13
$\begingroup$

Find all real polynomials $p(x)$ that satisfy $\sin( p(x) ) = p( \sin(x) )$.

Is there an easy way to prove this?

$\endgroup$
0

3 Answers 3

13
$\begingroup$

Hint:

  • $p(\sin(x))$ is periodic.

I hope this helps ;-)

$\endgroup$
8
$\begingroup$

From $\sin(p(0))=p(0)$, we get $p(0)=0$. Therefore $\sin(p(2k\pi)) = p(\sin(2k\pi)) = p(0) = 0$ for every integer $k$. In turn, $\cos(p(2k\pi))=\pm1$ and \begin{align*} &\sin(p(x)) = p(\sin(x))\\ \Rightarrow\ &p'(x) \cos(p(x)) = p'(\sin(x)) \cos(x)\\ \Rightarrow\ &p'(2k\pi) = \pm p'(0). \end{align*} Hence $p'(2k\pi)$ is bounded for all integers $k$ and $\deg(p)$ is at most $1$. Since $p(0)=0$, we have $p(x)=ax$. It remains to show that $a\in\{-1,0,1\}$. That should be easy.

$\endgroup$
0
$\begingroup$

Assume that $p\in\Bbb R[x]\space\land\space\deg (p)\ge 2$. May assume that $\lim\limits_{x\to \infty} p(x) = \infty$. Then there exists an interval $[c,\infty)$ such that $p\colon [c, \infty)\to [d, \infty)$ is bijective.

Consider the values $n\pi$ ($n\ge n_0$) in the interval $[d, \infty)$ and their preimages $x_n \in [c, \infty)$.

Now $p(\sin x_n) = \sin(p(x_n) ) = \sin (n\pi) = 0$

But we have $$x_n \to \infty \ \ \textrm{and} \\ (x_{n+1}-x_n) \to 0$$

(the last limit because $\lim\limits_{x\to \infty} p'(x) = \infty$ ), and so $(\sin x_n)_{n\ge n_0}$ is dense in $[-1,1]$.

We conclude $p\equiv 0$ on $[-1,1]$, contradiction.

Therefore, $p$ is of degree $\le 1$.

Now use $\sin(p(0)) = p(\sin 0) = p(0)$, and this implies $p(0) = 0$.

We are left with $p(x) = a x$, so $\sin a x = a \sin x$. Taking the derivative we get $a \cos a x = a \cos x$. We see from here that the solutions are $a =-1, 0, 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .