# Is a function periodic $f(x) = \cos (x) +\cos(x^2)$

This one I gave my students today, nobody solve it.

Is a following function periodic $f:\mathbb{R}\to\mathbb{R}$ $$f(x) = \cos (x) +\cos(x^2)$$

If someone is interested I can show a solution later.

• Quite a beautiful function... Feb 16 '18 at 15:39
• $f(ix)=\cos(ix)+\cos(x^2)$ would also be periodic.
– user530891
Feb 16 '18 at 15:52
• @TheSimpliFire yeah, looking at it, I agree :P Sep 30 '18 at 15:53

Without using the derivative, the equation $$f(x) = f(0)$$ has only one solution. Indeed, if $$\cos(x)+\cos(x^2)=2$$ then $$\cos(x) = 1 = \cos(x^2)$$ so there exists $$p,q \in \mathbb{Z}$$ such that $$x = 2p \pi$$ and $$x^2 = 2q \pi$$, so $$\pi = \frac{q}{2 p^2}$$. But $$\pi$$ is not rational ; absurd.
This may be overkill, but at least the same reasoning can prove the following : given any $$\beta$$-periodic function $$g$$ with $$\beta \in \mathbb{R} \backslash \mathbb{Q}$$ and such that $$g(x) \neq g(0)$$ for $$x \in ]0,\beta[$$, the function $$x \mapsto g(x)+g(x^2)$$ is not periodic. For instance, this is also true if $$g$$ is the Weierstrass function (if $$b\notin 2\mathbb{Z}$$, with the definition used by wiki).
If $f(x)=\cos x+\cos(x^{2})$ is periodic, then so is $f'(x) = -\sin x -2x\sin (x^{2})$, which is impossible since $f'(x)$ is not bounded. (For $x_{n} = \sqrt{(2n+1/2)\pi}$ ($n>0$), we have $f'(x_{n})=-\sin x_{n}-2x_{n}\leq 1-2\sqrt{2n\pi}$ which tends to $-\infty$ as $n\to \infty$.)
• @ChristianF You are right, but $\tan$ is not defined on the whole $\mathbb{R}$. Actually, any continuous periodic function $f(x)$ defined on $\mathbb{R}$ with period $T$ should be a bounded function, since it is bounded on the compact set $[0, T]$. Feb 17 '18 at 12:31