Half Range Fourier Series

**Rachel123** · January 29th 2013, 10:52 AM

If $f(x)$ is defined on $[0,L]$ then it can be extended to be either odd or even giving formulas
$f(x)=\sum_{n=1}^{\infty}b_n\sin\frac{\pi nx}{L}$ and
$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos\frac {\pi nx}{L}$ respectively
If $f(x)$ was not even or odd, is there another formula that can be used to express it?
I'm trying to get something of the form
$f(x)=\sum_{n=0}^{\infty}c_n\sin\frac{(2n+1)\pi x}{2L}$ , where $c_n$ is some constant expressed as an integral of $f$ , but I've no idea how to show this.

Thanks

**vincisonfire** · January 29th 2013, 11:37 AM

The Fourier basis functions $\sin$ and $\cos$ are eigenvectors of the operator $\nabla^2$ . Recall that eigenvectors satisfy $\nabla^2 u = -\lambda^2 u$ . Then $u = a\cos(\lambda x)+b\sin(\lambda x)$ . If you have (zero) Dirichlet boundary conditions, you get eigenvectors $u = \sin(\lambda x)$ where $\lambda = {\pi n\over L}$ . If you have (zero) Neumann boundary condition $u = \cos({\pi n\over L} x)$ are eigenvectors. You can also have periodic boundary conditions, which yields the traditional Fourier series.
Now you ask about $u = \sin\left({\pi (2n+1)\over 2L} x\right)$ eigenvectors. I think you can make it work with $u(0)=0$ and ${du\over dx}(L)=0$ .

**Rachel123** · January 29th 2013, 03:48 PM

I don't understand this, we have not yet covered eigenvectors relating to fourier series and Dirichlet boundary condition yet. Is there a simpler way to show this?

**hollywood** · January 30th 2013, 06:49 AM

I think you can just break it up into odd and even parts - so $f(x) = f_{odd}(x)+f_{even}(x)=$

$\sum_{n=1}^{\infty}b_n\sin\frac{\pi nx}{L} +\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos\frac {\pi nx}{L}=$

$\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos\frac {\pi nx}{L}+b_n\sin\frac{\pi nx}{L}$

And by the way, $f_{odd}(x)=\frac{f(x)-f(-x)}{2}$ and $f_{even}(x)=\frac{f(x)+f(-x)}{2}$ .

- Hollywood

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