factor said:
Homework Statement
I'm trying to find the sum of the infinite series:
1/n^4
using the Fourier series of x^2 on [-pi, pi] which I have as
PI^2/3 + 4*sum(n=1..infinity) (-1)^n/n^2*cos(nx)
Homework Equations
The Attempt at a Solution
So far all my attempts have been focused on finding a the series at various points in [-pi, pi] and seeing if I can add all the series together to cancel out the terms I don't need or leave the series I want with another series that I know the sum for. However, so far all these attempts have been unsuccessful and I think that since most of these equations seem fairly unwieldy that I'm just missing a small detail that would make this problem easily solvable. Anyone have any suggestion as to where to start on this problem so I can see if I can take it from there?
Let's take a step back and look at the theory of Fourier series again. When you expand a function like x^2 in a Fourier series, this is analogous as expanding a vector in terms of basis vectors in linear algebra. It is more convenient to work with exp(i n x) as the basis functions, but let's instead work with the set of functions:
e0(x) = 1/sqrt(2)
en(x) = cos(n x) for n >=1
Define an inner product as follws:
(f,g) = 1/pi Integral from -pi to pi of f(x)g(x)dx
We don't need to take the complex conjugate of h if we're dealing with real functions. It is easy to check that the en are orthonormal.
You can now expand any function f(x) as:
f(x) = Sum over n of (f,en) en(x)
If I take the series you obtained:
PI^2/3 + 4*sum(n=1..infinity) (-1)^n/n^2*cos(nx)
I can make the above form explicit:
x^2 = sqrt(2) Pi^2/3 eo(x) + sum(n=1..infinity) 4 (-1)^n/n^2 en(x)
You can now ask: What is the norm of the function x^2 considered as a vector of the Hilbert space? Of course, you can simply compute the inner product of x^2 with itself, to obtain:
(x^2,x^2) = 1/pi Integral from -pi to pi of x^4 dx =
2/5 pi^4
But, of course, if we have that f = sum over n of cn en(x)
then the squared norm will also be the sum of the squares of the cn. This is analogous to Pythagoras' theorem, you can see this by writing
(c0 e0 + c1 e1 + c2 e2..., c0 e0 + c1 e1 + c2 e2+ ..)
using linearity of the inner product and using the fact that (en,em) = 0 if n and m are different while it is equal to 1 if they are equal, you obtain the result.
So, the squared norm of x^2 is also given as:
2 Pi^4/9 + 16*desired summation.
Equating this to 2/5 pi^4 gives:
Desired summation = pi^4/90