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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
 

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

Why are you replying to your own posts?

It's easier to post further, and I want to make more explicit

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.

Why are you only whining instead of showing some cool images from these formulaes?

Image for Kompassbrot kind 2, I hope to can post

KIND 3, but I not sure that have this shape

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.

Hehe, the colours are nice. If 'll have time maybe i'll digitalise it.

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".

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

You are so amazing. Hand drawn fractals deserve a beer! :beer:
:D

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

I'm founded a beta formula, but my first beta formula was been mistake because I used the condition dV/dc=0 where V=z[1]*z[2] formula mentioned up.
The true condition for border of mandel cubic perturbation fc(z)= z^3+k*z^2+c, is G=1 and for to solve (c) from kompassbrot equation is dG/dc=0 where G = |3*z[1]^2+2k*z[1]^2|*|3*z[2]^2+2k*z[2]^2| what become 9(V^2)+6kvV+4(k^2)V = (3k+9c)*v-2*(k^2)*V+9+6kc what must derivating with (c).
Then dG/dc = (3k+9c)*(dv/dc)+9v-2*(k^2)*(dV/dc)+6k =0, an equation what must solving with more carefully.
I keep trying to solving this and keep working. The result coming soon only on this topic.
I keep verifying each stage of this calculus for avoiding any mistakes.
The result will be next beta formula for KOMPASSBROT KIND 2 fractal.
When I will found the right formula for KOMPASSBROT KIND 2, this my fractal from my handraw will be rendered with my ULTRAFRACTAL

I hope. Don't forget then to upload a formula in UF site;) Kind of strange genial research.

New difficulties was occured at searching formula for all kind kompassbrots:
If kompassbrot kind 3 and above need to solve algebrical equation degree five. I learnt that equation degree 5 and above certain has no solution in radicals due the Abel-Ruffini theorem at whick I found the demonstration, after I tried in vain 9 years to find a solution in radicals of algebrical equation degree 5, plus that I founded a new way of demonstration of the theorem of Abel-Ruffini just now, that equation z^5+pz^3+qz^2+rz+s =0 if has 3 real roots and yet two complex conjugates roots as solutions, neither combination z[1;2;3;4;5]=(w^k)*(A^(1/5))+(w^2k)*(B^(1/5))+(w^3k)*(C^(1/5))+(w^4k)*(D^(1/5)) can't give only three real numbers and two complex roots, where w=cos(TT/5)+i*sin(TT/5)=
(-1+sqrt(5))/2+i*(sqrt(10+2*sqrt(5))/4) and k=solution number.
 So I will limiting at kompassbrot kind 2, and I won't can to render on ULTRAFRACTAL the Kompassbrot kind over 2.
New beta formulas for Kompassbrot kind 2 coming soon.
Another beta formula c=(-k^3-9*k)/9+(1/9)*(((k*k-18)^2)*(11*k^5+36*k^3-324*k+sqrt(12*k^4-39*k^2+96+12*c*k^3+54*kc+81*c^2))/32)^(1/3)+
(1/9)*(((k*k-18)^2)*(11*k^5+36*k^3-324*k-sqrt(12*k^4-39*k^2+96+12*c*k^3+54*kc+81*c^2))/32)^(1/3).
whick unfortunethly isn't right formula, I tested allready on ULTRAFRACTAL. So I must have much patience for find the right formula. I have a fully journal with the formulas.