Title: Appearances can be deceptive
Post by: element90 on May 11, 2015, 12:29:17 PM
The following formula produces something that appears to be a standard Mandelbrot as part of a larger fractal. It turns out not to be.
The initial value of z is zero.
(https://copy.com/wGjLJT3wcuXHth9c)
(https://copy.com/rAmEEOSnztiPGY7u)
(https://copy.com/gxcc84vY2fG0wfuU)
(https://copy.com/jm118zuXKDcsZIaB)
Clearly zero is a critical point but I don't see how it is derived:
![f'(z) = \left( \frac{g(z)}{h(z)} \right)' = \frac{g'(z)h(z)-g(z)h'(z)}{[h(z)]^2}](/cgi-bin/mimetex.cgi?f'(z) = \left( \frac{g(z)}{h(z)} \right)' = \frac{g'(z)h(z)-g(z)h'(z)}{[h(z)]^2})
 = z^2+c)
 = 2z)
 = \frac{1}{z^6}+\frac{1}{c})
 = -\frac{6}{z^7})
![f'(z) = \frac{2z\left( \frac{1}{z^6}+\frac{1}{c} \right)-\left( z^2+c \right)\left( -\frac{6}{z^7} \right)}{[\frac{1}{z^6}+\frac{1}{c}]^2} = 0](/cgi-bin/mimetex.cgi?f'(z) = \frac{2z\left( \frac{1}{z^6}+\frac{1}{c} \right)-\left( z^2+c \right)\left( -\frac{6}{z^7} \right)}{[\frac{1}{z^6}+\frac{1}{c}]^2} = 0)
Multiplying both sides by h(z) squared simplifies the formula to:
 = 0)
The polynomial can be solved for zero and will give location dependent solutions and hence critical points. How can zero be a solution, as it must be if it is a critical point?
The initial value of z is zero.
(https://copy.com/wGjLJT3wcuXHth9c)
(https://copy.com/rAmEEOSnztiPGY7u)
(https://copy.com/gxcc84vY2fG0wfuU)
(https://copy.com/jm118zuXKDcsZIaB)
Clearly zero is a critical point but I don't see how it is derived:
Multiplying both sides by h(z) squared simplifies the formula to:
The polynomial can be solved for zero and will give location dependent solutions and hence critical points. How can zero be a solution, as it must be if it is a critical point?
Title: Re: Appearances can be deceptive
Post by: lkmitch on May 11, 2015, 05:59:06 PM
Two things: 1) a critical point is a spot where the derivative = 0 or fails to exist. In your last equation, the 1/z^7 factor is what makes 0 a critical point, since you'd be dividing by zero. 2) If you rewrite the first term of f(z) as a rational function (that is, one numerator polynomial and one denominator polynomial), you'll see in its derivative that every term has some extra z factors in the numerator, so the net effect is that z=0 would cause the derivative to be zero.
Title: Re: Appearances can be deceptive
Post by: Adam Majewski on May 11, 2015, 06:02:41 PM
IMHO one has to compute critical point for every point c
(%i1) f:(z^2+c)/(z^{-6} +c^{-1}) +c+7;
(%o1) (z^2+c)/(z^{−6}+c^{−1})+c+7
(%i2) d:diff(f,z,1);
(%o2) (2*z)/(z^{−6}+c^{−1})−({−6}*z^{−6}−1*(z^2+c))/(z^{−6}+c^{−1})^2
(%i3) solve(d=0);
solve: more unknowns than equations.
Unknowns given :
[c,z]
Equations given:
errexp1
-- an error. To debug this try: debugmode(true);
(%i4) expand(d);
(%o4) −({−6}*z^({−6}+1))/(z^(2*{−6})+2*c^{−1}*z^{−6}+c^(2*{−1}))−({−6}*c*z^{−6}−1)/(z^(2*{−6})+2*c^{−1}*z^{−6}+c^(2*{−1}))+(2*z)/(z^{−6}+c^{−1})
Compare https://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/qpolynomials
HTH
(%i1) f:(z^2+c)/(z^{-6} +c^{-1}) +c+7;
(%o1) (z^2+c)/(z^{−6}+c^{−1})+c+7
(%i2) d:diff(f,z,1);
(%o2) (2*z)/(z^{−6}+c^{−1})−({−6}*z^{−6}−1*(z^2+c))/(z^{−6}+c^{−1})^2
(%i3) solve(d=0);
solve: more unknowns than equations.
Unknowns given :
[c,z]
Equations given:
errexp1
-- an error. To debug this try: debugmode(true);
(%i4) expand(d);
(%o4) −({−6}*z^({−6}+1))/(z^(2*{−6})+2*c^{−1}*z^{−6}+c^(2*{−1}))−({−6}*c*z^{−6}−1)/(z^(2*{−6})+2*c^{−1}*z^{−6}+c^(2*{−1}))+(2*z)/(z^{−6}+c^{−1})
Compare https://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/qpolynomials
HTH
Title: Re: Appearances can be deceptive
Post by: element90 on May 11, 2015, 08:42:24 PM
Quote
IMHO one has to compute critical point for every point c
I know that's why I stated:
Quote
The polynomial can be solved for zero and will give location dependent solutions
i.e. the critical points have to determined for each location, I already do this in Neptune and in the new version of Saturn.
Quote
1) a critical point is a spot where the derivative = 0 or fails to exist.
I clearly know about the solutions to f'(z) = 0 but what does "fails to exist" mean?
Quote
2) If you rewrite the first term of f(z) as a rational function (that is, one numerator polynomial and one denominator polynomial), you'll see in its derivative that every term has some extra z factors in the numerator, so the net effect is that z=0 would cause the derivative to be zero.
I know how to do that.