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The Lyapunov exponent for logistic map $x_{n+1}=f(x_n) $ where f is given by $f(x)=bx(1-x)$ is given by: $$\lambda = \frac{1}{n}\sum^n_i log(b(1-2x_i))$$ where the index $i$ refers to $i$th iterate

http://demonstrations.wolfram.com/LyapunovExponentsForTheLogisticMap/

I read that $\lambda$ is constant for a fixed value of $b$. I want to ask why does it not depend on the seed value say $x_1$, since for any value of $b$, I can start with any value of $x_1$ between 0 and 1?

zed111
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    the way you define $\lambda$ doesn't make sense. What's $n$ ? is that an infinite sum ? shouldn't there be a limit somewhere ? – mercio Aug 30 '14 at 07:44
  • http://demonstrations.wolfram.com/LyapunovExponentsForTheLogisticMap/ – zed111 Aug 30 '14 at 08:00
  • How is the link related to the comment? – Did Aug 30 '14 at 08:02
  • It is the link which has the formula for Lyapunov Exponent – zed111 Aug 30 '14 at 08:03
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    The way I've seen it defined definitely includes taking a limit on $n$. The orbit of any one seed looks pretty much like the orbit of any other, on average over the long run, so when you take that limit, it doesn't matter what you chose as $x_1$. Well, you have to avoid periodic points, but with finite precision machines you're not likely to hit one. – Gerry Myerson Aug 30 '14 at 09:05

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