**The Logistic Map for a>4**

adapted from section 1.6 of

We now consider the behavior of

For the remainder of this section, we will usually drop the subscript *a*
and write *F* instead of . As before, all of the interesting dynamics
of *F* occur in the unit interval *I = [0,1]*. Note that, since *a > 4*, the
maximum value of *F* is larger than one. Hence certain points leave
*I* after one iteration of *F*. Denote the set of such points by .
Clearly, is an open interval centered at 1/2 and has the property that,
if then *F(x) > 1*, so
and . is the set of points which
immediately escape from *I*. All other points in *I* remain in *I* after one
iteration of *F*.

Let If , then , , and so, as before, . Inductively, let . That is,

so that consists of all points which escape from *I* at the
iteration. As above, if *x* lies in , it
follows that the orbit of *x* tends eventually to . Since we
know the ultimate fate of any point which lies in the , it
remains only to analyze the behavior of those points which never
escape from *I*, i.e., the set of points which lie in

Let us denote this set by *A*. Our first question is: what precisely is this
set of points? To understand *A*, we describe more carefully its recursive
construction.

Since is an open interval centered at *1/2*, consists of two
closed intervals, on the left and on the right.

Note that F maps both and , monotonically onto *I*; *F* is
increasing on and decreasing on .
Since , there are a pair of open intervals, one in
and one in , which are mapped into by *F*.
Therefore this pair of intervals is precisely the set .

Now consider . This set consists of 4 closed
intervals and *F* maps each of them monotonically onto either or .
Consequently maps each of them onto 1. Thus, each of
the four intervals in contains an open subinterval which
is mapped by onto . Therefore, points in these intervals escape
from I upon the third iteration of *F*. This is the set we called . For
later use, we observe that is alternately increasing and decreasing on
these four intervals. It follows that the graph of must therefore
have two ``humps''.

Continuing in this manner we note two facts. First, consists of disjoint open intervals. Hence consists of closed intervals since

Secondly, maps each of these closed intervals monotonically onto
*I*. In fact, the graph of is alternately increasing and
decreasing on these intervals. Thus the graph of has exactly
humps on *I*, and it follows that the graph of crosses the line
*y = x* at least times. This implies that has at least
fixed points or, equivalently, *Per(F)* consists of points in *I*.
Clearly, the structure of *A* is much more complicated when *a > 4* than
when *a < 3*.

The construction of *A* is reminiscent of the construction of the Middle
Thirds Cantor set: *A* is obtained by successively removing open intervals
from the "middles" of a set of closed intervals. See section 4.1 of
Alligood, Sauer, & Yorke and/or
Neal Carothers' Cantor Set web
pages for more
details.

**Definition:**
A set is a Cantor set if it is a closed, totally disconnected, and
perfect subset of an interval. A set is totally disconnected if it contains
no intervals; a set is perfect if every point in it is an accumulation point
or limit point of other points in the set.

**Example: The Middle-Thirds Cantor Set.**
This is the classical example of a Cantor set. Start with *[0,1]* but
remove the open "middle third," i.e. the interval . Next, remove the middle thirds of the resulting intervals,
that is, the pair of intervals and . Continue removing middle thirds in this fashion;
note that open intervals are removed at the stage of this
process. Thus, this procedure is entirely analogous to our construction
above.

**Remark.** The Middle-Thirds Cantor Set is an example of a fractal.
Intuitively, a fractal is a set which is self-similar under magnification.
In the Middle-Thirds Cantor Set, suppose we look only at those points which
lie in the left-hand interval . Under a microscope which
magnifies this interval by a factor of three, the "piece" of the Cantor set
in looks exactly like the original set. More precisely,
the linear map *L(x) = 3x* maps the portion of the Cantor set in homeomorphically onto the entire set.

This process does not stop at the first level: one may magnify any piece of the Cantor set contained in an interval by a factor of to obtain the original set.

To guarantee that our set *A* is a Cantor set, we need an additional
hypothesis on *a*. Suppose *a* is large enough so that *| F'(x)| > 1 * for
all . The reader may check that
suffices. Hence, for these values of *a*, there exists such
that for all . By the chain rule, it
follows that as well. We claim that *A* contains
no intervals. Indeed, if this were so, we could choose two distinct point
*x* and *y* in *A* with the closed interval . Notice that
for all . Choose *n* so
that . By the Mean Value Theorem, it then follows that
, which implies that at least one
of or lies outside of *I*. This is a contradiction, and so
*A* is totally disconnected.

Since *A* is a nested intersection of closed intervals, it is closed. We
now prove that *A* is perfect. First note that any endpoint of an is
in *A*: indeed, such points are eventually mapped to the fixed point at 0,
and so they stay in *I* under iteration. Now if a point were
isolated, every near point near *p* must leave *I* under iteration of *F*.
Such points must belong to some . Either there is a sequence of
endpoints of the converging to *p*, or else all points in a deleted
neighborhood of *p* are mapped out of I by some power of F. In the former
case, we are done as the endpoints of the map to 0 and hence are in
*A*. In the latter, we may assume that maps *p* to 0 and all other
points in a neighborhood of *p* into the negative real axis. But then
has a maximum at *p* so that . By the chain rule, we must
have for some *i < n*. Hence , and so
is not in *I*, contradicting the fact that .

Hence we have proved

**Theorem.** If , then *A* is a Cantor set.

**Remark**. The theorem is true for *a > 4*, but the proof is more
delicate. Essentially, all we need is that for each , there is an
*N* such that for all , we have . Then almost
exactly the same proof applies.

We have now succeeded in understanding the gross behavior of orbits of
when *a > 4*. Either a point tends to under iteration of
,or else its entire orbit lies in a Cantor set *A*.
For points , we see the same complicated type of behavior that we
have seen for *a=4*: there are periodic points of all periods, but even
more: we can symbolically specify a behavior by a string of Rs and Ls, and
such an orbit necessarily exists. We will return to this issue later.

Fri Jan 31 00:06:41 EST 1997