Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10 5/3/83; site teklabs.UUCP
Path: utzoo!linus!decvax!tektronix!tekchips!teklabs!stevens
From: stevens@teklabs.UUCP
Newsgroups: net.math
Subject: Iteration problem solved
Message-ID: <2517@teklabs.UUCP>
Date: Fri, 30-Sep-83 12:30:05 EDT
Article-I.D.: teklabs.2517
Posted: Fri Sep 30 12:30:05 1983
Date-Received: Sun, 2-Oct-83 02:27:17 EDT
Organization: Tektronix, Beaverton OR
Lines: 28


	I like this kind of problem,  let's get off this probability thing.

Problem:  Given Y(n) = Y(n-1) - X(n-1)
		X(n) = X(n-1) - X(n-1)*X(n-1)/Y(n-1)

	  Find X(1) that maximizes X(8) when Y(1) = 4

OK, lets let X(1) = a for now.  Through some calculations,

	Y(1) = 4		X(1) = a
	Y(2) = 4-a 		X(2) = a(4-a)/4
	Y(3) = (4-a)^2/4	X(3) = a(4-a)^2/16
	Y(4) = (4-a)^3/16	X(4) = a(4-a)^3/64

	Y(n) = (4-a)^(n-1)/2^n  X(n) = a(4-a)^(n-1)/4^(n-1)

So to find the maximum of X(n) for any n, we take the derivative (sp?) with
respect to a. We get

	dx/da = 0 = [a(n-1)(4-a)^(n-2) + (4-a)^(n-1)]

or simply       a(n-2) + 4 = 0      When n=8, a = -2/3

There's probably a more theoretical way of solving this, but this works, so
I use it.  By the way, a = -2/3 is a MINIMUM and not a maximum.  I wish I
knew how to find the maximum that was asked for in the problem
-- 
Steve Silberberg