Possible Error in Calculating Fourier Series for sin2x?

[sarideli18]

Homework Statement

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

Homework Equations

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?

 

[jbunniii]

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

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.

I found Ao=0 even though I came up with a Fourier cosine series. There must be something wrong.

If you show your calculation for $a_0$, you will get feedback regarding what is wrong, if anything.

And is that possible that I come up with a Fourier cosine series which converges to sin2x??

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.

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?

Your instinct is correct for the function sin(2x). However, once again, be careful: your function is NOT simply sin(2x).

 

[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.

 

[jbunniii]

Well, for Ao;

Ao=1/2pi ∫sin2x dx (from -pi to 0)

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

OK, that looks right.

For Bn;

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

When I do the integral, again I get 0.

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.

 

[sarideli18]

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.

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?

 

[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.

 

[sarideli18]

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.

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.

 

[LCKurtz]

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.

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)?

 

[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.