For what $p(x) \in \mathbb{R}[x]$ is the function $\sin(p(x))$ periodic?
It seems obvious to me that all linear polynomials satisfy the condition, but I can not prove it and I do not know what other functions can satisfy it.
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$\begingroup$ Hello and welcome to Maths Stack Exchange. Could you show us your thoughts about the question? $\endgroup$– WyllichCommented Sep 21, 2017 at 14:29
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$\begingroup$ it seems obvious to me that all linear polynomials satisfy the condition, but I can not prove it and do not know what other functions can satisfy it $\endgroup$– Luk17Commented Sep 21, 2017 at 14:30
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1$\begingroup$ @Luk17 why is it obvious. Have you tried using $p(x)=x^2$ as an experiment? What about $p(x)=\sin(x)$? $\endgroup$– Stella BidermanCommented Sep 21, 2017 at 14:35
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$\begingroup$ for linear and periodic $p(x)$ $\endgroup$– VasiliCommented Sep 21, 2017 at 14:36
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$\begingroup$ I tried using $p(x)=x^2$ and it isn't periodic. $\endgroup$– Luk17Commented Sep 21, 2017 at 14:40
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1 Answer
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The key idea is the following :
Let $f \, : \, \mathbb{R} \, \rightarrow \, \mathbb{R}$ be a periodic function with period $T$.
If $f$ is differentiable on $\mathbb{R}$, then $f'$ is periodic with period $T$.
If $\forall t \in \mathbb{R}, \; f(t) = \sin\big( P(t) \big)$ with $P \in \mathbb{R}[X]$, $f$ is differentiable on $\mathbb{R}$ and: $$\forall t \in \mathbb{R}, \; f'(t) = P'(t) \cos\big( P(t) \big). $$
Therefore, $f'$ is bounded on $\mathbb{R}$ if and only if $P'$ is. This implies that $\deg(P) \leq 1$.
Conversely, if $\deg(P) \leq 1$, $P'$ is bounded on $\mathbb{R}$.
