Partial sum of Fourier Coefficients

[sristi89]

Homework Statement

f (x) = 0 -pi<x<0

x^2 0<x<pi

Find the Fourier series and use it to show that

(pi^2)/6=1+1/2^2+1/3^2+...

Homework Equations

N/A

The Attempt at a Solution

I was able to find the Fourier series and my answer matched with the back of the book. But I don't understand how I am supposed to use it to prove the expansion of pi^2/6 . I tried plugging in (pi^2)/6 for f(x) but that didn't work out.


Thanks!

 

[nicksauce]

What do you find to be the Fourier series for f(x)?

 

[Dick]

Just to echo nicksauce, if you found the Fourier series as you said, what is it?

 

[sristi89]

I didn't want to type out the Fourier series since I am really new to latex but here it is

f(x) =pi^2/6 +\Sigma { (2(-1)^n/n^2) cos nx + ((-1)^(n+1) pi/n + 2/(pi * n^3) [(-1)^n-1] sin nx }

I hope that made sense and that I didn't make any typos.

 

[sristi89]

P.S: and n goes from 1 to infinity

 

[Dick]

Thanks. Since the mean value of the function over the interval is pi^2/6, I was hoping I could just say put x=0. But you've got that alternating sign and factor of 2 in the cosine part. Are you sure of the series? Otherwise, I'll have to work it out. Didn't want to do that.

 

[sristi89]

I am pretty sure about the series. I verified with the back of the book. We were also supposed to show pi^2/12=1-1/2^2+ etc. I plugged in x=o and was able to prove it.

For the pi^2/6, I tried setting x to be pi/2 so that the cosine term goes away but that didn't work out.

 

[Dick]

It looks to me like you should be putting x=pi. That turns cos(nx) into (-1)^n makes the sin(nx) vanish. I'm trying to check your series. I get something a lot like it. But I can't seem to get all the parts quite right. Guess I'm not very good at this...

 

[Dick]

Doh. I've been being stupid. Regarded as a periodic function f(x) is discontinuous at x=pi. The series doesn't converge to f(pi). It converges to (f(pi)+f(-pi))/2. Or pi^2/2. Do you know why? Now try x=pi in your series.