Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site alice.UUCP Path: utzoo!watmath!clyde!burl!ulysses!allegra!alice!ark From: ark@alice.UUCP (Andrew Koenig) Newsgroups: net.puzzle Subject: Re: x to the x to the x ... (ANSWER/SPOILER) Message-ID: <3681@alice.UUCP> Date: Wed, 1-May-85 11:46:45 EDT Article-I.D.: alice.3681 Posted: Wed May 1 11:46:45 1985 Date-Received: Thu, 2-May-85 01:59:43 EDT References: <7219@watdaisy.UUCP> Organization: Bell Labs, Murray Hill Lines: 42 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 0x^x^x^x...approaches infinity for x>1. > I suppose that it is possible that there exists some non-real solution. >Jeff Sonntag Greg Rawlins attempts to explain this with some puzzled hand-waving: > 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. In fact, the explanation is simpler than that: Jeff's arithmetic is wrong. He is mistakenly assuming that exponentiation is left-associative, where it is really right-associative. Thus: x = 1.414... x^x = 1.632... x^x^x = 1.761... x^x^x^x = 1.841... The series really does converge.