Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site rtech.ARPA Path: utzoo!watmath!clyde!cbosgd!ihnp4!houxm!vax135!cornell!uw-beaver!tektronix!zehntel!vlsvax1!qantel!dual!unisoft!mtxinu!rtech!training From: training@rtech.ARPA (Training account) Newsgroups: net.puzzle Subject: Re: x to the x to the x ... (ANSWER/SPOILER) Message-ID: <347@rtech.ARPA> Date: Fri, 3-May-85 12:32:20 EDT Article-I.D.: rtech.347 Posted: Fri May 3 12:32:20 1985 Date-Received: Sat, 11-May-85 00:29:14 EDT References: <748@whuxlm.UUCP> <351@gitpyr.UUCP> <821@mhuxt.UUCP> Organization: Relational Technology, Berkeley CA Lines: 51 > > >x^x^x^x... = 2 > > This problem is solved as follows: > > I couldn't figure out exactly what was wrong with the derivation, but > Scott's answer: > > (1/2) > > 6. x = 2 > > Is easily demonstrated to be wrong, with the help of my trusty TI-58. > 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^x^x^x...approaches infinity for x>1. Nothing was wrong with Scott's answer; you parenthesized incorrectly. You computed the answer to ((((x^x)^x)^x)^x...) which does grow without bound. If, however, you parenthesize as (x^(x^(x^(x...)))), and substitute radical-2 for x, the answer is indeed 2. There's a nice intuitive proof that this is true: suppose we raise radical-2 to the second power. The answer is 2. So if we raise radical-2 to the power of any number LESS than 2, the result will be less than 2. And if we raise radical-2 to that new power, the result must therefore be less than 2. And if we raise... you get the idea. Here's another problem: x*x*x*x*x...=2. This is the same as x*(x*x*x*x*x...)=2, so by substituting, we get x*2=2. Therefore, x = 1. But substituting back in the original equation, we get 1*1*1*1*1*1*1=2, so 1 = 2. What's wrong here? Robert Orenstein Relational Technology