bugman


« Reply #15 on: December 06, 2009, 07:24:19 PM » 

Perhaps a better title for this thread would have been "Triplex Arithmetic"?



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Paolo Bonzini
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« Reply #16 on: December 06, 2009, 09:25:33 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.
That would be really cool. Good luck! It's more than only using math till it works
LOL :) I hope it's not a totally accurate definition of what I'm doing (though it's close). Note however that j^2=1, it's just not clear from the Cartesian formulas. My impression is that any time you move away from the trig definition you are complicating your life with these numbers. It may be more accurate to first define as much as possible on spherical coordinates (since even things like e^x and ln x seem to work), and then define the Cartesian formulas more or less qualitatively (e.g. does the cartesian multiplication preserve modulus? what about the angles?) etc. The bare minimum that you need to get the moduli of polynomials i.e. to draw fractals. Having done this, extensions to >3D are probably much less complex, ehm, complicated. However, even though with fractals things seems to mostly work, a serious treatment of the topic should also understand why...



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bugman


« Reply #17 on: December 06, 2009, 10:06:36 PM » 

Here is the solution for the exponential function:


« Last Edit: December 16, 2009, 07:06:01 AM by bugman »

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Paolo Bonzini
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« Reply #18 on: December 06, 2009, 10:09:58 PM » 

OK, I think I found the exponential function, but I'm not sure how to check it yet...
Can you explain the derivation? Looks pretty different from mine...



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zee
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« Reply #19 on: December 06, 2009, 11:10:28 PM » 

Note however that j^2=1, it's just not clear from the Cartesian formulas.
Why j^2 must be 1? Okay, this would explain calling it hypercomplex but I think that can't work to get those equations! (and I tried a lot of things) In my definition it's something else BECAUSE of the imaginary IN the imaginary! If you calculate j^2 with my definition you'll get the 1 but not alone... and we should remember that if j is the 3rd dimension, every number has an own 3rd direction! So are we talking about or about ? I think I have to ask: Did you read and understood what I wrote down? And do you see that the 3rd dimension must be something more complex than a definition of a basis such as quarternions?


« Last Edit: December 06, 2009, 11:16:03 PM by zee »

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David Makin


« Reply #20 on: December 06, 2009, 11:38:56 PM » 

Perhaps someone reading this thread can clear up an issue related to the nontrig calculations.
I'm using a mix of complex and real to calculate the formula, in the case below for the "sine" version (first iteration initial value of magn is zero):
r = cabs(zri) magn = sqrt(magn)^@mpwr zjk = r + flip(zj) if r>0.0 zri = (zri/r)^@mpwr endif if (ph=cabs(zjk))>0.0 zjk = (zjk/ph)^@mpwr endif zj = cj  magn*imag(zjk) zri = magn*real(zjk)*zri + cri magn = zri + sqr(zj)
where zri, zjk, and cri are complex and the rest are real. cabs(zri) is sqrt(x^2 + y^2) where zri is (x+i*y) and zri is just x^2+y^2
You'll note that when zri is (0,0) and when zjk is (0,0) then I assume the sine/cosine calculation done by raising the normalised complex value to a power should return (0,0) i.e. it leaves the values unchanged. It would be equally valid to return (1,0)  would that be a better option mathematically speaking ?



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Paolo Bonzini
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« Reply #21 on: December 07, 2009, 12:33:42 AM » 

Note however that j^2=1, it's just not clear from the Cartesian formulas.
Why j^2 must be 1? [...] are we talking about or about ? Sorry, I was talking about one particular j, the one that is not associated to any i (because it has 0+0i as the complex part). In spherical coordinates (0,0,1) is (1,0,pi/2) so j^2=(1,0,pi) which becomes (1,0,0) when converted back to cartesian. I understood your definition and I liked it, still it does not explain (AFAICT) what happens when the complex parts are zero, hence my observation above. If you can explain that, that'd be great. You also have a bit of handwaving in how the j's cancel with z's (z's are reals, so I didn't really understand that part), but that's fine for now. I still think that it is better to milk as much as possible from the spherical definitions first (including an exponential and logarithm), though. Cartesian representation is turning out to be much more tricky. Perhaps someone reading this thread can clear up an issue related to the nontrig calculations. You'll note that when zri is (0,0) and when zjk is (0,0) then I assume the sine/cosine calculation done by raising the normalised complex value to a power should return (0,0) i.e. it leaves the values unchanged. It would be equally valid to return (1,0)  would that be a better option mathematically speaking ?
I don't know for the signreversed calculation, but I gave above the correct formulas for zri and/or zjk being (0,0): for (0,0,z1)*(0,0,z2), the result is (z1 z2,0,0).
for (0,0,z1)*(x2,y2,z2), I got (z1 z2 x2 / rho2, z1 z2 y2 / rho2, z1 rho2) where rho2 is sqrt(x2^2+y2^2).
The third special case is trivial due to commutativity. It should be easy to reconcile the difference between these definitions and the signreversal. BTW, I encountered some inconsistencies here, so I suggest that you compute z^n directly rather than through repeated multiplication.


« Last Edit: December 07, 2009, 01:19:06 AM by Paolo Bonzini »

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zee
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« Reply #22 on: December 07, 2009, 12:51:33 AM » 

In spherical coordinates (0,0,1) is (1,0,pi/2) so j^2=(1,0,pi) which becomes (1,0,0) when converted back to cartesian.
How can you follow this? Is everybody convinced that the equations (by wikipedia and bugman) for triplex multilplication are correct?So it's easy to show that j^2 can't be 1: like in other hypercomplex numbers. But the "from heaven falling" rule for Triplex multiplication says that the first component have to be: This is something other I'll work to understand that problem in the origins. I just started yesterday to think about that equations. Maybe I should write my program on and try to compare cartesian values with the spherical, too. But I didn't occupy myself very much with simulating 3d mandelbulbs after understanding that it works. I've got an 3d array with for example 1 and 0 and don't know how to get the surface with triangles instead of showing voxels. (I'm creating an .obj file and render with Bryce). I'm sorry, this is not the topic but I think I'll create a topic or read something about that and I hope you can help me a lot if I'm welcome in this amazing active community!


« Last Edit: December 07, 2009, 01:04:56 AM by zee »

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Paolo Bonzini
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« Reply #23 on: December 07, 2009, 01:13:46 AM » 

Is everybody convinced that the equations (by wikipedia and bugman) for triplex multilplication are correct?
Yes, they are (at least as correct as they can be, given the issue with range of elevation that I mentioned earlier). But if the x and y is zero, they are undefined and you need a different definition. BTW on wikipedia there's no reference to the cartesian formulas (luckily; they would be likely copied without all these details...). But the "from heaven falling" rule for Triplex multiplication says that the first component have to be: (x^2  y^2) (1  \frac{z^2}{r^2})
Basically, the (x^2  y^2) * 1 factor is eliminated. But (x^2  y^2) / (r^2) is 0 / 0, so you have to find its value in some way. I couldn't compute it with limits, but I could compute it with trigonometry and its value is 1. So you get that the first component of (0,0,1)^2 is z^2 = 1. The formulas do not fall from heaven. They come from the spherical coordinate definition of powers and multiplications, and can be derived from there. Try doing that on paper (it's not totally trivial, but since you know what the result must look like... ;) ). Note that I've written some quick Maxima functions, but no 3D stuff. I'm looking at it only from a theoretic point of view. I think we can continue discussing here in this topic.


« Last Edit: December 07, 2009, 01:42:31 AM by Paolo Bonzini »

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David Makin


« Reply #24 on: December 07, 2009, 01:47:21 AM » 

Perhaps someone reading this thread can clear up an issue related to the nontrig calculations. You'll note that when zri is (0,0) and when zjk is (0,0) then I assume the sine/cosine calculation done by raising the normalised complex value to a power should return (0,0) i.e. it leaves the values unchanged. It would be equally valid to return (1,0)  would that be a better option mathematically speaking ?
I don't know for the signreversed calculation, but I gave above the correct formulas for zri and/or zjk being (0,0): I think I deduce from that, that the answer is yes I should use (1,0) instead of (0,0) when normalising (0,0)  I guess the answer was in the question anyway since if it's normalised then the magnitude *has* to be 1



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Paolo Bonzini
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« Reply #25 on: December 07, 2009, 01:54:05 AM » 

Perhaps someone reading this thread can clear up an issue related to the nontrig calculations. You'll note that when zri is (0,0) and when zjk is (0,0) then I assume the sine/cosine calculation done by raising the normalised complex value to a power should return (0,0) i.e. it leaves the values unchanged. It would be equally valid to return (1,0)  would that be a better option mathematically speaking ?
I think I deduce from that, that the answer is yes I should use (1,0) instead of (0,0) when normalising (0,0)  I guess the answer was in the question anyway since if it's normalised then the magnitude *has* to be 1 I'm not sure. You should probably use an ifthenelse and use totally different formulas when zri and/or zjk are zero, I don't think normalising would work.



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zee
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« Reply #26 on: December 07, 2009, 02:01:00 AM » 

There is an error in my PDF because that equations: http://www.fractalforums.com/3dfractalgeneration/true3dmandlebrottypefractal/msg8680/#msg8680there r is only the length of the (x,y) vector but in Wikipedia and in this thread bugman wrote it's the length of the (x,y,z) vector. I wanted so compare the results by myself but I crashed my program But please tell me what equation is correct? I think I did it without the z^2 and got the same result like with the trigonometric equations...


« Last Edit: December 07, 2009, 02:19:00 AM by zee »

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David Makin


« Reply #27 on: December 07, 2009, 02:04:03 AM » 

Perhaps someone reading this thread can clear up an issue related to the nontrig calculations. You'll note that when zri is (0,0) and when zjk is (0,0) then I assume the sine/cosine calculation done by raising the normalised complex value to a power should return (0,0) i.e. it leaves the values unchanged. It would be equally valid to return (1,0)  would that be a better option mathematically speaking ?
I think I deduce from that, that the answer is yes I should use (1,0) instead of (0,0) when normalising (0,0)  I guess the answer was in the question anyway since if it's normalised then the magnitude *has* to be 1 I'm not sure. You should probably use an ifthenelse and use totally different formulas when zri and/or zjk are zero, I don't think normalising would work. Actually I just noticed that according to your version above it looks like I should have zri/mag(zri) as (1,0) when zri is (0,0)


« Last Edit: December 07, 2009, 02:06:17 AM by David Makin »

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David Makin


« Reply #28 on: December 07, 2009, 02:10:33 AM » 

There is an error in my PDF because that equations: http://www.fractalforums.com/3dfractalgeneration/true3dmandlebrottypefractal/msg8680/#msg8680there r is only the length of the (x,y) vector but in Wikipedia and in this thread bugman wrote it's the length of the (x,y,z) vector. I wanted so compare the results by myself but I crashed my program But please tell me what equation is correct? I think I did it without the z^2 and got the same result like with the trigonometric equations BUT if there is no z^2 you'll get a sigularity at (0,0,1) and I think your simulations say something others... There are 2 "r's" in the link you quoted, one is in the trig version and is sqrt(x^2+y^2+z^2) (that's plain R) and the other is used in the nontrig versions of the formula and that's sqrt(x^2 + y^2) (that's Rxy).



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zee
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« Reply #29 on: December 07, 2009, 02:13:36 AM » 



« Last Edit: December 07, 2009, 02:28:29 AM by zee »

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