Q:

Somewhat easy chaos theory question/summation

Show that for $1 <\mu< 3$ and $\mu\neq2$, the Lyapunov exponent of the logistic map $F_\mu(x)=\mu x(1-x)$ is given by $\lambda(x)=\ln|2-\mu|$.
My work:
We have $F_\mu(x(k))=\mu x(k)(1-x(k))$. We take $|F'_\mu(x(k))|=|\mu-2\mu x(k)|$
So we have: $$\lambda(x_0)=\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\ln|\mu-2\mu x(k)|$$
This is: $$\lim_{n\to\infty}\frac1n\cdot\left(\ln|\mu-2\mu x(0)|+\ln|\mu-2\mu x(1)|+\ln|\mu-2\mu x(2)|+\ln|\mu-2\mu x(3)|+\dots\right)$$
I'm stuck evaluating the sum and I'm not sure where to go from there. Keep in mind that $x(k) $ is the $k$th iterate of the map $f$. 

A:

You don't need to evaluate this sum exactly. There is a theorem that states that for a map with a fixed point $x^*$ that is a basin of attraction 
$$\lambda=\ln |f'(x^*)|$$
In your case, for $1<\mu<3$, the fixed point of the logistic map $x^*=0$ is unstable and repulsive, while the fixed point $x^*=1-\frac{1}{\mu}$ is attractive.
Evaluating at the fixed point we obtain the quoted result:
$$\lambda=\ln|\mu(1-2(1-\frac{1}{\mu}))|=\ln|2-\mu|$$

