The Problem.
Part I. A study of the logistic map f(x) = a x ( 1 - x ).

Do all computations with the parameter a as an arbitrary number between 0 and 4 inclusive unless indicated otherwise and concentrate on values of x in [0, 1].

1) Verify that for all 1 <= a <= 4, if x0 > 1 or x0 < 0, then the orbit of x0 diverges to negative infinity.

2) Compute the fixed points of f(x).

3) Compute the slope of f(x) at each fixed point.

4) Verify by example that fixed points xbar repel when |f'(xbar)| >1 and attract when |f'(xbar)| <1. Further, verify by example that if f'(xbar) < 0, then orbits cycle near xbar and when f'(xbar) > 0, orbits staircase near xbar.

5) Find the a-value (bifurcation value) for which the "trivial" fixed point switches from attracting to repelling.

6) Find the a-value for which the "nontrivial" fixed points switches from attracting to repelling.

7) Graph f^2(x) = f(f(x)) along with x for various values of a between 0 and 4. Can you tell for what a-value two period-2 points of f(x) emerge?

8) Determine formulas for the period-2 points of f(x) (i.e., the new fixed points of f^2(x) which are not fixed points of f(x).)

9) Verify by example that periodic points xbar repel when |(f^2)'(xbar)| > 1 and attract when |(f^2)'(xbar)| < 1.

10) Find the a-value(s) for which the period-2 points switch(s) from attracting to repelling.

11) Let a = 3.55 and plot f^3(x), f^4(x), f^5(x), f^6(x), f^7(x), f^8(x) along with the replacement line y = x. List how many periodic points f(x) has of every period less than or equal to 8. Do you think that f(x) has any higher periodic orbits (when a = 3.55?) Why or why not?