# Thread: Fourier Sine Series Exapnsion

1. ## Fourier Sine Series Exapnsion

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

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

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

$\displaystyle $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

$\displaystyle $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?

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

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

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

$\displaystyle $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

3. Thanks, for pointing out my mistake. I got the correct answer now.