As stated in the question, I want to find the topological entropy of the logistic map on the interval $ [0,1]$ for a “nice” range of the parameter $ \mu$ , namely $ \mu \in (1,3)$ . I think the fact that $ f:[0,1] \to [0,1]$ is a very important additional condition here which simplifies things.

I’ve tried something, which I’m not sure is the right way to approach the problem, but I’ll outline it here anyway.

I know a theorem that states $ h_{top}(f) = h_{top}(f|_{NW(f)})$ , where $ NW(f)$ is the set of non-wandering points of $ f$ , so I wanted to find that set. By drawing a lot of little pictures, I concluded that for $ x \notin \{0,1\}$ , we should have $ \lim_{n\to \infty} f^{n}(x) = 1-\frac{1}{\mu}$ , which is the other fixed point of $ f$ . Also, the convergence seems fairly straightforward (i.e. it gets closer with every iteration), so I somehow got it into my head that I should have $ NW(f) = \{0, 1- \frac{1}{\mu}$ .

To confirm this, Wikipedia says:

By varying the parameter r, the following behavior is observed:

`With r between 0 and 1, the population will eventually die, independent of the initial population. With r between 1 and 2, the population will quickly approach the value r − 1/r, independent of the initial population. With r between 2 and 3, the population will also eventually approach the same value r − 1/r, but first will fluctuate around that value for some time. The rate of convergence is linear, except for r = 3, when it is dramatically slow, less than linear (see Bifurcation memory). `

However, I haven’t been able to find a proof of these claims. Can anyone show me how to prove this, or give me a reference where the proof is clearly written out?

Also, if there is an easier way of finding the topological entropy (again, I emphasize that $ f:[0,1] \to [0,1]$ ; I’ve lost a lot of time reading about Mandelbrot sets by conjugating $ f$ to $ g(z) = z^2 + c$ and looking at formulas for the entropy of $ g$ which exist, but with domains $ \mathbb{C}$ or some variant), I’d be very happy to hear it.