Why Do Recursive Rational Functions Attract and Repel?

Date: 07/10/2008 at 10:07:50
From: Mike
Subject: Why do recursive rational functions attract and repel

Given a recursive linear rational polynomial e.g. f(x) = (x-a)/(x-b). 
f(x) = x has 2 real roots, chose x|x<>b, x<>either root.  Recursively 
apply f(x) and the function will converge to one of the roots no 
matter how close you are to the other root.
 
Why does the function have a preference to one of the roots?  How can
you determine the attractor root other than trial and error?  Where
can I find out more about recursive attractors?  (Excuse my “recursive
attractors” term.  I do not know the language.)

A fun example is the function f(x) = (x+1)/x which has the golden
ratio (1+sqrt(5))/2 as an attractor and (1-sqrt(5))/2 as a repeller.

f(x)-x = 0 is obviously quadratic and the graph suggests nothing of an 
attractor quality.  I noticed eigenvectors [A]X = X and more obviously
differential equations tend to have attractor-like solutions.  It is
confusing why a simple 1st order rational polynomial would exhibit
such behavior.

Another strange property I've noticed is when f(x) = x has no real 
solutions and a and b are prime then f(x) tends to act like a one 
dimensional fractal generator.  If you recursively evaluate f(x) = x 
lets say 1E6 times and plot the results, the repetitive pattern 
repeats itself as you zoom in on any cycle of the repetition.

Date: 07/19/2008 at 16:03:07
From: Doctor Vogler
Subject: Re: Why do recursive rational functions attract and repel

Hi Mike,

Thanks for writing to Dr. Math.  That's a very good question.  I think
you would very much like reading the book "A First Course in Chaotic
Dynamical Systems" by Robert L. Devaney, which discusses iterates of
continuous functions, attracting and repelling fixed points and
cycles, and other topics in fractals and chaos theory, also known as
discrete dynamical systems.  It's a great book and easy reading.

The short answer is that a fixed point of a function, that is, a value
of x where

  f(x) = x

is attracting (nearby values come closer to it when you iterate f) when

  abs(f'(x)) < 1

and it is repelling (nearby values go farther away when you iterate f)
when

  abs(f'(x)) > 1.

When abs(f'(x)) = 1, some very strange behavior can happen, including
weak attracting, weak repelling, oscillation, loops, and other funny
things.

In your case, f(x) = (x-a)/(x-b), and f'(x) = (a-b)/(x-b)^2.  You can
check that if you evaluate f'(x) at the two roots of f(x) = x, then
the product of the two f'(x) values is 1.  Therefore, if one of them
has absolute value < 1, then the other has absolute value > 1.  That
is, one is attracting and one is repelling.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/