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From: percus@acf4.UUCP (Allon G. Percus)
Newsgroups: net.math
Subject: Re: REPEATED ROOTS
Message-ID: <920006@acf4.UUCP>
Date: Tue, 25-Jun-85 10:10:00 EDT
Article-I.D.: acf4.920006
Posted: Tue Jun 25 10:10:00 1985
Date-Received: Sat, 29-Jun-85 00:36:49 EDT
References: <15187@watmath.UUCP>
Organization: New York University
Lines: 91

> 	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
> 
> 
> which of course ( here's were I get a bit suspicious ) can be rewritten as
> 
> EQ 3
> 	
> 	2 + x = x^2
> 
> which can then be rearranged and solved as follows
> 
> EQ 3a
> 	x^2 - x - 2 = 0
> 
> EQ 3b
> 	( x - 2 ) ( x + 1 ) = 0
> 
> EQ 3c
> 	x = 2, x = -1

> QUESTIONS:
> 
> 1) Is EQ 3 a legitimate expression to derive from EQ 2 ?

Absolutely.  You're just manipulating an identity.

> 2) If so, why are there two different answers produced, at first glance
>    I would expect the equation to generate a quadratic with repeated
>    roots as the positive square root is well defined.

There is only one answer produced.  You specified that "sqrt" means
the "positive square root", so, plugging x=-1 into it:

     sqrt(2+sqrt(2+sqrt(2+...)))=-1
     sqrt(2+x)=-1

The positive square root of 2+x can most certainly NOT be -1!!!
Therefore, x=-1 must be discarded as an extraneous root.

> 3) If EQ 3 is legitimate are these equations legitimate ?
> 
> EQ 4
> 	4 + 4( sqrt( 2+ sqrt( 2 + sqrt( 2 + .... )))) + 2 + sqrt( 2 +..)))
> 	= x^4
> 
> ( square of equation 2 )
> 
> EQ 5
> 	6 + 5x = x^4
> 
> EQ 6
> 	
> 	x^4 - 5x - 6 = 0
> 
> which has solutions
> 
> EQ 7
> 	x = -1
> 	x = 2
> 	x = -1/2 + i*sqrt( 11 )
> 	x = -1/2 - i*sqrt( 11 )

Equations 4, 5, and 6 are absolutely correct.  However, in #7, we
must again discard the negative root as extraneous, and the same
can be shown to hold for the complex roots.

> 4) If the answer to 3 is yes then can we continue this process 
> 	    indefinitely and if so:
> 	    a) if we are considering the 2^n power will all the solutions
> 	       of 2^n-1 power be included ?
> 	    b) are there infinitely many solutions ?

If you go on to 2^n, as n -> inf, you'll get an infinite number of
roots, but again, you'll have to discard all but one as extraneous:
except for x=2, all roots will either be negative or complex.  If
I have some time later on, I'll provide a semi-rigorous proof of this.
As for your first part of the question, I'll have to think about it.

METHOD 2, I agree, is somewhat more elegant than METHOD 1, but I'm
afraid I can't look at any more now.

I find this a VERY interesting problem.
                                         A. G. Percus
                                  (ARPA) percus@acf4
                                   (NYU) percus.acf4
                                  (UUCP) ...!ihnp4!cmcl2!acf4!percus