# Fourier series of sin2x

1. Oct 14, 2012

### sarideli18

1. The problem statement, all variables and given/known data

Fx= sin2x for -pi<x<0 and 0 for 0<x<pi. Compute the coefficients of the Fourier series.

2. Relevant equations

3. The attempt at a solution

I found Ao=0 even though I came up with a Fourier cosine series. There must be something wrong. And is that possible that I come up with a fourier cosine series which converges to sin2x?? Before I began doing the problem, I thought An will be equal to zero and Bn will be zero for n≠2 which is intuitively correct. So, anyone has any idea?

2. Oct 14, 2012

### jbunniii

First thing to note is that this function is neither even nor odd. Therefore the Fourier series will contain both cosines and sines. You will not end up with a cosine series or a sine series.
If you show your calculation for $a_0$, you will get feedback regarding what is wrong, if anything.
Two things to note: first, your function is not simply sin(2x). It is only defined that way for $-\pi < x < 0$, not for $0 < x < \pi$. Second, you will only end up with a cosine series if your function is even, which is not the case here.
Your instinct is correct for the function sin(2x). However, once again, be careful: your function is NOT simply sin(2x).

3. Oct 14, 2012

### sarideli18

Well, for Ao;

Ao=1/2pi ∫sin2x dx (from -pi to 0)

When I do the integral, I get cos2x and cos0-cos2pi=0.

For Bn;

Bn=1/pi ∫sin2x sinnx dx (from -pi to 0)

When I do the integral, again I get 0. So, the only coefficient i have is An. So, the fourier series of fx becomes fx~ƩAn cosnx which seems incorrect. I checked my algebra maybe 10 times but there is no mistake.

4. Oct 14, 2012

### jbunniii

OK, that looks right.

That is not true for every $n$. For example, if $n = 2$ then the integrand is $\sin^2(2x)$, which certainly does not integrate to zero.

5. Oct 14, 2012

### sarideli18

I did the integral for n=2 and I got Bn=1/2. I should have realized that since I had an n-2 in the denominator. Obviously what I found for n≠2 would not work for n=2. Thank you for that. However, for n=2, An=0 and since A0=0, I just have 1/2*sin2x which does not converge to sin2x. Is that because f(x)=sin2x for -pi<x<0 rather than -pi<x<pi or is the result incorrect?

6. Oct 14, 2012

### LCKurtz

Show us your calculations for $a_n$. You shouldn't be getting them all zero. Putting n=2 aside, the other a values come out every other one 0 and non-zero.

7. Oct 14, 2012

### sarideli18

Well, yes you are right. An=4/π(n+2)(n-2) for n is odd and An=0 for n is even. However, upon plugging A0=0 A2=0, B2=1/2, and n=2 I have;

F(x) ~ A0 + ƩAncosnx + ƩBnsinnx

F(x) = 1/2 sin2x. But F(x) must be sin2x for -pi<x<0. I checked my algebra for An but could not find any mistake at all.

8. Oct 14, 2012

### LCKurtz

How do you know you have a mistake? Have you plotted the graph of your series with, say, 10 non-zero terms to see if it is beginning to look like F(x)?

9. Oct 15, 2012

### Ibix

A slightly belated observation - A0 is the average value of F(x). Since the left-hand half of your function is a complete cycle of a sine wave and the right hand half is flat zero, the average value has to be zero - as your maths is also showing you.