I have recently started to study Fourier series and I have the following function:
$ f(x)= \begin{cases} 1&\text{if}\,& 0\leq x\leq \pi \\ -1&\text{if}\,& \pi<x\leq 2\pi\\ \end{cases} $
Where:
$a_0=0$ ,$a_n=0$ ,$b_n=\frac {2}{\pi}$$(\frac {1-(-1)^n}{n}$)
Given that even $b_n$s are equal to zero, I understand that I should have two choices for, somehow, including the odd ones:
1)
Let $b_{2n+1}=\frac {2}{\pi}$$(\frac {2}{2n+1}$)
And so:
$f(x)=\sum^\infty_0\frac {2}{\pi}$$(\frac {2}{2n+1})sin((2n+1)x) $
1)
Let $b_{2n-1}=\frac {2}{\pi}$$(\frac {2}{2n-1}$)
And so:
$f(x)=\sum^\infty_1\frac {2}{\pi}$$(\frac {2}{2n-1})sin((2n-1)x) $
Now, I don't grasp the reasoning of starting the sum of the Fourier series on 0 or 1 on either case depending on how I want to express the odd ones, and neither the whole reasoning of "including" the odd ones. I am sure that it's something basic and it may even go further than series and sums.
Any help is welcome :)
The function $$1-(-1)^n=\begin{cases} 2& \text{for odd } n\\ 0&\text{otherwise}\end{cases}$$ So you can emit all even $b_n$. As to the sum starting at $0$ and $1$ being different, you just have to ensure that every odd number is included. So when $n=0$ we get $2n+1=1$ whereas for $2n-1$ we need to start at $n=1$ for $2n-1=1$.