How Do You Derive the Fourier Series for Laplace's Equation Solutions?

[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.

 

[statdad]

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

 

[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?

 

[statdad]

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

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

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

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

 

[Somefantastik]

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