Although the algebra is hopeless, the geometric interpretation for the location of periodic orbits is straightforward. As we see in Figure 2.10(a), the location of the period one orbits is given by the intersection of the graphs y = f(x) and y = x.
Figure 2.10: First and second iterates of the quadratic map ( 
).
Figure 2.11: Third and fourth iterates of the quadratic map ( ).
The latter equation is simply a straight line
passing through the origin with slope +1. In the case of the quadratic
map, 
is an inverted parabola also passing through the origin. These two graphs
can intersect at two distinct points, giving rise to two distinct period one
orbits. One of these orbits is always at the origin and the other's exact
location depends on the height of the quadratic map, that is, the specific value of
in the quadratic map.
To find the location of the period two orbit we need to plot y = x and
. The
graph shown in Figure 2.10(b) shows three points of intersection in
addition to the origin. The middle point (the open circle) is the
period one orbit found above. The two remaining intersection points are the
two points belonging to a single period two orbit. A dashed line indicates
where these period two points sit on the original quadratic map (the two dark
circles), and the simple graphical construction of section 2.3 should
convince the reader that this is, in fact, a period two orbit.
The story for higher-order orbits is the same
(see Fig. 2.11(a) and (b)).
The graph of the third iterate,
, shows eight points of intersection with the straight line.
Not all eight intersection points are elements of a period three orbit.
Two
of these points are just the pair of period one orbits. The remaining six points
consist of a pair of period three orbits. The graph for the period four
orbits shows sixteen points of intersection.
Again, not all the intersection points are part of a period four
orbit. Two intersection points are from the pair of period one
orbits, and two are from the period two orbit. That leaves twelve remaining
points of intersection,
each of which is
part of some period four orbit. Since there are twelve remaining
points, there must be three (12 points / 4 points per orbit)
distinct period four orbits.
The number of intersection points
of 
depends on .
If 
and 
,
there are only two intersection points: the two distinct period one orbits.
In dramatic
contrast, if 
, then it is easy to show that there will be 
intersection points, and counting arguments like those just illustrated allow us
to determine how many of these intersection points are new periodic points of
period n [4]. One
fundamental question is: how can a system as
simple as the quadratic map change from having only two to having an infinite
number
of periodic orbits?
Like many aspects of the quadratic map,
the answers are surprising.
Before we tackle this problem,
let's resume our analysis of the period one and period two orbits.