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Let $f(x)=4x(1-x)$.

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)

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).

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)

EDIT: from W. Zudilin answer, it seems that the question is open in full generality. But maybe there is a chance of finding just one rational $r$ such that $f^n(r)$, or $Frac({2^n\over \pi} Arcsin\ r ) $, is dense in [0,1] ?

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Or, in other words, when is arcsin(p/q) a normal number in base 2? – Ian Morris Jun 9 '10 at 12:22
Sorry, that's not quite the same property. My bad. – Ian Morris Jun 9 '10 at 12:24
There was an Arizona Winter School earlier this year on arithmetic dynamics which might contain some results relevant to your query. Here is a link to Joe Silverman's lecture For more details you should look in Silverman's book "Arithmetic Dynamics" – Victor Miller Jun 9 '10 at 13:19
@Victor. Thanks for the reference. – coudy Jun 9 '10 at 15:04
Do you know about the theory of the logistic map? – Victor Miller Jun 9 '10 at 16:00

Ian Morris has it essentially correct in his comment. If you solve the equation $x = (\sin \pi y/2)^2$ for $y$, then the orbit of $y \in \mathbb{R}/\mathbb{Z}$ under $y \mapsto 2y$ is dense if and only if every finite binary string appears in the binary expansion of $y$. Now, this is weaker than being normal in base 2, 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.

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Well, Wadim makes a similar remark to mine in a comment to his answer. – Greg Kuperberg Jun 9 '10 at 14:22
My 2 cents: Personally, I would prefer the term "2-generic" instead of "topologically 2-normal" or "2-dense." – François G. Dorais Jun 9 '10 at 14:37
Francois: Sure. In a cursory search, I did not find any accepted name. – Greg Kuperberg Jun 9 '10 at 14:44
Greg, clearly this condition is the same as normality. Just add different prefixes to the string. – Oleg Eroshkin Jun 9 '10 at 15:29
Oleg: It is certainly possible to make a number in binary that has every finite string, but in which those strings are not equidistributed. You can just insert rapidly lengthening clumps of 0s. – Greg Kuperberg Jun 9 '10 at 17:50

If I understand your question correctly, you ask about the trajectory of $\lbrace\xi\alpha^n\rbrace$ for $\xi\in\mathbb Q$ and $\alpha=2$.

Here is the abstract of [A. Dubickas, ON THE FRACTIONAL PARTS OF THE NATURAL POWERS OF A FIXED NUMBER, Siberian Mathematical Journal 47 (2006) 879-–882]:

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 $\alpha$.

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.

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So, $\lbrace 2^n\xi\rbrace$ for $\xi$ transcendental. The article contains some historical background including your problem (Arturas is an expert in this area!): your wish is known for almost all $\xi$ only (with reference H. Weyl's 1916 article). There are some explicit constructions of transcendental numbers $\xi$ which satisfy the uniform distribution but these are "Liouville-type" constructions. Nothing is proved for your very concrete numbers $\xi=\asin(r)$, sorry! Your problem is open for almost 100 years. (And this means it's really hard.) – Wadim Zudilin Jun 9 '10 at 13:23
PS: Should be $\xi=\operatorname{asin}(r)$. Why there is no way to edit comments?! – Wadim Zudilin Jun 9 '10 at 13:24
Editing comments would be useful. You can copy the comment, make a new corrected comment from it, then delete the original comment. – Gerald Edgar Jun 9 '10 at 14:30
Division by $\pi$ doesn't make the things easier or harder. The problem is open... – Wadim Zudilin Jun 9 '10 at 23:13

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