# Fourier coefficients

1. Dec 4, 2012

### fluidistic

1. The problem statement, all variables and given/known data
I've reached a relation but then I need to obtain the coefficients $A_m$ and $B_m$'s, those are the only unknowns.
Here's the expression: $\sum _{m=0}^\infty a^m [A_m \cos (m \theta ) + B_m \sin (m \theta )]=T_0\sin ^3 \theta$.

2. Relevant equations
Fourier coefficients.
Result is probably $B_1=\frac{3T_0}{4a}$ and $B_3=-\frac{T_0}{4a^3}$.
All the other coefficients being 0.

3. The attempt at a solution
So I am not really sure. I have something of the form $\sum _{m=0}^\infty a^m [A_m \cos (m \theta ) + B_m \sin (m \theta )]=f(\theta)$.
So in order to get let's say the $A_m$ first, I guess I must multiplicate both sides by $cos (m \theta )$ and integrate from ... 0 to $2 \pi$?
I'd get $\sum _{m=0}^\infty a^m A_m \int _0^{2\pi } \cos ^2 (m \theta ) d \theta + \sum _{m=0}^\infty a^m B_m \int _0^{2\pi } \sin (m \theta ) \cos (m\theta ) d \theta=\int _0 ^{2\pi } f(\theta ) d \theta$. Where an integral vanishes.
I'd be left with $\sum _{m=0}^\infty a^m A_m \int _0^{2\pi } \cos ^2 (m \theta ) d \theta =\int _0 ^{2\pi } f(\theta ) d \theta$.
I don't really know how to evaluate those integrals nor how to isolate $A_m$.

2. Dec 4, 2012

### vela

Staff Emeritus
You can't multiply by $\cos m\theta$ because $m$ is your dummy index. Also, you forgot to multiply the righthand side. You can multiply by $\cos n\theta$ and then do the integrals. Only the term with n=m will survive.

But for this problem, it's probably better to use trig identities to express $\sin^3 \theta$ in terms of sines and cosines without exponents.