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$x_{n+1}=4x_n(1-x_n)$ I already proved that for $x_n\subset [0,1]$, $x_n=sin^2(2\pi y_n)$

with $y_{n+1}=\begin{cases}2y_n & 0 \le y_n < 0.5 \\vee 2y_n -1 & 0.5 \le y_n < 1 \end{cases}$

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

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.

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  • $\begingroup$ Not really a research question --- the kind of thing I teach in my 3rd-year undergrad class on discrete dynamical systems --- would fit better on math.stackexchange.com $\endgroup$ Nov 11 '12 at 5:54
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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.

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 smallest period.)

For instance, $w=001001001\dots$ is of period 3.

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.

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  • $\begingroup$ Thanks four your answer, so I can turn x into an m-periodic point by setting $sin^2(2\pi 2^m y)=sin^2(2\pi y)$ ?? $\endgroup$
    – Leitz
    Nov 11 '12 at 12:26
  • $\begingroup$ The word $(10^{m-1})^\infty$ is $m$-periodic for $\sigma$, whence $x=0.5/(1-2^{-m})$ is $m$-periodic for $T$. If your formula is correct I haven't checked it), then $\sin^2(\pi/(1-2^{-m}))$ is your guy. $\endgroup$ Nov 11 '12 at 12:56
  • $\begingroup$ The word $(10^{m-1})^\infty$ is arbitrary, right? $\endgroup$
    – Leitz
    Nov 11 '12 at 13:05
  • $\begingroup$ Right. Note however that for this word we know for sure that it is of period $m$ and not of any $k<m$. $\endgroup$ Nov 11 '12 at 13:13

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