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I am attempting to work through the following problem. I have no problem with part (a) and have included it only for context.

(a) Verify that $u_n = \sin^2(2^n)$ is a solution of the map $u_{n+1}=4u_n(1-u_n)$.

(b) Calculate the Lyapunov exponent (LE) of that solution.

I have the beginnings of a working of part (b) but I can't find any trig examples of calculating the LE in any notes. What I have so far is:

$L = \lim_{N\to\infty} \frac{1}{N} \sum\limits_{n=1}^N \ln|4\cos(2^{n+1})| \\ = \lim_{N\to\infty} \frac{1}{N} \left[ 2N\ln(2) + \ln|\cos(2^2)\cos(2^3)\dots \cos(2^{N+1})| \right] $

I could argue, due to $\cos$ being bounded, that $L \leq 2\ln(2)+1$ but that doesn't tell me if $L$ is negative or not. Maybe I'm overcomplicating. Any help would be very much appreciated.

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1 Answer 1

The following identity can be established by induction and/or the identity $\sin(2x)=2\sin(x)\cos(x)$:

$$\prod_{j=0}^{k-1}\cos(2^jx)=\frac{\sin(2^kx)}{2^k\sin(x)}.$$

Can you take it from here?

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