Aperiodicity of $\sin[2x+\cos(\sqrt{2}x)]$

Question

Is the function $f(x)=\sin[2x+\cos(\sqrt{2} \cdot x)]$ periodic?

I don't know how to prove that this function is not periodic. Can you help me out?

Answer 1

First, some intuition. Here's a graph of $f(x)=\sin(2x+\cos(\sqrt{2} \cdot x))$:

We have to prove that the function is aperiodic. From the graph it appears that something stronger is true: even if we focus on any specific $y$-value $y_0$, the values of $x$ where $f(x) = y_0$ will not be evenly spaced. So let's pick $y_0 = 0$ for convenience.

Towards a contradiction, let's also assume the function is periodic, and it has period $a$. Then let $r$ be any root, where $f(r) = 0$. We have $$ \cdots = f(r-2a) = f(r-a) = f(r) = f(r+a) = f(r+2a) = \cdots = 0. $$ In order for all these to be zero, we need for all $k \ge 1$: \begin{align*} 2r + \cos(\sqrt{2} \cdot r) &\in \pi \mathbb{Z} \tag{1} \\ 2(r+ka) + \cos(\sqrt{2} \cdot (r+ka)) &\in \pi \mathbb{Z} \tag{2} \\ 2(r-ka) + \cos(\sqrt{2} \cdot (r-ka)) &\in \pi \mathbb{Z} \tag{3} \\ \end{align*} where $\pi \mathbb{Z}$ is the set of real numbers that are $\pi$ times an integer. Adding (2) and (3) we get \begin{align*} 4r + \cos(\sqrt{2} \cdot (r+ka)) + \cos(\sqrt{2} \cdot (r-ka)) &\in \pi \mathbb{Z}, \end{align*} and this can be simplified: note $\cos(x + y) + \cos(x - y) = 2 \cos x \cos y$. So $$ 4r + 2\cos(\sqrt{2} \cdot r) \cos(\sqrt{2} \cdot ka) \in \pi \mathbb{Z}. \tag{4} $$ Now we can subtract two times (1) to get $$ 2 \cos(\sqrt{2} \cdot r) [\cos(\sqrt{2} \cdot ka) - 1] \in \pi \mathbb{Z}. $$ But $\cos$ is always between $-1$ and $1$, so this expression is between $-4$ and $4$ (it's the product of $2$, something between $-1$ and $1$, and something between $-2$ and $2$). Therefore this value is $-\pi, 0,$ or $\pi$. And it must be for all $k$. In particular it is either $0$, $\pi$, or $-\pi$ for infinitely many $k$. This gives us a couple of cases. Either (i) $\cos(\sqrt{2} \cdot r) = 0$; or (ii) $\cos(\sqrt{2} \cdot ka)$ attains the same value for infinitely many $k$.

• Case (i). We know $\sqrt{2} \cdot r - \frac{\pi}{2} \in \pi \mathbb{Z}$. So $(2 \sqrt{2})r \in \pi \mathbb{Z}$. At the same time, we go back to equation (1), and we get that $2r \in \pi \mathbb{Z}$. If $r \ne 0$ then this implies that the ratio $\frac{(2 \sqrt{2})r}{r} = 2 \sqrt{2}$ is rational, contradiction (because $\sqrt{2}$ is irrational). So then $r = 0$. But then $\cos(\sqrt{2} \cdot r) = 1$, not $0$.

• Case (ii). If $\cos x = \cos y$, what can we say about $x$ and $y$? Either $x$ and $y$ differ by a multiple of $2 \pi$, or $x$ and $-y$ differ by a multiple of $2 \pi$ (prove this). But if we have infinitely many such values, we must find two that differ by a multiple of $2 \pi$. So we have $k_1$ and $k_2$ such that $\sqrt{2} \cdot k_1 a - \sqrt{2} \cdot k_2 a \in 2 \pi \mathbb{Z}$. So $a (k_1 - k_2) \cdot \frac{\sqrt{2}}{\pi} \in \mathbb{Q}$, so $a$ is some rational number times $\frac{\pi}{\sqrt{2}}$.

  This gets us very close to our contradiction. Go back to equations (2) and (3), and plug in values of $k$ such that $\sqrt{2} k a$ is a multiple of $2 \pi$, so it can be ignored. We then have that $$ 2r + 2ka + \cos(\sqrt{2} \cdot r) \in \pi \mathbb{Z} $$ for all such values of $k$. Subtracting two such values $k_3$ and $k_4$, $2(k_3 - k_4) a \in \pi \mathbb{Z}$, so $a$ is $\pi$ times a rational number.

  But now $a$ is a rational number times $\pi$ and a rational number times $\frac{\pi}{\sqrt{2}}$. Since $a$ is nonzero (it's a period), this means that $\sqrt{2}$ is rational, contradiction.