Appearances can be deceptive

[element90]

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.






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?

[lkmitch]

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.

[Adam Majewski]

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

[element90]

IMHO one has to compute critical point for every point c

I know that's why I stated:

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.

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?

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.