Yes, but if the Fourier series of f(x) = ao/2 + sum over n from 1 to infinity (an*cos(n*x) + bn*sin(nx)) then if ao, an, and bn go to zero, you are left with simply zero as the Fourier series. Shouldn't this method still work?
Sorry to gravedig, but considering this is the first result for the fourier series of sin[x]. I had to correct it.
A fourier series is a series that has a period equal to part of a qualified f(x) from -L to L or 0 to 2l.
Yes, the fourier series of Sin[x] with L = 2n Pi where n is an element of Z is simply Sin[x].
But the fourier series of Sin[x] with L =/= 2n Pi is NOT Sin[x].
Look at the graph of Sin[x] from 0 to Pi/2. If you wanted a function that repeated that period, Sin[x] would no-where near work.
The first few terms of that fourier series would be: (4 Cos[1/2] Sin[x])/(-1 + \[Pi]^2) + (
8 Cos Sin[2 x])/(-4 + \[Pi]^2) + (
12 Cos[3/2] Sin[3 x])/(-9 + \[Pi]^2)
According to mathematica. (It's slightly incorrect, since the outcome of the periods are off. Working on it)