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Author Topic: Splitting up the exponent into a triplex  (Read 1778 times)
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cbuchner1
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« on: December 13, 2009, 03:58:47 PM »

The Mandelbulb uses a single exponent that applies to norm of Z and angles phi and theta likewise.

Z^p

Did you guys realize that it is perfectly valid to split this exponent into a triplex and define an
operation

Z^(p_norm, p_phi, p_theta) ?

If you want the Mandelbrot to live in the x,y plane then keep p_norm and p_phi at 2 and choose a different p_theta, e.g. 3 or 4.

The results will be interesting. For animations this gives your parameter space three degrees of freedom.

Christian
« Last Edit: December 13, 2009, 04:12:44 PM by cbuchner1 » Logged
David Makin
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« Reply #1 on: December 13, 2009, 04:16:36 PM »

That's basically the same as my suggestion of adding "self-rotation", many of my 2D complex formulas for UF have it built-in as an extra option.

I haven't tried splitting the power into a triplex on the Mandelbulb yet but I'm guessing the results will be very similar to what happens with "sel-rotation" on a complex fractal:
For a given magnitude power increasing either of the angle powers will result in increased chaos in the fractal and reducing them will reduce the chaos.
The effect of increasing the chaos is especially useful for rendering strange attractors from formulas - the more extra rotation involved the greater the chance of strange attraction occurring. For example adding self-rotation to the Henon gives the usual attractors a higher dimension and increases the areas of strange attraction overall.

I'm still interested in the "correct" z^p for the triplex where p is in the true "triplex" form (I have a method where p is complex though whether it's strictly "correct" or not I'm not sure).
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kram1032
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« Reply #2 on: December 13, 2009, 05:02:22 PM »

Afaik, in a simplified form where the signs might change, to raise a quaternion over a quaternion, one first takes the log of a quaternion a,b,c,d, multiplies the results with a quaternion e,f,g,h and takes the exponential of it...

With an a bit more complex forumla, you'll always get the right results...

Shouldn't triplex^triplex work more or less the same, just taking care of the special restrictions of triplex algebra?
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David Makin
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« Reply #3 on: December 13, 2009, 05:26:15 PM »

Afaik, in a simplified form where the signs might change, to raise a quaternion over a quaternion, one first takes the log of a quaternion a,b,c,d, multiplies the results with a quaternion e,f,g,h and takes the exponential of it...

With an a bit more complex forumla, you'll always get the right results...

Shouldn't triplex^triplex work more or less the same, just taking care of the special restrictions of triplex algebra?

Yes it should, but I'm not so sure about getting log(triplex) or exp(triplex) smiley
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kram1032
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« Reply #4 on: December 13, 2009, 06:12:32 PM »

ah, I see...
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Paolo Bonzini
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« Reply #5 on: December 13, 2009, 07:26:53 PM »

I'm still interested in the "correct" z^p for the triplex where p is in the true "triplex" form (I have a method where p is complex though whether it's strictly "correct" or not I'm not sure).

The correct method is to do q(z)^(p/2) i conj(q(z))^(p/2), see my thing for how q is computed.  I'm pretty sure you'd come out with the same formulas as Paul, since they work.

As to multiplication, I'd just say that multiplication of a triplex by a triplex is "fake", but if you really want to do it it would be q(z)^(1/2) w conj(q(z))^(1/2).
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Paolo Bonzini
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« Reply #6 on: December 13, 2009, 07:30:13 PM »

Z^p

Did you guys realize that it is perfectly valid to split this exponent into a triplex and define an
operation

Z^(p_norm, p_phi, p_theta) ?

No, it's not valid.  You don't have (rho1, phi1, theta1)^(rho2, phi2, theta2) = (rho1^phi1, phi1*phi2, theta1*theta2).  You have to define triplex exponentiation and logarithm, multiply one by the logarithm of the other and then do an exponentiation.

I'm fine with defining formulas inspired by modifications to 4D math (like David did), but if you want to stick with 3D, going to quaternions and back is the only way that makes sense from a mathematical point of view.  Everything else is chemtrails.
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David Makin
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« Reply #7 on: December 13, 2009, 09:09:32 PM »

I'm still interested in the "correct" z^p for the triplex where p is in the true "triplex" form (I have a method where p is complex though whether it's strictly "correct" or not I'm not sure).

The correct method is to do q(z)^(p/2) i conj(q(z))^(p/2), see my thing for how q is computed.  I'm pretty sure you'd come out with the same formulas as Paul, since they work.

As to multiplication, I'd just say that multiplication of a triplex by a triplex is "fake", but if you really want to do it it would be q(z)^(1/2) w conj(q(z))^(1/2).

I wasn't aware Paul had expanded to formulas where the power is either complex or triplex ?
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David Makin
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« Reply #8 on: December 13, 2009, 09:18:18 PM »

Z^p

Did you guys realize that it is perfectly valid to split this exponent into a triplex and define an
operation

Z^(p_norm, p_phi, p_theta) ?

No, it's not valid.  You don't have (rho1, phi1, theta1)^(rho2, phi2, theta2) = (rho1^phi1, phi1*phi2, theta1*theta2).  You have to define triplex exponentiation and logarithm, multiply one by the logarithm of the other and then do an exponentiation.

I'm fine with defining formulas inspired by modifications to 4D math (like David did), but if you want to stick with 3D, going to quaternions and back is the only way that makes sense from a mathematical point of view.  Everything else is chemtrails.

Hi Paolo. I agree it doesn't make sense mathematically speaking i.e. it's not "valid", however from the perspective of creating images then any formula is OK if it produces interesting results and using a multiplier for theta and/or phi that doesn't match the power for the magnitude is perfectly valid for that smiley
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cbuchner1
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« Reply #9 on: December 13, 2009, 09:27:41 PM »

No, it's not valid.

It has been quite an achievement that you have been able to generalize the Mandelbulb formula by expressing it with quaternion mathematics.  I am looking forward to reading your completed article.

The suggested modification to the Zˆp operator probably will not fit into any quaternion concept at all. But I am not even claiming this operator means "raising Z to the power of p" anymore. It's just an arbitrarily defined operator to be applied to Z. Whether mathematically valid or not - I don't care much.

But I do care about rendering interesting looking fractals - and the result of having a triplex of exponents does fact in give some new and interesting shapes (I tested it today in a GPU based raymarcher). The DE method still converges on it, so it produces images that may be interesting. Snowflake-Mandelbrots, anyone?

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Paolo Bonzini
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« Reply #10 on: December 13, 2009, 09:43:44 PM »

I'm still interested in the "correct" z^p for the triplex where p is in the true "triplex" form (I have a method where p is complex though whether it's strictly "correct" or not I'm not sure).

The correct method is to do q(z)^(p/2) i conj(q(z))^(p/2), see my thing for how q is computed.  I'm pretty sure you'd come out with the same formulas as Paul, since they work.

I wasn't aware Paul had expanded to formulas where the power is either complex or triplex ?

Doh, sorry I read where z is the true "triplex" form (probably I inferred from the fact that you write your triplex numbers as complex+real, dunno).  Read below.

It has been quite an achievement that you have been able to generalize the Mandelbulb formula by expressing it with quaternion mathematics.  I am looking forward to reading your completed article.
It's been finished since yesterday morning. :-)  Thanks!

The suggested modification to the Zˆp operator probably will not fit into any quaternion concept at all.

Oh, actually it does.  If you want to rotate theta and phi by different angles, the quaternion formulation just says "make Q(z) a bit more weird then usual, that's fine".  That's different from defining exponentiation though, it's just a different definition of the rotation (which is what it is).

Otherwise, you can raise a quaternion to a quaternion, so you can represent the exponent triplex as xi+yj+zk and do Q(z)^(p/2) i conj(Q(z))^(p/2).  I honestly have no idea what to expect, that would be equivalent to a z=z^n+c 2D mandelbrot with complex n.

What I am wary about is still talking about triplex numbers as algebras, because they sound awfully pseudomathematic to me.  Beware, it's not a kind of Not-Invented-Here syndrome.  If you're talking about expressing the iteration on R^3 -> R^3 that's fine with me; if you have a z^2 formula on triplex that gives a cool picture, that's great and I'll gladly help converting it to Q(v) if possible.  However, if you want to define an algebra and seek what are the properties of addition and multiplication, I don't trust it very much anymore.  It's not needed anymore.

I hope what I meant is clear.  (I was in a hurry before).
« Last Edit: December 13, 2009, 10:35:29 PM by Paolo Bonzini » Logged
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