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I have a family of rational maps $f$ ( the Blaschke fraction ) with one complex parameter $\rho$:

$$f(z) = \rho z^2 \frac{z-3}{1-3z}$$

I want to find $\rho$ such that map $f$ has a parabolic period 3 cycle on the unit circle.

So $\rho$ should be a solution of system of equations:

$$ \begin{cases}f^3(z_0) = z_0 \\ |(f^3)'(z_0)| = 1 \end{cases}$$

where

  • $f^3(z) = f(f(f(z)0)))$ is 3-rd iteration of function f
  • $(f^3)'(z_0)$ is the first derivative of $f^3$ with respect to $z$ at $z_0$
  • $z_0$ is the point from period 3 parabolic cycle

but computing derivative of iterated function is hard

The image of dynamic plane with Julia set:

enter image description here

It is fig 3 from paper : Near Parabolic Renormalization for Unicritical Holomorphic Maps by Arnaud Chéritat

I know that $\rho$ has a modulus equal to one (complex number):

$$ |\rho| = 1 $$

My first try was easy solution ( point on the unit circle with angle in turns = 1/3)

$$ \rho = e^{2 \pi i / 3} = -0.5 +0.8660254037844386 i$$

Then the period 3 cycle is on the unit cirlce but the cycle is not parabolic and the image is different.

Q. How can I find such $\rho$ using algebra or numerical root finding ? How can I solve above system of equations ?

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    $\begingroup$ does "a parabolic 3-cycle on the unit circle" mean $|\rho|=1$ or $|z_0|=1$ (or both?)? $\endgroup$
    – Claude
    Jun 18, 2022 at 14:32
  • $\begingroup$ @Claude Good question. I do not know the answer. $\endgroup$
    – Adam
    Jun 20, 2022 at 16:18

1 Answer 1

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It is an approximation of the answer using brute force and visual analysis.

First analyze the map f

  • find critical points
  • compute Mandelbrot set
  • find parameter $\rho$ : point on the unit circle which is parabolic point : cusp or a root

So there are 3 critical points :

  • 2 finite critical points : z=1.0 i z= 0.0
  • infinity

Here is parameter plane ( rho plane) full set

Here zoom around cusp point ( black) on the unit circle ( white)

zoom

So $\rho \approx -0.6170144002709304 +0.7869518599370003*I$

Here is dynamic plane ( z plane ) with Julia and Fatou sets

enter image description here

There are 3 Fatou components:

  • (white = 255) basin of attraction to infinity
  • ( light gray = 195) basin of attraction to fixed point z=0 ( superattracting) with critical point z=0
  • (dark gray = 135) basin of parabolic period 3 cycle ( critical point z= 1)

Increasing precision of $\rho$ value : there are two 3 point cycle. Choose rho such that 2 cycles join into one cycle.

$$\rho = e^{2 \pi i t}$$

t is changing from 0.3558289901601067 to 0.3558800000000000

Final rho (t) = e^(2piti) is rho = -0.6172665900123702 +0.7867540637673888I with turn(rho) =0.3558800000000000 and radius = 1.0000000000000000

0.3558289901601067 enter image description here enter image description here enter image description here

On the last image critical orbit goss to limit cycle from both sides . I think that it is numerical error . Rho value seem to be good

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