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I've just written the final exam in my introductory course to functional analysis (2nd year bachelor degree). I felt quite well prepared but nevertheless found the exam pretty challenging in the timeframe of two hours. I'd appreciate any comments about what you think about it! The exam can be found here: http://www.math.lmu.de/~michel/SS12_FA_Final_Test_01.pdf

In particular I'd appreciate any hints how to solve 6 ii), which states:

Use the fourier-series of $f(x)=\begin{cases}x^2 &\mbox{ for } x\in[0,\pi] \\ (2\pi-x)^2 &\mbox{ for } x\in (\pi,2\pi ]\end{cases} $ to calculate $\sum_{n\in\mathbb{N}}\frac{(-1)^{n-1}}{n^2}$

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    @Bolek Most likely you used Parseval to get i). The idea for ii) is to use the convergence of the Fourier series itself. It's actually a fairly routine exercise, certainly compared to other questions on that exam :).2012-07-16

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I would find the exam 'challenging' to do in 2 hours. The idea part is fine for me, the computation too demanding for me in that time.

The Fourier series for $f$ are $\hat{f}_0 = \frac{\pi^2}{3}$, $\hat{f}_n = 2 \frac{(-1)^n}{n^2}$, for $n \neq 0$. Then since $f$ is Lipschitz at $0$, we have $f(0) = \sum_n \hat{f}_n$. Since $f(0) = 0$, we get $\hat{f}_0 = - \sum_{n \neq 0} \hat{f}_n$. Since $\hat{f}_n = \hat{f}_{-n}$, we have $\hat{f}_0 = - 2\sum_{n > 0} \hat{f}_n$, from which it follows that $ \sum_{n > 0} \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}.$

(I used wxmaxima to do the fairly elementary integration, I tried by hand but made too many mistakes!)