1. ## fourier series

Hey everyone, I need a little bit of help with this question so any help would be much appreciated.

Find the Fourier series of the function and discuss the divergence of it.

$\displaystyle f(x)=\begin{cases} 1 \mbox{ for } |x| \le \frac{\pi}{2} \\ 0\mbox{ for } \frac{\pi}{2} < |x| < \pi \end{cases}$

Thanks for any help everyone.

2. ## Re: fourier series

Actually obtaining the series shouldn't be that bad. For $\displaystyle n\geq 1$

$\displaystyle a_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x)cos(nx)dx$

$\displaystyle a_n = \frac{1}{\pi}\Big (\int_{-\pi}^{-\frac{\pi}{2}} 0\cdot cos(nx)dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1\cdot cos(nx)dx + \int_{\frac{\pi}{2}}^\pi 0\cdot cos(nx)dx \Big )$

$\displaystyle a_n = \frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} cos(nx)dx$

And you'll get something very similar for $\displaystyle b_n$

Can you take it from here?

3. ## Re: fourier series

oh yes, stupid me.

I get $\displaystyle b_n = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} sin(nx)dx$

But what about the divergence of the Fourier series?

4. ## Re: fourier series

hi,
first the function must be periodic to be developed in Fourier series.
$\displaystyle f\left ( x \right )= \frac{a_{0}}{2}+\sum_{n=1}^{\infty }\left ( a_{n} cos\left ( nx \right )+b_{n}sin\left ( nx \right )\right )$
I consider that the function has a period of 2pi look the attachement. And it's clear that the function is even thus the (bn) coefficients are null.
$\displaystyle a_{0}=\frac{1}{\pi }\int_{-\pi }^{\pi }f\left ( x \right )dx$
$\displaystyle a_{0}=1$
$\displaystyle a_{n}=\frac{1}{\pi }\int_{-\pi }^{\pi }f\left ( x \right )cos\left ( nx \right )dx$
$\displaystyle a_{n}= sinc\left ( \frac{n}{2} \right )$

and about the convergence: The serie converge to f(x) where f(x) is continous
and diverge where f(x) is discontinous
this is from the lejeune-dirichlet theorem:

5. ## Re: fourier series

Thanks salim, get it now.