For any polynomial map $f$ we can define the filled Julia $K$ to be closure of the complement of $ \Omega$ in $\mathbb{C}$ of the basin of infinity
$$\Omega = \{z \in \mathbb{C}; f^{\circ n}(z)\rightarrow \infty \}.$$
Thus the boundary of $K$ is what is often called the Julia set. Now there are examples like $f=z^2$ for which the filled Julia set is connected and locally connected.
Lets consider iterations of the form
$$f(z) = z^2 + c \times z^5$$
for complex $z$ and a given positive real $c>0$
For what values $c$ is the Filled Julia set connected with a single component ??
I know that it is connected with a single component for $c=0.23$ but has infinitely many connected components for $c=0.22$
Where is the crossover value of $c$ to go from a single component to infinitely many ?
Does that value have a closed form ?
Do I need to solve a quintic or compute an infinite sum ?
I know -or at least more or less know- that if one finite critical point of a finite number of iterations escapes to infinity, then the filled-in Julia set consists of infinitely many components.
But Im not able to apply it.
I guess I only need to study the zero's of $f'(z)$ to see if they go to infinity ?
But how can I be certain if they will or will not ?
And especially the boundary cross-over value for $c$ ITSELF ; does it have one component or infinitely many ?
edit
Using the comment of Claude, I will use his computation.
I will try to make a little logical conclusion.
For critical points,
$$f′(z)=2z+5cz^4=z(2+5cz^3)=0$$
has solutions
$$z=0$$
,
$$z= \sqrt[3]{\frac{−2}{5c}}$$
(1 real, 2 complex (1 conjugate pair)).
Now lets iterate once with this real number.
( 0 is trivial so we try the other real value )
$$z = \sqrt[3]{\frac{−2}{5c}}$$
$$z^2 = \sqrt[3]{\frac{4}{25c^2}}$$
$$z^5 = \sqrt[3]{\frac{−32}{3125c^2}}$$
$$z^2 + c z^5$$
then gives us
$$\sqrt[3]{\frac{4}{25c^2}}+\sqrt[3]{\frac{−32}{3125c^2}}$$
so we solve that by setting it equal to $0$ or the starting value $z$ above :
$$\sqrt[3]{\frac{4}{25c^2}}+\sqrt[3]{\frac{−32}{3125c^2}}=0$$
no real solutions.
Further
$$\sqrt[3]{\frac{4}{25c^2}}+\sqrt[3]{\frac{−32}{3125c^2}} = \sqrt[3]{\frac{−2}{5c}}$$
This has 1 real solution :
$$c = – \frac{54}{625}$$
although this is actually negative and I asked about real $c>0$, we can conclude
$$f(z) = z^2 – \frac{54}{625} \times z^5$$
has a filled Julia set with exactly one connected component.
However this is also true of
$$f(z) = z^2 – \frac{54}{624} \times z^5$$
and
$$f(z) = z^2 – \frac{54}{626} \times z^5$$
So nothing special.
The question remains open.
Maybe we should use derivatives again or maybe we should use geometric function theory.
And maybe we should use a higher number of iterations before solving equations.
But that might get more complicated.
I have not considered the non-real solutions $c$ or $z_0$ since the question was about real $c$ but maybe it will help ? I dont know.
edit2
$$f(z) = z^2 – 0.07 \times z^5$$
has one component.
and
$$f(z) = z^2 – 0.03 \times z^5$$
has infinitely many components.
so $c$ around $0.2$ is not the only real interesting value.
Again, I have no closed form for this boundary value of $c$ between $-0.03$ and $-0.07$.
Nor do I know if the boundary values around $-0.07$ and $0.2$ are related by an algebraic equation.
It seems there are 5 intervals for $c$ more or less like this
$[-oo,-0.05[$ has one component.
$[-0.05,0[$ has infinitely many components.
$0$ has one component.
$]0,0.225[$ has infinitely many components.
$[0.225,+oo]$ has one component.
Which probably makes sense for a degree $5$ iteration.
Best Answer
Conclusion
The special point is at $$c = 0.2294487977025897\ldots$$ satisfying $$f_c(f_c(z_0(c))) = f_c(z_0(c))$$ where $$z_0(c) = \sqrt[3]{-\frac{2}{5c}}$$ is the non-zero real critical point. Its filled-in Julia set has a single component.
Finding the point
Script for (wx)Maxima computer algebra system:
Being more explicit:
gives the equation $$5^{35/3}c^{5/3}-5^{10}\cdot3\cdot2^{2/3}c-5^5\cdot3^4\cdot2^{8/3} = 0$$
whose real root $c \in [0.2, 0.3]$ seems to match the solution found in the previous code block.
So, yes, you need to solve a quintic (a polynomial in $x = c^{1/3}$).
Finding the number of components
Consider the long-term behaviour of (one of) the other non-zero complex conjugate pair of critical points to know how many components the Julia set has for this specific $c$. Using SageMath I found that the other non-zero critical point $z_1(c)$ is pre-periodic, with $$f_c(z_1(c)) = f_c(f_c(f_c(z_1(c))))$$ so it remains bounded. As no critical points escape to infinity, the Julia set for this $c$ value has a single component.
Script for SageMath:
The point is in the right place
From the left
Arbitrarily close nearby values of $c$ less than the special point give filled-in Julia sets with infinitely many components.
Considering small real perturbations in $c$ near the special point and non-zero real critical point, the difference in $z$ has the opposite sign to the difference in $c$. Now considering small real perturbations in $z$ near the eventually fixed point in the limit of the perturbation in $c$ going to $0$, if the perturbation in $z$ is initially positive, then it grows without bound, eventually escaping towards infinity.
Continuing with SageMath:
From the right
It remains to be shown that arbitrarily close nearby real values of $c$ greater than the special point give filled-in Julia sets with a single component.
If we have $|z_{n+1}| < |z_n| < \epsilon$, then iterations are certainly bounded as $n \to \infty$. Substituting, we get $$|\epsilon^2 + c \epsilon^5| < \epsilon^2 + |c| \epsilon^5 < \epsilon $$
And if $|c| > \frac{2}{5\epsilon^3}$ then $z_0 < \epsilon$. So $$\frac{2}{5\epsilon^3} < |c| < \frac{1 - \epsilon^3}{\epsilon^4}$$ whence $$5 \epsilon^3 + 2 \epsilon < 5,$$ that is $$\epsilon < \epsilon_{\text{max}} \approx 0.86755924139011\ldots$$ and $$|c| > R_{\text{min}} \approx 0.6125796570109\ldots$$
That still leaves a big gap of positive real $c$ that haven't been proven to remain bounded.
Visualisation
Here's the "Mandelbrot set" plot of the $c$ plane with the proven-bounded part in red, points that definitely escaped to infinity in white (these have filled-in Julia sets with infinitely many components, the rest have a single component), and the remainder in black:
Here's a zoom-in of the previous image with a grid overlaid in green (spacing 0.1 units, the axes are slightly thicker); the special $c$ is at the right-most corner of the white region:
Here's an animation of the filled in Julia set (black) for the special $c \pm \frac{1}{10}$.
Parameter plane GLSL fragment shader source code snippet for FragM (fork of Fragmentarium):
Filled-in Julia set GLSL fragment shader source code snippet for FragM:
Reproducibility
Try the SageMath scripts online here. Today the output is:
I used the Debian package for
maxima
, version 5.46.0-8 (not the latest version available in the Bookworm repositories).I used https://sagecell.sagemath.org as interface to SageMath.
Images were rendered with Fragmentarium 2.5.7.221026 Digilantism with my example files (include path:
Examples/Include;Examples/Claude
).