I missed my class where we went over the fourier series and am having extreme issues with this homework question.
$f(x) = f(x + 1) = f(x + \sqrt{2})$
Is there anyone who could be kind enough as to give me an explaination to this question?
I am able to do the fourier analysis when we are given a simple function such as $f(x) = x$, but the issue is that in this equation $f(x)$ is completely arbitrary.
Thanks-
There's a simple proof for this not involving Fourier Series: Notice that for every $n,m\in\mathbb{Z}$, we have $f(n+m\sqrt{2})=f(0)$. Thus, $f$ is constant in the subgroup $G=\mathbb{Z}+\sqrt{2}\mathbb{Z}$ of $\mathbb{R}$. Since every subgroup of $\mathbb{R}$ is either dense or of the form $k\mathbb{Z}$ for some $k\in\mathbb{R}$ (see Subgroup of $\mathbb{R}$ either dense or has a least positive element?), and $G$ is not of the latter form, then $G$ is dense in $\mathbb{R}$.
Therefore, $f$ is constant in a dense subset of $\mathbb{R}$, so $f$ is constant, by continuity.
(I am giving it a try here. I request the fellow readers to kindly check the sanctity!)
As the question explicitly asked for Fourier series, the time period is $T < \infty$.
Now $$c_n = \frac{1}{T}\int_0^Tf(t) e^{-2\pi int/T}dt$$
but $$c_n = \frac{1}{T}\int_0^Tf(t+1) e^{-2\pi int/T}\, dt=\left[ \frac{1}{T}\int_1^{T+1}f(t) e^{-2\pi int/T} \, dt \right]e^{2\pi in/T}$$ (Equation 1).
Similarly by putting $f(t+\sqrt2)$ in place of $f(t)$, we can get $$c_n =\left[ \frac{1}{T}\int_{\sqrt 2}^{T+\sqrt 2}f(t) e^{-2\pi int/T} \, dt \right]e^{-2 \sqrt 2\pi in/T}$$ (Equation 2).
The terms in the brackets (in the last two equations) are the same as we are integrating over 1 time period in both cases, and they are equal to $c_n$.
Now substracting eq. 1 from eq. 2, and replacing the bracket term with $c_n$ we can get
$$c_n(e^{-2 \sqrt 2\pi in/T} - e^{2\pi in/T}) = 0.$$ Now clearly the term in the bracket can not be equal to zero, hence $c_n$ must be zero for all $n\neq 0$. For $n=0, c_0$ is non zero.
So after all this, we can see that there is no fluctuating component in the Fourier Series expansion of the given function, and hence it has to be constant!
Note: I have used substitution method to get equation 1 & 2.
