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Author Topic: KOMPASBROT fractals  (Read 1047 times)
Description: Mandelbrot research
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hgjf2
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« on: July 11, 2013, 01:51:42 PM »

Is a new kind of KOMPASSBROT fractals?
The classic KOMPASSBROT set is only the kind 1 , this whick have formula fc(z) = z^3+c*(z^2)+((4*c*c*c+18*c)/27), when (z) going from 0
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hgjf2
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« Reply #1 on: July 11, 2013, 02:09:52 PM »

Is a new kind of KOMPASSBROT fractals?
The classic KOMPASSBROT set is only the kind 1 , this whick have formula fc(z) = z^3+c*(z^2)+((4*c*c*c+18*c)/27), when (z) going from 0

Yes exist. But have a formula hard.
Kind 1 represent the simetry point for cubic MANDELBROT sets with formulas fc(z) = z^3+k*z^2+c with (z) going from 0 , but why (z) can going from -2k/3 , because the Julia sets with cubic formula have two attraction basins, as example Julia set with have at base a formula z^3+z^2+0.1+0.2i have compact zones at {0} and {-2/3+0i} zone, and not considering the connectivity a this Julia sets.

If we try to search the conjunction point for the cubic Mandelbrot sets at 2nd minibrot/bulb we find more points but hard to render coordinates because request to solve 6th kind algebric equation like as
{z(c)|z^6+z^5+(3c+1)*(z^4)+(2c+1)*(z^3)+(3c^2+3c+1)*(z^2)+(c^2+2c+1)*z+(c^3+2*c^2+c+1)=0}
where z(c) is a implicit function. This equation is for 3rd bulb/minibrot attraction basin for classic Mandelbrot set,
the point (z_1=z_2*z_2+c;z_2=z_3*z_3+c;z_3=z_1*z_1+c).
So if try to determining equivalent of the Kompassbrot for a function like fc(fc(z)) where fc(z) = z^3+k*z^2+c
 
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hgjf2
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« Reply #2 on: July 11, 2013, 03:02:37 PM »

Yes exist. But have a formula hard.
Kind 1 represent the simetry point for cubic MANDELBROT sets with formulas fc(z) = z^3+k*z^2+c with (z) going from 0 , but why (z) can going from -2k/3 , because the Julia sets with cubic formula have two attraction basins, as example Julia set with have at base a formula z^3+z^2+0.1+0.2i have compact zones at {0} and {-2/3+0i} zone, and not considering the connectivity a this Julia sets.

If we try to search the conjunction point for the cubic Mandelbrot sets at 2nd minibrot/bulb we find more points but hard to render coordinates because request to solve 6th kind algebric equation like as
{z(c)|z^6+z^5+(3c+1)*(z^4)+(2c+1)*(z^3)+(3c^2+3c+1)*(z^2)+(c^2+2c+1)*z+(c^3+2*c^2+c+1)=0}
this equation is (((z^2+c)^2+c)^2+c-z)/(z^2+c-z)
where z(c) is a implicit function. This equation is for 3rd bulb/minibrot attraction basin for classic Mandelbrot set,
the point (z_1=z_2*z_2+c;z_2=z_3*z_3+c;z_3=z_1*z_1+c).
So if try to determining equivalent of the Kompassbrot for a function like fc(fc(z)) where fc(z) = z^3+k*z^2+c
 

So if searching the point(z_1=z_2*z_2*z_2+k*z_2*z_2+c;z_2=z_1*z_1*z_1+k*z_1*z_1+c) for z^3+k*z^2+c finding at point
{z(c)|z^6+2k*(z^5)+(k*k+1)*(z^4)+(2*k+2*c)*(z^3)+(k*k+2*k*c+1)*(z^2)+(k+c)*z+(c*c+k*c+1)} where this 6th degree polynom is just
((z^3+k*z^2+c)^3+k*(z^3+k*z^3)^2+c-z)/(z^3+k*z^2+c-z).
If we searching the derivative of z(c) from (c) then find kind 2 from Kompassbrot.
For 3rd kind from Kompassbrot must same derivative but for [(fc(fc(fc(z))))-z]/(fc(z)-z) and be very difficult and huger formula
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matsoljare
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« Reply #3 on: July 11, 2013, 09:00:48 PM »

Why are you replying to your own posts?
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hgjf2
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« Reply #4 on: July 11, 2013, 10:37:46 PM »

Why are you replying to your own posts?
It's easier to post further, and I want to make more explicit
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kram1032
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« Reply #5 on: July 12, 2013, 02:16:04 PM »

but why do you quote yourself? - Or more specifically, why do you quote each post in entirety? Especially when there is no post between those and the new one.
Quotes are for giving context to older messages as well as to specific parts of single messages. Fully quoting the previous message is kinda pointless.
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Kalles Fraktaler
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« Reply #6 on: July 12, 2013, 02:34:23 PM »

Why are you only whining instead of showing some cool images from these formulaes?
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hgjf2
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« Reply #7 on: July 13, 2013, 10:32:59 AM »

Image for Kompassbrot kind 2, I hope to can post


* kompass.JPG (44.06 KB, 964x681 - viewed 98 times.)
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hgjf2
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« Reply #8 on: July 13, 2013, 10:34:12 AM »

KIND 3, but I not sure that have this shape


* kompass_.JPG (28.61 KB, 944x694 - viewed 84 times.)
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hgjf2
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« Reply #9 on: July 13, 2013, 10:42:59 AM »

The pictures have rendered very manual, because yet I'm not found formula for this Mandelbrot set, probabily the formula contain certain large cubic roots, I know that need to solve a 6th degree algebric equation. I looked that cubic Mandelbrot sets have and twin minibrots like fc(z)= z^2+c^2 or
fc(z) = z^2 + (c+k)^2 where k =a+bi a static complex number, but this centre of almost simetry from those twins minibrots generating the Kompassbrot kind 2 from this picture if the twins have 2-bulb periodicity (like z^2-1 JULIA at classic Mandelbrot set) and kind 3 if the twins have 3-bulb periodicity.
Explains can finding at first picture with details.
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Alef
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« Reply #10 on: August 15, 2013, 03:41:12 PM »

Image for Kompassbrot kind 2, I hope to can post

Hehe, the colours are nice. If 'll have time maybe i'll digitalise it.
« Last Edit: August 15, 2013, 04:25:29 PM by Alef » Logged

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hgjf2
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« Reply #11 on: December 05, 2015, 04:44:38 PM »

Finally I'm solved this equation F=z^6+2k*z^5+(k^2+1)*z^4+(2k+2c)*z^3+(k^2+2kc+1)*z^2+(k+c)*z+(c^2+kc+1)=0. This using formulas
z[1]=z[2]^3+k*z[2]^2+c;
z[2]=z[1]^3+k*z[1]^2+c;
z[3]=z[4]^3+k*z[4]^2+c;
z[4]=z[3]^3+k*z[3]^2+c;
z[5]=z[6]^3+k*z[6]^2+c;
z[6]=z[5]^3+k*z[5]^2+c;
Having the formulas: v[1]=z[1]+z[2];v[2]=z[3]+z[4];v[3]=z[5]+z[6]; V[1]=z[1]*z[2];V[2]=z[3]*z[4];V[3]=z[5]*z[6];
Where V[1]=(c-v[1])/(v[1]+k) and V[2]=(c-v[2])/(v[2]+k) and V[3]=(c-v[3])/(v[3]+k) but and V[1]=v[1]^2+k*v[1]+1 and V[2]=v[2]^2+k*v[2]+1 and V[3]=v[3]^2+k*v[3]+1.
So will give any v^3+2k*v^2+(k^2+2)*v+(k-c)=0 where naming
v[1;2;3]= [-4k+(w^n)*sqrt[3](-20*k^3+36*k-108*c+4*sqrt(21*k^6-18*k^4-351*k^2+864+270*(k^3)*c-486*k*c+729*c^2) +
(w^2n)*sqrt[3](-20*k^3+36*k-108*c-4*sqrt(21*k^6-18*k^4-351*k^2+864+270*(k^3)*c-486*k*c+729*c^2)]/6
where w=(-1+i*sqrt(3))/2 is on root of the equation z^3-1=0.
Sqrt[3] is the cubic root: sqrt[3](z)=z^(1/3).

So the fractal formula for this compassbrot kind 2 is work in progress and this fractal is at beta testing.

This fractal will have the formula fc(z)=z^3+k*z^2+g(k), where g(k) is the solution of equation {c|dV/dc=0} where dV/dc is the partial derivative of V on direction "c".
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hgjf2
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« Reply #12 on: December 05, 2015, 04:51:14 PM »

I'm forget to say that z[1;3;5]+z[2;4;6]=[-4k+
(w^n)*sqrt[3](-20*k^3+36*k-108*c+4*sqrt(21*k^6-18*k^4-351*k^2+864+270*(k^3)*c-486*k*c+729*c^2) +
(w^2n)*sqrt[3](-20*k^3+36*k-108*c-4*sqrt(21*k^6-18*k^4-351*k^2+864+270*(k^3)*c-486*k*c+729*c^2)]/6
If we solving the system of equations:
 z[1]+z[2]=v with z[1]*z[2]=v^2+kv+1. then reaching at z[2]*(v-z[2])=v^2+kv+1 and z[2]^2-v*z[2]+(v^2+kv+1)=0 and
z[2]=(v+sqrt(v^2-4*v^2-4kv-4))/2=(v+sqrt(-3*v^2-4k*v-4))/2.

The final formula will coming soon only on FRACTALFORUMS
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DarkBeam
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Fragments of the fractal -like the tip of it


« Reply #13 on: December 05, 2015, 05:40:52 PM »

You are so amazing. Hand drawn fractals deserve a beer! A Beer Cup
cheesy
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No sweat, guardian of wisdom!
hgjf2
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« Reply #14 on: December 06, 2015, 08:48:52 AM »

If I will find the final formula of KOMPASSBROT kind 2 then I will can to render on PC with FRACTALEXPLORER or ULTRAFRACTAL this example on whick I made hand drawn
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