Consider the cubic functions f_c(z)=z^3+c.
Show that f_c has a super attractive fixed point.
Show that f_i and f_{-i} have super-attractive orbits of period 2.
Find a value of c so that f_c has a super-attractive orbit of period 3.
Illustrate the above phenomena using our fun polynomial Julia set plotter…
So this is the multibrot set for z^3 + c, so I think we want to keep our values of c inside this set, or else we might not have an attractive point at all, but I am not 100% sure of this.
Anyway at c = 0 we we have f_0(z) = z^3, this means we have a fixed point at z = 0 cause
f_0(0) = 0. Now let’s look at the derivative of f_0(z),
f'(z) = 3z^2 so |f'(0)| = 0, then we have a super attractive fixed point. Like in the case of z^2 we looked at in class, when c is close to 0 it will behave like c=0, so even though f_{0.1}(z) = z^3 + 0.1, does not have a derivative of value 0 it does have one very close to zero. That is for the fixed point p \approx 0.10103140329933635, f'(p) \approx 0.03062203335789946 so not technically supper attractive, but I assume this behaves very similarly to a super attractive fixed point. I am not sure if f_c(z) = z^3 + c has a super attractive fixed point for all values of c, but it does have at least one.
Lets compare the filled Julia sets for each of the functions we defined
f(z) = z^3
f(z) = z^3 + 0.1
This all looks great!
So this is quite rare actually. Typically, for each component, you’ve got one super-attractive orbit. There’s only one component where that orbit has length one, though.
I bet someone else can deal with f_{\pm i}. If you’re interested in searching for a value of c such that f_c has a super-attractive orbit of period 3, I’d recommend looking at your picture near the upper right.
