# Topological entropy of logistic map $f(x) = \mu x (1-x)$, $f:[0,1] \to [0,1]$ for $\mu \in (1,3)$

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.

• Under the claims, the topological entropy is $0$ by the Bowen-Dinaburg definition. – Bullet51 May 6 at 8:16
• Could you elaborate on why it's so? That's the definition I'm using. But I don't know if $NW(f) = \{0, 1- \frac{1}{\mu}$. – Matija Sreckovic May 6 at 9:09
• As to the convergence to $1-1/\mu$: it is indeed true for $0<x<1$ and $1<\mu\le 3$. Note this is the range for which $f_\mu$ has $2$ fixed points and no other $2$-periodic point. This makes the discussion easy. – Pietro Majer May 6 at 9:32
• Sorry if I'm asking an obvious question, but is there a theorem that states that, for example, $f_{\mu}^{n}(x)$ must converge to a periodic point? Why is it important that $f_{\mu}$ has no other $2-$periodic point? – Matija Sreckovic May 6 at 9:49
• @Matija Sreckovic If $0< \mu \le 3$, $f_\mu^{(2)}$ maps the interval $J:= [1/2, 1-1/\mu]$ to itself, and it is increasing. Therefore for any $x \in J$, $f_\mu^{(2n)}(x)$ converges monotonically to $1-1/\mu$. But then also $f(f_\mu^{(2n)}(x))=f_\mu^{(2n+1)}(x)$. So also $f_\mu^{(n)}(x)$. To generalize to any $x\in (0,1)$ you need to check that after some iterations $f_\mu^{(n)}(x)$ enters in $J$, which is easily seen. – Pietro Majer May 6 at 15:51