 # Is a function periodic f(x) = \cos (x) +\cos(x^2) This one Deven Livingston 2022-04-24 Answered
Is a function periodic $f\left(x\right)=\mathrm{cos}\left(x\right)+\mathrm{cos}\left({x}^{2}\right)$
This one I gave my students today, nobody solve it.
Is a following function periodic $f:\mathbb{R}\to \mathbb{R}$
$f\left(x\right)=\mathrm{cos}\left(x\right)+\mathrm{cos}\left({x}^{2}\right)$
If someone is interested I can show a solution later.
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Without using the derivative, the equation $f\left(x\right)=f\left(0\right)$ has only one solution. Indeed, if $\mathrm{cos}\left(x\right)+\mathrm{cos}\left({x}^{2}\right)=2$ then $\mathrm{cos}\left(x\right)=1=\mathrm{cos}\left({x}^{2}\right)$ 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 ; absqrt.
This may be overkill, but at least the same reasoning can prove the following : given any $\beta$-periodic function g with $\beta \in \mathbb{R}\mathrm{\setminus }\mathbb{Q}$ and such that $g\left(x\right)\ne g\left(0\right)$ for $x\in \right]0,\beta \left[$, the function $x↦g\left(x\right)+g\left({x}^{2}\right)$ is not periodic. For instance, this is also true if g is the Weierstrass function (if $b\notin 2\mathrm{ℤ}$, with the definition used by wiki).

###### Not exactly what you’re looking for? kayuukor9c
If $f\left(x\right)=\mathrm{cos}x+\mathrm{cos}\left({x}^{2}\right)$ is periodic, then so is ${f}^{\prime }\left(x\right)=-\mathrm{sin}x-2x\mathrm{sin}\left({x}^{2}\right)$, which is impossible since f'(x) is not bounded. (For , we have ${f}^{\prime }\left({x}_{n}\right)=-{\mathrm{sin}x}_{n}-2{x}_{n}\le 1-2\sqrt{2n\pi }$ which tends to $-\mathrm{\infty }$ as $n\to \mathrm{\infty }$.)