Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site linus.UUCP Path: utzoo!watmath!clyde!burl!ulysses!unc!mcnc!decvax!linus!meister From: meister@linus.UUCP (Phillip W. Servita) Newsgroups: net.math Subject: Re: REPEATED ROOTS Message-ID: <441@linus.UUCP> Date: Wed, 26-Jun-85 21:20:55 EDT Article-I.D.: linus.441 Posted: Wed Jun 26 21:20:55 1985 Date-Received: Sat, 29-Jun-85 00:37:49 EDT References: <15187@watmath.UUCP> <920006@acf4.UUCP> Reply-To: meister@linus.UUCP (Philip W. Servita) Organization: The MITRE Coporation, Bedford, MA Lines: 41 Summary: In article <920006@acf4.UUCP> percus@acf4.UUCP (Allon G. Percus) writes: >> The first solution which came to mind arose from the observation, >> which I now suspect to be invalid, that if we strip off the first radical >> ( that is we square both sides of the equation ) we get: >> >> EQ 2 >> 2 + sqrt( 2 + sqrt( 2 + sqrt( 2 + ....)))) = x^2 by now everybody knows the problem we are talking about here. when i was back in high school, i discovered this by accident. however, i took it a little further than people seem to be taking it here. take the general function: y = f(x) = sqrt(x + sqrt(x + sqrt(x + sqrt(x + ...)))) hence, 2 y = x + sqrt(x + sqrt(x + sqrt(x + sqrt(x + ...)))) or: 2 2 y = x + y , so y - y - x = 0, with solutions easily given by the quadratic formula: y = (1 +/- sqrt(1 + 4x))/2 for x = 2, this gives values of 2 and -1, giving rise to the extraneous root discussion. what i noticed, is that: TRY PLUGGING IN -1/4 for X! this gives a repeated root of 1/2! hence sqrt(-1/4 + sqrt(-1/4 + ...)) = 1/2. this one just defies all intuition. enjoy. -the venn buddhist