The coefficients of a Fourier series on $[-L, L]$

Question

Suppose we have $f : [-L, L] \rightarrow \mathbb{R}$ is a periodic function with period $2L$. We also have $g(x) = f(\frac{Lx}{\pi})$ and $$g(x) = \dfrac{a_0}{2} + \sum_{n=1}^\infty {\big(a_n\cos{nx} + b_n\sin{nx}\big)}$$ which is Fourier series for $g(x)$. Then $$a_n = \dfrac{1}{\pi} \int_{-\pi}^{\pi} {g(x)\cos{nx}\,dx} = \dfrac{1}{\pi} \int_{-\pi}^{\pi} {f(\frac{Lx}{\pi}}){\cos{nx}\,dx} = \frac{1}{L} \int_{-L}^L {f(t)\cos{\frac{\pi nt}{L}}\,dt}$$ $$b_n = \frac{1}{L} \int_{-L}^L {f(t)\sin{\frac{\pi nt}{L}}\,dt}$$ Now, $$f(x) = g(\frac{\pi x}{L}) = \dfrac{a_0}{2} + \sum_{n=1}^\infty {\big(a'_n\cos{\frac{\pi nx}{L}} + b'_n\sin{\frac{\pi nx}{L}}\big)}$$ Now the question. As I understand from the book, $a'_n$ and $a_n$, $b'_n$ and $b_n$ equal, but why? shouldn't $x$ be replaced with $\frac{\pi x}{L}$ in the representation of the coefficients for the Fourier series of $g(x)$?

Answer 1

No, let $\frac{\pi x}{L}=y$. Then

\begin{align*} f(x)&=g(\frac{\pi x}{L}) = g(y) = \frac{a_0}{2}+\sum_{n=1}^\infty(a_n \cos(ny)+b_n\sin(ny)\\ &=\frac{a_0}{2}+\sum_{n=1}^\infty(a_n \cos(n\frac{\pi x}{L})+b_n \sin(n\frac{\pi x}{L})) \end{align*}