# parseval's theorem

• Jan 19th 2009, 05:15 PM
PvtBillPilgrim
parseval's theorem
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.
• Jan 20th 2009, 12:10 AM
Opalg
Try the function f(x) = x^2 on the interval (–π,π). You should find that $\displaystyle a_0 = 2\pi^2/3$, and $\displaystyle a_n = 4(-1)^n/n^2$ when n≠0. Then Parseval's theorem gives $\displaystyle \sum_{n=1}^\infty\frac1{n^4} = \frac{\pi^4}{90}$.
• Jan 21st 2009, 01:49 PM
PvtBillPilgrim
Thank you very much. This does work.

What function do I pick for the summation from n =1 to infinity of (1/n^6)? If I try a Fourier cosine series with x^3, it seems I get the oscillation problem again.

Any help appreciated.
• Jan 22nd 2009, 11:23 AM
PvtBillPilgrim
Does anyone know a function to use, please? I can't get 1/n^3 in the series with easy to use terms in the numerator.
• Jan 22nd 2009, 11:47 AM
Mathstud28
Quote:

Originally Posted by PvtBillPilgrim
Does anyone know a function to use, please? I can't get 1/n^3 in the series with easy to use terms in the numerator.

If you can you should let someone know...because then you would have found $\displaystyle \zeta(3)$ (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
• Jan 22nd 2009, 12:07 PM
PvtBillPilgrim
Well, I want to determine the summation 1/n^6. How do I do this? I believe it's pi^6/(956).
• Jan 22nd 2009, 12:34 PM
Opalg
Quote:

Originally Posted by Mathstud28
If you can you should let someone know...because then you would have found $\displaystyle \zeta(3)$ (Apery's constant) whose irrationality was only found a couple of years ago...the actual value is unknown.

... and yet it clearly states here that "The Riemann zeta function $\displaystyle \zeta(2n)$ 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 $\displaystyle (-1)^n/n^3$. That would get round the problem of having to avoid anything that looks like the Apéry constant, yet still allows you to get $\displaystyle {\textstyle\sum} 1/n^6$ using Parseval's theorem. Have you tried the function $\displaystyle |x^3|$? (That's just a guess — I have no idea whether it will work.)

Quote:

Originally Posted by PvtBillPilgrim
Well, I want to determine the summation 1/n^6. How do I do this? I believe it's pi^6/(956).

Actually $\displaystyle \pi^6/945$ (see the above link).
• May 2nd 2010, 12:55 AM
magus
Quote:

Originally Posted by Opalg
... and yet it clearly states here that "The Riemann zeta function $\displaystyle \zeta(2n)$ 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 $\displaystyle (-1)^n/n^3$. That would get round the problem of having to avoid anything that looks like the Apéry constant, yet still allows you to get $\displaystyle {\textstyle\sum} 1/n^6$ using Parseval's theorem. Have you tried the function $\displaystyle |x^3|$? (That's just a guess — I have no idea whether it will work.)

Actually $\displaystyle \pi^6/945$ (see the above link).

My text on PDEs that I'm working on has a similar problem but gives the Fourier expansion:

$\displaystyle \dfrac{x}{2} = \sum_{n=1}^{inf}\dfrac{(-1)^{n+1}}{n}\sin{nx}$

I think I know how to start but I'd also like some guidance.

EDIT: N/M That was just to find $\displaystyle \zeta(2)$ You need Bernoulli numbers for a generic formula.