You might want to look at the help screens below
for this module.
The screen below shows how the TI-92 can find the equilibrium points for a logistic
model of the form
p(n + 1) = f(p(n))
where
f(p) = a(1 - b * p) * p.
Notice the answers
p = 0
and
a - 1
p = -----
ab
involve the constants a and b.
The same procedure solve can be used to find 2-cycles as shown in the screen below.
Notice, however, that because there are four solutions and some of the solutions are very long they don't all appear on the screen. Use the cursor pad to move the cursor up and then right to see the second and third solutions as shown in the screen below.
Finally, use the cursor pad to see the last of the solutions as shown in the screen below.
Now we have all four solutions of the equation f(f(p)) = p.
p = 0
a - 1
p = -----
ab
Sqrt(a - 3) Sqrt(a + 1) a + 1
p = ----------------------- + -----
2ab 2ab
-Sqrt(a - 3) Sqrt(a + 1) a + 1
p = ------------------------ + ---------
2ab 2ab
The first two solutions are equilibrium points, so we are really interested in the last two. The next step is to save these two solutions as shown in the screens below.
Next we calculate the derivative gp(p) of the function f(f(p)) as shown in the screen below.
We want to find out for which values of a, gp evaluated at the two 2-cycle points is equal to 1 or -1. The following screen shows how to do that.
Because this is a complicated equation, the TI-92 used numerical methods to solve it and did not find all the solutions. Nonetheless it did find the solutions needed in the browser window.
Finally we evaluate gp(p) at one 2-cycle point and one value of a between 3 and 3.44948 and another value of a between 3.44949 and 4 using a typical value of b as shown in the screen below.
