Why is a Fourier series the same on $[-3,3]$ and on $[-\pi,\pi]$?

Question

I have calculated the Fourier series in Mathematica for a function on an interval $[-3,3]$ and on an interval $[-\pi,\pi]$ and they are exactly the same.

The function is:

\begin{equation} f(\xi)=\begin{cases} 2, \ \ \ \ -2\le \xi<-1 \\ 1, \ \ \ \ -1\le \xi<0 \\ 2, \ \ \ \ 0\le \xi<1 \\ 3, \ \ \ \ 1\le \xi\le2, \end{cases} \end{equation}

They seem scaled, but have not changed at all.

The core of this lies in the Fourier coefficients. Can someone explain how the Fourier coefficients become the same on $[-3,3]$ and on an interval $[-\pi,\pi]$?

The coefficients are:

$$\alpha_0=\frac{1}{2a}\int_{-a}^a f(\xi)d\xi$$

$$\alpha_k=\frac{1}{2a}\int_{-a}^a f(\xi)\cos\omega k\xi d\xi$$

$$\beta_k=\frac{1}{2a}\int_{-a}^a f(\xi)\sin\omega k\xi d\xi$$

Where $\omega=\frac{n\pi}{a}$

Answer 1

It is obvious since once a periodic function can be represented as a linear combination of monotone harmonics, its scaled version can be represented as a linear combination of scaled monotone harmonics, with the same scale factor and linear combination coefficients. In mathematical language, if $$ f(x)=\sum_{n=0}^\infty a_n\cos \frac{2n\pi}{T}x+b_n\sin \frac{2n\pi}{T}x $$ and $$ f(kx)=\sum_{n=0}^\infty c_n\cos \frac{2n\pi}{T/k}x+d_n\sin \frac{2n\pi}{T/k}x, $$ then $$ { c_0=\frac{1}{T/k}\int_0^{T/k} f(kx)dx=\frac{1}{T}\int_0^{T} f(x)dx=a_0, \\ c_{n\ge 1}=\frac{2}{T/k}\int_0^{T/k} f(kx)\cos \frac{2n\pi}{T/k}x dx=\frac{2}{T}\int_0^{T} f(x)\cos \frac{2n\pi}{T}x dx=a_n. } $$ similarly, $d_n=b_n.$