If f(x) = (c-x)x and x(k+1) = f(x(k)) [iterative formula]. How do i prove that if 1<c<2 and 0<x(0)<c that ...

x(k) tends tcwards to c-1(where c/2 corresponds to c-1). To do this I have found the derivative of f(x) and found f'(x) = 0....which is c/2 (which corresponds to c-1). If this is correct, then what relevance do the boundary counditions have? ( 1%26lt;c%26lt;2 , 0%26lt;x(0)%26lt;c )

If f(x) = (c-x)x and x(k+1) = f(x(k)) [iterative formula]. How do i prove that if 1%26lt;c%26lt;2 and 0%26lt;x(0)%26lt;c that ...
This is the famous logistic map. It has two fixed points: x=0 and x=c-1. When c%26lt;1 the series converges to zero. When 0%26lt;c%26lt;1, it converges to c-1, and when c%26gt;1 it shows period doubling, and becomes chaotic somewhere for c%26gt;3.5 or so. For c%26gt;4, the series blows up. I think the limit 2 in the problem must have been an error (nothing interesting there that I know of).


Here follows a simple mathematica program that shows those effects. You should be able to write your own on a hand calculator:





namax = 100; da =


4/namax; nmax = 1000; nlast = 20; seta = {}; setal = {}; Do[{a =


na da; x = .1;


setx = {{0, x}};


Do[{x = a x (1 - x), setx = Append[setx, x]}, {n, 1, nmax}];


Do[seta = Append[seta, {a, Part[setx, nmax - k]}], {k, 0,


nlast}]}, {na, 0, namax}];


ListPlot[seta, PlotRange -%26gt; All, Axes -%26gt; {False, True},


Frame -%26gt; True]
Reply:If the iteration is convergent as k-%26gt;∞, then x(k+1) = x(k) = x.


x = (c-x)x


=%26gt; x^2 + (1-c)x = 0


=%26gt; x(x+1-c) = 0


Since x ≠ 0, we must have


(x+1-c) = 0


So, x = c-1 as k-%26gt;∞

brenda song