Logistic Map Analysis

The numerical results of the last section raise many tantalizing questions. Let s try to answer a few of the more straightforward ones. 

Flip bifurcations are often associated with period-doubling. In the logistic map, the flip bifurcation at r = 3 does indeed spawn a 2-cycle, as shown in the next example. 

Solution A 2-cycle exists if and only if there are two points p and q such that f p = q and f(,q)= p. Equivalently, such a p must satisfy f(f(p))= p, where /(x) = rjc(l-x). Hence p is a fixed point of the second-iterate map 

To find p and q, we need to solve for the points where the graph intersects the diagonal, i.e., we need to solve the fourth-degree equation / (x) = x. That sounds hard until you realize that the fixed points x = 0 and x -1 - -J- are trivial solutions of this equation. (They satisfy /(x ) = x, so / (x ) = x automatically.) After factoring out the fixed points, the problem reduces to solving a quadratic equation. 

We outline the algebra involved in the rest of the solution. Expansion of the equation / (x)-x = 0 gives r x(l-x)[l - rx(l-x)]-x =0. After factoring out X and X - (1 - -t) by long division, and solving the resulting quadratic equation, we obtain a pair of roots 

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