= Solution
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> gives
$$
I_0\longrightarrow I_1,
\qquad I_1\longrightarrow I_0,
\qquad I_1\longrightarrow I_1
$$
for 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 as
$$
3\prec5\prec7\prec\cdots
\prec2\cdot3\prec2\cdot5\prec\cdots
\prec2^2\cdot3\prec\cdots
\prec\cdots\prec2^3\prec2^2\prec2\prec1,
$$
and says that a cycle of a given period forces cycles of every period to its right.
Suppose $N=2^rq$ 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
$$
\boxed{N\text{ not a power of two and an }N\text{-cycle}\implies F\text{ is Glendinning-chaotic}.}
$$
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 $\mu>1+\sqrt8$. The stated facts alone give no conclusion for the intermediate range:
$$
\boxed{\mu<\mu_\infty:\ ext{no ten-cycle};
\qquad \mu>1+\sqrt8:\ ext{a ten-cycle exists}.}
$$
Back to article page
