# Period of 3 implies chaos

Let $f(x)$ be a continuous function from $\mathbb{R}\rightarrow\mathbb{R}$.

Let's denote $k$-times repeated application of the function, $f(f(f(...f(x)...)))$ as $f^{(k)}(x)$.

Let's call any $x$ a periodic point with period $n$ if $f^{(n)}(x)=x$.

Is it true that if a point with period 3 exists, then points with all possible periods exist?

In other words is it true that

$$\exists x:f^{(3)}(x)=x\Rightarrow \forall n>0 \exists y:f^{(n)}(y)=y$$

and if so, why?

-

I found Chaos: An Introduction to Dynamical Systems by Kathleen T. Alligood, Tim D. Sauer, and James A. Yorke, to be a sensible text. A small section of the text contains a brief explanation of Sharkovsky's Theorem.

In short, Sharkovsky uses this unique ordering of the natural numbers in such a way as to ensure that for each natural number n, the existence of a period-n point implies the existence of periods containing all the periods of orders higher than n. In my opinion, the text provides a reasonable simplification of this ordering and gently walks one through a decipherable step by step proof.

-

I believe you want Sharkovsky's Theorem.

A reasonable explanation is in:

Kaplan, Harvey. "A cartoon-assisted proof of Sarkowskiĭ's theorem." Amer. J. Phys. 55 (1987), no. 11, 1023–1032. MR917121 DOI:10.1119/1.14928"

-
Thanks... do you have the proof as well? –  KalEl Aug 21 '10 at 0:08
I linked a fairly good explanation in an article. It looks like this is covered in a lot of "dynamics" or "chaos" textbooks too, so you might just check the library for a readable one. I get the feeling that Harvey Kaplan felt most of the explanations were too scary, and so wrote this reasonable one. It's been 23 years, so I would assume the textbooks are better now. –  Jack Schmidt Aug 21 '10 at 0:13

See the Monthly paper of that exact title Period three implies chaos - which made the JSTOR All-Stars list as the third most frequently accessed Monthly article. See also A simple proof of Sharkovsky's theorem and this paper for some history on Sharkovsky's theorem.

-

This is (a special case of) a famous theorem, and of course there are many published proofs. However, knowing the statement, you might be able to work out a proof yourself.

The simple key fact for the proof is:

Lemma. Suppose that $f$ is a continuous function on the compact interval $I=[a,b]$ such that the image of $f$ contains $I$. Then $f$ has a fixed point.

(Hint: intermediate value theorem.)

Now suppose you have a period 3 orbit. First of all think about how the three points $x_1<x_2<x_3$ of the orbit can be permuted by the function - you will notice that you can assume without loss of generality that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$. Consider the two intervals $I_1=[x_1,x_2]$ and $I_2=[x_2,x_3]$. What do we know about the image of these intervals under $f$? Already you should see that $f$ must have a fixed point. With a little bit more work, you can see that $f$ has periodic points of every period $n$. (Show that there is a subinterval $J$ of $I_2$ such that $f^{n-1}(J)= I_1$ and $f^j(J)\subset I_2$ for $1\leq j<n-1$.)

-

This blog entry has a nice intro to the theorem, although no proof.

http://divisbyzero.com/2008/12/18/sharkovskys-theorem/

-