Strange Mandelbrot Set

[Endemyon]

Hello Everybody !
I tried a formula wich obtains interesting results:
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15275

It's obtained from iteration of
Zn+1=(Zn^3+c)*(Zn^2+c)
I assume it's possible to do every formula's of type:
Zn+1=(Zn^p+c)*(Zn^q+c) with  p,q>1 integers
because the two parts of the equation are mandelbrot's formulas.
Maybe it's even possible to multiply it with other terms of type (Zn^k+c)

It's possible to obtained corresponding julia sets just like the normal way, but they're not totally symetric. It seems like a composition of the two sets !
Does anyone is familiar with this kind of thing and tell me more about it ?


Friendly, a fractal fellow

Edit: A corresponding Julia's set:
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15278
And it appears to me that i might have misplaced this topic, could a great asministrator deplace it to the appropriate area ?

[cKleinhuis]

hey there, nice meeting you as well, i move it to the math section, what you did there is a kind of hybrid fractal,
and yes, basically this is one form of hybridisation, combining existing formulas to new ones, you connected the two
z^2 and z^3 mandelbrot equation with the multiplication operation

[Rychveldir]

In your Formula



Is c a constant? And did you use at all? Or was it something like


or


I also like exploring generalized versions of the mandelbrot set using polynomials of the form



They can generate really interesting effects.

[cKleinhuis]

i think the "c" is just the z0 value in this case
but anyways, all of the combinations should provide interesting results

[element90]

The difficulty with the formula

z = (z^3 + c)(z^2 + c)

is that the critical points can't be easily determined except for the point at zero.

The formula expands to

z = z^5 +cz^3 + cz^2 + c^2

so the critical points are the solutions of

f'(z) = 5z^4 + 3cz^2 + 2cz = 0

so

z(5z^3 + 3cz^2 + c) = 0

which implies that there is a critical point at 0, the remaining 3 critical values have to calculated for each value of c (the location in the complex plane).

[kram1032]

Since that's a Polynomial of degree four, you can still solve for the generic critical point, however the expressions are rather complex:
http://www.wolframalpha.com/input/?i=5z%5E4%2B3+c+z%5E2+%2B+2+c+z+%3D+0

[element90]

For higher level polynomials from a programming point of view it is even easier, the coefficients can be used to produce the companion matrix and the eigenvalues are the roots, provided you have the routines to do it.. It just has to be done for each location. Implementing this in a script with a program such as Ultra Fractal won't be easy.

I looked at

z = (z^2 + c)(z^2 + c)

so

f'(z) = 4z^3 + 2cz = 0

z(4z^2 + 2c) = 0

which has three critical points 0, -sqrt(-c) and sqrt(-c) which can be plugged into a UF or Gnofract4d script which it turns out produces three identical pictures which isn't surprising as the pictures are properly formed M4 Mandelbrots.

[element90]

The formula

z = (z^2 + c)(z^4 + c)

has somewhat easier critical points to determine, for the values see the code below in the Gnofract4d script.

The critical points are solutions of

f'(z) = 6z^5 + 4cz^3 + 2cz = 0
z(6z^4 + 4cz^2 + 2c) = 0

which gives zero and the square roots of the roots of a quadratic.

Code:

M2.M4 {
; M2 Mandelbrot multiplied by an M4 Mandelbrot
init:
;        z = 0
        z0 = #pixel
;        z = sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12)
;        z = sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12)
        z = -sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12)
;        z = -sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12)
loop:
        z2 = z*z
	z = (z2 + #pixel)*(z2*z2 + #pixel)
bailout:
	@bailfunc(z) < @bailout
default:
float param bailout
	default = 4.0
endparam
float func bailfunc
	default = cmag
endfunc
}

Critical point 0


https://copy.com/hqSr3VYbJBuO

Critical point z = sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12) and  sqrt((-z0 + sqrt(z0*z0 - 24*z0))/12)


https://copy.com/kqLZT9DwdGFJ

Critical point z = sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12) and  -sqrt((-z0 - sqrt(z0*z0 - 24*z0))/12)


https://copy.com/UMOwFuqWj8ed

Combined


https://copy.com/BGUDDqTrGMt7

What usually happens is that areas in one critical point picture that lack Mandelbrot buds are provided with them from an other critical point picture. Oddly the areas at the top and the bottom have areas where this isn't so, assuming I haven't made a mistake.

[jdebord]

There are analytical formulas for polynomials up to degree 4. See for instance:

http://dlmf.nist.gov/1.11#iii

Since we are looking for the roots of the derivative, this applies for polynomials up to degree 5.

[Endemyon]

Hey guys, thanks for your answers !
I can't resist to show you this julia set, obtained by Zn+1=(Zn^2+c)*(Zn^3+c)*(Zn^4+c) with C=0,7937+i0,0904
http://www.fractalforums.com/index.php?action=gallery;sa=view;id=15802

I don't understand why you need to find the roots of critical points (if it's not to have a completely connex Julia set ) ?

It's possible to have an approximation for higher degrees with the newton's method, but I think you don't learn something new 

[jdebord]

The critical point is required only for the Mandelbrot set, not for the Julia set.

The Julia set iteration always start with the pixel coordinates.

[Endemyon]

The critical point is required only for the Mandelbrot set, not for the Julia set.
The Julia set iteration always start with the pixel coordinates.

If you do a Mandelbrot set for the equation Zn+1=Zn^2+C (the classic).  C  is the point which parcours the pixel coordinates and Z0=0.
So you don't need a critical point

So that doesn't not explain me why it's needed, is it from the software you use?

Now, why I was talking about critical point for a julia set:
If you do a julia set for the equation Zn+1=Zn^2+C.  C is the chosen constant and Z0 is the point which is the pixel coordinates.

If you try do a julia set where C is the critical point of mandelbrot set, you obtain a  julia set where all converging parts are connected. ( for any n)

 

[jdebord]

If f(z) denotes the iterated formula, the critical points are the roots of the equation f'(z) = 0

For the classical Mandelbrot set, f(z) = z^2 + c and f'(z) = 2z.

The equation is 2z = 0, hence z = 0

So, z0 = 0 is a critical point (the only one, in this case).

[TheRedshiftRider]

I tried something else, I tried to divide the two parts of the basic formula and the results are nice:



(z^10+c)/(z^2+c)


(z^10+c)/(z^8+c)

[jdebord]

Nice pictures, especially the second one !

I have tried to find a formula for the derivatives, assuming p > q > 1 :







The formula for f'(c) is needed if you want to apply the distance estimator method (here z' is the value of the derivative at the previous iteration).

The formula for f'(z) shows that 0 is always a critical point, but the other critical points depend on c and need to solve a polynomial.

I assume that the pictures are for the critical point zero ?