Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2(pesnta.1.3) 9/5/84; site epicen.UUCP
Path: utzoo!watmath!clyde!bonnie!akgua!whuxlm!harpo!decvax!decwrl!pyramid!nsc!cadtec!csi!epicen!jbuck
From: jbuck@epicen.UUCP (Joe Buck)
Newsgroups: net.math
Subject: Re: Is the Mandelbrot set a fiction??
Message-ID: <221@epicen.UUCP>
Date: Tue, 24-Sep-85 22:32:34 EDT
Article-I.D.: epicen.221
Posted: Tue Sep 24 22:32:34 1985
Date-Received: Sat, 28-Sep-85 07:02:25 EDT
References: <418@aero.ARPA> <646@petsd.UUCP>
Organization: Entropic Processing, Inc., Cupertino, CA
Lines: 27
Summary: It's real enough

> From: cjh@petsd.UUCP (Chris Henrich)
> Date: 17 Sep 85 14:26:31 GMT
> 
>      By the way, I think that article was mistaken in stating
> that if the value of z ever got to where |z| > 2, it was sure
> to go to "infinity".  It is fairly easy to show that if
>           |z| > |c| + 1
> then the sequence will go to infinity.

I don't understand why you think discovering a different bound invalidates
the one given in Scientific American. To show the result given there, first
prove that the set is bounded by the circle |c| = 2. To show this, study
the sequence z'=z*z+c for |c|>2; it grows without bound since |c^2|=|c|^2
is at least 4 and c*2+c must have a strictly larger magnitude than c.

The slowest growth is when c is real and negative; in fact, it's not hard
to show that all real, negative c's in [-2,0] are in the set. It's only 
slightly trickier to see that no point on the positive real axis is in
the set (each term is at least c^2 greater in magnitude than the previous
term, so the sequence goes to infinity).

Next, assuming that |c|<2, it's straightforward to see that once |z|>2,
the sequence grows without bound.
-- 
Joe Buck				|  Entropic Processing, Inc.
UUCP: {ucbvax,ihnp4}!dual!epicen!jbuck  |  10011 N. Foothill Blvd.
ARPA: dual!epicen!jbuck@BERKELEY.ARPA   |  Cupertino, CA 95014


Brought to you by Super Global Mega Corp .com