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From: bs@faron.UUCP (Robert D. Silverman)
Newsgroups: net.puzzle
Subject: Re: x to the x to the x ... (maybe SPOILER)
Message-ID: <292@faron.UUCP>
Date: Tue, 30-Apr-85 16:16:56 EDT
Article-I.D.: faron.292
Posted: Tue Apr 30 16:16:56 1985
Date-Received: Wed, 1-May-85 05:42:44 EDT
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Organization: The MITRE Coporation, Bedford, MA
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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

*** REPLACE THIS LINE WITH YOUR MESSAGE ***

Hey people, 2^(1/2) happens to be the right answer and while the formal
derivations are more or less correct noone has bothered to show whether
the infinite exponentiation converged for x = 2^(1/2). The question
of convergence appeared on the Putnam exam a few years back (~ 10).
The sequence converges if x < e^(1/e)  and x > e^(-1/e). 

The derivation x^x^x^x ... = 2  IF (and that's a big IF) it converges
then let y = x^x^x... = 2   leads immediately to x^2 = 2.
 
As it also happens (for complex variable freaks) that it still 
conveges as long as x*conjugate(x) falls in the above bounds.