Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.PCS 1/10/84; site hocsj.UUCP Path: utzoo!linus!decvax!harpo!whuxle!spuxll!abnjh!u1100a!pyuxn!pyuxww!gamma!ulysses!mhuxj!houxm!hogpc!pegasus!hocsj!ecl 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