Q:

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?

A:

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"

A:

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.

A:

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.

