Binary shift map

Binary shift map) Show that the binary shift map x , = 2x (modi) has sensitive dependence on initial conditions, infinitely many periodic and aperiodic orbits, and a dense orbit. (Hint Redo Exercises 10.3.7 and 10.3.8, but write x as a binary number, not a decimal.)... 

Exact solutions for the logistic map with r = 4) The previous exercise shows that the orbits of the binary shift map can be wild. Now we are going to see that this same wildness occurs in the logistic map when r = 4. 

Show that the binary shift map x +, = 2x (mod 1) has countably many periodic orbits and uncountably many aperiodic orbits. 

The name shift map derives from the following observation Let xo be a number in the unit interval [0,1]. An example for xq in binary notation is... 

Thus, applying the shift map is equivalent to scanning through the binary digits of the seed xq. 

Since the binary digit sequence in ro can have any arbitrary complexity imaginable, we expect that the shift map for a generic seed xq produces the most complex orbits imaginable, in other words, the shift map is fully chaotic. 

The calculation of fixed points and periodic obits for the shift map is straightforward. There are exactly two fixed points. Since for a fixed point of period 1 a single shift of the binary point to the right has to reproduce the seed, there are only two possibilities ... 

If we check the periodic orbits, especially their representation in binary notation, it is obvious that the seeds for the periodic orbits of the shift map are the rational numbers in the unit interval [0,1]. Since the rationals are dense in the irrationals, we obtain a second important result the periodic orbits of the shift map axe dense. 

It is not a diflficult matter to prove sensitivity of the shift map. Since X and y are not identical, the binary expansion of their difference has at least one binary digit 1 in some binary position after the binary point. Iteration of the shift map will bring this 1 closer and closer to... 

This binary decimal representation of xq should make obvious the reason why this map is named the Bernoulli shift. If xq < 1/2, then a = 0 if xq > 1/2 then ai = 1. Thus... 

In other words, a single application of the map / to the point Xq discards the first digit and shifts to the left all of the remaining digits in the binary decimal expansion of Xq. In this way, the iterate is given by Xn = an+iCtn+2 ... 

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