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 Author Topic: Triplex Algebra  (Read 28080 times) Description: Triplex Algebra 0 Members and 1 Guest are viewing this topic.
bugman
Conqueror

Posts: 122

 « on: December 04, 2009, 11:22:16 PM »

Some people have requested for a more complete definition of the triplex. Unfortunately, it doesn't exactly form a complete algebra, but I will try to summarize the most consistent formulas here. I will only focus on the positive z-component formula here, because that is the most consistent formula.
 TriplexArithmetic.gif (12.39 KB, 675x651 - viewed 1319 times.) « Last Edit: December 16, 2009, 07:06:55 AM by bugman » Logged
bugman
Conqueror

Posts: 122

 « Reply #1 on: December 04, 2009, 11:24:18 PM »

The nth root of the power formula has n² unique valid roots (branches) if n is an integer. If n is not an integer, then depending on the values of {x, y, z}, some of the roots will not be valid.

Here is some C++ code in case you have difficulty following the math formulas below:

//Arbitrary integer roots, allows for n² branches
//this code is written for ease of understanding and is not intended to be efficient
//all unique valid roots can be found using all possible combinations of:
// ktheta=0,1,...abs(n)-1
// kphi=0,1,...abs(n)-1
triplex TriplexRoot(triplex p, int n, int ktheta, int kphi) {
int k1=(abs(n)-(p.z<0?0:1))/4, k2=(3*abs(n)+(p.z<0?4:2))/4, dk=0;
if(abs(n)%2==0 && kphi>k1 && kphi<k2) dk=sign(p.z)*(abs(n)%4==0?-1:1);
double r=sqrt(sqr(p.x)+sqr(p.y)+sqr(p.z));
double theta=(atan2(p.y,p.x)+(2*ktheta+dk)*pi)/n;
double phi=(asin(p.z/r)+(2*kphi-dk)*pi)/n;
return pow(r,1.0/n)*triplex(cos(theta)*cos(phi),sin(theta)*cos(phi),sin(phi));
}

Also, here is some Mathematica code in case that is easier for you to understand:
TriplexRoot[ {x_, y_, z_}, n_, ktheta_, kphi_] := Module[ {dk = If[ Mod[ n, 2] == 0 && Abs[ n] < 4kphi + If[ z < 0, 0, 1] <= 3Abs[ n], Sign[ z]If[ Mod[ n, 4] == 0, -1, 1], 0], r = Sqrt[ x^2 + y^2 + z^2],theta, phi}, theta = (ArcTan[ x, y] + (2ktheta + dk)Pi)/n; phi = (ArcSin[ z/r] + (2kphi - dk)Pi)/n; r^(1/n){Cos[ theta] Cos[ phi], Sin[ theta]Cos[ phi], Sin[ phi]}];
 Roots.gif (8.67 KB, 644x439 - viewed 941 times.) « Last Edit: December 17, 2009, 12:27:59 AM by bugman » Logged
David Makin
Global Moderator
Fractal Senior

Posts: 2269

 « Reply #2 on: December 04, 2009, 11:53:59 PM »

Thanks Paul !
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Safarist

Posts: 85

 « Reply #3 on: December 05, 2009, 04:37:47 AM »

Have you tried defining transcendental functions by their Taylor series, such as $e^z=(1,0,0)+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$ where $z$ is a triplex, to see if that makes any kind of sense?
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bugman
Conqueror

Posts: 122

 « Reply #4 on: December 05, 2009, 06:42:38 AM »

Have you tried defining transcendental functions by their Taylor series, such as <Quoted Image Removed> where <Quoted Image Removed> is a triplex, to see if that makes any kind of sense?

That's an interesting idea, but it looks difficult. The triplex power function is rather complicated. I'll think about it and see if I come up with any ideas.
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Paolo Bonzini
Guest
 « Reply #5 on: December 05, 2009, 08:43:25 PM »

FWIW, the triplex numbers do not even form an alternative algebra, and there is no distributivity either.  They only have power associativity (which is no big news) .
 « Last Edit: December 05, 2009, 11:29:18 PM by Paolo Bonzini » Logged
Paolo Bonzini
Guest
 « Reply #6 on: December 05, 2009, 11:39:07 PM »

Another little thing, it is enough that one of the three factors be complex (z=0) to have associativity (it is obvious that one is enough, since there is commutativity).

Also, it is enough that a is complex to have a(b+c) = ab+ac.

This may make it a bit easier to manipulate power series...
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zee
Guest
 « Reply #7 on: December 06, 2009, 01:47:50 AM »

Thank you for this definition! I think you added this to wikipedia?! (http://en.wikipedia.org/wiki/Hypercomplex_number#Three-dimensional_complex_numbers_based_on_spherical_coordinates)

But I can't see why the third dimension can be called hypercomplex.
In the normal Mandelbrot you have the simple multiplication z² in complex and only because of i² = -1 you get the mandelbrot iteration very simple; its the property that complex numbers can do a rotation and rescaling.

So I think the 3rd dimension can't be a second complex dimension, and the multiplication and result shows that a second complex number like j will not be used (and makes no sense) because the result is more than multiplying 2 hypercomplex numbers (which are sums)! In 2D it works, but try to multiply quarternions - you can't get terms with x y and z in one component. This is impossible by squaring.

I see that the 2D mandelbrot equations are still in there... But it looks like that the "triplex" is something that can't be understand in calling it hypercomplex... there must be somthing more/other...
 « Last Edit: December 06, 2009, 07:43:49 AM by zee » Logged
David Makin
Global Moderator
Fractal Senior

Posts: 2269

 « Reply #8 on: December 06, 2009, 02:29:04 AM »

@bugman: There's a mistake on the Wikipedia page - you have the 3rd value for the multiply as sin(theta2+theta2).
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Paolo Bonzini
Guest
 « Reply #9 on: December 06, 2009, 03:43:57 AM »

I fixed the mistake in Wikipedia.  BTW, I think it can be called a hypercomplex number if the third axis is taken to have j^2=0 (or maybe not because unlike dual numbers, here you have to count z^2 when you compute the norm; these numbers are strange...).  You can also say that (x,-y,-z) is the conjugate.

Also, here is a closed form expression for powers of the triplex numbers based on Chebyshev polynomials Tn(x) and Un-1(x).  These give $T_n(x) = \cos(n\ \arccos\ x)$, and $\sin\ \arccos\ x\ \cdot\ U_{n-1}(x) = \sin(n\ \arccos\ x)$ without explicitly computing the trigonometric functions, so they are very handy for this kind of transformation.

I initially assume that x^2+y^2+z^2 = 1. I express the angles in terms of arccos and use the Chebyshev polynomials:

$\rho_{xy} = \sqrt{x^2+y^2}$
$\theta = \arccos\ \frac{x}{\rho_{xy}},\ \phi = \arccos\ \rho_{xy}$

$
(x,y,z)^n = (\cos\ n\theta\ \cos\ n\phi,\ \sin\ n\theta\ \cos\ n\phi,\ \sin n\phi) \\
\, = (T_n(\frac{x}{\rho_{xy}}) T_n(\rho_{xy}),\ \frac y{\rho_{xy}} U_{n-1}(\frac{x}{\rho_{xy}}) T_n(\rho_{xy}),\ z U_{n-1}(\rho_{xy})) \\
\, = (T_n(\frac{x}{\rho_{xy}}) T_n(\sqrt{1-z^2}),\ \frac y{\rho_{xy}} U_{n-1}(\frac{x}{\rho_{xy}}) T_n(\sqrt{1-z^2}),\ z U_{n-1}(\sqrt{1-z^2))
$

From this formula, the various pieces in http://www.fractalforums.com/index.php?action=dlattach;topic=742.0;attach=436;image should be more or less identifiable.

Since (x,y,z) does not usually have unary modulus, you should divide the arguments by $\rho$ and multiply the results by $\rho^n$.  Most of the divisions cancel, and what you get is:

$
(x,y,z)^n = (\rho^n T_n(\frac{x}{\rho_{xy}}) T_n(\sqrt{1-\frac{z^2}{\rho^2}}),\ \rho^n \frac y{\rho_{xy}} U_{n-1}(\frac{x}{\rho_{xy}}) T_n(\sqrt{1-\frac{z^2}{\rho^2}}),\ \rho^{n-1} z U_{n-1}(\sqrt{1-\frac{z^2}{\rho^2}}))$

 « Last Edit: December 06, 2009, 03:48:58 AM by Paolo Bonzini » Logged
bugman
Conqueror

Posts: 122

 « Reply #10 on: December 06, 2009, 06:05:09 AM »

Disclaimer: I'm not a mathematician, and I can't say exactly what constitutes an "alternative algebra" or what constitutes a "hypercomplex" number. I also haven't added anything to Wikipedia, but I am impressed with Paolo's contributions on this topic.

There are other ways of defining multiplication and power operators for the "triplex", and I am not entirely convinced that the method I outlined here is the best way. For example, I was wondering if applying the rotational matrix about the y-axis was really the most logical choice. I have been experimenting with some other rotations, but so far it hasn't produced a very good looking 3D Mandelbrot set. Of course, the most natural method of applying a single rotation about the axis that is perpendicular to both the x-axis and the point in question, reduces to the quaternion method with w = 0, which isn't very interesting.
 « Last Edit: December 06, 2009, 06:46:42 PM by bugman » Logged
zee
Guest
 « Reply #11 on: December 06, 2009, 07:38:03 AM »

I think I got it!

There are maybe some small problems because this is not a full definition, but there is enought defined to derive the non-trigometric equation.
The 3rd dimension j is a very complicated thing because i*j = -1 but j² is something else...
 « Last Edit: December 22, 2009, 09:02:48 PM by zee » Logged
Paolo Bonzini
Guest
 « Reply #12 on: December 06, 2009, 11:14:02 AM »

Heh, I don't really think it is a great contribution, at least I don't know really enough math to move it ahead.  You can read on Wikipedia for alternative algebra, it's not hard.  Hypercomplex is a bit over my head though. :-)  I am not a mathematician either, I just dabble more in paper than in graphics.

Anyway, here is my little conjecture:

$e^{(x,y,z)}=e^x e^{(0,y,0)} e^{(0,0,z)} = e^x (\cos\ y,\ \sin\ y,0)\ (\cos\ z,\ 0,\ \sin\ z) = e^x (\cos\ y\ \cos\ z,\ \sin\ y\ \cos\ z,\ \sin\ z)$

It's not totally out of thin air.  Since $e^x$ and $e^{(0,y,0)}$ are complex, the multiplications in $e^x e^{(0,y,0)} e^{(0,0,z)}$ are associative, so you can actually sum the exponents to $e^{(x,y,z)}$. Still a long way from a proof though...

Of course you could just read the equalities from right to left and turn the conjecture into a definition but proving the above (especially $e^{(0,0,z)}$) from the power series would be really nice...
 « Last Edit: December 06, 2009, 10:43:48 PM by Paolo Bonzini » Logged
Paolo Bonzini
Guest
 « Reply #13 on: December 06, 2009, 04:51:27 PM »

And one more thing.  I said above that (0,0,1)^2=0, but that's not true.  I was trusting the Cartesian formulas too much, but they are clearly undefined if one of the rho_xy is zero, and the simplification is not at all trivial.

In particular:

for (0,0,z1)*(0,0,z2), the result is (-z1 z2,0,0).  This is particularly important because it basically gives the required proof for the value of e^(0,0,z).  The idea is that powers alternate between (0,0,(-1)^n z^2n+1) and ((-1)^n z^2n+2,0,0), so that the series can be split to sines and cosines.

for (0,0,z1)*(x2,y2,z2), I got $\left(-z_1 z_2 x_2 / \rho_2, -z_1 z_2 y_2 / \rho_2, z_1 \rho_2 + z_2 \rho_1\right)$ but this does not really make sense to me. For example it gives (0,0,2)*(0,1,1) = (0,-2,2) but it should be (0,1,1)^3 = (0,2,2) according to the trig formula.  Apparently elevation's full range must be considered between (-pi,+pi) because the (usually redundant) range matters when the elevation is multiplied by n.  So each triplex has two spherical representations, for example (1,3 pi/2, 3 pi/4) and (1,pi/2,pi/4), and bad things happen when the cartesian formulas exchange one with the other.

It may mean that we have abandoned of power associativity too (bad!), or maybe we're just complicating the formulas unnecessarily and the real underlying space is 4D, such as $(r\ \cos\ \vartheta,\ r\ \sin\ \vartheta, r\ \cos\ \varphi,\ r\ \sin\ \varphi)$. I honestly have no idea, and need to work it out a bit more on paper.  Funny though that there is an exponential operator, but not a good definition of multiplication in cartesian coordinates!

As far as fractals are concerned, it's not too bad because there is a definition of power that can be used for Mandelbrot and Newton fractals at least!  Also, the results seem to vary in sign only if you change cartesian vs. trigonometric computation of powers, so it is fine if the non-trig formulas are used for drawing!

(After this detour, the third special case is trivial from commutativity).
 « Last Edit: December 06, 2009, 10:42:51 PM by Paolo Bonzini » Logged
zee
Guest
 « Reply #14 on: December 06, 2009, 07:07:21 PM »

I'm not a mathematician, too, but a theoretical physicist and I want to write my master thesis about that. I'll try to define the whole set mathematically.

I think my solution is okay (providet that the equations I derived are correct) because its more than only using math till it works; There is an idea that says the 3rd basis is not only a hypercomplex basis but it's the (normalized) rotation of the 1st and 2nd coordinate (the 2d complex number) because it has to know the first rotation.

I think that makes much sense!