More Fourier series

Markov

Member
1) Compute Fourier series of $f(x)=\left\{\begin{array}{cl}\sin x,&\sin x\ge0,\\0,&\sin x<0.\end{array}\right.$

2) Consider the $2\pi-$periodic function $f(x)=\left\{\begin{array}{cl}1,&\text{ if }\phantom{-}0\le x<\pi,\\0,&\text{ if }-\pi\le x<0.\end{array}\right.$ Use Parseval's identity to prove that $\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}=\frac{\pi^2}8.$

Attempts:

1) How can I get the bounds to compute the Fourier series?

2) I think this is straightforward, I just need t compute $a_n$ and $b_n$ by writing for example $a_n=\displaystyle\frac1\pi\int_{-\pi}^{\pi}f(x)\cos (nx)\,dx=\frac{1}{\pi }\left( {\int_{ - \pi }^0 {f(x)\cos (nx)\,dx} + \int_0^\pi {f(x)\sin (nx)\,dx} } \right),$ and the same for $b_n,$ the rest is using $\displaystyle\frac{{a_0^2}}{2} + \sum\limits_{n = 1}^\infty {\left( {a_n^2 + b_n^2} \right)} = \frac{1}{\pi }\int_{ - \pi }^\pi {f{{(x)}^2}\,dx} ,$ but I have a problem here, when calculating the integral of the last formula, I'd need to split it up into two integrals but the fact that $f(x)=1$ for $0\le x<\pi$ doesn't imply that $f(x)^2=1,$ doesn't?

chisigma

Well-known member
1) Compute Fourier series of $f(x)=\left\{\begin{array}{cl}\sin x,&\sin x\ge0,\\0,&\sin x<0.\end{array}\right.$

If f(x) is $2 \pi$ periodic You have...

$\displaystyle a_{n}= \frac{1}{\pi}\ \int_{0}^{\pi} \sin x\ \cos nx\ dx = \frac{1}{\pi} \frac{-1+(-1)^{n+1}}{n^{2}-1}$

$\displaystyle b_{n}= \frac{1}{\pi}\ \int_{0}^{\pi} \sin x\ \sin nx\ dx = \left\{\begin{array}{cl}\frac{1}{2},&n=1\\0,&n \ne 1\end{array}\right.$

Kind regards

$\chi$ $\sigma$

Last edited:
• Markov

Markov

Member
Okay but why $0\le x<\pi$ ?

chisigma

Well-known member
Okay but why $0\le x<\pi$ ?
For $- \pi< x<0$ [or, that is the same, for $-1< \sin x<0$...] is $f(x)=0$...

Kind regards

$\chi$ $\sigma$