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I need help with this problem. I'm asked if it is possible to find $\sum_{n=0}^\infty \frac{1}{n^2}$ with the Fourier series of $f(x)=1-|x|$ if $|x|\leq 1$ and $f(x+2)=f(x)$. I tried to do the Fourier series, but I didn't ebd up with the correct result.

Fourier series definition: $f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos\left(\frac{n\pi x}{L}\right)+\sum_{n=1}^\infty bn\sin\left(\frac{n\pi x}{L}\right)$

I found that:

$a_0=\frac{1}{L}\int_{-L}^{L}f(x)dx=\int_{-1}^1(1-|x|)dx=1 $ since $|x|=x $ in $[0,L]$.

$a_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos(\frac{n\pi x}{L}) dx=\int_{-1}^1(1-|x|)\cos(\frac{n\pi x}{1})dx=\frac{2-2\cos\left({\pi}n\right)}{{\pi}^2n^2}=\frac{2-2(-1)^n}{{\pi}^2n^2}$

$b_n=\frac{1}{L}\int_{-L}^{L}f(x)\sin(\frac{n\pi x}{L}) dx=\int_{-1}^1(1-|x|)\sin(\frac{n\pi x}{1})dx=0$

I know that $\sum_{n=0}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$, so I guess that the answer is no. Am I correct or what am I doing wrong?

davidllerenav
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    Check the result for $b_n$: $1-|x|$ is an even function, so I would expect all of them to be $0$. Then recall also $\cos(\pi n)=(-1)^n$, and also it isn't the end of the world if you can first calculate something similar (e.g. $1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots$) because that can be used easily to find the original sum you're after. –  Jan 26 '20 at 09:24
  • Oh yes, you're right. So $bn=0$ and that leaves the fourier series like $f (x)=1+\sum_{n=1}^\infty \frac {2(-1)^{n+1}-2}{\pi^2 n^2} \sin (n\pi x) $, right? – davidllerenav Jan 26 '20 at 09:45
  • $\cos$, not $\sin$, and then I believe this might do: substitute $x=1$ because the Fourier series converges pointwise. –  Jan 26 '20 at 09:49
  • It looks like your Integration by substituion contains at least a mistake and i struggle to see how your definition of fourier coefficients leads to the fourier series you gave. Please rewrite the question using this definition. –  Jan 26 '20 at 14:29
  • If you find the fourier coefficients to be of the shape $\frac{1}{n}$ we may use Parsevals identity to connect the sum of the squares of the fourier coefficients to the integral over the square of the function. –  Jan 26 '20 at 14:40
  • @StinkingBishop but isn't $a_n$ multiplied by $\cos(\frac{n\pi x}{L})$ in the fourier series? – davidllerenav Jan 26 '20 at 16:19
  • @PeterSheldrick The error is in the integral of $b_n$, right? I have just made an edit to correct that mistake and to write the fourier series definition and the coefficients. I haven't seen Parsevals identity in class, how do I use it? – davidllerenav Jan 26 '20 at 16:40
  • https://www.wolframalpha.com/input/?i=integral_%28-1%29%5E%28%2B1%29+%281+-+x%29+sin%283+%CF%80+x+%2F+1%29+dx –  Jan 26 '20 at 17:10
  • @PeterSheldrick But if you write $|x|$ instead of $x$ then you get $0$ as the answer. Am I supposed to use $x$ or $|x|$? – davidllerenav Jan 26 '20 at 17:35
  • Let us continue this discussion in chat. –  Jan 26 '20 at 18:04

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