Logistic Map closed form for $x_{n+1}=rx_n(1-x_n), n\in\mathbb N_0$? $x_{n+1}=rx_n(1-x_n), n\in\mathbb N_0.$ 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? 
 A: 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 $S^k((a_n)_{n\geqslant1})=(a_{n+k})_{n\geqslant1}$.
