# Fourier Sine Series Exapnsion

• July 14th 2010, 04:37 PM
dats13
Fourier Sine Series Exapnsion
The question ask to expand the following function of period $2L$ on the interval $(0,L)$ as a Fourier sine series.

$$f(x)=sin(\frac{\pi}{2L}x)$
$

This function is periodic on a period of $4L$ and thus the coefficients are given by

$$b_{n}=\frac{1}{2L}\int_{0}^{2L}sin(\frac{\pi x}{2L})sin(\frac{n\pi x}{2L})dx$
$

Upon evaluating the integral, I split the sine terms into two seperate cosine terms using a trig identity. After integration I am unable to get a single cosine as given in the correct answer

$$f(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\frac{ncos(n\ pi)}{1-4n^{2}}sin(\frac{n\pi}{L}x)$
$

Have I made an error or am I on the correct track?
• July 14th 2010, 09:47 PM
CaptainBlack
Quote:

Originally Posted by dats13
The question ask to expand the following function of period $2L$ on the interval $(0,L)$ as a Fourier sine series.

$$f(x)=sin(\frac{\pi}{2L}x)$
$

This function is periodic on a period of $4L$ and thus the coefficients are given by

$$b_{n}=\frac{1}{2L}\int_{0}^{2L}sin(\frac{\pi x}{2L})sin(\frac{n\pi x}{2L})dx$
$

Because the integrand is not periodic with period 2L you have to integrate over -L,L here, also you divide by 2L in the second sin term and you should not it should just be L. You could also do this the way shown in the MathWorld page.

CB
• July 17th 2010, 12:45 PM
dats13
Thanks, for pointing out my mistake. I got the correct answer now.