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Author Topic: Attempt at Unitary Mandelbrot Set  (Read 1868 times)
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kram1032
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« Reply #30 on: May 15, 2013, 09:33:18 PM »

Ok, so w,x,y and z never change.
What changes is \phi
You already tried out the first method and a variation of it.

What's left is:
\phi \to arg(e^{\phi+\psi})

or


\phi \to arg(r e^{\phi} + s e^{\psi})

Where \phi is the phase of z and \psi the phase of c.

And yeah, it's not all that unlikely that all of them will basically look like revolution surfaces. After those initial experiments, it might be a nice idea to just randomly mess around with signs or factors, to get slightly different formulae that might be able to produce far more interesting dynamics and behaviours. Just try to apply such an additional phase factor to other kinds of 3D M-Sets that already are known to produce interesting things. Maybe, some new and interesting things emerge.
Also to be tried are things like higher powers or other functions. I already provided generic multiplication and division, and except for the phase factor, addition is well defined too.
« Last Edit: May 15, 2013, 09:37:48 PM by kram1032 » Logged
Alef
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« Reply #31 on: May 16, 2013, 03:19:25 PM »

Ok, so w,x,y and z never change.
What changes is <Quoted Image Removed>
You already tried out the first method and a variation of it.

What's left is:
<Quoted Image Removed>

or


<Quoted Image Removed>

Where <Quoted Image Removed> is the phase of z and <Quoted Image Removed> the phase of c.


OK. I will try.
Then this is:  huh?


Ф (0) = some number.
Ksi (n) = Ф (n-1)
Ф (n+1) = arg(e^(Ф (n)+Ksi(n) )  )

Am I right, Ksi is the Ф of previous iteartion? Or they are unrelated whatever picked numbers?

But then arg is deffined as angle of complex argument. But if Ф and Ksi are real numbers, then e^(Ф+ Ksi) is real number and so there is no complex angle. Ahh, forgot what was complex powers, but complex powers alsou resulted in real value.

Then I understand Ф (n+1) = arg ( e^Ф (n) + i*e^Ksi(n) )  huh?

Preferably could you explain it in plain text becouse in my phone latex aren't visible.


But I think revolution should be becouse of formula quadratic polynomial part. I think, changing quadratic parts could result in not revolution.
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kram1032
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« Reply #32 on: May 16, 2013, 10:29:47 PM »

I should have been more careful with consistency, but I'll roll with this now:

Z was made up of w,x,y,z,\phi
C was made up of a,b,c,d,\psi

So the \psi here isn't related at all to the previous iteration of \phi. It's a constant phase, just like a,b,c and d are constant components of C.
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jehovajah
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« Reply #33 on: May 17, 2013, 09:50:28 AM »

Hooray for Alef!

I do not know where we would be if Alef did not take an interest in visualizing these weird concepts. Thanks man!

These equations are Grassmann Forms, but very few would know that.
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Alef
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« Reply #34 on: May 17, 2013, 01:01:22 PM »

Hooray for Alef!

I do not know where we would be if Alef did not take an interest in visualizing these weird concepts. Thanks man!

These equations are Grassmann Forms, but very few would know that.
Thanks  cheesy

I should have been more careful with consistency, but I'll roll with this now:

Z was made up of w,x,y,z,<Quoted Image Removed>
C was made up of a,b,c,d,<Quoted Image Removed>

So the <Quoted Image Removed> here isn't related at all to the previous iteration of <Quoted Image Removed>. It's a constant phase, just like a,b,c and d are constant components of C.

Well. But if Ksi and Fi are real numbers, how equation -> =e^(Ф (n)+Ksi(n) ) would produce complex number. arg or atan2 calculates complex angle, but without complex number there are no complex angle to calculate. Shouldn't it be Ф (n+1) = arg(e^(Ф (n)+i*Ksi(n) )  ) huh? Or does Ksi are imaginary number?
« Last Edit: May 17, 2013, 01:06:12 PM by Alef » Logged

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kram1032
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Posts: 1863


« Reply #35 on: May 17, 2013, 04:59:29 PM »

Oh, I'm sorry, I forgot an i in those exponential equations


\phi \to arg(e^{i\left(\phi+ \psi\right)})

or


\phi \to arg(r e^{i \phi} + s e^{i \psi})

Now it should make more sense, I hope.
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