Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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


Can you take it from here?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.