Dynamics on a circle

**godelproof** · January 1st 2012, 01:46 AM

Consider a circle $S$ and a function $f: S \mapsto S$ given by $f(\theta)=2 \theta \ (mod\ 2 \pi)$ where $\theta \in [0,2 \pi)$ .

Question: Given any open sets $U \in S$ and $V \in S$ , how can we show that $\exists\ k$ such that ${f}^{k}(U)\cap V\neq \emptyset$ ?

The problem comes from An Introduction to Chaotic Dynamical Systems by Devaney, Example 8.6. There's no explanation there though.

**godelproof** · January 1st 2012, 11:57 PM

Consider another problem:

${a}_{i}(x):=ix \ (mod\ 1)$ , for $i=1,2,3,...$

Now choose any irrational $x \in [0,1]$ , prove that the series $\{{a}_{i}(x)\}$ is dense in $[0,1]$ .

There's a nice proof by mapping ${a}_{i}(x)$ to a circle of unit length, instead of considering mod 1 (In the same book by Devaney). But is there a more direct or simpler method? Because this seems like some classical problem.

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Edit: If, say, ${a}_{i}(x):={2}^{i}x \ (mod\ 1)$ , then the series no longer needs to be dense in $[0,1]$ for irrational x. So linearality is the key.

**emakarov** · January 2nd 2012, 01:15 AM

Originally Posted by godelproof

Question: Given any open sets $U \in S$ and $V \in S$ , how can we show that $\exists\ k$ such that ${f}^{k}(U)\cap V\neq \emptyset$ ?

Do you mean $U\subseteq S$ and $V\subseteq S$ ?It is necessary that U and V are nonempty. Then U contains an interval. Repeatedly applying f to this interval inflates it so that eventually it covers the whole circle.

**emakarov** · January 2nd 2012, 02:37 AM

Originally Posted by godelproof

There's a nice proof by mapping ${a}_{i}(x)$ to a circle of unit length, instead of considering mod 1 (In the same book by Devaney). But is there a more direct or simpler method? Because this seems like some classical problem.

I would not consider dealing with a unit circle to be a different method. Considering a circle vs numbers mod 1 is just a reformulation of the problem.

This is indeed a classic problem. A stronger statement that $\{a_i(x)\}$ is not only dense but uniform is the equidistribution theorem. The original fact follows from the observation that $\inf\{m\cdot1+n\cdot x\mid m\cdot1+n\cdot x>0\mbox{ and }m,n\in\mathbb{Z}\}=0$ since this set does not have a minimum; otherwise, the minimum would be a common divisor of 1 and x and they would be commensurate. (Of course, the absence of a minimum by itself does not imply that the $\inf$ is 0; another simple step is needed.) See also this thread and this post. See also this statement on cut-the-knot (set a = b = 0). The fact also follows from the Dirichlet's approximation theorem.

**godelproof** · January 2nd 2012, 03:28 AM

Originally Posted by emakarov

Do you mean $U\subseteq S$ and $V\subseteq S$ ?It is necessary that U and V are nonempty. Then U contains an interval. Repeatedly applying f to this interval inflates it so that eventually it covers the whole circle.

But why must repeated application of $f$ inflates ANY interval to cover the whole circle? This does not seem obvious to me. Say the interval is $({x}_{1},{x}_{2})$ where ${x}_{1},{x}_{2} \in (2 \pi/{2}^{n+1},2 \pi/({2}^{n+1}-1))$ , then the interval does not expand to cover $S$ in $n$ applications of $f$ or less. But at the $(n+1)$ th application, we may again have ${f}^{n+1}({x}_{1}),{f}^{n+1}({x}_{2}) \in (2 \pi/{2}^{n+1},2 \pi/({2}^{n+1}-1))$ , which we started out with. Maybe I'm missing something obvious, but I need more explanation here, please.

**emakarov** · January 2nd 2012, 09:02 AM

Originally Posted by godelproof

But why must repeated application of $f$ inflates ANY interval to cover the whole circle?

This is because $f$ transforms the whole interval and not only the endpoints.

Consider the following picture.

Even though $f^n(x_1)=2^nx_1\mbox{ mod }2\pi$ and $f^n(x_2)=2^nx_2\mbox{ mod }2\pi$ are quite close, the interval $(2^nx_1, 2^nx_2)$ is mapped to $[0,2\pi]$ by the modulo function. The interval $(f^n(x_1),f_n(x_2))$ covers the whole circle as soon as $2^nx_2-2^nx_1>2\pi$ .

**godelproof** · January 2nd 2012, 10:27 PM

Thank you. BTW, what software did you use to plot the picture? It's neat.

**godelproof** · January 3rd 2012, 05:38 AM

Originally Posted by emakarov

I would not consider dealing with a unit circle to be a different method. Considering a circle vs numbers mod 1 is just a reformulation of the problem.

This is indeed a classic problem. A stronger statement that $\{a_i(x)\}$ is not only dense but uniform is the equidistribution theorem. The original fact follows from the observation that $\inf\{m\cdot1+n\cdot x\mid m\cdot1+n\cdot x>0\mbox{ and }m,n\in\mathbb{Z}\}=0$ since this set does not have a minimum; otherwise, the minimum would be a common divisor of 1 and x and they would be commensurate. (Of course, the absence of a minimum by itself does not imply that the $\inf$ is 0; another simple step is needed.) See also this thread and this post. See also this statement on cut-the-knot (set a = b = 0). The fact also follows from the Dirichlet's approximation theorem.

Can we find a real function $T: \mathbb{R} \mapsto \mathbb{R}$ , such that the sequence { $T(\theta)\ mod \ 1, {T}^{2}(\theta)\ mod \ 1, {T}^{3}(\theta)\ mod \ 1,...$ } is dense in $[0,1]$ whenever $\theta \in [0,1]$ is irrational?

**emakarov** · January 3rd 2012, 10:59 AM

Originally Posted by godelproof

BTW, what software did you use to plot the picture? It's neat.

I used PGF/TikZ (on CTAN and on sourceforge.net).

Originally Posted by godelproof

Can we find a real function $T: \mathbb{R} \mapsto \mathbb{R}$ , such that the sequence { $T(\theta)\ mod \ 1, {T}^{2}(\theta)\ mod \ 1, {T}^{3}(\theta)\ mod \ 1,...$ } is dense in $[0,1]$ whenever $\theta \in [0,1]$ is irrational?

This seems plausible, but I am not a specialist in this area. Maybe you can make a new post (probably in the Calculus/Topology section) to attract attention to this question.

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