Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Z Values in the Mandelbrot Set

Date: 04/27/2006 at 22:16:40
From: Louise
Subject: Mandelbrot set

Could you please explain why for the Mandelbrot set the modulus for 
the resulting z value must remain less than 2?  Why is 2 crucial?



Date: 04/29/2006 at 11:28:39
From: Doctor Vogler
Subject: Re: Mandelbrot set

Hi Louise,

Thanks for writing to Dr. Math.  That's a good question.  You'll 
recall that the Mandelbrot set is defined to be those complex numbers 
c such that if you start with z = 0 and then change z to

  z^2 + c

and repeat, then all of the values that you get will remain small 
(will stay in a bounded set).  That is, the absolute values won't grow
to infinity.  So we want to prove that if the absolute value of c is
bigger than 2, then these iterates will ALWAYS grow to infinity.

Suppose that |c| > 2.  Then the first iterations are

  z(0) = 0
  z(1) = c
  z(2) = c^2 + c.

Consider that if |z| >= |c| > 2, then the triangle inequality gives

  |z^2 + c| + |-c| >= |z^2|

and therefore

  |z^2 + c| >= |z|^2 - |c|  (since |-c| = |c| and |z^2| = |z|^2)
            >= |z|^2 - |z|  (since -|c| >= -|z|)
            =  |z|(|z| - 1) (distribution law)
            >= |z|(|c| - 1) (since |z| >= |c|)

so that if we pick some (real) number k between 1 and |c| - 1 (which
is bigger than 1), then that means that

  |z^2 + c| >= k|z|,

and therefore we find that the absolute value of the new iteration is
at least k times the absolute value of the last iteration.  Well, we
already have

  |z(1)| = |c| > 2

after only one iteration, which means that after n iterations, we will
have

  |z(n)| >= (k^n)|c|

and since k > 1, the right side grows to infinity, which means that
the left side grows to infinity.  Therefore, the number c does not
belong to the Mandelbrot set.

Actually, all we really needed in this proof was |z| >= |c| and
|z| > 2 (since we can pick k to be between 1 and |z| - 1).  That means
that if |z| >= |c| and |z| > 2, then n iterations later, we will have

  |z| >= (k^n)|c|.

This is why we can stop as soon as we get |z| > 2; when |z| > 2, we
know that the iterates will keep growing to infinity.  Of course, 
that's not the *only* time they go to infinity, because they go to
infinity when c is any point outside of the Mandelbrot set.  But it is
worth noting that there is one point c with |c| = 2 that is contained
in the Mandelbrot set.  Since you have to get equality in all of the
inequalities above, there is only one point that does it, and that
point is c = -2.  That's why we can't use a number smaller than 2.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to 
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Imaginary/Complex Numbers
High School Fractals
High School Imaginary/Complex Numbers
High School Sets

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/