Solving Fourier Coefficients for A_m & B_m

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SUMMARY

The discussion focuses on solving for the Fourier coefficients \(A_m\) and \(B_m\) in the equation \(\sum _{m=0}^\infty a^m [A_m \cos (m \theta ) + B_m \sin (m \theta )]=T_0\sin ^3 \theta\). The derived results indicate that \(B_1=\frac{3T_0}{4a}\) and \(B_3=-\frac{T_0}{4a^3}\), with all other coefficients equating to zero. Participants suggest using trigonometric identities to simplify \(\sin^3 \theta\) into a form suitable for Fourier analysis, emphasizing the importance of integrating with respect to \(\cos(m\theta)\) and \(\sin(m\theta)\) to isolate the coefficients.

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Homework Statement


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##.

Homework 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.

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##.
 
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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.
 

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