The logistic family is defined by f_{\lambda}(z) = \lambda z(1-z).
Sketch several Julia sets from this family using our fun polynomial Julia set plotter. Maybe \lambda=2, 3, 3.2, and 1+0.5i would be decent starting values.
Find a value of \lambda such that f_{\lambda} has a super attractive orbit of period 3.
So I have done part one:
\lambda = 2
\lambda = 3
\lambda = 3.2
\lambda = 1 + 0.5i
\lambda = i
I think \lambda = i has period four. After doing a composite of f(f(z)) = -iz^4 - 2z^3 + iz^2 + z^2 - z, then setting that equal to z and solving using by factoring out the original fixed points and solving the quadradic I found two more fixed points at
z = \dfrac{-(1-i) \pm \sqrt{4+2i}}{2}
then I used the following code to check and see if these had a period of 4, which is what I guessed they must be from the picture
def iteration_function(gamma, z): return gamma*z*(1-z)
gamma = 1j
z_0 = (-1+1j + (4-2j)**(1/2))/2
period_check = 10
for i in range(period_check):
print(z_0)
z_0 = iteration_function(gamma, z_0)
and I got this result:
(0.5290855136357462+0.2570658641216772j)
(0.014953785392391838+0.31523689139316946j)
(-0.3058089217498531+0.11410446739005904j)
(-0.1838927956688495-0.3863081888932918j)
(0.5283867745840066-0.06847533916173948j)
There might be some rounding error going on but it seems like the real part came back to almost the same value after 4 iterates. Still its not really close on the imaginary part.
I know to find a super attractive fixed point of orbit 3, we want to find F(z) = f(f(f(z))), then solve the system
F(z) = z
|F'(z)| = 0
but this is easier said than done, not sure if I am missing some trick, but it seems like you would want to find the period 2 orbits, then use them to factor down F(z).
So after some thought and talking with Mark, I think I got an answer. We need to check and see what the super attractive fixed point of the logistic family is
f(z) = \gamma z (1-z)
f'(z) = \gamma - 2\gamma z
so
f'(z) = 0, when z = \tfrac{1}{2}
So, what \gamma can we have that will give us 0.5 as a fixed point. f(1/2) = \gamma (1/2)(1-1/2) gives use \gamma = 2, so all super attractive orbits should have a factor of two.
Now I wrote some code that will help us find our \gamma
from sympy import *
c = Symbol('c')
def g(z): return c*z*(1-z)
def g_3(z): return g(g(g(z)))
expand(g_3(1/2))
this finds the F(z) = f^3(z) for the value of a super attractive fixed point of our original function,
then we just need to find the zeros,
from numpy import roots
root_list = roots([-0.00390625,0.03125,-0.0625,-0.0625,0.25,0,0,-1/2])
print(root_list)
which produced the results
[ 3.83187406+0. j 2.55267525+0.95945597j 2.55267525-0.95945597j
2. +0. j -1.83187406+0. j -0.55267525+0.95945597j
-0.55267525-0.95945597j]
so I just used the value \gamma = 2.55267525+0.95945597i and found there did seem to be some 3 orbits and when we look at the picture we can confirm there are three branches
