Computing Fourier Series for a Piecewise Function

[Knissp]

Homework Statement

Find the Fourier series for y(x)=\begin{cases}<br /> A\sin(\frac{2\pi x}{L}) &amp; 0\leq x\leq\frac{L}{2}\\<br /> 0 &amp; \frac{L}{2}\leq x\leq L\end{cases}

Homework Equations

B_{n}=\frac{2}{L}\int_{0}^{L}y(x)\sin(\frac{n\pi x}{L})dx

The Attempt at a Solution

B_{n}=\frac{2}{L}\int_{0}^{L/2}A\sin(\frac{2\pi x}{L})\sin(\frac{n\pi x}{L})dx

=\frac{2}{L}\int_{0}^{\frac{L}{2}}A\sin(\frac{\pi x}{L/2})\sin(\frac{(n/2)\pi x}{L/2})dx

=\frac{1}{p}\int_{0}^{p}A\sin(\frac{\pi x}{p})\sin(\frac{\frac{n}{2}\pi x}{p})dx

=\begin{cases}<br /> 0 &amp; \frac{n}{2}=1\\<br /> \frac{A}{2} &amp; \frac{n}{2}\in\mathbb{Z}\text{ and }\frac{n}{2}\neq1\end{cases}

So this takes care of the even values of n, but I'm not sure what to do when n is odd.

=\frac{1}{p}\int_{0}^{p}A\sin(\frac{\pi x}{p})\sin(\frac{(\frac{2k+1}{2})\pi x}{p})dx

=\frac{1}{p}\int_{0}^{p}A\sin(\frac{\pi x}{p})\sin(\frac{(k\pi x+\frac{1}{2}\pi x}{p})dx

=\frac{1}{p}\int_{0}^{p}A\sin(\frac{\pi x}{p})[\sin(\frac{k\pi x}{p})\cos(\frac{\frac{1}{2}\pi x}{p})+\cos(\frac{k\pi x}{p})\sin(\frac{\frac{1}{2}\pi x}{p})]dx

I'm not really sure if this is going anywhere. The final answer should be B_{n}=\frac{4A(-1)^{\frac{n+1}{2}}}{\pi(n^{2}-4)}, where n is odd.

Any ideas? Thank you.

 

[jbunniii]

B_{n}=\frac{2}{L}\int_{0}^{L/2}A\sin(\frac{2\pi x}{L})\sin(\frac{n\pi x}{L})dx

Try using a trig identity at this point.

\sin(a) \sin (b) = ?

 

[Knissp]

Thanks, that worked perfectly!