About the tent map and eventually fix points

Question

I need some help with this question:

Given the tent map, $T:[0,1]\to[0,1]$ defined as $T(x)=2x$ if $0\leq x\leq\frac{1}{2}$ and $T(x)=2-2x$ if $\frac{1}{2}\leq x\leq1$,

I'm trying to check that the points $\displaystyle\frac{s}{2^m}$ ($m,s\in\mathbb{N}$, with $0<s\leq2^m$) are eventually fixed points for $T$ (that is, to check if it exists some natural $n$ verifying $\displaystyle T^n(\frac{s}{2^m})=p$, where $p$ is a fixed point of $T$).

I tried computing $T^2$, $T^3$...but this seems to get me nowhere.

Thanks a lot for any help!

Answer 1

Suppose that $x_0 = \frac s{2^n}$ where $s$ is odd and $n$ is positive. Then $x_{1} = T(x_0)$ is either $\frac s{2^{n-1}}$ or $2 - \frac s{2^{n-1}} = \frac{2^{n} - s}{2^{n-1}}$, both of which are also in lowest terms because the numerator in each case is odd. So the power of 2 in the denominator must decrease by exactly 1 in each iteration, while the numerator always remains odd.

After exactly $n$ iterations the denominator has decreased to $2^0 = 1$, while the numerator $s$ is still odd. So $x_{n}$ is an odd integer. But the range of the function is $[0,1]$, so $x_{n} = 1$, and thus $x_{n+1} = 0$, which is a fixed point.