Mathematics marathon

[y3k9]

Ok, this was quite interesting on aops, and I learned quite recently that there's a lot math people like me on the forum, I am starting the mathematics marathon.

The basic idea is that you go on post an interesting problem and solve the one posted above you. It's quite simple. However, on aops there's marathons for multiple skill levels, and I don't want multiple threads so just say for what level the problem is intended and solve only those that are of your level. In this thread abstract math is strongly encouraged. Here are rules:

1. Have an included solution to the question you answered and bold the answer.
2. Make use of spoilers, make sure to have a separate spoiler for both your answer and solution.
3. If your really into the REALLY abstract math title your questions "ABSTRACT"
4. Try to post multiple problems if possible.
5. Frequent this thread, and help someone if they answer your question wrong.

That's about it, I'll go first:
Level: intermediate (8th grade)
Problem 1:

Spoiler

For what value x is \( 2+x-x^2 = 0 \)

As of 31/12/10:
This thread seems to be doing good, might I ask however that you either box or bold your answer in a solution to a problem as well as number your problems. It was intention for each problem to be numbered, so please number the next problem you write.

 

[aronpm]

Answer TO PROBLEM 1:

Spoiler

Divide by -1 to get \( x^2 - x - 2 = 0 \) and factorize to get \( (x - 2)(x + 1) = 0 \). Therefore either \( x - 2 = 0 \) (so \( x = 2 \)) or \( x + 1 = 0 \) (\( x = -1 \))

IDK why you asked for me to write a proof to that question since my hypothetical dog could do it in his sleep.

Problem 2:

Spoiler

Provide a unique proof of Fermat's Last Theorem

 

[Christopher Mowla]

Provide a unique proof of Fermat's Last Theorem

You have a very good sense of humor, my friend.

Level: intermediate (8th grade)

Sorry, but in the math world, basic algebra is not intermediate: it is knowledge that should already be second nature for you to have any hope to solve other problems.

Not that I really expect anyone to solve this problem, but here is an "easy" problem: (not abstract, just ingenuity):

Spoiler

Find
\( \frac{d}{dx}\left[ \frac{1}{\sqrt{\sin ^{2}\left( x \right)-1}} \right] \)

 

[Rinfiyks]

Answer:

Spoiler

\( \sin^2(x)+\cos^2(x) = 1 \)
\( \sin^2(x) - 1 = -\cos^2(x) \)
So it becomes
\( \frac{d}{dx} \left[ \frac{1}{\sqrt{-\cos^2(x)}} \right] \)
Which... er... I thought that you'd have designed it so the minus sign isn't in the square root!
I can't do it

 

[TimMc]

Sorry, but in the math world, basic algebra is not intermediate: it is knowledge that should already be second nature for you to have any hope to solve other problems.

They didn't say:

I hereby declare the following problem to be of an intermediate level of difficulty in the world of mathematics.

They might very well have said:

The average 8th grade student could solve this problem because it's likely to be a part of their curriculum.

Hooray for Maths!

Tim.

 

[Sa967St]

Answer to cmowla's problem:

Spoiler

Still not too sure if it's right.

Problem:

Spoiler

Find the derivative of:

 

[Christopher Mowla]

Answer to Sa967St's problem

Spoiler

\( \frac{d}{dx}\left[ \sqrt{x^{2}+9}+2\ln \left( x+\sqrt{x^{2}+9} \right) \right] \)

\( =\frac{d}{dx}\left[ \left( x^{2}+9 \right)^{1/2}+2\ln \left( x+\left( x^{2}+9 \right)^{1/2} \right) \right] \)

\( =\frac{1}{2}\left( x^{2}+9 \right)^{-1/2}\left( 2x \right)+2\frac{1+\frac{1}{2}\left( x^{2}+9 \right)^{-1/2}\left( 2x \right)}{x+\left( x^{2}+9 \right)^{1/2}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+2\frac{1+\frac{x}{\sqrt{x^{2}+9}}}{x+\sqrt{x^{2}+9}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+2\frac{\frac{\sqrt{x^{2}+9}}{\sqrt{x^{2}+9}}+\frac{x}{\sqrt{x^{2}+9}}}{x+\sqrt{x^{2}+9}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+2\frac{\sqrt{x^{2}+9}+x}{\sqrt{x^{2}+9}}\frac{1}{x+\sqrt{x^{2}+9}} \)

\( =\frac{x}{\sqrt{x^{2}+9}}+\frac{2}{\sqrt{x^{2}+9}} \)

\( =\frac{x+2}{\sqrt{x^{2}+9}} \)

Problem:

Spoiler

The same one I already posted

 

[Sa967St]

Cmowla got mine correct.
I edited my answer to his question, but I'm not too sure if it's correct.

 

[Christopher Mowla]

Still not too sure if it's right.

It's not. I noticed that, where I posted the problem, latex makes the superscript 2 so small. Try again.

This is the solution to check your answer:

Spoiler

\( \frac{d}{dx}\left[ \frac{1}{\sqrt{\sin ^{2}\left( x \right)-1}} \right] \)

Does not exist for real numbers. The function is not even defined on its own domain. (Which is a problem because we need a function to be continuous at least somewhere in order to calculate the derivative at that x-value). See its graph. Notice that the orange is the imaginary portion, and the blue is the real portion.

\( \text{Reason:} \)

\( \sin ^{2}\left( x \right)\le 1\text{ for all }x\in \left( -\infty ,\infty \right). \)

\( \text{Hence, if }\sin ^{2}\left( x \right)=1,\text{ that is, if }x=\frac{n\pi }{2}\text{ for }n\text{ being non-zero integers,} \)

\( \text{then we have }\frac{d}{dx}\left[ \frac{1}{\sqrt{1-1}} \right]=\frac{d}{dx}\left[ \frac{1}{\sqrt{0}} \right]. \)

\( \text{If sin}^{2}\left( x \right)<1,\text{ then we have a negative in the square root}\text{.} \)

 

[Rinfiyks]

But it still has a solution in complex numbers. I thought that was what we were supposed to find?

 

[keemy]

Answer to cmowla

Spoiler

\( (-cos^2(x))^{\frac{-1}{2}} \)
(i guess this is a complex valued function?)
use chain rule
\( \frac{-2cos(x)sin(x)}{2(-cos^2(x))^{\frac{3}{2}}} \)
\( \frac{cos(x)sin(x)}{(cos^2(x))^{\frac{3}{2}}} \)

Problem:

Spoiler

Prove:\( \tbinom {2n}{n} = \sum_{k=0}^{n} {\tbinom {n}{k}}^2 \)

 

[Whippopotamus]

Hi, I'm doing As Level Maths at the minute, and I have a fair interest in maths, and I would like to pose this question, its just something I noticed when I was bored in a physics lesson:

Spoiler

Why is it that:
sin cos cos sin (n)
sin cos cos sin cos cos cos sin (n)
sin cos cos sin cos cos cos sin cos cos cos cos sin (n)

Firstly, for any value of n, tend towards the same value, and secondly, why do they tend towards a more specific value as you work down the list (as the number of sin and cos functions are increased in that pattern)?

P.S Im not 100% sure of my answer...

 

[Christopher Mowla]

But it still has a solution in complex numbers. I thought that was what we were supposed to find?

You are correct. We can express the solution as a complex function if we wanted to.

Spoiler

\( \frac{d}{dx}\left[ \frac{1}{\sqrt{\sin ^{2}\left( x \right)-1}} \right]=\frac{d}{dx}\left[ \frac{1}{\sqrt{-\left( 1-\sin ^{2}\left( x \right) \right)}} \right] \)

\( =\frac{1}{i}\frac{d}{dx}\left[ \frac{1}{\sqrt{1-\sin ^{2}\left( x \right)}} \right]=\frac{1}{i}\frac{d}{dx}\left[ \left( 1-\sin ^{2}\left( x \right) \right)^{-1/2} \right] \)

\( =\frac{1}{i}\left( -\frac{1}{2}\left( 1-\sin ^{2}\left( x \right) \right)^{-3/2}\left( -2\sin \left( x \right)\cos \left( x \right) \right) \right) \)

\( =\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left( 1-\sin ^{2}\left( x \right) \right)^{3/2}} \)

\( =\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left( \cos ^{2}\left( x \right) \right)^{3/2}}=\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left| \cos ^{3}\left( x \right) \right|} \)

\( =\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left( \cos ^{2}\left( x \right) \right)^{3/2}}=\frac{1}{i}\frac{\sin \left( x \right)\cos \left( x \right)}{\left| \cos ^{3}\left( x \right) \right|} \)

\( =\frac{1}{i}\tan \left( x \right)\left| \frac{1}{\cos \left( x \right)} \right|\times \left( \frac{i}{i} \right)=-i\tan \left( x \right)\left| \frac{1}{\cos \left( x \right)} \right| \)

\( (\text{For }x\ne \frac{n\pi }{2},\text{ where }n\text{ is any non-zero integer)}\text{.} \)

EDIT:

(i guess this is a complex valued function?)

Yes. See my spoiler.

 

[macky]

Answer to keemy

Spoiler

Rewrite as
\( \binom{2n}{n} = \sum_{k = 0}^n \binom{n}{k}\binom{n}{n-k}. \)
Partition of set of 2n objects into two sets of n. Then this says that one chooses n objects by choosing k from the first set and n-k from the second, where k runs over 0 through n.

[edit]
Can I not center displays? No align environment?

[edit]
A problem:

Spoiler

Prove that
\( \frac{a_3^2}{a_3^2 + a_4^2 + a_5^2 - a_3a_4 - a_4a_5} \)
is maximized for real \( a_3, a_4, a_5 \) when \( a_3 = 3a_5, a_4 = 2a_5 \).

I'll reveal the source later.

 

[vcuber13]

Answer to cmowla's problem:

Spoiler

Still not too sure if it's right.

woulnt it be the square root of the sinx -1 cubed?

 

[Sa967St]

woulnt it be the square root of the sinx -1 cubed?

Squared, but the ^2 wasn't there when I answered it:

I noticed that, where I posted the problem, latex makes the superscript 2 so small.

 

[vcuber13]

i mean wouldn't n^(3/2) be sqrt(n^3) not cuberoot(n^2)?

 

[Sa967St]

i mean wouldn't n^(3/2) be sqrt(n^3) not cuberoot(n^2)?

Yes. I always make dumb mistakes.
Still wasn't correct since there are no real answers as cmwola explained.

 

[keemy]

ya the latex on SS is kind of annoying.

answer to macky

Spoiler

Prove that
k going to just let x=a2, y=a4, z=a5.
plugging in we find we want to prove max is 3/2. So set up inequality.


\( \frac {x^2}{x^2+y^2+z^2-xy-yz} \leq \frac {3}{2} \)
\( 2x^2 \leq 3x^2 +3 y^2 +3z^2 -3xy - 3yz \)
\( (x-y)(x-2y) + (y-z)(y-2z) +z^2 \geq 0 \)

so the only way this could be false would be if \( y<x<2y \) and/or \( z<y<2z \) (so one or both of first terms are negative) note that the least the first 2 terms could be (in said range) are when \( x=1.5y \), \( y=1.5z \) but plugging in gets the \( LHS=.1875z^2 \) which clearly can't be negative

I feel like I missed something...

problem uhh I think there is still one left unanswered also don't want to find one now.

 

[macky]

That works. Simpler:

Spoiler

\( \frac{a_3^2}{a_3^2 + a_4^2 + a_5^2 - a_3a_4 - a_4a_5} \le \frac32 \)
\( \Leftrightarrow 2a_3^2 \le 3a_3^2 + 3a_4^2 + 3a_5^2 - 3a_3a_4 - 3a_4a_5 \)
\( \Leftrightarrow 0 \le \frac14(4a_3^2 + 9a_4^2 - 12a_3a_4) + \frac34(a_4^2 + 4a_5^2 - 4a_4a_5) \)
\( \Leftrightarrow 0 \le \frac14(2a_3 - 3a_4)^2 + \frac34(a_4 - 2a_5)^2 \)

And this proof easily generalizes to more variables.

This was part of Exercise 21.14 from Fulton and Harris's A First Course in Representation Theory. These maximization problems come up when ruling out Dynkin diagrams of simple Lie algebras (the above corresponds to a branch of length 3). For example, they can be used to show that if a node with three branches with length \( p, q, r \), then \( 1/p + 1/q + 1/r \le 1 \). This rules out all diagrams of this kind except \( D_5, E_6, E_7, E_8 \).

Also, there's no good reason why the index starts at 3.