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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.

nonuser
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2 Answers2

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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 $\cos$.

ajotatxe
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    The “if” part isn't quite true (it requires that the integral of $f'$ over one period is zero), but “only if” is true, which is all you need in this case. – Hans Lundmark Feb 17 '18 at 12:39
  • Yeah, I was thinking about functions defined over $\Bbb R$, which is not the case. Fixed. – ajotatxe Feb 17 '18 at 13:32
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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 $$n\pi=\dfrac{m\pi}{\sqrt2}$$

Ataulfo
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