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Author Topic: Question about Mandelbrot orbits..  (Read 2107 times)
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Posts: 24


« on: September 30, 2010, 07:10:08 AM »

Something hasn't really been sitting right with me lately..

I've been looking at visualized Mandelbrot orbits, and noticed that some of the orbit points for starting points that are within the set appear to be outside of the set.  For instance, you can see the orbits with this applet: http://math.hws.edu/xJava/MB/.  You have to go to Tools, then select "Mouse Shows Orbit".  From there, if you hold the left mouse button down on any area on the Mandelbrot set, it will show the orbit of iterations. 

Now, if you go to the first main circle to the left of the main cardiod, and then to the very next circle to the left of that, you will notice that part of the orbit goes to the right of the set which normally we think of as not being in the set at all.  If you move your cursor around in this circle you will notice that this orbit point reflects the position in the circle, it is as if there is an invisible circle there which encloses all of the possible orbit points from the circle where the cursor is. 

My question is, if these points are part of the orbit of points that are within the Mandelbrot set, then why don't they show up as part of the set?  Wouldn't starting an iteration with one of these points result in the exact same orbit pattern produced by the second circle to the left of the main cardiod?  I feel like I'm missing something here...
Posts: 24


« Reply #1 on: September 30, 2010, 07:28:37 AM »

Uhh, I think I found the answer to my own question.   head banging wall

I seem to have forgotten that we keep adding the original starting point (c) to every iteration.  This would explain why starting at one of the orbit points for one orbit doesn't produce the same orbit pattern.
Fractal Senior
Posts: 7044

formerly known as 'Trifox'

« Reply #2 on: September 30, 2010, 10:31:43 AM »

the bailout check ( if the point moves to infinity ) is at a certain radius ( 2 ), so any point within this radius can be inside or outside the set,
it does not matters if it is "outside the set", because everything with a contracting behaviour ( |z|<1 as an example ) will stay within that
circle, and values greater than the circle ( e.g. 3 ) will expand soon to infinity, because of the large grow it takes when iterating z^2, adding
the constant factor then *could* move it back to a radius <2, but this will most likely NOT happen

hello and welcome to the forums wink


divide and conquer - iterate and rule - chaos is No random!
David Makin
Global Moderator
Fractal Senior
Posts: 2286

Makin' Magic Fractals
« Reply #3 on: September 30, 2010, 03:07:24 PM »

Just to add to what Trifox said, for the standard Mandelbrot Set (z^2+c) (and for all standard z^2+c Julias I think?) it's actually been proved that if at any time in an orbit the magnitude of z exceeds 2 then the original point for that orbit is definitely "outside" i.e. the attractor for the point is infinity.
Of course this threshold value will change depending on the fractal formula being iterated.
Also for visualisation and colouring purposes it's generally best to use a larger bailout value because many outside colouring methods (in particular "smooth" ones) rely on the behaviour of the orbits at higher magnitudes - for instance the idea that as z gets large then z^2+c is approximately equal to z^2.
I wrote a bit more on orbits and attractors here:

The meaning and purpose of life is to give life purpose and meaning.

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Fractal Molossus
Posts: 795

« Reply #4 on: September 30, 2010, 03:54:49 PM »

Nice, thank you for this link smiley

often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
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