Fourier Series of $f=x$ in $[0,2\pi)$

Question

I have to find the Fourier Series of $f=x$ in $[0,2\pi)$, I already know that $g=x$ in $[-\pi,\pi)$ is $$\sum_{n=1}^\infty \frac{(-1)^n (-2) }{n} \sin(nx) $$ How does the difference in the region change me, since in both regions the period is $2\pi$?

Answer 1

If you know that $$ x= \sum_{n=1}^{\infty}\frac{(-1)^n(-2)}{n}\sin(nx),\;\;\; -\pi < x < \pi, $$ then $$ x-\pi=\sum_{n=1}^{\infty}\frac{(-1)^n(-2)}{n}\sin(n(x-\pi)), \;\;\; 0 < x < 2\pi. $$

Using \begin{align} \sin(n(x-\pi))& =\sin(nx)\cos(-n\pi)+\cos(nx)\sin(-n\pi)\\ &=(-1)^n\sin(nx) \end{align} gives $$ x = \pi-\sum_{n=1}^{\infty}\frac{2}{n}\sin(nx),\;\;\; 0 < x < 2\pi. $$