What is the Fourier series for a periodic function over [-\pi,\pi]?

[TFM]

Homework Statement

Find the Fourier series corresponding to the following functions that are periodic over the
interval [-\pi,\pi]

f(x) = 1, -\pi/2 < x< \pi/2; f(x) otherwise.

Homework Equations

Fourier Series:

f(x) = \frac{1}{2}a_0 + \sum^\infty_{n=1}a_n cos\frac{2*\pi*n*x}{l} + \sum^\infty_{n=1} b_n sin\frac{2*\pi*n*x}{l}

\frac{1}{l}\int^{l/2}_{-l/2}f(x) dx

a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos frac{2*\pi*n*x}{l}dx

a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin frac{2*\pi*n*x}{l}dx

The Attempt at a Solution

So far I have:

a_0 = 1

a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2}

b_n = -\frac{1}{\pi n}[cos(nx)]^{\pi}_{-\pi}

But I am not sure what to do now. I seem to be mainly confused about the n's

TFM

 

[TFM]

My equations didn't come out quite right...Sorry. Should be:


<br /> a_0 = \frac{1}{l}\int^{l/2}_{-l/2}f(x) dx <br />

<br /> a_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) cos \frac{2*\pi*n*x}{l} dx <br />

<br /> b_n = \frac{1}{l}\int^{l/2}_{-l/2}f(x) sin \frac{2*\pi*n*x}{l}dx <br />

TFM

 

[dirk_mec1]

Note that:

\sin( n \cdot \pi /2)= 1 if n = 1,3,5...
\sin( n \cdot \pi /2) = 0 if n = 2,4,6,...and

\cos( n \cdot \pi ) = (-1)^{n}

 

[TFM]

So for a_n

<br /> a_n = \frac{1}{\pi n}[sin(nx)]^{\pi/2}_{-\pi/2} <br />

n = 1,3,5...

so

a_n = \frac{1}{\pi 1}[sin(1x)]^{\pi/2}_{-\pi/2},

\frac{1}{\pi * 3}[sin(3x)]^{\pi/2}_{-\pi/2},

\frac{1}{\pi * 5}[sin(5x)]^{\pi/2}_{-\pi/2} ...


and

<br /> \cos( n \cdot \pi ) = (-1)^{n} <br />

cos of n pi always = -1? (as long as pi is a whole number)

TFM