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I'm having a few questions regarding the following problem:

Calculate the Fourier series of $f(t)=|t|$ in $[-\pi, \pi)$ and then prove with $$\sum_{k=-n}^n |ck^2| = \frac{1}{2\pi}\int_0^\pi{|f(x)|^2}\,\mathrm dx$$ that $$\sum_{k=0}^{+\infty} \frac{1}{(2k+1)^4}=\frac{\pi^4}{96}.$$

I have calculated the Fourier series: $$f(t)=\pi+\sum_{k=0}^n -\frac{4}{\pi k}\cos(kx)$$ but I don't see any connection to the second part of the problem.

What have I missed or done wrong?

rubik
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m79
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    Your Fourier series is wrong. One way to tell is that the convergence is too slow: $O(1/k)$ decay of the coefficients corresponds to something that's discontinuous, while the periodic extension of $|t|$ is continuous (and is almost $C^1$). It ought to have decay on order $1/k^2$ as a result. –  Jul 28 '16 at 08:30
  • You are right, it should be k^2 and it should $ ak=-\frac{4((-1)^n-1)}{pik^2}$ – m79 Jul 28 '16 at 10:11

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$$\begin{align*}&c_n=\frac1\pi\int_{-\pi}^\pi |t|e^{-int}\,dt=\frac1\pi\int_0^\pi t\left(e^{int}+e^{-int}\right)dt=\frac2\pi\int_0^\pi t\cos nt\,dt\\{}\\ &=\overbrace{\left.\frac{2t}{\pi n}\sin nt\right|_0^\pi}^{=0}-\frac2{\pi n}\int_0^\pi\sin nt\,dt=\left.\frac2{\pi n^2}\cos nt\right|_0^\pi=\\{}\\&= \frac2{\pi n^2}\left[(-1)^n-1\right]=\begin{cases}\;\;\;\;\;0,&n\;\text{even}\\{}\\-\cfrac4{\pi n^2},&n\;\text{odd}\end{cases}\end{align*}$$

and finally (observe the limits of the integral! It is not zero in the lower limit as you can see here)

$$2\sum_{n=0}^\infty\frac{16}{\pi^2(2n+1)^4}=\color{red}{\sum_{-\infty}^\infty|a_n|^2=\frac1{2\pi}\int_{-\pi}^\pi t^2\,dt}=\frac1{6\pi}2\pi^3=\frac{\pi^2}3\implies$$

$$\sum_{n=0}^\infty\frac1{(2n+1)^4}=\frac{\pi^4}{96}$$

DonAntonio
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  • Why are you using the real coefficient an instead of ck=an/2? – m79 Jul 28 '16 at 10:19
  • Thank you for this calculation, you are correct and I see know that I have failed to take into account the behaviour of cos(n*pi). In your last line of calculation, you wrote (2n+1)^2, actually it should be (2n+1)^4. – m79 Jul 28 '16 at 10:27
  • @m79 What "real coefficient"? I used complex Fourier series and then I used $$\cos nt=\frac12\left(e^{int}+e^{-int}\right);...$$ – DonAntonio Jul 28 '16 at 11:05