I am interested in the topological entropy of the map: $T_s(x) = min\{ sx, s(1-x)\}$, the tent map with slopes $\pm s$ and peak at $x = \frac{1}{2}$. For $s \in [1, 2]$.
When $s = 2$, this is the standard tent map and the topological entropy is log(2).
I do not need an exact answer, I am just wondering if either
a) The topological entropy is log(s)
b) The topological entropy is NOT log(s).
I have found one source that claims (without clear source) that the answer is log(s). I am inclined to believe it is probably NOT log(s), but I cannot find a clear argument for why this is the case. If you could provide any direction or intuition for either argument, that would be greatly appreciated.
Thank you in advance.
It doesn't seem to me that $[0, 1/s] \cup [1-1/s, 1]$ is a Markov partition in this case because $T_s$ is not monotone on $[1-1/s, 1]$. Wouldn't the point $x = 1/2$ also have to be included in the partition?
– curiousaboutthings Sep 13 '16 at 13:16