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 '16 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 '16 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 $2\pi T$. $\endgroup$ – Fimpellizieri Sep 7 '16 at 19:21
  • 1
    $\begingroup$ You could try addition theorems for $\cos(a+b)$ $\endgroup$ – Kaligule Sep 7 '16 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 '16 at 19:24

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 '17 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.