Computing Fourier Series for Odd Functions

[kiwifruit]

Homework Statement

f(t)= -1 if -∏ < t ≤ 0
1 if 0 < t ≤ ∏

f(t+2∏) = f(t)

question asks to compute first 3 non-zero terms in Fourier series expansion of f(t)

Homework Equations

The Attempt at a Solution

since this is an odd function i used the Fourier sine series formula

f(t)=
∞
Ʃ (bn) sin(nwt)
n=1

(bn)= (2/L)*
L
∫ f(t)sin(nwt)
0
this is just integral from 0 to L cause i don't know how to use the subscipts on the forum

i got L=∏ since the period,T=2∏
w=1

so my (bn)=(-2/n∏) [cos(n∏)-1]
so as a refult my Fourier expansion becomes (bn) sin(nwt)
and i get (-2/n∏) [cos(n∏)sin(nt) - sin(nt)]

and whatever n value i get cos(n∏)=1 so it will be 0 for every n value. I am pretty sure i did something wrong here since the answer is
4/∏ [sin(t) + 1/3sin(3t) + 1/5sin(5t)]

 

[I like Serena]

Welcome to PF, kiwifruit!

You have (bn)=(-2/n∏) [cos(n∏)-1].

Let's try to fill in a couple of values for n.
What do you get for n=1?
Since I do not get zero.

 

[kiwifruit]

thank you. i got it now. i confused cos∏=1 when it should be -1

 

[I like Serena]

Cheers!