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.
Best Answer
Consider the $n$th 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,
$$\frac{4}{\pi}\sum_{k=1}^{\infty}\frac{sin(2k-1)(\frac{\pi}{2})}{2k-1}=f(\frac{\pi}{2})$$
$$\implies \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{2k-1}=1$$
$$ \implies1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots = \frac{\pi}{4}$$