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I would like to show that

$B(x)= 2x$ if $\ 0\leq x\leq 1/2$

$B(x)=2x-1$ if $\ 1/2 \leq x \leq 1$

is chaotic on [0,1]. I used the symbolic dynamics; any suggestions please?

I briefly recall some definitions from "An Introduction to Chaotic Dynamical System" of R. L. Devaney:

Let V be a set. $f: V \rightarrow V$ is said to be chaotic on V if

1) $f$ has sensitive dependence on initial condition, i.e. there exists $\delta>0$ such that, for any $x\in V$ and any neighborhood $N$ of $x$, there exists $y\in N$ and $n\geq 0$ such that $|f^{n}(x)-f^{n}(y)|>\delta$.

2) $f$ is topologically transitive, i.e. for any pair of open sets $J,I\subset V$ there exists $k>0$ such that $f^{k}(J)\cap I\neq 0$.

3) periodic points are dense in V.

Now, let $f:A\rightarrow A$ and $g:B\rightarrow B$ be two maps. $f$ and $g$ are said to be topologically conjugate it there exists a homeomorphism $h:A\rightarrow B$ such that, $h\circ f=g \circ h$. Mappings which are topologically conjugate are completely equivalent in terms of their dynamics.

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It's not hard to prove that the map $g(\theta)=2\theta$ is chaotic on the circle ($\theta$ measures the angle). Then the map $h(x)=2\pi x$ from the unit interval to the circle conjugates the Baker's map to $g$. Then there's a theorem that says that maps that are conjugate are both, or neither, chaotic.

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I know this theorem. But: $B\circ h(x)=4\pi x$ if $0\leq 2\pi x \leq 1/2$, and $B\circ h(x)=4\pi x-1$ if $1/2\leq 2\pi x \leq 1$, while $h\circ g=4 \pi \theta$. It seems to me that $B\circ h\neq h\circ g$. Where am I wrong? –  Mark Sep 4 '12 at 15:40
    
I'm not sure whether this is a difference in mathematics or just in notation, but I'd write $h\circ B=g\circ h$. I think you ought to be getting $4\pi x-2\pi$ (not $4\pi x-1$) which, in the circle, is the same as $4\pi x$. –  Gerry Myerson Sep 4 '12 at 22:40
    
Thanks, I realized the mistake I had made​​. Can I ask you two things? 1) To prove that the map is chaotic in $[0,1]$, $h$ must be: $[0,\frac{1}{2\pi}]\rightarrow [0,1]$ is this so? 2) –  Mark Sep 5 '12 at 6:50
    
2) The itinerary of x is a sequence $S(x)=s_0 s_1 s_2 ...$ where $s_j = 0$ if $B_n (x)\in [0,1/2]$, $s_j = 1$ if $B_n (x)\in [1/2,1]$. The shift map $\sigma : \Sigma_2 \rightarrow \Sigma_2$ is given by $\sigma(s_0 s_1 s_2 ...)=(s_1 s_2 s_3 ...)$. I know that the logistic map is topologically conjugate to the shift map and the logistic map is chaotic: hence the shift map is chaotic. If I show that B (x) is topologically conjugate to the shift map, I can also say that B (x) is chaotic? –  Mark Sep 5 '12 at 7:02
    
I was thinking of $h$ as a map from $[0,1)$ to $[0,2\pi)$. Concerning the logistic map, I think it's easier to prove the shift map is chaotic directly from the definitions, and then use topological conjugacy to prove the logistic map is chaotic, rather than the other way around. But, in any event, sure, if you get $B$ conjugate to shift, and you have a proof that shift is chaotic, then you have proved $B$ is chaotic. –  Gerry Myerson Sep 5 '12 at 7:38
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