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Path: utzoo!watmath!ljdickey
From: ljdickey@watmath.UUCP (Lee Dickey)
Newsgroups: net.math
Subject: Re: Square Roots give Powers (Solution)
Message-ID: <9165@watmath.UUCP>
Date: Tue, 25-Sep-84 23:53:54 EDT
Article-I.D.: watmath.9165
Posted: Tue Sep 25 23:53:54 1984
Date-Received: Thu, 27-Sep-84 03:29:53 EDT
References: <9092@watmath.UUCP>
Organization: U of Waterloo, Ontario
Lines: 88

Six people responded to my article "Square Roots give Powers". 
They are:
  Mark Manasse	=  ihnp4!gargoyle!msm
  Dave Clark	=  ihnp4!houxm!homxa!wdc
  Albert Algava	=  ihnp4!houxm!hound!5143ama	
  Kevin Martin	=  kpmartin
  		=  allegra!hplabs!hplabsb!ayanoglu
  Roy Haas	=  allegra!ulysses!gamma!exodus!rwh
All of them had correct methods of justifying the result I mentioned.
Kevin even suggested an improvement.  A proof was mentioned by Roy,
and one is included below.  Several of the arguments are summarized 
here:
=================================================================
We want to show that
	(1+y*(x^(1/(2^n))-1))^(2^n)	~=  x^y
Let M=2^n.  What we want to show is:
	(1+y*(x^(1/M)-1))^M		~=  x^y
Now, let  a = x^(1/M)-1.  Then what we are looking at is:
	(1+y*a)^M			~=  ((1+a)^M)^y
But, ((1+a)^M)^y = ((1+a)^y)^M.  So, taking power 1/M, we get:
	(1+y*a)				~=  (1+a)^y
This last approximation is well known as a generalation of the
binomial expansion.  Accuracy improves for "a" small, and here,
"a" gets small quickly as "n" gets large.  Kevin Martin points 
out that the identity can be sharpened by taking three (or more) 
terms on the left side instead of just two.  For instance,
	(1+ y*a + .5*y*(y-1)*a^2)	~= (1+a)^y
is an improvement.
=================================================================
Here is a short proof that the process really does converge:

The proof is one that uses limits, along the lines of those proofs
that are used in elementary calculus.  Recall the algorithm:

    Enter the number  x.
    Press the "square root" key  N  times.
    Subtract 1.
    Multiply by  y.
    Add 1.
    Press the "square" key  N  times.

Pressing the SQRT key  N  times gives      
    (((x^.5)^.5)...)^.5 = x^(1/(2^N)).  
To simplify, I write M=2^N.
Thus
    (((x^.5)^.5)...)^.5 = x^(1/(2^N)) = x^(1/M).
Combining the rest of the steps, and calling the result  Z, we get:
    Z = (1+y*(x^(1/M) - 1))^M.
The problem is to show that for N (and hence M) large enough, this
expression is approximately x^y.  That is, that

    Z --> x^y,  as  M --> infinity.

Now,
    ln Z =  M * ln ( 1+y*(x^(1/M) - 1) )

                ln ( 1+y*(x^(1/M) - 1) )
             =  --------------------------------------
    		1/M

The above quotient satisfies the conditions of one version 
of L'Hospital's rule.  Thus, differentiating both numerator 
and denominator with respect to M gives
    limit ln Z = limit W
where
         (1 / ( 1+y*(x^(1/M) - 1) )) * y * (ln x) * (x^(1/M)) * (-1/M^2)
    W = -----------------------------------------------------------------
    		(-1/M^2)

      =  (ln x^y) * (x^(1/M)) / ( 1+y*(x^(1/M) - 1) )

But, as M-->infinity,  (1/M)-->0, and  x^(1/M)-->1.  And thus,
    W-->(ln x^y) * (1) / (1+y*(1 - 1) )
        = ln x^y.
Hence,
    limit ln Z = ln x^y, and
      limit Z  = x^y.
    					Q.E.D.
=================================================================
How many times should I press the SQRT key?  Here is a simple
heuristic:  For numbers "x" bigger than "1", successive square roots 
get closer and closer to 1.  Because the machine truncates, you do not
want to press too many times, because eventually the some successor
could become 1.  So watch the display.  The numbers become the form
	1.0000xxxx
A good time to stop is when there are about as many zeros as there are
digits "x" to the right of the zeros.