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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

753.0. "bounded but nonperiodic Mandelbrot Set members?" by PSW::WINALSKI (Paul S. Winalski) Thu Aug 20 1987 14:11

Consider the function:

                 2
	F (Z) = Z  + C
         C       

for complex Z and C.  Now consider the infinite series:

                                      
	F (0), F (F (0)), F (F (F (0))), ...
         C      C  C       C  C  C

The Mandelbrot Set is the set of C for which this series remains bounded.

There are some values of C for which the series is periodic, that is, there
exist m and n such that:

         m       n
	F (0) = F (0)
         C       C

C = 2 is an example.  My question is, are there C for which the series is
bounded but not periodic?

--PSW
T.RTitleUserPersonal
Name
DateLines
753.1SureCHOVAX::YOUNGBack from the Shadows Again,Thu Aug 20 1987 20:011
    Any N where ( -.25 > N > 0 )
753.2CLT::GILBERTBuilderThu Aug 20 1987 21:258
        m       n
    If F (0) = F (0), for any m <> n, then C must be algebraic.
        C       C

    If you can find a trancendental C such that the F series remains
    bounded, then the F series cannot be periodic.  And such C are easy
    to find (based on some common knowledge of the set -- any line segment
    wholly in the set provides a multitude of such C).
753.3PSW::WINALSKIPaul S. WinalskiThu Aug 20 1987 21:386
RE: .0

Oops.  I meant to say that C = -2 is an example where the series is bounded and
periodic.  C = 2 is not even bounded.

--PSW
753.4BEING::POSTPISCHILAlways mount a scratch monkey.Fri Aug 21 1987 14:4314
    Re .1:
    
    > Any N where ( -.25 > N > 0 )
    
    There aren't many positive numbers less than -.25.  :-)
    
    Why stop there, though?  What is wrong with rational c's such that -1 <
    c < 0?  They are not periodic because the number of digits after the
    radix point (c can be represented with a finite number of digits in
    some radix, since it is rational) keeps increasing, and they are
    bounded because -1 < z^2+c < 1 as long as -1 < z < 1.
    
    
    				-- edp 
753.5ooopsCHOVAX::YOUNGBack from the Shadows Again,Fri Aug 21 1987 16:394
    Re.4:
    
    You are right on both counts.  I bollixed up a sign when I was
    desk checking it.
753.6ENGINE::ROTHMon Aug 24 1987 14:2420
    All the periodic points will lie on the boundry of the Mandelbrot
    set.  Points exterior to the set diverge, and points interior converge
    to a fixed point inside the set.

    I've forgotten transformation to use, but you can map the cardioid
    shaped main boundry onto a pair of circles - then points at rational
    angular points around the circles will be periodic.  But so will
    points at rational points around the sprouts attached to the main
    circle, and so on.  Points at irrational angles will exhibit ergodic
    motion on the boundry.  I forget the expression, but you can also
    use number theory to get the points where the filaments fork, and
    also points whose angles are transcendental (Liouville numbers) have
    some interesting properties.

    If you read French, I can send you Hubbard and Douady's paper where
    these results were explained and proved.  I have not looked at this
    subject in a long time, but understood it fairly well when I was
    hacking with it...

    - Jim
753.7ENGINE::ROTHMon Aug 24 1987 14:278
753.8AITG::DERAMODaniel V. {AITG,ZFC}:: D'EramoSun Aug 13 1989 12:5614
        What exactly are you saying in .6 and .7?
        
        Is there a countable set {a0, a1, a2, ...} such that for
        any interior point c, the sequence 0, c, c^2 + c, ...,
        converges to one of the an?
        
        Or, for any interior point c, the set {0, c, c^2 + c, ...}
        has countably many limit points?
        
        Or yet another interpretation?
        
        Makes me wish that I read French.
        
        Dan