Show that
$$x(\pi - x)= \frac{\pi^2}{6}-\sum_{k=1}^{\infty} \frac{\cos(2kx)}{k^2}$$
for $ 0<x<\pi$
My idea:
I've defined the periodic function $$f(x) = 0 \text{ if } x \in [- \pi, 0) \text{ and }$$ $$f(x)=x(\pi-x) \text{ if } x \in [0, \pi)$$
and make the Fourier Series of f.But, imediatly we have
$$a_o = \frac{1}{\pi} \int_0^{\pi} x(\pi -x) dx = \frac{\pi^2}{6}$$
So I am seeing a problem since we should use $ a_o/2. $Also,
$$b_n = \frac{1}{\pi} \int_0^{\pi} x(\pi-x) sin(nx) dx = \frac{-n \pi \sin(n \pi) - 2 cos(n \pi) +2}{n^3 \pi} = \frac{4}{n^3 \pi} $$
If $ n = 2k+1$. So the series term with $\sin(n x) $ cannot be zero. Since the f is continuous over $ (0, \pi)$, I thought the series should converge tl the given function.What am I doing wrong?
P.s: for the cosine part, I've obtained exactly the series part given an answer. P.S.S: we usethat the fourier series is
$$SF_f = \frac{a_o}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) + b_n \sin(nx)$$
Edit:
I've found this searching about Fourier series. Since the function f is even on $(0,\pi$), does the theorem holds or it must be even in $R$?
Edit 2:
$$f(x) = -x(\pi+x) \text{ if } x \in [- \pi, 0) \text{ and }$$ $$f(x)=x(\pi-x) \text{ if } x \in [0, \pi)$$
Is even and the equality, now, holds.
Target function
$$ f(x) = x \left(\pi - x \right) $$
To reproduce the answer, use an even reflection of the function: $$ \begin{align} r(x) &= x \left(\pi - x \right), \quad x\ge 0, \\ l(x) &= -x \left(\pi + x \right), \quad x < 0. \end{align} $$
Fourier amplitudes
$$ \begin{align} % a_{0} &= \frac{1}{\pi} \left( \int_{-\pi}^{0} l (x) \, dx + \int_{0}^{\pi} r (x) \, dx \right) = \frac{\pi ^3}{3} \\[5pt] % a_{k} &= \frac{1}{\pi} \left( \int_{-\pi}^{0} l (x) \, dx + \int_{0}^{\pi} r (x) \, dx \right) = -\frac{2 \left((-1)^k+1\right)}{k^2} \\[5pt] % b_{k} &= \frac{1}{\pi} \left( \int_{-\pi}^{0} l (x) \, dx + \int_{0}^{\pi} r (x) \, dx \right) = 0 % % \end{align} $$
The first few terms in the series look like $$ \begin{array}{rr} k & a_{k} \\\hline 1 & 0 \\ 2 & -1 \\ 3 & 0 \\ 4 & -\frac{1}{4} \\ 5 & 0 \\ 6 & -\frac{1}{9} \\ \end{array} $$
Fourier Series
$$ \begin{align} f(x) &= \frac{a_{0}}{2} + \sum_{k=1}^{\infty} \frac{-2 \left((-1)^k+1\right)}{k^2}\cos \left( kx \right) \\[3pt] &= \frac{\pi^{3}}{6} - \sum_{k=1}^{\infty} k^{-2} \cos \left( 2kx \right) \end{align} $$
Building the approximation