Find Fourier sine series

Question

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!

How did you extend the range? If you just let $f(x)=x^2$ over $(-2,2]$, that is even, and then the $b_n$ will be zero. What you want to do (or what you did?) is to extend the function so that it is odd. $f(x)=x^2$ when $0<x\le2$ and $f(x)=-x^2$ when $-2<x\le0.$ — mjw, Nov 10 '20 at 16:34
I see what you mean, I thought I needed a symmetric range .. If I extend t0 $-2<x\leq0$ would my period still be 4? — Aline Bellangero, Nov 10 '20 at 16:42
The range is fine. The function should be "extended" so that it is odd. Also, not sure what you mean by extend to $0<x\le2.$ That's the original range of the function, right? — mjw, Nov 10 '20 at 16:43
It was a typo, I've updated my attempt. I see if n is odd I get $\frac{8}{\pi n} - \frac{32}{(\pi n)^3}$ and if n is even I get $\frac{-8}{\pi n}$. How can I show this fourier series converges to $0$? @mjw — Aline Bellangero, Nov 10 '20 at 19:22

mjw · Accepted Answer · 2020-11-11T17:06:00.297

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 $x_0=-2.$

Find Fourier sine series

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