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From: chuck@dartvax.UUCP (Chuck Simmons)
Newsgroups: net.puzzle
Subject: Re: x to the x to the x ... (maybe SPOILER)
Message-ID: <2981@dartvax.UUCP>
Date: Sun, 28-Apr-85 08:59:13 EDT
Article-I.D.: dartvax.2981
Posted: Sun Apr 28 08:59:13 1985
Date-Received: Tue, 30-Apr-85 03:13:28 EDT
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> > x^x^x^x... = 2
> > 
> Let zeta be x^x^x^x^..., and re-write the left-hand side as
> 
> 	x^zeta = 2	Note also that x^zeta = zeta, and therefore that
> 			2 = zeta.  Rewrite as
> 	x^2 = 2
> 
> Simple, no?  Not done yet:  there are two answers, plus and minus root-2.
> 
> It (seems to me that it) can't be plus root-2, because trying to do the
> "series" by hand leads to an unbounded value.  So it must be minus root-2.
> However, I am not comfortable with that, either, because I don't remember
> my complex arithmetic well enough to work out whether or not it works (I
> knew it - I've been programming in C too long!  I had to look in a FORTRAN
> manual just to convince myself that log and exponent operations are OK on
> negative quantities.  All that math in college has just poured out my ears!).
> 
> Well, I tried.
> -- 
> John Woods, Charles River Data Systems, Framingham MA, (617) 626-1101

You better go back and dust off your math books.  First, minus root 2
is not a really nice answer because you have to leave the real numbers
and get into the complex numbers.  Secondly, "by hand" calculations show
that plus root 2 converges to 2 fairly quickly.

I wrote a real simple pl1 program to do my by hand calculations for
both the positive and negative square roots.  (Arn't languages with
complex numbers wonderful?)  (Oh bummer!  I suddenly remember that last
summer I found a few bugs in our complex number runtime package exponentiation
routines.  So I have no "by hand" calculations for negative square roots.)

Your by-hand calculations should use the following algorithm:

    result = root2;
    do forever;
        result = root2 ** result;
    end;

Here, 'result' will quickly converge to 2.0.  (We should be able to
prove we have a monotonic-increasing bounded sequence.  Any suggestions?)

Possibly your by-hand calculations used the similar algorithm:

    result = root2;
    do forever;
        result = result ** root2;
    end;

Here 'result' rapidly diverges.  However, this algorithm solves
for ((((...^root2)^root2)^root2)^root2)^root2; which is not what we want.

-- Chuck Simmons