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.

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.


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!

  • $\begingroup$ Doesn't $T = 2\pi $ work? $\cos (\alpha x + 2\pi T + \cos (x + 2\pi)) = \cos (\alpha x + \cos x)$? $\endgroup$
    – Zain Patel
    Sep 7, 2016 at 19:20
  • 1
    $\begingroup$ Try writing $f(x) = \cos(\alpha x) \cos (\cos x) - \sin(\alpha x)\sin(\cos x)$. $\endgroup$
    – Umberto P.
    Sep 7, 2016 at 19:20
  • 1
    $\begingroup$ @ZainPatel Notice we'd have $f(x+2\pi)=\cos(\alpha x + 2\pi\alpha + \cos(x))$; ie, that's $2\pi\alpha$ rather than $2\pi T$. $\endgroup$ Sep 7, 2016 at 19:21
  • 1
    $\begingroup$ You could try addition theorems for $\cos(a+b)$ $\endgroup$
    – Kaligule
    Sep 7, 2016 at 19:23
  • $\begingroup$ @ZainPatel, thanks. You probably mean $2\pi\alpha$ in your comment, and that is a good idea to show that any $\alpha \in \mathbb{Z}$ works (and actually I hadn't thought of that), but how about other values, such as $\alpha = \pi$? $\endgroup$
    – Pedro A
    Sep 7, 2016 at 19:24

1 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 $\alpha \in \mathbb Q$.

  • $\begingroup$ Impressive, thank you very much and sorry to take so long to give feedback. $\endgroup$
    – Pedro A
    Oct 14, 2017 at 17:25

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