Plane filling Julia sets

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Offline knighty

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« Reply #30 on: September 23, 2019, 03:11:49 PM »
That last picture is gorgeous! Are those power2 minibrots in the 2nd picture?

Here are some evaldraw scripts that I've written to compute pre-periodic points for f(z)=1+c/z▓. Few of them are actually periodic points (like -1) and the complex ones come in conjugate pairs. I haven't filtered them out.


Offline gerrit

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« Reply #31 on: September 23, 2019, 08:00:57 PM »
That last picture is gorgeous! Are those power2 minibrots in the 2nd picture?
Yes, they are always there except for 1+c/z^n which only has power n^2 minibrots for reasons I don't understand.

Offline gerrit

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« Reply #32 on: September 29, 2019, 08:41:25 AM »
Parameter space and a space filling Julia set of 1/"burning ship":
\( z \leftarrow 1/z\\z \leftarrow bship(z) + c \)

Offline gerrit

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« Reply #33 on: October 08, 2019, 04:18:40 AM »
Another 1/burningShip Julia set.

Offline marcm200

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« Reply #34 on: October 08, 2019, 09:20:07 AM »
Very nice - and very 3D looking! How did that 3D effect in the lower left green objects arise?

Offline hgjf2

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« Reply #35 on: October 12, 2019, 10:23:41 AM »
Plane filling Julia set is a key for to can solve the Collatz's conjecture: His function is the polynom P=(1+cos*pi*z)/2+z*((7+5*cos*pi*z)/4).
The Julia set JP(z)={z|P(P(P...P(z)...))<>infinity} filling the space and has reough five attraction basins as {4;2;1} and {0} and {-1;-2}.

Offline gerrit

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« Reply #36 on: October 12, 2019, 10:35:10 PM »
Plane filling Julia set is a key for to can solve the Collatz's conjecture: His function is the polynom P=(1+cos*pi*z)/2+z*((7+5*cos*pi*z)/4).
The Julia set JP(z)={z|P(P(P...P(z)...))<>infinity} filling the space and has reough five attraction basins as {4;2;1} and {0} and {-1;-2}.
That function, which I think maybe you got from http://yozh.org/2012/01/12/the_collatz_fractal/, though a match requires an assumption of several typos,  does not agree with https://en.wikipedia.org/wiki/Collatz_conjecture#Iterating_on_real_or_complex_numbers.

As I understand it these functions, neither of which are polynomials, are supposed to be analytic continuations of a function on the integers and as an analytic continuation is unique only one can be right.

???

PS The second function on the WIKI page ".5*z*cos(#pi*z/2)^2 +(3*z+1)*sin(#pi*z/2)^2/2", critical point
z= -1.357974913200654158349384242675, has an interesting Julia set. First image.

Each of the "crosses" has a different filled Julia set Kf in it, some of them looking a bit like Mandelbrot minis, but never "really", and each branch of those crosses has sub-crosses, each with a different Kf. Example is second image. (Some black artifacts due to overflow of the exponential in the sin/cos, a common problem with this kind of functions.)
« Last Edit: October 12, 2019, 11:08:47 PM by gerrit »

Offline gerrit

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« Reply #37 on: October 12, 2019, 10:39:04 PM »
Very nice - and very 3D looking! How did that 3D effect in the lower left green objects arise?
Just a side effect of the orbit trap.

Offline mrrudewords

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« Reply #38 on: October 14, 2019, 10:44:13 AM »
Some black artifacts due to overflow of the exponential in the sin/cos, a common problem with this kind of functions.

Reduce the bailout to 100(ish) and they should get filled.
Z = Z2 + C (obvs)

Online pauldelbrot

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« Reply #39 on: October 14, 2019, 04:59:40 PM »
as an analytic continuation is unique only one can be right.

Analytic "tyre puncture patches" are unique, but analytic interpolations of functions on the integers are emphatically not unique.

To see this, note that \( a \sin \frac{x}{\pi} \) is zero when x is an integer, is nonzero elsewhere, is analytic, and is nonunique due to the free parameter a. Furthermore, given a function f(x) passing through specific values on the integers, so equal to a specific (not necessarily all-zero) function when restricted to the integers, \( f(x) + a \sin \frac{x}{\pi} \) then describes a whole family of functions that are unequal to one another and share the same restriction to the integers, with the original recovered for the value a = 0, so if an analytic such function f exists it too is nonunique, since the pointwise sum of two analytic functions is analytic, thus proving: if g is a function on the integers, either there is no analytic function f whose restriction to integers is g, or there are infinitely many distinct such functions. (Indeed, since sums and scalar multiples of analytic functions are analytic, the analytic functions that restrict to g form a vector space with at least one dimension, with \( f(x) + a \sin \frac{x}{\pi} \) being a nonzero vector in that space when \( a \neq 0 \).)

Online pauldelbrot

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« Reply #40 on: October 14, 2019, 05:07:27 PM »
Should be an infinite-dimensional vector space, in fact: just consider harmonics. Any sum of terms of the form \( a \sin \frac{kx}{\pi} \) where the ks are positive integers will be zero on the integers and analytic if the sum is finite (if it's an infinite series, it need not be convergent and even if convergent need not be analytic; the triangle and sawtooth waves can be constructed this way and are not analytic, or in the sawtooth's case even continuous). The terms for a fixed (e.g. 1) and k a positive integer give a set of basis vectors, and the vector space consisting of all vectors with finite support in the span of those basis vectors is a subspace of the vector space of all analytic functions that are zero on the integers (maybe not a proper subspace though -- can every periodic analytic function be Fourier decomposed to a finite sum of sine waves?).  So given a function g on the integers the vector space of analytic interpolations of g should be likewise infinite-dimensional, to say nothing of merely infinite.

Offline gerrit

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« Reply #41 on: Yesterday at 03:18:32 AM »
>>Some black artifacts due to overflow of the exponential in the sin/cos, a common problem with this kind of functions.
Reduce the bailout to 100(ish) and they should get filled.
Works. Can you explain why? How can those regions never reach 1000000 if they truly escape, but do reach 100? Does some underflow happen on the way to 1000000? I expected I had to raise escape radius to fix it.

For exp(z) this does not help, which I understand, but the cos(z) I don't understand.

Below clean parameter space image of that Collatz fractal (formula + c=pixel) with escape radius 100.

Offline gerrit

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« Reply #42 on: Yesterday at 03:20:24 AM »
The Julia set of the Collatz formula can be seen fully when inverting z0 = 1/pixel. You can see the Mandelbrot-like but not really shapes.
Unlike Julia sets I know, this one has an infinite variety of different filled Julia shapes.

Offline gerrit

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« Reply #43 on: Yesterday at 03:22:37 AM »
Analytic "tyre puncture patches" are unique, but analytic interpolations of functions on the integers are emphatically not unique.

To see this, note that \( a \sin \frac{x}{\pi} \) is zero when x is an integer, is nonzero elsewhere, is analytic, and is nonunique due to the free parameter a. Furthermore, given a function f(x) passing through specific values on the integers, so equal to a specific (not necessarily all-zero) function when restricted to the integers, \( f(x) + a \sin \frac{x}{\pi} \) then describes a whole family of functions that are unequal to one another and share the same restriction to the integers, with the original recovered for the value a = 0, so if an analytic such function f exists it too is nonunique, since the pointwise sum of two analytic functions is analytic, thus proving: if g is a function on the integers, either there is no analytic function f whose restriction to integers is g, or there are infinitely many distinct such functions. (Indeed, since sums and scalar multiples of analytic functions are analytic, the analytic functions that restrict to g form a vector space with at least one dimension, with \( f(x) + a \sin \frac{x}{\pi} \) being a nonzero vector in that space when \( a \neq 0 \).)
Right, sin(k/pi) is zero for all integers, good explanation. I somehow thought the gamma function was the unique analytic extension of the factorial, for most of my life, duh!


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