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I have to prove that the function $F(x) = 8x^{4} - 8x^{2} + 1$ is chaotic. I would like to use the definition of a chaotic function which is:

Let $F$ be $F: V$ -> $V$.

1) Sensitive dependance on initial conditions: it exists $\delta > 0$ such that it exists $y \in N$ and $n \ge 0$ such that $|f_{n}(x) - f_{n}(y)| > \delta$ for any $x \in V$ and any neighborhood $N$ of $x$.

2) Topologically transitive: it exists $k > 0$ such that $f_{k}(J) \cap I \ne 0$ for any pair of open sets $J$ and $J \subset V$.

3) Periodic points are dense in $V$.

I wanted to try the following method:

Find $G$ and $\phi$ such that $F o \phi = \phi o G$, prove that G is chaotic and then as a matter of fact that F is chaotic.

Here, Use the definition of "topologically conjugate" to prove that $F(x)=4x^3-3x$ is chaotic., you can see the method for the function $F(x) = 4x^{3} - 3x$.

The problem I have is to find a function $G$.

Thanks for reading,

Sign009

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Dust009
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  • You may be familiar with the dictum that period three implies chaos. Is your Question about a proof using "the definition of a chaotic function" or about one using the method outlined in the last part of your post? – hardmath Jan 22 '15 at 12:42
  • My question is about having a hint on what is G so I can prove that G is chaotic using the definition of a chaotic function and hence F is chaotic :) – Dust009 Jan 22 '15 at 12:47
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    @hardmath The problem with applying such a general theorem is that it gives you very little information on the chaotic domain. Take $f(x)=3.83 x(1-x)$, for example. This function has an orbit of period 3 - an attractive orbit, in fact, so it's easy to find. As a result the dynamics look rather tame. The chaotic domain is just a Cantor set. The function in question, however, is chaotic on an entire interval so it's characteristics are quite different. – Mark McClure Jan 22 '15 at 13:18

1 Answers1

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Let's take a look at the two graphs you mention.

enter image description here

It looks to me like the first is conjugate to $\theta \mapsto 4\theta \mod 2\pi$ while the second is conjugate to $\theta \mapsto 3\theta \mod 2\pi$. In fact, the conjugacy functions are both given by $\varphi(\theta) = \cos(n\theta)$ for $n=3,4$. This is because these are both Chebyshev polynomials and, in that context, their defining characteristic is that $T_n(\cos(\theta))=\cos(n\theta))$. Of course, $\theta \mapsto n\theta \mod 2\pi$ is well known to be chaotic.

Mark McClure
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