FS4

Find the Fourier series for $f(x) = \begin{cases}1 & -1 \le x < 0\\ \frac{1}{2} & x = 0\\x&0<x\le 1\end{cases}$ on $[-1,1]\,$

The general Fourier series on $[-L,L]\,$ is:

$f(x) = \frac{a_0}{2} + \sum_{k=1}^\infty a_k \, \cos(\frac{k\pi x}{L}) + b_k \, \sin(\frac{k\pi x}{L})\,$

$a_n = \frac{1}{L} \int_{-L}^L f(x) \cos(\frac{n\pi x}{L}) \, dx, \,\,\, n=0,1,2,...\,$

$b_n = \frac{1}{L} \int_{-L}^L f(x) \sin(\frac{n\pi x}{L}) \, dx, \,\,\, n=1,2,3,...\,$

In the present problem,

$a_n = \int_{-1}^1 f(x) \cos(n\pi x) \, dx = \int_{-1}^0 f(x) \cos(n\pi x)\,dx + \int_0^1 f(x) \cos(n\pi x)\,dx\,$

$= \int_{-1}^0 \cos(n\pi x)\,dx + \int_0^1 x \cos(n\pi x) \, dx\,$

$= \left[\frac{\sin(n\pi x)}{n\pi}\right]_{-1}^0 + \left[\frac{x \sin(n\pi x)}{n\pi} + \frac{\cos(n\pi x)}{n^2 \pi^2}\right]_0^1\,$

$= \frac{\cos(n\pi)-1}{n^2 \pi^2} = \frac{(-1)^n-1}{n^2\pi^2}\,\,\,n=1,2,3,...\,$

$= \int_{-1}^0 \,dx + \int_0^1 x\,dx = \frac{3}{2}\,$

$b_n = \int_{-1}^1 f(x) \sin(n\pi x)\,dx = \int_{-1}^0 f(x) \sin(n\pi x)\,dx + \int_0^1 f(x) \sin(n\pi x)\,dx\,$

$= \int_{-1}^0 \sin(n\pi x)\,dx + \int_0^1 x\,\sin(n\pi x)\,dx\,$

$= \left[-\,\frac{\cos(n\pi x)}{n\pi}\right]_{-1}^0 + \left[ -\, \frac{x\,\cos(n\pi x)}{n\pi} + \frac{\sin(n\pi x)}{n^2\pi^2}\right]_0^1\,$

$= -\,\frac{1}{n\pi}\,$

So the Fourier series is:

$f(x) = \frac{3}{4} + \sum_{n=1}^\infty \frac{(-1)^n-1}{n^2\pi^2} \cos(n\pi x) - \frac{1}{n\pi} \sin(n\pi x)\,$

Setting $x = 0$ gives $\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ...\,$