Eventually an Orbit?

Probability Level 5

Let $g : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be an injective map. If $g$ has periodic orbits, will $g$ also have 'eventually periodic' points?

Background:

Some functions $f(x)$ display what are known as 'periodic orbits,' which is where for some initial point $x_{0},$ repeated application of $f$ eventually maps back to $x_{0}.$

More specifically, a point $x_{0}$ is a part of a period- $k$ orbit if $f^{k}(x_{0}) = x_{0}.$ The points $f(x_{0}), \ f\big(f(x_{0})\big), \ f\big(f(f(x_{0}))\big),\ ..., \ f^{k-1}(x_{0})$ are also part of this period- $k$ orbit. Points which are a part of some periodic orbit are known as periodic points.

A point $x_{1}$ is said to be 'eventually periodic' on $f$ if $x_{1}$ is not part of any periodic orbit of $f,$ but for some $j,$ $f^{j}(x_{1})$ is part of a periodic orbit of $f.$ The point begins outside of an orbit but eventually maps to one.

As an explicit example, the function $f(x)=4x(1-x)$ is a 1-D map. $f(0)=0,$ so that $x_{0}=0$ is a period-1 orbit. The point $x_{1}=\frac{1}{2}$ is not a periodic point of $f,$ and satisfies $f(x_{1})=1, f^{2}(x_{1})=f(1)=0,$ which is a periodic point of $f.$ Hence, $x_{1}=\frac{1}{2}$ is an eventually periodic point.

Details and Assumptions:

It may be helpful to review what an injection is.
Here, $f^{k}(x)$ does not mean the $k^\text{th}$ derivative of $f.$
The map $g$ does not need to be continuous.

Always Sometimes Never

2 solutions

Brandon Monsen
Sep 27, 2018

No knowledge on anything besides what an injection is and what is given in the problem are needed!

if $x_{0}$ is a periodic point of period $k$ , then $y_{0} = g^{k-1}(x_{0})$ satisfies $g(y_{0})=x_{0}$ . We also know that $y_{0}$ is part of the same periodic orbit, since $g^{k-1}(x_{0})=y_{0}$ and $g(y_{0})=x_{0}$ imply that $g^{k-1}(g(y_{0})=y_{0}$ or that $g^{k}(y_{0})=y_{0}$ so that $y_{0}$ is a part of the same period- $k$ orbit.

Now, we use the fact that $g$ is injective so that if $g(z)=x_{0}$ , and $g(y_{0})=x_{0}$ , then we know that $z=y_{0}$ . Hence, no point outside of our orbit can map to $x_{0}$ , and our answer is $\boxed{\text{NEVER}}$

Typical Beam
Oct 6, 2018

Suppose some point $a$ is eventually periodic. Then for some $j,f^j(a)$ belongs to a periodic orbit, that is, for some $k$ , $f^{j+k}(a)=f_j(a)$ . But $f$ is injective, so this forces $f^k(a)=a$ which cannot be, as $a$ does not belong to any periodic orbit. Thus $f$ has no eventually periodic points.

Eventually an Orbit?

2 solutions

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