I'm trying to find the Fourier Coefficients for $ x\cdot(2\pi-x) \rightarrow 0 \le x \le 2\pi $.
This task is an easy one..
$$ a_0=\frac{4}{3}\pi^2 $$
$$ a_n = -\frac{4}{n^2} $$
$$ b_n = 0 $$
So where is my problem? The task is an exercise from a book and the solution says that f(x) was even so $b_n = 0$ but I'm not able to prove that.
A function is even when $ f(-x) = f(x) $. Usually I would do something like f(-3) = f(3). But this doesn't work here.
For example: $ f(3)=9,85; f(-3)=-27,85 $.
Furthermore when I plot the function it doesn't look very symmetric either.
Best Answer
Fourier series applies to periodic functions. When we are given the function on a specific interval, it is implied that we are dealing with a function whose period equals the length of the interval [in your case, $2\pi$] and the restriction of this periodic function to the given interval is the function given [in your case, $x(2\pi -x)$ on [$0, 2\pi$]]. Check your textbook if you have one.
As I said, if you $2\pi$-periodically continue the function $f$ to $F$ on the whole axis $\mathbb R$, then clearly this $F$ is even [by observing the graph, or check the equation $F(x) = F(-x)$ for all $x$].
UPDATE
Compute the expression of the function $F$. Example. For $x \in [-2\pi, 0]$, $x + 2\pi \in[0, 2\pi]$. Since $F(x) = F(x+2\pi)$ for $x \in [-2\pi, 0]$, we have $$ F(x) = (x +2\pi) (2\pi - (x + 2\pi)) = -x (x+2\pi). $$