Why do the basis functions of a Fourier series have only integer multiples of the frequency?

Question

We are learning about Fourier series expansion, and we were given this equation: $$x(t)=\sum_{n=-\infty}^{\infty}a_n e^{j2\pi n f_0 t}$$ where $f_0$ is the frequency of the function $x(t)$

I only know a little linear algebra, but I understand that this equation (sort of...?) says the set of complex exponentials: $$\{ e^{j2\pi n f_0 t}:n\in\mathbb{Z}\}$$ forms an orthonormal basis of the vector space in which $x(t)$ resides. And I get that this means that we can represent $x(t)$ as a linear combination of the basis functions of its overarching vector space.

But I am very confused as to why complex exponentials with only integer multiples, $2\pi n f_0 t$, of the function's frequency suffice to form a basis for this space--why not the set of complex exponentials of ALL frequencies, i.e. $n\in\mathbb{R}$?

Said another way--Why does it only take the set of all complex exponentials with integer multiples of $f_0$ to express the $x(t)$?.

Another, related question--what exactly is the vector space that $x(t)$ is a member of? Is it tied directly to $f_0$?

Can anyone shed some light on this?

Answer 1

You only need integer multiples of a single base frequency because $x(t)$ is assumed to be periodic with period $\frac1{f_0}.$

If $x(t)$ weren't assumed to be periodic, this wouldn't work (and you'd need to try to move to something like the Fourier integral).

Answer 2

lets assume x(t) is some periodic function that repeats after $T_0$, then

x(t) = x(t+ n$T_0$)

Lets introduce new function with different time period

g(t) = g(t + n (p/q$T_0$)) $\qquad$ { p and q are integers and q > 1 }

p/q is some fraction of $T_0$, then its assume g(t) contributes its part along linear combination of complex exponential basis functions in x(t)'s

x(t) = $\sum_{k=-\infty}^\infty a_k e^{j 2\pi k f_0 t} + g(t)$

x(t + m$T_0$) = $\quad$above summation remains as it is $\quad$ + $\quad$g(t + m$T_0$),

g(t + m$T_0$) will not always give the same value unless $\rightarrow$ m = multiples of q

It is guarantee that g(t) gives same value after every m=k(q$T_0$), then to synchronise with

$\rightarrow \quad$ x(t)'s $T_0 \quad$ and $\quad$ g(t)'s q$T_0$

q had to be 1, g(t)'s time period had to be at least $T_0$

And further $T_0$ can be made generalised to expression n$T_0$ {n $\in$ I}

So we can expand g(t) into its Fourier series and it will smoothly and beautifully merge into the above summation series.

Thus x(t) cannot be made out of other frequencies else its Time period would not be the same and signal would be something else.