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From: rwh@exodus.UUCP (Roy Haas)
Newsgroups: net.math
Subject: Re: Square Roots give Powers
Message-ID: <221@exodus.UUCP>
Date: Fri, 21-Sep-84 09:39:45 EDT
Article-I.D.: exodus.221
Posted: Fri Sep 21 09:39:45 1984
Date-Received: Tue, 25-Sep-84 20:49:57 EDT
References: <9092@watmath.UUCP>
Organization: Bell Communications Research, Inc., Murray Hill, NJ
Lines: 13

Since the limit

	lim [ y( x^(1/2n) + 1 ]^2n = x^y
	n-> inf

holds (use L'Hospital's Rule), your algorithm works for "large n".

The rate of convergence depends on how large x is, since repeated 
rooting drives x^1/2n to 1.  It also depends on how large y is,
since y(x^1/2n - 1) must approach 0. Note that the algorithm is
only useful as an approximatiion for  y where the fractional
part has an infinite binary expansion (otherwise you can compute x^y
exactly in a finite number of steps).