Knowing that $f(\theta)=\pi - \theta$ defined on $[0,2\pi]$ has the fourier series $\sum \limits _{n=1}^{\infty}\frac{2\sin(n\theta)}{n}$, then find the coefficients of $g(\theta)=\pi-\theta$ defined on $[0,\pi]$.
Well, I know that $f(\theta)=\sum \limits _{n=1}^{\infty}a_n\sin(n\theta)$ with $2/n=a_n=\frac{1}{\pi}\int _0 ^{2\pi}f(\theta)\sin(n\theta)d\theta$. I am trying to find a relation between $a'_n = \frac{2}{\pi}\int _0^{\pi}f(\theta)\sin(2n\theta)d\theta$ and $a_n$... I tried to change variables, like if $x=2\theta$ then $$a'_n = \frac{2}{\pi}\int _0^{2\pi}f(x)\sin(4nx)dx$$ but I couldnt get anywhere... if anyone could help me! Also, if there is a general formula relating the Fourier coefficients of an integrable function with periodic extensions with different periodicities, I would like to know how to get to that relation.
Through a few transformations of $f$, $\displaystyle g(\theta)=\frac{f(2\theta)+π}{2}$.
I hope that makes it simpler!
You can just substitute $2n$ for $n$, divide by $2$, and add $\displaystyle \frac{π}{2}$.