test for the supposed winner of the taffy pulling contest
Roger Bagula
In the 2d Mandelbrot set the Pc(n) polynomials have solutions that
give clear cyclic
Julia bulbs and the set has nice clear lapin Julia sets that show
the cycles.
So you should in 2d have three variable 'invariants'
as polynomials that give solutions whose Julias are countable cycles
by the n level of the polynomial.
A 3d equivalent to a Douady Rabbit Julia set would be the first step.
References;
The Beauty of Fractals page 43,page 58.
Complex Dynamics page 129-130.
So you have to find the 3d polynomials,
solve them and look at the Julias.
The phase locking behavior of self-similar sets is why
I say that the true Mandelbrot set in 3d should have
some sort on antenna.
The posting of boasts of some sort of solution
without giving a derivation and clear
mathematical formula is why the mathematicians call
these "dumb' fractal artists"?
If something is true mathematically it is subject to
intimate examination by everybody
and isn't a 'secret formula'.
So mocking the authority figures just gets you in deeper
and makes them laugh harder at the ignorant fractal artist clowns?
Pictures of a pretty set posted to the Internet...
Roger Bagula
Mandelbrot algorithm:
Clear[g, gg, a, p, q, r, x, y, z, x0, y0, z0]
(* definition of a 3d Mandelbrot algorithm*)
p[x_, y_, z_] = 2*y*x - x^2 + x0
q[x_, y_, z_] = 2*z*y - y^2 + y0
r[x_, y_, z_] = 2*x*z - z^2 + z0
g[x_, y_, z_] = {p[x, y, z], q[x, y, z], r[x, y, z]}
(* definition of recursive 3d polynomials*)
gg[0] := {x0, y0, z0}
gg[1] := g[x0, y0, z0]
gg[n_] := gg[n] = g[gg[n - 1][[1]], gg[n - 1][[2]], gg[n - 1][[3]]]
a = Table[Table[ExpandAll[gg[n][[i]]] == 0, {i, 1, 3}], {n, 0, 3}]
First level solution:
Solve[a[[1]], {x0, y0, z0}]
{{x0 -> -1, y0 -> -1, z0 -> -1},
{x0 -> 0, y0 -> 0, z0 -> 0},
{x0 -> 0, y0 -> 0, z0 -> 1},
{x0 -> 0, y0 -> 1, z0 -> 0},
{x0 -> 0, y0 -> 3, z0 -> 1},
{x0 -> 1, y0 -> 0, z0 -> 0},
{x0 -> 1, y0 -> 0, z0 -> 3},
{x0 -> 3, y0 -> 1, z0 -> 0}}
Second level solution:
Solve[a[[2]], {x0, y0, z0}]
{{x0 -> -1, y0 -> -1, z0 -> -1},
{x0 -> 0, y0 -> 0, z0 -> 0},
{x0 -> 0, y0 -> 0, z0 -> 1},
{x0 -> 0, y0 -> 1, z0 -> 0},
{x0 -> 0, y0 -> 3, z0 -> 1},
{x0 -> 1, y0 -> 0, z0 -> 0},
{x0 -> 1, y0 -> 0, z0 -> 3},
{x0 -> 3, y0 -> 1, z0 -> 0}}
The third level hasn't yet come down ( is calculating).
That is how the Julia root points for the cycles if they exist for a
3d Mandelbrot algorithm are calculated
There is another way to look at these as implicit
curves in product. In 2d it gives another representation of the
Mandelbrot set:
http://www.mathematica-users.org/webMathematica/wiki/wiki.jsp?pageName=User:Roger_Bagula
In the 3d case each polynomial set has a 3d implicit surface
associated with it.
They are no simple curves in the higher iterations.
Respectfully, Roger L. Bagula
> 11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
> http://www.google.com/profiles/Roger.Bagula
> alternative email: roger....@gmail.com
David Makin
> If something is true mathematically it is subject to
> intimate examination by everybody
> and isn't a 'secret formula'.
Here's Paul Nylander's latest suggestion - you'll see why he thinks
this is "better" :)
http://www.fractalforums.com/3d-fractal-generation/true-3d-mandlebrot-type-fractal/msg8680/#msg8680
Or, I prefer the compact UF version with ztemp, zri and cri complex,
r, zj, cj and @mpwr real though @mpwr could be complex:
ztemp = ((r=cabs(zri)) + flip(zj))^@mpwr
zri = real(ztemp)*(zri/r)^@mpwr + cri
zj = imag(ztemp) + cj
It'd be interesting if someone could add the 3rd dimension to the
power/degree but that's beyond my maths )
Roger Bagula
> On 16 Nov, 19:05, Roger Bagula <roger.bag...@gmail.com> wrote:
>
> > If something is true mathematically it is subject to
> > intimate examination by everybody
> > and isn't a 'secret formula'.
>
> Here's Paul Nylander's latest suggestion - you'll see why he thinks
> this is "better" :)
>
>
> Or, I prefer the compact UF version with ztemp, zri and cri complex,
> r, zj, cj and @mpwr real though @mpwr could be complex:
>
> ztemp = ((r=cabs(zri)) + flip(zj))^@mpwr
> zri = real(ztemp)*(zri/r)^@mpwr + cri
> zj = imag(ztemp) + cj
>
> It'd be interesting if someone could add the 3rd dimension to the
> power/degree but that's beyond my maths )
I looked at the stuff; your recent movie has one with
maybe too many things sticking out?
If you can write it out in the form:
x'=f1(x,y,z)+x0
y'=f2(x,y,z)+y0
z'=f2(x,y,z)+z0
we might be able to get a Douady's rabbit
equivalent for the cycles test of a Mandelbrot set.
David Makin
> I looked at the stuff; your recent movie has one with
> maybe too many things sticking out?
Which one ?
David Makin
Note that some are using a simplistic numerically based pseudo-
distance estimate routine which is essentially based on the smooth
iteration density alone. This means that it's definition of "solid"
when based on a distance estimate threshold is not quite the same as
that based on an analytical distance estimation threshold, but in
general it's pretty close :)
David Makin
With respect that to the cycle test I'll leave that to you or others
more mathematically qualified than myself ;)
Besides I can't really follow Mathematica code and have no access to
it.
David Makin
> Or, I prefer the compact UF version with ztemp, zri and cri complex,
> r, zj, cj and @mpwr real though @mpwr could be complex:
>
> ztemp = ((r=cabs(zri)) + flip(zj))^@mpwr
> zri = real(ztemp)*(zri/r)^@mpwr + cri
> zj = imag(ztemp) + cj
>
> It'd be interesting if someone could add the 3rd dimension to the
> power/degree but that's beyond my maths )
Here's an idea:
ztemp = ((r=cabs(zri)) + flip(zj))^pri
zri = real(ztemp)*(zri/r)^pri + cri
zj = imag(ztemp)^pj + cj
Roger Bagula
From Mandelbrot sets in 2d the Feigenbaum set
which comes from the Lapin form:
f(x,c)=z^2/(c*z^2+1)
Feigenbaum:
f(x,c)=c*z^2+1
is the toral inverse of the Mandelbrot:
m(1/z,c)=1/z^2+c
z^2*(1/z^2+c)
The point of this is that the toral inverse transform on
{x,y,z} is ( inversion around the unit sphere) :
{x',y',z'}={x,y,z}/(x^2+y^2+z^2)
So if you apply that in 3d you should get a set
very like the first,
if the set is really "Mandelbrot set" like.
The PC constants/roots should be similar as well.
Respectfully, Roger L. Bagula
11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
http://www.google.com/profiles/Roger.Bagula
Roger Bagula
Mathematica below:
Douady's rabbit is at ( found by observation not solving...).
c={x0,y0,z0}={-.0122561,0.744862,0}
definition from Paul Nylander is:
{x,y,z}'= (x^2 - y^2)(1 - z^2/(x^2 + y^2)) + x0,2*x*y(1 - z^2/(x^2 +
y^2)) + y0, -2*z*Sqrt[x^2 + y^2] + z0}
The only real problem with this algorithm is one of scale as it is
nearly all inside the unit sphere instead of inside
the sphere of radius 2.
A better scaling may be:
{x,y,z}'={(x^2 - y^2 - z^2) /(1 - z^2)+x0, 2 x y+y0, -2 *(x^2 + y^2)^
(3/2)* z/(1 - z^2)+z0}
which seems to give the same sort of Mandelbulb , but larger
to match the PC polynomial roots.
\!\(Clear[g, gg, a, p, q, r, x, y, z, x0, y0,
z0]\[IndentingNewLine] (*definition\ of\ a\ 3
d\ Mandelbrot\ Mandelbulb\ algorithm*) \n
\(p[x_, y_, z_] = \((x^2\ - \ y^2)\) \((1\ - \
z^2/\((x^2\ + \ y^2)\))\) + x0;\)\n
q[x_, y_, z_] = \ 2*x*y \((1\ - z^2/\((x^2\ + \ y^2)\))\) + y0; \n
r[x_, y_, z_] = \(-2\)*z*Sqrt[x^2\ + \ y^2] + z0; \n
\(g[x_, y_, z_] = {p[x, y, z], q[x, y, z], r[x, y, z]};\)\n
(*definition\
of\ recursive\ 3 d\ polynomials*) \n
gg[0] := {x0, y0, z0}\n
gg[1] := g[x0, y0, z0]\n
gg[n_] := \(gg[n] = g[\(gg[n -
1]\)[\([\)\(1\)\(]\)], \(gg[n - 1]\)[\([\)\(2\)\(]
\)], \
\(gg[n - 1]\)[\([\)\(3\)\(]\)]]\)\n
\(a = Table[Table[FullSimplify[\(gg[n]\)[\([\)\(i\)\(]\)]] ==
0, {i, 1, 3}], {n, 0,
*) \n
NSolve[a[\([\)\(1\)\(]\)], {x0, y0, z0}]\n
{{x0 -> 0, y0 -> 0, z0 -> 0}}\[IndentingNewLine] (*Second\ level\ \
\(\(solution\)\(:\)\)*) \n
NSolve[a[\([\)\(2\)\(]\)], {x0, y0, z0}]\n
{{z0 -> 0, y0 -> 0,
x0 -> \(-1\)}, {z0 -> \(-\(\[ImaginaryI]\/2\)\), y0 -> 0,
x0 -> \(-\(1\/2\)\)}, {z0 -> \[ImaginaryI]\/2, y0 ->
0, x0 -> \(-\(1\/2\)\)}, {z0 -> \(-\(\@3\/2\)\), y0 -> 0, x0
-> \
1\/2}, {z0 -> \@3\/2, y0 -> 0, x0 -> 1\/2}}\n
NSolve[a[\([\)\(3\)\(]\)], {x0, y0, z0}]\)