TITLE: converging to cosine by iteration
QUESTION [6 upvotes]: In what sense (if at all) does the iteration $x \mapsto 2x^2 - 1$ converge to $\cos 2^n x$ in the unit interval [-1,1]?
One might try to plot in Mathematica:
y = x;
Table[y = 2 y^2 - 1; y, {k, 1, 5}]
Plot[%, {x, -1, 1}]

And you get an some plots resembling cosines of increasing frequency.  Can this be made rigorous?

This is related to a similar question I asked.

REPLY [4 votes]: The recurrence $x_{n+1}=2x_n^2-1$ with $x_0$ in $[-1,1]$ can be translated to $x_0=\cos \alpha_0$ and $\cos\alpha_{n+1}=\cos 2\alpha_n=...=\cos 2^n \alpha_0$.
You cannot say that the iteration converges to $\cos 2^n \alpha_0$, because convergence would mean taking the limit as $n \to \infty$, and for $\alpha_0 \neq 0$ that limit may not exist.
Still, the recurrence means doubling the argument in a cosine function.