Chaotic behavior of the logistic map at $r=4$.

Question

With Sarkovskii's theorem I want to conclude the chaotic behavior of the logistic map $f(x)=r \cdot x(1-x)$. I can't find a value of $x$ which leads to a periodic three orbit. Does anyone know a value of $x$ that leads to a periodic three orbit? If I found such starting point of the iteration, I could use Sarkovskii's theorem to argue the chaotic behavior of the logistic map at $r=4$.

The following pictures (generated with Mathematica) show the graphic iterations of the logistic map for different starting points:

Iteration of the logistic map with starting point $x=0.50000$:

Iteration of the logistic map with starting point $x=0.50001$:

Answer 1

It is well-known that a parametrization of $x_n=\sin^2(\pi \theta_n)$ transforms the dynamic to $\theta_{n+1}=2\theta_n\bmod 1$, the destructive left-shift of the binary digit sequence of $\theta_n$, removing the leading digit in every step. For a period-3 cycle you now need a rational number with a period-3 binary digit sequence like $\theta_0=1/7$, thus take $$x_0=\sin^2(\pi/7).$$

This logic also predicts that using double precision floating point numbers the cycle will deteriorate in about 50 iterations, as then the encoding bits of the initial value are used up, and the remaining contents in $x_n$ will be random bits produced by the floating point truncation errors during the computation.

Answer 2

To build up on the Lutz Lehmann's answer, the map $\Phi:\theta\mapsto \sin^2(\pi\theta)$ exhibits the logistic map $L:x\mapsto 4x(1-x)$ as a $2:1$ factor of the doubling map $E: \theta\mapsto 2\theta \mod1$. This latter map has (exactly) $8-2=6$ points of period $3$, hence the logistic map $L$ has at least $3$ points of period $3$, that is, at least one $3$-cycle ($\dagger$). Here is a cobweb diagram exhibiting the cycle given by the semiconjugacy $\Phi$:

(The interactive graph is available at: https://www.desmos.com/calculator/ucr4tj7zdq)

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($\dagger$) In an earlier version of this answer I had claimed that the logistic map $L$ has exactly one $3$-cycle; as David pointed out in a comment this is false. The true statement is that $L$ has exactly two $3$-cycles:

(The first cycle is the one discussed already. The interactive graph is available at: https://www.desmos.com/calculator/mk3qfhibgb)

The error in the previous version of my answer stemmed from assuming that any $3$-periodic point of $L$ must be the image of a $3$-periodic point of the doubling map $E$ under the semiconjugacy $\Phi$. In reality $(\dagger\dagger)$ we have that if $\theta$ is a point with

$$\Phi\circ E(\theta) = \Phi(\theta), \quad\quad(\star)$$

then $\Phi(\theta)$ is a fixed point of $f^3$, and indeed any fixed point of $f^3$ is of the form $\Phi(\theta)$ for such a $\theta$. Expanding $(\star)$, for numerical values of initial conditions for $3$-cycles $\theta$ is required to solve one of the two analytic equations

$$\sum_{n\in\mathbb{Z}_{\geq0}}\dfrac{(-1)^n(8^{2n+1}\pm 1)\pi^{2n+1}}{(2n+1)!}\theta^{2n+1}=0.$$

There are various methods that approximate zeros of analytic equations; but for the sake of this discussion it seems it's sufficient to refer to graphs (of concatenations); indeed the interactive graph at https://www.desmos.com/calculator/mwbq9gpc5d does count the distinct initial conditions for $3$-cycles accurately. (Once the existence of a cycle is established, one can also use the cobweb diagrams to find numerical values.)

$(\dagger\dagger)$ More generally if $X,Y$ are compact metric spaces and $f:X\to X$, $g:Y\to Y$, $\Phi:X\to Y$ are continuous maps with $\Phi$ onto and $\Phi\circ f=g\circ \Phi$, then

$$\{x\in X | \Phi\circ f(x)=\Phi(x)\} = \Phi^{-1}(\text{Fix}(g)).$$