Imitations of classical Julia set using Newton's method

[giann_tee]

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.

http://www.chiark.greenend.org.uk/~sgtatham/Newton/

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:

dsolve('z-f/Df=z^2-z+c','z')

ans =

C1*exp(-1/(c-1)^(1/2)*atan((z-1)/(c-1)^(1/2)))

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

int(-1/(-z+z^2+c),z)

ans =

-2/(4*c-1)^(1/2)*atan((2*z-1)/(4*c-1)^(1/2))

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

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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]
roots(p)

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:

dsolve('z-f/Df=z^5-z+c','z')

ans =

C1*exp(Int(-1/(-2*z+z^5+c),z))

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

int(-1/(-2*z+z^5+c),z)

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?

 

[jvicens]

Thank you for sharing your exploration of fractals derived from Newton's method. It is always exciting to see people delving into the world of mathematics and discovering new insights and connections.

To answer your first question, the equation that Simon Tatham presents in the "imitations" section of his website is a special case of the general form of the differential equation z_new = z - f(z)/f'(z). In this case, the function f(z) is defined as f(z) = z^2 + c, where c is a complex number. This makes the equation z_new = z - f(z)/f'(z) equivalent to the iteration z_new = z - (z^2 + c)/2z, which is the same as the iteration for the Julia set J(z^2+c).

To solve this differential equation, Simon Tatham uses the technique of partial fractions, which is a common method for solving rational functions. In this case, the rational function is -dz/(z^2-z+c), and by using partial fractions, Simon is able to find the solution f(z) = (z-a)^(1/(b-a)) * (z-b)^(1/(a-b)), where a and b are the roots of z^2-z+c=0. This solution is then substituted into the equation z_new = z - f(z)/f'(z), which simplifies to the iteration for the Julia set J(z^2+c).

To answer your second question, the output from the Matlab and Mathematica software may seem unreadable because they are giving the general solution to the differential equation. This means that the solution contains a constant, C1, which can take on any value. However, in the case of the Julia set J(z^2+c), we are only interested in a specific solution, which is the one with C1=0. This is the solution that Simon Tatham presents on his website.

In general, these programs for mathematics work very well, but sometimes the output may seem confusing or unreadable because they give the general solution rather than a specific one. To read the output, you can try substituting different values for the constants to see how the solution changes. In this case, substituting C1=0 will give you the specific solution for the Julia set J(z^2+c).

I hope this helps to clarify the solution to the differential equation and the output from the software. Keep