Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site watmath.UUCP 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.