Part I. The Logistic Family: f(x) = a x ( 1 - x ).

1) Verify that orbits of x0 > 1 or x0 < 0 diverge to negative infinity (for a in (0, 4].)

Note that if x0 > 1, then f(x0) < 0. Thus, it suffices to show that orbits of x0<0 tend to negative infinity. But a x (1 - x) - x = x ( a - 1 - a x ) < 0 for all x < 0 implying that
f(x) < x for all x < 0 (since a >= 1.)
So, all orbits must tend to negative infinity.

2) Compute the fixed points of f(x) = a x (1 - x).

Input := 

f[x_] = a x ( 1 - x )

Output =

a (1 - x) x

Input := 

fps = Solve[f[t]==t, t]

Output =

                 -1 + a
{{t -> 0}, {t -> ------}}
                   a

Input := 

fp1 = t /. fps[[1]]

Output =

0

Input := 

fp2 = t /. fps[[2]]

Output =

-1 + a
------
  a

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

Input := 

f'[fp1]

Output =

a

Input := 

Simplify[f'[fp2]]

Output =

2 - a

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.

For the trivial fixed point, we'll compare the functions with a = 0.5 and a = 1.5. In each of these cases, we see a staircase either towards or away from the fixed point. For the nontrivial fixed point, we'll compare the function with a = 2.7 to see that spirals occur near the fixed point when the slope is negative.

a = 0.5

a = 1.5

a = 2.7.

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

Since f'(0) = a, by inspection a = 1 is the desired bifurcation value.

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

Since f'((a-1)/a) = 2 - a, a = 3 by inspection.
NOTE: We do not choose a = 1 since the fixed point a-1/a is really the trivial fixed point for that value of a.

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).)

Input := 

a = .;

Input := 

pps = Solve[h[a, x]==x, x]

Output =

                   1 - a
{{x -> 0}, {x -> -(-----)}, 
                     a
 
                                     2  2      2
        -((-1 - a) a) - Sqrt[(-1 - a)  a  - 4 a  (1 + a)]
  {x -> -------------------------------------------------}, 
                                 2
                              2 a
 
                                     2  2      2
        -((-1 - a) a) + Sqrt[(-1 - a)  a  - 4 a  (1 + a)]
  {x -> -------------------------------------------------}}
                                 2
                              2 a

Input := 

pp1 = Simplify[x /. pps[[3]]]
pp2 = Simplify[x /. pps[[4]]]

Output =

     2         2              2
a + a  - Sqrt[a  (-3 - 2 a + a )]
---------------------------------
                 2
              2 a

Output =

     2         2              2
a + a  + Sqrt[a  (-3 - 2 a + a )]
---------------------------------
                 2
              2 a

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

We contrast the cases of a = 3.2 in which |(f^2)'(x)| < 1 and a = 3.5 in which |(f^2)'(x)| > 1. In the first, we see that f^2 has an attracting fixed point corresponding to an attracting period-2 orbit. In the second, we see that f^2 has a repelling fixed point (about which there is cycling) and the corresponding period-2 orbit is no longer attracting. The derivatives (which had to be the same) are given for the period-2 map at the period-2 points.

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?