gclark@utcsri.UUCP (Graeme Clark) (03/20/86)
[Balanced meal for line eater goes here.]
As you may know, the Mandelbrot set is the set of complex numbers c
such that the sequence
2 3
0, f(0), f (0), f (0), ...
k
is not bounded, where f(z) = z^2+c, and f (z) = f(f(f(...f(z)...)))
k
(k applications of f). It is claimed that if |f (0)| > 2 for some
k, then the seqence blows up (in absolute value) to infinity. Can
someone show a proof for this?
Graeme Clark -- Dept. of Computer Science, Univ. of Toronto, Canada M5S 1A4
{allegra,cornell,decvax,ihnp4,linus,utzoo}!utcsri!gclarklarsen@fisher.UUCP (Michael Larsen) (03/28/86)
> [Balanced meal for line eater goes here.] > > As you may know, the Mandelbrot set is the set of complex numbers c > such that the sequence > > 2 3 > 0, f(0), f (0), f (0), ... > k > is not bounded, where f(z) = z^2+c, and f (z) = f(f(f(...f(z)...))) > k > (k applications of f). It is claimed that if |f (0)| > 2 for some > k, then the seqence blows up (in absolute value) to infinity. Can > someone show a proof for this? > > Graeme Clark -- Dept. of Computer Science, Univ. of Toronto, Canada M5S 1A4 > {allegra,cornell,decvax,ihnp4,linus,utzoo}!utcsri!gclark Let x(0) = 0, x(k + 1) = x(k) ^ 2 - a. Denote the absolute value function abs. Then Theorem: If abs(x(n)) > 2 for some n, then lim abs(x(k)) = inf. Lemma: If x(k) > (1 + sqrt(1 + 4*abs(a))) / 2, then lim abs(x(k)) = inf. Proof: The above inequality implies x(k) > e + (1 + sqrt(1 + 4*abs(a))) / 2 for some e > 0. By the triangle inequality, abs(x(k + 1)) = abs(x(k) ^ 2 - a) >= abs(x(k)) ^ 2 - abs(a) > abs(x(k)) + e*sqrt(1 + 4*abs(a)). By induction, abs(x(k + N)) > abs(x(k)) + N*e. The lemma follows. Proof: If abs(x(n)) = b > 2 for some n, then max(b, abs(x(1))) >= (2*b + abs(x(1))) / 3 > (4 + abs(a)) / 3 >= (1 + sqrt(1 + 4*abs(a))) / 2. The theorem follows immediately from the lemma.