Chaos Task 1

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Task 7: Solutions

Fix a = 4. Set x_k = sin² py_k (for each k Î \mathbbN), and substitute in the logistic equation, x_n+1 = ax_n (1-x_n).

Part 1. Show that this simplifies to sin² py_n+1 = sin² 2 py_n.

We use the trigonometric identities 1- sin² A = cos² A and the `Double Angle Fomula', sin2A = 2 ·sinA ·cosA. So...

sin² 2 py_n+1

x_n+1

4 ·x_n ·(1-x_n)

4 ·sin² 2 py_n ·(1-sin² 2 py_n)

4 ·sin²2 py_n ·cos² 2 2 2 py_n

(2 ·sin 2 py_n ·cos2 py_n)²

sin² 2 ·(2 py_n)

It follows that y_n+1 = 2ⁿ y₁ mod 1, and to calculate y_n it is sufficient to shift the binary string of y₁ to the left n places, and drop the integer part.

Part 2. Use the above to show that y_n, and hence x_n, is chaotic - if y₁ and y˘₁ are in [0,1], then y_n - y˘_n = 2ⁿ (y₁ - y˘_{1˘₁) mod 1.}

This is clear: y_n - y˘_n = 2ⁿ y₁ - 2ⁿ y˘₁ mod 1 = 2ⁿ(y₁ - y˘₁) mod 1.

[Further, show that given a and b in [0,1], with a < b, then, provided the seed y₁ has an infinite binary expansion, there is an N such that a < y_N < b. (Hint: write a and b in their binary expansion, and note that a < b, means that the binary digits of a and b are equal up to some nth term, where a has binary digit 0 and b has binary digit 1.)]

NOTE: The question is not stated carefully enough. There are seeds with infinite expansion with finite orbit, and hence are not ergodic. The condition the seed must satisfy is the following:

...the seed has an expansion such that:
for any finite sequence of 0's and 1's, there is a segment of the seeds expansion equal to the given finite sequence.

Let us note that only countably many seeds fail to have this property. Thus if we were to pick a seed `at random' it would almost certainly (probability 1) have the required property, and so the population sequence it generates (y_n) (and (x_n)), would (and (x_n)), would be ergodic.

Following the hint, and assuming the seed y₁ = 0. t₁ t₂Ľ has the property above, we show (y_n) (and (x_n)) is ergodic: given a < b in [0,1], there is an N such that a < y_N < b. Write a = 0. a₁ a_Ľ and b = 0. b₁ b₂Ľ. As the hint says there is an N so that that a_i = b_i for i < N, but a_N = 0 and b_N = 1. By the given property of the seed, there is an M such that t_M+1 t_M+2 Ľt_M+N-1 = a₁ a₂ Ľa_N-1. Hence after M iterations we get to y_M which starts 0. a₁ a₂ Ľa_N-1. Therefore y_M is between a and b, as required.

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