### Parabolic Julia sets

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#### Adam Majewski #### Re: Parabolic Julia sets

« Reply #30 on: June 19, 2020, 07:27:32 PM »
My question with approximative c values always is, how accurate is the image? If the exact c value has irrational imaginary and/or real part and is on the boundary of the Mset, any finite precision approximations can be sometimes inside the Mandelbrot set or outside
Yes, it is a good question.
In other words, why when I take a number ( internal angle = floating point number) I draw    Siegel disc  not parabolic Julia set?
I think that is math question ( number theory- how fast irrational number is aproximated by rational numbers)

#### Adam Majewski #### Re: Parabolic Julia sets

« Reply #31 on: June 19, 2020, 09:34:39 PM »
Not for the TSA, my current record is still just slightly above 100. I think I have to wait for a new computer with more memory and the parallelization I'm thinking of, working out (separating the plane into tupels of subregions, pairwise not sharing a one-step forward or backward iterate bounding box).
Can you post 100 ?

#### marcm200 #### Re: Parabolic Julia sets

« Reply #32 on: June 21, 2020, 10:01:29 AM »
Can you post 100 ?

My current record holder: cycle of length 130: posts #134 and #135 at: https://fractalforums.org/fractal-mathematics-and-new-theories/28/julia-sets-true-shape-and-escape-time/2725/msg19173#msg19173

A new one with a cycle of length 105: https://fractalforums.org/fractal-mathematics-and-new-theories/28/julia-sets-true-shape-and-escape-time/2725/msg22335#msg22335
(I posted it in one of my other, TSA threads as the set is not parabolic. My apologies for this redirecting.)

#### Adam Majewski #### Re: Parabolic Julia sets

« Reply #33 on: June 21, 2020, 10:12:14 AM »
My current record holder: cycle of length 130: posts #134 and #135 at: https://fractalforums.org/fractal-mathematics-and-new-theories/28/julia-sets-true-shape-and-escape-time/2725/msg19173#msg19173

A new one with a cycle of length 105: https://fractalforums.org/fractal-mathematics-and-new-theories/28/julia-sets-true-shape-and-escape-time/2725/msg22335#msg22335
(I posted it in one of my other, TSA threads as the set is not parabolic. My apologies for this redirecting.)

Cool. Period 3 looks nice but period 105 looks like need more time to compute (:-))

My record is only 30
https://commons.wikimedia.org/wiki/File:Parabolic_Julia_set_for_internal_angle_1_over_30.png

but I'm not happy with it : see some distorion of Julia near alfa

look also lins to periods 377 and 987

http://maths.nju.edu.cn/~yangfei/gallary-2.html

https://www.math.univ-toulouse.fr/~cheritat/GalII/galery.html

#### marcm200 #### Re: Parabolic Julia sets

« Reply #34 on: July 01, 2020, 08:19:59 AM »
"The next best thing: Not all going out"

To detect interior for any Julia set with the TSA, the algorithm must find a collection of pixels that keep to themselves in terms of bounding boxes only intersecting with collection pixels. While this currently cannot be found for parabolic sets, I tried the next best thing: Finding regions in the plane where only a few bounding boxes of pixels from within land fully outside. A parabolic fix point might then lie there, or one might be deep within a Siegel disc.

Here is the first example for (z^2-0.75)^2-0.75 having 2 parabolic fix points (z=0.5, z=-0.5) and one repelling one (z=1.5). It is a 2-iterate version of the classical z^2-0.75 - but this did not give a positive result in the algorithm below.

Technical details / Image description
• The complex 2-square was tiled în two hierarchical steps. First into 2^14 x 2^14 pixels, whose bounding boxes were to be analyzed. Secondly in groups of 2^N x 2^N pixels which define a region. The main image below has N=4.
• If the bounding box of a pixel lands fully outside that region, the pixel is colored white, otherwise black.
• Grid lines (gray) do not shadow underlying pixels as the regions were automatically cropped and placed in a distance of 2 pixels to one another so there's room for circumferencing lines.
• Red circumferenced pixels contain a definite repelling fix point,
• Blue circumferences a region where a parabolic or attracting fix point can lie.
• All calculations were done with a sufficient numbertype or with the kv IA library, so rounding errors were accounted for and locations and bounding boxes are accurate.
• As all IA-computed fix point regions here resulted in point-"intervals" falling onto a grid point in sufficiently fine regioning and therefore belonging to 4 regions, it was kind of arbitrary which harbouring region was marked blue or red. I chose the one where the fix point is lower and left.

The right column shows the cropped fix points. The bottom depicts a series of different secondary region sizes.

One parabolic point shows black pixels, the second one does not (yet?) at the current level. Interestingly the repelling fix point also shows some black pixels, so not all points in that vicinity escape rapidly and with large steps.

The algorithm here only allows for a very weak statement: "If black is present, the union of all black covers at least one parabolic fix point and part of its immediate basin." But it gives a hint where the parabolic point is - without the use of guiding data (the whole 2-square was analyzed).

I hope I can detect (part of) a Siegel disc next.

[/list]
« Last Edit: July 01, 2020, 08:36:00 AM by marcm200 »

#### marcm200 #### Re: Parabolic Julia sets

« Reply #35 on: July 02, 2020, 08:26:09 AM »
"A Cremer number"

Result
Between 10.099999999999 and 10.1 there is a Cremer number.

The quadratic formula $$z^2+e^{2i\dot\pi \cdot \alpha}\cdot z$$ has a Siegel disc if $$\alpha$$ is a Brjuno number, it is a Cremer set otherwise. Almost all irrationals are Brjuno.

In the article below I found a way of computing a Cremer number, at least in principle:

Code: [Select]
EF LeeThe structure and topology of the Brjuno numbersTopology proceedings, vol 24, 1999
Proposition 1.1 gives an example that is defined recursively:

There is a non-Brjuno number $$\alpha$$ which is approximated by converging rationals (based on continued fractions):

$\frac{P_{2k}}{Q_{2k}} < \alpha < \frac{P_{2k+1}}{Q_{2k+1}}$

with

$P_n = a_n\cdot P_{n-1} + P_{n-2}\\ Q_n = a_n\cdot Q_{n-1} + Q_{n-2}\\$

at starting conditions

$$P_0=a_0, P_1=a_1\cdot a_0+1, Q_0=1, Q_1=a_1$$

and the main Cremer-producing formulae

$a_0=10\\ a_1=10\\ a_n=Q_{n-1}^{Q_{n-1}}$

I wrote a small maxima program that could in principle perform the iteration, but after P2 memory error already occurs - (calls to cremer_init(10,10) followed by several cremer_iterate()).

Code: [Select]
cremer_init(a0,a1):=( listan:[a1,a0], listpn:[a0*a1+1,a0], listqn:[a1,1], approx:[listpn/listqn,listpn/listqn])$cremer_iterate():=( an:listqn^listqn, /* very fast growth */ listan:append([an],listan), tmpqn:an*listqn+listqn, tmppn:an*listpn+listpn, listqn:append([tmpqn],listqn), listpn:append([tmppn],listpn), approx:append([tmppn/tmpqn],approx))$
The first 3 rationals are (indices 2,1,0):

$\frac{1010000000010}{100000000001}, \frac{101}{10}, 10$

numerically: 10.099999999999, 10.1, 10.0.

The difference of the first two rationals (lying on different sides of alpha) is smaller than 10-11. I will compute a suitable c value for z^2+c to see if it fits into a complex interval of my usual width 2^-25 to draw some images next.

To arrive at a narrower interval (I intend to get a final c interval of width 2^-48), a higher number of iterations is required:
(The following are just sketches of ideas, so might turn out not to be feasible, but I like this area in mathematics)

• Number type As the fraction contains mainly zeros (like a Liouville number),  implementing a specific number type that works with: non-zero-digit followed by a lot of zeros, then another non-zero-digit, etc.
• an = 10^something I think if a_n is larger than in the definition above, the number the series converges to is also a Cremer number (fast growing denominators Q are a characteristic of non-Brjuno numbers). E.g. taking a value that is the next greater power of 10 than Q(n-1)^Q(n-1), eventually reaching a power-tower of 10^10^10^.., where only the height of the tower must be stored.
• constant^Q Maybe it is possible to use a constant base for exponentation in the a_n formula, which would make the numbers much smaller.

Does anyone have other ways of arriving at a specific Cremer number? Or articles with different existence proofs?

#### Adam Majewski #### Re: Parabolic Julia sets

« Reply #36 on: July 02, 2020, 08:03:50 PM »
"A Cremer number"

Does anyone have other ways of arriving at a specific Cremer number? Or articles with different existence proofs?

Thank you for that post. It is very interesting

Quote
"A Cremer internal angle is obtained as 0.10001000000000000010000... with fast growing 0-gaps, but any finite approximation is parabolic." ( Wolf Jung )

IMHO the problem is : if you use floating point number with fixed precision then it is a rational number ( so it gives parabolic Julia set)
https://en.wikibooks.org/wiki/Fractals/Mathematics/sequences#sequence_from_Siegel_disk_to_Leau-Fatou_flower

#### marcm200 #### Re: Parabolic Julia sets

« Reply #37 on: July 02, 2020, 08:28:12 PM »
@Adam: Thanks, "fast growing zeros" - yes, that's what I got from the paper as well. I think it's not restricted to zeros, as long as the denominators grow exponentially fast, so the Brjuno sum diverges.

As for Wolf Jung's statement: I am not computing a rational approximation of an angle to finally produce an image. I plan on using two rational P/Q-values, one from each side of the converging inequality as an interval and with that I compute the e^... term via interval arithmetics, so the final image will hopefully be informative of some sort. Naturally, this interval angle will include rational ones, so this will be - in the sense of Figueiredo, an intersection of underlying sets and neither merely a parabolic one nor a pure Cremer one.

It will still be possible to get some information about the Cremer case, e.g. using the SCB algorithm I posted a while ago as long as the orbit (again with IA and complex intervals as parameters) is inside one pixel (=square in the plane).

I'm currently computing the Golden mean Siegel set as a test for that angle enclosure calculation (should be finished by tomorrow).

#### marcm200 #### Re: Parabolic Julia sets

« Reply #38 on: July 03, 2020, 07:39:41 AM »
"The golden mean Siegel set"

Combining the output of the TSA and the containment algorithm (post #34) found part of the Siegel disc in z^2+c with c=([-13104378..-13104377] + i*[-19689335..-19689334]) * 2^-25) which is a conjugated form of the classical Golden mean set (details below). The image shows the 2-square with a pixel representing complex squares of width 2^-10 (downscaled from L17).

The statement here is: "The union of all orange pixels cover part of the Siegel disc that is present in the filled-in Julia set of the exact value for c after conjugating the Golden mean angle (g) in z2+z*e2*i*pi*g into the form z2+c."

• white (from TSA) the underlying square completely lies in the exterior of all individual Julia sets of the interval seed.
• blue, red artificially introduced to circumference an analyzed region for the containment algorithm. Blue indicates the underlying sets can have a parabolic fix point (and as c includes at least one Siegel seed, there is one), red means all individual sets have a repelling fix-point therein.
• orange (from containment) the bounding box of the underlying square lands at least partially again in the circumferenced marked region.
• gray (TSA) covers the union of all Julia set boundaries and their interior (if present).

Currently the enclosed regions are fixed in position. On my todo-list is a sliding window, possibly of oval or octagonal shape and using a c width of 2^-48. I hope that way I can discover a larger portion of the Siegel disc.

Technical details
• Golden mean Siegel set: $$z^2+z\cdot e^{2i\pi \alpha}$$ with $$\alpha = \frac{\sqrt{5}+1}{2}$$
• All calculations for alpha, the factor in front of the z-term and the conjugation were done via IA using the kv library.
• The resulting z^2+([a0..a1]+i*[b0..b1])*z set was affinely conjugated to the form z^2+c.
• The TSA and the containment algorithm work with a sufficient number type to allow for error-free computations of the bounding box.

#### marcm200 #### Re: Parabolic Julia sets

« Reply #39 on: Yesterday at 08:49:01 AM »
"Cremer number 2"

The last Cremer number produced a final Julia seed interval of width 7*2^-25, a bit too large for my liking. So I tried to calculate another one.

$a_0 = 2 \\ a_1 = 2 \\ a_n=2^{Q_{n-1}}$

also converges to a Cremer number (proof in the zipped file).

The maxima code to compute the convergents:

Code: [Select]
cremer_init(a0,a1):=( listan:[a1,a0], listpn:[a0*a1+1,a0], listqn:[a1,1], approx:[listpn/listqn,listpn/listqn])$cremer_iterate():=( an:2^listqn, /* very fast growth */ listan:append([an],listan), tmpqn:an*listqn+listqn, tmppn:an*listpn+listpn, listqn:append([tmpqn],listqn), listpn:append([tmppn],listpn), approx:append([tmppn/tmpqn],approx))$denom2N(a,b,N) := (/* converts two rationals a <= b comprising an interval into two rationals *//* cdl <= cdr whose denominator is 2^N with cdl <= a <= b <= cdr */ if a < b then left:a else left:b, if a > b then right:a else right:b, left2:left * 2^N, /* attn: floating point operations might be used internally */ cdl:floor(left2), /* make sure the following verification works solely on rationals */ if (integerp(cdl) = false) then print("Error 1"), /* verify that it is valid */ /* the check then works again on rationals */ if ( (cd1 / (2^N)) > left) then print("Error 2"), if (cdl <= (2^63 - 1)) then print("          cdl usable as int64_t") else print("          ATTN: cdl larger than int64_t"), if (cdl <= (2^52 - 1)) then print("          cdl usable as double") else print("          ATTN: cdl might need more precision than double"), right2:right * 2^N, if mod(right2,2^N) = 0 then cdr:floor(right2) else cdr:1+floor(right2), if (integerp(cdr) = false) then print("Error 3"), /* and check if a valid enclosement */ if ( (cdr / (2^N)) < right) then print("Error 4"), if (cdr <= (2^63 - 1)) then print("          cdr usable as int64_t") else print("          ATTN: cdr larger than int64_t"), if (cdr <= (2^52 - 1)) then print("          cdr usable as double") else print("          ATTN: cdr might need more precision than double"), display(cdl), display(cdr) )\$
using

cremer_init(2,2);
cremer_iterate();
cremer_iterate();
cremer_iterate();
denom2N(approx,approx,48);

With that definition I could compute P5/Q5 (values in the zipped file), the difference to P4/Q4 being smaller than 10-1300. I guess that's why Brjuno numbers are said to be badly approximated by rationals - this one is the opposite, P4/Q4 = 11269/4610 is already pretty close.

Using those rationals (outward converted to a denominator of 2^48) to compute a final c interval (after conjugation) for z^2+c resulted in:

$c=( [-186176550019915~..-186176550019910]~+~i*[93330855667149~..93330855667158] )~*~2^{-48}$

which has only a width of 9 * 2^-48.

Next steps now: Image generation, proof extension to obtain further values and trying to formalize the proof to be verifiable by an automated proof checker system.

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