True 3D mandelbrot fractal (search for the holy grail continues)

[stigomaster]

But i has a square root in C,

[fracmonk]

Yes, it can be expressed like that, but again, that's not necessary either. We'd only be plotting coefficents of j and ij just as we do for Re and Im, so that f(v)->v^2+g, v=x+yi+zj+wij, g=a+bi+cj+dij.  We get a set of 8 rotating powers of j, just as we have 4 of i.  It's the relationship between j and i and real that counts.  Note that when j and ij dims are zero, we have the standard M-set in two dims.  Preserving that, and going to a dimensional extension that is meaningful to the function is what appealed to me.  Has anyone done this yet, only I did not recognize it?

(came back to fix above, may as well add:)
j^2=i, j^3=ij, j^4=ii=-1, j^5=-j, j^6=-i, j^7=-ij, j^8=j^0=-ii=1, so that Re=j^8-j^4 values, Im=j^2-j^6 values, sqrt(i)dim=j^1-j^5 values, and i*sqrt(i)dim=j^3-j^7 values.  That's how the coefficients are processed.  Any nonzero value for c or d above will contribute to x+y+z+w under iteration, though of course we are only interested in eventual fates of the coordinates for now, and can only look at (at most) 3-d slices.

Hope I put it down right this time.

[jehovajah]

For<Quoted Image Removed> I designed a transform based on the <Quoted Image Removed> transform in the following way:

(a,b,c) vx (d,e,f)  

(ad - be - cf, ae + bd, af +cd, bf + ce) using <Quoted Image Removed> as a manipulation /construction space.

This i reduce to

(A,B,C,D) =(ad - be - cf, ae + bd, af +cd, bf + ce)

Using pairs from the construction bracket in the <Quoted Image Removed> transform i obtain 6 building blocks

1 (AA - BB, 2AB)

2 (AA - CC, 2AC)

3 (-CC - BB, 2BC)

4(AA + DD, 2AD)

5 (DD - BB, 2BD)

6 (DD - CC, 2CD)

THE unary OPERATORS i and j are used to inform the manipulations so that i² = j² = -1 and (ij)² =+1.


Now my intention was to rotate the planes xy, xz, yz by this construction and i assumed that was what was happening until i rechecked the construction principles. The yz plane is not the same as the other two planes with the unary operators i and j operating on the axes. Under the <Quoted Image Removed> transform the yz plane is sheared to the xij plane whatever that is. It may be a vortex surface.

so the first constructed transform is mistaken in two counts. The manipulations were faulty and i will show the correct manipulations; but the design was mistaken as it was not tranforming to a map of geometrical space.


The expansions are as follows for the right handed form

AA = (ad)² + (be)² + (cf)² - 2abde - 2acdf + 2bcef

BB = (ae)² + 2abde + (bd)²

CC = (af)² + 2acdf + (cd)²

2AB = 2(a)²de + 2ab(d)² - 2ab(e)² - 2(b)²de - 2acef - 2bcdf

2AC = 2(a)²df + 2ac(d)² - 2ac(f)² - 2(c)²df - 2abef - 2bcde

2BC = 2(a)²ef + 2bc(d)² + 2abdf + 2acde

Now the first posted construction was based on combining blocks 1,2,3

supposedly giving

<Quoted Image Removed> *{AA - BB + AA - CC, 2AB + BB - CC,  2AC + 2BC}

=> {AA - BB/2 - CC/2, AB + BB/2 - CC/2, AC + BC }.                                                    


 [ in fact it should be {AA - BB/2 - CC/2, AB - BB/2 - CC/2, AC + BC } due to an error in the original formulation of block 3]

So clearly (when i expand it) my original manipulations were wrongly copied from page to page to screen.

But now i realise i have not combined like with like and so have to construct the following transform from blocks 1and 2 which i fear will be even less interesting than my mistaken one

{AA -BB/2 - CC/2. AB, AC}

Contrary to  predicted it is the corrected expansion that is  less interesting or possibly more! The image sculpted here is thin slice in the yz  plane seen at an offset angle. This may mean that the x(ij) resultant of the yz  plane rotation under vx is in fact a 3d operator of sorts, like the two planar right triangles in the twinbee nylander formulae. The important thing here and there is the orthogonal or otherwise linking of the two planar operators so as to span geometrical space. When i used the i23 linking, dave produced a solid object and so did i. (see the second image.) This seems to be the result of the mistakes.

By the way the formulae do not need to be fully expanded to draw them. i think i can use the building blocks and their definition to   draw them i the latest quasz by terry.

[fracmonk]

Continuing 181, if anyone's interested, the consequent formulation is:

x part: x^2-y^2-2zw+a+
y part: (2xy+z^2-w^2+b)i+
z part: (2xz-2yw+c)j+
w part: (2xw+2yz+d)ij

[jehovajah]

AA = (ad)² + (be)² + (cf)² - 2abde - 2acdf + 2bcef

BB = (ae)² + 2abde + (bd)²

CC = (af)² + 2acdf + (cd)²

2AB = 2(a)²de + 2ab(d)² - 2ab(e)² - 2(b)²de - 2acef - 2bcdf

2AC = 2(a)²df + 2ac(d)² - 2ac(f)² - 2(c)²df - 2abef - 2bcde

2BC = 2(a)²ef + 2bc(d)² + 2abdf + 2acde

The left handed formulation requires the following

DD = (bf)² + 2bcef + (ce)²

2AD =  2abdf + 2acde - 2bc(e)² - 2bc(f)² - 2(b)²ef   - 2(c)²ef

2BD =2ac(e)² + 2(b)²df + 2abef + 2bcde

2CD = 2ab(f)² + 2(c)²de + 2bcdf + 2acef


Now from the 6 blocks i can construct 3 linked formulations abbreviated as followa

{AA - BB/2 - CC/2, AB, AC}                           RIGHT HANDED

{(AA + DD - BB - CC)/2, BC + AD}                 A MIRROR PLANE  at <Quoted Image Removed> to the xy plane but here plotted in xy to have a look

{DD - BB/2 - CC/2, CD, BD}                          LEFT HANDED


THIS STRUCTURE i think might be interesting even if the mandelbrots are not. Particularly the plane as it may contain reflections of details in the brots not visible in the 3d brots.

Any way is any one interested as i am in seeing what this system looks like? I will expand it and find the vx for (x, y, z) if you are.

These 3 formulae produce linked but different shapes. There is no plane as guessed but a curious linking surface, which apparently changes orientation with the order of the BC product.

[jehovajah]

@fracmonk

using this:
wx= x#*x#-y#*y#-2*z2*w
wy=(2*x#*y#+z1-w^2)*i
wz=(2*x#*z2-2*y#*w)*(1+i)/1.142
ww=(2*x#*w+2*y#*z2)*i*(1+i)/1.142

i get these. Do not know if it is what you are after. A julia and 2 mandy's.
z1=z^2 and z2=z
However have had to use quad c to get this as your extension of the polynomial numerals in i cannot be rounded off in 3d. You just get a "sausage".

[fracmonk]

I'm not familiar with the code, and can't be sure if it executes as intended.  If you wrote your own from scratch, you'd keep the four parts separate (as you did) but treat as real computationally, only *assuming* the r,i,j,+ij after each component, and all values would go where they belong and change in each step appropriately.  You got interesting results, but I'd guess they're just artifacts.  I'll print it, take it with me, & try to decipher.  Best I can do, for now.

[fracmonk]

jehovajah-  Familiar with FractInt?  It's what I use.  If you know it, it automates complex math so you can use a single variable which it treats as complex.  Stayed up late "fooling" it into treating my variables as 1 dimensional real.  Then followed my own advice left to you in last post and did 2-d slices of some of the 6 possible combinations:  a by b with c+d as set constants, etc.   It's good, and if you get your code right, you will have the glory (and it is glorious!) of doing its 6(?)  3-d index sets first.  Good luck.  Later.

[jehovajah]

So adding to Karls original analysis:
Power-Reducing/Half Angle Formulas

Product-to-Sum Formulas

For the mandelbrot in 2d

z= x+ iy=rcosø+irsinø

z²=r²(cos²ø-sin²ø+2icosøsinø)

                  =r²cos2ø+ir²sin2ø


For the 3d mandelbrot


z= x+ iy+jz=rsinΩcosø+irsinΩsinø+jrcosΩ

z²=r²sin²Ωcos²ø-r²sin²Ωsin²ø-r²cos²Ω+i2r²sin²Ωsinøcosø+j2r²sinΩcosΩcosø+ijr²sinΩcosΩsinø+jir²sinΩcosΩsinø

                 =r²(+i+j+ij+ji)
                 =r²(+i+j+ij+ji)

So doubling the angles is only part of the transformation if this analysis is right. Please check and point out my mistakes as it is easy to rectify.

I thought i would derive the alternative cosine formula.

z= x+ iy+jz=rcos¥cosø+ircos¥sinø+jrsin¥

z²=r²cos²¥cos²ø-r²cos²¥sin²ø-r²sin²¥+i2r²cos²¥sinøcosø+j2r²cos¥sin¥cosø+ijr²sin¥cos¥sinø+jir²sin¥cosYsinø
                  =r²(+i+j+ij+ji)

With some initial sculpures of the sinΩ version.
And one i like by just squaring

z=rsinΩcosø+irsinΩsinø+jrcosΩ.

[jehovajah]

I'm not familiar with the code, and can't be sure if it executes as intended.  If you wrote your own from scratch, you'd keep the four parts separate (as you did) but treat as real computationally, only *assuming* the r,i,j,+ij after each component, and all values would go where they belong and change in each step appropriately.  You got interesting results, but I'd guess they're just artifacts.  I'll print it, take it with me, & try to decipher.  Best I can do, for now.

Hiya fracmonk.

I am using QuaSZ by Terry Gintz which does not completely match with FractInt but is close.

The description you gave of the extension of the polynomial numerals in i i combined with the post that explained the square root of i as a polynomial numeral.  That is where the revision of your formula came from.

Now I am reminded of Kujonai and his mod 3 sign formulation, but what I cannot follow at preset is how you algebraically link these systems to the Cartesian or polar coordinate system.  Your advice and explanation do not at present help me to do that.  When you have a thought like this sometimes you are on your own until you can get others to play with the ideas the way you do.  I had a go because your formulation was suggestive of a way to realize it.  I assumed in one attempt that the formulation was attached to the quaternion math structure but with the coefficients being modified by the square root of i.  So by mistake I added quad c and got the sculptures posted.  I then corrected c to reflect the polynomial basis you outlined and got the "sausages" which I did not post.  On analysis my assumption would produce sausage results because your definition of the basis is in terms of i and that represents only 2 dimensions in quaternion math.

Hope I have explained clearly where I am up to with your polynomial numeral extension.

my second assumption would be to use your formulation as coefficients for r,i,j,k with the sqrt(i) function and see how that plays out.  What do you think?   

[fracmonk]

hey, jehovajah!  O.k., first to correct last msg- 4  3-d index sets, assuming unused dim param=0.  If u remember early diy M-set programming advice, it was given in real math- newx=x*x-y*y+a, new y=2*x*y+b, and the important part is that there is no "i" to be seen anywhere.  So, your generator's core iterator should *actually* be made to carry out the following for each iteration:

x=y=z=w=o (to start)  2 or 3 parts of your constant are auto, 2 or 1 are fixed, initially

newx=(x*x)-(y*y)-(2*z*w)+a
newy=(2*x*y)+(z*z)-(w*w)+b
newz=(2*x*z)-(2*y*w)+c
neww=(2*x*w)+(2*y*z)+d

I made the bailout test high, and just added them all together for it. Crude, but effective.
in 2-d:
For any given a+b, c by d exhibits origin symmetry.
There is x-axis sym for b by d for any a when c=0
   "     "     "        "     "  b by c  ''     "    "    "    d=0
   "     "     "        "     "  a by d  "     "    b    "    c=0
   "     "     "        "     "  a by c  "     "     "    "    d=0

Time!  All for now. Sorry.
 (time passes...)
Let me explain my situation:
I get all internet access from a public library computer.  Very limited time, purposely disabled drives, and lots of competition for equipment.  Very frustrating!  Got limits?  Always have to rush, and don't mean to be terse. Please forgive that. Thanx                               

[fracmonk]

Jehovajah!  Should have mentioned earlier this system is not quat, or any other similar scheme.  If i've got it right, it's a field.  Why give up field properties if you don't have to?  Pretty sure it's associative, commutative, distributive, etc.  Early on, stigomaster pointed out that sqrts of i are in C, but more than that, they're on the unit circle @ 45 deg angles to 1,-1,i, +-i.
C is the smallest closed subfield of a set of dimensions 2^n.  It is n=1, and the environment I propose here (n=2) is in turn a subfield of an 8-D system n=3, which would pare down the unit circle further into 22.5 deg arcs, etc.  You are very good at polar, I suspect, + I hope that offers insight.  I suck at it, + look at it from p.o.v. of algebra.  I mentioned that it's the relationships betw the dims that count.  Let's square a number: (distributively)

(i+j+ij)^2=-1+ij-j+ij+i-1-j-1-i=-3-2j+2ij    Notice in this case that real was zero before this particular number was squared, but now there's a real part in the product, and i terms canceled.  If still lost, refer back to the earlier table of 8 powers of j.  I'm deadly sure you can figure it out.

Also, I made a small mistake indicating symmetries: where it said "xaxis", that only referred to my 2-d rendering.  They have a "butterfly's symmetry" like in M.  No other plane in right angle orientation to C has simple connectedness. M on C is the only one.  Accept no substitutes.  I'll try to stay in touch but have pressing matters in personal life.  Later.

[jehovajah]

hey, jehovajah!  O.k., first to correct last msg- 4  3-d index sets, assuming unused dim param=0.  If u remember early diy M-set programming advice, it was given in real math- newx=x*x-y*y+a, new y=2*x*y+b, and the important part is that there is no "i" to be seen anywhere.  So, your generator's core iterator should *actually* be made to carry out the following for each iteration:

x=y=z=w=o (to start)  2 or 3 parts of your constant are auto, 2 or 1 are fixed, initially

newx=(x*x)-(y*y)-(2*z*w)+a
newy=(2*x*y)+(z*z)-(w*w)+b
newz=(2*x*z)-(2*y*w)+c
neww=(2*x*w)+(2*y*z)+d

I made the bailout test high, and just added them all together for it. Crude, but effective.
in 2-d:
For any given a+b, c by d exhibits origin symmetry.
There is x-axis sym for b by d for any a when c=0
   "     "     "        "     "  b by c  ''     "    "    "    d=0
   "     "     "        "     "  a by d  "     "    b    "    c=0
   "     "     "        "     "  a by c  "     "     "    "    d=0

Time!  All for now. Sorry.
 (time passes...)
Let me explain my situation:
I get all internet access from a public library computer.  Very limited time, purposely disabled drives, and lots of competition for equipment.  Very frustrating!  Got limits?  Always have to rush, and don't mean to be terse. Please forgive that. Thanx                              

Thanks for that. So i did assumption three which was my first thought that the coefficients are in fact the extensions for the four orientations i⁰₀ i¹₀ j₀ ij₀ where these are totally the same basis as i₀ v₀ j¹₀ k₀ as described in http://www.fractalforums.com/complex-numbers/polynomial-rotations/.

Quasz as Terry has updated it makes this a simple thing to do using the rfun..rend bracket.

I am touched by your situation and wish you feelings of gratitude along with opportunity to pursue your dreams  at every available instance resulting in joy to you and all around you.

I am attempting to post to the gallery and cut down on attachments as this uses up too much of Trifox's precious and appreciated space, but i think it is appropriate to attach these images for you to look at now,and if they are what you are finding to gallery them later. A julia and 2 mandelbulbs.
pixelsculpt mandy

xyz view of ordinary mandy

[fracmonk]

Jehovajah- went to link in your last msg to print it & take it away for study. At a very quick glance, I have doubts about this being the same animal (for instance, nothing about rotation in these), but will read, just the same.  I could not leave without adding a couple thoughts (& there are always more):
I've done a survey of 2-d "sliced mandelbread" studies, in which one param was zero, another fixed for each picture, and two became screen coordinates.  The issue I was trying to bring to bear in this discussion is about what is the most meaningful and appropriate extension of C not just for M, but anything complex.  I think this might be it.  It might be helpful to do a 3-d visualization that includes a+b recognizable cross-section in one of the butterfly symmetricals I listed.  Look for:

1. Any 3-d visualizations for a non-escaping value of the unseen component should be in one piece.
2. Is condition 1 necessary?
3. Does anything in 4-d extend beyond |2|?  It shouldn't, & my studies don't suggest it.

There IS a sound and provable way to do division in this system, and though it's not relevant to M-set generation in it, I think the question of field properties hinges on it, and that will go later to appropriateness arguments.

Thanx for the pix. I will look for coincidence between things in them and my own.

Later!

[jehovajah]

I recognised something like the complexified quaterinions in these images and refer you to the follwing,copied from Terry gintz Quasz manual for mac.

9.2 Tutorial
  An Introduction To CQuat Fractals By Terry W. Gintz
In the process of exploring all possible extensions to a fractal generator of this type, I
considered using discrete modifications of the standard quaternion algebra to discover new
and exciting images.  The author of Fractal Ecstasy [6] produced variations of the
Mandelbrot set by altering the discrete complex algebra of z2+c.  The extension of this to
quad algebra was intriguing.  There was also the possibility of different forms of quad algebra
besides quaternion or hypercomplex types.
Having modeled 3D fractals with complexified octonion algebra, as described in Charles
Muses' non-distributive algebra [7], it was natural to speculate on what shapes a
"complexified" quaternion algebra would produce.  Would it be something that was between
the images produced with hypercomplex and quaternion algebra?  Quaternion shapes tend to
be composed of mainly rounded lines, and hypercomplex shapes are mainly square (see
Figures 1 and 2.)
  i   j  k
i -1 k -j
j k -1 -i
k -j -i 1
Table 1 Hypercomplex variable multiplication rules
   i  j   k
i -1 k -j
j -k -1 i
k j -i -1
Table 2 Quaternion variable multiplication rules
In both quaternion and hypercomplex algebra, i2=-1.  The hypercomplex rules provide for one
real variable, two complex variables, (i and j) and one variable that Charles Muses refers to as
countercomplex (k), since k*k = 1.  It would appear from this that k = 1, but the rules in Table
1 show that k has complex characteristics.  In quaternion algebra there is one real variable
and three complex variables.  In hypercomplex algebra, unlike quaternion algebra, the
commutative law holds; that is, reversing the order of multiplication doesn't change the
product.  The basics of quaternion and hypercomplex algebra are covered in Appendix B of
Fractal Creations [8].  One other concept important to non-distributive algebra is the idea of
a "ring".  There is one ring in quaternion and hypercomplex algebra (i,j,k). (There are seven
rings in octonion algebra.)  If you start anywhere in this ring and proceed to multiply three
variables in a loop, backwards or forwards, you get the same number, 1 for hypercomplex,
and 1 or -1 for quaternion, depending on the direction you follow on the ring.  The latter
emphasizes the non-commutative nature of quaternions.  E.g. : using quaternion rules, i*j*k =
k*k = -1, but k*j*i = -i*i = 1.
For "complexified" quaternion algebra, the following rules were conceived:
  i    j   k
i -1 -k  -j
j -k   1   i
k -j   i   1
Table 3 CQuat variable multiplication rules
Note that there are two countercomplex variables here, (j and k).  The commutative law holds
like in hypercomplex algebra, and the "ring" equals -1 in either direction.  Multiplying two
identical quad numbers together, (x+yi+zj+wk)(x+yi+zj+wk) according to the rules of the

complexified multiplication table, combining terms and adding the complex constant, the
following iterative formula was derived for the "complexified" quaternion set, q2+c:
x -> x*x - y*y + z*z + w*w + cx
y  -> 2.0*x*y + 2.0*w*z + cy
z -> 2.0*x*z - 2.0*w*y + cz
w -> 2.0*x*w - 2.0*y*z + cw

I do think Terry would be best able to direct you on how to use quasz to get what you want.

By the way the link to polynomial rotations is to an unfinished peice of work.