2

Consider $f(x):=\lvert x\rvert, x\in [-\pi,\pi]$. Then the Fourier series is $$ f(x)=\frac{\pi}{2}-\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\cos((2n-1)x)}{(2n-1)^2}. $$

Now my task is to write down the related Parseval equation.

The general Parseval equation is $$ \frac{a_0^2}{2}+\sum_{n=1}^{\infty}(a_n^2+b_n^2)=\frac{1}{\pi}\int_{-\pi}^{\pi}\lvert f(x)\rvert^2\, dx, $$ so here it is $$ \frac{\pi^2}{2}+\frac{16}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^4}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\ dx. $$

Is that the hole task or what?!

Asaf Karagila
  • 393,674
  • 1
    it is correct; note that choosing $x=0$ in the Fourier series expansion you get a numerical estimate of the series for $a_n=\frac{1}{(2n-1)^2}$. With Parseval you get a num. estimate for the series of $b_n=\frac{1}{(2n-1)^4}$. – Avitus Nov 18 '13 at 11:56
  • I wonder why this task is rated with so much points on my recent worksheet... –  Nov 18 '13 at 11:58
  • probably because you are supposed to understand the Parseval formula and being able to prove it ;) – Avitus Nov 18 '13 at 12:26
  • Another task is: Caclulate the series of the sums of the inverse of the quadrat of the even natural numbers... do you understand what is meant? –  Nov 18 '13 at 12:58
  • Let $c_n= \frac{1}{(2n)^2}$ be the sequence of the inverse-of the square-of even natural numbers. Compute $\sum_{n\geq 1}c_n$. Can you produce a function $f$ s.t. its Fourier series evaluated, for example at $x=0$ gives you such series? I think that you are supposed to solve the exercise using Fourier Series. – Avitus Nov 18 '13 at 13:01
  • 1
    You have already the series of the inverse of the square of odd integers...using $\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}$ (famous formula) you could probably be able to compute the series for $c_n$ given in the above comment... – Avitus Nov 18 '13 at 13:04

1 Answers1

3

You got

$$0=f(0)=\frac\pi2-\frac4\pi\sum_{n=1}^\infty\frac1{(2n-1)^2}$$

and as commented by Avitus you have a numerical value for the above series. Now check that

$$\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n)^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}=\frac14\sum_{n=1}^\infty\frac1{n^2}+\sum_{n=1}^\infty\frac1{(2n-1)^2}\implies\ldots$$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • okay, then the searched sum is $\frac{\pi^2}{24}$. –  Nov 18 '13 at 13:18
  • Nop...Check again. You have $$\frac34\sum_{n=1}^\infty\frac1{n^2}=\sum_{n=1}^\infty\frac1{(2n-1)^2}\ldots$$ – DonAntonio Nov 18 '13 at 13:28
  • Not right? I have again the same. It is $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{\pi^2}{8}$. Then $\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{8}\Leftrightarrow \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{24}$. –  Nov 18 '13 at 14:04
  • Try again: $$\frac43\cdot\frac{\pi^2}8=\frac{\pi^2}6\ldots\ldots$$ – DonAntonio Nov 18 '13 at 14:22
  • Sorry, I do not see your point! My aim is to get $\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$. So I divide by 3, getting $\frac{\pi^2}{24}$. –  Nov 18 '13 at 14:28
  • But you are calculating $\sum_{n=1}^{\infty}\frac{1}{n^2}$. –  Nov 18 '13 at 14:35
  • Ok, now I got it: you want the sum of of the squares of the even natural numbers...and then yes: you get that. I missed the even thing from the beginning. – DonAntonio Nov 18 '13 at 14:41
  • No problem at all. You helped me a lot. Thanks to you and of course to Avitus! –  Nov 18 '13 at 14:46