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    (Original post by Flauta)
    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?
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    (Original post by henpen)
    Which part don't you understand?
    The transformation of the initial integral into


    EDIT: Oh wait, it's the maclaurin series for \dfrac{1}{1-x} right?

    Yeah I understand it all now thanks. That's a really cool way of doing it!
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    Problem 387**

    Find (as a power series) the function that satisfies

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

    Spoiler:
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    http://math.uchicago.edu/~chonoles/e...tricseries.pdf, also use the Fibonacci numbers' generating function.
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    Problem 387***

    Evaluate the following infinite sum

    \displaystyle \sum_{r=0}^{\infty} \dfrac{4r}{3^{r+1}}

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    Problem 388***

    Here's another one

    Evaluate

    \displaystyle \sum_{r=1}^{\infty} \dfrac{(-1)^r}{r^2}

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    Solution 388

    \displaystyle \sum_{n\geq 0} \frac{4n}{3^{n+1}} = \sum_{n\geq 1} \frac{4n}{3^{n+1}}

    |x| < 1:

    \displaystyle\begin{aligned} \sum_{n\geq 1} x^n = \frac{x}{1-x} \implies \sum_{n\geq 1} n x^{n-1} & = \frac{d}{dx} \frac{x}{1-x} \\ & = \frac{1}{(x-1)^2} \\ \implies \sum_{n\geq 1} 4n x^{n+1} & = \frac{4x^2}{(x-1)^2} \\ \implies \sum_{n\geq 0} \frac{4n}{3^{n+1}} & = \frac{4x^2}{(x-1)^2} \bigg|_{x=\frac{1}{3}} = 1

    Solution 389
    It is very well known that \displaystyle\sum_{n\geq 1} \frac{1}{n^2} = \frac{\pi^2}{6} (unless that was the intention of the question).

    \displaystyle\begin{aligned} \sum_{n\geq 1} \frac{(-1)^n}{n^2} & = \sum_{n\geq 1} \frac{1}{(2n)^2} - \sum_{n\geq 1} \frac{1}{(2n-1)^2} \\ & = \frac{\pi^2}{24} - \left( \sum_{n\geq 1} \frac{1}{n^2} -  \sum_{n\geq 1} \frac{1}{(2n)^2} \right) \\ & = -\frac{\pi^2}{12}
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    Alternative:

    Solution 389

    Consider \displaystyle f(z)= \frac{1}{z^{2} \sin \pi z}. Choose some beautiful contour C_{n}; for example, one the most popular - square with vertices at \displaystyle \pm (n+\frac{1}{2}) \pm i(n+\frac{1}{2}) will do. We can easily (indeed, all we need is to bound \sin \pi z below) see that \displaystyle \int_{C_{n}} f \to 0 as n \to \infty.

    Our function has simple poles at z=n \in \mathbb{Z}, n \not= 0 and 3-pole at z=0.
    For n \not= 0, we have \displaystyle \text{Res}(f,n) = \frac{(-1)^{n}}{\pi n^{2}}, which is obvious.

    We turn to the case z=0. We note \displaystyle \sin \pi z = \sum_{0}^{\infty} (-1)^{k}\frac{(z \pi)^{2k+1}}{(2k+1)!} and \displaystyle f(z) = z^{-3}\frac{1}{\frac{\sin z \pi}{z}}. Hence, \displaystyle \text{Res}(f,0) = \frac{\pi}{6}.

    Whence, \displaystyle \frac{1}{\pi}\sum_{n  \in \mathbb{Z} | n \not= 0} \frac{(-1)^{n}}{n^{2}} + \frac{\pi}{6}= 0.

    Problem 390***

    Evaluate

    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}.
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    (Original post by Mladenov)

    Problem 390***

    Evaluate

    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}.
    Does that converge?

    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}=\sum_{N \ge 2} c_N \frac{(-1)^{N}}{N^{2}}

    \displaystyle c_N= \sum_{a+b=N, a,b>0} ab =\binom{N+1}{3}

    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}=\sum_{N \ge 2} \frac{(N+1)N(N-1)}{6} \frac{(-1)^{N}}{N^{2}}

    \displaystyle =\sum_{N \ge 2} \frac{N^2-1}{6} \frac{(-1)^{N}}{N}=\frac{1}{6}\sum_{N \ge 2} \left(N- \frac{1}{N} \right) (-1)^N

    ,which I think is divergent.
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    (Original post by henpen)
    Spoiler:
    Show
    Does that converge?

    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}=\sum_{N \ge 2} c_N \frac{(-1)^{N}}{N^{2}}

    \displaystyle c_N= \sum_{a+b=N, a,b>0} ab =\binom{N+1}{3}

    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}=\sum_{N \ge 2} \frac{(N+1)N(N-1)}{6} \frac{(-1)^{N}}{N^{2}}

    \displaystyle =\sum_{N \ge 2} \frac{N^2-1}{6} \frac{(-1)^{N}}{N}=\frac{1}{6}\sum_{N \ge 2} \left(N- \frac{1}{N} \right) (-1)^N

    ,which I think is divergent.
    It converges. How do you justify the first line?
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    (Original post by Mladenov)
    It converges. How do you justify the first line?
    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}

    \displaystyle \large = \begin{matrix}\frac{1 \cdot 1(-1)^{1+1}}{(1+1)^{2}} &+\frac{2 \cdot 1(-1)^{1+2}}{(2+1)^{2}}  &+\frac{3 \cdot 1(-1)^{1+3}}{(3+1)^{2}}  &+\frac{4 \cdot 1(-1)^{1+4}}{(4+1)^{2}}+\cdots \\ 

\frac{2 \cdot 1(-1)^{1+2}}{(1+2)^{2}} & +\frac{2 \cdot 2(-1)^{2+2}}{(2+2)^{2}} & +\cdots  & \\ 

\frac{3 \cdot 1(-1)^{1+3}}{(1+3)^{2}} & +\cdots &  & \\ 

 &  &  & 

\end{matrix}
    , then, summing diagonally,
     \displaystyle= \sum_{N \ge 2} \left(\sum_{a+b=N ,a,b>0} ab \right) \frac{(-1)^N}{N^2}.
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    I hope none of you have seen this question before

    Problem 390**

    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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    (Original post by henpen)
    Spoiler:
    Show
    \displaystyle \sum_{m \ge 1} \sum_{n \ge 1} \frac{nm(-1)^{m+n}}{(n+m)^{2}}

    \displaystyle \large = \begin{matrix}\frac{1 \cdot 1(-1)^{1+1}}{(1+1)^{2}} &+\frac{2 \cdot 1(-1)^{1+2}}{(2+1)^{2}}  &+\frac{3 \cdot 1(-1)^{1+3}}{(3+1)^{2}}  &+\frac{4 \cdot 1(-1)^{1+4}}{(4+1)^{2}}+\cdots \\ 

\frac{2 \cdot 1(-1)^{1+2}}{(1+2)^{2}} & +\frac{2 \cdot 2(-1)^{2+2}}{(2+2)^{2}} & +\cdots  & \\ 

\frac{3 \cdot 1(-1)^{1+3}}{(1+3)^{2}} & +\cdots &  & \\ 

 &  &  & 

\end{matrix}
    , then, summing diagonally,
     \displaystyle= \sum_{N \ge 2} \left(\sum_{a+b=N ,a,b>0} ab \right) \frac{(-1)^N}{N^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.
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    (Original post by Flauta)
    Problem 386**/***

    

Find \displaystyle\int^{2\pi}_0 \frac{1}{5-3\sin x} dx
    (Original post by henpen)
    Problem 387**

    Find (as a power series) the function that satisfies

    f''(x)+f'(x)=f(x)+1.
    Spoiler:
    Show
    There seem to be two problems of the same number. Unless I'm missing a very clever link between the two
    (Original post by Mladenov)
    Can you guys amend your problems/solutions' numeration (+1), starting from Henpen's differential equation.
    Edited

    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}
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    (Original post by henpen)
    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.
    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
    $ k =\frac{a^n+b^n}{(ab)^{n-1}+1}\in\mathbb{N})\Rightarrow \exists c\in\mathbb{N}\; (k = c^n) $
    Which wouldn't be possible in root flipping as the power is not 2.
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    (Original post by Khallil)
    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 &lt; x &lt; 2\pi} + \underbrace{\displaystyle \int_{0}^{\infty} \dfrac{dt}{\left(t-\frac{3}{5}\right)^2 + \left(\frac{4}{5} \right)^2}}_{0 &lt; x &lt; \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
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    (Original post by Flauta)
    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
    All of the thanks goes to you for posting such great questions! (May I ask where you get them from?)

    Also, have a look at LoTF's solution to problem 10. That substitution is another one of my favourites!

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    (Original post by Llewellyn)
    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
    $ k =\frac{a^n+b^n}{(ab)^{n-1}+1}\in\mathbb{N})\Rightarrow \exists c\in\mathbb{N}\; (k = c^n) $
    Which wouldn't be possible in root flipping as the power is not 2.
    It's regarded by some to be the hardest IMO question ever. Don't know whether that's substantiated by the marks competitors achieved though. It's a niche trick so not too unreasonable that you won't think of it in a pressured situation.
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    (Original post by Mladenov)
    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.
    Thanks. It's a shame I didn't notice that, it's one of the few results concerning convergence I know of.
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    (Original post by Arieisit)
    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.



    Posted from TSR Mobile

    Solution 391

    Colour the plane black, then choose a point P on none of the lines such that with each new line, switch the colour of all the regions that are on the same side of the new line as P. If n lines coincide, only perform this operation once.

    I assume by adjacent you mean share an edge (rather than a vertex). If so, by passing across an edge you are passing over a line, but the colours on either side of the line must be different due to the operation we used to create it, therefore the adjacent regions have different colours.

    This easily generalises to any number of dimensions.
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    Problem 392**

    For n \ge 0 define a sequence {u_n} by u_0=u_1=u_2=1 and

    \displaystyle \det\begin{pmatrix}u_{n} &u_{n+1} \\ 

 u_{n+2}& u_{n+3}

\end{pmatrix}=n!.

    Prove that for all n, u_n \in \mathbb{Z}.
 
 
 
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