Is a following function periodic $f(x) =\tan x +\tan (\sqrt{2}x)$

Question

Motivated with $\cos(x)+\cos(x\sqrt{2})$ is not periodic I found this question interesting:

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

The approach in that link can not help. Any ideas? Also from a graph I draw in Geogebra I can not rule out it is not periodic.

Answer 1

It's easy to show that a differentiable function $f$ is periodic only if $f'$ is periodic.

So

$$f'(x)=1+\tan^2(x)+\sqrt 2+\sqrt 2\tan^2(\sqrt 2 x)$$

If $f'$ has a period $T>0$ then

$$1+\sqrt 2=1+\tan^2(T)+\sqrt 2+\sqrt 2\tan^2(\sqrt 2T)$$

This implies that $\tan T$ and $\tan(\sqrt 2 T)$ are both $0$, which is impossible for the same reason as for $\cos$.

Answer 2

HINT.-$\tan(x)$ has period $\pi$ and $\tan(\sqrt2 x)$ has period $\dfrac{\pi}{\sqrt2}$. How to display a period of $\tan(x)+\tan(\sqrt2 x)$ with the irrational $\sqrt2$ in place? You need for this two integers $n,m$ such that $$n\pi=\dfrac{m\pi}{\sqrt2}$$