I was given the problem to find the Fourier Sine Series for $f(x) = 1 + x $ on the unit interval. I took suggested steps from my lecture but I'm not sure if I did it correctly.
According to my lecture the function has to be odd and also periodic. Therefore I extended it according to the notes: $F:[-1,1]$ $$F(x) = \begin{cases} f(x) & \text{if $0\lt x \leq L$} \\0 & \text{if $x =0 \lor x=L$}\\ -f(-x) & \text{if $-L \leq x \lt 0$} \end{cases}$$ with $F(x+2L) = F(x)$
which is for functions defined on $[0,L]$. Then since $ L = 1$ my function has the form
$$F(x) = \begin{cases} 1+x & \text{if $0\lt x \leq 1$} \\0 & \text{if $x =0 \lor x=1$}\\ x-1 & \text{if $-1 \leq x \lt 0$} \end{cases}$$
And so I should obtain the Fourier Sine Series : $$\sum_{n=1}^\infty b_nsin(n\pi x) $$ with $b_n =2\int_{0}^1(1+x)sin(n\pi x)dx$ but I think $F(x+2) \neq F(x)$.
Any suggestions as to how I go from here?
Best Answer
To make $F$ odd and periodic of period $2$ from $f$ defined on $[0,1],$ you can define:
$$F(x)=\begin{cases}0&x\in\mathbb Z\\ f(x-2m)&x\in (2m,2m+1),\,m\in \mathbb Z \\ -f(2m-x)&x\in(2m-1,2m),\, m\in\mathbb Z \end{cases}$$
In particular, $F(x)=f(x)$ only on $(0,1),$ not when $x=0$ or $x=1.$ These are necessary to make $F$ odd and period $2.$
$F(0)=0$ is obviously necessary because $$F(0)=F(-0)=-F(0).$$
$F(1)=0$ is necessary for $$F(1)=F(-1+2)=F(-1)=-F(1).$$
So periodicity and oddness require $F(m)=0$ for all integers $m.$
However, you can change $F(0)$ and $F(1)$ because it doesn’t affect the integrals.
You can avoid all this work if you know that $F$ only has to satisfy oddness and periodicity “almost everywhere.” But that requires some measure theory.
Now, that’s a lot of work, but we don’t need to do anything with it to compute the coefficients. That’s because, as you’ve noted:
$$b_n =\int_{-1}^1 F(x)\sin n\pi x\,dx=2\int_0^1 f(x)\sin n\pi x\,dx$$
For $f(x)=x+1,$ this integral is easy with integration by parts, with $u=x+1, dv=\sin n\pi x\,dx.$