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For a real number $\beta >1$, the beta transformation is the interval mapon $[0,1)$. Each interval of continuity is expanding with slope $\beta$.

A continuous interval map is chaotic in Glendinning's sense when some positive iterate has a horseshoe for an interval map.

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and disjoint open intervals $K_0,K_1\subseteq J$ such thatIterating inverse branches produces full two-symbol itinerary dynamics.

A continuous interval map has positive topological entropy if and only if some positive iterate has a horseshoe for an interval map.

An interval map has a horseshoe for an interval map when two disjoint subintervals are each mapped across a common interval. It is Glendinning-chaotic when one of its positive iterates has such a horseshoe.

Let the points of the three-cycle be $x_1<x_2<x_3$ and put $I_0=[x_1,x_2]$, $I_1=[x_2,x_3]$. There are two possible cyclic orders. If $x_1\mapsto x_2\mapsto x_3\mapsto x_1$, the intermediate value theorem givesfor the interval covering relation. Thus there are two closed covering walks of length two based at $I_1$. In the reverse cyclic order there are instead the arrows $I_0\to I_0$, $I_0\to I_1$ and $I_1\to I_0$, again giving two closed walks of length two. The horseshoe from two closed covering walks therefore shows that $F^2$ has a horseshoe in either case, so $F$ is chaotic.

The Sharkovsky theorem orders the positive integers asand says that a cycle of a given period forces cycles of every period to its right.

Suppose $N=2^{r}q$ with $q>1$ odd. Then $F^{2^r}$ has a $q$-cycle. The Sharkovsky theorem gives that map a six-cycle, and squaring it produces a three-cycle. Hence $F^{2^{r+1}}$ has a three-cycle and an iterate of $F$ has a horseshoe. Thus

A horseshoe contains the symbolic dynamics of the Bernoulli shift. Periodic binary words give cycles of every symbolic period; after translating from an iterate back to $F$, this supplies cycles of many periods that are not powers of two.

For the logistic map, a ten-cycle would imply chaos because $10$ is not a power of two. Hence there is no ten-cycle for $\mu <{\mu }_{\infty }$. A three-cycle forces every period, so a ten-cycle exists for $\mu >1+\sqrt{8}$. The stated facts alone give no conclusion for the intermediate range:

An interval map $F:I\to I$ has a horseshoe when there are an open interval $J\subseteq I$ and two disjoint open intervals $K_0,K_1\subseteq J$ such that

For $0<\mu <1$, all orbits converge to one fixed point, and for $\mu =1$ every orbit is eventually fixed or period two, so no iterate has a horseshoe. For the symmetric tent map with slope magnitude $\mu >1$, the topological entropy is $\log \mu >0$. The interval-map positive-entropy horseshoe theorem says that a continuous interval map has positive topological entropy exactly when some iterate has a horseshoe. HenceFor $\mu \ge \sqrt{2}$ this already follows from part (iii); for $1<\mu <\sqrt{2}$, a higher iterate supplies the horseshoe.

An interval map $F:I\to I$ has a horseshoe if there are two closed subintervals $J_0,J_1\subseteq I$ with disjoint interiors such thatIt is chaotic in Glendinning's sense if some positive iterate $F^n$ has a horseshoe.

Suppose $a<b<c$ are the points of a 3-cycle and put $J_0=[a,b]$, $J_1=[b,c]$. There are two possible cyclic orders. Ifthen the intermediate value theorem givesConsequently both $F^2(J_0)$ and $F^2(J_1)$ contain $J_0\cup J_1$. If insteadthenand again both second images contain $J_0\cup J_1$. Thus in either cyclic order $F^2$ has a horseshoe.

A point $x$ is periodic for an interval map $F$ when $F^n(x)=x$ for some positive integer $n$. Its least such $n$ is its period, and the finite setis its periodic orbit.