Logistic map periodic point - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:32:51Z http://mathoverflow.net/feeds/question/112050 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112050/logistic-map-periodic-point Logistic map periodic point Leitz 2012-11-11T01:41:55Z 2012-11-11T02:46:17Z <p>$x_{n+1}=4x_n(1-x_n)$ I already proved that for $x_n\subset [0,1]$, $x_n=sin^2(2\pi y_n)$</p> <p>with $y_{n+1}=\begin{cases}2y_n &amp; 0 \le y_n &lt; 0.5 \vee 2y_n -1 &amp; 0.5 \le y_n &lt; 1 \end{cases}$</p> <p>Now I would like to prove that for an arbitrary number $m\in\mathbb N$ there exists an $x\in [0,1]$ of the recursion with period lenght m</p> <p>I think it can be shown using the fact that, if I write $y_n$ in the binary system as $y_n=\sum_{k=1}^\infty a_{k,n}2^{-k}$ the recursion for $y_n$ equivalent to $a_{k,n+1}=a_{k+1,n}$ is, but I dont know how. </p> http://mathoverflow.net/questions/112050/logistic-map-periodic-point/112054#112054 Answer by Nikita Sidorov for Logistic map periodic point Nikita Sidorov 2012-11-11T02:46:17Z 2012-11-11T02:46:17Z <p>Essentially you have proved that the logistic map is conjugate to the doubling map $Tx=2x\bmod 1$. Now, $T$ is in turn conjugate to the shift map $\sigma:\Sigma\to\Sigma$, where $\Sigma$ is the space of infinite 0-1 words. </p> <p>More precisely, if $$\pi(w_1,w_2,\dots)=\sum_{n=1}^\infty w_n2^{-n},$$ then you have $$\pi \sigma = T\pi.$$ Since $\pi$ is 1-1, except for a countable set of finite words, you can just take any word $u$ of length $m$ and the corresponding infinite word $w=uuuu\dots$. Clearly, $\sigma^m(w)=w$, i.e., $w$ is $\sigma$-periodic of period $m$. (With a little effort you can make this the <em>smallest</em> period.) </p> <p>For instance, $w=001001001\dots$ is of period 3. </p> <p>Now take $x:=\pi(w)$; it is $T$-periodic of period $m$. Finally, use you conjugate map to turn $x$ into an $m$-periodic point for the logistic map. </p>