1. ## 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.

2. Try the function f(x) = x^2 on the interval (–π,π). You should find that $a_0 = 2\pi^2/3$, and $a_n = 4(-1)^n/n^2$ when n≠0. Then Parseval's theorem gives $\sum_{n=1}^\infty\frac1{n^4} = \frac{\pi^4}{90}$.

3. 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.

4. 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.

5. 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 $\zeta(3)$ (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

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

7. Originally Posted by Mathstud28
If you can you should let someone know...because then you would have found $\zeta(3)$ (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 $\zeta(2n)$ 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 $(-1)^n/n^3$. That would get round the problem of having to avoid anything that looks like the Apéry constant, yet still allows you to get ${\textstyle\sum} 1/n^6$ using Parseval's theorem. Have you tried the function $|x^3|$? (That's just a guess — I have no idea whether it will work.)

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

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

Actually $\pi^6/945$ (see the above link).
My text on PDEs that I'm working on has a similar problem but gives the Fourier expansion:

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

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

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

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# using fourier series prove that summation n=1 to infinity 1Ã·n^4=Î ^2/90

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