Finding summation when given a fourier expansion of a function

[Jncik]

Homework Statement

I have a function

f(x) = x^2/4 for |x|<π

I have the Fourier series of this function which is

and I need to prove that

The Attempt at a Solution

I tried to use dirichlet for x = 0 but I get -pi^2/3

 

[Disconnected]

Have you tried Parsaval's theorem?

 

[Jncik]

if I use parseval won't I get 1/n^4 instead of 1/n^2?

 

[Dick]

If you put in x=0 you get an alternating series. And I don't know how you got -pi^2/3. I get -pi^2/12 for the alternating series. Try putting x=pi if you don't want the series to alternate. Which you don't.

 

[Jncik]

actually for x = 0 I said that since we want 1/n^2

instead of (-1)^n

I will have to use (-1)^2n

hence I will get a 1/4 in the right side of the equation and a -pi^2/12 in the left side

hence -4pi^2/12 will be the answer..

how did you find yours?

as for x = pi

indeed, it's not continuous in this point hence

I will have

(pi^2/4 + 0)/2 = pi^2/8

pi^2/8 = pi^2/12 + 1/4 * x => x = pi^2/6

indeed it's correct...

but can you please explain me how you found your answer?

thanks in advance

 

[Dick]

actually for x = 0 I said that since we want 1/n^2

instead of (-1)^n

I will have to use (-1)^2n

hence I will get a 1/4 in the right side of the equation and a -pi^2/12 in the left side

hence -4pi^2/12 will be the answer..

how did you find yours?

as for x = pi

indeed, it's not continuous in this point hence

I will have

(pi^2/4 + 0)/2 = pi^2/8

pi^2/8 = pi^2/12 + 1/4 * x => x = pi^2/6

indeed it's correct...

but can you please explain me how you found your answer?

thanks in advance

Ok, now I see what you are thinking. And it's wrong. You can't just change to n to 2n. If you do that then you are only summing over the even terms in the Fourier series. So it's not equal to x^2/4 anymore. The way to do this is to notice cos(pi*n)=(-1)^n. Put x=pi to cancel the other (-1)^n. And why do you think the function is discontinuous anywhere? You are doing the Fourier expansion on [-pi,pi], aren't you??

 

[Jncik]

i think that we assume that it's periodic... but I'm not sure...

yes I was wrong

but if I put x = pi and cancel them out won't the result be -pi^2/12?

isn't this wrong?

 

[Dick]

i think that we assume that it's periodic... but I'm not sure...

yes I was wrong

but if I put x = pi and cancel them out won't the result be -pi^2/12?

isn't this wrong?

Certainly you assume it's periodic. That's what Fourier series are about. And yes, it's wrong. Could you explain why you think the answer is -pi^2/12?? Like, show your steps to get to that answer?

 

[Jncik]

Oh I just made a silly mistake

I have

pi^2/4 = pi^2/12 + sum

sum = pi^2/4 - pi^2/12 = pi^2/6

instead of pi^2/4 I put 0, I got confused from what I said in my previous reply..

I understand it now, thanks a lot for your help