Re: Feigenbaumnumber Message #2 Posted by Julian Miranda (Spain) on 30 May 2004, 1:22 p.m., in response to message #1 by Tizedes Csaba [Hungary]
I don't have the program. From the book "Chaos. Making a New Science" by James Gleick, I can tell you the equations he was working with.
First he worked with x(t+1)=r·x·[1x(t)]. A quadratic equation. y=rxrx^2.
You start with a value for "r" and a value for "x" and the result is the next value for "x" if you keep iterating finally you get always the same value. With some pairs of values you end in an alternation between two values. He searched for the values of "r" that produced this duplication.
Samples of values r=2.7 x=0.02 that ends in 0.6296. r=3.5 x1=0.850 x2=0.3828 x3=0.8269 x4=0.5009, in this case there is a double duplication so we end with four x values.
Then he repeated the process with the following equation
x(t+1)=r·sin PI·x(t). Where PI is 3.14159... as you may supposse.
I hope that can help you a little.
