The question: Show that if $ x= \frac{k}{2^{n}}$ where k and n are positive integers with $ 0 < \frac{k}{2^{n}} <1 $, then x is eventually a fixed point of the tent map.
My Attempt:
If you recall, the tent map is this function:
$$T(x) = \begin{cases} 2x & 0<x< 1/2\\ 2(1-x) & 1/2\leq x\leq 1 \\ \end{cases} $$
I tried proving this by induction. For the base case, I can ket n=1 which would mean that $x = \frac{k}{2}$. From the immediate value theorem, since the tent map is a continuous map, it means for any real number, r, between f(a) and f(b), there exists a real number, c, such that f(c) = r. Since if we pick $x = \frac{k}{2}$ between the interval f(a) and f(b), we could pick a value in the domain which we can make x a fixed point.
Am I really on the right track so far? If I am, could I possibly use this theorem to prove for the induction step?
Thanks a lot. I really appreciate your help.
Consider $x = \sum_{k = 1}^{\infty} \omega_{k}(x) 2^{-k}, \omega_{k} \in \{ 0, 1 \}$, i.e. let $\omega_{k}$ be the $k$th binary digit of $x$. Then \begin{align*} T(x) & = \begin{cases} \sum_{k = 1}^{\infty} \omega_{k + 1}(x)2^{-k} & (\omega_{1}(x) = 0), \\ 1 - \sum_{k = 1}^{\infty} \omega_{k + 1}(x)2^{-k} & (\omega_{1}(x) = 1). \end{cases} \end{align*} Note, however, that $1 - \sum_{k = 1}^{\infty} \omega_{k + 1}(x)2^{-k} = \sum_{k = 1}^{\infty} \overline{\omega_{k + 1}(x)}2^{-k}$, where $1 = \overline{0}, 0 = \overline{1}$. To confirm this, note that $\sum_{k = 1}^{\infty} (\omega_{k}(x) + \overline{\omega_{k}(x)})2^{-k} = \sum_{k = 1}^{\infty} (1) 2^{-k} = 1$.
Now if $x = K/2^{N}$ for some $K \in \{0, 1, \ldots, 2^{N} \}$, then for $k > N$, we have that $\omega_{k}(x) = 0$. This means that $T^{N}(x)$ is either $\sum_{k = 1}^{\infty} (0) 2^{-k} = 0$ or $\sum_{k = 1}^{\infty} (1)2^{-k} = 1$. In either case, $T^{N + 1}(x) = 0$.