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- Newsgroups: sci.fractals
- Path: sparky!uunet!mnemosyne.cs.du.edu!nyx!lmitchel
- From: lmitchel@nyx.cs.du.edu (lloyd mitchell)
- Subject: Re: Newton's Method
- Message-ID: <1992Nov9.113335.8516@mnemosyne.cs.du.edu>
- Sender: usenet@mnemosyne.cs.du.edu (netnews admin account)
- Organization: University of Denver, Dept. of Math & Comp. Sci.
- Date: Mon, 9 Nov 92 11:33:35 GMT
- Lines: 33
-
- In article <1992Nov8.221326.11267@gagme.chi.il.us> modus@gagme.chi.il.us (Patrick M. Kane) writes:
- >I recently read the book FRACTAL PROGRAMMING IN PASCAL and have a question
- >relating to the generation of Newton's Method fractals. The following code
- >fragment iterates the equation z^3-1=0 :
- >xsquare= x * x
- >ysquare = y * y
- >denom = 3 * ((xsquare - ysquare) * (xsquare - yswaure) - 4 * xsquare *
- >ysquare)
- >IF denom = 0 then denom = 0.0000001
- >x = (2/3) * x + (xsquare - ysquare) / denom
- >y = (2/3) * y - (2 * x * y) / denom
- >
- >This does, of course, work beautifully. However, with my high school calculus
- >skills, I am unable to understand WHY. Could someone please e-mail or post a
- >step by step explanation of each part of the above fragment? Any help is
- >appreciated.
- >
- >Patrick Kane
-
- If we use z for x+iy, then Newton's method for z^3 - 1 = 0 would be:
-
- z = z - f/f', where f = z^3 - 1 and f' is the derivative of f with respect
- to z, or 3z^2 in this case. Thus,
-
- z = z - (z^3 - 1)/3z^2 = z - z^3/3z^2 + 1/3z^2 = z - z/3 + 1/3z^2,
- z = 2/3*z + 1/3z^2.
-
- So the first term in your x and y equations is the 2/3*z term. To get the
- 1/3z^2 term, your program uses z^2/(3*|z^4|), which is the same thing. So
- 'denom' is 3 times the magnitude of z^4. xsquare - ysquare is the real
- part of z^2, and 2*x*y is the imaginary part.
-
- Kerry Mitchell
-