# F. S. Expansion

1. Sep 16, 2008

### Somefantastik

I'm supposed to derive this monster!

$$\frac{1}{2} + \frac{2}{\pi} \sum^{\infty}_{k = 1}\frac{1}{2k-1}sin(2k-1)x = \left\{^{0 \ for \ -\pi < x < 0}_{1 \ for \ 0<x<\pi}$$

I don't even know where to start right now. And no examples to work from. Can anyone get me started?

the Chapter is on Fourier Expansions for solutions to Laplace's Equation.

Any direction at all would be really appreciated.

2. Sep 16, 2008

### statdad

What have you tried? use the formulas for the coefficients in a fourier expansion and be careful with your integrations.

3. Sep 16, 2008

### Somefantastik

I don't understand. What am I to integrate? If I integrate the coefficients, what is my f(x)? What does that get me when I integrate the coefficients other than just a compacted form?

4. Sep 16, 2008

### statdad

If you have a periodic function $$f$$ on $$[-\pi, \pi]$$, then setting

\begin{align*} a_n = \frac 1 \pi \int_{-\pi}^{\pi} \cos{(nx)} f(x) \, dx \\ b_n = \frac 1 \pi \int_{-\pi}^{\pi} \sin{(nx)} f(x) \, dx \end{align*}

are the Fourier coefficients of the function $$f$$. With them you have
the formal representation

$$f(x) \sim \frac{a_0} 2 + \sum_{n=1}^\infty {\left(a_n \cos(nx) + b_n \sin(nx)}$$

The conditions that show when the series actually converges to $$f$$ are varied, and
should be given in your text.

Looking at your first post, it seems that in your case the function

$$f(x) = \begin{cases} & 0 \text{ if } -\pi < x < 0\\ & 1 \text{ if } 0 < x < \pi \end{cases}$$

5. Sep 18, 2008

### Somefantastik

Yes, that worked. It all fell together pretty nicely. Seems like a pretty trivial problem. Thanks for your help :)

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