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The abstract for "Quartic Julia sets including any two copies of quadratic Julia sets" in Discrete & Continuous Dynamical Systems - A,36,4,2103,2112,2015-9-1, by Koh Katagata, states

[...] for any two quadratic Julia sets, there exists a quartic polynomial whose Julia set includes copies of the two quadratic Julia sets.

I am struggling with the construction, particularly as I have not found a copy of the paper to read. For example, I choose two quadratic polynomials $f_1(z) = z^2-2$ and $f_2(z) = z^2-1$, and let the quartic be $F(z) = z^4 + az^3 + bz^2 + cz + d$ with $a,b,c,d$ to be determined with $F$ behaving like $f_1$ near $z_1$ and like $f_2$ near $z_2$.

The quadratic Julia sets look like this:

Julia set -2 Julia set -1

In particular I've tried "behaving like" to mean something like affine conjugacy, and end up with things like $F(z_k + \lambda_k z) = z_k + \lambda_k f_k(z) + O(z^3)$ where $k \in\{1,2\}$. I get (using Maxima): $$\lambda_{1}^2\,b+3\,z_{1}\,\lambda_{1}^2\,a+6\,z_{1}^2 \,\lambda_{1}^2 = \lambda_1\\ \lambda_{1}\,c+2\,z_{1}\,\lambda_{1}\,b+3\,z_{1}^2\,\lambda_{1}\,a+4\,z_{1}^3\,\lambda_{1} = 0 \\ d+z_{1}\,c+z_{1}^2\,b+z_{1}^3\,a+z_{1}^4=-2\,\lambda_{1}+z_{1}$$ and $$\lambda_{2}^2\,b+3\,z_{2}\,\lambda_{2}^2\,a+6\,z_{2}^2\,\lambda_{2}^2 = \lambda_{2}\\ \lambda_{2}\,c+2\,z_{2}\,\lambda_{2}\,b+3\,z_{2}^2\,\lambda_{2}\,a+4\,z_{2}^3\,\lambda_{2} = 0\\ d+z_{2}\,c+z _{2}^2\,b+z_{2}^3\,a+z_{2}^4=-\lambda_{2}+z_{2}$$

Which is 6 nonlinear equations in 8 unknowns. Arbitrarily setting $z_1 = -1, z_2 = 1, \lambda_k \ne 0 $ gives $$ (b - 3 a + 6) \lambda_1 = 1 \\ (c - 2 b + 3 a - 4) \lambda_1 = 0 \\ d - c + b - a + 2 + 2\lambda_1 = 0\\ (b + 3 a + 6) \lambda_2 = 1 \\ (c + 2 b + 3 a + 4) \lambda_2 = 0 \\ d + c + b + a + \lambda_2 = 0 \\ $$ which has three solutions according to WolframAlpha. Their Julia sets look like this:

Julia set A Julia set B Julia set C

of which the middle one looks most like what I expected, albeit somewhat distorted on the right. It has the $\lambda_k$ about a factor of 2 apart, the other two have $\lambda_k$ about a factor -35 apart.

Is this the right way to construct the quartic?

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I have not seen the paper either but I might try something like the following. First, I believe that the general quartic is affinely conjugate to a function of the form $$ f(z) = z^4-\frac{4}{3}(a+b) z^3 + 2abz^2 + c. $$ The form of those coefficients are chosen so that the critical points are $a$, $b$, and $0$.

Now, the critical orbits dictate the structure. Thus, we might choose the coefficients so that the orbits of $a$ and $b$ behave like the critical orbits of our target quadratics and the orbit of zero diverges. That last critical orbit ensures that we have some separation between the others.

For example, suppose we want a quartic Julia set that contains "copies" of the Julia sets of $z^2$ and $z^2-1$. Since the critical orbit of $z^2$ is fixed and the critical orbit of $z^2-1$ has period two, we could write down the system $$ \begin{align} f(a) &= a \\ f(f(b)) &= b \\ f(0) &= 2.2. \end{align} $$ That gives us three equations in the three unknowns $a$, $b$, and $c$. Solving the system numerically with Mathematica, I found 64 solution, most of which do not yield anything nice. Many of the solutions yield something like what we want, though. In particular, $$ f(z) = z^4+0.689832 z^3-2.91154 z^2+2.2 $$ looks about like what we expect:

enter image description here

enter image description here

enter image description here

It's pretty clear in that last picture that immediate basin of attraction of the super-attractive fixed point is not a perfect circle, as is the case for $z^2$. I seriously doubt, though, that the author of the paper shows that the "copies" are exact geometric images. Perhaps, they mean image under another type of function?

To generate your example for $z^2-1$ and $z^2-2$, we'd use the fact that the critical orbit of $z^2-2$ lands on a fixed point after two iterations. Thus, our equations look like so: $$ \begin{align} f(f(f(a))) &= f(f(a)) \\ f(f(b)) &= b \\ f(0) &= 2.2. \end{align} $$ Now, the first equation is much higher order than before and I wasn't able to solve the system in a reasonable amount of time.

To deal with Julia sets that have attractive but not super-attractive orbits, simply add a pair of equations of the form $$ \begin{align} f(z) &= z \\ f'(z) &= \lambda. \end{align} $$ The point $z$ is then an unknown that also needs to be solved for.

For example, suppose we wish to find a quartic Julia set with a "copy" of the Julia set for $z^2-1/2$ (which has an attractive fixed point with multiplier $1-\sqrt{3}$) and a "copy" of Douady's rabbit (which has a super attractive orbit of period 3). We could then set up and solve the system $$ \begin{align} f(z) &= z \\ f'(z) &= 1-\sqrt{3} \\ f(f(f(0))) &= 0 \\ f(b) &= 2. \end{align} $$ This solution has 306 solutions, one of which yielding the quartic $$ f(z) = z^{4} +\left(5.07076+0.05647 i\right)z^{3} +\left(6.02721+0.166772 i\right)z^{2} - \left(0.00824741+0.130767 i\right) $$ whose Julia set looks like so:

enter image description here

Even though the first two equations involve just the fixed point, that fixed point must attract at least one critical point.

One thing I do not know how to deal with is find a quartic that contains copies of a couple of totally disconnected Julia sets.

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  • $\begingroup$ doi.org/10.1088/1361-6544/aa6c01 "Entire functions whose Julia sets include any finitely many copies of quadratic Julia sets" Koh Katagata 2017 uses A-B kneading sequences, Green's functions, quasi-conformal surgery, ..., and has a figure thus: mathr.co.uk/tmp/figure-3.png corresponding to example 5.2: $f(z) = (0.802 379 + 0.755 898i) + (0.155 968 − 1.747 14i)z^3 + (0.547 987 + 0.605 057i)z^4 + (0.125 851 + 0.047 9252i)z^5 − (0.548 184 − 0.430 263i)z^6$. but the main result defers to the paper mentioned in the question... $\endgroup$ – Claude Mar 26 at 6:46
  • $\begingroup$ How could you generalize your method to non-superattracting orbits? For example $z^2-\frac{1}{2}$ which is tends to a fixed point with multiplier $1-\sqrt{3}$. Would it make sense to use the multiplier too, solving $4w(w-a)(w-b)=1-\sqrt{3}$ along with $f(w)=w$ (where $f$ is the quartic)? $\endgroup$ – Claude Mar 26 at 7:45
  • $\begingroup$ @Claude Yes, your suggestion for finding a copy of the Julia set for a function like $z^2-1/2$ is reasonable. I've edited my answer to account for that. $\endgroup$ – Mark McClure Mar 26 at 15:36

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