# Difficulty of functional analysis exam

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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Which Fourier series did you find for $f$? – Davide Giraudo Jul 16 '12 at 20:13
I dont know anymore. Something like $\frac{\sqrt{\pi}(-1)^n}{n^2}$ multiplied by some constant and something different for $n=0$. The result I got for i) was $\frac{\pi^4}{90}$ – Bolek Jul 16 '12 at 20:15
I think it is only the last part of the question which is really appropriate for this site. To help those who want to help you, could you reproduce question 6ii) in the text of the question itself? – Pete L. Clark Jul 16 '12 at 20:16
sure. Sorry that the rest is not relevant. Just wanted to get a feeling wheater it is only me who finds this exam difficult! – Bolek Jul 16 '12 at 20:18
I believe the difficulty of an exam is not something to be discussed here, after all we can't make a fair judgement about a particular course at a particular university, not knowing anything about the level, grading, etc.. Rather, ask your fellow students about their experience. In any case, could you show your attempts at that particular exercise? – wildildildlife Jul 16 '12 at 21:04

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 $$\sum_{n > 0} \frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12}.$$