of finite order, then $g^m = e$ for some $m \in \mathbb{Z}^+$

correct?

Also for the 2nd part, is this right?

Counterexample: Let the group G be $<\mathbb{Z}, +>$ . Let H be the infinite subset $\mathbb{Z}^+$ which is closed under +. However the identity element 0 of $\mathbb{Z}$ is not in $\mathbb{Z}^+$ thus H is not a subgroup of G.

Also just a new question, show that a group with only a finite number of subgroups must be finite.

I've made a start on this question but I just don't quite know if it's heading in the right direction lol.

Assume that a group G is of finite order. Let $g_1$ be an element in G.

Consider the subgroup $<g_1>$ of G. If $<g_1>$ is of infinite order, then it is isomorphic to $\mathbb{Z}$ and thus has an infinite number of subgroups (Since all subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ for $n \in \mathbb{Z}^+$ ) and thus G has an infinite number of subgroups, contradicting our assumption that G is of finite order.

So we can assume that $<g_1>$ is of finite order. If $<g_1>$ is of finite order then it is isomorphic to $\mathbb{Z}_n$

Thus if $G=<g_1>$ then G is isomorphic to $\mathbb{Z}_n$ and since $\mathbb{Z}_n$ has a finite number of subgroups, G must also has a finite number of subgroups.

Now this is where I am stuck. First, what if $g_1$ is not a generator for G? And also I just "assumed" $\mathbb{Z}_n$ has a finite number of subgroups, how can you actually prove that $\mathbb{Z}_n$ has a finite number of subgroups?

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 09, 2011, 01:24:07 pm

Quote

Then 1. and 2. are not true anymore right? But we can adjust it to:

1. If G is of infinite order, then g^m \neq e \ \forall m \in \mathbb{Z}^+

2. If G is of finite order, then g^m = e for some m \in \mathbb{Z}^+

correct?

2 is true I verified it in previous post. 1 is not true I showed you a counterexample in the previous post.

Quote

Also for the 2nd part, is this right?

Counterexample: Let the group G be <\mathbb{Z}, +>. Let H be the infinite subset \mathbb{Z}^+ which is closed under +. However the identity element 0 of \mathbb{Z} is not in \mathbb{Z}^+ thus H is not a subgroup of G.

yeah.

As for the next question. Decent start.

Here is a hint: You are right, what if $g_1$ was not a generator? then there is some $g_2$ not in $<g_1>$ , so $<g_2>$ would be a DIFFERENT subgroup. Now what if there was a $g_3$ that was not in $<g_1>$ and not in $<g_2>$ etc. etc. (can this argument go on forever?), what happens when it stops?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 09, 2011, 05:23:27 pm

Cheers

but what counter example? You mean

Quote

$ (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z}) \times$ ... (infinite product). It is an infinite group but every element is of order 2.

? what do you mean every element is of order 2? you didn't show that an element g in that group can produce $g^m = e$ for some $m \in \mathbb{Z}^+$

Yeah for the other question, i thought about that as well, but look if we picked a g2 not in g1, it would certainly produce another subgroup, however there could also be elements IN <g_1> that could also produce another subgroup. eg consider Z_18, look at <1>, if we picked 2 from <1> and formed <2> we make another subgroup? Or is that already taken into consideration because like i said <g_1> is isomorphic to some Z_n and thus it has a finite number of subgroups. (is that reasoning right?) so if we had <1> for Z_18, then the other subgroups for <1> is already "taken care" of since we know Z_18 has a finite number of subgroups. So the question remains how do you show Z_n has a finite number of subgroups?

Now continuing on with the question, that means if we pick a g2 not in <g1> and form <g2>, how do you know <g2> would be a different subgroup? How do you define 'different'?

like in my previous example for Z_18, if we had <1>, then <2> is also a different subgroup, ie, <1> = {0, 1, 2, 3, ..., 17} and <2> = {0, 2, 4, ..., 16} theyre both different subgroups of Z_18 yet $2 \in <1>$

So if we assume $G = \{g1, g2, g3, g4, \cdots \}$ I don't get what happens if we continue this argument, can't see it clearly...

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 09, 2011, 09:24:31 pm

Quote

? what do you mean every element is of order 2? you didn't show that an element g in that group can produce g^m = e for some m \in \mathbb{Z}^+

Yes every element is of order 2, if we take any element such as $a=(a_1,a_2,a_3...)$ then, $a+a=(a_1+a_1,a_2+a_2,..)=(0,0,...)$ .

As for the question, it is essential that you know what it means for two subgroups to be different in order to understand the statement "finitely many subgroups". Two subgroups are said to be the SAME if they are the same set.

Anyway you can do it like I suggested before but I realised there is an easier way: As there are only finitely many subgroups of $G$ , there are only finitely many cyclic subgroups of $G$ . Now as we saw before each cyclic subgroup of $G$ must be finite (for if it was infinite then it has infinitely many subgroups), hence G is a union of finitely many finite cyclic subgroups (each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself). Hence G is finite.

Title: Re: TT's Maths Thread
Post by: /0 on May 10, 2011, 01:24:20 am

Suppose $G$ has $n$ subgroups. If each subgroup is finite and the largest of them has $m$ elements (since you proved the subgroups must be finite) then $G$ has $\leq nm$ elements.
Suppose there was an element of $G$ which was not part of a subgroup. Then it would not be in its own cyclic group, which is impossible.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 10, 2011, 01:53:14 am

Quote from: kamil9876 on May 09, 2011, 09:24:31 pm

Quote
? what do you mean every element is of order 2? you didn't show that an element g in that group can produce g^m = e for some m \in \mathbb{Z}^+

Yes every element is of order 2, if we take any element such as $a=(a_1,a_2,a_3...)$ then, $a+a=(a_1+a_1,a_2+a_2,..)=(0,0,...)$ .

As for the question, it is essential that you know what it means for two subgroups to be different in order to understand the statement "finitely many subgroups". Two subgroups are said to be the SAME if they are the same set.

Anyway you can do it like I suggested before but I realised there is an easier way: As there are only finitely many subgroups of $G$ , there are only finitely many cyclic subgroups of $G$ . Now as we saw before each cyclic subgroup of $G$ must be finite (for if it was infinite then it has infinitely many subgroups), hence G is a union of finitely many finite cyclic subgroups (each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself). Hence G is finite.

Thanks, well i am still not quite sure how to complete the method before, can you answer the queries i have about it? They are the following:

When we said that <g1> = G, then we are done, is that because <g1> is isomorphic to $\mathbb{Z}_n$ where n is the order of <g1> and since $\mathbb{Z}_n$ has a finite number of subgroups, then so does G? But how can we actually prove that $\mathbb{Z}_n$ has a finite number of subgroups?

So if we consider a g2 not in <g1> then <g2> will be another DIFFERENT subgroup, how can you show <g2> is a DIFFERENT subgroup, ie <g1> and <g2> are not the same set?
Is it because if we assume <g1> and <g2> are the same set then g2 must be in <g1> contradicting the fact that we assumed g2 is not in <g1>?

We know that both <g1> and <g2> are finite sets (because of their isomorphisms to $\mathbb{Z}_n$ for some n). However I am still a bit lost on how to complete the proof.

First why don't we also consider elements IN <g1>? Since these elements could also produce a different subgroup, ie, consider the subgroup <2> of $\mathbb{Z}_{18}$ . If we consider the element 6 in <2>, then <6> is a different subgroup as well...

Second, I don't see how picking g1, g2, g3... and then forming cyclic groups out of these elements suffices to show G is finite.

Can you please explain the loose ended questions above?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 10, 2011, 04:05:54 am

Quote from: kamil9876 on May 09, 2011, 09:24:31 pm

there is an easier way: As there are only finitely many subgroups of $G$ , there are only finitely many cyclic subgroups of $G$ . Now as we saw before each cyclic subgroup of $G$ must be finite (for if it was infinite then it has infinitely many subgroups), hence G is a union of finitely many finite cyclic subgroups (each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself). Hence G is finite.

Also for your new method, i have a few questions, how does there being finitely many subgroups of $G$ imply that there are finitely many cyclic subgroups of $G$ ?

Also why is G a union of the finitely many finite cyclic subgroups? How do you know that each element of G is in some subgroup? Aren't you basing this on the assumption that each element of G is in at least one subgroup and the subgroup have to be of finite order. What if the subgroup is not of finite order? (you are assuming that G has finitely many subgroups, but that doesn't imply each subgroup has a finite order) What if an element of G is not in any of its subgroups?

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 10, 2011, 04:34:57 pm

Quote

how does there being finitely many subgroups of G imply that there are finitely many cyclic subgroups of G?

The cyclic subgroups are subgroups. A subcollection of a finite collection of objects is finite, that is obvious?

Quote

How do you know that each element of G is in some subgroup

Answer:

Quote

(each element is in some subgroup (possibly more than one), ie the cyclic subgroup generated by itself)

AKA $g \in <g>$ so each element is in some cyclic subgroup.

Quote

What if the subgroup is not of finite order?

If the CYCLIC subgroup $<g>$ is not of finite order then it has infinitely many subgroups $<g^k>$ k=1,2,3,4.. (as you showed here

Quote

If <g_1> is of infinite order, then it is isomorphic to \mathbb{Z} and thus has an infinite number of subgroups

Title: Re: TT's Maths Thread
Post by: TrueTears on May 16, 2011, 06:17:12 am

[IMG]http://img713.imageshack.us/img713/8/ps9s.jpg[/img]

how do i do part a) and b)?

Cheers

Title: Re: TT's Maths Thread
Post by: Mao on May 16, 2011, 11:02:22 am

a) the set of all solutions is a vector space if it is closed under addition and scalar multiplication (taking into account the n-degrees of freedom from constants of integration)

that is, if $\textbf{y}$ is a solution, then $a\textbf{y}$ must also be a solution.

$\frac{d(a\textbf{y})}{dt} = a\frac{d\textbf{y}}{dt} = a(A\textbf{y}) = A(a\textbf{y})$ , satisfied

if $\textbf{y}_1$ and $\textbf{y}_2$ are solutions, then $\textbf{y}_1+\textbf{y}_2$ must also be a solution.

$\frac{d(\textbf{y}_1+\textbf{y}_2)}{dt} = \frac{d\textbf{y}_1}{dt} + \frac{d\textbf{y}_2}{dt} = A\textbf{y}_1 + A \textbf{y}_2 = A(\textbf{y}_1+\textbf{y}_2)$ , satisfied

$\textbf{y}$ is a subspace.

b) $A\textbf{y} = c_1 \lambda_1 e^{\lambda_1 t} \textbf{v}_1 + c_2 \lambda_2 e^{\lambda_2 t} \textbf{v}_2 \cdots$ , since v1, v2,... are eigenvectors of A.
differentiating normally, $\frac{d}{dt} \textbf{y} = c_1 \lambda_1 e^{\lambda_1 t} \textbf{v}_1 + c_2 \lambda_2 e^{\lambda_2 t} \textbf{v}_2 \cdots$

They are the same, verified.

Title: Re: TT's Maths Thread
Post by: /0 on May 16, 2011, 11:07:35 am

a)
You need to show that $\mathbf{0}$ solves the DE.
You need to show that if $\mathbf{y}_1$ and $\mathbf{y}_2$ solve the DE, then so does $a\mathbf{y}_1+b\mathbf{y}_2$ (adding equations etc.)

b)
Suppose that $\mathbf{y}=\mathbf{e}^{\lambda_1 t}\mathbf{v}_1$ solves the DE.

Then $\lambda_1 \mathbf{e}^{\lambda_1 t}\mathbf{v}_1= A\mathbf{e}^{\lambda_1 t}\mathbf{v}_1$

$(A-\lambda_1 I)\mathbf{e}^{\lambda_1 t}\mathbf{v}_1=0$ .

$(A-\lambda_1 I)\mathbf{v}_1=0$ .

So $\lambda_1$ must be an eigenvalue and $\mathbf{v}_1$ must be an eigenvector of $A$ . Then by part a), you take linear combinations of them.

(Beaten!)

Title: Re: TT's Maths Thread
Post by: TrueTears on May 23, 2011, 05:29:52 pm

[IMG]http://img860.imageshack.us/img860/9325/18467514.jpg[/img]

How would you do these 2 questions?

Cheers

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 23, 2011, 05:44:38 pm

In fact every homomorphism from $\mathbb{Z}$ to any group is uniquely determined by the image of $1$ .

Prove that $\phi(x)=6x$ (where by $6x$ I of course mean that element of $\mathbb{Z}/10\mathbb{Z}$ ).

You can prove it for x>0 by noticing that $x=1+1...+1$ $x$ times and so $\phi(x)=\phi(1+1+...+1)$ , now use the homomorphism property. For x<0 the proof is similair you just have to notice that $\phi(-1)=-6$ . Now it should be clear.

As for the first question notice that $48=8*2*3*1$ So you should take $\mathbb{Z}_8 \times G_2 \times G_3 \times G_4$ where $G_2$ is a subgroup of $\mathbb{Z}_4$ with 2 elements, $G_3$ is a subgrpup of $\mathbb{Z}_9$ with three elements and $G_4$ is a subgroup of $\mathhbb{Z}_7$ with 1 element. Can you find such groups?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 23, 2011, 06:00:06 pm

Quote from: kamil9876 on May 23, 2011, 05:44:38 pm

In fact every homomorphism from $\mathbb{Z}$ to any group is uniquely determined by the image of $1$ .

Prove that $\phi(x)=6x$ (where by $6x$ I of course mean that element of $\mathbb{Z}/10\mathbb{Z}$ ).

You can prove it for x>0 by noticing that $x=1+1...+1$ $x$ times and so $\phi(x)=\phi(1+1+...+1)$ , now use the homomorphism property. For x<0 the proof is similair you just have to notice that $\phi(-1)=-6$ . Now it should be clear.

As for the first question notice that $48=8*2*3*1$ So you should take $\mathbb{Z}_8 \times G_2 \times G_3 \times G_4$ where $G_2$ is a subgroup of $\mathbb{Z}_4$ with 2 elements, $G_3$ is a subgrpup of $\mathbb{Z}_9$ with three elements and $G_4$ is a subgroup of $\mathhbb{Z}_7$ with 1 element. Can you find such groups?

Not really sure on that homomorphism question, I'm a bit vague in this area, can you finish off the gaps in your working?

Also for the subgroup question for order 48, LCM(8*2*3*1) is not 48 so you can't really do that?

Also what about the other question? (the isomorphic one?)

Thanks.

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 23, 2011, 06:50:26 pm

You can take G_2={0,2}, G_3={0,3,6}, G_4={0} Viola.

What's not clear? The point is $\phi(x)=6x$ is what you have to really prove and then you can answer i and ii immediately.

As for the question I missed I think you should use the chineese remainder theorem to decompose it into a product of prime power groups and then it may be easier to analyse.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 23, 2011, 07:01:02 pm

Thanks.

Well what I mean is how did you get $\phi(x)=6x$ from the $\phi(1) = 6$ ? And how do you prove it for the case x<0?

Quote

As for the question I missed I think you should use the chineese remainder theorem to decompose it into a product of prime power groups and then it may be easier to analyse.

Can you show me? I've never used or heard of this technique of CRT to decompose...

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 23, 2011, 07:17:19 pm

for $x>0$ , we have $\phi(x)=\phi(1+1+1...+1)=\phi(1)+\phi(1)....+\phi(1)=6+6+6...+6=6x$

$0=\phi(0)=\phi(-x)+\phi(x)$ so we have $\phi(-x)=\phi(x)$ , hence $\phi(x)=6x$ is verified for negative $x$ as well.

As for the other one:

$\mathbb{Z}_{126}=\mathbb{Z}_{2*3^2*7}=\mathbb{Z}_2\mathbb{Z}_{3^2} \times \mathhbb{Z}_{7}$

Title: Re: TT's Maths Thread
Post by: TrueTears on May 23, 2011, 08:38:44 pm

Quote from: kamil9876 on May 23, 2011, 07:17:19 pm

for $x>0$ , we have $\phi(x)=\phi(1+1+1...+1)=\phi(1)+\phi(1)....+\phi(1)=6+6+6...+6=6x$

$0=\phi(0)=\phi(-x)+\phi(x)$ so we have $\phi(-x)=\phi(x)$ , hence $\phi(x)=6x$ is verified for negative $x$ as well.

As for the other one:

$\mathbb{Z}_{126}=\mathbb{Z}_{3^2*7^2}=\mathbb{Z}_{3^2} \times \mathhbb{Z}_{7^2}$ (because $3^2$ and $7^2$ have no common factor) decompose the others in a similair fashion.

Thanks for that, but it should be $\phi(-x) = -\phi(x)$ right?

Also is this right for ii?

$Ker(\phi) = \{x \in \mathbb{Z} | \phi(x) = 0\}$ where 0 is the identity element in $\mathbb{Z}_{10}$

Thus we have $6x \equiv 0 \pmod{10} \implies x = 10k$ where $k \in \mathbb{Z}$

Also $\mathbb{Z}_{4} \times \mathbb{Z}_{126}$

The factorisation you got for 126 is wrong right... should be:

$\mathbb{Z}_4 = \mathbb{Z}_{2^2}$ and $\mathbb{Z}_{126} = \mathbb{Z}_{23^27}$ which means it is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$

So $\mathbb{Z}_{4} \times \mathbb{Z}_{126}$ is isomorphic to $\mathbb{Z}_{2^2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$

For $\mathbb{Z}_6 \times \mathbb{Z}_{84}$

$\mathbb{Z}_6$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3$ and $\mathbb{Z}_{84}$ is isomorphic to $\mathbb{Z}_{2^2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{7}$

So $\mathbb{Z}_6 \times \mathbb{Z}_{84}$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_{2^2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{7}$

Now what?

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 23, 2011, 08:48:33 pm

Quote

Thanks for that, but it should be \phi(-x) = -\phi(x) right?

yes sorry.

No it is not x=10k, because x=5 is in the kernel but it is not of that form. Think about it, when is 6x a multiple of 10? (ie when is 2*3*x a multiple of 2*5 ?)

Now the beauty of that decomposition is that you can see that the first group has an element of order 9 (the element (0,0,1,0) in $\mathbb{Z}_{2^2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$ ). But you can argue that the second group does not and hence that means they are NOT isomorphic.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 23, 2011, 08:56:53 pm

Quote from: kamil9876 on May 23, 2011, 08:48:33 pm

Quote
Thanks for that, but it should be \phi(-x) = -\phi(x) right?

yes sorry.

No it is not x=10k, because x=5 is in the kernel but it is not of that form. Think about it, when is 6x a multiple of 10? (ie when is 2*3*x a multiple of 2*5 ?)

Now the beauty of that decomposition is that you can see that the first group has an element of order 9 (the element (0,0,1,0) in $\mathbb{Z}_{2^2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{7}$ ). But you can argue that the second group does not and hence that means they are NOT isomorphic.

oh shit opps, i mean x = 5k since lcm(6, 10) = 30, thats better yeah?

yeah check my edited post, i realised that lol

Title: Re: TT's Maths Thread
Post by: TrueTears on May 23, 2011, 11:06:15 pm

[IMG]http://img716.imageshack.us/img716/9218/4bcz.jpg[/img]

Any ideas on these kamil?

I'm a bit lost on the first one however the second one I have a rough starting, not sure where to go with it though.

for a homomorphism $\phi: G \to G'$ , then we know $Ker(\phi)$ is a normal subgroup of G, by Lagrange's Theorem $|Ker(\phi)|$ divides $|G|$ .

This is the only thing I can think of which relates the order of G, but then i'm kinda stuck to prove what's required.

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 23, 2011, 11:45:20 pm

To prove the first one it depends on what you know, are you aware of what the Sign of a permutation is?

As for the second one, try to use the first isomorphism theorem.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 23, 2011, 11:52:08 pm

Nope, I haven't heard of what a sign of a permutation is... I know most about them, transpositions, even/odd and etc etc

Yeah I've thought about the first isomorphism theorem too but how is it of use here? can you show me by finishing the proof with it? i dont get how to apply it

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 24, 2011, 12:06:56 am

So suppose that $h \in A_n$ and $g \in S_n$ . Suppose we have written $g=c_1c_2..c_m$ (product of m cycles). Then $g^{-1}=c_m^{-1}....c_1^{-1}$ .

Thus:

$ghg^{-1}=c_1...c_m h c_m^{-1}...c_1^{-1}$

So you see, $ghg^{-1}$ has been written as a product of 2m+(number of cycles in h) cycles which is an even number A_n is closed under conjugation (I'm assuming u already know it is a group).

Title: Re: TT's Maths Thread
Post by: TrueTears on May 24, 2011, 12:51:49 am

Quote from: kamil9876 on May 24, 2011, 12:06:56 am

So suppose that $h \in A_n$ and $g \in S_n$ . Suppose we have written $g=c_1c_2..c_m$ (product of m cycles). Then $g^{-1}=c_m^{-1}....c_1^{-1}$ .

Thus:

$ghg^{-1}=c_1...c_m h c_m^{-1}...c_1^{-1}$

So you see, $ghg^{-1}$ has been written as a product of 2m+(number of cycles in h) cycles which is an even number A_n is closed under conjugation (I'm assuming u already know it is a group).

Thanks, I don't quite get what you mean by "...A_n is closed under conjugation (I'm assuming u already know it is a group). "? What is conjugation? and wat r u assuming to be a group..?

And also how have you shown A_n is a normal subgroup of S_n? Shouldn't you prove $gA_ng^{-1} = A_n \ \forall \ g \in S_n$ ? Why did you just pick an element from A_n....?

Also your last sentence doesn't seem to make sense lol

Quote

So you see, $ghg^{-1}$ has been written as a product of 2m+(number of cycles in h) cycles which is an even number A_n is closed under conjugation

there should be a word b/w number and A_n?

What about the other Q btw?

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 24, 2011, 04:28:06 pm

it should be "is an even number, and so A_n is closed..."

Conjugation means taking conjugates, ie $ghg^{-1}$ is a conjugate of $h$ . In order to show that a subgroup $h$ is normal it suffices to show that $g \in G, H\in H \implies ghg^{-1} \in H$ , that is what I meant by closed under conjugation. Of course you also have to show that it is a subgroup, but I'm assuming that u already know A_n is a subgroup.

(see here http://en.wikipedia.org/wiki/Normal_subgroup#Definitions, I used that definition of normal subgroup, there are other definitions such as yours but this one is the easiest to use when showing that something is normal)

Title: Re: TT's Maths Thread
Post by: TrueTears on May 24, 2011, 05:42:13 pm

Ahh okay thanks for that, never seen that definition before haha

Also for the 2nd question, i know that the first isomorphism theorem is that the image of $\phi$ is isomorphic to the quotient group $G / Ker(\phi)$

but i dono what to do with it, can u show me

oh nvm i got it lol

Title: Re: TT's Maths Thread
Post by: TrueTears on May 24, 2011, 08:39:09 pm

And just a final proof question: Let n be an odd integer and let G be an abelian group of order 2n. Prove that G has exactly one element of order 2.

Okay so what i started off with was let this element by g. Then we can create <g> and by lagrange's theorem we know that |G|/|<g>| = some constant. So (2n)/|<g>| = some constant, so how do i show that |<g>| is 2 and prove uniqueness?

Cheers

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 24, 2011, 09:25:09 pm

Well you can't really show that |<g>|=2 as g is arbitrary and so may be an element that is not of order 2. I guess this can be solved a number of ways depending on what theorems are available to you. Are we allowed to use the structure theorem for finitely generated abelian groups?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 24, 2011, 10:58:55 pm

Quote

Are we allowed to use the structure theorem for finitely generated abelian groups?

wats that? i should know it, maybe i just dono the theorem name, what's the theorem about?

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 24, 2011, 11:28:35 pm

google it. I think you've seen it before. The fact that every finitely generated abelian group is a product of $\mathbb{Z}$ and quotients $\mathbb{Z}_m$

Title: Re: TT's Maths Thread
Post by: TrueTears on May 25, 2011, 12:20:38 am

ohhh yeah i know that, how does that play any role here though...?

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 25, 2011, 07:01:37 pm

So by the structure theorem we can assume that:

$G=\mathbb{Z}_{d_1} \times \mathbb{Z}_{d_2}... \times \mathbb{Z}_{d_m}$

We know that $2n=d_1d_2..d_m$ it follows that one of the $d_i=2$ while all the others are odd, suppose $d_1=2$ and all others are odd. So

$G=\mathbb{Z}_{2} \times \mathbb{Z}_{d_2}... \times \mathbb{Z}_{d_m}$

Can you now find an element of order 2 here? should be very easy.

Now try to prove uniqueness by proving that no other element is of order 2, how would such an element look like?

Title: Re: TT's Maths Thread
Post by: TrueTears on May 25, 2011, 07:37:42 pm

oh i think i get it now, an element of order 2 is (1, 0, 0, ..., 0)

now since all the other d_i's are odd, then any other element would have an order of at least 3, ie d_2 =>3, so say if d_2 was 3, then an element would look like (0, 1, 0, 0, ..., 0) which would have an order of at least 3.

Title: Re: TT's Maths Thread
Post by: kamil9876 on May 25, 2011, 07:50:07 pm

In general, let p be a prime and n a number not having p is a factor. Then every abelian group of order pn has a unique element of order p.

Title: Re: TT's Maths Thread
Post by: TrueTears on May 25, 2011, 07:57:28 pm

alright, nice, that's a really nice result lol

Title: Re: TT's Maths Thread
Post by: TrueTears on June 03, 2011, 04:11:44 am

Prove that for any group G, and any element g in G that gG = G

So if we pick g = e, then eG = G, but the question says g is ANY element of G, how can i show that gG = G for some g not the identity?

Title: Re: TT's Maths Thread
Post by: /0 on June 03, 2011, 08:53:49 am

$gG \subseteq G$ :
Let $h \in G$ , then $gh \in G$ for all $h \in G$ by closure. Then, since $h$ was arbitrary, $gG \subseteq G$ .

$G \subseteq gG$ :
For any $a \in G$ , does there exist $b \in G$ such that $gb = a$ ? By construction, yes: $b= g^{-1}a$ . Hence, for every element in $G$ , that element exists in $gG$ , so $G \subseteq gG$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on June 03, 2011, 05:41:22 pm

Quote from: /0 on June 03, 2011, 08:53:49 am

$gG \subseteq G$ :
Let $h \in G$ , then $gh \in G$ for all $h \in G$ by closure. Then, since $h$ was arbitrary, $gG \subseteq G$ .

$G \subseteq gG$ :
For any $a \in G$ , does there exist $b \in G$ such that $gb = a$ ? By construction, yes: $b= g^{-1}a$ . Hence, for every element in $G$ , that element exists in $gG$ , so $G \subseteq gG$ .

Cheers man

Title: Re: TT's Maths Thread
Post by: TrueTears on June 04, 2011, 03:16:33 am

How do i show that any two left cosets of a subgroup H of a group G is either disjoint or equal?

I mean it's so trivial but how do you actually prove it?

I kinda get it but i don't think my proof is that complete:

consider g1H and g2H, clearly g1H n g2H = null set or g1H n g2H = {some non empty set}

Now we just have to show that if g1H n g2H = {some non empty set} then g1H = g2H

so then what...?

Also how is the symmetric group on 3 letters S_3 isomorphic to Z_6?

~~Well i know that it is isomorphic if S_3 is cyclic, however how can i show it is (or is not?)?~~

Well actually nvm it's not isomorphic since S_3 is not abelian while Z_6 is, however i'd still like to know: is S_3 cyclic? is there a systematic way of doing this rather than testing all 6 elements of S_3 and then raising them to integral powers to see if they generate S_3 or not?

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 04, 2011, 11:29:40 am

If you already know it's not isomorphic to Z_6 then it cannot be cyclic.

==============================================================

if they have a non-empty intersection, that means we have $h_1,h_2 \in H$ s.t

$g_1h_1=g_2h_2$

and so

$g_2^{-1}g_1=h_2h_1^{-1} \in H$ .

But any element in $g_1H$ can be written as $g_1h=g_2(g_2^{-1}}g_1h)$ but by closure the thing inside the bracket is in $H$ . hence $g_1H \subset g_2H$ . You can get the other inclusion with the same argument.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 04, 2011, 06:48:07 pm

cheers kamil, actually i found an even easier way, consider an element x in the non empty set, then $x \in g1H$ so xH = g1H. But $x \in g2H$ so xH = g2H. Thus g1H=xH=g2H

this is a valid argument right?

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 04, 2011, 06:58:50 pm

You should justify why does $x \in g_1 H$ imply $xH=g_1H$ ? I think you may be using circular reasoning (ie. using the result we are actually trying to prove) but I cannot say because I don't know your reason.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 04, 2011, 07:02:23 pm

Quote from: kamil9876 on June 04, 2011, 06:58:50 pm

You should justify why does $x \in g_1 H$ imply $xH=g_1H$ ? I think you may be using circular reasoning (ie. using the result we are actually trying to prove) but I cannot say because I don't know your reason.

it's trivial to prove isn't it? but i guess the reasoning would be like, x would be of the g1h for some h in H, so xH = g1hH = g1(hH) = g1H

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 04, 2011, 07:05:55 pm

yep that's fine.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 06, 2011, 06:02:28 am

[IMG]http://img838.imageshack.us/img838/6016/finalquestions.jpg[/img]

Not sure how to do the first question...

Second question, is this right? D_8 has an order of 2*8 = 16 so it can't be isomorphic to Z_8 which only has an order of 8.

I read in my book that D_n has an order of 2n, but how do i prove this?

Third question, is my reasoning for this question right?

Since both are cyclic groups of finite order, then G and H are both isomorphic to Z_2003, thus they are isomorphic to each other.

Fourth question, again can someone please check my reasoning.

None of the 3 sentences are true, in other words all of the D_i's for i = 10, 8, 6 are not isomorphic to their claimed counterparts, since their orders aren't the same.

So again it comes back to the question, how do i show D_n has order 2n?

Fifth question, dw i figured it out

lol dw about this question

7th question: i kinda know how to do this question but dono how to complete it.

basically we just have to show that any right coset of G has the same number of elements of H, the subgroup.

so we can define a one-to-one, onto function mapping from H to Hx where x is an element of G.

So define phi:H -> Hx, phi(h) = hx for all h in H.

to show 1-to-1, assume phi(h1) = phi(h2) -> h1x = h2x -> h1=h2

to show onto, assume for some y in Hx we must show there exists some h in H such that hx = y. But by the definition of right coset, if y is in Hx then there has to exist some h such that hx = y.

So all right cosets of H in G has the same number of elements.

Not sure how to do this question, help please :D

i) ive already done

ii) just to confirm that's just the same Q as as q 7 right, except its for left cosets.

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 06, 2011, 01:20:07 pm

First question: C'mon I thought you liked combinatorics! What is a binary operation? if is a function of two variables f(a,b). there are $m^2$ different pairs (a,b) and we must assign to each some element (m choices). So how many possible different assignments?

Second: Actually your using a different definition of $D_8$ , (their $D_8$ is what your textbook would call $D_4$ ), ie there are $8$ symmetries of the square so they must be referring to what you would call $D_4$ . So a simple cardinality check won't suffice. (Hint, Z_8 has an element of order 8, does any symmetry of the square have such an order ?)

Third: Yes in general any two cyclic groups of the same finite cardinality are isomorphic, i guess they want you to prove this? Find an isomorphism which maps the generator from one to the generator of the other.

cbf... this is enough for you to chew for now anyway.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 06, 2011, 05:17:12 pm

Q1: I get what you mean, but I dono how do the q, a binary operation on a set S is just a function mapping SxS into S. How to work out how many binary operations?

Q2: O i see... well then my Q4 is wrong too, so in general how do i show that these aren't isomorphic? First how do you show that D_4 doesn't have an element of order 8? And how did you know to pick order 8? I would have never thought of that...

Same with question 4, how do i do them?

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 06, 2011, 05:36:04 pm

Q1: i pretty much spelt it out for you, it's combinatorics.

Q2: Well a group with 8 elements is isomorphic to Z_8 iff it has an element of order 8 so it's always the perfect strategy actually. To show D_4 doesn't have an element of order 8 just work out the order of each symmetry in D_4: every rotation is a multiply of 2pi/4 ie $2k\pi/4$ but if you multiply that by 4 you get $2k\pi$ and so every rotation has order at most 4. Every reflection has order 2. Hence no element is of order 8.

For the first part of q4, the same strategy works.

for the second part just do a simply cardinality check.

as for the third one, try to find an isomorphism. (hint: label the vertices 1,2,3 and try to create a map from there)

Title: Re: TT's Maths Thread
Post by: TrueTears on June 06, 2011, 06:35:40 pm

Quote from: kamil9876 on June 06, 2011, 05:36:04 pm

Q1: i pretty much spelt it out for you, it's combinatorics.

Q2: Well a group with 8 elements is isomorphic to Z_8 iff it has an element of order 8 so it's always the perfect strategy actually. To show D_4 doesn't have an element of order 8 just work out the order of each symmetry in D_4: every rotation is a multiply of 2pi/4 ie $2k\pi/4$ but if you multiply that by 4 you get $2k\pi$ and so every rotation has order at most 4. Every reflection has order 2. Hence no element is of order 8.

For the first part of q4, the same strategy works.

for the second part just do a simply cardinality check.

as for the third one, try to find an isomorphism. (hint: label the vertices 1,2,3 and try to create a map from there)

oh i see, so its just m^(m^2) right?

for Q 2 does that mean for ANY dihedral group, the maximum order of any rotation permutation is equal to n where n corresponds to the regular n-gon that D_n is acting upon. And also for ANY dihedral group, any reflection permutation has order 2, since if you reflected it "twice" then you just end up with the original again.

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 06, 2011, 06:56:44 pm

yes.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 06, 2011, 11:37:31 pm

cheers i get it now, so how about this question, how many elements of order 7 does S_4 have?

Title: Re: TT's Maths Thread
Post by: TrueTears on June 20, 2011, 06:40:43 am

[IMG]http://img94.imageshack.us/img94/3196/idempotent.jpg[/img]

I can do i) but just a bit stuck on ii).

my working for i) is: E^2-E=0 so that means $\lambda^2-\lambda$ is an eigenvalue of E^2-E=0, since the zero vector only has an eigenvalue of 0, then $\lambda^2-\lambda=0$ and so \lambda = 0 or 1

cheers :)

[IMG]http://img64.imageshack.us/img64/5682/adjacencymatrix.jpg[/img]

Denote the vertices of G G1, G2, G3, does the 2 in the matrix just means there's 2 paths of length 1 from G2 to G1? ie, look like this:

[IMG]http://img4.imageshack.us/img4/9931/graphgq.jpg[/img]

thx

[IMG]http://img854.imageshack.us/img854/1174/eigenvaluesqfrom2008.jpg[/img]

I sorta have an idea but dono how to finish this Q

If we let x be a eigenvector of $B^TB$ corresponding to $\lambda$ , then we have $B^TBx = \lambda x$

Now just playing around: $B^TBx.x = x. ((B^TB)^Tx)=x.(B^TB)x= x.(\lambda x)$

but dono how to get to the result $\lambda \ge 0$

[IMG]http://img718.imageshack.us/img718/1669/gramschmitprocess.png[/img]

Just stuck on part c)

my answer to part b) is $\hat{u_1} = (\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}), \hat{u_2} = (\frac{1}{3}, -\frac{2}{3}, -\frac{2}{3}), \hat{u_3} = (\frac{2}{3}, \frac{2}{3}, -\frac{1}{3})$

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 20, 2011, 07:43:20 pm

Here is a solution but it may use theorems you are not familiar with:

$X(X-1)$ is an annhilator polynomial for $E$ hence the minimal polynomial is either X,(X-1) or X(X-1). Since 0 is not an eigenvalue that means that the minimal polynomial does not have 0 as a root and hence it must be X-1 so E=I.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 20, 2011, 08:10:32 pm

Quote from: kamil9876 on June 20, 2011, 07:43:20 pm

Here is a solution but it may use theorems you are not familiar with:

$X(X-1)$ is an annhilator polynomial for $E$ hence the minimal polynomial is either X,(X-1) or X(X-1). Since 0 is not an eigenvalue that means that the minimal polynomial does not have 0 as a root and hence it must be X-1 so E=I.

lol i dont know what you mean, do you have a more elementary method?

Title: Re: TT's Maths Thread
Post by: TrueTears on June 21, 2011, 12:45:21 am

Actually for my question 4 (the gram schmidt question)

is this way of doing it correct? although it seems a bit too long, i was wondering if theres any shorter ways?

So let $o = (o_1, o_2, o_3)$ be a vector in $R^3$ , then $proj_W o = <o, \hat{u_1}>\hat{u_1} + <o, \hat{u_2}>\hat{u_2}$

so then we can basically sub in the respective components for o, compute the respective inner products etc etc and we get:

$proj_W o = (\frac{5}{9}o_1-\frac{4}{9}o+\frac{2}{9}o_3, -\frac{4}{9}o_1+\frac{5}{9}o_2+\frac{2}{9}o_3, \frac{2}{9}o_1+\frac{2}{9}o_2+\frac{8}{9}o_3)$

and so.. the matrix is pretty obvious from that, but yeah that took alot of computation, any slicker ways?

cheers

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 21, 2011, 10:15:57 am

Quote from: TrueTears on June 20, 2011, 08:10:32 pm

Quote from: kamil9876 on June 20, 2011, 07:43:20 pm
Here is a solution but it may use theorems you are not familiar with:

$X(X-1)$ is an annhilator polynomial for $E$ hence the minimal polynomial is either X,(X-1) or X(X-1). Since 0 is not an eigenvalue that means that the minimal polynomial does not have 0 as a root and hence it must be X-1 so E=I.
lol i dont know what you mean, do you have a more elementary method?

Well a bit hard to say because in some courses the fact that the roots of the minimal polynomial are exactly the eigenvalues is proven pretty early on. What can I use? Jordan Normal Form?

As for the eigenvalue question:
$ x^TB^TBx=\lambda x^Tx$

$(Bx)^T(Bx)=\lambda x^Tx$

$||Bx||^2=\lambda||x||^2$ where $||.||$ denotes the usual length.

hence the result. (I am assuming the matrices have real number coefficients?)

I think your third one didn't really take that much computation?

Title: Re: TT's Maths Thread
Post by: zzdfa on June 23, 2011, 09:02:57 pm

Another way to obtain the projection matrix is to observe that -every- projection matrix is orthogonally similar to a diagonal matrix with only 1s or 0s on the diagonal (bc projection is just eliminating components of the vector in some orthonormal basis - think geometrically)

So you found your orthonomal basis {u1,u2,u3}. Let U be the matrix with columns u1,u2,u3, then we should have P= U D U^T where D=[100,010,000] (because you want to project onto the span of the first 2 vectors).

how'd you go with the exam today?

Title: Re: TT's Maths Thread
Post by: zzdfa on June 23, 2011, 09:04:05 pm

edit: double post

Title: Re: TT's Maths Thread
Post by: TrueTears on June 24, 2011, 08:24:35 am

haha yeah, thanks guys i got them in the end anyway, yeah exam was easy as today :)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 13, 2011, 10:40:33 pm

Hey guys, I have a few questions that I've been stuck on for a few days, these questions have no solutions provided so I was wondering if anyone can check them for me. The first 3 questions I have completed with full working however I'm not entirely sure if they are correct, the last 4 I have no clue how to solve :3 Any help would be appreciated!

(http://img140.imageshack.us/img140/4136/q21f.jpg)

So first we need to work out the amount of each installment, $R$ , thus $1000000=Ra_{25:0.08} \implies R = 93678.779$

Now X purchases the annuity on 1 Jan 1986, so we need to work out the price that X bought the annuity for.

Since he purchased it on 1 Jan 1986 and the annuity expires at the end of 2005, the annuity has a maturity left of 19 years.

But since X pays a tax rate of 0.45 on the interest portion of each payment of R, then we need to split up R each period to find the interest payment and capital payment so we can account for the tax.

To derive an expression for the interest payment and capital payment of R each period we let L be the lump sum borrowed at the start of an arbitrary period, $n$ be the number of periods and $i$ be the effective interest rate per period. Then for the first payment period, the loan outstanding at the start of the period is L, the interest payment for the first period is $iL = i(Ra_{n:i}) = iR\left(\frac{1-v^n}{i}\right) = R(1-v^n)$ where $v = (1+i)^{-1}$ . Then clearly the capital component paid at the end of the first period is $Rv^n$ since $R(1-v^n) +Rv^n = R$ , we can show in general that for payment period k, the interest component paid at the end of the period is $R(1-v^{n-k+1})$ and the capital component paid at the end of the period is $Rv^{n-k+1}$

Now back to the question: let $u = (1+0.09)^{-1}$ and $v = (1+0.08)^{-1}$

The capital payments of each period starting from 1986 until the end of 2005 when discounted back to 1986 form the cash flow stream:

$u(Rv^{19}) + u^2(Rv^{18}) + \cdots + u^{19}(Rv)$

The interest payments of each period starting from 1986 until the end of 2005 when discounted back to 1986 form the cash flow stream: (Note we take into consideration the tax rate of 45% which is (1-0.45) = 0.55 of the interest payments)

$u(0.55R(1-v^{19})) + u^2(0.55R(1-v^{18})) + \cdots + u^{19}(0.55R(1-v))$

Now we need to combine both of these cash flow streams and form one single expression to evaluate it.

Let us match together the $v^{k}$ components:

$u(Rv^{19}) + u(0.55R(1-v^{19})) = 0.45uRv^{19} + 0.55uR$

$u^2(Rv^{18}) + u^2(0.55R(1-v^{18})) = 0.45u^2Rv^{18} +0.55u^2R$

And so on... so let us now group together the expressions with 0.55 as the coefficient:

$0.55uR + 0.55u^2R + \cdots + 0.55u^{19}R = 0.55R(u+u^2+ \cdots + u^{19}) \cdots [1]$

Now group together the expressions with 0.45 as the coefficient:

$0.45uRv^{19} + 0.45u^2Rv^{18} + \cdots + 0.45u^{19}Rv = 0.45R(uv^{19} + u^2v^{18} + \cdots + u^{19}v) \cdots [2]$

$[1] = 0.55Ra_{19:0.09} = 0.55(93678.779)\left(\frac{1-(1.09)^{-19}}{0.09}\right) = 461139.703695$

$[2] = 0.45R\left(uv^{19}\left(\frac{1-(uv^{-1})^{19}}{1-uv^{-1}}\right)\right) = 156912.68$

Thus $[1]+[2] = 618052.3837$

Now let the lump sum paid on April 1995 be L, L must satisfy the following equation of value: (Let $q = 1.1^{-1}$ )

$\left[0.55R(q+q^2+\cdots+q^9)\right] + \left[0.45R(qv^{19}+q^2v^{18} + \cdots + q^9v^{11}\right] + Lq^{9.25} = 618052.3837$

$0.55R\left(\frac{1-1.1^{-9}}{0.1}\right) + 0.45R\left[qv^{19}\left(\frac{1-(qv^{-1})^9}{1-qv^{-1}}\right)\right] + Lq^{9.25} = 618052.3837$

Solving the above yields $L = 596411.93$

(http://img823.imageshack.us/img823/7391/q24f.jpg)

First we need to work out the amount of each installment paid in advance from 1 Jan 1985 to 2005 (20 years).

Let each installment be R.

First note $(1+0.1) = (1-d)^{-1}$ where d is the effective annual rate of discount.

$d = \frac{1}{11}$

$10000=R\ddot{a}_{20:0.1} \implies 10000=R\left(\frac{1-1.1^{-20}}{d}\right) \implies R = 1067.81477$

Now we need to find on 1 Jan 1990 the total outstanding capital left to be repaid.

This is given by $R\ddot{a}_{15:0.1} = R\left(\frac{1-1.1^{-15}}{d}\right) = 8934.07$

Now from 1 Jan 1990, if B were to increase the yield to 0.12, then we need to find the new installments per period, these new R is given by:

(Note $(1+0.12) = (1-d)^{-1} \implies d = \frac{3}{28}$ )

$8934.07 = R\ddot{a}_{15:0.12} \implies 8934.07 = R\left(\frac{1-1.12^{-15}}{\frac{3}{28}}\right) \implies R = 1171.19499$

Now let A's final payment be L, L must satisfy the following equation of value:

$R\ddot{a}_{5:0.12} + L(1.12)^{-5}= 8934.07$

$\implies R\left(\frac{1-1.12^{-5}}{\frac{3}{28}}\right) + L(1.12)^{-5} = 8934.07$

$\implies L = 7411.61448$

(http://img18.imageshack.us/img18/6838/q27n.jpg)

Let $C_n$ be the capital component paid in arrear for the nth period and $I_n$ be the interest component paid in arrear for the nth period. Let $L_0$ be the loan outstanding.

From the question $C_1+(1-0.3)I_1 = 5000$

But $I_1 = 0.1L_0 \implies C_1 +0.07L_0 = 5000$

Likewise, $C_2+0.7I_2=5000$

But $I_2 =0.1L_1 = 0.1(L_0 - C_1)$ [This is obvious as $L_1$ represent the loan outstanding at the end of the first period, which is equal to the loan outstanding at the beginning of the period minus the capital paid in the first period]

So $C_2+0.07(L_0-C_1) = 5000$

Furthering this pattern, $C_3+0.7I_3 = 5000$

But $I_3 = 0.1L_2 = 0.1(L_1-C_2) = 0.1(L_0-C_1-C_2)$

Hence $C_3+0.07(L_0-C_1-C_2)=5000$

So we have:

$C_1 +0.07L_0 = 5000 \cdots [1]$

$C_2+0.07(L_0-C_1) = 5000 \cdots [2]$

$C_3+0.07(L_0-C_1-C_2)=5000 \cdots [3]$

From [1] we have $C_1 = 5000 - 0.07L_0$

So [2] becomes $C_2+0.07(L_0 - 5000 +0.07L_0) = 5000 \implies C_2 + 0.07L_0 - 0.07(5000) + 0.07^2L_0 = 5000 \implies C_2 = 5000+0.07(5000) - (0.07+0.07^2)L_0 \implies C_2 = 1.07(5000-0.07L_0) = 1.07C_1$

In [3] we have:

$C_3 = 5000- 0.07(L_0-C_1-C_2) \implies C_3 = 0.07C_1+0.07(1.07)C_1+C_1 \implies C_3 = 1.07C_1 + 0.07(1.07)C_1 = 1.07^2C_1$

This forms a recurrence relation and we can show that $C_n = 1.07^{n-1}C_1$

Now we need to find what $L_0$ is. Note that if we assume the person who bought this decreasing annuity does not pay tax, then we should have:

$\sum_{i=1}^{10} \left(C_i+I_i\right)v^i = L_0$ where $v = 1.1^{-1}$

$\implies \sum_{i=1}^{10} \left(C_i + \frac{1}{0.7}(5000-C_i)\right)v^i=L_0 \implies \sum_{i=1}^{10} \left(1.07^{i-1}C_1 + \frac{1}{0.7}(5000-1.07^{i-1}C_1)\right)v^i=L_0 \implies \sum_{i=1}^{10} \left(1.07^{i-1}(5000-0.07L_0) + \frac{1}{0.7}(5000-1.07^{i-1}(5000-0.07L_0))\right)v^i=L_0$

Now we can solve for $L_0, L_0 = 35117.0977$

The price that someone would pay for this decreasing annuity if they do not pay tax is given by:

$\sum_{i=1}^{10} \left(1.07^{i-1}(5000-0.07L_0) + \frac{1}{0.7}(5000-1.07^{i-1}(5000-0.07L_0))\right)v^i$ where $v^i = 1.08^{-i}$

The price is 38252.909

(http://img714.imageshack.us/img714/9620/q20n.jpg)

(http://img64.imageshack.us/img64/8656/q22q.jpg)

(http://img819.imageshack.us/img819/3273/q23m.jpg)

(http://img638.imageshack.us/img638/8881/q26tc.jpg)

Title: Re: TT's Maths Thread
Post by: TrueTears on June 26, 2012, 08:18:59 pm

I'm trying to show that given $A_1, A_2, A_3$ and $A_4$ are independent (ie, they are all mutually independent and pairwise independent) then this implies that $P(A_1 \cup A_2 | A_3 \cup A_4) = P(A_1 \cup A_2)$ . And in general I claim that if we have a string of independent $A_i$ 's for i = 1 to n, then $P(A_1 * A_2 * \cdots * A_i | A_{i+1} * \cdots * A_n) = P(A_1 * A_2 * \cdots * A_i)$ where * can be $\cup$ or $\cap$

So firstly, to prove my initial claim this is what I have so far:

$P(A_1 \cup A_2 | A_3 \cup A_4) = \frac{P[(A_1 \cup A_2) \cap (A_3 \cup A_4)]}{P(A_3 \cup A_4)}$

$=\frac{P\left([(A_1 \cup A_2) \cap A_3] \cup [(A_1 \cup A_2) \cap A_4]\right)}{P(A_3) + P(A_4) - P(A_3)P(A_4)}$

$=\frac{P(A_1 \cup A_2)P(A_3) + P(A_1 \cup A_2)P(A_4) - P(A_1 \cup A_2)}{P(A_3) + P(A_4) - P(A_3)P(A_4)}$

$= \frac{P(A_1 \cup A_2)\left[P(A_3) + P(A_4) - 1\right]}{P(A_3) + P(A_4) - P(A_3)P(A_4)}$

This is where I'm stuck as I want the $P(A_3)+P(A_4) + ...$ to cancel but something must have went wrong, but I can't see what?

Any ideas kamil, ahmad?

Title: Re: TT's Maths Thread
Post by: dcc on June 26, 2012, 08:37:42 pm

If you take A3 and A4 to be something like 'get heads' then your last line is 0

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 26, 2012, 08:38:57 pm

I'm pretty sure you didn't expand out the numerator correctly. Instead of $-P(A_1 \cup A_2)$ you should have $-P((A_1 \cup A_2)\cap A_3 \cap A_4)$

Title: Re: TT's Maths Thread
Post by: TrueTears on June 26, 2012, 08:42:11 pm

Quote from: kamil9876 on June 26, 2012, 08:38:57 pm

I'm pretty sure you didn't expand out the numerator correctly. Instead of $-P(A_1 \cup A_2)$ you should have $-P((A_1 \cup A_2)\cap A_3 \cap A_4)$

Yes I think so too, but I just used the principle of inclusion and exclusion on:

$P\left([(A_1 \cup A_2) \cap A_3] \cup [(A_1 \cup A_2) \cap A_4]\right)$

To get $P([(A_1 \cup A_2) \cap A_3]) + P([(A_1 \cup A_2) \cap A_4]) - P([(A_1 \cup A_2) \cap A_3] \cap [(A_1 \cup A_2) \cap A_4])$

I think the following is wrong, but why?

$P([(A_1 \cup A_2) \cap A_3] \cap [(A_1 \cup A_2) \cap A_4]) = P[(A_1 \cup A_2)]$ ? Since that's the intersection.

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 26, 2012, 08:43:59 pm

Nope, you forgot the $A_3$ and $A_4$ . Note that taking intersections is an associative operation

Title: Re: TT's Maths Thread
Post by: TrueTears on June 26, 2012, 08:49:25 pm

Ah stupid me, what a stupid mistake, thanks it works out now.

Title: Re: TT's Maths Thread
Post by: TrueTears on June 27, 2012, 02:51:01 am

Just wanna verify my solution to this:

Let $X$ be a Discrete Uniform random variable over the interval $[-n, n], n \ge 1$ . What is the distribution of $|X|$ ?

The PMF of the Discrete Uniform RV is given by: $p_X(x) = \frac{1}{n-m+1}, x \in [m, n]$

Let $Y$ be the RV such that $Y = |X|$

Thus $p_Y(y) = \sum_{\{x | |x| = y\}}p_X(x)$

Note that in this case $x \in [-n, n]$ , thus $p_X(x) = \frac{1}{2n+1}$

$p_Y(0) = \sum_{\{x | |x| = 0\}}p_X(x) = \sum_{\{x | x = 0\}}p_X(x) = \frac{1}{2n+1}$ since $0 \in [-n, n]$ as $n \ge 1$

$p_Y(1) = \sum_{\{x | |x| = 1\}}p_X(x) = \sum_{\{x | x = -1, 1\}}p_X(x) = p_X(-1) + p_X(1) = \frac{2}{2n+1}$

$p_Y(2) = \sum_{\{x | |x| = 1\}}p_X(x) = \sum_{\{x | x = -2, 2\}}p_X(x) = p_X(-2) + p_X(2) = \frac{2}{2n+1}$

$p_Y(n) = \sum_{\{x | |x| = 1\}}p_X(x) = \sum_{\{x | x = -n, n\}}p_X(x) = p_X(-n) + p_X(n) = \frac{2}{2n+1}$

Thus the PMF of $Y = |X|$ is given by:

$p_Y(y) = \frac{1}{2n+1}$ if $y = 0$ and $p_Y(y) = \frac{2}{2n+1}$ if $y = 1, 2, \cdots, n$

Thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on June 28, 2012, 08:59:42 pm

(http://img36.imageshack.us/img36/7396/questionaboutdice.jpg)

I was wondering in order to compute E(S) using the method they described, is it really as tedious as the following? (Obviously I could just apply the definition of $E(S) = \sum_{s = 2}^{12} s \times p_S(s)$ straight away, but here I wanted to try their exercise)

So in order to compute E(S) using the method they described we have to work out E(S|M=y) first which is given by $E(S|M=y) = \sum_{s = 2}^{12} s \times p_{S|M}(s|y) = \sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right)$

Now we have to do that for $y = 1, \cdots, 6$ : which means we have to manually find $\sum_{s = 2}^{12} s \left(\frac{p_{S,M}(s,y)}{p_{M}(y)}\right)$ for $y = 1, \cdots, 6$

Then we have to manually find $p_M(y)$ for $y = 1, \cdots, 6$

Then after all that sub back into $E(S) = \sum_{y=1}^{6} p_M(y) E(S|M=y)$

I did all of the above and ended up with so much working, is there any shortcuts rather than applying the definitions of each expression directly?

Title: Re: TT's Maths Thread
Post by: Mao on June 30, 2012, 03:59:37 pm

Quote from: TrueTears on June 28, 2012, 08:59:42 pm

I did all of the above and ended up with so much working, is there any shortcuts rather than applying the definitions of each expression directly?

Mathematica. :P

Title: Re: TT's Maths Thread
Post by: TrueTears on July 05, 2012, 12:30:51 am

(http://img84.imageshack.us/img84/3668/bias2.jpg)

If the picture is too small, please view attached pic.

I'm just reading through the proof of the biased-ness of OLS estimates if an important independent variable is left out and I'm just stuck at the very end, basically the mathematical part that confuses me is the part circled in red.

When we take the expectation of $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ how does it simply end as the same thing?

Ie, Why is $E\left[\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}\right] = \frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ ?

Since $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ isn't a constant as it depends on $n$ so it's like a random variable which outputs different values as $n$ changes, so why does it follow the rule that $E(c) = c$ where $c$ is a constant?

(http://img201.imageshack.us/img201/8966/assumptionsols.jpg)

I don't get why the part boxed in red.

Note that X, Y and U are all random variables, how is $E(\beta_1 X|X) = \beta_1 X$ ?

Shouldn't it be $E(\beta_1 X|X) = \beta_1E(X|X)$ since $\beta_1$ is just a constant, then $E(X|X) = \sum_{i=1}^n x_i P(X=x_i|X=x_i) = \sum_{i=1}^n x_i since P(X=x_i|X=x_i) = 1$

Thus $E(\beta_1 X|X) = \beta_1 \left(\sum_{i=1}^n x_i\right) \neq \beta_1 X$

Thanks :)

Title: Re: TT's Maths Thread
Post by: TrueTears on August 17, 2012, 03:55:54 am

(http://img253.imageshack.us/img253/7306/moments.jpg)

This is a pretty weird question... because:

$E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k$

But the limit: $\lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right]$ is undefined?

How am I meant to compute the MGF then?

Thanks

Title: Re: TT's Maths Thread
Post by: Mao on August 17, 2012, 05:37:02 am

This could be a complete misinterpretation

$M(t) = \lim_{k\to \infty} \left[ \frac{e^{k(t-1)}}{t-1} \right] - \frac{1}{t-1} = \frac{1}{1-t} \left( 1 - \lim_{k\to \infty} e^{k(t-1)}\right)$

This latter limit has two branches:

1. for $t<1$ , $e^{k(t-1)} \to e^{-\infty} = 0$ , thus $M(t) = \frac{1}{1-t}$ for $t<1$

2. for $t=1$ , $M(t=1)=0$

3. For $t>1$ , the limit diverges, thus $M(t)$ diverges.

No idea what the 'first three moments' are referring to.

Title: Re: TT's Maths Thread
Post by: TrueTears on August 17, 2012, 07:23:56 pm

Oh right, thanks for that Mao, yup I realised that we only needed the t<1 branch, thanks again :)

Title: Re: TT's Maths Thread
Post by: TrueTears on September 26, 2012, 04:40:16 pm

If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$ ? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?

Title: Re: TT's Maths Thread
Post by: golden on September 26, 2012, 04:51:36 pm

You forgot the squared for the (x-mean) part I think.

Title: Re: TT's Maths Thread
Post by: TrueTears on September 26, 2012, 05:04:30 pm

yeah typo'ed that lol

anyways so i've tried chebyshev's inequality on:

$\lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) \ge \lim_{n \to \infty} 1 - \frac{V(S_n^2)}{t^2}$

to somehow arrive at:

$\lim_{n \to \infty} P(|S^2 - \sigma^2| \le \epsilon) = 1$ by working backwards but can't seem to get anywhere :\

Actually I think I got it, but needed a piece of info from wiki

$V[s^2] = \sigma^4 \left( \frac{2}{n-1} + \frac{\kappa}{n} \right)$

so $\lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) \ge \lim_{n \to \infty} 1 - \frac{V(S_n^2)}{t^2} \implies \lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) \ge \lim_{n \to \infty} 1 - \frac{\sigma^4 \left( \frac{2}{n-1} + \frac{\kappa}{n} \right)}{t^2} \implies \lim_{n \to \infty} P(|S_n^2 - \sigma^2| \le t) =1$ and just simply let $t = \epsilon$ .

so the q is how to show $V[s^2] = \sigma^4 \left( \frac{2}{n-1} + \frac{\kappa}{n} \right)$ ? lol

Title: Re: TT's Maths Thread
Post by: golden on September 26, 2012, 05:23:45 pm

I don't know what's going on but here's a link:
http://www.ma.utexas.edu/users/mks/M358KInstr/SampleSDPf.pdf

The part with:
We will prove that the sample variance, S2 (not MOSqD) is an unbiased estimator of the
population variance 𝜎...

Whether it is related I wouldn't know lol.

Edit: oh you seem to have it and it doesn't seem related to this haha.

Title: Re: TT's Maths Thread
Post by: TrueTears on September 26, 2012, 05:26:58 pm

Quote from: golden on September 26, 2012, 05:23:45 pm

I don't know what's going on but here's a link:
http://www.ma.utexas.edu/users/mks/M358KInstr/SampleSDPf.pdf

The part with:
We will prove that the sample variance, S2 (not MOSqD) is an unbiased estimator of the
population variance 𝜎...

Whether it is related I wouldn't know lol.

Edit: oh you seem to have it and it doesn't seem related to this haha.

cheers for that :) actually that's for another statement $E(S^2) = \sigma^2$ which is a nice result, but I was after consistency of S^2 :P

Title: Re: TT's Maths Thread
Post by: Mao on September 27, 2012, 04:27:24 pm

Quote from: TrueTears on September 26, 2012, 04:40:16 pm

If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$ ? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?

Is the population a continuous distribution, or a large set of discrete points?

Title: Re: TT's Maths Thread
Post by: TrueTears on September 27, 2012, 04:31:43 pm

Quote from: Mao on September 27, 2012, 04:27:24 pm

Quote from: TrueTears on September 26, 2012, 04:40:16 pm
If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$ ? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?

Is the population a continuous distribution, or a large set of discrete points?

It can be any, discrete or continuous, however I am interested in the asymptotic properties

Title: Re: TT's Maths Thread
Post by: TrueTears on September 30, 2012, 10:46:48 pm

Does anyone know the exact definition of conditional moment generating functions?

In other words, can anyone confirm whether this definition is correct or not?

Given the conditional mgf of a random variable X|Y is $m_{X|Y}(t) = f(Y;t)$ then the marginal mgf of X is given by $m_{X}(t) = E_{Y}(f(Y))$

Title: Re: TT's Maths Thread
Post by: Jenny_2108 on September 30, 2012, 10:59:34 pm

Quote from: TrueTears on September 30, 2012, 10:46:48 pm

Does anyone know the exact definition of conditional moment generating functions?

In other words, can anyone confirm whether this definition is correct or not?

Given the conditional mgf of a random variable X|Y is $m_{X|Y}(t) = f(Y;t)$ then the marginal mgf of X is given by $m_{X}(t) = E_{Y}(f(Y))$

Hey TT, see part 3, page 20 of this link: conditional moment generating functions Hope it helps :P

Title: Re: TT's Maths Thread
Post by: TrueTears on September 30, 2012, 11:01:02 pm

Thanks Ennjy! Although my query is a bit different from that haha but dw all good I asked Ahmad and sorted it out :)

Title: Re: TT's Maths Thread
Post by: Jenny_2108 on September 30, 2012, 11:05:56 pm

Quote from: TrueTears on September 30, 2012, 11:01:02 pm

Thanks Ennjy! Although my query is a bit different from that haha but dw all good I asked Ahmad and sorted it out :)

LOL sr, I don't study uni though :-\

Title: Re: TT's Maths Thread
Post by: Mao on October 03, 2012, 03:32:30 pm

Quote from: TrueTears on September 27, 2012, 04:31:43 pm

Quote from: Mao on September 27, 2012, 04:27:24 pm
Quote from: TrueTears on September 26, 2012, 04:40:16 pm
If we let $X_1, X_2, \cdots, X_n$ be independent and identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ then the weak law of large number states that $\bar{X_n} \rightarrow^p \mu$ and I can prove this part, however does $S^2 \rightarrow^p \sigma^2$ ? Where $S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2$ the sample variance? If so, how to prove it?

Is the population a continuous distribution, or a large set of discrete points?
It can be any, discrete or continuous, however I am interested in the asymptotic properties

Okay. I will just look at the continuous case, because that's easy. By definition:

$\sigma^2 = \int_{-\infty}^{\infty} x^2 p(x)\, dx - \mu^2$

Now, we have the sample $X_1, X_2, \cdots, X_n$ . By definition:

$\bar{X^2} = \frac{1}{n} \sum_{i=1}^n X_i^2$ and $S^2 = \frac{n}{n-1} \left( \bar{X^2} - (\bar{X_n})^2\right)$

As $n\to\infty$ , $\frac{n}{n-1} \to 1$ , $\bar{X_n}\to \mu$ . We can approximate $\frac{1}{n} \sum_{i=1}^n X_i^2$ by binning and summing over histograms. As $n\to\infty$ , the frequencies will tend towards some discretised form of $p(x)$ ; also as we decrease the bin width, the approximation improves. Thus in the limit, $\bar{X^2} \to \int_{-\infty}^{\infty} x^2 p(x)\, dx$

Thus $\bar{X^2} \to \sigma^2$ as $n\to \infty$ .

I wouldn't say this is a truly rigorous proof a pure mathematician would embrace, but hey, it gets there.

Title: Re: TT's Maths Thread
Post by: TrueTears on October 03, 2012, 03:47:35 pm

Nice! Thanks Mao :)

Actually the other day I did something very similar to what you just did, but just a tiny bit more rigorous towards the end of the proof:

$S^2_n = \frac{n}{n-1}\left(\frac{1}{n}\sum_{i=1}^n Y_i^2 -\bar{Y_n}^2\right)$

$E(Y_i^2) = \mu_2^'$ and $V(Y_i^2) = \mu_4^'$ $- (\mu_2^{'})^{2} < \infty$

By LLN: $\frac{1}{n}\sum_{i=1}^n Y_i^2 \rightarrow^{p} \mu_2^'$

And obviously $\bar{Y_n} \rightarrow^{p} \mu$

Now since $g(x)=x^2$ is continuous, then $\bar{Y_n}^2 \rightarrow^{p} \mu^2$

Thus $\frac{1}{n}\sum_{i=1}^n Y_i^2 - \bar{Y_n}^2 \rightarrow^{p} \mu_2^{'} - \mu^2 = \sigma^2$ since $n\to\infty, \frac{n}{n-1} \to 1$

:D

Title: Re: TT's Maths Thread
Post by: Mao on October 03, 2012, 05:44:58 pm

Quote from: TrueTears on October 03, 2012, 03:47:35 pm

Nice! Thanks Mao :)

Actually the other day I did something very similar to what you just did, but just a tiny bit more rigorous towards the end of the proof:

$S^2_n = \frac{n}{n-1}\left(\frac{1}{n}\sum_{i=1}^n Y_i^2 -\bar{Y_n}^2\right)$

$E(Y_i^2) = \mu_2^'$ and $V(Y_i^2) = \mu_4^'$ $- (\mu_2^{'})^{2} < \infty$

By LLN: $\frac{1}{n}\sum_{i=1}^n Y_i^2 \rightarrow^{p} \mu_2^'$

And obviously $\bar{Y_n} \rightarrow^{p} \mu$

Now since $g(x)=x^2$ is continuous, then $\bar{Y_n}^2 \rightarrow^{p} \mu^2$

Thus $\frac{1}{n}\sum_{i=1}^n Y_i^2 - \bar{Y_n}^2 \rightarrow^{p} \mu_2^{'} - \mu^2 = \sigma^2$ since $n\to\infty, \frac{n}{n-1} \to 1$

:D

I got lost in the alphabet soup and the p and the subscripts and the dashes. But yes. Great!

Title: Re: TT's Maths Thread
Post by: TrueTears on December 03, 2012, 10:26:52 pm

Ok here's a funny ODE to solve:

$xy'' + (1-2x)y' + (x-1)y = 0$

clearly a straight forward power series substitution won't work here since we have a regular singularity at x = 0

so try the frobenius method by expanding around x = 0.

Assume $y = \sum_{m=0}^{\infty} a_mx^{m+r}$ is a solution where $r$ is some constant.

So we have $y' = \sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1}$ and $y'' = \sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}$

Put this back in:

$x \left(\sum_{m=0}^{\infty}(m+r)(m+r-1) a_mx^{m+r-2}\right) + (1-2x) \left(\sum_{m=0}^{\infty} (m+r)a_mx^{m+r-1}\right) + (x-1) \left(\sum_{m=0}^{\infty} a_mx^{m+r}\right) = 0$

after some algebra and stuff:

$\sum_{m=0}^{\infty} (m+r)^2 a_m x^{m+r-1} - \sum_{m=0}^{\infty} [2(m+r)+1] a_m x^{m+r} + \sum_{m=0}^{\infty} a_m x^{m+r+1} = 0$

clearly lowest term is $x^{r-1}$ with it's coefficient as $r^2a_0$ hence $r^2a_0 = 0$

Now $a_0 \neq 0$ , so $r^2 = 0 \implies r = 0$

Now we find the coefficients of the term $x^s$ where s is some constant, this gives:

$(s+1)^2a_{s+1} x^s - (2s+1) a_s x^s + a_{s-1} x^s = 0$

rearranging gives:

$a_{s+1} = \frac{(2s+1)a_s - a_{s-1}}{(s+1)^2}$ for s = 1, 2, etc

Thus we found a recurrence relationship with $a_0$ and $a_1$ as arbitrary initial values.

A bit of playing around quickly shows that:

$a_2 = \frac{3a_1 - a_0}{4}$

$a_3 = \frac{11a_1 - 5a_0}{36}$

$a_4 = \frac{25a_1 - 13a_0}{288}$

Thus we have one of the solutions to be $y = a_0 + a_1x + (\frac{3a_1 - a_0}{4})x^2 + (\frac{11a_1 - 5a_0}{36})x^3 + (\frac{25a_1 - 13a_0}{288})x^4 + ...$

However because a_0 and a_1 are arbitrary, let us pick.... $a_0 = a_1 = 1$ , now magically we have:

$y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+ ... = e^x$

So $y = e^x$ is one of the basis for the general solution of this ode.

I was wondering, since $a_0$ and $a_1$ are arbitrary, then would ANY $a_0$ ( $\neq 0$ ) and $a_1$ work? Say $a_0 = 4$ and $a_1 = 3$ which then implies that there is an "infinite" number of different basis for the general solution of this ode? pretty cool ~

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 03, 2012, 11:12:07 pm

Quote from: TrueTears on December 03, 2012, 10:26:52 pm

which then implies that there is an "infinite" number of different basis for the general solution of this ode? pretty cool ~

Pardon my ignorance, but every (non-zero) vector space has an infinite number of different bases, so this shouldn't be so surprising :P Secondly doesn't your computation show that the solution space is (at most?) 2-dimensional ?

Title: Re: TT's Maths Thread
Post by: TrueTears on December 03, 2012, 11:15:13 pm

well it's just cool because by plugging in some random values for the initial values of the recurrence relationship you can condense the entire infinite sum down into a function, i wouldn't have thought of that if it wasn't for a hint in my book, in general, how do i know what values to put in for the initial values? intuitive guessing (and hoping it is some expansion of a function) or just plug any random values in?

Title: Re: TT's Maths Thread
Post by: kamil9876 on December 04, 2012, 03:55:06 pm

I'm wondering, is the solution space a 1-dimensional or 2-dimensional vector space? You found $e^x$ as a solution but is there some solution that isn't a scalar multiple of it? e.g $a_1=0, a_0=1$

Title: Re: TT's Maths Thread
Post by: TrueTears on December 04, 2012, 04:37:45 pm

there is another solution, i just didnt bother finding it, it's quite easy to find by reduction of order

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2012, 05:43:34 pm

(http://img515.imageshack.us/img515/7160/oddeven.jpg)

Now in the example of transforming $f(x) = e^{-kx}$ for x>0 and k>0, they assumed it was an even function, however f(x) is not even, how then can you apply the fourier integral for even functions on it?

Is it because we are kinda doing a "half-range expansion", ie, f(x) can be manipulated into an even function (by simply just reflecting it in the y axis), so this then allows us to apply the fourier integral for even functions but then it is only valid over the domain of x>0 and k>0?

Title: Re: TT's Maths Thread
Post by: Planck's constant on December 06, 2012, 06:42:47 pm

TT, Kreyszig discusses the half-range expansion issue in the Fourier series section. (section 11.3) Both an odd-periodic and even-periodic extension is possible.
The argument carries over to Fourier Intregrals (with period -> infinity)

Title: Re: TT's Maths Thread
Post by: TrueTears on December 06, 2012, 06:43:32 pm

Quote from: argonaut on December 06, 2012, 06:42:47 pm

TT, Kreyszig discusses the issue in the Fourier series section. (section 11.3) Both an odd-periodic and even-periodic extension is possible.
The argument carries over to Fourier Intregrals (with period -> infinity)

sweet! wanted to confirm that, cheers man

Title: Re: TT's Maths Thread
Post by: FlorianK on December 07, 2012, 08:58:51 am

Quote from: argonaut on December 06, 2012, 06:42:47 pm

TT, Kreyszig discusses the half-range expansion issue in the Fourier series section. (section 11.3) Both an odd-periodic and even-periodic extension is possible.
The argument carries over to Fourier Intregrals (with period -> infinity)

OMG!!! ;D ;D ;D ;D ;D ;D ;D

Title: Re: TT's Maths Thread
Post by: Planck's constant on December 07, 2012, 12:50:47 pm

Quote from: FlorianK on December 07, 2012, 08:58:51 am

OMG!!! ;D ;D ;D ;D ;D ;D ;D

EDIT. Sorry Floriank, but I belatedly realised that you specifically highlighted the word 'Kreyszig', and I have now made the German connection.
Has this guy been giving you nightmares? :)
Which makes my question below to TT even more pertinent :

PS. TT what do you think of the Kreyszig text? It seems like people either love it or hate it.

Title: Re: TT's Maths Thread
Post by: TrueTears on December 07, 2012, 08:42:43 pm

Hmm since it's a book for engineers I can't comment much on it basically I just flip through it and skim sections here and there, although I had a look at its introductory ODE section, pretty good and solid however definitely comes from an engineers perspective lol. Other than that, the sections on laplace and fourier are good enough for me, haven't really had a look at the other sections.

Title: Re: TT's Maths Thread
Post by: TrueTears on March 03, 2013, 06:42:55 pm

(http://img547.imageshack.us/img547/9872/optimization.jpg)

Cbf typing it again on AN as the latex codes are different, if the picture is too small, original sauce: http://math.stackexchange.com/questions/319254/conversion-of-primal-problem-into-dual-optimization-problem#319254

Title: Re: TT's Maths Thread
Post by: Dedicated on March 03, 2013, 06:50:42 pm

Glancing over the posts and there is some awesome maths in here despite not understanding it. Hope I can reach this level soon through lots of reading.

Title: Re: TT's Maths Thread
Post by: TrueTears on March 16, 2013, 11:32:15 pm

question attached

Title: Re: TT's Maths Thread
Post by: #1procrastinator on March 17, 2013, 09:58:24 am

What area of math is this? OO

Title: Re: TT's Maths Thread
Post by: TrueTears on June 27, 2013, 06:01:49 pm

If X and Y are independent random variables (on a well defined probability space), then are f(X) and g(Y) also independent for any given real functions f and g? If so, any ideas how to go about proving it (non-measure theory methods?)

Title: Re: TT's Maths Thread
Post by: kamil9876 on June 28, 2013, 08:03:50 am

Hrmm, what if $f,g$ are both the zero function? Then both $f \circ X$ and $g \circ Y$ are the same random variable.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 11, 2013, 04:40:06 am

(http://img829.imageshack.us/img829/9229/vx5e.jpg)

Thanks :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on July 11, 2013, 11:59:27 am

No, you can't take $P=C$ . For example let's look at 1 by 1 matrices i.e real numbers

$2=\sqrt{2}\sqrt{2}$ but does $\frac{1}{2}=\sqrt{2}\sqrt{2}$ ?

The relationship between P and C is as follows:

$(CC^{T})^{-1}=(C^{T})^{-1}C^{-1}=(C^{-1})^TC^{-1}$

So in fact we can take $P=C^{-1}$ .

Title: Re: TT's Maths Thread
Post by: TrueTears on July 11, 2013, 04:56:36 pm

thanks kamil :)

Title: Re: TT's Maths Thread
Post by: TrueTears on July 13, 2013, 07:54:33 pm

Let $\mathbf{X}$ be a n by k matrix with full column rank, that is, rank k, hence $\mathbf{X'X}$ has full rank k and is invertible. Let $\mathbf{R}$ be a m by k matrix with full row rank, that is, rank m. Show that $\mathbf{R}(\mathbf{X'X})^{-1} \mathbf{R'}$ is of rank m.

EDIT: Actually just show that $\mathbf{R}(\mathbf{X'X})^{-1} \mathbf{R'}$ is a positive definite matrix.

kamil? :P

Title: Re: TT's Maths Thread
Post by: kamil9876 on July 13, 2013, 10:28:36 pm

Ok so showing that it is positive definite will already imply it has rull rank $m$ .

Now we know that $X'X$ is positive definite. Thus we can by the theorems in your posts above find an invertible square matrix $C$ such that $C'C=(X'X)^{-1}$ . Now:

$RC'CR'=(RC')(RC')'=AA'$ where $A=RC'$ .

So it only remains to check that $A$ is of rank $m$ but this follows because C is an invertible square matrix and R is of rank m.

Further Question:
It's interesting to see if the statement about the rank holds in other fields (where of course the notion of positive definite doesn't necessarily exist).

Edit: Actually it doesn't in the field with two elements $\mathbb{F}_2$ . Take $X=[1 , 1]^T$ , then $X^TX=1+1=0$ :o

Title: Re: TT's Maths Thread
Post by: TrueTears on July 13, 2013, 10:38:20 pm

ahhh thanks kamil, i just managed to prove it as well, although your way is much easier, i just used the definition of positive definiteness:

let $\mathbf{z}$ be a m by 1 matrix. So we need to show $\mathbf{z'R(X'X)^{-1}R'z} >0$ , since $\mathbf{X'X}$ is PD, then we can decompose $\mathbf{(X'X)}^{-1} = \mathbf{P'P}$ , thus we have $\mathbf{(R'z)'P'P(R'z)}$ , now let $\mathbf{v} = \mathbf{R'z}$ and we have $\mathbf{v'P'Pv} = v.v >0$

:D

Title: Re: TT's Maths Thread
Post by: TrueTears on July 14, 2013, 01:12:58 am

How do we define independence for random vectors?

Eg, let $\mathbf{e}$ be a random n by 1 vector and $\mathbf{b}$ be a random k by 1 vector (where n does not necessarily have to equal to k), then how do we define independence between $\mathbf{e}$ and $\mathbf{b}$ ?

For example, if we take just 2 random variables, X, Y then X and Y are independent iff f(x,y) = f(x)f(y) where f(x,y) characterizes their joint pdf and f(x), f(y) are their marginal pdfs. If we adapt this onto the random vector case, we have $f(\mathbf{e}, \mathbf{b}) = f(\mathbf{e}) f(\mathbf{b})$ , examining the RHS, we see that $f(\mathbf{e})$ is simply the joint distribution of $e_1, e_2, ..., e_n$ and likewise, $f(\mathbf{b})$ is the joint distribution of $b_1, b_2, ..., b_k$ , but then what is $f(\mathbf{e}, \mathbf{b})$ ? How is it defined?

Continuing on, for a more special case, consider when $\mathbf{e}$ is jointly multivariate normal, ie, $\mathbf{e} \sim N(\mathbf{\mu_e}, \mathbf{\Sigma_e})$ and $\mathbf{b}$ is also jointly multivariate normal, ie, $\mathbf{b} \sim N(\mathbf{\mu_b}, \mathbf{\Sigma_b})$ . Now in the special case of when two jointly normally distributed variables X and Y, a sufficient condition for independence is when cov(X,Y) = 0 where cov(.) denotes the covariance between X and Y. How then, do we generalise that into the random vector case? Ie, can we say that $\mathbf{e}$ and $\mathbf{b}$ are independent iff $cov(\mathbf{e}, \mathbf{b}) = 0$ ? But then we would have to show $\mathbf{e}$ and $\mathbf{b}$ are jointly normally distributed... but this doesn't really make sense since $\mathbf{e}$ and $\mathbf{b}$ are already itself jointly normally distributed, so then wouldn't we be talking about the 'joint' distribution of two joint distributions?

EDIT: Thanks Ahmad for clarifying on IRC :)

Title: Re: TT's Maths Thread
Post by: kamil9876 on July 14, 2013, 12:10:41 pm

^ In the middle of a Saturday night :o

Title: Re: TT's Maths Thread
Post by: TrueTears on July 14, 2013, 04:53:42 pm

Let $\mathbf{M}$ be a n by n symmetric, idempotent matrix with rank (hence its trace as well) equal to n-k < n (hence $\mathbf{M}$ is singular). Since $\mathbf{M}$ can be written as $\mathbf{M} = \mathbf{MM'}$ then $\mathbf{M}$ is positive semi definite and hence can be decomposed as $\mathbf{M} = \mathbf{P}\mathbf{P'}$ where $\mathbf{P}$ is a n by n-k matrix with full column rank (that is with rank n-k). Show that $\mathbf{PP'} = \mathbf{M} \iff \mathbf{P'P} = \mathbf{I_{n-k}}$ where $\mathbf{I_{n-k}}$ denotes the (n-k) by (n-k) identity matrix.

Title: Re: TT's Maths Thread
Post by: kamil9876 on July 14, 2013, 05:32:01 pm

If you are familiar with Jordan Normal Form, then it's possible to solve it with that. (I mean how else did you know that the trace=rank for idempotent matrices?)

Title: Re: TT's Maths Thread
Post by: TrueTears on July 14, 2013, 05:34:10 pm

lol I was given the latter fact.

Title: Re: TT's Maths Thread
Post by: TrueTears on July 18, 2013, 12:23:54 am

Anyone wanna do a bit of algebra soup? nothing difficult, just annoying lol

Inverted Gamma pdf is: $f(x; \alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{-\alpha - 1}\exp\left(-\frac{\beta}{x}\right)$

My equation:

$f(\sigma; ?, ?) = \frac{2}{\Gamma\left(\frac{v}{2}\right)}\left(\frac{v\hat{\sigma}^2}{2}\right)^{\frac{v}{2}} \frac{1}{\sigma^{v+1}}\exp\left[\frac{-v\hat{\sigma}^2}{2\sigma^2}\right]$

Clearly $\alpha = \frac{v}{2}$ , what's $\beta$ ?

nvm finally got it lol

ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: TrueTears on November 12, 2009, 05:01:58 pm