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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/