Is this the odd expansion of the function $f(x)=x$ if $0\leq x<1$ and $f(x)=1$ if $1\leq x<2$?

Question

We have the function $f(x)=\left\{\begin{matrix} x &, 0\leq x<1 \\ 1 &, 1\leq x<2 \end{matrix}\right.$

I want to compute the Fourier sinus series with period 4.

Is the odd expansion of the function the following?

$h(x)=\left\{\begin{matrix} -1 &, -2<x<-1 \\ x& , -1<x<0 \\ 0 & , x=0,2 \\ x &, 0\leq x<1 \\ 1 &, 1\leq x<2 \end{matrix}\right.$

The graph of that function is this:

I think that it is wrong, since a periodic function must be the same for each period. But in this case the line segment $y=x$ is getting higher at each interval. Can that be?

Answer 1

Your extension function $h$ is right, and the extended graph is

now compute the Fourier series!.

Answer 2

In another way, differentiate $f(x)$ and subtract the continuous component of $1/2$ to get a (odd) square wave, period $2$: suppose you can get easily the sine transform of it.