Past exam of the topics in applied mathematics course of the University of Oxford / 2010 / B / 5 / iii

In the two-symbol shift representation, $\boxed{\dots 000.000\dots }$ is fixed under a shift and codes the fixed point labelled $0$. The constant all-ones sequence supplies the analogous example. The dot only marks the present position; shifting it advances one iterate. The assertion uses the faithful orbit coding assumed in the question.

Use the primitive repeating word $\boxed{\dots 001001.001001\dots }$. One shift changes the sequence and three shifts return it, so the least period is three. A word such as $000$ repeated would instead have least period one; primitivity is what distinguishes a true three-cycle from a periodic notation for a fixed point.

The sequence $\boxed{\dots 000010000.00000\dots }$ has exactly one $1$ and otherwise zero entries. Under forward shifts that exceptional symbol moves out of every fixed observation window; the same happens under backward shifts in the other direction. Thus both temporal limits are the all-zero fixed sequence, while the orbit itself is nonconstant: it is homoclinic to that fixed point. The bi-infinite coding specifies a chosen full orbit even when the original interval map is not invertible.

Choose $\boxed{\dots 00000.11111\dots }$, with zeros on the entire left tail and ones on the entire right tail. Forward shifts converge in the symbolic metric to the all-one fixed point, while backward shifts converge to the all-zero fixed point. Its two limiting invariant states differ, so it is a heteroclinic orbit. Reversing the two tails reverses these asymptotic roles.

One explicit future itinerary is obtained by concatenating all binary words, ordered first by length and then lexicographically:Extend it arbitrarily to negative indices, for example by zeros. Every finite binary block occurs in its forward itinerary, and indeed occurs arbitrarily far out as a subword of longer words. Its shift orbit is therefore dense in the full two-symbol shift space. It is not eventually periodic: arbitrarily long runs of zeros and ones occur, together with all other finite words.

The full shift has sensitive dependence because sequences agreeing on an arbitrarily large initial window can differ just beyond it, and shifting brings that disagreement into view. Thus this itinerary supplies a dense aperiodic orbit in the chaotic symbolic system assumed by the question. A merely nonrepeating sequence would not be enough: restricted aperiodic sequences can have much simpler, nonchaotic orbit closures. Nor does symbolic description alone assign a numerical Lyapunov exponent without information about the underlying map and its coding.