Tent map: show that $x$ is a periodic point IFF it is a rational number of the form $\frac {m}{p}$ where $m$ is even and $p$ is odd

Question

Let us consider the tent map: $f [0,1] \rightarrow [0,1]$ where $f(x) = 2x$ if $0\leq x \leq \frac{1}{2}$ and $f(x) = 2(1-x)$ if $\frac{1}{2}\leq x \leq 1$.

Show that $x$ is a periodic point IFF it is a rational number of the form $\frac {m}{p}$ where $m$ is even and $p$ is odd.

I have a proof of this here, but it seems overly complicated with Euler's functions and such. So I was wondering whether any of you can provide a simpler proof that is more natural and avoids using specialized functions.

Answer 1

There is a shorter proof if you are willing to concede certain facts about binary expansions.

Let $D\colon[0,1]\to[0,1]$ be the doubling map: $$ D(x) \;=\; \begin{cases}2x & \text{if }x < 1/2, \\ 2x-1 & \text{if }x \geq 1/2.\end{cases} $$ Then $D$ acts as a bitwise left shift on the binary digits of a number. In particular, a point is periodic under $D$ if and only if its binary expansion is repeating, i.e. if and only if it is a rational number with odd denominator. (We will assume this bit of knowledge about binary expansions.)

Now, observe that $T\circ D = T \circ T$, where $T$ is the tent map. (This is a semiconjugacy between $T$ and $D$.) Then $T\circ D^n = T^{n+1}$ for all $n$, which makes it easy to classify periodic points for $T$.

Theorem. Let $p\in[0,1]$. Then $p$ is periodic under $T$ if and only if $p$ is a rational number with even numerator and odd denominator.

Proof: Suppose first that $p$ is a fraction with odd denominator and even numerator. Then $p/2$ is a fraction with odd denominator, so $D^n(p/2) = p/2$ for some $n$. Then $$ T^n(p) \;=\; T^{n+1}(p/2) \;=\; T(D^n(p/2)) \;=\; T(p/2) \;=\; p $$ so $p$ is periodic under $T$.

For the converse, suppose that $p$ is periodic under $T$. Then $T^n(p) = p$ for some $n$. Since $D(p/2) = p$, it follows that $$ T(D^n(p/2)) = T(D^{n-1}(p)) = T^n(p) = p = T(p/2). $$ Therefore, either $D^n(p/2) = p/2$, or $D^n(p/2) = 1-(p/2)$.

• If $D^n(p/2) = p/2$, then $p/2$ is periodic under $D$. Thus $p/2$ is a fraction with odd denominator, so $p$ is a fraction with odd denominator and even numerator.

• If $D^n(p/2) = 1-(p/2)$, then $D^{2n}(p/2) = p/2$ (since $D(1-x) = 1-D(x)$ for all $x$), so again $p/2$ is periodic under $D$, and hence $p$ is a fraction with odd denominator and even numerator.

Q.E.D.

Of course, this sweeps most of the number theory details under the rug of binary expansions, but even if you aren't content to assume the required fact, it's probably slightly easier to sort through the number theory of binary expansions and then apply the results to tent map than to work through everything at once.