Mandelbrot and the road forward

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Offline DaveFash

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« on: February 22, 2018, 06:01:05 PM »
I jumped out of the box. I took Mandelbrot's recursive 2D equation, using 1 imaginary plane; and expanded it to 3D with 2 imaginary planes [1 didn't do it]. The two imaginary planes forced me to invent new math rules, like i x i' = -1, and required a visual basic computation to implement the new rules. But, I think I just stumbled into a fascinating new view of fractal configurations.

I have submitted my findings to the university that issued my math degree for consideration toward a graduate thesis -- but we'll see where that leads. In the meantime, I am in awe of the visual products that my little pc is turning out ( see MandelFash.com - listen to the short intro video and then checkout the images at the very bottom of the web page). I am more than happy to share the technique, even the VB code that generates the 'babies'; I have even made the source data files available for download, if you have graphing capabilities. (see sample data file attached)

Thanks for your time,
Dave Fashenpour.


Offline utak3r

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« Reply #1 on: February 22, 2018, 09:50:42 PM »
That looks interesting and promising, but I've only read it quickly. I think that the method of rather abstract maths enhancing is promising to discover new things, as pure cutouts of real quaternions ain't so interesting  :embarrass:  good luck with this job  :thumbs:

Offline DaveFash

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« Reply #2 on: June 15, 2018, 11:02:14 PM »
 :yes: UPDATE - big breakthrough in 3D Mandelbrot.  Please see latest results at: http://MandelFash.com

Offline 3DickUlus

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« Reply #3 on: June 16, 2018, 02:31:20 AM »
quote from your site... (spelling corrected btw)
Quote
We would like to see Math professionals scrutinize OUR MATHEMATICS.
post the source code if you want some input
Resistance is fertile... you will be illuminated!

https://en.wikibooks.org/wiki/Fractals/fragmentarium

Offline gerrit

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« Reply #4 on: June 16, 2018, 08:51:04 AM »
I get for the 3D M-iteration using your \( i' \)
\( x +iy+i'z \leftarrow (x+iy+i'z)^2 + c_x+ic_y+i'c_z = x^2-y^2-z^2-2yz +c_x+i(2xy+c_y)+i'(2xz+c_z). \)
So using normal real numbers it's
\( x \leftarrow x^2-y^2-z^2-2yz + c_x \)
\( y \leftarrow 2xy + c_y \)
\( z \leftarrow 2xz +c_z \)

I tried it in UF, taking \( (c_x,c_y,c_z) \) as various planes through the origin but all it seems to produce is rotated and translated normal Mandelbrots. I think this means there is a linear change of variables that will eliminate one of them and leave you with just the 2D M-set.
Am I missing something?

Offline 3DickUlus

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« Reply #5 on: June 16, 2018, 06:40:41 PM »
@gerrit did you watch the videos on his site? the images there lead me to the same conclusion, 2D translated.

Offline gerrit

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« Reply #6 on: June 16, 2018, 09:05:46 PM »
@gerrit did you watch the videos on his site? the images there lead me to the same conclusion, 2D translated.
I sort of skimmed them.
This may explain why: https://math.stackexchange.com/questions/751106/complex-number-with-3-dimensions

Offline FractalDave

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« Reply #7 on: August 06, 2018, 09:59:56 PM »
Yep - nothing special, been tried before....

IMO the more interesting extension of complex to multiple dimensions (that goes to all powers of 2 i.e, 4D, 8D etc.) is:

2D: x+iy with x and y real, giving x^2-y^2+2xyi when squared
4D: x+jy with x and y complex giving x^2-y^2+2xyj when squared
8D: Same again but using x+yk with x and y as 4D above etc.

The 4D version is the same (fractal-wise) as what's often referred to historically as plain "hypercomplex" but more accurately termed bi-complex i.e. it's the type that produces the squarry-brots and julias.
The 4D version fails the field test due (only??) to lack of a true division algebra because it's possible for two non-zero numbers to produce zero on multiplication. Personally I like this more than standard quaternions because their fractals are so boring and I dislike anything not commutative :)


For multiplication for 2D complex (a+ib)(c+id) = (ac-bd)+i(ad+bc) it's obvious that if any single values of real a or b and c or d are non-zero then the result will be non-zero, also that if a and b or c and d are non-zero and either of the other pair is non-zero then the result is non-zero.
This also applies to the 4D version.
For complex we can also say that if all 4 real parts (a,b,c,d) are non zero then the result cannot be zero because for ac-bd to be zero then the signs of ac and bd must be the same i.e. for zero if a and c are the same sign then b and d must also be the same sign and if a and c are opposite signs then b and d are also opposite signs therefore in any case ad+bc would be non-zero - i.e. only multiplying by (0+0i) produces a zero result.
Unfortunately for the 4D case this last part fails thus allowing the possibility of zero being returned on multiplying two non-zero 4D values hence the extension to 4D is not a true field.

The 8D version probably fails further tests for a field - but then again so do octonions.

BTW Based on the definition of complex division at Wolfram Mathworld one would think that the division algebra for the 4D version fails completely because the divisor c^2+d^2 could be zero for certain non-zero complex c and d, but the correct version of the divisor is (UF) |c|+|d| i.e. cabs(c)^2+cabs(d)^2 or x^2+y^2+z^2+w^2 if c is x+iy and d is z+iw.
« Last Edit: August 07, 2018, 12:46:11 AM by FractalDave, Reason: Additions »
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