The Proof is Trivial!

I'm not going to pretend to understand anything you've written there, but that is the correct answer! Guess there are more than 2 ways of doing it

Which part don't you understand?

Problem 390***

Evaluate

$\displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}$.

It converges. How do you justify the first line?

Problem 387**

Find (as a power series) the function that satisfies

$f''(x)+f'(x)=f(x)+1.$

Problem 384***

It seems I incorrectly starred this one. Here's a better rewording:

Prove that if $a,b,\frac{a^2+b^2}{1+ab} \in \mathbb{Z}$ , then $\frac{a^2+b^2}{1+ab}$ is a perfect square.

There seem to be two problems of the same number. Unless I'm missing a very clever link between the two

Solution 386**/***

$I = \displaystyle \int_{0}^{2\pi} \frac{1}{5-3\sin x}\ dx$

I'm going with the Weierstrass sub as I've seen an alternative method used already

$\text{Let} \ t = \tan \left( \dfrac{x}{2} \right)$

$\begin{aligned} \sin x & = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \\ & = 2 \left( \frac{t}{\sqrt{1+t^2}} \right) \left( \frac{1}{\sqrt{1+t^2}} \right) \\ & = \dfrac{2}{1+t^2} \end{aligned}$

$\begin{aligned} dx & = 2 \cos^2 \left(\frac{x}{2} \right)\ dt \\ & = \dfrac{2}{1+t^2}\ dt \end{aligned}$

$x=0 \Rightarrow t = 0, \ x=2\pi \Rightarrow t = 0$

However, we have to account for the discontinuity at $x=\pi$ by splitting the integral in two:

$\frac{5}{2} I = \displaystyle \int_{-\infty}^{0} \dfrac{dt}{t^2 - \frac{6}{5}t +1} + \displaystyle \int_{0}^{\infty} \dfrac{dt}{t^2 - \frac{6}{5}t +1}$

By completing the square, we get a more familiar integral:

$\frac{5}{2} I = \underbrace{\displaystyle \int_{-\infty}^{0} \dfrac{dt}{\left(t-\frac{3}{5}\right)^2 + \left(\frac{4}{5} \right)^2}}_{\pi < x < 2\pi} + \underbrace{\displaystyle \int_{0}^{\infty} \dfrac{dt}{\left(t-\frac{3}{5}\right)^2 + \left(\frac{4}{5} \right)^2}}_{0 < x < \pi}$

$\frac{5}{2} I = \left[ \frac{5}{4} \arctan \left( \frac{5 \left( t-\frac{3}{5} \right)}{4} \right) \right]_{-\infty}^{0} + \left[ \frac{5}{4} \arctan \left( \frac{5 \left( t-\frac{3}{5} \right)}{4} \right) \right]_{0}^{\infty}$

$\frac{1}{2} I = \frac{1}{4} \left( \frac{\pi}{2} - \arctan \frac{-3}{4} \right) + \frac{1}{4} \left( \arctan \frac{-3}{4} - \left(\frac{-\pi}{2} \right) \right)$

$\therefore \ I = \dfrac{\pi}{2}$

That substitution is just, words can't even describe. Thanks for posting another method Amazing how many different ways the same problem can be evaluated, maths is perfect

I am almost certain I've seen this before and that the solution that went with it was absolutely brilliant****.
But actually I think this is manageable with root flipping:

If I set the fraction as p then
$a^2 + b^2 -pab - p = 0$
If a=b then obviously a=b=p=1.
Otherwise we assume a>b and take the equation as a quadratic in a. Ignoring the trivial points, there are roots a1 and a2, where a1 = a. Also a1 + a2 = pb and a1a2 = b^2 - p hold (vieta).

The idea now becomes to show $a_2 \geq 0$. So we assume a2 < 0 and so b^2 - p < 0 so p>b^2. Now a1 > pb >b^3 and so ab+1>b^4+1 which is a contradiction to the initial expression.

Because we have shown the holds are consistent, we have created a new solution a,b with a < b. We can repeat this until one (b) is 0. Since p does not change we reduce it to p = a^2 / 1 which is a perfect square.

(This is of course not the full solution you would want to give, it is just an outline and in my opinion Vieta jumping/ root flipping means this problem is not that hard for the IMO).

**** - I am pretty sure the solution used an infinite series to derive further expressions through induction. This may also work to show that

Which wouldn't be possible in root flipping as the power is not 2.

This is, however, true only when the series converges absolutely, which is not the case. Many trivial examples exist, showing that you can rearrange only absolutely convergent series. By the way, Jordan did exactly the same thing, and was convinced that only absolutely convergent series exist.

I hope none of you have seen this question before

Problem 391**

A number of straight lines are drawn in a plane, dividing it into regions. Show that each region may be colored either red or black in such a way that no two neighboring regions have the same color.



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