Print Page - Uni Maths Questions

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: vcestudent94 on March 14, 2013, 12:43:42 pm

Title: Uni Maths Questions
Post by: vcestudent94 on March 14, 2013, 12:43:42 pm

Can someone help me with this sequence limit question?
Lim n-->infinity (3^n - 4^n)/(3*n^2+4^n+7)

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on March 14, 2013, 05:01:12 pm

divide numerator and denominator by 4^n

Title: Re: Uni Maths Questions
Post by: vcestudent94 on March 14, 2013, 05:43:33 pm

Thanks. How about this sequence:
(3^n+2!)/(5^n-n!)

And also, how would you use the comparison test to determine if sums like this one is convergent or not?
sum( (sqrt(n) - 1)/(n^2 + 1) )

Title: Re: Uni Maths Questions
Post by: vcestudent94 on March 15, 2013, 11:18:55 pm

Bump ::)

Title: Re: Uni Maths Questions
Post by: Nagisa on March 16, 2013, 01:39:05 am

Quote from: vcestudent94 on March 14, 2013, 05:43:33 pm

And also, how would you use the comparison test to determine if sums like this one is convergent or not?
sum( (sqrt(n) - 1)/(n^2 + 1) )

yes, this is how i would do it

$\sum^n_{n=1} \ \frac{\sqrt{n} - 1}{n^2 + 1}$

the comparison test states that if you have two series (say, $a_n$ and $b_n$ ). if $b_n$ is convergent and $a_n \le b_n$ then $a_n$ is also convergent.
conversely, if $b_n$ diverges and $a_n \ge b_n$ then $a_n$ is also divergent.

so this means in this case we can look at the dominant parts of the series which are $\sqrt{n}$ on top and $n^2$ on bottom.

so we should choose our $b_n$ to be $\frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}$

now we have $\frac{\sqrt{n} - 1}{n^2 + 1} < \frac{1}{n^{3/2}}$ (since now we have $b_n$ as a 'p series' (1/[p^n]). a p series converges if n > 1 and diverges if n < 1)

so $b_n$ converges and via the comparison test, so does $a_n$

Title: Re: Uni Maths Questions
Post by: vcestudent94 on March 16, 2013, 06:17:12 pm

Thanks! I get it now.. would you happen to know how to approach this question:

Lim n->infinity [(3^n+2!)/(5^n-n!)]

Title: Re: Uni Maths Questions
Post by: Nagisa on March 16, 2013, 06:26:10 pm

do you mean show convergence or divergence?

Title: Re: Uni Maths Questions
Post by: kamil9876 on March 16, 2013, 06:56:42 pm

I'm wondering if there is a typo in there since why write $2!$ instead of just $2$ ?

Anyhow, assuming this is what you meant, then divide top and bottom by 3^n and you see that you get $\frac{1+2.3^{-n}}{(5/3)^n - n!3^{-n}}$ .

The numerator goes to $1$ and so if you can show the denominator $\to -\infty$ then you have shown that it goes to $0$ . It isn't hard to see that if $a_n=n!3^{-n}$ and $b_n=(5/3)^n$ then $a_n$ grows much faster than $b_n$ and $b_n-a_n \to -\infty$ the point is that $a_{n+1}/a_n=(n+1)/3$ while $b_{n+1}/b_n=5/3$ so from this you can show that eventually $a_n$ will be larger than $b_n$ and from then on will grow much faster and hence the difference goes to infinity. (As an exercise you can turn this into a more precise argument).

Title: Re: Uni Maths Questions
Post by: vcestudent94 on March 16, 2013, 07:06:30 pm

Yes its a typo. Sorry, I meant 2n! instead of just 2!

Title: Re: Uni Maths Questions
Post by: kamil9876 on March 17, 2013, 01:21:34 pm

Just before I solve an irrelevant problem again, is that (2n)! or n! ? In either case you should multiply by $\frac{1/n!}{1/n!}$ and then use the fact that $a^n/n! \to 0$ for each fixed $a \in \mathbb{R}$ . So the limit is either $-2$ or $-\infty$ depending on whether you meant $2n!$ or $(2n)!$ respectively

Title: Re: Uni Maths Questions
Post by: nubs on March 17, 2013, 06:23:38 pm

How would you use the sandwich theorem to find the limit of (3^n+1)^(1/n) and n!/(n^n)?

Title: Re: Uni Maths Questions
Post by: kamil9876 on March 18, 2013, 11:34:11 am

$(3^n)^{1/n} \leq (3^n+1)^{1/n} \leq (3^n+3^n)^{1/n} =3.2^{1/n} \to 3$

$0 \leq \frac{n!}{n^n}=\frac{1}{n} \frac{2}{n} \ldots \frac{n}{n} \leq \frac{1}{n}$

Title: Re: Uni Maths Questions
Post by: vcestudent94 on March 27, 2013, 10:01:10 pm

Find the 576th derivative of e^(4t)*cos(4t) ?

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on March 27, 2013, 10:10:57 pm

Express it as $\mathrm{Re}\left( \mathrm{e}^{4(1+i)t}\right)$

Title: Re: Uni Maths Questions
Post by: Jeggz on March 27, 2013, 10:19:05 pm

$\frac{d^5^7^6}{dt^576}{$ $e^4^t Re(e^4^i^t)$

$= \frac{d^5^7^6}{dt^576}{$ $Re(e^4^t * e^4^i^t)$

$= \frac{d^5^7^6}{dt^576}{\mathrm {Re}\left( \mathrm{e}^{4(1+i)t}\right)$

$=Re (\frac{d^5^7^6}{dt^576}{\left( \mathrm{e}^{4(1+i)t}\right))$

Then just differentiate from there, hope that helped :)

EDIT: Beaten :P

Title: Re: Uni Maths Questions
Post by: vcestudent94 on March 27, 2013, 10:32:31 pm

Thanks for the method guys...
For the second step, its unintuitive to me how you can just chuck the e^(4t) into the Re(...) ?

Title: Re: Uni Maths Questions
Post by: Jeggz on March 27, 2013, 10:35:57 pm

I think you just have to commit these two into memory.
1. $cos(t) = Re(e^i^t)$
2. $sin(t) = Im(e^i^t)$

Title: Re: Uni Maths Questions
Post by: vcestudent94 on March 27, 2013, 10:45:54 pm

I get it now. I just needed to stare for a while because I've never come across that technique before. Thanks for the help!

Title: Re: Uni Maths Questions
Post by: b^3 on March 27, 2013, 10:55:09 pm

If it makes it easier to remember, you know that
$e^{i \theta}=\cos(\theta) + i \sin(\theta)$
Now how would we get $\cos(\theta)$ out of that? We would take the real part of $e^{i \theta}$ , $\text{Re}(e^{i \theta})=\cos(\theta)$
How would we get $\sin{\theta}$ out of that? We would take the imagainary part of $e^{i \theta}$ , $\text{Im}(e^{i \theta})=\sin(\theta)$

Title: Re: Uni Maths Questions
Post by: Mao on March 27, 2013, 11:44:40 pm

Quote from: vcestudent94 on March 27, 2013, 10:32:31 pm

Thanks for the method guys...
For the second step, its unintuitive to me how you can just chuck the e^(4t) into the Re(...) ?

Consider two complex numbers, $u,v\in \mathbb{C}$ , where $u=\Re(u)$ (its imaginary component is zero)

Then, multiplication by $u$ works like a 'scalar' multiplication for vectors: $\Re (u v) = \Re(u)\times \Re(v)$ , $\Im(u v) = \Re(u) \times \Im(v)$ .
A similar technique applies if $u = \Im(u) i$ .

Title: Re: Uni Maths Questions
Post by: Jeggz on April 13, 2013, 09:25:45 pm

Can someone please explain when and why it is okay to interchange columns/rows in row reduction?
And also I would really appreciate some help reducing the matrix below to find the inverse (it's meant to have the identity matrix next to it, if that makes sense) - i keep going around and round in circles :-\ Thanks in advance!!

2   1   -1
3   -1   1
-1   0   2

Title: Re: Uni Maths Questions
Post by: b^3 on April 13, 2013, 10:44:12 pm

The way I justify it is that, what row reduction normally leads into is solving systems of linear equations. Since our rows are representing the coefficients of variables in a linear equation, if we swap two rows, we are not changing the system of equations. e.g. $\begin{bmatrix}a & b & c \\ d & e & f \end{bmatrix}$ would correspond to the same system of equations as the matrix $\begin{bmatrix}d & e & f \\ a & b & c \end{bmatrix}$ , that is by swapping rows, we haven't changed the solution set.

As to finding the inverse matrix, we try to make the left into the identity matrix, leaving the right as the inverse matrix.

$\begin{alignedat}{1} & \left[\begin{array}{ccc|} 2 & 1 & -1 \\ 3 & -1 & 1 \\ -1 & 0 & 2 \end{array}\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\end{alignedat} $
Firstly we can swap rows 1 and 3, so that we get a nice 1 in our 1,1 entry.
$\begin{alignedat}{1}r_{1}\leftrightarrow r_{3} & \left[\begin{array}{ccc|} -1 & 0 & 2 \\ 3 & -1 & 1 \\ 2 & 1 & -1 \end{array}\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] \\ r_{1}\rightarrow-\left(r_{1}\right) & \left[\begin{array}{ccc|} 1 & 0 & -2 \\ 3 & -1 & 1 \\ 2 & 1 & -1 \end{array}\begin{array}{ccc} 0 & 0 & -1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right] \end{alignedat} $
Now we want to form a pivot in the 2,2 entry, so to do this we need to clear the 2,1 entry out, so we can take a multiple of row 1 from row 2 to do this. We can also clear out the 3,1 entry by taking away a mutiple of row 1 from row 3.
$\begin{alignedat}{1}\begin{array}{c} r_{2}\rightarrow r_{2}-3r_{1} \\ r_{3}\rightarrow r_{3}-2r_{1} \end{array} & \left[\begin{array}{ccc|} 1 & 0 & -2 \\ 0 & -1 & 7 \\ 0 & 1 & 3 \end{array}\begin{array}{ccc} 0 & 0 & -1 \\ 0 & 1 & 3 \\ 1 & 0 & 2 \end{array}\right]\end{alignedat} $
Next we need to clear out the 3,2 entry to make the lower triangle of the matrix be all zeros. We can clear out this entry by taking away a mutiple of row 2 from row 3.
$\begin{alignedat}{1}r_{3}\rightarrow r_{3}+r_{2} & \left[\begin{array}{ccc|} 1 & 0 & -2 \\ 0 & -1 & 7 \\ 0 & 0 & 10 \end{array}\begin{array}{ccc} 0 & 0 & -1 \\ 0 & 1 & 3 \\ 1 & 1 & 5 \end{array}\right] \\ \begin{array}{c} r_{2}\rightarrow-\left(r_{2}\right) \\ r_{3}\rightarrow\frac{1}{10}\left(r_{3}\right)\vphantom{\int_{a}^{b}} \end{array} & \left[\begin{array}{ccc|} 1 & 0 & -2 \\ 0 & 1 & -7 \\ 0 & 0 & 1 \end{array}\begin{array}{ccc} 0 & 0 & -1 \\ 0 & -1 & -3 \\ \frac{1}{10}\vphantom{\int_{a}^{b}} & \frac{1}{10} & \frac{1}{2} \end{array}\right] \end{alignedat} $
Next we need to start working back upwards, to clear out the top right triangle, forming the identity matrix. So we first start by working on the entry 2,3. If we take a mutiple of row one away, we will affect our zeros we formed, but if we are to take a mutiple of row 3 away from row 2, then we will preserve these zeros, and be able to form another zero.
$\begin{alignedat}{1}r_{2}\rightarrow r_{2}+7r_{3} & \left[\begin{array}{ccc|} 1 & 0 & -2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\begin{array}{ccc} 0 & 0 & -1 \\ \frac{7}{10}\vphantom{\int_{a}^{b}} & -\frac{3}{10} & \frac{1}{2} \\ \frac{1}{10}\vphantom{\int_{a}^{b}} & \frac{1}{10} & \frac{1}{2} \end{array}\right]\end{alignedat} $
Now we work on the 1,3 entry. We can add a mutiple of row three away from row one without affecting our other zeros.
$\begin{alignedat}{1}r_{1}\rightarrow r_{1}+2r_{3} & \left[\begin{array}{ccc|} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\begin{array}{ccc} \frac{1}{5}\vphantom{\int_{a}^{b}} & \frac{1}{5} & 0 \\ \frac{7}{10}\vphantom{\int_{a}^{b}} & -\frac{3}{10} & \frac{1}{2} \\ \frac{1}{10}\vphantom{\int_{a}^{b}} & \frac{1}{10} & \frac{1}{2} \end{array}\right]\end{alignedat} $
Now normally we wouldn't be finished yet, but since our 1,2 entry is what we want it to be, we have formed the identity matrix, so the matrix on the right is our inverse matrix. If we did need to work on the 1,2 entry, then we could take a mutiple of row 2 away from row 1 without affecting the other zeros.
$A^{-1} = \begin{bmatrix}\frac{1}{5}\vphantom{\int_{a}^{b}} & \frac{1}{5} & 0 \\ \frac{7}{10}\vphantom{\int_{a}^{b}} & -\frac{3}{10} & \frac{1}{2} \\ \frac{1}{10}\vphantom{\int_{a}^{b}} & \frac{1}{10} & \frac{1}{2} \end{bmatrix} = \frac{1}{10}\begin{bmatrix}2 & 2 & 0 \\ 7 & -3 & 5 \\ 1 & 1 & 5 \end{bmatrix} $

Wow, that turned into a long post, anyways, hope that makes sense and hope it helps :)

Title: Re: Uni Maths Questions
Post by: Jeggz on April 14, 2013, 08:56:19 am

You are a legend b^3! Thankyouu :)

Title: Re: Uni Maths Questions
Post by: Deleted User on April 15, 2013, 08:06:58 pm

Which one of the following is a vector space and could you explain me why?

The set of all (real) polynomials with positive coefficients.

The set of all (real valued) continuous functions with the property that the function is 0 at every integer (eg. f(x)=sin(pi*x)).

Thank you

Title: Re: Uni Maths Questions
Post by: kamil9876 on April 15, 2013, 08:18:40 pm

I'm assuming you are asking whether they are vector spaces over $\mathbb{R}$ with the obvious operations?

For polynomials with positive coefficients: No. It isn't closed under scalar multiplication. E.g $x$ is in your supposed vector space, but $(-1).x=-x$ is not.

The continous functions which vanish at the integers form a vector subspace of the vector space of all real valued continous functions (I'm assuming you already know this is a vector space?). Here is a proof:

Non-empty: The zero function, it vanishes all real numbers hence in particular at integers too.

Addition: If f,g vanish at the integers then for all integers n we have (f+g)(n)=f(n)+g(n)=0+0=0

Scalar multiplication: if a is a real number and f vanishes on the integers then we have: (af)(n)=af(n)=a.0=0

Title: Re: Uni Maths Questions
Post by: Deleted User on April 15, 2013, 09:20:34 pm

Thank you!

Generally, how do we prove that something is closed under addition/multiplication? Like for 2a - 3b + 5c = 0, how do we show that this is a subspace?

Title: Re: Uni Maths Questions
Post by: kamil9876 on April 15, 2013, 10:54:13 pm

So in other words, you want to know how does one show that the set of all (a,b,c) such that 2a-3b+5c=0 is closed under addition? Well let us call it S. Now suppose $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are in $S$ , we want to show that $(a_1+a_1,b_1+b_2,c_1+c_2)$ is in $S$ , so let us see if the following is zero:

$2(a_1+a_2)-3(b_1+b_2)+5(c_1+c_2)=2a_1-3b_1+5c_1+2a_2-3b_2+5c_2=0+0=0$ since $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are in $S$ . Hence this shows S is closed under addition and a similair argument works for scalar multiplication.

In general notice that this shows that the solution of any system of the form $\lambda_1x_1 + \ldots\lambda_n x_n=0$ is a subspace of $\mathbb{R}^n$ .

Aside: Of course, if you know matrices then this shouldn't also be so surprising, any such equation can be written as Ax=0 for some row matrix A where x is the column vector of x_i's. Then if x=v and x=w are solutions, then by standard matrix properties we have $A(v+w)=Av+Aw=0+0=0$ so x=v+w is also a solution

Title: Re: Uni Maths Questions
Post by: Deleted User on April 16, 2013, 08:02:23 pm

Thanks! You're a life saver!

Title: Re: Uni Maths Questions
Post by: Deleted User on April 17, 2013, 12:48:32 pm

Okay for the following problems, I know how to identify which is a real vector space with the usual operations or wihch is not, but I just don't know how to prove it in writing! Could you help me out? Thank you.

a) Set of real polynomials of any degree

b) Set of real polynomials of degree =< n.

c) Set of real polynomials of degree exactly n.

Title: Re: Uni Maths Questions
Post by: humph on April 18, 2013, 06:57:26 am

Quote from: Deleted User on April 17, 2013, 12:48:32 pm

Okay for the following problems, I know how to identify which is a real vector space with the usual operations or wihch is not, but I just don't know how to prove it in writing! Could you help me out? Thank you.

a) Set of real polynomials of any degree

b) Set of real polynomials of degree =< n.

c) Set of real polynomials of degree exactly n.

To prove it in writing: if you think it is a vector space, show that it satisfies the desired properties. For example, if
$f(x) = a_m x^m + a_{m - 1} x^{m - 1} + \cdots + a_0$
and
$g(x) = b_n x^n + b_{n - 1} x^{n - 1} + \cdots + b_0$
are polynomials of any degree (assuming without loss of generality that $m \geq n$ ), then
$(f + g)(x) = a_m x^m + a_{m - 1} x^{m - 1} + \cdots + a_{n + 1} x^{n + 1} + (a_n + b_n) x^n + (a_{n - 1} + b_{n - 1}) x^{n - 1} + \cdots + (a_0 + b_0)$
is clearly also a polynomial of some degree.
If, on the other hand, you don't think it's a vector space, simply give an example where a vector space property fails (e.g. when $f + g$ is not in the original space).

Title: Re: Uni Maths Questions
Post by: Deleted User on April 18, 2013, 10:20:01 am

Thank you. That was what I was after

Title: Re: Uni Maths Questions
Post by: Sach1_K on April 18, 2013, 10:58:40 am

I don't exactly know where to post this. Can you guys help me with it?
Thanks in advance
- Attached question

Title: Re: Uni Maths Questions
Post by: Deleted User on April 18, 2013, 02:13:19 pm

How do I find a basis of a subspace?

Eg. What is a basis for this: {(-1,2,0,4), (3,1,-1,2), (-5,3,1,6), (7,0,-2,0)} R^4

Thank you

Title: Re: Uni Maths Questions
Post by: Alwin on April 18, 2013, 04:35:16 pm

For the basis of a subset, put the vectors you have in matrix form:

$ \begin{pmatrix} -1 & 3 & -5 & 7 \\ 2 & 1 & 3 & 0 \\ 0 & -1 & 1 & -2 \\ 4 & 2 & 6 & 0 \\ \end{pmatrix} $

Next, you find the row reduced echelon form, I'll assume that you know how to do this already:

$ \begin{pmatrix} 1 & 0 & -2/7 & 0 \\ 0 & 1 & -1/7 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $

So the column space is clearly the first two rows of the original martix

Thus, a basis is:

$\{(-1,2,0,4), (3,1,-1,2)\}$

Hope that helps!

Title: Re: Uni Maths Questions
Post by: Sach1_K on April 19, 2013, 05:43:32 pm

Quote from: Sach1_K on April 18, 2013, 10:58:40 am

I don't exactly know where to post this. Can you guys help me with it?
Thanks in advance
- Attached question

Don't worry, I got it now. :)

Title: Re: Uni Maths Questions
Post by: Deleted User on April 20, 2013, 04:48:15 pm

Quote from: Alwin on April 18, 2013, 04:35:16 pm

For the basis of a subset, put the vectors you have in matrix form:

$ \begin{pmatrix} -1 & 3 & -5 & 7 \\ 2 & 1 & 3 & 0 \\ 0 & -1 & 1 & -2 \\ 4 & 2 & 6 & 0 \\ \end{pmatrix} $

Next, you find the row reduced echelon form, I'll assume that you know how to do this already:

$ \begin{pmatrix} 1 & 0 & -2/7 & 0 \\ 0 & 1 & -1/7 & 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} $

So the column space is clearly the first two rows of the original martix

Thus, a basis is:

$\{(-1,2,0,4), (3,1,-1,2)\}$

Hope that helps!

Thanks. Does this method work for all subspaces?

Title: Re: Uni Maths Questions
Post by: Deleted User on April 20, 2013, 04:52:07 pm

How do I show that the set (a,b,a-b,a+b) is closed under addition and scalar multiplication (R^4)?

Title: Re: Uni Maths Questions
Post by: Alwin on April 22, 2013, 11:25:49 am

Quote from: Deleted User on April 20, 2013, 04:48:15 pm

Thanks. Does this method work for all subspaces?

Yes, this called The Column Method of finding the basis for the span of a set of vectors given.

Quote from: Deleted User on April 20, 2013, 04:52:07 pm

How do I show that the set (a,b,a-b,a+b) is closed under addition and scalar multiplication (R^4)?

Addition:
You can sub in values to test it:
$a=1, b=1: u=(1,1,0,2) a=2, b=1: v=(2,1,1,3)$
$u+v=(3,2,1,5)$ which is in the set of $(a,b,a-b,a+b)$ since $3-2=1$ and $3+2=5$

For a more general proof,
let $u=(u_1, u_2, u_1-u_2, u_1 +u_2)$ and $v=(v_1, v_2, v_1-v_2, v_1 +v_2)$
$u+v=(u_1+v_1, u_2+v_2, u_1-u_2+v_1-v_2, u_1 +u_2+ v_1 +v_2)$
$u+v=(u_1+v_1, u_2+v_2, (u_1+v_1)-(u_2+v_2), (u_1+v_1)+(u_2+v_2))$ by manipulation
Hence, the set is closed under addition

Scalar multiplication:
Again, you can sub in values to test it:
$a=1, b=1: u=(1,1,0,2)$
$3u=(3,3,0,6)$ which is in the set of $(a,b,a-b,a+b)$ since $3-3=0$ and $3+3=6$

For a more general proof,
let $u=(u_1, u_2, u_1-u_2, u_1 +u_2)$
$\lambda u =(\lambda u_1,\lambda u_2, \lambda(u_1-u_2), \lambda(u_1 +u_2))$
$\lambda u =(\lambda u_1,\lambda u_2, \lambda u_1-\lambda u_2, \lambda u_1 +\lambda u_2)$
Clearly the scalar, $\lambda u$ is also in the set $(a,b,a-b,a+b)$

I think this proves it, but there may be a more elegant method. Refer to your textbook if you have one

Title: Re: Uni Maths Questions
Post by: Deleted User on April 23, 2013, 10:34:09 am

How do I put the set of vectors {(2 1; 0 0), (0 0; 2 1), (3 -1; 0 0), (0 0; 3 1)} into a matrix with each vector as a column? Thanks

Title: Re: Uni Maths Questions
Post by: Alwin on April 23, 2013, 05:15:31 pm

Quote from: Deleted User on April 23, 2013, 10:34:09 am

How do I put the set of vectors {(2 1; 0 0), (0 0; 2 1), (3 -1; 0 0), (0 0; 3 1)} into a matrix with each vector as a column? Thanks

I'm not sure, because of your sparsely spaced semi-colons but is this what you mean?

$ \begin{pmatrix} 2 & 0 & 3 & 0 \\ 1 & 0 & -1& 0 \\ 0 & 2 & 0 & 3 \\ 0 & 1 & 0 &1 \end{pmatrix} $

If not, let me know

Title: Re: Uni Maths Questions
Post by: Deleted User on April 24, 2013, 10:42:18 am

I meant that eg. (2 1; 0 0) =
2 1
0 0

Title: Re: Uni Maths Questions
Post by: Deleted User on April 24, 2013, 11:49:16 am

Ok this is what I meant.

How do I write this set as a matrix? { $\Big(\begin{matrix}2 & 1\\0 & 0\end{matrix}\Big)$ , $\Big(\begin{matrix}0 & 0\\2 & 1\end{matrix}\Big)$ , $\Big(\begin{matrix}3 & -1\\0 & 0\end{matrix}\Big)$ , $\Big(\begin{matrix}0 & 0\\3 & 1\end{matrix}\Big)$ } (M_2,2.)

Title: Re: Uni Maths Questions
Post by: mark_alec on April 24, 2013, 01:00:52 pm

Consider the basis made up of {(1,0;0,0), (0,1;0,0), (0,0;1,0),(0,0;0,1)}.

Title: Re: Uni Maths Questions
Post by: Deleted User on April 24, 2013, 01:07:27 pm

I don't understand

Title: Re: Uni Maths Questions
Post by: Yendall on April 29, 2013, 09:03:41 am

Q1: Which ordered pairs need to be added to the relation $p = \left\{(a,a)\right\}$ on the set $X = \left\{a,b,c,d\right\}$ to create the symmetric closure $p^s$ of $p$ ?

This is a really simple question, is this relation already symmetrically closed due to the fact that the symmetry of (a,a) is (a,a) ? Thanks.

Q2: Which ordered pairs need to be added to the relation $p = \left\{(a,a)\right\}$ on the set $X = \left\{a,b,c,d\right\}$ to create the equivalence relation $p^*$ generated by p?

Reflexivity: add $(b,b) (c,c)$ and $(d,d)$
Symmetry: add nothing?
Transitivity: Would I also add nothing?

This is where i'm confused. If the rule defines transitivity as $(x,y) \in p, (y,z) \in p \Rightarrow (x,z) \in p$ , if there aren't any combinations of (x,z) in the set, do I need to add ordered pairs so that there are? If so, how many pairs is enough to make the entire set transitive?

Thanks for any help guize.

Title: Re: Uni Maths Questions
Post by: kamil9876 on April 29, 2013, 10:30:02 pm

I agree with both of your answers (not sure what a "symmetric closure" is but the name suggests that it is the smallest relation contain your relation that is symmetric, if that's the case then YES).

Title: Re: Uni Maths Questions
Post by: Deleted User on May 04, 2013, 08:58:08 pm

How do I do this question?

Find a basis for the solution space of the following system in five variables:
-x1 - x2 -3*x4 - x5 = 0
3*x1 + 3*x2 - x3 -3*x4 + x5 = 0

The answer gives me a 3x5 matrix but I don't know how this is possible if there are only 2 rows in the system?

Title: Re: Uni Maths Questions
Post by: Jeggz on May 05, 2013, 07:41:09 pm

I'm having a bit of trouble with this question -
Find the distance from the point (1,0,0) to the line through the points (1,2,0) and (-1,1,1)
Any help would really be appreciated! :)

Title: Re: Uni Maths Questions
Post by: Will T on May 05, 2013, 07:59:36 pm

Find the equation of the line in Parametric form:
$(1,2,0)-(-1,1,1)=(2,1,-1)$
$X(t)=t(2,1,-1)+(1,2,0)$
Any point on the line $X(t)$ has co-ordinates $(2t+1,t+2,-t)$
From the distance formula, the distance between the line and a point will be:
$\sqrt{((2t+1)-(1))^2+((t+2)-(0))^2+((-t)-(0))^2}$
$=\sqrt{4t^2+t^2+4t+4+t^2}$
$=\sqrt{6t^2+4t+4}$
$=\sqrt{6(t^2+\frac{2t}{3}+\frac{2}{3})}$
$=\sqrt{6(t^2+\frac{2t}{3}+\frac{1}{9}-\frac{1}{9}+\frac{2}{3})}$
$=\sqrt{6((t+\frac{1}{3})^2+\frac{6}{9}-\frac{1}{9})}$
$=\sqrt{6(t+\frac{1}{3})^2+6(\frac{5}{9})}$
$=\sqrt{6(t+\frac{1}{3})^2+\frac{10}{3}}$
Therefore, the minimum occurs when $t=-\frac{1}{3}$ and is $\sqrt{\frac{10}{3}}$
Or something like that....

Title: Re: Uni Maths Questions
Post by: Alwin on May 05, 2013, 08:28:45 pm

Quote from: Deleted User on May 04, 2013, 08:58:08 pm

How do I do this question?

Find a basis for the solution space of the following system in five variables:
-x1 - x2 -3*x4 - x5 = 0
3*x1 + 3*x2 - x3 -3*x4 + x5 = 0

The answer gives me a 3x5 matrix but I don't know how this is possible if there are only 2 rows in the system?

Think of the kernal space (ie solve the system of equations) ;)
You have 2 equations and 5 equations. Hence, 3 parameters required which gives you your 3 rows. Your 5 columns are a result of the 5 variables

Quote from: Jeggz on May 05, 2013, 07:41:09 pm

I'm having a bit of trouble with this question -
Find the distance from the point (1,0,0) to the line through the points (1,2,0) and (-1,1,1)
Any help would really be appreciated! :)

@Will T, your method is perfectly acceptable. However, a more elegant solution that requires less algebraic manipulation is as follows:

$\text{direction vector of line}=v=(2,1,-1)$
$\text{vector to point }(1,0,0) \text{ from } (1,2,0) = u=(0,-2,0)$

$d=\frac{|u\times v|}{|v|}$

$d=\frac{|(0,-2,0)\times (2,1,-1)|}{|(2,1,-1)|}$
$d=\frac{|(2, 0, 4)|}{|(2,1,-1)|}$
$d=\frac{2 \sqrt{5}}{\sqrt{6}}$
$d=\sqrt{\frac{10}{3}}\text{ units}$

I find this method better because Will T's method requires a lot more work.

If you are interested in the proof of this formula, it is a generalisation of projections of vectors, ie the vector method of finding distance rather than the algebraic method.

Hope it helps!

Title: Re: Uni Maths Questions
Post by: Phy124 on May 06, 2013, 01:31:09 am

I'll get back to you on this later (if someone else doesn't beforehand) but in the mean time you've made an error in this line:

Quote from: Will T on May 05, 2013, 10:33:29 pm

$\frac{\partial f}{\partial y}=2(4y+x+1)-4(-2y-x-2)=16y+6x+12=0$

Title: Re: Uni Maths Questions
Post by: Deleted User on May 07, 2013, 04:04:09 pm

How do I show whether this is a linear transformation (or not)?
F: R^3 -> R^2, F(x,y,z) = (0,2x+y)

I know I have to prove F(u+v) = F(u) + F(v) and F(au)= aF(u) but idk how to go about doing this

Title: Re: Uni Maths Questions
Post by: mark_alec on May 07, 2013, 04:19:24 pm

F(a+x, b+y, c+z) = (0, 2a+2x+b+y) = (0, 2a+b) + (0, 2x+y) = F(a,b,c) + F(x,y,z)
F(cx, cy, cz) = (0, 2cx+cy) = c(0, 2x+y) = cF(x,y,z)

Title: Re: Uni Maths Questions
Post by: Deleted User on May 07, 2013, 09:37:08 pm

Thanks man!

Title: Re: Uni Maths Questions
Post by: Deleted User on May 07, 2013, 09:38:09 pm

How do a find a single matrix that gives this linear transformation: reflects about y-axis, then expands by factor 5 in x-direction, then reflects about y=x? Thanks

Title: Re: Uni Maths Questions
Post by: kamil9876 on May 08, 2013, 06:04:24 pm

So doing B then A is equivalent to doing AB (because B takes v to Bv, then A takes Bv to A(Bv)=(AB)v)

So if you know the matrix of each of those transformations you can just multiply them together.

Another Way:

The image of (1,0) is the first column. The image of (0,1) is the second column. So let us see how (1,0) travels:

(1,0) -> (1,0) -> (5,0) -> (0,5)

(0,1) -> (0,-1) -> (0,-1) -> (-1,0)

So the matrix is:

0 -1
5 0

Title: Re: Uni Maths Questions
Post by: Jeggz on May 09, 2013, 07:45:17 pm

Find the point of intersection of the line 2x=y-1=z with the plane x+2y-z=2. What is the angle between the line and the plane?
I'm stuck guys, any help would really be appreciated :)

Title: Re: Uni Maths Questions
Post by: brightsky on May 09, 2013, 08:12:03 pm

line has equation r = (0,1,0) + t*(1/2,1,1), where t E R
plane has equation r.(1,2,-1) = 2
sub the line into the plane:
(0,1,0).(1,2,-1) + t(1/2,1,1).(1,2,-1) = 2
2 + t(1/2+2-1) = 2
2 + t(3/2) = 2
t = 0
so the point of intersection is (0,1,0).

the line has direction vector d = (1/2,1,1). the plane has normal vector n = (1,2,-1).
so the cosine of the acute angle between d and n is given by:
absolute value(d.n/mod(d)mod(n)) = (1/2 + 2 - 1)/(3/2)(sqrt(6)) = sqrt(6)/6
so the acute angle is arccos (sqrt(6)/6).
now we require the angle between the line and the plane, which is, then: pi/2 - arccos(sqrt(6)/6)

Title: Re: Uni Maths Questions
Post by: Deleted User on May 09, 2013, 09:58:38 pm

Hi guys, what do these bases mean: B= {e1,e2,e3} and B'={E^(ij)|i=1,2;j=1,2}? Something to do with 2x2 matrices?

Title: Re: Uni Maths Questions
Post by: Alwin on May 10, 2013, 09:36:19 pm

I have a question and its really bugging me >:(

Quote

prove that if all the rows of a matrix add up to the same number k, then k is an eigenvalue of this matrix. Describe one possible eigenvector corresponding to this special eigenvalue.

Okay, I got the second part, the eigenvector is a constant vector eg $[1\text{ }1\cdot \cdot \cdot1]^T$ but any ideas about proving k is an eigenvalue??

Thanks guys!

Title: Re: Uni Maths Questions
Post by: kamil9876 on May 10, 2013, 10:35:21 pm

So just evaluate Av where v is the column vector you have provided us. Each entry is in fact a sum of entries in a row (think about the definition of matrix multiplication.

Title: Re: Uni Maths Questions
Post by: Alwin on May 10, 2013, 10:43:24 pm

Quote from: kamil9876 on May 10, 2013, 10:35:21 pm

So just evaluate Av where v is the column vector you have provided us. Each entry is in fact a sum of entries in a row (think about the definition of matrix multiplication.

no no I was supposed to prove it first THEN HENCE find v, not the other way around. but dont worry, it's kool. I mucked around with determinants and the characteristic polynomial and got it eventually. it's not a very slick method tho.. Kamil9876 if u have a suggestion as to the proof (without using v first!) I'd love to hear it! or anyone else for that matter haha. thnx.

Title: Re: Uni Maths Questions
Post by: Mao on May 10, 2013, 11:59:17 pm

Quote from: Alwin on May 10, 2013, 10:43:24 pm

no no I was supposed to prove it first THEN HENCE find v, not the other way around. but dont worry, it's kool. I mucked around with determinants and the characteristic polynomial and got it eventually. it's not a very slick method tho.. Kamil9876 if u have a suggestion as to the proof (without using v first!) I'd love to hear it! or anyone else for that matter haha. thnx.

How is demonstrating that k is an eigenvalue not enough proof? Not every mathematical proof has to be a direct proof. This is an example of a proof by construction, which is equally valid. http://en.wikipedia.org/wiki/Proof_by_construction

Title: Re: Uni Maths Questions
Post by: Alwin on May 11, 2013, 08:36:43 am

Quote from: Mao on May 10, 2013, 11:59:17 pm

How is demonstrating that k is an eigenvalue not enough proof? Not every mathematical proof has to be a direct proof. This is an example of a proof by construction, which is equally valid. http://en.wikipedia.org/wiki/Proof_by_construction

That's the thing. I know proof by construction, but how do you propose to prove it here without using the matrix v ?

It's like trying to solve a quadratic like the following:

$x^2 - 5x - 6 =0$

I construct the answer to be 2,3 Then,
$(x-2)(x-3)=0$
$x^2 - 5x - 6 =0$

Oh, so my construction is correct!!
-.-

Seriously, you almost always find the eigenvalues first then the eigenvector. So explain please how to use proof by construction to prove:

Quote

that if all the rows of a matrix add up to the same number k, then k is an eigenvalue of this matrix.

Sorry, it's not that I'm being uncooperative, it's just that I have this massive ugly looking proof for it and it seems wayy to long...

EDIT: Don't worry guys, I got it.

Title: Re: Uni Maths Questions
Post by: Mao on May 12, 2013, 02:08:01 am

Quote

So explain please how to use proof by construction

Let A be a square matrix where each row adds up to k.

It follows $A\tilde{1} = k \tilde{1}$ , where $\tilde{1}$ is a column vector with each element being 1.

Therefore, k is an eigenvalue to A.

QED.

Title: Re: Uni Maths Questions
Post by: Yendall on May 12, 2013, 06:31:09 pm

Let $P(x,y,z)$ be the predicate $xy > z$ . If the domain of interpretation is $D = \{1,2\left\}$ , is the following true or false? There exist $x, y$ and $z$ such that $P(x,y,z)$ is true.

When the universe of discourse has only two variables, how do I determine Z?

I might be wrong, but wouldn't it be via assumption? Like the pattern 1,2... would step to 3?

Title: Re: Uni Maths Questions
Post by: kamil9876 on May 13, 2013, 12:16:39 am

Quote

When the universe of discourse has only two variables, how do I determine Z?

But of course the $x,y,z$ do not have to be distinct! So you want to find $x,y,z \in \{1,2\}$ such that $xy>z$ . Take $x=2, y=1, z=1$ and you see that indeed $P(2,1,1)$ is true. Hence the statement is true: there do exist such $x,y,z$ .

Title: Re: Uni Maths Questions
Post by: Yendall on May 13, 2013, 08:08:38 am

Quote from: kamil9876 on May 13, 2013, 12:16:39 am

But of course the $x,y,z$ do not have to be distinct! So you want to find $x,y,z \in \{1,2\}$ such that $xy>z$ . Take $x=2, y=1, z=1$ and you see that indeed $P(2,1,1)$ is true. Hence the statement is true: there do exist such $x,y,z$ .

Ohhhhh! Thank you, makes so much more sense.

Title: Re: Uni Maths Questions
Post by: Deleted User on May 13, 2013, 08:29:04 am

Let S: P2 -> P3 be defined as follows. For each p(x) = a2*x^2 + a1*x + a0, define S(p) = 1/3*a2*x^3 + 1/2*a1*x^2 + a0*x. Find the matrix A that represents S with respect to the bases B={1,x,x^2} and B'={1,x,x^2,x^3}.

How do I do this question thanks!!!!

Title: Re: Uni Maths Questions
Post by: MJRomeo81 on May 17, 2013, 11:58:16 pm

Hey guys, is anyone in the mood for tackling some recurrence relations?If someone could provide a detailed solution that would be great.

http://i.imgur.com/tdJPbn8.png

Title: Re: Uni Maths Questions
Post by: b^3 on May 18, 2013, 12:23:43 am

Not going to stick to it exactly, but give you enough so that you should be able to do the question.

4. This is a homogeneous recurrence relation since we don't have any terms that don't involve S(something).

Charateristic Equation

$\begin{alignedat}{1}x^{2} & =2x+8 \\ x^{2}-2x-8 & =0 \\ \left(x-4\right)\left(x+2\right) & =0 \\ x & \in\left\{ -2,4\right\} \end{alignedat}$

This means our general solution will be of the for

$\begin{alignedat}{1}S\left(n\right) & =a\cdot\left(-2\right)^{n}+b\cdot4^{n}\end{alignedat}$

Then the complete solution will arise when we fit the two initial conditions to the situation.
$\begin{alignedat}{1}S\left(0\right) & =1 \\ a\cdot\left(-2\right)^{0}+b\cdot4^{0} & =1 \\ a+b & =1 \\ b & =1-a....[1] \\ S\left(1\right) & =16 \\ a\cdot\left(-2\right)^{1}+b\cdot4^{1} & =16 \\-2a+4b & =16 \\ 2b-a & =8....[2] \\ \text{Substitute [1] into [2]} \\ 2\left(1-a\right)-a & =8 \\ 2-3a & =8 \\ 3a & =-6 \\ \implies a & =-2 \\ \text{substitute into [1]} \\ b & =1-\left(-2\right) \\ \implies b & =3 \\ \implies S\left(n\right) & =-2\cdot\left(-2\right)^{n}+3\cdot4^{n} \\ \therefore S\left(n\right) & =\left(-2\right)^{n+1}+3\cdot4^{n},\: n\geq0 \end{alignedat}$

Can't quite get 5. to work, may give it another shot tomorrow, or someone else might be able to help.

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on May 18, 2013, 01:19:37 am

Solution to Q5 attached.

Title: Re: Uni Maths Questions
Post by: Deleted User on May 18, 2013, 10:23:33 pm

How do I find the eigenvector of a 2x2 matrix?
So for
3 -2
2 -2

I've found that the eigenvalues are -2 and 1.

Then, for eigenvalue -2, I solve
|5 -2||x| = |0|
|2 0||y| |0|

So 5x-2y=0 and 2x=0. How do I solve for the eigenvector from here? What do I let equal to the parameter?

Thanks

Title: Re: Uni Maths Questions
Post by: Alwin on May 18, 2013, 10:48:34 pm

Quote from: Deleted User on May 18, 2013, 10:23:33 pm

How do I find the eigenvector of a 2x2 matrix?
So for
3 -2
2 -2

I've found that the eigenvalues are -2 and 1.

You may want to check that working. I get 2 and -1

Quote from: Deleted User on May 18, 2013, 10:23:33 pm

So 5x-2y=0 and 2x=0. How do I solve for the eigenvector from here? What do I let equal to the parameter?
Note these equations would also be incorrect
Thanks

And for your question, yes you do let a parameter, say t, equal y. Then let t equal a convenient number. t=1 is always pretty good!

Title: Re: Uni Maths Questions
Post by: Deleted User on May 19, 2013, 02:21:33 pm

Ok thanks. Does it matter whether I let x or y be the parameter?

Title: Re: Uni Maths Questions
Post by: Alwin on May 19, 2013, 03:06:53 pm

Quote from: Deleted User on May 19, 2013, 02:21:33 pm

Ok thanks. Does it matter whether I let x or y be the parameter?

Sorry, I don't mean to spoon feed you, but because of your initial mistake with the eigenvalues, I'll just write out how I would do this question.

$\text{let } A = \begin{bmatrix} 3 & -2 \\ 2 & -2 \end{bmatrix}$

$det(A-\lambda I)= \begin{bmatrix} 3-\lambda & -2 \\ 2 & -2-\lambda \end{bmatrix}= 0$

$(3-\lambda)(-2-\lambda) + 4 = 0$

$\lambda^2-\lambda-2 = 0$

$(\lambda-2) (\lambda+1) = 0$

$\therefore \lambda_1 = -1, \lambda_2 = 2$

$\lambda_1 = -1,$

$ \begin{bmatrix} 3-(-1) & -2 \\ 2 & -2-(-1) \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Solving gives:

$\begin{bmatrix} 1 & -\frac{1}{2} & | & 0\\ 0 & 0 & | & 0 \end{bmatrix}$

$\text{let t be the parameter where } y=t \text{ since y is the free variable},$

$\text{hence: }$

$ \begin{bmatrix} x \\ y \end{bmatrix} = t \begin{bmatrix} \frac{t}{2} \\ t \end{bmatrix} = t \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix}$

$\text{let } t=2, (\text{you can choose any value of t, except 0!})$

$\therefore \begin{bmatrix} 1 \\ 2 \end{bmatrix} \text{ is an eigenvector corresponding to } \lamdba_1 = -1$

Have a go at finding the other eigenvector corresponding to $\lambda_2=2$

Here's what you should get for the second eigenvector (see spoiler):

Spoiler

$\begin{bmatrix} 2 \\ 1 \end{bmatrix}$

Hope that clears up a few things! :D

Title: Re: Uni Maths Questions
Post by: Deleted User on May 19, 2013, 03:23:38 pm

Sweet thanks man!

Title: Re: Uni Maths Questions
Post by: Deleted User on May 19, 2013, 03:43:14 pm

Also, how do I find the image of a transformation?

For example, a linear transformation is given by
T(|x|) = |x+2y|
(|y|) | -y |
|x - y |

The standard matrix is
|1 2|
|0 -1|
|1 -1|

I'm not sure how to find the image matrix?

Title: Re: Uni Maths Questions
Post by: kamil9876 on May 19, 2013, 11:19:03 pm

It depends what sort of form you want it in. I'm not sure what an "image matrix" is. One way is to say that the image is the span of {(1,0,1),(2,-1,-1)}. In fact, this is a basis. In general the image is the span of the columns (think why) so if you want to find a basis you can find it the usual way.

Title: Re: Uni Maths Questions
Post by: Deleted User on May 20, 2013, 09:45:24 pm

I'm trying to find an eigenvector.

I've got the RREF of this matrix:
1 0 -36
0 0 1
0 0 0

What do I let as the parameter? Is it 'y' because there's no leading 1 in its column? If so, is the eigenvector
0
1
0
?

Title: Re: Uni Maths Questions
Post by: b^3 on May 20, 2013, 10:21:50 pm

Since there is no pivot in the $y$ column, ' $y$ is free', so you let $y=\alpha$ (well any parameter) where $\alpha \in \mathbb{R}$ and then solve each row. In this case you'd get $z=0$ , $x=0$ and $y=\alpha$ .
That is $\alpha\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}$ Where $\alpha \in \mathbb{R}$

Title: Re: Uni Maths Questions
Post by: Deleted User on May 21, 2013, 06:54:49 pm

Thank you that's what I thought.

Title: Re: Uni Maths Questions
Post by: Jeggz on May 25, 2013, 11:14:26 pm

Determine the dimensions of a rectangular box open at the top, having Volume V and possessing the least surface area?
I'm a bit stuck guys and any help would be appreciated :)

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on May 25, 2013, 11:36:11 pm

$V=xyz\iff z=\frac{V}{xy}$ , $A=xy+2yz+2xz=xy+\frac{2V}{x}+\frac{2V}{y}$ . Minimise $A$ by finding when both partial derivatives are zero.

Title: Re: Uni Maths Questions
Post by: Alwin on June 02, 2013, 11:05:50 pm

Quote from: Jeggz on June 02, 2013, 10:43:19 pm

It's just a maths question..but I'm trying to find the dimensions of a rectangular box which is open at the top, having volume V and possessing the least surface area. My question to you however is... if it's a rectangular box does that mean that the side faces are both equal? Do you get what I'm saying ? Like would it still work if I wrote down the volume as not V=xyz, but V=x^2z? Do you get me?

For a rectangular box, there are 3 "different" faces, top (same as bottom), front (same as back) and left (same as front). What you propose with a volume x^2 z is a box with a square face, because the faces can only be x by x dimensions or x by z dimension. Using V=xyz is more generic.

PS: sorry didn't pm reply, it's just I saw ur post^ so decided to add it here.

Title: Re: Uni Maths Questions
Post by: M-D on June 03, 2013, 10:48:12 am

i have a question in which i must sketch the graph of a function without the use of a calculator.

$f(x)=\frac{1+x^2}{1-x^2}$

i need to find where it is concave up and down.

$f''(x)=\frac{-4(3x^2+1)}{(x^2+1)^3}$

the graph is concave up when $f''(x)>0$

how can i find out manually over which interval $f''(x)=\frac{-4(3x^2+1)}{(x^2+1)^3}>0$ and $f''(x)=\frac{-4(3x^2+1)}{(x^2+1)^3}<0$ for concave down

thank you for your help

Title: Re: Uni Maths Questions
Post by: b^3 on June 03, 2013, 11:07:10 am

You may want to double check your double derivative :P
$\begin{alignedat}{1}f''\left(x\right) & =\frac{-4\left(3x^{2}+1\right)}{\left(x^{2}-1\right)^{3}}\end{alignedat} $
Now something to take note of here is that, $x^{2}$ is always positive or zero, and thus $3x^{2}+1$ is always positive. So our numerator $-4\left(3x^{2}+1\right)$ is always negative(-ve/-ve=+ve). Now we will get $f''(x)>0$ if the denominator is negative, that is
$\begin{alignedat}{1}\left(x^{2}+1\right)^{3}< & 0 \\ x^{2}+1 & <0 \\ -1<x<1 \end{alignedat} $
(do a quick sketch of the parabola if needed).
Now for $f''(x)<0$ , we need the denominator to be positive (-ve/+ve=-ve)
$\begin{alignedat}{1}\left(x^{2}+1\right)^{3}> & 0 \\ x^{2}+1 & >0 \\ x<-1\text{ or }x>1 \end{alignedat} $

Anyways, hope that helps :)

Title: Re: Uni Maths Questions
Post by: M-D on June 03, 2013, 12:07:47 pm

thanks b^3. you're amazing

Title: Re: Uni Maths Questions
Post by: M-D on June 03, 2013, 02:35:54 pm

i have a question similar to my previous one in which i must sketch the graph of a function without the use of a calculator.

$f(x)=\frac{x^3}{x^2+1}$

i need to find where it is concave up and down.

$f''(x)=\frac{-2x(x^2+3)}{(x^2+1)^3}$

the graph is concave up when $f''(x)>0$

how can i find out manually over which interval $f''(x)=\frac{-2x(x^2+3)}{(x^2+1)^3}$ and $f''(x)=\frac{-2x(x^2+3)}{(x^2+1)^3}$ for concave down

also is there an easier method for solving inequalities like these?

thank you for your help

Title: Re: Uni Maths Questions
Post by: Phy124 on June 03, 2013, 04:22:06 pm

Do the same for this question as b^3 did for the previous.

Firstly check over your double derivative working out because I'm pretty sure you should have $f''(x) = \frac{-2x(x^2-3)}{(x^2+1)^3}$

The whole term will be positive when both numerator and the denominator are positive or both the numerator and the denominator are negative.

$-2x(x^2-3)<0$ for all $x\in (-\sqrt{3},0) \cup (\sqrt{3},\infty)$ (Try sketching the graph, you can note that it is a negative cubic with x-intercepts at $x=-\sqrt{3},0,\sqrt{3}$ )

$(x^2+1)^3$ is never less than zero. (You know that $x^2+1 > 0$ for all $x \in \mathbb{R}$ and that the cubed part will not change this)

Hence both are never less than zero at the same time.

We will now focus on when both are positive.

$-2x(x^2-3)>0$ for all $x\in (-\infty,-\sqrt{3}) \cup (0,\sqrt{3})$

$(x^2+1)^3>0$ for all $x \in \mathbb{R}$

Therefore the whole term is positive for all $x \in (-\infty,-\sqrt{3}) \cup (0,\sqrt{3})$ (the intersection of the two domains)

The whole term will be negative when either of the numerator and the denominator are negative and the other isn't.

As the denominator is never negative, we don't need to find the condition of numerator = positive and denominator = negative. Instead, just the condition of the numerator being negative and the denominator being positive.

As shown before these were:

$-2x(x^2-3)<0$ for all $x\in (-\sqrt{3},0) \cup (\sqrt{3},\infty)$

$(x^2+1)^3>0$ for all $x \in \mathbb{R}$

Therefore the whole term is negative for all $x\in (-\sqrt{3},0) \cup (\sqrt{3},\infty)$ (the intersection of the two domains)

Therefore we have:

$f''(x)>0$ for all $x \in (-\infty,-\sqrt{3}) \cup (0,\sqrt{3})$

$f''(x)<0$ for all $x\in (-\sqrt{3},0) \cup (\sqrt{3},\infty)$

Title: Re: Uni Maths Questions
Post by: Deleted User on June 03, 2013, 09:24:24 pm

How do I do this question?

Find all solutions in mod 2 of the following linear system:
x1 + x3 + x5 = 1
x1 + x3 + x4 + x5 = 0
x4 + x5 = 0

Express your answer as a finite list.

So I've put the system into an augmented matrix and reduced it to RREF:
1 0 1 0 1 | 1
0 0 0 1 0 | 1
0 0 0 0 1 | 1

How do I find the solutions from here? Thanks

Title: Re: Uni Maths Questions
Post by: Mao on June 04, 2013, 12:21:26 am

The full RREF in $\mathbb{Z}_2$ is

Code: [Select]

1	0	1	0	0	0
0	0	0	1	0	1
0	0	0	0	1	1

x1=t, x2=s, x3=t, x4=1, x5=1, $t,s\in \mathbb{Z}_2$

Title: Re: Uni Maths Questions
Post by: M-D on June 04, 2013, 03:39:55 pm

how can i express the following in Cartesian form a+ib:

$\frac{cos(theta)+i sin(theta)}{cos(phi)+i sin(phi)}$

the answer in the solutions is $cos(theta - phi)+i sin(theta- phi)$

thanks

Title: Re: Uni Maths Questions
Post by: mark_alec on June 04, 2013, 03:44:13 pm

Since this is university maths, I assume you have come across the identity:
$e^{ix} = \cos{x}+i\sin{x}$ .

Title: Re: Uni Maths Questions
Post by: Deleted User on June 04, 2013, 03:46:32 pm

Quote from: Mao on June 04, 2013, 12:21:26 am

That RREF isn't right. It should be
Code: [Select]
1 0 1 0 0 0 0 0 0 1 0 -1 0 0 0 0 1 1
The answer is x1=t, x3=-t, x4=-1, x5=1, $t\in \mathbb{R}$

I thought in mod 2 there are only two numbers 0 and 1?

Title: Re: Uni Maths Questions
Post by: M-D on June 04, 2013, 04:46:37 pm

Quote from: mark_alec on June 04, 2013, 03:44:13 pm

Since this is university maths, I assume you have come across the identity:
$e^{ix} = \cos{x}+i\sin{x}$ .

could you show me the working out please. thanks

Title: Re: Uni Maths Questions
Post by: mark_alec on June 04, 2013, 04:47:47 pm

Use that identity and the usual index law:
$e^x/e^y = e^{x-y}$

Title: Re: Uni Maths Questions
Post by: Alwin on June 04, 2013, 05:00:03 pm

Quote from: M-D on June 04, 2013, 04:46:37 pm

Quote from: mark_alec on June 04, 2013, 03:44:13 pm
Since this is university maths, I assume you have come across the identity:
$e^{ix} = \cos{x}+i\sin{x}$ .
could you show me the working out please. thanks

Hmm, M-D I'm not sure if you're doing Uni Maths, or UMEP. Perhaps clarifying would help :)

Anyways, from Complex Numbers in the UMEP course:

To Prove:
$e^{ix} = \cos(x) + i\sin(x)$

Let:
$f(x) = ( \cos(x) + i\sin(x) ) e^{-ix}$

Hence:
$f'(x) = ( -\sin(x) + i\cos(x) ) e^{-ix} - i( \cos(x) + i\sin(x) ) e^{-ix}$

$f'(x) = -\sin(x)e^{-ix} + i\cos(x)e^{-ix} - i\cos(x)e^{-ix} - i^2\sin(x)e^{-ix}$

$f'(x) = -\sin(x)e^{-ix} + i\cos(x)e^{-ix} - i\cos(x)e^{-ix} + \sin(x)e^{-ix}$

$\therefore f'(x) = 0$

This is a constant function, since gradient = 0

Also,
$f(0) = ( \cos(0) + i\sin(0) ) e^{0} = 1$

Thus:
$f(x) = ( \cos(x) + i\sin(x) ) e^{-ix} = 1$

$( \cos(x) + i\sin(x) ) e^{-ix}e^{ix} = e^{ix}$

$\cos(x) + i\sin(x) = e^{ix}$

As Required :)

Title: Re: Uni Maths Questions
Post by: Jeggz on June 04, 2013, 05:06:51 pm

Quote from: Alwin on June 04, 2013, 05:00:03 pm

could you show me the working out please. thanks

Hmm, M-D I'm not sure if you're doing Uni Maths, or UMEP. Perhaps clarifying would help :)

There's a difference? ???

Title: Re: Uni Maths Questions
Post by: Alwin on June 04, 2013, 05:12:30 pm

Quote from: Jeggz on June 04, 2013, 05:06:51 pm

There's a difference? ???

Hmm, not really. Just that our course, MAST10013 UMEP Mathematics for High Achieving Students is actually only a 1 semester subject at Melbourn University haha.

Just wasn't sure what Maths M-D does in uni, like pure maths and statistics-based maths would vary by a LOT :P

EDIT: Deleted 2nd post, just added it here:

Quote from: M-D on June 04, 2013, 03:39:55 pm

how can i express the following in Cartesian form a+ib:

$\frac{cos(theta)+i sin(theta)}{cos(phi)+i sin(phi)}$

the answer in the solutions is $cos(theta - phi)+i sin(theta- phi)$

thanks

AHAHAHAHAHAHA SORRY M-D I did't actually answer your question.

So, you have the identity:
$e^{ix} = \cos{x}+i\sin{x}$ See my other post, Re: Uni Maths Questions, if you still want the formal proof.

Continuing, so you now have:
$\frac{\cos(\theta)+i \sin(\theta)}{\cos(\phi)+i \sin(\phi)}$

$= \frac{e^{\theta}}{e^{\phi}}$

$= e^{(\theta - \phi)}$

$=\cos{(\theta-\phi)} + i \sin{(\theta-\phi)}$

TADAAAA :D

Title: Re: Uni Maths Questions
Post by: M-D on June 04, 2013, 07:16:25 pm

thanks Alwin. how to do get the theta and phi symbols in Latex?

Title: Re: Uni Maths Questions
Post by: Lasercookie on June 04, 2013, 07:31:09 pm

Quote from: M-D on June 04, 2013, 07:16:25 pm

thanks Alwin. how to do get the theta and phi symbols in Latex?

\phi \Phi
\theta \Theta

The case of the first letter alternates between upper or lowercase. You'd be using the lowercase ones in this case.

If you click 'quote post' on a post with latex, you can look through the latex code they used that way too. (Or right click -> open image in new tab and read the latex code from the URL).

Title: Re: Uni Maths Questions
Post by: Mao on June 04, 2013, 11:47:05 pm

Quote from: Deleted User on June 04, 2013, 03:46:32 pm

I thought in mod 2 there are only two numbers 0 and 1?

Oops, I didn't even read the question. I'll update the answer.

Title: Re: Uni Maths Questions
Post by: Mao on June 04, 2013, 11:57:44 pm

Quote from: Alwin on June 04, 2013, 05:12:30 pm

So, you have the identity:
$e^{ix} = \cos{x}+i\sin{x}$ See my other post, Re: Uni Maths Questions, if you still want the formal proof.

Continuing, so you now have:
$\frac{\cos(\theta)+i \sin(\theta)}{\cos(\phi)+i \sin(\phi)}$

$= \frac{e^{\theta}}{e^{\phi}}$

$= e^{(\theta - \phi)}$

$=\cos{(\theta-\phi)} + i \sin{(\theta-\phi)}$

TADAAAA :D

At this level, it's probably more intuitive to stay away from $\exp(ix)$ :

$\begin{align} \frac{\cos(\theta)+i \sin(\theta)}{\cos(\phi)+i \sin(\phi)} = & \frac{\cos(\theta)+i \sin(\theta)}{\cos(\phi)+i \sin(\phi)} \frac{\cos(\phi)-i \sin(\phi)}{\cos(\phi)-i \sin(\phi)} \\ = & \frac{(\cos(\theta)+i \sin(\theta))(\cos(\phi)-i \sin(\phi))}{\cos^2(\phi)+\sin^2(\phi)} \\ = & \cos(\theta)\cos(\phi) + \sin(\theta)\sin(\phi) + i ( \cos(\phi)\sin(\theta)-\cos(\theta)\sin(\phi)) \\ = & \cos(\theta-\phi) + i \sin(\theta-\phi) \end{align}$

Title: Re: Uni Maths Questions
Post by: Deleted User on June 06, 2013, 09:43:06 pm

Thanks Mao!

Could someone help me with this question as well?

Without having to find the linear transformation matrix, how do I calculate the dimensions of the image and kernel for the linear transformation:

R: R^3->R^3 by reflection in the plane x+y+z=1 ?

So I know that dim(Ker) = nullity of R and dim(Im) = rank of R but I don't know whether these facts are of any use in this question?

Title: Re: Uni Maths Questions
Post by: M-D on June 07, 2013, 10:12:49 am

find all the solutions for $z^8 -256=0$

here is what i have done:

$z^8= 256$

$z=re^\(i\theta$

$256=256e^\(i0$

$therefore: r^8=256, r=+- 2$

$8\theta=0+2k\pi$ , k is an element of z

$let k=0, z= +-2$
$let k=1, z=+-2e^\(i(\pi/4)$
$let k=-1, z= +-2e^\(i(-\pi/4)$
$let k=-2, z= +-2e^\(i(-\pi/2)$

these are the solutions in polar form (are they correct?).also to convert them to Cartesian form i use the Argand diagram, however, what i don't understand is that here the modulus is 2 and -2 but the modulus must always be positive. i really don't know what to do

Title: Re: Uni Maths Questions
Post by: brightsky on June 07, 2013, 04:09:12 pm

not exactly sure what you did there, but here's how i would go about it:

z^8 = 256
z^8 = 256 cis(0)
z = (256 cis (2kpi))^(1/8)
= 2 cis (2kpi/8)
= 2 cis(kpi/4)
so you want your argument to be in (-pi,pi]
-pi<kpi/4 =< pi
-1 < k/4 =< 1
-4 < k/4 =< 4
so k = -3, -2, -1, 0, 1, 2, 3, 4
so z = 2cis(-3pi/4), 2cis(-pi/2), 2cis(-pi/4), 2cis(0), 2cis(pi/4), 2cis(pi/2), 2cis(3pi/4), 2cis(pi)

if this is uni maths, then just replace rcis(t) with r*e^(i*t). but yeah, modulus = distance from origin so it is strictly non-negative.

Title: Re: Uni Maths Questions
Post by: kamil9876 on June 07, 2013, 08:54:29 pm

Quote from: Deleted User on June 06, 2013, 09:43:06 pm

Thanks Mao!

Could someone help me with this question as well?

Without having to find the linear transformation matrix, how do I calculate the dimensions of the image and kernel for the linear transformation:

R: R^3->R^3 by reflection in the plane x+y+z=1 ?

So I know that dim(Ker) = nullity of R and dim(Im) = rank of R but I don't know whether these facts are of any use in this question?

But this transformation is not linear. Is it supposed to be x+y+z=0 perhaps?

Title: Re: Uni Maths Questions
Post by: Deleted User on June 07, 2013, 10:12:15 pm

Hmm... Not Sure. It's from an exam paper and it says that it's x+y+z=1.

Title: Re: Uni Maths Questions
Post by: M-D on June 08, 2013, 12:19:43 pm

could someone please help with integrating this. i have done as much as i could.

$\int\frac{4x^3+42x+1}{x^2+10} dx$

after applying long divison i get: $\int4x dx +\int\frac{2x+1}{x^2+10} dx$

i don't know to continue from here with the second integral. the first is easy. In the answers their is $arctan$ so i think the numerator and denominatos would have to be changed somehow. please help. thanks :)

Title: Re: Uni Maths Questions
Post by: brightsky on June 08, 2013, 01:25:35 pm

split the second integral up. (2x+1)/(x^2+10) = 2x/(x^2+10) + 1/(x^2+10). you can integrate the first fraction using substitution (you will get ln(something)). the arctan arises when you integrate the second fraction.

Title: Re: Uni Maths Questions
Post by: M-D on June 08, 2013, 03:34:33 pm

thanks brightsky. :)

Title: Re: Uni Maths Questions
Post by: Deleted User on June 10, 2013, 10:36:25 am

Hey could someone help me out with this question?

Find the point closest to the origin (0,0,0) on the surface given by the graph of the function z=x^2 - y + 1.

No idea where to start...

Title: Re: Uni Maths Questions
Post by: b^3 on June 10, 2013, 05:05:29 pm

I feel like I'm about to go about doing this the long way, but anyways.
The distance between the origin and any point on the surface is given by
$\begin{alignedat}{1}|d| & =\sqrt{\left(x-0\right)^{2}+\left(y-0\right)^{2}+\left(z-0\right)^{2}} \\ & =\sqrt{x^{2}+y^{2}+\left(x^{2}-y+1\right)^{2}} \\ & =\sqrt{x^{2}+y^{2}+x^{4}-x^{2}y+x^{2}-x^{2}y+y^{2}-y+x^{2}-y+1} \\ & =\sqrt{x^{4}+3x^{2}+2y^{2}-2x^{2}y-2y+1} \end{alignedat}$
Now we want to minimize that, which we can achieve by minimising whats under the square root, so lets let $f\left(x,y\right)=x^{4}+3x^{2}+2y^{2}-2x^{2}y-2y+1$ .
Now to find a minimum, we will take the partial derivatives with respect to $x$ and $y$ and set to zero. (For there to be a local minimum, the partial derivatives in both directions has to be zero).
$\begin{alignedat}{1}\frac{\partial f}{\partial x} & =4x^{3}+6x-4xy \\ \frac{\partial f}{\partial x} & =0 \\ 2x\left(2x^{2}-2y+3\right) & =0 \\ x=0 & \text{ or }2x^{2}-2y+3=0....[1] \\ \frac{\partial f}{\partial y} & =4y-2x^{2}-2 \\ \frac{\partial f}{\partial y} & =0 \\ 4y-2x^{2}-2 & =0 \\ 2y & =\frac{2x^{2}+2}{2} \\ 2y & =x^{2}+1....[2] \\ \text{subtitute [2] into [1]} \\ 2x^{2}-x^{2}-1+3 & =0 \\ x^{2}+2 & =0 \\ x^{2} & =-2 \\ \text{no solution} \end{alignedat} $
So the only solution we have so far is $x=0$ , which makes our partial derivative with respect to $x$ zero, but we still need $\frac{\partial f}{\partial y}$ to be zero as well, so we can substitute $x=0$ into [2].
$\begin{alignedat}{1}[2] \\ x=0 \\ 2y & =0+1 \\ y & =\frac{1}{2} \\ x=0,\: y=\frac{1}{2} \\ z & =x^{2}-y+1 \\ & =0^{2}-\frac{1}{2}+1 \\ & =\frac{1}{2} \\ \left(0,\frac{1}{2},\frac{1}{2}\right) \end{alignedat} $
We probably should verify that this distance is a minimum (just for the sake of it :P)
$\begin{alignedat}{1}\Delta & =\begin{vmatrix}f_{xx}\left(a,b\right) & f_{xy}\left(a,b\right) \\ f_{yx}\left(a,b\right) & f_{yy}\left(a,b\right) \end{vmatrix} \\ & =\begin{vmatrix}12x^{2}+6-4x & -4x \\ -4x & 4 \end{vmatrix} \\ & =8\left(6x^{2}+3-2x\right)+16x^{2} \\ x=0,\: y=\frac{1}{2} \\ \Delta & =8\left(3\right) \\ & =24 \\ f_{xx}\left(0,\frac{1}{2}\right) & =24 \end{alignedat} $
As $\Delta>0$ and $f_{xx}\left(0,\frac{1}{2}\right)>0$ , $\left(0,\frac{1}{2}\right)$ is a local minimum.
That is the point on the surface that is closest to the origin is $\left(0,\frac{1}{2},\frac{1}{2}\right)$

Title: Re: Uni Maths Questions
Post by: Deleted User on June 10, 2013, 06:43:21 pm

Thanks!!!

Title: Re: Uni Maths Questions
Post by: Deleted User on June 10, 2013, 09:36:10 pm

Another question if you guys don't mind.

Consider C[-1,1] = space of continuous functions defined on [-1,1] with the inner product
<f(x),g(x)> = antidiff (f(x)*g(x) dx) from -1 to 1.

Use the Gram-Schmidt procedure to find an orthonormal basis for the subspace of C[1,1] spanned by 1 and x^2.

I'm used to dealing with vectors but not so much with functions so could you guys show me how to find the first element of the basis?

u1 = v1/magnitude of v1 so if v1 = 1, wouldn't u1 just be = 1? Why is the answer u1 = 1/sqrt(2)?

Thanks

Title: Re: Uni Maths Questions
Post by: mark_alec on June 14, 2013, 12:22:30 am

Because <v1,v1> = |v1|^2 and its value is equal to 2.

Title: Re: Uni Maths Questions
Post by: M-D on June 14, 2013, 10:28:58 am

i have a question which say: solve the following separable differential equations whose right hand side depends only on $y$ .

1) $\frac{dy}{dx}=\frac{1}{y^2}$

2) $\frac{dy}{dx}=1+y^2$

how would i go about solving this. it does not really look like a separable diff. equation because it is only with respect to $y$ . thanks

Title: Re: Uni Maths Questions
Post by: Alwin on June 14, 2013, 12:30:50 pm

Quote from: M-D on June 14, 2013, 10:28:58 am

i have a question which say: solve the following separable differential equations whose right hand side depends only on $y$ .

1) $\frac{dy}{dx}=\frac{1}{y^2}$

2) $\frac{dy}{dx}=1+y^2$

how would i go about solving this. it does not really look like a separable diff. equation because it is only with respect to $y$ . thanks

Hint:

$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$

So you can 'flip' both sides and then you can integrate in terms of y

Title: Re: Uni Maths Questions
Post by: M-D on June 14, 2013, 07:17:45 pm

thank you very much Alwin. I have another question relating to implicit differentiation and points where the tangent line to a curve is vertical. Here's the question:

Consider the curve given by the equation: $x^4+xy+y^4=3$

(a) Find the equation of the tangent line to the curve at the point $(1,1)$

(b) Find all the points on the curve where the tangent line is vertical.

Solution to (a)

$\frac{dy}{dx}=\frac{-4x^3-y}{x+4y^3}$

let $x=1$ and $y=1$

$\frac{dy}{dx}=-1$

equation of the line is $y=2-x$

how should go about part (b)

Title: Re: Uni Maths Questions
Post by: brightsky on June 14, 2013, 07:25:32 pm

let the denominator of the derivative = 0. that is, let x + 4y^3 = 0 => x = 4y^3. sub this into the original equation and solve to get the values of x for which tangent is vertical.

Title: Re: Uni Maths Questions
Post by: M-D on June 15, 2013, 06:51:46 pm

could someone please help with these two questions:

1) Use the complex exponential to express $sin^3\theta$ in terms of $sin\theta$ and $sin3\theta$

2) Find all the roots of the polynomial: $P(z)=z^6-9z^3+8$

thanks. i really appreciate your help

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on June 15, 2013, 07:21:39 pm

1) Expand out $\left(\frac{1}{2i}\left(\exp(\mathrm{i}\theta)-\exp(-\mathrm{i}\theta)\right)\right)^3$ , try and get it in the form $\frac{a}{2i}\left(\exp(\mathrm{i}\theta)-\exp(-\mathrm{i}\theta)\right)+\frac{b}{2i}\left(\exp(3\mathrm{i}\theta)-\exp(-3\mathrm{i}\theta)\right)+c$ , which is equal to $a\sin(\theta)+b\sin(3\theta)+c$ .

2) Let $a=z^3$ , solve resulting quadratic.

Title: Re: Uni Maths Questions
Post by: b^3 on June 15, 2013, 07:29:07 pm

EDIT: Technically used the complex exponential but this might not be the way you were meant to do it.

Quote from: M-D on June 15, 2013, 06:51:46 pm

1) Use the complex exponential to express $sin^3\theta$ in terms of $sin\theta$ and $sin3\theta$

Firstly we can use the complex exponential to obtain the result below, then we can find an equivalent term by expanding using the binomial theorem.

Spoiler

$\begin{alignedat}{1}e^{i\theta} & =\cos\left(\theta\right)+i\sin\left(\theta\right) \\ \left(\cos\left(\theta\right)+i\sin\left(\theta\right)\right)^{3} & =\left(e^{i\theta}\right)^{3} \\ & =e^{3i\theta} \\ & =\cos\left(3\theta\right)+i\sin\left(3\theta\right) \\ \left(\cos\left(\theta\right)+i\sin\left(\theta\right)\right)^{3} & =\dbinom{3}{0}\left(\cos\theta\right)^{3}\left(i\sin\theta\right)^{0}+\dbinom{3}{1}\left(\cos\theta\right)^{2}\left(i\sin\theta\right)^{1}+\dbinom{3}{2}\left(\cos\theta\right)^{1}\left(i\sin\theta\right)^{2}+\dbinom{3}{3}\left(\cos\theta\right)^{0}\left(i\sin\theta\right)^{3} \\ & =\cos^{3}\theta+3i\cos^{2}\theta\sin\theta+3i^{2}\cos\theta\sin^{2}\theta+i^{3}\sin^{3}\theta \\ & =\cos^{3}\theta+3i\cos^{2}\theta\sin\theta-3\cos\theta\sin^{2}\theta-i\sin^{3}\theta \\ & =\left(\cos^{3}\theta-3\cos\theta\sin^{2}\theta\right)+i\left(3\cos^{2}\theta\sin\theta-\sin^{3}\theta\right) \end{alignedat}$

Then we can equate this to what we found previously, since we want to be dealing with $\sin^{3}\theta$ we can just equate the imaginary part.

Spoiler

$\begin{alignedat}{1}\cos\left(3\theta\right)+i\sin\left(3\theta\right) & =\left(\cos^{3}\theta-3\cos\theta\sin^{2}\theta\right)+i\left(3\cos^{2}\theta\sin\theta-\sin^{3}\theta\right) \\ \sin\left(3\theta\right) & =3\cos^{2}\left(\theta\right)\sin\left(\theta\right)-\sin^{3}\left(\theta\right) \\ \sin\left(3\theta\right) & =3\left(1-\sin^{2}\left(\theta\right)\right)\sin\left(\theta\right) \\ \sin\left(3\theta\right) & =3\sin\left(\theta\right)-3\sin^{3}\left(\theta\right)-\sin^{3}\left(\theta\right) \\ 4\sin^{3}\left(\theta\right) & =3\sin\left(\theta\right)-\sin\left(3\theta\right) \\ \sin^{3}\left(\theta\right) & =\frac{3}{4}\sin\left(\theta\right)-\frac{1}{4}\sin\left(3\theta\right) \end{alignedat}$

Hope that helps :)

Title: Re: Uni Maths Questions
Post by: Deleted User on June 15, 2013, 08:56:50 pm

Hi guys. Could someone explain to me how the quadric x^2+8xy+7y^2=9 is equivalent to the hyperbola u^2 - v^2/9 =1? Thanks

Title: Re: Uni Maths Questions
Post by: M-D on June 15, 2013, 10:01:22 pm

Quote from: b^3 on June 15, 2013, 07:29:07 pm

Hope that helps :)

thanks b^3. :D

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on June 15, 2013, 10:58:03 pm

Quote from: Deleted User on June 15, 2013, 08:56:50 pm

Hi guys. Could someone explain to me how the quadric x^2+8xy+7y^2=9 is equivalent to the hyperbola u^2 - v^2/9 =1? Thanks

How much do you know about rotations and stuff (for getting rid of the xy term)? I don't wanna explain stuff you already know. :P

Title: Re: Uni Maths Questions
Post by: Deleted User on June 15, 2013, 11:09:11 pm

I have no idea. I probably should considering I have an exam on it in 4 days but yeah, do you mind explaining it from the basics? Thanks!

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on June 15, 2013, 11:34:18 pm

Do you know how to rotate a curve about the origin using the 2D rotation matrix?

Title: Re: Uni Maths Questions
Post by: Deleted User on June 17, 2013, 04:22:48 pm

Is it this matrix?
|cosθ -sinθ |
|sinθ cosθ|

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on June 17, 2013, 07:17:38 pm

Yes. Is there anything in your notes about removing the $xy$ term by rotation?

Title: Re: Uni Maths Questions
Post by: Deleted User on June 17, 2013, 09:34:52 pm

Nope

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on June 17, 2013, 10:16:49 pm

I don't know why you'd be expected to be able to do that question then. But just to give you a decent hint, try to rotate the graph of $x^2+8xy+7y^2=9$ clockwise about the origin by an angle of $\arctan(2)$ .

Title: Re: Uni Maths Questions
Post by: Deleted User on June 17, 2013, 10:31:04 pm

Wow thanks! So why do I use the angle arctan(2)? And why do I have to use u and v instead of x and y?

Title: Re: Uni Maths Questions
Post by: BubbleWrapMan on June 18, 2013, 12:16:42 am

This has a decent explanation of rotations (courtesy of $b^{3}$ ): http://math.sci.ccny.cuny.edu/document/show/2685

This method gives you something like $-\frac{1}{2}\arctan\left(\frac{4}{3}\right)$ for your rotation angle, which I think is equal to $-\arctan\left(\frac{1}{2}\right)$ by some double angle formula. If you use this angle you get a hyperbola with a top half and a bottom half (as opposed to left and right halves) so you can add $\frac{\pi}{2}$ to this angle and rotate it further to get the type of hyperbola you want. You can do this because $\cot\left(2\left(\theta+\frac{\pi}{2}\right)\right)=\cot(2\theta)$ , so this angle still works and gives you a better result.

$u$ and $v$ are really different coordinates to $x$ and $y$ , but you can express them both in terms of $x$ and $y$ . I think that's what the question would be asking you for: something like $u=ax+by$ and $v=cx+dy$ .

Title: Re: Uni Maths Questions
Post by: Deleted User on June 18, 2013, 10:20:29 am

Sweet thanks. Yeah I'm pretty sure tht we didn't cover this in our lectures

Title: Re: Uni Maths Questions
Post by: FlorianK on June 20, 2013, 06:59:39 am

Hey, how would you go about solving the integral:
$\int_{0}^{\pi} (x\cdot sin(x))^{2} dx$

Title: Re: Uni Maths Questions
Post by: b^3 on June 20, 2013, 08:22:34 am

Change the $\sin^{2}(x)$ into $\frac{1-\cos(2x)}{2}$ , then expand out the integral. The first term is easy to int, the second you can use integration by parts.

Title: Re: Uni Maths Questions
Post by: FlorianK on June 21, 2013, 12:37:35 am

Quote from: b^3 on June 20, 2013, 08:22:34 am

Change the $\sin^{2}(x)$ into $\frac{1-\cos(2x)}{2}$ , then expand out the integral. The first term is easy to int, the second you can use integration by parts.

thx :)
Is there any webpage with a list of all those identities?

Title: Re: Uni Maths Questions
Post by: appianway on June 21, 2013, 12:49:10 am

Derive them with Euler's equation :P

Title: Re: Uni Maths Questions
Post by: mark_alec on June 21, 2013, 01:15:10 am

Quote from: FlorianK on June 21, 2013, 12:37:35 am

Is there any webpage with a list of all those identities?

Wikipedia -> List of trigonometry identities

Title: Re: Uni Maths Questions
Post by: FlorianK on June 21, 2013, 05:55:50 am

Can you check my working?

$\int x^{2} \cdot sin^{2}(x) dx$
$sin^{2}(x)=\frac{1}{2}\cdot(1-cos(2x)$
$\int x^{2} \cdot sin^{2}(x) dx = \frac{1}{2} \int x^{2} dx - \frac{1}{2}\int x^{2} \cdot cos(2x) dx$
$\int x^{2} \cdot cos(2x)dx = \frac{x^{2}}{2} \cdot sin(2x) - \int x \cdot sin(2x) dx$
$\int x \cdot sin(2x) dx = \frac{-x}{2} \cdot cos(2x) + \int \frac{1}{2} \cdot cos(2x) dx$
$\int \frac{1}{2} cos(2x)dx = \frac{1}{4} \cdot sin(2x)$
$\int x \cdot sin(2x) dx = \frac{-x}{2} \cdot cos(2x) + \frac{1}{4} \cdot sin(2x)$
$\int x^{2} \cdot cos(2x)dx = \frac{x^{2}}{2} \cdot sin(2x) + \frac{x}{2} \cdot cos(2x) - \frac{1}{4} \cdot sin(2x)$
$\int x^{2} \cdot sin^{2}(x) dx = \frac{1}{2} \int x^{2} dx - \frac{1}{2}\int x^{2} \cdot cos(2x) dx = \frac{1}{6} x^{3} - \frac{x^{2}}{4} \cdot sin(2x) - \frac{x}{4} \cdot cos(2x) + \frac{1}{8} \cdot sin(2x)$
$\int_{0}^{\pi} x^{2} \cdot sin^{2}(x) dx = [\frac{1}{6} x^{3} - \frac{x^{2}}{4} \cdot sin(2x) - \frac{x}{4} \cdot cos(2x) + \frac{1}{8} \cdot sin(2x)]^{\pi}_{0} = \frac{\pi^{3}}{6}-\frac{\pi}{4}$

Is there an easier way?

Title: Re: Uni Maths Questions
Post by: b^3 on June 21, 2013, 12:23:38 pm

Probably would have not worked on the integrals on different lines, but yeah it all works out.

Title: Re: Uni Maths Questions
Post by: b^3 on June 21, 2013, 08:32:40 pm

Quote from: Deleted User on June 15, 2013, 08:56:50 pm

Hi guys. Could someone explain to me how the quadric x^2+8xy+7y^2=9 is equivalent to the hyperbola u^2 - v^2/9 =1? Thanks

Something I've just come across while cramming for my linear algebra exam. Going to try and explain the theory first, but once we actually get to working it out it's not that long.
Since the xy term has disappeared we want to orthogonalise the curve, so we will need the quadratic form for the curve. The quadratic form for a curve is given by
$Q_{A}\left(\underset{\sim}{x}\right)&=&\underset{\sim}{x}^{T}A\underset{\sim}{x}$
where $A$ is the matrix of the coefficients of the curve, that is the coefficients of powers of $x$ go in the first column, starting with the highest power, then the coefficients of the $y$ terms starting with the highest power go in the second column again starting with the highest power and working downwards.
Now if we try to orthogonalise the matrix by using $A=PDP^{T}$ where $P$ is the matrix that is formed by taking the nullspace of $A$ corresponding to each eigenvector and making that the columns of $P$ and $D$ is the matrix with the eigenvalues on it's diagonal, and all other entries $0$ . This means that
$\begin{alignedat}{1}\underset{\sim}{x}^{T}A\underset{\sim}{x} & =\underset{\sim}{x}^{T}PDP^{T}\underset{\sim}{x} \\ & =\left(P^{T}\underset{\sim}{x}\right)^{T}D\left(P^{T}\underset{\sim}{x}\right) \\ & =Q\left(\underset{\sim}{x}\right) \\ \text{If }\underset{\sim}{x}=P\underset{\sim}{y} \\ \underset{\sim}{y} & =P^{-1}\underset{\sim}{x} \\ & =P^{T}\underset{\sim}{x}\:\left(\text{as the matrix is a diagonal matrix}\right) \\ \implies Q\left(\underset{\sim}{x}\right) & =\underset{\sim}{y}^{T}D\underset{\sim}{y} \\ & =\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\dots+\lambda_{n}y_{n}^{2} \end{alignedat}$
Which means we can take the eigenvalues as the coefficients of the squared terms. Now this will become useful later on.

Any conics can be written in the form $ax_{1}^{2}+2bx_{1}x_{2}+cx_{2}^{2}+dx_{1}+ex_{2}+f&=&0$ .
Now if $d=e=0$ then we can write it in the below form:
$\begin{bmatrix}x_{1} & x_{2}\end{bmatrix}\begin{bmatrix}a & b \\ b & c \end{bmatrix}\begin{bmatrix}x_{1} \\ x_{2} \end{bmatrix}+f&=&0$ .
Now if we divide by $-f$ this will give $\text{quadratic form}&=&1$ .
Now we have our quadratic from, we can rewrite the quadratic form in such a way that the cross terms vanish, obtaining a form called standard position. This has the form below:
$\begin{bmatrix}y_{1} & y_{2}\end{bmatrix}\begin{bmatrix}\lambda_{1} & 0 \\ 0 & \lambda_{2} \end{bmatrix}\begin{bmatrix}y_{1} \\ y_{2} \end{bmatrix}&=&1$
Where $\lambda_{1}$ and $\lambda_{2}$ are the eigenvalues of the coefficient matrix $A$ !
Expanding out the above gives $\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}&=&1$
Which we can see that if both the eigenvalues are positive, we will get an ellipse, if one is positive and one is negative we get a hyperbola and if both are negative we get no graph.

So we can take the coefficient matrix $A$ and the eigenvalues will be the coefficients on $x^{2}$ and $y^{2}$ when we orthogonalise the matrix. (How cool is that! :D )

Now to the actual working.
So firstly we need to get the equation into the right form.
$\begin{alignedat}{1}x^{2}+8xy+7y^{2} & =9 \\ \frac{1}{9}x^{2}+\frac{8}{9}xy+\frac{7}{9}y^{2} & =1 \end{alignedat}$
Next we find the matrix $A$ which is the coefficient matrix. The coefficients for $xy$ will be halfed, as we need to account for if we expanded the matrix back out. In our case this is
$\begin{alignedat}{1}A & =\begin{bmatrix}\frac{1}{9}\vphantom{\int_{a}^{b}} & \frac{4}{9} \\ \frac{4}{9}\vphantom{\int_{a}^{b}} & \frac{7}{9} \end{bmatrix} \\ & =\frac{1}{9}\begin{bmatrix}1 & 4 \\ 4 & 7 \end{bmatrix} \end{alignedat}$

Now we need to find the eigenvalues of $A$ , to form the matrix $D$ .
$\begin{alignedat}{1}\det\left(\lambda I-A\right) & =\begin{vmatrix}\lambda-\frac{1}{9} & -\frac{4}{9} \\ -\frac{4}{9} & \lambda-\frac{7}{9} \end{vmatrix} \\ & =\left(\lambda-\frac{1}{9}\right)\left(\lambda-\frac{7}{9}\right)-\frac{16}{81} \\ & =\frac{1}{9}\left(9\lambda^{2}-8\lambda-1\right) \\ & =\frac{1}{9}\left(9\lambda+1\right)\left(\lambda-1\right) \\ & =0 \\ \implies\lambda_{1} & =1 \\ \lambda_{2} & =-\frac{1}{9} \\ u^{2}-\frac{1}{9}v^{2} & =1 \end{alignedat}$
Which is what we wanted to get to. Hope that helps (and makes sense) :D ....now back to cramming :P (banning myself from AN for a few days to get work done).

Title: Re: Uni Maths Questions
Post by: dpagan on July 20, 2013, 01:48:59 pm

Hi, this is a vector spaces question which we just started:

Is there anything flawed with the comment, if a Matrix has an inverse (invertible) it is a spanning set AND independent thus it is a basis.

If however det=0, parameter(s) needed thus it is not the aforementioned.

Is that correct?

Also, what exactly is a subspace and why is it needed?

Thanks

Title: Re: Uni Maths Questions
Post by: Will T on July 22, 2013, 08:57:34 pm

Need help with a question:
Let A be any m x n matrix and let V be the set of vectors x \in R such that Ax=0
Prove that V is a subspace of R^n

Title: Re: Uni Maths Questions
Post by: kamil9876 on July 22, 2013, 10:04:43 pm

Use the properties of matrix multiplication: $A(x+y)=Ax+Ay$ along with the property that $A(cx)=c(Ax)$ for $c \in R$ .

Title: Re: Uni Maths Questions
Post by: Jeggz on July 25, 2013, 05:29:49 pm

Hey can someone please help me with this question?
Suppose T: v-->W is a linear tranformation and a bijection. Prove that T inverse is also linear.
Thanks :)

Title: Re: Uni Maths Questions
Post by: kamil9876 on July 25, 2013, 06:38:47 pm

So we must show that $T^{-1}w+T^{-1}w'=T^{-1}(w+w')$ for $w,w' \in W$

Now $T(T^{-1}w+T^{-1}w')=T(T^{-1}w)+T(T^{-1}w')=w+w'$ as $T$ is linear.

On the other hand we also have $T(T^{-1}(w+w'))=w+w'$ by the definition of inverse.

Thus we have $T(T^{-1}w+T^{-1}w')=T(T^{-1}(w+w'))$ . Now since $T$ is bijective it is also injective and so $T^{-1}w+T^{-1}w'=T^{-1}(w+w')$ .

(Remember, injective means $Tx=Ty$ implies $x=y$ )

I leave you to do the exact same thing to check the scalar multiplication condition. :)

Title: Re: Uni Maths Questions
Post by: Jeggz on July 25, 2013, 07:46:17 pm

Quote from: kamil9876 on July 25, 2013, 06:38:47 pm

So we must show that $T^{-1}w+T^{-1}w'=T^{-1}(w+w')$ for $w,w' \in W$

Now $T(T^{-1}w+T^{-1}w')=T(T^{-1}w)+T(T^{-1}w')=w+w'$ as $T$ is linear.

On the other hand we also have $T(T^{-1}(w+w'))=w+w'$ by the definition of inverse.

Thus we have $T(T^{-1}w+T^{-1}w')=T(T^{-1}(w+w'))$ . Now since $T$ is bijective it is also injective and so $T^{-1}w+T^{-1}w'=T^{-1}(w+w')$ .

(Remember, injective means $Tx=Ty$ implies $x=y$ )

I leave you to do the exact same thing to check the scalar multiplication condition. :)

Thanks a heap mate!! :)

Title: Re: Uni Maths Questions
Post by: vcestudent94 on August 15, 2013, 09:54:46 pm

Can someone please check if I did this proof right? I've never done a 'formal' proof before so just need some reassurance to tackle the rest of the questions as they're all pretty similar. Cheers.

(http://i43.tinypic.com/5lz95j.jpg)

Edit: Sorry about the big picture, it wouldn't let me attach it.

Title: Re: Uni Maths Questions
Post by: BigAl on August 16, 2013, 05:53:37 pm

Can someone check the validity of this proof and answer my questions below?

Title: Re: Uni Maths Questions
Post by: BigAl on August 16, 2013, 05:59:27 pm

there is a typo up there
delta=sqrt(x^2+(y-1)^2, not -(y-1)^2

Title: Re: Uni Maths Questions
Post by: humph on August 17, 2013, 05:48:53 am

Here's how I would write up my proof:

For $(x,y) \neq (0,1)$ and using shifted polar coordinates $(x,y) = (r\cos \theta, r\sin\theta + 1)$ with $r \in (0,\infty)$ , $\theta \in [0,2\pi)$ , so that $r = \sqrt{x^2 + (y - 1)^2}$ , we can write
$f(x,y) = r^2 \sin \theta \cos^3 \theta (r\sin \theta + 1),$
and so by the fact that $-1 \leq \sin \theta \leq 1$ and $-1 \leq \cos \theta \leq 1$ for all $\theta \in [0,2\pi)$ , together with the triangle inequality, we have that
$|f(x,y) - 0| \leq r^2(r + 1) \leq 2r^2$
for $0 < r < 1$ .
So for any $\varepsilon > 0$ , we let
$\delta = \min\left\{\sqrt{\frac{\varepsilon}{2}},\frac{1}{\sqrt{2}}\right\},$
and it follows by the previous argument that
$0 < \sqrt{x^2 + (y - 1)^2} < \delta$
implies that
$|f(x,y) - 0| < \varepsilon$
for all $(x,y) \neq (0,1)$ . We therefore conclude that
$\lim_{(x,y) \to (0,1)} f(x,y) = 0.$

Title: Re: Uni Maths Questions
Post by: BigAl on August 17, 2013, 12:50:56 pm

Quote from: humph on August 17, 2013, 05:48:53 am

Here's how I would write up my proof:

For $(x,y) \neq (0,1)$ and using shifted polar coordinates $(x,y) = (r\cos \theta, r\sin\theta + 1)$ with $r \in (0,\infty)$ , $\theta \in [0,2\pi)$ , so that $r = \sqrt{x^2 + (y - 1)^2}$ , we can write
$f(x,y) = r^2 \sin \theta \cos^3 \theta (r\sin \theta + 1),$
and so by the fact that $-1 \leq \sin \theta \leq 1$ and $-1 \leq \cos \theta \leq 1$ for all $\theta \in [0,2\pi)$ , together with the triangle inequality, we have that
$|f(x,y) - 0| \leq r^2(r + 1) \leq 2r^2$
for $0 < r < 1$ .
So for any $\varepsilon > 0$ , we let
$\delta = \min\left\{\sqrt{\frac{\varepsilon}{2}},\frac{1}{\sqrt{2}}\right\},$
and it follows by the previous argument that
$0 < \sqrt{x^2 + (y - 1)^2} < \delta$
implies that
$|f(x,y) - 0| < \varepsilon$
for all $(x,y) \neq (0,1)$ . We therefore conclude that
$\lim_{(x,y) \to (0,1)} f(x,y) = 0.$

Thanks for formal maths...I finally end up with the same result as yours after realising that 2r^2 is much more proper function than 2r^3 for the values between 0 and 1. Thanks again

Title: Re: Uni Maths Questions
Post by: Jeggz on October 20, 2013, 11:46:02 am

I would really appreciate help with this question guys! Thanks in advance :)

Consider the Inner product <f,g> = $\int_a^b \! f(t)g(t) \, \mathrm{d}t.$

Let $f(x)= x^2$ and define D(a,b) = $||f(x)-(ax+b)||^2$ .
Minimise the function D and sketch both f(x) and the solution ax+b over [0,1]

Title: Re: Uni Maths Questions
Post by: stolenclay on October 20, 2013, 03:46:50 pm

Are you sure that's the right question? Because it doesn't seem to have a solution that you could get by hand...

Are you sure you don't mean $\langle f,g \rangle = \int_0^1 f \left ( t \right )g \left ( t \right ) \, \mathrm{d}t$ ?

Spoiler

If you do mean that, then:
$D(a,b) = \langle f\left ( x \right )-\left ( ax+b \right ),f\left ( x \right )-\left ( ax+b \right ) \rangle = \int_0^1 \left [x^{2}-\left ( ax+b \right ) \right ]^{2} \, \mathrm{d}t$
$D(a,b) = \frac{a^2}{3}+a b-\frac{a}{2}+b^2-\frac{2 b}{3}+\frac{1}{5}$
To minimise that, you could try partial derivatives. (Interestingly, it's also $3 \left(\frac{a}{3}+\frac{b}{2}-\frac{1}{4}\right)^2+\frac{1}{4} \left(b+\frac{1}{6}\right)^2+\frac{1}{180}$ if you try and complete the square, but it's probably easier to use partial derivatives.)
Anyway, you should get that the minimum of $D(a,b)$ is $\frac{1}{180}$ , with $a = 1 \$ and $b = -\frac{1}{6}$ .

Hopefully I didn't solve the wrong question.

Title: Re: Uni Maths Questions
Post by: Jeggz on October 20, 2013, 06:43:10 pm

OMG YES I DID MEAN THAT!
Thank you so much legend :)

Title: Re: Uni Maths Questions
Post by: BigAl on October 22, 2013, 08:06:51 pm

Just a general question. How do we find potential field such that F=Gradf ...isn't it just integratating component by component then adding them up?

Title: Re: Uni Maths Questions
Post by: rife168 on October 22, 2013, 10:37:11 pm

Quote from: BigAl on October 22, 2013, 08:06:51 pm

Just a general question. How do we find potential field such that F=Gradf ...isn't it just integratating component by component then adding them up?

Yeah that's pretty much it, once you have determined that it is conservative and well defined and all that business...
Then when you integrate it's the case that you have 'constant' functions instead of the usual '+C' as in the case of single variables.

See here:
http://mathinsight.org/conservative_vector_field_find_potential

ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: vcestudent94 on March 14, 2013, 12:43:42 pm