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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...

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

Nice, thank you for this link