Author Topic: TT's Maths Thread (Read 146707 times) Tweet Share

Over9000 · « **Reply #45 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.

TrueTears · « **Reply #46 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.

TrueTears · « **Reply #47 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 $y = -x$

kamil9876 · « **Reply #48 on:** November 13, 2009, 11:41:52 pm »

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

humph · « **Reply #49 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

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.

kamil9876 · « **Reply #50 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

plus it's a good way to spend time and make the most of uni i realised

TrueTears · « **Reply #51 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

humph · « **Reply #52 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 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.

TrueTears · « **Reply #53 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 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

TrueTears · « **Reply #54 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.

kamil9876 · « **Reply #55 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$

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

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

TrueTears · « **Reply #56 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

TrueTears · « **Reply #57 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.

dcc · « **Reply #58 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.

TrueTears · « **Reply #59 on:** November 14, 2009, 09:57:48 pm »

Thanks a lot dcc.

Can you show me induction method?

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