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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.

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    $\begingroup$ Quite a beautiful function... $\endgroup$
    – TheSimpliFire
    Feb 16, 2018 at 15:39
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    $\begingroup$ $f(ix)=\cos(ix)+\cos(x^2)$ would also be periodic. $\endgroup$
    – user530891
    Feb 16, 2018 at 15:52
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    $\begingroup$ @TheSimpliFire yeah, looking at it, I agree :P $\endgroup$
    – Mr Pie
    Sep 30, 2018 at 15:53

2 Answers 2

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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).

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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 $n\to \infty$.)

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  • $\begingroup$ But tan is unbounded and periodic function. I don't understand your argument. $\endgroup$
    – nonuser
    Feb 17, 2018 at 6:46
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    $\begingroup$ @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 $[0, T]$. $\endgroup$
    – Seewoo Lee
    Feb 17, 2018 at 12:31

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