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Author Topic: Hidden Mandelbrot  (Read 1404 times)
Description: Why does it appear ???
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Moresque
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Posts: 5


« on: May 02, 2013, 11:32:02 PM »

Hi all,

I just discovered a new formula to make appear a Mandelbrot, and I don't success in explaining why...

--
Z0 = c

Zn+1 = Zn+c▓*(Zn)^(-1)-1

--

And for these using Xaos : z+(c^2)*(z^(-1))-1     -> It will automatically change the initialization point in c.

Any idea ?
« Last Edit: May 03, 2013, 10:40:15 PM by Moresque » Logged
Tglad
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« Reply #1 on: May 03, 2013, 04:28:54 AM »

Because the mandelbrot set is universal.
(https://www.google.com.au/search?q=the+mandelbrot+set+is+universal)

My interpretation is that the gradient of the mandelbrot set transform is linear, so any formula that you come up with will have small sections where the transform gradient is approximately linear, so in those sections it looks like the mandelbrot set. It is a bit like the fact that any smooth curve looks like a plane if you zoom in close enough.
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Moresque
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Posts: 5


« Reply #2 on: May 03, 2013, 01:25:13 PM »

I see what you mean, and I know well the universality of the Mandelbrot set (the combination of a big panel of Julia sets), but how do you explain that with very similar formula (for exemple z▓ + c -1), there is no Mandelbrot ?

I tried a lot of similar fractals, and the formula I quote is the only one wich gives a perfect Mandelbrot set...
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lkmitch
Fractal Lover
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Posts: 238



« Reply #3 on: May 03, 2013, 06:02:02 PM »

Hi all,

I just discovered a new formula to make appear a Mandelbrot, and I don't success in explaining why...

--
Z0 = 0

Zn+1 = Zn+c▓*(Zn)^(-1)-1

--

And for these using Xaos : z+(c^2)*(z^(-1))-1

Any idea ?

I don't understand how this formula works.  If you begin with z = 0, then the first iteration should give you a divide-by-zero error with the z^(-1).

Other formulas will give perfect Mandelbrot sets, if you start with z = a critical point.  z = 0 is used with the standard Mandelbrot set because 0 is the critical point of f(z) = z^2 + c.  If you change the function, like f(z) = z^2 + z + c, then you'll need a different starting point.
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Alef
Fractal Supremo
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Posts: 1174



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« Reply #4 on: May 03, 2013, 06:26:39 PM »

Probably started with a Z(0) = pixel. And fractal software could have no error message when there is division by 0.
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fractal catalisator
Moresque
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Posts: 5


« Reply #5 on: May 03, 2013, 10:26:37 PM »

I don't understand how this formula works.  If you begin with z = 0, then the first iteration should give you a divide-by-zero error with the z^(-1).

Other formulas will give perfect Mandelbrot sets, if you start with z = a critical point.  z = 0 is used with the standard Mandelbrot set because 0 is the critical point of f(z) = z^2 + c.  If you change the function, like f(z) = z^2 + z + c, then you'll need a different starting point.

You're right... I was thinking that XaoS was able to know the limit of the sequence even with a divide-by-zero, but when I typed 0 in user initialization, the fractal was different (but it does exist).

So I think that the software automatically fixes a critical point with the formula, but I can't find it...
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Moresque
Forums Newbie
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Posts: 5


« Reply #6 on: May 03, 2013, 10:39:13 PM »

I found it...

The starting point is c...

So I changed the formula in first post.
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