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Author Topic: Triplex Algebra  (Read 26172 times)
Description: Triplex Algebra
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kram1032
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« Reply #90 on: December 30, 2009, 11:36:31 PM »

BradC:
Which sum did you use for ln?

I'm not sure but Wolfram Alpha gives me several sums where only one seems to be fully defined and that one is pretty complex.... - a triplex one might actually require an even further extension of the definition and thus a whole new sum...

Also, what exactly does mathematica return if you throw exp(ln(a,b,c)) on it and simplify the results?
Or the other way round, ln(exp(a,b,c))?
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BradC
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« Reply #91 on: January 01, 2010, 09:45:46 PM »

BradC:
Which sum did you use for ln?

I used the series expansion at z=1 since it seems like a pretty simple one. See the second paragraph in http://www.fractalforums.com/theory/triplex-algebra/msg9962/#msg9962 for details.

Also, what exactly does mathematica return if you throw exp(ln(a,b,c)) on it and simplify the results?
Or the other way round, ln(exp(a,b,c))?

For both exp(ln(z)) and ln(exp(z)), it simplifies to something that's still really complicated - a few pages long when written out.
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kram1032
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« Reply #92 on: January 04, 2010, 03:29:55 AM »

outch... lol

I'm not sure for mathematica as I don't have it but in case of wolfram alpha I noticed that for very complex results, I can simplify them quite some bits by hand (if you're willing to go through pages of maths-code and search for such spots lol) and then, if I put that hand-simplified code back into wolfram alpha, it magically simplifies that even furhter although it didn't find a simpler solution before....

log definitely is quite complex if you go out of the bounds of R+ so you might need to take care to use the correct sum (if you're unlucky, even find a totally new one) to correctly define the log.... smiley

look here (scroll down, click more)
http://www.wolframalpha.com/input/?i=series+ln%28x%2By*i%29
you'll see that only two series expansions of log are defined for the whole complex plane directly. - and both need the log function themselves...
maybe there are others, wolfram alpha isn't aware of... or the definitions aren't as strict as it claims them to be...
but I'm afraid not  cry
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BradC
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« Reply #93 on: January 05, 2010, 11:22:07 PM »

\cos (x,y,z)=\left(\sin (x) \sinh (y) \sinh (t x) \sin (t y)+\cos (x) \cosh (y) \cosh (t x) \cos (t y),\cos (x) \cosh (y) \sinh (t x) \sin (t y)-\sin (x) \sinh (y) \cosh (t x) \cos (t y),-\sinh (z) \sin \left(\sqrt{x^2+y^2}\right)\right) where t=\frac{z}{\sqrt{x^2+y^2}}
\sin (x,y,z)=\left(\sin (x) \cosh (y) \cosh (t x) \cos (t y)-\cos (x) \sinh (y) \sinh (t x) \sin (t y),\cos (x) \sinh (y) \cosh (t x) \cos (t y)+\sin (x) \cosh (y) \sinh (t x) \sin (t y),\sinh (z) \cos \left(\sqrt{x^2+y^2}\right)\right) where t=\frac{z}{\sqrt{x^2+y^2}}

I derived these formulas using Mathematica, and tested them with 1000 random points (-4<x<4,-4<y<4,-4<z<4) by comparing them with the series truncated after 32 terms. Agreement was always at least ~12 digits.

When z=0, these formulas match those for complex sin and cos: \cos (x,y,0)=(\cos (x) \cosh (y),-\sin (x) \sinh (y),0) and \sin (x,y,0)=(\sin (x) \cosh (y),\cos (x) \sinh (y),0).
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kram1032
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« Reply #94 on: January 05, 2010, 11:43:50 PM »

nice, so now we can see a kind of triplex sine and/or cosine bulb cheesy
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Timeroot
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« Reply #95 on: January 19, 2010, 04:32:13 AM »

How about when you square them and add? Do they equal 1? That would seem like an important property to me.

I was thinking that maybe the Triplex sine and cosine could be defined using some patterns in other functions. First off,
Cosh(x)=\frac{e^x + e^{-x}}{2} and Cos(z)=\frac{e^{i z} + e^{-i z}}{2}, so maybe we could say Cost(q)=\frac{e^{q*(0,0,1)} + e^{q*(0,0,-1)}}{2}. This wouldn't exactly give us a cosine for triplex numbers, but it might give us new function for any real (complex? triplex?) number. It may act identically to the regular cosine, but if so, it would give us clues to how the exponential function works. Once we know that, we could
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« Reply #96 on: January 20, 2010, 03:09:23 AM »

Okay, I was thinking some more, and I was just wondering what the agreed upon defintion(s) or exponentiation. This thread is tooo looong and messed up. As far as I can tell, if A and B are two three-dimensional numbers, then A^B=Result, where Result.Radius=A.Radius^B.X, Result.Theta=A.Theta*B.X + B.Theta, and Result.Phi=A.Theta*B.X + B.Theta. From what I hear, multiplication is commutative and associative, and is done by multiplying radii and adding angles. Addition is commutative and associative, and you add x, y, and z. In that case, cos should be defined just fine by using the regular imaginary-exponent formula above.

I'm wondering how feasible it is do define a derivative on the triplex volume. If one can, then that could give an alternative (or possibly confirmative) definition of the exponential function / trigonometric functions. It could even be used to define something like the Airy/Bessel functions... not that they'd be used very often in fractal formulas.
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David Makin
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« Reply #97 on: January 20, 2010, 02:16:10 PM »

Okay, I was thinking some more, and I was just wondering what the agreed upon defintion(s) or exponentiation. This thread is tooo looong and messed up. As far as I can tell, if A and B are two three-dimensional numbers, then A^B=Result, where Result.Radius=A.Radius^B.X, Result.Theta=A.Theta*B.X + B.Theta, and Result.Phi=A.Theta*B.X + B.Theta. From what I hear, multiplication is commutative and associative, and is done by multiplying radii and adding angles. Addition is commutative and associative, and you add x, y, and z. In that case, cos should be defined just fine by using the regular imaginary-exponent formula above.

I'm wondering how feasible it is do define a derivative on the triplex volume. If one can, then that could give an alternative (or possibly confirmative) definition of the exponential function / trigonometric functions. It could even be used to define something like the Airy/Bessel functions... not that they'd be used very often in fractal formulas.

Ermm - I think you'll find that multiplication is commutative but neither associative nor distributive.
Also the multiply and divide are not inverses of each other and apparently neither are log() and exp().
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Paolo Bonzini
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« Reply #98 on: January 20, 2010, 03:44:40 PM »

Sorry for pointing to my paper again, but Timeroot is somewhat correct about the multiplication.

Addition on cartesian coordinates works great, indeed you can only express it on cartesian coordinates (maybe there is a closed form on quaternions, but I haven't looked for it yet).

Multiplication on cartesian coordinates is not associative or distributive *because first and foremost it makes no sense*.  Multiplication on spherical coordinates (i.e. multiply radius, sum angles) or triplex-expressed-as-quaternion is commutative and associative, not sure about distributive.

Exponentiation on cartesian coordinates gives the correct values only when you compute the formula in advance, because in that case you are doing a "hidden" conversion (to quaternions or spherical coordinates) and back.

So, for Mandelbulbs you can take the shortcut, but otherwise you need to be explicit about your domain being a point (elevation in -pi/2...pi/2) or a rotation (elevation in -pi...pi).
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David Makin
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« Reply #99 on: January 20, 2010, 04:13:38 PM »

Multiplication on cartesian coordinates is not associative or distributive *because first and foremost it makes no sense*.  Multiplication on spherical coordinates (i.e. multiply radius, sum angles) or triplex-expressed-as-quaternion is commutative and associative, not sure about distributive.

I'm afraid you've lost me there.

For the triplex maths as applied in the Mandelbulb, the following are *not* true for all values:

a*(b*c) = a*b*c = (a*b)*c

a*(b+c) = a*b + a*c

a*(b/a) = b

exp(ln(a)) = a

If for your quaternion form any of these *do* hold true then it's not the same mathematical form as that used for the original Mandelbulb - I'm not saying it can't produce similar fractals though.

Note that I think however that the above equalities obviously do hold true if in a, b and c the "j" component is zero i.e. we're reduced to complex numbers.
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David Makin
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« Reply #100 on: January 20, 2010, 04:34:48 PM »

Multiplication on cartesian coordinates is not associative or distributive *because first and foremost it makes no sense*.  Multiplication on spherical coordinates (i.e. multiply radius, sum angles) or triplex-expressed-as-quaternion is commutative and associative, not sure about distributive.

I'm afraid you've lost me there.

For the triplex maths as applied in the Mandelbulb, the following are *not* true for all values:

a*(b*c) = a*b*c = (a*b)*c

a*(b+c) = a*b + a*c

a*(b/a) = b

exp(ln(a)) = a

If for your quaternion form any of these *do* hold true then it's not the same mathematical form as that used for the original Mandelbulb - I'm not saying it can't produce similar fractals though.

Note that I think however that the above equalities obviously do hold true if in a, b and c the "j" component is zero i.e. we're reduced to complex numbers.


I should add that even:

(a^2)*a = a^3

does not hold true for all values.
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David Makin
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« Reply #101 on: January 21, 2010, 12:06:26 AM »

Multiplication on cartesian coordinates is not associative or distributive *because first and foremost it makes no sense*.  Multiplication on spherical coordinates (i.e. multiply radius, sum angles) or triplex-expressed-as-quaternion is commutative and associative, not sure about distributive.

I'm afraid you've lost me there.

For the triplex maths as applied in the Mandelbulb, the following are *not* true for all values:

a*(b*c) = a*b*c = (a*b)*c

a*(b+c) = a*b + a*c

a*(b/a) = b

exp(ln(a)) = a

If for your quaternion form any of these *do* hold true then it's not the same mathematical form as that used for the original Mandelbulb - I'm not saying it can't produce similar fractals though.

Note that I think however that the above equalities obviously do hold true if in a, b and c the "j" component is zero i.e. we're reduced to complex numbers.


I should add that even:

(a^2)*a = a^3

does not hold true for all values.


Just been thinking about it and realised that the Paolo is correct with respect to treating the triplex in polar terms when dealing solely with multiplication/division since if you convert all the terms to polar first and then perform the calculations then associativity works and a*(b/a)=a *but* as soon as you need to include addition/subtraction then things break down because however you do the calculations (in 3 dimensions only) the distributive law does not apply.

I'm just wondering if the derivations done in this thread for the higher functions using Mathematica etc. were done in such a way that Mathematica took into account that the distributive law does not apply ?
Or were they all done using Paolo's quaternion form ?

An obvious problem is something as basic as the derivation of the derivative of y = x^2 where:

y+dy = (x+dx)^2
y + dy = x^2 + 2*x*dx + dx^2
x^2 + dy = x^2 +2*x*dx + dx^2
dy = 2*x*dx + dx^2
dy/dx = 2*x + dx
So as dx->0:
Dy/Dx = 2*x

The problem being that for the triplex you cannot expand (x+dx)^2 as x^2 + 2*x*dx + dx^2 for all x/dx.

This may be part of the reason why the analytical DE seems to break down in certain areas.
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Timeroot
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« Reply #102 on: January 21, 2010, 02:03:31 AM »

Okay, so it *is* associative, since you're keeping it in spherical coordinates and only using addition/multiplication. Also, the division/multiplication inverse *does* work, agreed? I can understand, though, why the distributive property doesn't work. I'm curious as to how exactly exp(a) is defined - you could use taylor series, or the exponentiation formula I suggested with A=2.71..., or exp(x)=e^x * (cos x,sin y,0) * (cos z, 0, sin z) [or WHATEVER that was on the first page].... it seems there is an agreed upon definition for Mandelbulb purposes, can someone please tell me what it is?

With regard to derivatives, I think it would be interesting how exactly (x+dx)^2 can be expanded [that is, if it can be].

EDIT:
 Forgive me if I'm being a pain with having all this explained, but why doesn't (z*z)*z=z^3? I'd think that z^3=Result, where Result.Radius=z.Radius^3, Result.Theta=z.Theta*3, and Result.Phi=z.Phi*3. z*z=ResultA, where ResultA.Radius=z.Radius*z.Radius, ResultA.Theta=z.Theta + z.Theta, and and ResultA.Phi=z.Phi + z.Phi. Then (z*z)*z=ResultA*z=Result B, where ResultB.Radius=z.Radius * ResultA.Radius = (substituting here) z.Radius * z.Radius * z.Radius = z.Radius^3. ResultB.Theta=z.Theta + ResultA.Theta = z.Theta + z.Theta + z.Theta = z.Theta*3. Same goes for Result.Phi

Clearly, z^3=Result should equal (z*z)*z=ResultB. Can somebody explain to me why it is not?
« Last Edit: January 21, 2010, 02:15:10 AM by Timeroot » Logged

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David Makin
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« Reply #103 on: January 21, 2010, 02:59:14 AM »

Okay, so it *is* associative, since you're keeping it in spherical coordinates and only using addition/multiplication. Also, the division/multiplication inverse *does* work, agreed? I can understand, though, why the distributive property doesn't work. I'm curious as to how exactly exp(a) is defined - you could use taylor series, or the exponentiation formula I suggested with A=2.71..., or exp(x)=e^x * (cos x,sin y,0) * (cos z, 0, sin z) [or WHATEVER that was on the first page].... it seems there is an agreed upon definition for Mandelbulb purposes, can someone please tell me what it is?

With regard to derivatives, I think it would be interesting how exactly (x+dx)^2 can be expanded [that is, if it can be].

EDIT:
 Forgive me if I'm being a pain with having all this explained, but why doesn't (z*z)*z=z^3? I'd think that z^3=Result, where Result.Radius=z.Radius^3, Result.Theta=z.Theta*3, and Result.Phi=z.Phi*3. z*z=ResultA, where ResultA.Radius=z.Radius*z.Radius, ResultA.Theta=z.Theta + z.Theta, and and ResultA.Phi=z.Phi + z.Phi. Then (z*z)*z=ResultA*z=Result B, where ResultB.Radius=z.Radius * ResultA.Radius = (substituting here) z.Radius * z.Radius * z.Radius = z.Radius^3. ResultB.Theta=z.Theta + ResultA.Theta = z.Theta + z.Theta + z.Theta = z.Theta*3. Same goes for Result.Phi

Clearly, z^3=Result should equal (z*z)*z=ResultB. Can somebody explain to me why it is not?

I guess it's my fault for not following Paolo's comments well enough - I was describing the issue if you calculate z^2 getting the result in cartesian form then back to polar to multiply by the original z giving z^2*z in cartesian form at the end or using the fully cartesian calculation to get z^2 and then say polar and back for z^2*z.
These do not result in the same value as doing the calculation of z^3 in polar form and only converting to cartesian at the end (at least not for all z).
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Paolo Bonzini
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« Reply #104 on: January 21, 2010, 10:18:40 AM »

Okay, so it *is* associative, since you're keeping it in spherical coordinates and only using addition/multiplication.
Only using multiplication/division, actually.  Addition requires cartesian coordinates, so it is associative but not distributive.

Also, the division/multiplication inverse *does* work, agreed? I can understand, though, why the distributive property doesn't work. I'm curious as to how exactly exp(a) is defined - you could use taylor series, or the exponentiation formula I suggested with A=2.71..., or exp(x)=e^x * (cos x,sin y,0) * (cos z, 0, sin z) [or WHATEVER that was on the first page].... it seems there is an agreed upon definition for Mandelbulb purposes, can someone please tell me what it is?
I have no idea.  The whatever was on the first page probably does not agree with the Taylor series, which is a problem.  Maybe quaternions could help, I don't have much time now.

Forgive me if I'm being a pain with having all this explained, but why doesn't (z*z)*z=z^3?
When you do z*z*z*z*z in cartesian form, every multiplication includes an implicit conversion to and from spherical coordinates.  It's a bit hard to visualize, but the conversion loses information because of the different range of the elevation vs. the y-axis rotation term.  Instead, you have to use specially crafted formulas for each exponent(*) that include exactly one conversion to spherical coordinates and one from, which is what is called z^n.

(*)I had a generic exponentiation formula expressed using Chebyshev polynomials, but I haven't looked at it for a while and I'm not sure I had no calculation errors since I did it on paper.
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