Does using a multiple of the fundamental period for a periodic function give the same fourier series?

Question

The fourier expansion of a "sufficiently nice" periodic function is given by $$a_0/2+\sum_{k=1}^\infty a_k \cos(\frac{2\pi}{T}kx)+\sum_{k=1}^\infty b_k \sin(\frac{2\pi}{T}kx)$$ where $a_k=\frac{2}{T}\int_0^T f(x)\cos(kx)dx$ and $b_k=\frac{2}{T}\int_0^T f(x)\sin(kx)dx$.

In pretty much every resource I've found, there is no mention of $T$ being the fundamental period. I want to know if it matters, as long as the function is $T$ periodic. In particular, I want to prove that if $T$ is the fundamental period of a function and $T'=nT$ where $n\in \mathbb{N}$, then the Fourier expansion obtained using $T'$ as the period is identical.

Unless I'm mistaken, this amounts to showing that $$\frac{2}{nT}\int_0^{nT}f(x)\cos\left(\frac{k}{n}\frac{2\pi }{T} x\right)=\frac{2}{nT}\int_0^{nT}f(x)\sin\left(\frac{k}{n}\frac{2\pi }{T} x\right)dx=0$$ if $k$ is not a multiple of $n$.

I am not sure whether what I want to prove is even true, and if it is, I am unable to prove it. I would appreciate any hints that will guide me further.

Answer 1

First, it's convenient to reframe this statement in terms of real sine/cosine integrals to a form in terms of complex exponential integrals. Namely, Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$ yields \begin{align} \frac{1}{nT}\int_0^{nT}f(x)e^{i 2\pi \frac{k}{nT} x}~dx&=\frac{1}{nT}\int_0^{nT}f(x)\cos\left(2\pi \frac{k}{nT} x\right)dx\\ &\hspace{1.5cm}+i\frac{1}{nT}\int_0^{nT}f(x)\sin\left(2\pi \frac{k}{nT} x\right)dx \end{align}

Note that the two real integrals considered originally are the real and imaginary parts of the present complex integral. Hence the original condition can be restated as

$$\frac{1}{nT}\int_0^{nT}f(x)e^{i 2\pi \frac{k}{nT} x}~dx=0$$ unless $k$ is a multiple of $n$ for all integer $k$ (including $k\leq 0$).

To prove this, we chop the interval $[0,nT]$ into $n$ domains of length $T$. The $j$th such interval is of the form $[(j-1)T,jT]$; we can reparametrize the integral for this contribution using $x'=x-(j-1)T$ to get $$\frac{1}{nT}\int_{0}^{T} f(x'+(j-1)T)e^{i 2\pi \frac{k}{nT} [x'+(j-1)T]}dx'=\frac{1}{nT}e^{i 2\pi \frac{k}{n}(j-1)}\int_{0}^{T} f(x')e^{i 2\pi \frac{k}{nT} x'}dx'$$ where the second equality follows from the $T$-periodicity of $f$. Summing these contributions yields

$$\frac{1}{n}\sum_{j=1}^n (e^{i 2\pi \frac{k}{n}})^{j-1}\cdot \frac{1}{T}\int_{0}^{T} f(x')e^{i 2\pi \frac{k}{nT} x'}dx'$$

The sum is known as a root of unity filter: it sums to zero unless $k/n$ is an integer, in which case $e^{i 2\pi (k/n)}=1$ and thus the sum reduces to $1/n\cdot n=1$. Hence the integral indeed vanishes unless $k$ is a multiple of $n$; in that case, we instead have $$\frac{1}{T}\int_{0}^{T} f(x')e^{i 2\pi \frac{k/n}{T} x'}dx'$$ which we identify as the $(k/n)$th complex Fourier coefficient of $f$ on the original domain.