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.

  • $\begingroup$ Periodic continuation. $\endgroup$ – xbh Sep 5 '18 at 14:34
  • $\begingroup$ Your title does not match your question. If $f (x)= x(2\pi - x) \; [x \in [0,2\pi]]$, then it is even. But for $x(\pi -x)$, it is not. $\endgroup$ – xbh Sep 5 '18 at 14:38
  • $\begingroup$ Sorry, typo... But how to mathematically prove this? Am I right that f(-x) = -f(x) doesn't work here or am I doing it wrong? $\endgroup$ – TimSch Sep 5 '18 at 14:41
  • $\begingroup$ $f$ is even iff $f(x) = f(-x)$ for all $x$, not $f(-x) = -f(x)$, which is $\color{red}{\mathrm{odd}}$. $\endgroup$ – xbh Sep 5 '18 at 14:43
  • $\begingroup$ oh my... yes you're right. But it doesn't solve my problem. I updated the question. $\endgroup$ – TimSch Sep 5 '18 at 14:47

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 $x$].


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 $$ F(x) = (x +2\pi) (2\pi - (x + 2\pi)) = -x (x+2\pi). $$

  • $\begingroup$ Thank you! It makes sense that when I periodically continue this parabola it's symmetric. But I'm stuck at the F(x) = F(-x). How can I represent that periodic function as a formula (because obviously using $ x(2 \Pi -x) $ is wrong). Sorry for being so pedantic but I'm feeling that I'm missing something here. $\endgroup$ – TimSch Sep 5 '18 at 16:14
  • $\begingroup$ See my update. The expression at other points could be derived similarly. $\endgroup$ – xbh Sep 5 '18 at 16:18
  • $\begingroup$ I just plotted your function and it seems not quite right. Here is the plot: wolframalpha.com/input/… I calculated it by myself and got f(x)=x(-x-2Pi). wolframalpha.com/input/… $\endgroup$ – TimSch Sep 5 '18 at 22:32
  • $\begingroup$ @TimSch Yes. I missed a minus sign. Thanks for warning. $\endgroup$ – xbh Sep 6 '18 at 0:24

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