Homework 2

Problem 1

• Determine the two points of the orbit as a function of r. [Solve f(f(x)) = x.]

• Determine the range of r for period two orbits. [Find conditions on r such that x is a real number in [0,1].]

• Do a stability analysis for the period two orbit. [As shown in class, [d/ dx] f(f(x))|_x0 = f¢(x₀)f¢(x₁) .]

We first find the period two orbit. [Note: This is also done at the site http://mathworld.wolfram.com/LogisticMap.html]

We now examine the stability of this orbit {x₀, x₁, x₀, x₁,x₀, ... }.

Other Problems of Interest

The following are solutions to unassigned problems this year. Problem 1.15 demonstrates Proof by Induction on n. That is, (1) Prove it true for n=0. (2) Assume it true for n=k and prove it true for n=k+1. Problem 1.17 looks at lengths of itineraries for the logistic map.

1.15

As per the hint, we will do a proof by induction on n. The statement to prove is

P_n: The nth point in the orbit generated by iterations of G(x) = 4x(1-x) is x_n = ¹/₂ - ¹/₂cos(2ⁿcos^-1(1-2x₀)).

Prove P₀ true.

1 2 - 1 2 cos(20cos-1(1-2x0)) = 1 2 - 1 2 cos(cos-1(1-2x0)) = 1 2 - 1 2 (1-2x0)) = x0.

So, P₀ is true

Prove P_k true Þ P_k+1 true.

We assume that

xk = 1 2 - 1 2 cos(2kcos-1(1-2x0)).

We want to conclude that

xk+1 = 1 2 - 1 2 cos(2k+1cos-1(1-2x0))

gives the same result as x_k+1 = G(x_k). So, we compute

xk+1 = G(xk) = 4xk(1-xk) = 4[ 1 2 - 1 2 cos(2kcos-1(1-2x0))](1-[ 1 2 - 1 2 cos(2kcos-1(1-2x0))]) = [1 - cos(2kcos-1(1-2x0))][1+cos(2kcos-1(1-2x0))] = 1 - cos2(2kcos-1(1-2x0)) = sin2(2kcos-1(1-2x0)) = 1 2 [1 -cos2(2(2kcos-1(1-2x0)))] = 1 2 - 1 2 cos(2k+1cos-1(1-2x0)).

Therefore we have proven that if P_k is true then P_k+1 true for any number k.

So, by the Principle of Mathematical Induction on n, we can now say that statement P_n is true for all n ³ 0. [Note, that this last sentence is needed to complete the proof.]

1.17

In this problem we are looking at the length of the itinerary interval for G(x) given by LLLLL... L has length [1-cos(p/2^k]/2. Thus, we need to find the range of x₀ such that 0 < Gⁿ(x₀) < 1/2, for 0 £ n < k.

0 < xn = Gn(x0) < 1 2 0 < 1 2 - 1 2 cos(2ncos-1(1-2x0)) < 1 2 0 < 1 -cos(2ncos-1(1-2x0)) < 1 1 > cos(2ncos-1(1-2x0)) > 0 - p 2 < 2ncos-1(1-2x0) < p 2 - p 2n+1 < cos-1(1-2x0) < p 2n+1 cos( p 2n+1 ) < 1-2x0 < 1 -cos( p 2n+1 ) > 2x0-1 > -1 1 -cos( p 2n+1 ) > 2x0 > 0 1 2 [1-cos( p 2n+1 )] > x0 > 0.

Therefore, letting k = n+1, we have shown that for x₀ Î (0, ¹/₂[1-cos([(p)/( 2^k)])]) 0 < x_n < ¹/₂, 0 £ n < k. The length of the interval is obviously ¹/₂[1 -cos([(p)/( 2^k)])].