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Path: utzoo!utcsri!garfield!robertj
From: robertj@garfield.UUCP (Robert Janes)
Newsgroups: net.math
Subject: REPEATED ROOTS
Message-ID: <3122@garfield.UUCP>
Date: Tue, 18-Jun-85 09:59:19 EDT
Article-I.D.: garfield.3122
Posted: Tue Jun 18 09:59:19 1985
Date-Received: Tue, 18-Jun-85 21:08:01 EDT
Distribution: net
Organization: Memorial U. of Nfld. C.S. Dept., St. John's
Lines: 129


	This is a small problem which has bothered me for some time and I
and I would like to hear some comments from various folks out there in
netland.

	The problem arises in considering the expression:

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

where sqrt is taken to mean the standard positive square root of a number
and there are an "infinite" number of these square roots nested one inside
another as shown.

	The problem is what is x ?

( METHOD I)

	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

:DONE:

QUESTIONS:

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

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.

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 )

question 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 ?


METHOD 2

	This is a very indirect approach to the problem:

	let a1 = pi/4

	then cos( a1 ) = sqrt( 2 )/2

	let ai = a(i-1)/2

	then by induction we can show that

	cos( ai ) = sqrt( 2 + sqrt( 2 + ....)))))..)/2

	where there are i sqrt's

	but in the limit as i->infinity ai = 0
	hence cos( ai ) = 1

	hence 2 = sqrt( 2 + sqrt( 2 +sqrt( 2 + .....)))))))

:DONE:

QUESTIONS:
question 1) Is this a legitimate method ?
question 2) Are there any other such constructions which will yield
	    similar results using trignometric functions
question 3) Can a similar series be constructed for each of the 
	    solutions given by METHOD 1 ?

_______________________________________________________________________________

	I would appreciate any comments on the above and any corrections 
appropos algebra or trigonometry.

	I suspect that since we are dealing with infinite sequences that
the results shown in section 1 are somewhat bogus as they make that great
leap into defining certain things to be x when it may not be rigourously
certain to be true.

	As you may have gathered I have a certain preference for METHOD 2
as it assumes very little in comparison to METHOD I and seems somewhat more
elegant.

						Robert Janes
						Memorial University of
						Newfoundland