Is $\cos(\alpha x + \cos(x))$ periodic?

Question

Consider the function $f: \mathbb{R} \to [-1, 1]$ defined as

$$f(x) = \cos(\alpha x + \cos(x))$$

What conditions must be placed on $\alpha \in \mathbb{R}$ such that the function $f$ is periodic?

First of all, I tried plotting some values on Wolfram|Alpha, and for all the values of $\alpha$ that I tested, it seems that any $\alpha$ works... But I couldn't prove it.

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My attempt:

We want to study $\alpha$ such that the following statement is true:

$$\exists \,\, T > 0 \quad \forall \,x \in \mathbb{R} \quad \cos(\alpha (x + T) + \cos(x + T)) = \cos(\alpha x + \cos(x))$$

I was able to show, with some trigonometric substitutions, that this statement is equivalent to the following statement:

$$\exists \,\, T > 0 \quad \forall \,x \in \mathbb{R} \quad \exists \,\, K \in \mathbb{Z} \quad \text{such that}$$ $$\sin(x + T) = \dfrac{\alpha T - K\pi}{\sin (T)} \quad \text{or} \quad \cos(x + T) = \dfrac{K\pi - \alpha(x + T)}{\cos (T)}$$

I couldn't make any progress after that, though.

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EDIT:

Inspired by a quick comment by @ZainPatel, I was actually able to show that all $\alpha \in \mathbb{Q}$ works! It's quite simple, I am surprised I didn't try this before.

Let $\alpha \in \mathbb{Q}$, $\alpha = \dfrac{p}{q}$. Then $T = 2q\pi$ works, since

$$f(x + 2q\pi) = \cos(\alpha (x + 2q\pi) + \cos(x + 2q\pi)) = \cos(\alpha x + 2p\pi + \cos(x)) = f(x)$$

The matter is still open for irrationals though!

Answer 1

As you already proven, each $\alpha \in \mathbb Q$ works.

We show that if $f$ is periodic, then $\alpha \in \mathbb Q$:

Let $T>0$ be so so that

$$f(x+T) =f(x) $$

$$\cos(\alpha x +\alpha T + \cos(x+T))=\cos(\alpha x + \cos(x))$$

This shows that $$-2 \sin\bigg(\frac{\alpha x +\alpha T + \cos(x+T)+ \alpha x + \cos(x)}{2}\bigg) \sin\bigg(\frac{\alpha T + \cos(x+T) - \cos(x)}{2} \bigg)=0$$

Let $$A:= \{ x | \sin\bigg(\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2}\bigg) =0 \} \,;$$ $$B:=\{ x| \sin\bigg(\frac{\alpha T + \cos(x+T) - \cos(x)}{2} \bigg) =0 \}$$

Then, the above shows that $A \cup B= \mathbb R$. Moreover, by continuity both sets are closed.

It follows from here that either $A$ or $B$ contains an interval.

Case 1: $A$ contains some interval $(a,b)$.

Since $$\sin\bigg(\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2}\bigg) =0$$ for all $x \in (a,b)$ we get that $$\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2} \in \{ k\pi |k \in \mathbb Z \} $$ for all $x \in (a,b)$.

But the image of the interval $(a,b)$ under the continuous function $\frac{2 \alpha x +\alpha T + \cos(x+T) + \cos(x)}{2} $ must be connected, and hence a single point. This implies that $\alpha = 0$.

Case 2: $B$ contains some interval $(a,b)$.

Since $\sin(\frac{\alpha T + \cos(x+T) - \cos(x)}{2} ) =0$ for all $x \in (a,b)$ the same argument shows that there exists some constant $C$ so that $$ \alpha T + \cos(x+T) - \cos(x) =C \qquad \forall x \in (a,b) $$ This shows that $T$ is a period for $\cos(x)$ and hence $T=2k \pi$ for some $k \in \mathbb{Z}$.

Now, for all $x \in (a,b)$ we have by the definition of $B$ $$\sin\bigg(\frac{\alpha 2 k \pi + \cos(x+2 k \pi) - \cos(x)}{2} \bigg) =0 $$

This gives $$\sin(\alpha k \pi ) =0 $$ from which is easy to derive that $\alpha \in \mathbb Q$.

Answer 2

$\newcommand{\boxedtext}[1]{\require{enclose}\enclose{box}{\textbf{#1.}}~~}$ Define $f_\alpha(x) = \cos(\alpha x + \cos(x))$. Below, I prove the following.

Claim: The function $f_\alpha$ satisfies the following periodicity property.
(I) If $\alpha \neq 0$, then $f_\alpha$ is $2\pi L$-periodic if and only if $L, \alpha L \in \mathbb{Z}$.
(II) If $\alpha = 0$, then $f_\alpha$ is $2\pi L$-periodic if and only if $2L \in \mathbb{Z}$.

I only prove (I) below since (II) is trivial. Of course, it implies $\alpha = 0$ or $\alpha \neq 0$ in which case $L, \alpha L \in \mathbb{Z}$, so $\alpha \in \mathbb{Q} \setminus\{0\}$.

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$\boxedtext{Proof of $"\implies"$}$ Define $\delta_{\pm} = \tfrac{1}{2\pi}(\cos(2\pi L) \pm 1)$. Periodicity shows that \begin{align*} f_\alpha(2\pi L) = f_\alpha(-2\pi L) &\iff 2\alpha L\stackrel{\rm(i)}{\in} \mathbb{Z} \quad \mbox{OR} \quad 4L \stackrel{\rm(ii)}{\in} 1 + 2\mathbb{Z} , \quad \mbox{and,} \\ f_\alpha(2\pi L) = f_\alpha(0) &\iff \alpha L + \delta_{-} \in \mathbb{Z} \quad \mbox{OR} \quad \alpha L + \delta_{+} \in \mathbb{Z}. \end{align*}

$\boxedtext{Case (ii)}$ Cannot hold. Indeed $f_\alpha(2\pi k L) = f_\alpha(0)$ at $k \in \{1,2,4\}$ shows that \begin{align*} p(u) \!=\! (\sin(2\pi u)\!\sin(2\pi u + 1))^{2} \!+\! (\sin(\pi u)\!\sin(\pi u - 1))^{\!2} + \!(\sin(\tfrac{\pi u + 1}{2})\!\sin(\tfrac{\pi u -1}{2}))^{\!2} \end{align*} vanishes at $u = 2\alpha L$. However, $p > 0$ on $\mathbb{R}$.

$\boxedtext{Case (i)}$ Now, $0\in\{\delta_+, \delta_{-}\}$, so $2L \in \mathbb{Z}$.

$\boxedtext{Completing the proof}$ Set $m =2L$. So far we showed that if $$ \cos(\alpha x + \cos(x)) = \cos(\alpha (x+m\pi) + \cos(x + \pi m)) \quad \mbox{for all}~x, $$ then $m, \alpha m \in \mathbb{Z}$. Hence, $\cos(\alpha m \pi + (-1)^m) = \cos(1)$, which implies $\alpha m \in 2\mathbb{Z}$. If $m$ is odd, \begin{align*} 0 & = \cos(\alpha \pi x/2 - \cos(\pi x/2)) - \cos(\alpha\pi x/2 + \cos(\pi x/2)) \\ &= 2 \sin(\alpha \pi x/2) \sin(\cos(\pi x/2)) \quad (\mbox{for all}~x) \end{align*} Thus $\alpha = 0$, as necessary.

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$\boxedtext{Proof of $"\impliedby"$}$ Fix $L, \alpha$ such that $L, \alpha L \in \mathbb{Z}$. Then, \begin{align*} f_\alpha(x + 2\pi L) &= \cos(\alpha x + \cos(x+ 2\pi L) + 2\alpha \pi L) = \cos(\alpha x + \cos(x)) = f_\alpha(x). \end{align*}