$x_{n+1}=4x_n(1-x_n)$ 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.

  • $\begingroup$ 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, 2012 at 5:54

1 Answer 1


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.

  • $\begingroup$ 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, 2012 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, 2012 at 12:56
  • $\begingroup$ The word $(10^{m-1})^\infty$ is arbitrary, right? $\endgroup$
    – Leitz
    Nov 11, 2012 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, 2012 at 13:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.