How can you find the Fourier series of Sin[x] with a period other than 2n Pi?

[misterme09]

I know it's trivial...but how do you find the Fourier series of sin(x) itself? I seem to get everything going to zero...

 

[DeadWolfe]

You should. The Fourier series of sin(x) is simply sin(x). This is like asking what the Taylor series of y=x^2 is.

 

[misterme09]

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?

 

[D H]

What ever makes you think all of the terms go to zero? One term does not, the b₁ term. You end up with sin(x).

 

[misterme09]

bn = (1/pi) int ( sin(x) * sin(n*x) dx) from -pi to pi becomes a multiple of sin (n*pi) which is always zero.

 

[nicksauce]

if n = 1, then it's the integral of sin^2x, which is certainly not zero over -pi to pi.

 

[misterme09]

That's true. Okay, but for bn where n isn't 1, bn is zero, correct?

 

[nicksauce]

That's true. Okay, but for bn where n isn't 1, bn is zero, correct?

Correct

 

[QuantumPhenom]

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[1] 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)