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Author Topic: pretty fractal from a noncommutative 2 element algebra  (Read 652 times)
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Khashishi
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Posts: 13


« on: July 14, 2017, 08:29:37 PM »

I found a nice looking fractal using a 2 element algebra. Here's the times table:
*ij
iij
j-ij
So, the product (a_1,b_1)(a_2,b_2) = (a_1 a_2 - a_2 b_1, b_1 b_2 + a_1 b_2)
I don't know what this algebra is called. It's certainly not the same as the complex numbers.
The formula is the classic Mandelbrot/Julia:
z=z^2+c
(Since z is only multiplied by itself, it doesn't matter if you exchange the result of i*j and j*i in the times table.)
It resembles the burning ship fractal, but I think it is fundamentally different.


* noncommutative mandel.jpg (23.78 KB, 640x480 - viewed 87 times.)

* noncommutative mandel zoom.jpg (93.73 KB, 640x480 - viewed 80 times.)

* noncommutative julia.jpg (77.1 KB, 640x480 - viewed 92 times.)
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tsl
Forums Newbie
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Posts: 5


« Reply #1 on: August 05, 2017, 03:48:10 PM »

How did you derive this multiplication formula from your table?

I've made a little bit different approach. Since complex number can be treated as matrix. Where matrix multiplication and additions represent complex operations. Square root of det (or absolute value of det) is complex absolute value.

Code:
(a+bi) -> [a -b
           b  a]

Having this, you can completely differently arrange a and b values in matrix. Or even use some functions or whatever. The only thing is that matrix should have non zero determinant for all values (but zero). For example, with matrix:

Code:
[a    (b-a)
(b+a) sinh(a)]

Iteration z->z^2+c gives such mandelbrot:


or such julia for some point


one more example with random matrix

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Khashishi
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Posts: 13


« Reply #2 on: August 11, 2017, 08:52:34 PM »

How did you derive this multiplication formula from your table?

Maybe it isn't clear from the notation I used, but there are two elements which I called "i" and "j". There is no element "1". The ordered pairs just means:
(a,b) = a*i + b*j
a and b are just real numbers.

so, (a1,b1)*(a2,b2) = (a1*i + b1*j)*(a2*i + b2*j)
Using FOIL
= a1*a2*i*i + a1*b2*i*j + b1*a2*j*i + b1*b2*j*j
Now using the table, i*i = i, i*j = j, j*i = -i, j*j = j
-> a1*a2*i + a1*b2*j - b1*a2*i + b1*b2*j
-> (a1*a2-b1*a2)*i + (a1*b2+b1*b2)*j
-> (a1*a2-b1*a2, a1*b2+b1*b2)
a1, a2, b1, b2 are just real numbers and commute, so they can be rearranged to what I have in the original post.
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