Try the function f(x) = x^2 on the interval (–π,π). You should find that , and when n≠0. Then Parseval's theorem gives .
How can I use Parseval's theorem to find the summation from n =1 to infinity of (1/n^4)?
I'm pretty sure I can calculate the Fourier cosine or sine series for a function; however, I'm not sure what this function should be. I tried f(x) = x, but I get something with (-1)^n - 1 in the numerator with n^2 in the denominator. Because it has this oscillating part, I can't solve the summation.
Tough to explain I guess, but does anyone have an idea what function to use and how you do it? I've been trying to get 1/n^2 in the denominator of the an terms (and then you square it), but I can't get it with just constants. I know the answer is pi/90. I would appreciate any help.
If you can you should let someone know...because then you would have found (Apery's constant) whose irrationality was only found a couple of years ago...the actual value is unknown.
Here is an article if you are interested
http://mathworld.wolfram.com/AperysConstant.html
... and yet it clearly states here that "The Riemann zeta function may be computed analytically for even n using either contour integration or Parseval's theorem with the appropriate Fourier series" (my underlining). Unfortunately, it doesn't tell you which Fourier series to use.
My guess is that there must be a function whose Fourier coefficients are . That would get round the problem of having to avoid anything that looks like the Apéry constant, yet still allows you to get using Parseval's theorem. Have you tried the function ? (That's just a guess — I have no idea whether it will work.)
Actually (see the above link).