TITLE: Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.
QUESTION [10 upvotes]: Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.

My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^n)=U_{k^n}(\sin(x))=U_k(u_{n-1})$. I thought of introducing a variant of Chebyshev polynomials $U_n(\sin(x))=\sin(nx)$. But it didn't help much. Does someone have a hint? Thanks. (here $\Bbb N = \{0,1,2,\ldots\}$.)

I needed this result for solving this exercice (which is a oral exam that was posed during the competitive exams of Polytechnique, #1 engineering school in France).

Show that for $f(z)=z^k$ with $k \ge 2$
$$
\exists \delta >0, \forall x,y \in \Bbb U, x \neq y \implies \exists n \in \Bbb N, |f^{(n)}(x)-f^{(n)}(y)| \ge \delta.
$$

Suppose $x \neq 0$ and $y/x=e^{i \theta}$,
\begin{align}
| f^{(n)}(x) - f^{(n)}(y) | &= |x^{k^n} - y^{k^n}|
=
 | 1- e^{i \theta k^n}|\\
&= | e^{-i \theta k^n/2} - e^{i \theta k^n/2}|
= |2i \sin(\theta k^n/2)| = 2|\sin (\theta k^n/2)|
\end{align}
So
$$
|f^{(n)}(x)-f^{(n)}(y)| \ge \delta \iff |\sin (\theta k^n/2)| \ge \delta/2
$$

REPLY [6 votes]: I only really see one way to solve this problem, though how you write the details can vary considerably. You asked for an "elegant" solution, so here's my arrangement.
Key Lemma: Let $y \in (0, 1)$ and $k \in \mathbb{Z}_{\geq 2}$. Then
$$\{yk^n \pmod{1} : n \in \mathbb{Z}_{\geq 0}\} \cap [1/k^2, 1-1/k^2] \neq \varnothing.$$
Proof: Consider the base $k$ expansion of $y$, so $y = 0{.}y_1 y_2 y_3{\ldots}$ and $yk^n \pmod{1} = 0{.}y_{n+1} y_{n+2}{\ldots}$. If the latter expression is not in $[1/k, 1-1/k]$, we must have $y_{n+1} = 0$ or $y_{n+1} = k-1$. So, we may suppose that $y_{n+1} \in \{0, k-1\}$ for all $n \in \mathbb{Z}_{\geq 0}$. Since $y \neq 0, 1$ by assumption, we have some $N$ where $y_{N+1} \neq y_{N+2}$. Hence $yk^N \pmod{1} = 0{.}y_{N+1} y_{N+2}{\ldots} = 0{.}0(k-1){\ldots}$ or $0{.}(k-1)0{\ldots}$. In the first case, $yk^N \pmod{1} \in [1/k^2, 1/k]$, while in the second case, $yk^N \pmod{1} \in [1-1/k, 1-1/k^2]$. $\Box$
Now set $\delta = \sin(\pi/k^2) > 0$. For all $x \in (0, \pi)$, by the key lemma applied to $y = x/\pi$, we have some $n \in \mathbb{Z}_{\geq 0}$ such that $xk^n \pmod{\pi} \in [\pi/k^2, \pi-\pi/k^2]$. From the unimodality and symmetry of $\sin$, it follows that $|\sin(xk^n)| \geq |\sin(\pi/k^2)| = \delta$.