Period Doubling -- The Road to Chaos -- Answers

1. These models have one nonzero equilibrium point

   q = (log a) / (log 2)
   

   By calculating f'(q) and solving the equation

   |f'(q)| = 1
   

   we see that the point at which this equlibrium point becomes repelling is a = e^2. This family of models exhibits exactly the same kind of period-doubling as the logistic model.

2. We study 3-cycles by examining the function

   g(p) = f(f(f(p)))
   

   and looking for places where the curve y = g(p) crosses the diagonal line y = p that are not equilibrium points. The figure below shows two crucial frames from a movie showing these graphs as a increases from 0 to 4. The graph of the function y = f(p) is shown in red and the graph of the function y = g(p) = f(f(f(p))) is shown in blue. In the lefthand frame the blue graph touches the diagonal line y = p in only one point and that point is an equilibrium point. In the righthand frame it touches the diagonal line in three pairs of additional points. These three pairs correspond to two 3-cycles. When these two 3-cycles first appear one of them is attracting and the other is repelling. This is because just before they appear the blue curve is tangent to the diagonal line and just after they appear the each new pair of points of intersection has one point whose slope is less than one in absolute value and another whose slope is bigger than one in absolute value.