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$x_{n+1}=rx_n(1-x_n), n\in\mathbb N_0.$ I am only interested in $r=4.$

On Wikipedia I found the following statement:

Every solution $(x_n)^\infty_{n=0}\subset [0,1]$ of the recursion can be written as $x_n=\sin^2(2\pi y_n)$, with $(y_n)^\infty_{n=0}\subset [0,1)$ that satifies $$y_{n+1}=\begin{cases}2y_n & 0 \le y_n < 0.5 \\2y_n -1 & 0.5 \le y_n < 1 \end{cases}$$

How can this be shown?

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up vote 1 down vote accepted

By putting together two facts:

  • If $x_0$ is in $[0,1]$, there exists $y_0$ in $[0,1]$ such that $x_0=\sin^2(2\pi y_0)$ (and one can even assume that $y_0$ is in $[0,\frac14]$).
  • If $x_n=\sin^2(2\pi y_n)$ for some $n\geqslant0$ and some $y_n$ then $$ x_{n+1}=4\sin^2(2\pi y_n)\cos^2(2\pi y_n)=\sin^2(4\pi y_n)=\sin^2(2\pi y_{n+1}), $$ for every $y_{n+1}$ such that $y_{n+1}-2y_n$ is in $\frac12\mathbb Z$, for example, $y_{n+1}=2y_n\pmod{1}$.

Edit: As mentioned in the comments, the logistic map $L:[0,1)\to[0,1)$, $x\mapsto4x(1-x)$, is conjugate to the shift $S:\{0,1\}^\mathbb N\to\{0,1\}^\mathbb N$, $(a_n)_{n\geqslant1}\mapsto(a_{n+1})_{n\geqslant1}$, through the binary expansion $B:\{0,1\}^\mathbb N\to[0,1)$, $(a_n)_{n\geqslant1}\mapsto\sum\limits_{n\geqslant1}a_n2^{-n}$. Simply put, this means that $$ L\circ B=B\circ S, $$ and has the consequence that, for every $k\geqslant0$, $L^k$ is encoded by the identity $$ L^k\circ B=B\circ S^k, $$ where $S^k:\{0,1\}^\mathbb N\to\{0,1\}^\mathbb N$ is the shift defined by $S^k((a_n)_{n\geqslant1})=(a_{n+k})_{n\geqslant1}$.

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Thank you for your answer. What about the first part of the statement, I mean why can $x_{n}$ exactly be written as $x_n=sin^2(2\pi y_n)$ – Montaigne Oct 31 '12 at 13:32
Because $x_0$ is in $[0,1]$, which is the image set of the function $y\mapsto\sin^2(2\pi y)$. – Did Oct 31 '12 at 13:35
Thank you. I was just thinking of the following: Assume I write $y_n$ in the binary numeral system: $y_n=\sum_{k=1}^\infty a_{k,n} \cdot 2^{-k}$ for $a_{k,n}\in\{0,1\}, k\in\mathbb N$, then the recurison equation for $y_n$ should be equivalent to the recursion $a_{k,n+1}=a_{k+1,n}$ for the sequence $(a_{k,n})\subset\{0,1\}$ – Montaigne Oct 31 '12 at 13:41
Bravo! You (re)discovered the fact that the logistic map $x\mapsto4x(1-x)$ on $[0,1)$ is conjugate to the shift $(a_n)_{n\geqslant0}\mapsto(a_{n+1})_{n\geqslant0}$ on $\{0,1\}^\mathbb N$. – Did Oct 31 '12 at 13:51
I would really appreciate it if you could explain me this equivalence a little bit more in detail. – Montaigne Oct 31 '12 at 13:57

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