Rational numbers with dense orbits in [0,1] under iteration by f(x)=4x(1-x) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:46:07Z http://mathoverflow.net/feeds/question/27573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27573/rational-numbers-with-dense-orbits-in-0-1-under-iteration-by-fx4x1-x Rational numbers with dense orbits in [0,1] under iteration by f(x)=4x(1-x) coudy 2010-06-09T12:17:11Z 2010-06-09T14:52:42Z <p>Let $f(x)=4x(1-x)$.</p> <p>For which rational numbers $r\in [0,1]$ is the sequence $f^n(r)$, $n\in \mathbb N$, dense in $[0,1]$ ? $(f^n(r)=f\circ f\circ ...\circ f(r)$ n times)</p> <p>I would be happy to find a single rational number with dense orbit in $[0,1]$, but my guess is that all rational numbers different from k/2^n should work. The numbers 1/3, 1/5, 1/10 are candidates (at least numerically).</p> <p>It is known that a.e. point x in [0,1] has a dense orbit (w.r.t the Lebesgue measure). This is shown by conjugating f to x->2x mod 1 with the map $x\rightarrow sin^2(\pi x/2)$. So the question can be rephrased as: for which rational r does ${1\over \pi} Arcsin(r)$ have a dense orbit in [0,1] under the action of x->2x mod 1 ? (this does not seem simpler though)</p> <p><strong>EDIT</strong>: from W. Zudilin answer, it seems that the question is open in full generality. But maybe there is a chance of finding just <em>one</em> rational $r$ such that $f^n(r)$, or $Frac({2^n\over \pi} Arcsin\ r )$, is dense in [0,1] ?</p> http://mathoverflow.net/questions/27573/rational-numbers-with-dense-orbits-in-0-1-under-iteration-by-fx4x1-x/27576#27576 Answer by Wadim Zudilin for Rational numbers with dense orbits in [0,1] under iteration by f(x)=4x(1-x) Wadim Zudilin 2010-06-09T12:47:10Z 2010-06-09T12:47:10Z <p>If I understand your question correctly, you ask about the trajectory of $\lbrace\xi\alpha^n\rbrace$ for $\xi\in\mathbb Q$ and $\alpha=2$.</p> <p>Here is the abstract of [A. Dubickas, ON THE FRACTIONAL PARTS OF THE NATURAL POWERS OF A FIXED NUMBER, <em>Siberian Mathematical Journal</em> <strong>47</strong> (2006) 879-–882]: </p> <blockquote> <p>Let $\xi\ne0$ and $\alpha > 1$ be reals. We prove that the fractional parts $\lbrace\xi\alpha^n\rbrace$, $n = 1, 2, 3, \dots$, take every value only finitely many times except for the case when $\alpha$ is the root of an integer: $\alpha = q^{1/d}$, where $q \ge 2$ and $d \ge 1$ are integers and $\xi$ is a rational factor of a nonnegative integer power of $\alpha$.</p> </blockquote> <p>It is not the complete answer to your question, but you will find all relevant references about the distrubution of the fractional parts. These problems are on the market for a long time, I even played this game myself some years ago.</p> http://mathoverflow.net/questions/27573/rational-numbers-with-dense-orbits-in-0-1-under-iteration-by-fx4x1-x/27580#27580 Answer by Greg Kuperberg for Rational numbers with dense orbits in [0,1] under iteration by f(x)=4x(1-x) Greg Kuperberg 2010-06-09T13:49:25Z 2010-06-09T14:10:20Z <p>Ian Morris has it essentially correct in his comment. If you solve the equation <code>$x = (\sin \pi y/2)^2$</code> for $y$, then the orbit of $y \in \mathbb{R}/\mathbb{Z}$ under <code>$y \mapsto 2y$</code> is dense if and only if every finite binary string appears in the binary expansion of $y$. Now, this is weaker than being <a href="http://en.wikipedia.org/wiki/Normal_number" rel="nofollow">normal in base 2</a>, because that requires that every binary string appears equally often, not just that it appears. Let's call such a number "topologically 2-normal" (or 2-dense could be another name), because the 2-normality condition is equivalent to saying that the orbit of $y$ is not just dense, but ergodic. My impression is that not much more is known about topologically normal numbers than about normal numbers. For instance, you can conjecture that any irrational algebraic number is topologically normal in base 2 (or in any other base), but it doesn't look like it is known. In any case, topological normality is the heart of the question.</p>