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Does $x_0=1/3$ lead to periodicity in the logistic map $x_{k+1}=4x_k(1-x_k)$?

I believe it does not, but this is equivalent to proving that $(2\pi)^{-1}\arcsin(\sqrt{1/3})$ is irrational. I am wondering if there are any known results. After all, $1/3$ is the most rudimentary seed that (I suspect) leads to non-periodicity in the fully chaotic logistic map. Or maybe not?

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  • $\begingroup$ I believe it does not lead to periodicity, but this nothing more than a belief. If it is periodic, the period must be gigantic (my guess). $\endgroup$
    – Vincent Granville
    Commented Feb 27, 2023 at 9:33
  • $\begingroup$ Thank you Loïc for noticing my typo. I corrected it. $\endgroup$
    – Vincent Granville
    Commented Feb 27, 2023 at 16:00

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The orbit of $1/3$ is infinite. You can show this via the $3$-adic valuation $\nu_3$.

Let us show by induction that $\nu_3(x_n) = -2^{n}$: We have $\nu_3(x_0) = \nu_3(1/3) = -1 = -2^0$.

Now $\nu_3(x_{n+1}) = \nu_3(4 x_{n} (1 - x_{n})) = \nu_3(4)+ \nu_3(x_n) + \nu_3(1 - x_n)$.

We have $\nu_3(4) = 0$, and since $\nu_3(x_n)$ is negative by induction, we have $\nu_3(1 - x_n) = \nu_3(x_n)$, so overall $\nu_3(x_{n+1}) = 2\nu_3(x_n) =-2^{n+1}$.

From this is is clear that the orbit must be infinite.

A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.

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    $\begingroup$ That's all correct; it can also be explained in more elementary (and general) fashion: if x is rational with odd denominator q, then 4x(1-x) has denominator q^2, so by induction the n-th iterate has denominator q^(2^n), and in particular x is never periodic once q>1. $\endgroup$
    – Noam D. Elkies
    Commented Feb 27, 2023 at 14:42
  • $\begingroup$ I may be navie, but how to ensure that $x_{lr}, l=1,2....$ for some $r$ (period) don't tend to an irrational solution of $f^{(r)}(x)=x$ where $f(x)=4x(1-x)$? $\endgroup$
    – Alapan Das
    Commented Feb 27, 2023 at 18:05
  • $\begingroup$ @AlapanDas Even if they do converge to such a solution, that doesn't mean that the sequence 'leads to' periodicity in the sense OP means — they're specifically asking (AFAICT) if 1/3 is a preperiodic point. $\endgroup$
    – Steven Stadnicki
    Commented Feb 27, 2023 at 20:15
  • $\begingroup$ @StevenStadnicki Oh, I see. Thank you for clearing my doubt. $\endgroup$
    – Alapan Das
    Commented Feb 28, 2023 at 4:02
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It is straightforward to show that $(2\pi)^{-1}\arcsin(\sqrt{1/3})$ is irrational. Indeed, if this equals a rational number $r\in\mathbb{Q}$, then $2\sin(2\pi r)=\sqrt{4/3}$. However, $2\sin(2\pi r)$ is a sum of two roots of unity, hence an algebraic integer, while $\sqrt{4/3}$ is not.

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