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From: ecl@hocsj.UUCP
Newsgroups: net.math
Subject: re: Square Roots Give Powers
Message-ID: <144@hocsj.UUCP>
Date: Sun, 30-Sep-84 12:38:14 EDT
Article-I.D.: hocsj.144
Posted: Sun Sep 30 12:38:14 1984
Date-Received: Wed, 3-Oct-84 05:48:00 EDT
Organization: AT&T Information Systems Labs, Holmdel NJ
Lines: 38

Subject: Re: Square Roots give Powers

An interesting sidelight of this problem is that x^(1/c)-1 is very
nearly a logarithm.

For large c

x^(1/c) ~= 1

x^((1/c) - 1) ~= 1/x

integrating both sides and approximating the integration constant we get

c*(x^(1/c) - 1) ~= ln(x)

x^(1/c) - 1 ~= ln(x)/ln(exp(c)) = log      x
                                     exp(c)

That is "log to the base exp(c) of x"
My TI calculator said to take the square root 11 times.  That says that
c = 2^11 = 2048.  This means that when you take the square root of a
number 11 times and subtract 1 you are really getting a close
approximation of log base (exp(2048)) of that number.  You are dealing
with logarithms of base approximately 1.7E38.  Who would have thought
that a logarit with such a humungous base would be so easy to
calculate?  Usually the algorithms I have see always have you perform
the inverse operation at the end.  Nobody usually deals with logarithms
of giant bases.  However if what I have found useful on my pocket CASIO
is to take square roots 11 times, subtract one, and divide by the easy
to remember constant 0.0011248.  This gives you the common logarithm to
about three decimal place accuracy.  Dividing by 0.0004883 gives you
the natural logarithm.  Where do these numbers come from?  Take 10,
take its square root 11 times and subtract 1 and you get 0.0011248.

					(Evelyn C. Leeper for)
					Mark R. Leeper
					...ihnp4!lznv!mrl