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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?

enter image description here

This is related to a similar question I asked.

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1 Answer

up vote 4 down vote accepted

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.

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Interestingly enough, you can use repeated doubling for numerically computing the cosine: start with some approximation like the Maclaurin series evaluated on an argument halved for the appropriate number of times, and then apply the doubling recursion the same number of times the original argument was halved. One could equivalently view this iteration with Chebyshev polynomials... –  J. M. Sep 15 '11 at 2:51
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