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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