Thread: fourier cosine series

1. fourier cosine series

Hi, this question is being a pain to me can someone please help.

a) Find the Fourier cosine series of $\displaystyle f(x)=|sinx|$ for interval $\displaystyle (-\pi, \pi)$.

b) Use it to find the sum

$\displaystyle \sum_{n \ge 1} \frac{1}{4n^2-1}$

Thanks

2. Re: fourier cosine series

Use the standard formulas:

\usepackage[usenames]{color}\gammacorrection{1.5}\usepackage[11pt]{extsizes}{\begin{aligned}f(x)&=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx,\quad x\in(-\pi,\pi).\\[7pt] a_0&=\frac{2}{\pi}\int\limits_0^\pi f(x)\,dx=\frac{2}{\pi}\int\limits_0^\pi\sin x\,dx=\left.{-\frac{2}{\pi}\cos x}\right|_0^\pi=-\frac{2}{\pi}(-1-1)=\frac{4}{\pi}.\\[5pt] a_n&=\frac{2}{\pi}\int\limits_0^\pi f(x)\cos nx\,dx= \frac{2}{\pi }\int\limits_0^\pi\sin x\cos nx\,dx=\frac{1}{\pi}\int\limits_0^\pi\Bigl[\sin(x+nx)+\sin(x-nx)\Bigr]dx=\\ &=\left.{-\frac{1}{\pi}\!\left[\frac{\cos(x+nx)}{1+n}+\frac{\cos(x-nx)}{1-n}\right]\right|_0^\pi= -\frac{1}{\pi}\!\left[\frac{-(-1)^n}{1+n}+\frac{-(-1)^n}{1-n}-\left(\frac{1}{1+n}+\frac{1}{1-n}\right)\right]=\\ &=-\frac{1}{\pi}\!\left[-\frac{2(-1)^n}{1-n^2}-\frac{2}{1-n^2}\right]=\frac{2}{\pi}\frac{1+(-1)^n}{1-n^2}= \begin{cases}0,&\text{if}\quad n=2k-1,\\[5pt]\dfrac{4}{\pi}\dfrac{1}{1-4k^2},&\text{if}\quad n=2k,\end{cases}k\in\mathbb{N}_1.\\[7pt] f(x)&=\frac{2}{\pi}+\frac{4}{\pi}\sum_{k=1}^\infty\frac{\cos(2kx)}{1-4k^2},\quad x\in(-\pi,\pi).\\[5pt]f(0)&=\frac{2}{\pi}+\frac{4}{\pi}\sum_{k=1}^\infty\frac{1}{1-4k^2}=0\quad\Rightarrow\quad\sum_{k=1}^\infty\frac{1}{4k^2-1}=\frac{2}{\pi}\div\frac{4}{\pi}=\frac{1}{2}.\end{aligned}}

3. Re: fourier cosine series

wow, thanks demath, I was just thrown a bit by the absolute sign. I understand it now.