I have to find a Fourier sine series of $f(x)=x^2$ for $0<x\leq2$. And show that converges to $0$ at $x=\pm 2$.
(UPDATE) My attempt was :
extended the function to $0<x\leq-2$ s.t. period is 2.
Then, $b_n=\frac{2}{2}\int_{-2}^{0}x^2sin\frac{\pi nx}{2}dx$
Solving this I got: $$-\frac{8}{\pi n}cos(\pi n)+\frac{16}{(\pi n)^3}cos(\pi n)-\frac{16}{(\pi n )^3}$$
$$=-\frac{8}{\pi n}(-1)^n+\frac{16}{(\pi n)^3}((-1)^n-1)$$
is this correct? how do I show this converges to zero?
Thanks!
Best Answer
If you define $f(x)$ to be odd over $-2<x\le 2$, and periodic, with period 4, then the Fourier sine series will converge to $f(x)$ at all points where $f$ is continous, and will converge to $\lim_{x\downarrow x_0} f(x) + \lim_{x\uparrow x_0}f(x)$ at points $x_0$ where $f(x)$ is discontinuous.
At $x_0=2$, the limit from the left is $2^2=4$ and the limit from the right is $-2^2=-4$ so the series converges to $4-4=0.$ Similarly at $x_0=-2.$