### Parabolic Julia sets

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#### Re: Parabolic Julia sets

« Reply #60 on: August 18, 2020, 10:47:12 AM »
What is the relation of the convergence radius of the Siegel linearization to external and internal radsiu of Siegel disc ? if any can you explain it in simple words ?

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#### Re: Parabolic Julia sets

« Reply #61 on: August 18, 2020, 11:49:11 AM »
There are three terms I know of:

1. the inner radius (of the Siegel disc) denotes a circle centered at the Siegel fix point that is definitely contained in the disc.

2. The convergence radius of the linearization (sometimes also called Siegel disc radius) denotes the radius of convergence of the power series of the function h that conjugates the polynomial to the rotation: h($$\lambda\cdot z$$) = f(h(z)). Sometimes one also uses the inverse function of h to describe the relation.

3. The conformal radius of a connected domain (in C) at a specific point z0 is the (smallest) distance of z0 to the boundary of the domain.

The main relation is:

The conformal radius and the radius of convergence are identical for (bounded) Siegel discs. As I'm only working with polynomials, that's the case here.

My goal is to find a guaranteed lower bound on the inner radius (1/4*conformal radius e.g.) so I can mark a region in the TSA as definite interior and then detect its preimages as interior as well.

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#### Re: Parabolic Julia sets

« Reply #62 on: August 22, 2020, 11:18:58 AM »
"Parabolic basin in base-6 numbers"

Summary
For the polynomial (with non-floating-point representable coefficients and periodic point coordinates) similar to post #48

$f(z)=-z^3 + \frac{4}{3}\cdot z - \frac{2}{27}$

with a constructed super-attracting fix-point at x=-2/3 and a parabolic (derivative f=+1) at 1/3, using a number representation to base-6 detected the parabolic basin (image upper left).

Long version

So far I only found parabolic periodic points when they lie on the grid lines, i.e. are dyadic fractions. Here I constructed a low-degree polynomial where the fix points are deliberately set on non-floating-point representable numbers.

The coefficients of the polynomial are not dyadic fractions, so I needed to use small intervals. In the classical Mandelbrot set case this would mean having seed values that lie outside the Mset and produce a Cantor set with no interior, so no chance of finding a parabolic cycle. But here, in the degree-3 case with two distinct critical points, "outside" the connectedness locus (assuming this set's parameter vector lies on the boundary) can mean: no critical point bounded - or one. So if the attracting basin's critical point turned escaping by using specific intervals, it might very well be that the parabolic basin could be detected.

Lower left is the classical output at a high level (binary numbers) with the attracting fix point still found.

This polynomial poses an inherent second obstruction that will hinder detection of the parabolic basin forever, independent of the use of coefficient intervals: Its parabolic point is not a dyadic fraction. Hence it lies in the interior of any pixel's underlying square. And a path leading to that pixel will lead to the exterior and the parabolic point, so the source pixel can never be judged interior.

There were two ways I could think of, trying to overcome this.

First, I used an extension of the fixed-point arithmetics number type from post #21, equipped now with an arbitrary choice of the base of the numbers represented. Here I chose base-6. Then the fix point at 1/3 is representable. And indeed, upper left, a base-6 TSA detects both basins (I currently cannot detect cycles directly, just interior).

Secondly, one could try an affine conjugation e.g. with z+1/3 to shift the parabolic point to the origin, i.e. analyzing the set f(z+1/3)-1/3. If this turns out to have representable coefficients at best, chances are the basin can be detected. And this was the case (lower right).

I'm currently trying to find/construct an example where the conjugation method will fail, so using a different number type is necessary.

Technical details
• Images are non-trustworthily downscaled for display purposes and cannot be used as a filter.
• A full degree-5 polynomial with recursive monotonicity check to tighten the bounding boxes as much as possible was used - and necessary.
• Base-6 was chosen to be able to represent the fix points and coefficients and the number 1/2 as this is used in the mean-value form interval evaluation of the component functions.
• Image size and range was adjusted appropriately to move the complex coordinates of the underlying pixel's square to base-6 representable numbers. Refinement then splits one pixel into a grid of 6x6.

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#### Re: Parabolic Julia sets

« Reply #63 on: September 15, 2020, 06:22:50 PM »
"Flowers, petals and Dv"

The polynomial z-z^5 has a parabolic fix-point with derivative 1 at the origin and 4 petals (upper row, image I, 2-square, escape time, images marked with PS are point-sampled, non-rounding controlled, double precision; images marked R are reliably computed, interval arithmetics, a pixel represents a complex square).

The point-clouds

$D_v(d) = \{ z \in C : |z^v - d| < d \}$

(v in the above case is 4) form flowers (upper row, image II).

If d is sufficiently small, these D_v sets can be taken as attracting petals (M Abate, Fatou flowers and parabolic curves, 2016. Equation 1.3). Unfortunately, the author did not state how one can arrive at or prove a value to be sufficiently small. Does anyone know of such a way? Maybe the proofs of the Leau-Fatou-Flower theorem provide a constructive method, but I haven't understood those to that extent.

For small enough petals, the polynomial maps a petal to itself (AF Beardon, Iteration of Rational Functions, theorem 6.5.4). So I thought the same might happen with the Dv sets and performed the following algorithm:

1. Calculate the flower set D_v for some value d that is definitely too large, here d=4 in a reliable manner (image II, black: pixel's square lies fully in the flower; gray, pixel's square straddles the flower's boundary; white, fully outside).

2. Iterate each black square A once under f (via IA) and judge the image f(A) whether it lies fully in the flower in general (it was not tested, if it is fully in the originating flower petal). If it is inside, A is kept (turned blue), otherwise it is discarded (set to white) (image III) and the value d judged as too large. For illustration purposes, the boundary of the flower is kept (gray).

3. Overlaying the judged image and the original parabolic one shows that it is indeed too large (IV, green pixels lie both in the flower and are (point-sampled using one corner) interior; red are within the flower, but lie in the point-sampled exterior), and the flower itself is way too large for the Fatou component it intersects with.

4. Choose a smaller d, e.g. d/2 and repeat this process until no pixel A is discarded. For d=0.25 this was the case (VI). Overlaying the flower after this f(A) test with the point-sampled version, shows it is fully in the interior (image VII).

I wonder, whether this is a general way to arrive at a sufficiently small d in the above petal sense, or at least one whose Dv set lies fully in the interior of the filled-in Julia set? Any proof ideas or counter-examples?

Starting from blue does not necessarily mean landing in blue, as f(A) might intersect with a gray pixel (so the pixel itself is not fully inside), but the part of f(A) intersecting with it, still is (even in axis-parallel IA overestimation). I think this will occur near the fix point.

Such a definite d value and its Dv set could then be used as a starting set of interior squares for the assisted TSA to detect (large?) portions of the parabolic interior.

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#### Re: Parabolic Julia sets

« Reply #64 on: September 15, 2020, 08:28:40 PM »
Nice.
It is a special case : here multiplier is a real ( non complex) so there is no spiralling around parabolic fixed point ( center of the flower).
If  multiplier of parabolic periodic point will be complex number then it will be harder to find petals.

I use sectors of the circle around parabolic point:

Code: [Select]
if (abs(z)<AR) { color=GiveColorOfInterior(x,y); break;}/* interior of Filled-in Julia set   */ 
where AR is a radius of circle around parabolic point is found by Trial and error method ( it must be inside imediate basin of attraction and choosed so small that width of exterior sectors will be smaller then pixel size)

In similar case f(z) = z+z^5
https://commons.wikimedia.org/wiki/File:Julia_set_z%2Bz%5E5.png

Code: [Select]
//all points of interior fall into fixed point z = 0// check at what angle = which quadrantunsigned charGiveColorOfInterior(double x, double y){        if (x > 0.0 && y > 0.0)                return 235;        if (x > 0.0 && y < 0.0)                return 225;        if (x < 0.0 && y > 0.0)                return 215;        //if (x < 0 && y < 0)                return 185;}
in f(z) = z^4 + z
https://commons.wikimedia.org/wiki/File:Julia_set_for_f(z)_%3D_z%5E4_%2B_z.png

Code: [Select]
unsigned char GiveColorOfInterior(double x, double y){ double angle; // all points tend to z=0 thru 3 petals // check to which sector / petal fall   angle=GiveTurn(x+y*I);   if (angle<1.0/3.0 ) return 235; // ()  if (angle<2.0/3.0)     return 225;       return 175; // 9/12 ; 11/12  }

For z^4 - z https://commons.wikimedia.org/wiki/File:Julia_set_for_f(z)_%3D_z%5E4_-_z.png
Code: [Select]
unsigned char GiveColorOfInterior(double x, double y){ double angle; //double offset = 1.0/(2.0*6.0);// all points tend to z=0 thru 6 petals // check to which sector / petal fall   angle=GiveTurn(x+y*I);   if (angle<1.0/12.0 || angle > (11.0/12.0)) return 235; // (11/12 ; 1/12)  if (angle<3.0/12.0)     return 225; // 1/12 ; 3/12  if (angle<5.0/12.0) return 215; // 3/12 ; 5/12  if (angle<7.0/12.0) return 205; // 5/12 ; 7/12  if (angle<9.0/12.0) return 195; // 7/12 ; 9/12//if (angle<11.0/12.0)      return 175; // 9/12 ; 11/12  }
« Last Edit: September 15, 2020, 09:03:39 PM by Adam Majewski, Reason: pixel »

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#### Re: Parabolic Julia sets

« Reply #65 on: September 16, 2020, 11:27:56 AM »
I use sectors of the circle around parabolic point:

Code: [Select]
if (abs(z)<AR) { color=GiveColorOfInterior(x,y); break;}/* interior of Filled-in Julia set   */ 
where AR is a radius of circle around parabolic point is found by Trial and error method ( it must be inside imediate basin of attraction and choosed so small that width of exterior sectors will be smaller then pixel size)

Is your check "being inside interior" done by point-sampling / escape time? Or do you have a mathematical proof that, once the sectors are found, they contain explicitely describable petals that are definitely part of the Julia set's interior?

As the parabolic fix point is in the Julia set itself and not the interior, any circle centered around it must intersect with the exterior, so you mentioning "it must be so small and inside the immediate basin" is an image-based criterion for illustration purposes rather than a mathematical one, or am I mistaken?

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#### Re: Parabolic Julia sets

« Reply #66 on: September 16, 2020, 05:28:57 PM »
First I really like what you are doing, Thx

Is your check "being inside interior" done by point-sampling / escape time? Or do you have a mathematical proof that, once the sectors are found, they contain explicitely describable petals that are definitely part of the Julia set's interior?
In my programs I check it by point sampling with escape time, so I do not have a math proof.
There are cases were I have a proof but not for radius but for angles describing sectors :
https://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/r_a_directions

Quote
As the parabolic fix point is in the Julia set itself and not the interior, any circle centered around it must intersect with the exterior, so you mentioning "it must be so small and inside the immediate basin" is an image-based criterion for illustration purposes rather than a mathematical one, or am I mistaken?

You are right. It is image based not mathemathical one.

I am very curious if you will find differences between my images and TSA images ?
We can make a challenge (:-))

It is a good chance for me to learn

btw. AFAIK there is no simple relation between polynomial and number of petals.
« Last Edit: September 16, 2020, 07:30:05 PM by Adam Majewski, Reason: petals »

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#### Re: Parabolic Julia sets

« Reply #67 on: September 17, 2020, 09:06:21 AM »
First I really like what you are doing,
Thanks,

I am very curious if you will find differences between my images and TSA images ?
We can make a challenge (:-))
You'd win. For the TSA I could only provide largely gray images with definite exterior, but no interior and maybe an impression of how many petals there could be (although this might be misleading, as gray pixels could vanish). And the Dv sets only work for the special cases z-z^n+... (and their affine conjugates).

btw. AFAIK there is no simple relation between polynomial and number of petals.
In the Abate paper I mentioned, there is a theorem 1.2, which, if I understand it correctly, states, that if you have a multiplier of e^2*i*pi*p/q for p,q coprime, the q-th iterate of f has multiplicity of a multiple of q, and hence a multipe of q attracting petals. As the (filled in) Julia sets for a polynom f and any of its iterates are identical, the Fatou components harbouring the petals for q-iterate-f are also existent in the f Julia set and attached to the parabolic point, so I guess it would be rather strange if some of them owned no longer attracting petals. So, the number of petals is a multiple of the denominator of the rational angle.

Quite interesting how powerful denominators are: Here they define the number of petals, and in the Brjuno number formula, they define whether the set is Cremer or Siegel (although I think the Brjuno condition could be reformulated using the numerator, as the quotients p_n/q_n approximate an irrational and are correlated).

---------------

Another piece of information on the Dv sets.

For points on the central line of an attracting petal (assuming the origin as a parabolic fix point), once close enough to the fix point, complex numbers there are directly attracted to it, i.e, |f(z)| < |z| for z close enough (M Abate, example 1.1). For the set I posted, z-z^5, in the positive real axis direction, the open unit interval would be such a "close enough". Every point therein gets mapped closer to the origin. And as the point z=1 is in the Julia set itself (it is mapped to the origin), no petal there can exceed distance 1 to the origin (the same holds for the other 3 coordinate axis directions due to symmetry).

So as the value d=0.25 I found with the 1-iteration approach has flowers not exceeding 1, I think it's worthwile scanning a bit more through the literature for a "small enough" d.

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#### Re: Parabolic Julia sets

« Reply #68 on: September 19, 2020, 05:58:31 PM »
"Thought experiment"

btw. AFAIK there is no simple relation between polynomial and number of petals.

I think one can symbolically calculate the number of attracting petals. I first tried it here with z^2+c for c lying on the main cardioid having a parabolic fix-point (see below).

The general idea after some reading (B Dozier, Classification and structure of periodic Fatou components, 2012, esp. chapter 3.3):

1. Bring f(z)=z^2+c for a parabolic c into g(z)=z^2+L*z by conjugation. This moves the sole fix point of f to the origin, hence L is the parabolic multiplier.

2. Determine the rational angle p/q (coprime) of L.

3. Calculate symbolically the q-fold composition of g(z), say G. This has the form (L^q)*z+a*z^n+higher order terms, where L^q equals +1 (rational angle rotation) and has the form of the Abate article with n being the lowest non-linear term with a non-zero coefficient (they use a minus sign, but as there are no restrictions on the coefficient a, especially not that it has to be a positive real value, using + or - is equally working).

4. The number of attracting petals is then n-1.

Below is a maxima script (not fully tested), the main function is attPetals( c value of z^2+c ) which returns the number of attracting petals - or a negative value in case of an error (non parabolic, not a fix-point or otherwise).

Note, that it only works if the provided c value is exact, so a numerical estimate will not have multiplier 1 and will return an error value. I'm not sure yet that relaxing this requirement and working with approximate values will provide the correct answer.

Code: [Select]
kill(all);numer:false;display2d:false;attPetals(c) := ( f:z^2+c, g:z^2+L*z, /* conjugate f to g */ h:expand( (subst(a*z+b,z,g) - b) / a ), sol:solve( [ coeff(h,z,2) = coeff(f,z,2), coeff(h,z,1) = coeff(f,z,1), coeff(h,z,0) = coeff(f,z,0) ],[a,b,L]), la:-1000, qdenom:-1, for i from 1 thru length(sol) do for k from 1 thru length(sol[i]) do if lhs(sol[i][k]) = L then ( cval:expand(rhs(sol[i][k])), if cabs(cval) = 1 then ( print("L",i), angle:carg(cval), so2:solve(angle=2*%pi*p/q,[p,q]), tp:-1000,tq:-1000, for n from 1 thru length(so2[1]) do ( if lhs(so2[1][n]) = p then tp:rhs(so2[1][n]), if lhs(so2[1][n]) = q then tq:rhs(so2[1][n]) ), print("angle p/q: ",tp," / ",tq), if denom(tq) # 1 then qdenom:-1 else ( qdenom:denom(tp/tq), R:cabs(cval), print("R=",R) ) R:cabs(cval), print("R=",R) ) ), if qdenom > 0 then ( F:R*exp(2*%i*%pi*tp/tq)*z+z^2, Ferg:F, /* compute composition q */ for i from 2 thru qdenom do Ferg:subst(F,z,Ferg), Ferg:expand(Ferg), /* find highest non-constant term */ err:0, petals:0, for i from 0 thru hipow(Ferg,z) do ( if i = 0 then if coeff(Ferg,z,i) # 0 then err:1, if i > 1 then if coeff(Ferg,z,i) # 0 then if petals = 0 then petals:i-1 ), if err # 0 then petals:-2, petals ) else err:-3);/* point of Mset */ap:attPetals( 1/4+1/2*%i );print("predicting ",ap," attracting petals for z^2+",cc)\$
The returned values were (image, red dot = fix point):

Code: [Select]
 petalsc=0.25 1c=-0.75 2c=0.25+0.5*i 4c=3/8+sqrt(3)/8*i 6`
I'm currently trying to extend this to parabolic cycles of period b > 1.

The general idea is (not fully proven): Instead of taking the polynomial f, one uses the b-fold composition of f, termed F. This breaks the cycle into b parabolic fixpoints of the same derivative as the b-cycle in f (and having the same Julia set). Then one of those parabolic fix points is translated to the origin via affine conjugation. This formula then has the form L*z+higher order terms. L is equal to the multiplier of the initial cycle. Then one determines the rational angle p/q of L and uses the q-fold composition of F and looks for the first non-linear occuring term. Its degree minus 1 is the number of attracting petals of that fix point and hence of every periodic point of the initial parabolic cycle in f.

« Last Edit: September 20, 2020, 09:36:10 AM by marcm200, Reason: code: error handling improved »

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#### Re: Parabolic Julia sets

« Reply #69 on: September 20, 2020, 08:56:28 AM »

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