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Imitations of classical Julia set using Newton's method

  1. Jan 10, 2009 #1
    I have been exploring the world of fractals derived from Newton's method for finding roots (solutions) to equations. The following page contains insights into some mathematics behind Newton's method.


    If you open the page and step into the topic of "imitations" you will see what I mean.

    Author mentions that we could try to generate Newton's fractal that looks like good old Julia set J(z^2+c) by solving a differential equation:

    z_new = z - f(z)/f'(z) = z^2 + c

    (all variables are complex numbers, function is holomorphic). We wonder, for which equation Newton's equation will produce the iteration of z^2+c. Simon Tatham gives the solution to that equation which I don't understand. He treats this like the simplest form of differential equation which can be solved in a single step - integration of left and right side:

    df/f = -dz / (z^2-z+c)

    Here I am rather puzzled - if I may explain. Simon says that he solved the integral to the right as an integral of a rational function (http://www.answers.com/topic/partial-fractions-in-integration). Somehow he arrives at this equation:

    f(z)= (z-a)^(1/(b-a)) * (z-b)^(1/(a-b))

    a, b are roots of z^2-z+c=0.

    (Please note that derivative of this function f can be simplified further for final calculation in computer using a perfectly well described method in the precious section of Simon's web site. That part is not a problem.)

    I used Matlab software to check on this work. Matlab treats the initial differential equation as homogeneous differential equation and gives:


    ans =


    Or, to follow Simon's story and compute integral of -dz/(z^2-z+c)


    ans =


    Therefore I conclude I don't know how to solve it Simon's way and I need you to explain to me!


    If you're still following the story....

    roots of z^2-z+c and some c are:

    c=complex(0.2, 0.3)
    p=[1 -1 c]

    So far, that was the Matlab way.


    Now if you are really brave and enlightened like me of course, you will type into Matlab some other differential equation to see how it works in general:


    ans =


    which means Matlab left this integral for us to solve...


    ans =

    -sum(1/(-2+5*_R^4)*log(z-_R),_R = RootOf(-2*_Z+_Z^5+c))

    which means that the output is now unreadable to me. Even though it seems like an error, "RootOf" is a legal way to mention roots of some polynomial. Mathematica also produces strange output for that integral: try here... http://integrals.wolfram.com/index.jsp

    My second question is therefore, why are these programs for mathematics like this? why don't they work well? How do I read output if its good?
  2. jcsd
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