# Fourier series of sin(x)

1. Sep 16, 2007

### 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...

2. Sep 16, 2007

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.

3. Sep 16, 2007

### 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?

4. Sep 16, 2007

### D H

Staff Emeritus
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).

5. Sep 16, 2007

### 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.

6. Sep 16, 2007

### nicksauce

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

7. Sep 16, 2007

### misterme09

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

8. Sep 17, 2007

### nicksauce

Correct

9. Aug 20, 2010

### 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)