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Disclaimer: I am only to work within the orthonormal system $\{e^{inx}\mid n\in\mathbb{Z}\}$! No cosine series/sine series straight off the bat.

I am working with the function $f(x)=\begin{cases} 0 & -\pi< x<0 \\\sin x & 0\leq x\leq \pi \end{cases}$

I am to show that $\sum_{-\infty}^{\infty}\frac{1}{4k^2-1}=0$ and $\sum_{-\infty}^{\infty}\frac{1}{(4k^2-1)^2}=\frac{\pi^2}{8}$

I have started off by finding the Fourier coefficients, which are:

$$c_1=-\frac{i}{4}, \hspace{0.5cm} c_{-1}=\frac{i}{4}, \hspace{0.5cm} c_0=\frac{1}{\pi},\hspace{0.5cm} c_{2k-1}=0, k\in\mathbb{Z}\setminus\{0,1\} , \hspace{0.5cm}c_{2k}=-\frac{1}{\pi}\frac{1}{4k^2-1}$$

I can see, that the above sums are quite close to summing over $(-\pi )c_{2k}$ and $(-\pi)^2c_{2k}^2$ respectively. I have found the Fourier series of $f$ to be

$$c_{-1}+c_0+c_1+\sum_{n=2}^{\infty}c_n(e^{inx}+e^{-inx})=c_0+2\sum_{n=2}^{\infty}c_n\cos(nx)$$

I have previously shown, that the Fourier series converges to f, so since $f(0)=0$, I evaluate the Fourier series in $0$ and get:

$$0=f(0)=\frac{1}{\pi}-\sum_{n=2}^\infty\frac{1}{\pi}\frac{1}{2k^2-1}$$

I feel like I might have miscalculated something, somewhere along the line or overseen something. However, I note that every uneven term of the series is $0$, which I am unsure of how to use.

As for the second sum, I am thinking that I should apply Parseval's Identity, seeing as it has something that is nearly my Fourier coefficients squared?

Bottom line is, I would like some hints or guidance to show the equalities presented at the top, using the Fourier series of $f$.

Thank you.

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  • $\begingroup$ Are you sure the following is correct? $\sum_{-\infty}^{\infty}\frac{1}{(4k^2-1)^2}=\frac{\pi^2}{8}$ $\endgroup$ – poyea Jun 6 '18 at 17:38
  • $\begingroup$ It is what is being asked of me to show, yes. A quick run through Maple shows that it is indeed correct. $\endgroup$ – user7924249 Jun 6 '18 at 17:41
  • $\begingroup$ I've got an answer that's very close, using the result you have. $\endgroup$ – poyea Jun 7 '18 at 18:57
  • $\begingroup$ A sketch (hope it's correct): For the first part, you write down the Fourier representation, and evaluate it at $x=0$. For the second part, it is indeed Parseval's Identity. The problem may be due to the way you compute the coefficient $c_{2k}$. $\endgroup$ – poyea Jun 7 '18 at 18:59

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