# For which $P \in \mathbb{R}[X]$ is $\sin \circ P$ a periodic function?

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.

• Hello and welcome to Maths Stack Exchange. Could you show us your thoughts about the question? Sep 21, 2017 at 14:29
• 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 Sep 21, 2017 at 14:30
• @Luk17 why is it obvious. Have you tried using $p(x)=x^2$ as an experiment? What about $p(x)=\sin(x)$? Sep 21, 2017 at 14:35
• for linear and periodic $p(x)$ Sep 21, 2017 at 14:36
• I tried using $p(x)=x^2$ and it isn't periodic. Sep 21, 2017 at 14:40

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 $$\mathbb{R}$$.