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From: gjerawlins@watdaisy.UUCP (Gregory J.E. Rawlins)
Newsgroups: net.puzzle
Subject: Re: x to the x to the x ... (ANSWER/SPOILER)
Message-ID: <7219@watdaisy.UUCP>
Date: Tue, 30-Apr-85 21:14:35 EDT
Article-I.D.: watdaisy.7219
Posted: Tue Apr 30 21:14:35 1985
Date-Received: Wed, 1-May-85 03:21:41 EDT
References: <748@whuxlm.UUCP> <351@gitpyr.UUCP> <821@mhuxt.UUCP>
Reply-To: gjerawlins@watdaisy.UUCP (Gregory J.E. Rawlins)
Organization: U of Waterloo, Ontario
Lines: 31

In article <821@mhuxt.UUCP> js2j@mhuxt.UUCP (sonntag) writes:
>Let x=2^(1/2).
>       x^x=1.632...
>      x^x^x=2.000...
>      x^x^x^x=2.665..., and the series just keeps on growing.  I suspect
>that the problem *has* no solution, and that x^x^x...=1 for  0<x<=1 and that
>x^x^x^x...approaches infinity for x>1.
>     I suppose that it is possible that there exists some non-real solution.
>Jeff Sonntag
    Unfortunately mathematics is not only stranger than we
imagine it's stranger than we *can* imagine. The sequence you
give does not converge because you leave off infinitely many
terms which *will* cause equality with 2.
    Let a=2^(1/2) Then   a^2 = 2
                         a^a^2 = 2
                         a^a^a^2 = 2
                         a^a^a^a^2 = 2
                         a^a^a^a^a^2 = 2
                         a^a^a^a^a^a^2 = 2
    ...to infinity. The problem is that unlike an (addition)
series of numbers we cannot separate the terms of the
exponentiations and stop after finitely many terms with a "good"
approximation to the answer. I confess that there are hard
problems here of convergence, but this is the essence of the
flaw. Is there a space in which we can prove (the normal)
epsilon-delta convergence??
    greg.

-- 
Gregory J.E. Rawlins, Department of Computer Science, U. Waterloo
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