0
$\begingroup$

Consider $f:[0,2\pi) \to [0,2\pi )$ given by $f(x) = (x + 1) \bmod 2\pi$ for all $x\in [0,2\pi )$, i.e. a shift map on the unit circle with anti-clockwise shift of $1$. Denote the sequence $\{ x_n \}$ for which $x_0=0$ and $x_n$ satisfies the recurrence relation $x_{n} = f(x_{n-1})$ for each $n\in\mathbb{N}$. Moreover, denote $$ Y_N = \bigcup_{i=0}^N (x_i-\delta , x_i+\delta ) \quad \forall N\in\mathbb{N} .$$ From Jacobi's theorem (see Devaney - An introduction to chaotic dynamical systems) it follows that $\{ x_n \}$ is dense in $[0, 2\pi)$. Thus for each $\delta >0$ there exists a smallest value $N_\delta\in\mathbb{N}$ such that $[0,1]\subset Y_{N_\delta}$.

$\textbf{Question}$ Is there a reference for (or an explanation of) an estimate for $N_\delta$ as $\delta \to 0$ and if so, what is this estimate? If an estimate isn't readily available, information concerning nontrivial upper or lower bounds on $N_\delta$ as $\delta \to 0$ would be appreciated.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ Since the irrationality measure of $\pi$ is finite (and in fact less than $8$), you can bound $N_\delta \le C\delta^{-9}$ -- find $x_m$ with $|x_m| < \delta$ or $|2\pi - x_m| < \delta$ (such an $m \le \frac{2\pi}{\delta}$ by the usual pigeonhole principle), then from irrationality measure we get $|x_m|, |2\pi - x_m| \gtrsim \delta^8$, so iterating only with multiples of $m$ we cover all of $[0, 1]$ (and all of $[0, 2\pi)$) in at most $\delta^{-9}$ steps. On the other hand, $N_\delta \ge \frac{1}{\delta}$ by pigeonhole again. Are these estimates good enough for you? $\endgroup$
    – Aleksei Kulikov
    Commented Mar 12 at 23:22
  • $\begingroup$ These estimates are substantially better than I had so thank you. If better is available I'd still like to know though. $\endgroup$
    – JCM
    Commented Mar 13 at 0:17
  • $\begingroup$ I'd still be quite interested to know if a larger lower bound (in terms of order of delta) could be established. I'm mostly asking out of curiosity given I attended a seminar in which this was a (relevant to me anyway) query which was unanswered. $\endgroup$
    – JCM
    Commented Mar 13 at 13:07
  • 1
    $\begingroup$ I think better than linear lower bounds are directly related to better estimates for the terms in the continued fraction for $\pi$ and since, as far as I know, humanity does not even know that they are not bounded, I don't think currently it is possible to get a better than linear lower bound. $\endgroup$
    – Aleksei Kulikov
    Commented Mar 13 at 18:56
  • $\begingroup$ Thanks. The upper bound is nonetheless informative. I wasn't familiar with irrationality measure prior to this exchange so learning a bit about that alone has been interesting. $\endgroup$
    – JCM
    Commented Mar 13 at 19:35

0

Reset to default

You must log in to answer this question.