🚀 go-pugleaf

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