Newsgroups: sci.math.num-analysis
From: pmontgom@cwi.nl (Peter L. Montgomery)
Subject: Re: Newton's method
Date: Wed, 22 Jul 1998 07:38:36 GMT

In article <35B4472E.78DB@scilab.uct.ac.za> jhncla02@scilab.uct.ac.za writes:
>I was wondering (this isn't a very highbrow question) whether anyone
>knows of an example of a function and a point (a nice rational number,
>if possible?) which, when Newton's method is used with that starting
>point, causes the parallelogram shape one always sees in texts showing
>where the method fails.

>What must happen is that x1 = x3 = ... = x(2n+1) and x2 = x4 = ... =
>x(2n). The method fails as the function is odd about the starting point,
>and the start point must be picked very precisely.

>I have a value for f(x)=x^3-x, but it is to be used to demonstrate a
>computer program with only finite accuracy, and after a bit it diverges
>owing to numerical inaccuracy. I've tried to solve for a point
>analytically with some odd functions, but it gets hairy.

      Modify your example to f(x) = x^3 - 5x.
If x1 = 1, then x_i = 1 for odd i and x_i = -1 for even i.

      For a cycle of length 3, try f(x) = x^3 - x + 1/2 with x1 = 1.

-- 
        Peter-Lawrence.Montgomery@cwi.nl    San Rafael, California