Fast way to find period-n points of a tent map?

Question

I need to find a period-3 orbit for a tent map defined by $$T(x)=\begin{cases}2x & 0\le x\le0.5 \\[2ex] 2-2x &0.5 < x \le 1 \end{cases}$$

So, what is a fast and efficient way of finding it? I only know how to do it by guessing-and-checking. For example, I picked a random point, $x = \frac{2}{7}$ and found that $$T\left(\frac{2}{7}\right)=\frac{4}{7}\\[4ex]T\left(\frac{4}{7}\right)=\frac{6}{7}\\[4ex]T\left(\frac{6}{7}\right)=\frac{2}{7}$$ which is a period-3 orbit. Is there any better way of finding a period-3 orbit , more specifically, a period-3 point?

Answer 1

Every real number $x$ in $[0,1]$ can be written (almost) uniquely as $$x=\frac12-\frac12\sum_{n=1}^\infty\frac1{2^n}b_n(x)$$ where, for each $n\geqslant1$, $b_n(x)$ is either $+1$ or $-1$. Then $$b_n(Tx)=b_1(x)b_{n+1}(x)$$ for every $n\geqslant1$ (can you check this?) hence $$b_n(T^kx)=b_k(x)b_{n+k}(x)$$ for every $n\geqslant1$ and $k\geqslant1$. In particular, $T^3x=x$ means that $$b_{n+3}(x)=b_3(x)b_n(x)$$ for every $n\geqslant1$. Such numbers $x$ are determined by the triple $(b_1(x),b_2(x),b_3(x))$ since then, $$b_{3n}(x)=b_3(x)^n\quad b_{3n+1}(x)=b_3(x)^nb_1(x)\quad b_{3n+2}(x)=b_3(x)^nb_2(x)$$ for every $n$ (can you check this?). Then, $$x=\frac12-\frac12\sum\limits_{n=0}^\infty\left(\frac1{2^{3n+1}}b_3(x)^nb_1(x)+\frac1{2^{3n+2}}b_3(x)^nb_2(x)+\frac1{2^{3n+3}}b_3(x)^{n+1}\right)$$ that is, $$x=\frac12-\frac12\left(\frac1{2}b_1(x)+\frac1{2^2}b_2(x)+\frac1{2^3}b_3(x)\right)\sum\limits_{n=0}^\infty\frac1{2^{3n}}b_3(x)^n$$ or, $$x=\frac12-\frac12\frac{4b_1(x)+2b_2(x)+b_3(x)}{8-b_3(x)}=1-\frac{4+2b_1(x)+b_2(x)}{8-b_3(x)}$$ Plugging each $(b_1(x),b_2(x),b_3(x))$ in $\{-1,+1\}^3$ into this yields that the solutions of the equation $T^3x=x$ are $$0\quad\frac27\quad\frac47\quad\frac67\quad\frac29\quad\frac49\quad\frac69\quad\frac89$$ thus the orbits of length $3$ are exactly $$\frac27\to\frac47\to\frac67\to\frac27\qquad\frac29\to\frac49\to\frac89\to\frac29$$ Edit: Likewise, the $2^k$ solutions of $T^kx=x$ are $$x=1-\frac1{2^k-b_k(x)}\left(2^{k-1}+\sum_{i=1}^{k-1}2^{k-1-i}b_i(x)\right)$$ or equivalently, $$x=\frac{2n}{2^{k}+s}$$ where $s=\pm1$ and $0\leqslant2n\leqslant 2^k+s$.

Answer 2

Split the domain in to L $=[0,0.5]$ and R $=(0.5,1]$. The section of the given periodic orbit $2/7\to4/7\to6/7$ can be represented by the sequence LRR (as $2/7$ is in L etc.). Sequences RLR and RRL are equivalent to LRR by just shifting along the periodic orbit.

For a sequence LRR, we start with an $x$ in L, so $T(x)=2x$, which we have specified is in R. Then $T^2(x)=T(2x)=2-2(2x)=2-4x$, which is also in R. So $T^3(x)=T(2-4x)=2-2(2-4x)=8x-2$. Now, for a period-3 orbit, $T^3(x)=x$, i.e. $8x-2=x$ which gives uniquely $x=2/7$.

Let's try LLR. For the first two iterations, $x$ just doubles, so $T^2(x)=4x$. Next, $4x$ is in R, so $T^3(x)=T(4x)=2-2(4x)=2-8x$. For a period-3 orbit, $T^3(x)=2-8x=x$ implies uniquely $x=2/9$, giving the periodic orbit $2/9\to 4/9\to 8/9\to 2/9\to\dots$

Applying the same procedure for RRR, we find uniquely $x=2/3$, but $T(2/3)=2/3$, so we have found a fixed point of $T$, rather than a period-3 orbit.

Finally, the last possible sequence, LLL gives $T^3(x)=8x$ which equals $x$ only at $x=0$.

(When we can label sections of the domain with letters, and identify trajectories by their letter sequence, the approach is called 'symbolic dynamics.')