Periodic Function Fourier Series: Proving with Trigonometric Equations

[icystrike]

Homework Statement

A periodic function of period 2\pi is defined by:
f(t)=\frac{t}{2} , 0<t<2\pi

Show that the trigonometric Fourier series of f(t) is given by:
f(t)=\frac{\pi}{2} - \sum_{n=1}^{\infty} \frac{1}{n}sin(nt)

Homework Equations

The Attempt at a Solution

I've gotten \frac{\pi}{4} for constant C instead of \frac{\pi}{2}

 

[hunt_mat]

What have you done so far, show us your working.

 

[icystrike]

Since function is odd, An = 0.

Bn = - \frac{1}{2\pi} \frac{2\pi}{n} = -\frac{1}{n}

Thats for now...

 

[hunt_mat]

My advice is to write:

<br /> f(t)=a_{0}+\sum_{n=-\infty}^{\infty}a_{n}\sin nt+b_{n}\cos nt<br />

 

[icystrike]

<br /> <br /> f(t)=\sum_{n=-\infty}^{\infty} \frac{(1)^{n+1}}{n}\sin nt<br /> <br />

 

[hunt_mat]

So you compute te A_n's and b_n is the usual way, what you have done with this?

 

[icystrike]

 

[hunt_mat]

Hmm, I am at a loss.