Fourier series and even/odd functions

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SUMMARY

The discussion centers on the application of Fourier series to solve a partial differential equation (PDE) involving the function u(rho, phi) = [A_n*cos(n*phi)+B_n*sin(n*phi)]*rho^n. The user questions the necessity of calculating the Fourier coefficient A_n for the odd function sin(phi/2), which is typically expected to yield only B_n. However, it is clarified that A_n cannot be discarded because the integration interval [0, 2π] does not satisfy the conditions for the integral of an even function multiplied by an odd function to vanish. Thus, both coefficients must be computed.

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[SOLVED] Fourier series and even/odd functions

Homework Statement


I found the solution to a PDE in this thread: https://www.physicsforums.com/showthread.php?t=224902 (not important)

The solution is the sum of u(rho, phi) = [A_n*cos(n*phi)+B_n*sin(n*phi)]*rho^n.

I have to find the general solution, where rho=c, so I equal rho = c, and I am told that u in this point equals sin(phi/2) when phi is between 0 and 2*pi.

I must find the Fourier-coefficients (since it is a Fourier-series).

My questions are:

Since sin(phi/2) is an ODD function, can I discard A_n and just find B_n? That is what I would do, but in the solutions in the back of my book they find A_n as well. Why is that?! The book even says that for an odd function, the Fourier-series only contains sine, so A_n can be discarded, but they still find it. Can you explain to me why A_n must be found as well?
 
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The integral of an even function times and odd function will generally vanish only if you are integrating over an interval symmetric around the origin, like [-L,L]. Your interval here is [0,2pi]. The A_n's don't automatically vanish.
 
Ahh, I see.

You have helped me very much lately. Thank you.
 

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