FFDiaz
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« on: December 28, 2011, 09:35:40 AM » 

Yesterday, in the Spanish section at this forum, I was speaking with Kali about this theory of numbers and I thought that perhaps someone more is interested in take a look. The part that might be more useful for the generation of 3D Fractals probably is the numeric cyclebasis, an extension of complex numbers using its cyclical properties with the powers,so the first cyclo basis is g=1, g^{2}=g=1 and the period is 1, the second is h=1, h^{3}=h=1 and the period is 2,the third one is i, i^{5}=i and the period is 4, the fourth one is j, j^{9}=j and the period is 8 and so on.
For a more detailed description see the attached article (isn't full 256 kb only). Greetings Paco Fdez.


« Last Edit: December 28, 2011, 11:40:33 PM by FFDiaz »

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DarkBeam
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Fragments of the fractal like the tip of it


« Reply #1 on: December 28, 2011, 10:45:08 PM » 

can you do a preliminar render of something? would be appreciated Luca



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FFDiaz
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Posts: 18


« Reply #2 on: December 28, 2011, 11:38:47 PM » 

Sorry Luca, I do not have the necessary software to represent it because I have, only works for conventional complex numbers, it would have to develop the code to do this and since the 1990s not code in C. I will try to do something in Basic but it is a bit crude. Precisely I joined this forum because I know that people very skilled in this type of software development there are and my take me much time to get uptodate and the short time available to me I need it for my theoretical research.
Thanks Paco Fdez.



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David Makin


« Reply #3 on: December 29, 2011, 12:25:45 AM » 

Basic or basicstyle pseudocode would I guess be fine for most of the programmers on the forum Certainly better that the "raw" math for me anyway



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DarkBeam
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Fragments of the fractal like the tip of it


« Reply #4 on: December 29, 2011, 06:25:46 AM » 

Ok I can try with Fragmentarium ... If it will look good will be implemented somewhere



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fractower
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« Reply #5 on: December 29, 2011, 06:58:58 AM » 

I think you can get there using complex numbers. Some periods have multiple solutions.
g^2 = g; g = 1 Special case I wish would go away.
h^3 = h; h = 1 i^5 = i; i = sqrt(1) j^7 = j; j = [(1)^(1/3), (1)^(2/3)] First multisolution. k^9 = k; k = [(1)^(1/4), (1)^(3/4)] Exclude sub harmonic.
x^n = x; x = [(1)^(m*/((n1)/2))] where m is a positive int less than (n1)/2 and shares no common factors with (n1)/2.



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DarkBeam
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Fragments of the fractal like the tip of it


« Reply #6 on: December 29, 2011, 09:24:29 AM » 

Reduced to the bone your paper proposes this law; float xx = real(zri) float yy = imag(zri) float zz = real(zjk) float ww = imag(zjk) zri = xx*xx + yy*yy + flip(2*xx*yy + zz*zz) + cri zjk = 2*zz*(xxyy) + 2*ww*flip(xxyy+zz) + cjk // I may have forgot some ww terms but they stay = 0 in 3D slices so... You should already know that all the quaternionstyle formulas have already been explored (well, almost... or in any case the result is known) The result is not so satisfying? Or yes? It depends on what you wished in the end. Also tried more recipes; // pacobrot B zri = xx*xx + yy*yy + flip(2*xx*yy  zz*zz) + cri zjk = 2*zz*(xxyy) + 2*ww*flip(xx+yyzz) + cjk // pacobrot B zri = xx*xx + yy*yy + flip(2*xx*yy  zz*zz) + cri zjk = 2*zz*(xxyy) + 2*ww*flip(xx+yyzz) + cjk // pacobrot C zri = xx*xx  yy*yy + flip(2*xx*yy + zz*zz) + cri zjk = 2*zz*(xx+yy) + 2*ww*flip(xx+yyzz) + cjk The quality is horrible because I used UF with brute force... Even the holy grail would look awful with that method. Anyway it is realistic at 70% ... So don't hope in miracles Hi!!!

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FFDiaz
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« Reply #7 on: December 29, 2011, 06:35:39 PM » 

I will respond in order, first to David Makin. Nothing further from my intention that insinuate that Basic is a little capable programming language, I was referring to what I am able to do with it. I apologize in advance because I'm not speaking English and it is easy to not me express correctly, so my sincere apologies, David. Greetings Paco Fdez.



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FFDiaz
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« Reply #8 on: December 29, 2011, 08:13:33 PM » 

Now to Luca or Darkbeam as you prefer. I I am excited to see the first results, particularly in option B. It has been something like to see a son for the first time. Many thanks in advance for your interest and work. I would like to tell you, since I do not know if this is the approach that you have given in your programming, which are my ideas on the graphical representation. If t is a number belonging to RC3 them t=x0·g+x1·h+y·i+z·j. x0 is in the positive part of the x axis and x1 is in the negative part of x axis. y is on the yaxis and z on the zaxis. All initial values x0, x1, y and z are positive. In other words, all the coefficients of c are positive in the iteration formula f(t)=t ^{2}+c. t^{2}=(x0^{2}+x1^{2})·g+(2·x0·x1+y^{2})·h+(2·y·(x0x1)+z^{2})·i+2·z·(x0x1+y)·j And would to check if diverges the iteration for every number of the space so defined we take as square of the module t^{2}=(x0x1)^{2}+y^{2}+z^{2} Again many thanks Luca. Greetings Paco Fdez.


« Last Edit: December 30, 2011, 03:35:57 PM by FFDiaz »

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FFDiaz
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Posts: 18


« Reply #9 on: December 29, 2011, 09:05:39 PM » 

And now Fractower. Is not convenient to think i like sqrt ( 1) or (1) ^{1/2} because this brings us to i ^{2} could be 1 or 1 depending on the properties of the powers that we use. So if we use the property (a·b) ^{n}=a ^{n}·b ^{n} let see: i^{2}=i·i=(1)^{1/2}·(1)^{1/2}=((1)·(1))^{1/2}=1^{1/2}=1 or if we use a ^{n}·a ^{m}=a ^{m+n}: i^{2}=i·i=(1)^{1/2}·(1)^{1/2}=(1)^{1/2+1/2}=(1)^{1}=1 Similarly it is not convenient to think of j=(1) ^{1/4}, k=(1) ^{1/8}... I think that the correct interpretation of the successive powers of the representatives numbers of each cyclobase is given in column 3 of the table that is on page 16 of my article. Many thanks for your interest. Greetings, Paco Fdez.


« Last Edit: December 31, 2011, 02:12:29 AM by FFDiaz »

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DarkBeam
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Fragments of the fractal like the tip of it


« Reply #10 on: December 29, 2011, 10:22:25 PM » 

Ahh! So you did not made a normal quaternion Please please express your formula in clear basic language using common funcs and symbols! I need that you call x like x, y like y... Use how many symbols as you please... I need a clear transcription to do the translation correctly No need to reply in a day but be clear por favor



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David Makin


« Reply #11 on: December 29, 2011, 11:05:01 PM » 

Ahh! So you did not made a normal quaternion Please please express your formula in clear basic language using common funcs and symbols! I need that you call x like x, y like y... Use how many symbols as you please... I need a clear transcription to do the translation correctly No need to reply in a day but be clear por favor Actually it's most definitely not a quaternion. The key is that the form concerned has *4* terms for an "observable" 3 dimensional system as I understand it. In terms of 4 separate "dimensions" addition and subtraction are as you would expect and the system is commutative and distributive but not associative(, nor does it have a unique inverse???) Anyway here's the standard multiplication table for up to 5 "dimensions" > == 4 "observable" dimensions, I give this in terms of r,i,j,k,l as that's how one would normally expect but under Paco's system r and i are not "separately observable" and in fact combine to a single numerical dimension (as in the magnitude calculation): r i j k l r r i j k l i i r j k l j j j i k l k k k k j l l l l l l k
For 3 observable dimensions then r,i,j,k are required, addition and subtraction are as you would expect, multiplication follows the above table and the magnitude xr+yi+zj+wk is sqrt((xy)^2+z^2+w^2) if I've understood correctly. Unfortunately as far as I can see there is no definition given for division  though I guess working out the appropriate matrix and inverting it would probably give a form to use though my math is not quite up to that. @Paco  have you a definition of division (for two 4D variables x and y in terms of your g,h,i,j or in terms of the above r,i,j,k) for an "observable" 3d system ?



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David Makin


« Reply #12 on: December 30, 2011, 04:19:42 AM » 

r i j k l r r i j k l i i r j k l j j j i k l k k k k j l l l l l l k So the square of (x, y, z, w) would be (x^2+y^2, 2*x*y+z^2, 2*x*z2*y*z+w^2, 2*x*w2*y*w+2*z*w)



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DarkBeam
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Fragments of the fractal like the tip of it


« Reply #13 on: December 30, 2011, 09:08:44 AM » 

Now to Luca or Darkbeam as you prefer. I I am excited to see the first results, particularly in option B. It has been something like to see a son for the first time. Many thanks in advance for your interest and work. I would like to tell you, since I do not know if this is the approach that you have given in your programming, which are my ideas on the graphical representation. If t is a number belonging to RC3 them t=x0·g+x1·h+y·i+z·j. x0 is in the positive part of the x axis and x1 is in the negative part of x axis. y is on the yaxis and z on the zaxis. All initial values x0, x1, y and z are positive. In other words, all the coefficients of c are positive in the iteration formula f(t)=t ^{2}+c. t^{2}=(x0^{2}+x1^{2})·g+(2·x0·x1+y^{2})·h+2·y·(x0x1)·i+2·z·(x0x1+y)·j And would to check if diverges the iteration for every number of the space so defined we take as square of the module t^{2}=(x0x1)^{2}+y^{2}+z^{2} Again many thanks Luca. Greetings Paco Fdez. hmmm, if I understand right x0 is x when x is lt 0 but when gt 0? 0? for x1 the same problem and y z should be normal? cx cy cz must be preprocessed with abs() Probably?



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FFDiaz
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Posts: 18


« Reply #14 on: December 30, 2011, 11:40:29 PM » 

Indeed, these numbers have nothing to do with quaternions.
i is the imaginary number conventional but j isn't the j of quaternions. j^2 is i not 1 as in the quaternions.
The real part of the number is constructed with the cycle bases g and h. The coefficient which multiplies to g is the numerical value of an observable of a positive nature (for example, the charge of proton) and the coefficient of h for a negative observable. Therefore, usually the coefficient of g or h is zero depending on the type of observable (this is important when it comes to the programming of the fractal).
The square of a number of RC3 is indeed as have written you David (I made a mistake and not wrote the last term of the coefficient of i, but I have now corrected). Indeed, as suspicions David, this is not probably a division algebra, but I have to study it in more detail the inverse of a number, I believe that it is not unique. But although this is very important to mathematicians, the division is not a necessary operation in physics that I am creating.
Luca in a while I will try to translate this into relationships that you can include in a program.
Greetings, Paco Fdez.


« Last Edit: December 31, 2011, 02:09:51 AM by FFDiaz »

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