Chaos theory
I would suppose the general problem of finding continuous roots is still
unsolved, however, given peoples' reaction to the problem on sci.math. I am
becoming more interested in chaos theory in addition to this problem. I am
only using James Gleik as a background (he writes Chaos for the general reader)
but this is all I need to know what I am trying to understand. This page is
a background on some phenomena and my advances so far. (My "advances" are most
certainly trivial but maybe they won't be so trivial if I keep going.) The
theory which was intended to solve f2 = g should be helpful in
this.
Consider the function f(x) = kx(1 - x), k > 1, which maps R to R. Obviously this
function is an upside-down parabola with zeros at 0 and 1, and fixed points
0 and (k - 1)/k. This simple looking function has interesting properties for
various k. Choose k < 3, 0 < x < 1, and iterate x by f repeatedly. The sequence
x, f(x), f2(x), f3(x), ... converges to a single value.
If k = 3, this sequence no longer converges to one value but has two
cluster-points, and fn(x) alternates from close to one value to
close to another. Increasing k, the cluster-points branch into 4 in number, 8,
16, 32, and at a point the values of the iteration become chaotic. Although
this behavior has probably already been explained in depth, I hope to use class
theory to help explain this phenomena. I state a simple theorem that links
this problem to class theory.
Theorem: Let f:R -> R be a continuous function, let x be a real number and
suppose for some n the ordered set (fkn(x), fkn + 1(x),
fkn + 2(x), ..., f(k + 1)n - 1(x)) converges as k ->
infinity. In essence, suppose x, when iterated, oscillates as explained in
the common example of kx(1 - x). Then f has a cycle of order n and is equal
to the limit of the above sequence.
Proof: Let m1 = lim fkn(x), m2 =
lim fkn + 1(x). It suffices to show that f(m1) =
m2; the rest of the theorem follows from renaming the cycle
elements. By a simple computation using theorems from elementary calculus
we find m2 = lim fkn + 1(x) = lim f(fkn(x))
= f(lim fkn(x)) = f(m1), proving our theorem. QED
Cycles in a continuous function which have x as in the theorem are called
attractive cycles and the ones that are not are called
unattractive cycles. Apparently there is one (interesting)
attractive cycle for each logistic equation kx(1 - x).
The existence of limit cycles, although a trivial result, leads to another
important question: How can we find the cycles of a continuous function
f:R -> R? After stumbling upon this question I have found that it is a
popular question from dynamical systems. It is of great interest to me
to research this question because many secrets of chaos can be unlocked
with its answer. I have decided to prove for myself the already
discovered result that the existence of a 3-cycle implies the
existence of cycles of all orders. This was proved by James Yorke in
1975 in his paper "Period Three Implies Chaos." The proof I discovered, as
a matter of fact, looks almost exactly like another proof at another web
page I will soon post. Right now, I'm trying to prove "Sarkovskii's theorem"
which states this.
Theorem: Order the natural numbers with relation > like this.
3 > 5 > 7 > ...
> 2.3 > 2.5 > 2.7 > ...
> 22.3 > 22.5 > 22.7 > ...
> ...
> 25 > 24 > 23 > 22 > 2 > 1
If a continuous function f:R -> R has a cycle of order m, then f has cycles
of order n for each m > n. Thus, period three implies period n is a
corollary. In addition, period 5 implies all cycles except possibly period
3.
The theorem is simpler than it looks. In order to prove
> 25 > 24 > 23 > 22 > 2 > 1
all we must do is show that a 4-cycle implies a 2-cycle. (Obviously any
function with a 2-cycle implies a fixed point or 1-cycle.) If f has an
8-cycle, f2 has a 4-cycle and thus has a 2-cycle, and by a
simple theorem of class theory, f must have a 4-cycle, so 8 > 4 > 2 > 1.
If f has a 16-cycle, then f2 has an 8-cycle, thus a 4-cycle, and
f has an 8-cycle, giving 16 > 8 > 4 > 2 > 1. By induction, we could show
> 25 > 24 > 23 > 22 > 2 > 1.
Similarly, we only have to prove the first two lines of the first part.
I'm not looking for a proof of Sarkovskii's theorem on the web. I'm trying to
prove it in a particular manner: using symbolic dynamics. (Laughter,
if this is a common way to prove it.) In this way, I want to generalize
Sarkovskii's order to the permutations. First, let me give an overview of
symbolic dynamics.
Let S = {a1, a2, ..., an} be a set of n
elements, called characters, and let X be the set of all
strings of the characters. (For example,
a1a1a3a2 is a string.) Let
there be a function on X called a replacement rule, which is induced
by a function from S onto X sending each character to a string. Taking an
initial string, called the initial state, and iterating based
on the replacement rules, you get a sequence of generations of
different strings. For example, let the characters be {a, b, c}
and the replacement rules be
a -> ab
b -> c
c -> b
Then if you start with ac, you get the sequence of iterations
Generation 0: ac
Generation 1: abb
Generation 2: abcc
Generation 3: abcbb
Generation 4: abcbcc
Generation 5: abcbcbb
...
The study of such sequences is, roughly, symbolic dynamics. (I only skimmed
a section of it on the web, so that might be a flawed definition. For our
purposes, we will refer to it as "symbolic dynamics" anyway.)
Definition: Let f:R -> R be a continuous function and suppose {x1,
..., xn} is an n-cycle of f,
x1 < ... < xn. Then f cyclically permutes the
indices of the variables with a permutation p in Sn. I call p
the cycle type of the cycle in f.
If X = Uk = 1inf Sk, then we can order X
with another Sarkovskii order. This will be more helpful to finding cycles
of a continuous function. I have been told that this has already been done,
and the order is extremely complicated. So what.
Definition: Let f:R -> R be a continuous function with cycle type p on
x1 < ... < xn. Let S = {(x1x2),
(x2x3), ..., (xn - 1xn)} be
a set of characters and define the replacement rule
(xixi + 1) ->
(xaxa + 1)(xa + 1xa + 2) ...
(xb - 1xb)
where a and b are min{f(xi), f(xi + 1)} and
max{f(xi), f(xi + 1)} respectively. The resulting
symbolic system is called the iterative system of the cycle type
p (independent of the function f). We often write (12) in place of
(x1x2), etc. The point of this iterative system is
that, taking (i i+1) as the initial state, the presence of (i i+1) in the
kth generation corresponds to a point x in (i, i + 1) such that
fk(x) = x. (This is yet to be proved or falsified in the case
that k is a multiple of n.) Showing that an (i i+1) exists in the nth
generation but it is not the spawn of any other (i i+1) in previous
generations dividing n can prove the existence of a cycle of order n.
Moreover, it may not only give Sarkovskii's order but also an order on
all permutations. The details are still being worked out.
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