Fourier Series of x^2-1 on Interval [-pi,pi]: Coefficient of cos3x Calculation

[aeroguy2008]

Determine the fourierseries of the function

f(x)=x^2−1

defined on the interval [-pi,pi]. What will the coefficient a/b be in front of the cos3x in the fourierseries?

I am not sure how to start this solving this problem. Pls help.

 

[nicksauce]

Well you use your formula for a_n and b_n. Did you try that?

 

[aeroguy2008]

I read that determining if the function is odd or even might be a good start.

 

[nicksauce]

True. So is this function even or odd or neither?

 

[aeroguy2008]

Its even if f(-x)=f(x) and odd if f(-x)=-x so in my case:

f(-x)=(-x)^2−1=x^2-1
so it must be even.

 

[aeroguy2008]

Hence I must use the cosine series on the interval -pi<x<pi. as

f(x) ~ a_0/2+ sum (n=1 to inf) a_n*cos((n*pi)/Pi)*x where:
a_0=2/pi*int(0 to pi) f(x)dx and
a_n=2/pi*int(0 to pi) f(x)cos(n*pi)/pi*x dx

But do I just start integrating from 0 to Pi to get a_0? Why not -Pi to Pi as indicated by the range of the problem? Is it because the integral will go to zero then?

 

[matt grime]

It's an even integrand. The integral from -K to K is twice that of the integral from 0 to K.

There are 2 things here.

1) Can you use the formulae you've been given?
2) Do you understand the formulae you've been given?

If you focus on 2), then 1) becomes easier.

 

[nicksauce]

You could integrate from 0 to Pi, or from -Pi to Pi and take out the factor of 2. Since f(x) is even, it'll be the same either way.

 

[aeroguy2008]

It's an even integrand. The integral from -K to K is twice that of the integral from 0 to K.

There are 2 things here.

1) Can you use the formulae you've been given?
2) Do you understand the formulae you've been given?

If you focus on 2), then 1) becomes easier.

Thanks. I did find this out. Since I integrated and what you guys are saying is surely correct.
But once I try to get a_n I end up with nasty partial integrations. I have to write back once I got my final results.

 

[HallsofIvy]

I have no idea what you mean by "partial integration". Basically you are integrating $(x^2- 1)cos(nx)$. That should be easy using integration by parts. Was that what you meant?

 

[dirk_mec1]

I have no idea what you mean by "partial integration". Basically you are integrating $(x^2- 1)cos(nx)$. That should be easy using integration by parts. Was that what you meant?

partial integration = IBP = integration by parts

 

[HallsofIvy]

So the whole problem is to integrate
$$\int_{-\pi}^{\pi} (x^2-1)cos(3x)dx$$

That should be easy using integration by parts (twice).

 

[aeroguy2008]

Yes dirk_mec1 and HallsofIvy, I was wrong...I was thinking of integration by parts. I have used IBP to get the solution of the integrand you specified and my solution matched with the one I got from a computer software, as attached. I can see that the cos(3x) term appears in my solution but the coefficient of (2/9)Pi in front of the cos(3x) term is incorrect. This gives me reason to question what I am doing wrong. I do understand the procedure of this problem, but I am a little confused. Can you pls point me in the right direction?

 

[aeroguy2008]

You could integrate from 0 to Pi, or from -Pi to Pi and take out the factor of 2. Since f(x) is even, it'll be the same either way.

Thanks that was helpful.

 

[nicksauce]

Yes dirk_mec1 and HallsofIvy, I was wrong...I was thinking of integration by parts. I have used IBP to get the solution of the integrand you specified and my solution matched with the one I got from a computer software, as attached. I can see that the cos(3x) term appears in my solution but the coefficient of (2/9)Pi in front of the cos(3x) term is incorrect. This gives me reason to question what I am doing wrong. I do understand the procedure of this problem, but I am a little confused. Can you pls point me in the right direction?

Well for now I'll assume your integration is correct, without checking it myself. The sin terms disappear because sin(0) = sin(n*pi) = 0. The x=0 boundary dissapears for the first term, because x=0, so for the whole integration, you have 2/9 * pi * cos(3pi) = -2pi/9. To get the a_3 term, you must multiply by 2/pi, giving you -4/9, not (2/9)Pi as you said.

 

[aeroguy2008]

Nick...thanks so much. What you said made perfect sense. I was disregarding the terms for n=3 and that's what I did wrong. So, now my answer is correct; a=-4 and b=9. Thus, This problem is solved:)