I want to prove $$\cos x = \frac{8}{\pi}\sum_n \frac{n\sin 2nx}{4n^2-1}\;x\in(0,2\pi)\;\;\;\;[1]$$
I have two questions regarding this:
$(1)$ How can I find a function $f$ such that the former series can be obtained using the Fourier series of $f$? I know the Fourier series will be given by
$$f(x)=a_0+\sum_n\left(a_n\cos\frac{n\pi x}{L}+b_n\sin\frac{n\pi x}{L}\right)$$
then should I asssume that the $\cos$ in the left hand side of $[1]$ is the $\cos$ inside the latter series?
Since no $\pi$ appear in the denominator of $[1]$, and $n\sin 2nx=n\sin nx\cos nx$ I thought that could $a_n=n\sin nx$ or $b_n =n\cos nx$ but these terms are given by definite integrals so no $x$ term can appear in them.
An alternative approach could be using $L=\pi/2$, then it must be $b_n=\frac{n}{4n^2-1}$, but this would imply $\frac{n}{4n^2-1}=\int_0^{\pi/2}f(x)\sin nx dx$, and I'm having some issues to get ride of the sine.
$(2)$ In the general case, given any series how can I proceed to find the function $f$?
Best Answer
You want the interval to be $0<x<\color{blue}{\pi}$, because then you can show $$ \cos x = \frac{8}{\pi}\sum_{n=1}^\infty \frac{n\sin(2nx)}{4n^2-1}, \quad x\in(0,\color{blue}{\pi}).\tag{1} $$ as follows.
In this case, the standard computation of Fourier sine coefficients for the series $$ \cos x=\sum_{n=1}^\infty b_n\sin(nx),\quad 0<x<\pi, $$ yields \begin{align} b_n={2\over \pi}\int_0^\pi \cos x\sin(nx)\,dx &= \begin{cases} 0, & n=1,\\ {2n(1+(-1)^n)\over \pi(n^2-1)}, &n=2,3,\dots, \end{cases} \\ &= \begin{cases} 0, & n=1,3,5,\dots\\ {4n\over \pi(n^2-1)}, &n=2,4,6,\dots \end{cases} \end{align} That is, $$ \cos x=\sum_{n\text{ even}}{4n\over \pi(n^2-1)}\sin(nx) %=\sum_{k=1}^\infty {4(2k)\over \pi((2k)^2-1)}\sin(2kx)\ \ \overset{\color{blue}{n=2k}}{=}\ \ \sum_{k=1}^\infty {8k\over \pi(4k^2-1)}\sin(2kx), $$ which is $(1)$.