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Question Find the Fourier series for the function defined by

$$f(x)=\begin{cases} -1, & -\pi\leq x\lt 0 \\ 0, & x=0\\ 1, & 0\lt x\leq \pi \end{cases}.$$

Tell whether the series is an expansion of $f(x)$. Hence deduce the value of the series $$1-\frac{1}{3}+ \frac{1}{5}- \frac{1}{7}+\cdots $$.

Effort: Fourier series of $f(x)$ is
$$\frac{a_0}{2}+\sum\limits_{n=1}^{\infty}(a_n\cos nx+ b_n\sin nx)$$ where

$a_0= \frac{1}{\pi}\int\limits_{-\pi}^{\pi} f(x) dx$,

$a_n= \frac{1}{\pi}\int\limits_{-\pi}^{\pi} f(x) \cos nx dx$

$b_n= \frac{1}{\pi}\int\limits_{-\pi}^{\pi} f(x) \sin nx dx$

Here the function is odd so Fourier coefficients, $a_n=0$ and we have after some calculations $b_n= \frac{2}{\pi}(1-(-1)^n$. Please help me to deduce the value of the last series.

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  • $\begingroup$ Is this a question you are asking, or a question someone is asking of you? It looks suspiciously like a homework exercise that you're just typing in without doing any thinking about it for yourself, and expecting random strangers on the internet to do your homework for you so you can avoid actually learning anything ... $\endgroup$ – Henning Makholm Sep 10 '16 at 16:09
  • $\begingroup$ The $b_n$ you found is false. $\endgroup$ – nicomezi Sep 10 '16 at 16:21
  • $\begingroup$ @nicomezi how did you find $b_n$ $\endgroup$ – user356595 Sep 10 '16 at 16:32
  • $\begingroup$ I am pretty sure you forgot to divide by $n$ when you have integrated $\sin(nx)$. $\endgroup$ – nicomezi Sep 10 '16 at 16:39
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Hint: Recall that a Fourier series decomposes $$ f(x) = \sum a_ne^{inx}. $$ So you need to find the $a_n$. What is their definition? Look it up and compute.

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Consider the nth Fourier coefficients of $f$ i.e. $\hat f(n)= \frac{1}{2 \pi} \int_{\pi}^{\pi} f(x) e^{-inx} dx$

Do a trivial computation to get, $\hat f(0) =0$ (since $f$ is odd!) and for $n \ne 0$, $\hat f(n)= \frac{1-(-1)^n}{i \pi n}$ .Thus only odd terms survive!

Then, $\sum_{n=-\infty, }^{+\infty} \hat{f}(n) e^{inx}= \sum_{n=-\infty, \text{n odd}}^{+\infty}\frac{1-(-1)^n}{i \pi n} e^{inx}$ . Now coupling $n$th and $-n$th term together get, $$\sum_{n=-\infty, }^{+\infty} \hat{f}(n) e^{inx}= \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{sin(2k-1)x}{2k-1} \ldots (1)$$

Consider any neighborhood of the form $(\frac{\pi}{2}-\delta, \frac{\pi}{2}+\delta)$ where $0 < \delta < \frac{\pi}{2}$ , then on this interval $f =1$ and hence Lipschitz. Thus at the point $x=\frac{\pi}{2}$, the Fourier series of $f$ converges to $f(\frac{\pi}{2})$ . Thus by (1) we get,

$$\frac{4}{\pi}\sum_{k=1}^{\infty}\frac{sin(2k-1)(\frac{\pi}{2})}{2k-1}=f(\frac{\pi}{2})$$

$$\implies \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}=1$$

$$ \implies1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \frac{\pi}{4}$$

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