# Thread: Separation of variables in Fourier Series

1. ## Separation of variables in Fourier Series

Problem: Find the Fourier Series of $\displaystyle f(x) = 0.05sin \pi x$, with the initial velocity $\displaystyle g(x)=0$, $\displaystyle c= \frac{1}{n}$, $\displaystyle L = 1$.

My solution: I use the separation of variables method, then I have:

$\displaystyle u(x,t)= \sum^{\infty}_{n=1}sin \frac{n \pi x}{L}[b_{n}cos( \frac {cn \pi}{L})t+ \hat{b_{n}} sin( \frac {cn \pi}{L})t]$

where

$\displaystyle b_{n}=\frac{2}{L} \int^{L}_{0}f(x)sin( \frac {cn \pi}{L})dx$
$\displaystyle \hat{b_{n}} = \frac{2}{cnx} \int^{L}_{0}g(x)sin( \frac {cn \pi}{L})dx$

Now I know that [tex]\hat{b_{n}}=0[tex] since $\displaystyle g(x)=0$, but when I solve for $\displaystyle b_{n}$, I have:

$\displaystyle b_{n}=2 \int^{1}_{0}0.05sin( \pi x)sin(n \pi x)dx$

I use the table of integrals, and I get bn = 0, is there something I'm doing wrong here? (Most likely)

K

2. What are you trying to do?

3. I'm trying to find the Fourier Series representation of this f(x), and I'm using the separation method to do it.

4. So you are trying to solve the wave equation.

Use the following facts,
$\displaystyle \int_0^L \sin \frac{\pi n x}{L}\sin \frac{\pi m x}{L} dx= L\delta_{nm}$

$\displaystyle \int_0^L \sin \frac{\pi nx}{L}\cos \frac{\pi mx}{L}dx=0$

$\displaystyle \int_0^L \cos \frac{\pi nx}{L}\cos \frac{\pi mx}{L} dx = L\delta_{nm}$