Finding the fourier series representation for a piecewise function

Question

Expand the given function in the appropriate Fourier series: $$\begin{align} f(x) = \begin{cases} x+1 &\mbox{if } -1 \leq x \leq 0 \\ x-1 &\mbox{if } 0 \leq x \lt 1 \end{cases} \end{align}$$

To my knowledge, the first step is to determine whether the function is even or odd and that's where I get stuck.

UPDATE AFTER DISCUSSION WITH MARIANO:

on $[-1,0]$, $f(-x)=-x+1 \neq -f(x)$

on $[0, 1]$, $f(-x)=-x-1 \neq -f(x)$

so now i'm at a loss as to how to continue...

http://en.wikipedia.org/wiki/Even_and_odd_functions#Odd_functions — Frank Epps, Jul 20 '13 at 03:08
No. Tell me what it means for a function to be odd. (Trust me: I already know the answer. The PhD committee made sure of that) — Mariano Suárez-Álvarez, Jul 20 '13 at 03:08
So: suppose that $x\in[0,1]$. What is $f(x)$ and what is $f(-x)$ for your specific function? — Mariano Suárez-Álvarez, Jul 20 '13 at 03:10
Yes it is, so it is odd on $[0,1]$, but what about $[-1,0]$ do we just ignore it? — Frank Epps, Jul 20 '13 at 03:13
No. Now suppose that $x$ is in $[-1,0]$. WHat is $f(x)$ and what is $f(-x)$? — Mariano Suárez-Álvarez, Jul 20 '13 at 03:13
$f(x)=x+1$ and $f(-x)=-x+1$ [PS: i made a mistake in finding $f(-x)$ for the earlier interval, so $f(-x) \neq f(x)$] — Frank Epps, Jul 20 '13 at 03:15
why would you want that $f(-x)$ and $f(x)$ be equal? We are trying to figure out if the function is odd at the moment. — Mariano Suárez-Álvarez, Jul 20 '13 at 03:16
I don't see where you think you made a mistake, really. In any case, you probably get the idea of how to check if the function is odd by now. Do it :-) — Mariano Suárez-Álvarez, Jul 20 '13 at 03:19
I'll leave it to someone else to walk you through the computation of the fourier series... — Mariano Suárez-Álvarez, Jul 20 '13 at 03:28
I understand how to proceed, was just confused about odd-even. ur a lifesaver!! — Frank Epps, Jul 20 '13 at 03:29
The FourierSeries command of the OrthogonalExpansions package of Maple produces $$\sum _{i=1}^{n}-2,{\frac {\sin \left( \pi ,ix \right) }{\pi ,i}} .$$ — user64494, Jul 20 '13 at 07:46
thank u sir, that's what i ended up getting in my calculations! — Frank Epps, Jul 20 '13 at 16:32

score 1 · Answer 1 · answered Jul 20 '13 at 22:10

It would be easier to determine that $f(x)=x+\operatorname{sign}x$ (sign function), where both terms are odd. Hence $f$ is odd, and we only worry about the sines; thus, integrate $f(x)\sin \pi n x$ over $[0,1]$ and multiply by $2$ to get the coefficient of $\sin \pi n x$ in the expansion. The key ingredient is
$$\int_0^1 (x-1)\,\sin \pi nx \,dx = -\frac{x-1}{\pi n}\cos \pi n x\bigg|_0^1 +\frac{1}{\pi n}\int_0^\pi \cos \pi nx \,dx = -\frac{1}{\pi n}+0 $$

Finding the fourier series representation for a piecewise function

1 Answers1