# Fractals/Rational

Iteration of complex rational functions[1][2][3]

Herman ring - image with c++ src code

# Examples

Published online: June 15, 2010 DOI: 10.1007/s10114-010-6611-3 Http://www.ActaMath.com

## Blaschke fraction

Fatou sets for Blaschke fraction f(z) = rho * z^2 * (z-3) over (1-3z)

Analysis of critical points:

kill(all);
remvalue(all);
display2d:false;
ratprint : false; /* remove "rat :replaced " */

rho : -0.6170144002709304 +0.7869518599370003*%i;

define(f(z), rho * z^2 * (z-3)/(1-3*z));

/* first derivativa wrt z */
define( d(z), diff(f(z),z,1));

/* hipow does not expand expr, so hipow (expr, x) and hipow (expand (expr, x)) may yield different results */
n : hipow(num(expand(f(z))),z);
m : hipow(denom(expand(f(z))),z);

/* check if infinity is a fixed point */
limit(f(z),z,infinity);

/* finite critical points */

s:solve(d(z)=0)$s : map(rhs,s)$
s : map('float,s)$s : map('rectform,s)$



So there are 3 critical points :

• 2 finite critical points : z=1.0 i z= 0.0
• infinity

Dynamical plane consist of 3 basins

• basin of attraction of fixed point z = infinity ( superattracting) with inf many componnets
• basin of attraction of fixed point z = 0 ( superattracting) with inf many componnets
• basin of parabolic period 3 cycle ( with z= 1 critical point)

## Finite Blaschke product

${\displaystyle B_{n}(z)}$  is a finite Blaschke product of degree n.[11] It is: [12]

• a rational function
• an analytic function on the open unit disc such that f can be extended to a continuous function on the closed unit disc that maps the unit circle to itself
• have no poles in the open unit disc
• In particular, if ƒ satisfies the condition above and has no zeros inside the unit circle, then ƒ is constant (this fact is also a consequence of the maximum principle for harmonic functions, applied to the harmonic function log(|ƒ(z)|)).
• the Blaschke products B are rational perturbations of the doubling map of the circle R(z) = z^2 (equivalently given by θ → 2θ (mod 1)).
• a finite Blaschke product may be uniquely described by the set of its critical points
• rational map which fix a disc = which takes the closed unit disc D to itself
• their iteration theory can be analyzed from the point of view of Fuchsian groups.
• polynomials’ in the hyperbolic plane = hyperbolic polynomial
• A finite Blaschke product, restricted to the unit circle, is a smooth covering map
• the unit disk D, the unit circle ∂D and the complement of the closed unit disk C\D are all completely invariant sets for B

${\displaystyle B_{n}(z)=\zeta \prod _{i=1}^{n}\left({{z-a_{i}} \over {1-{\overline {a_{i}}}z}}\right)^{m_{i}}}$

where

• ${\displaystyle \zeta }$  is a unimodular constant. It is a point which lies on the unit circle: ${\displaystyle |\zeta |=1}$
• ${\displaystyle m_{i}}$  is the multiplicity of the zero ${\displaystyle a_{i}}$
• ${\displaystyle (a_{i})_{i=1}^{n}}$  is a finite sequence of n points in the open unit disc ${\displaystyle |a_{i}|<1}$

${\displaystyle (a_{i})_{i=1}^{n}=(a_{1},a_{2},\ldots ,a_{n}).}$

The building blocks[13] of Blaschke products are Mobius transformations of the form

${\displaystyle b_{k}(z)=e^{i\theta k}{\frac {a_{k}-z}{1-{\overline {a_{k}}}z}}}$

where

• ak ∈ D := {z ∈ C|, |z| < 1}
• θk ∈ R.

A finite (infinite) Blaschke product has the form

${\displaystyle w=B(z)=\prod _{k=1}^{n}b_{k}(z)}$

Examples:

### classification

There is a classification of finite Blaschke products in analogy with Möbius transformations.[14]

• B is elliptic if the Denjoy-Wolff point z0 of B lies in D. |B' (z0)| < 1.
• B is hyperbolic if the Denjoy-Wolff point z0 of B lies on ∂D and B'(z0) < 1,
• B is parabolic if the Denjoy-Wolff point z0 of B lies on ∂D and B'(z0) = 1,

The Denjoy-Wolff point of B is a unique z0 ∈ D such that ${\displaystyle B^{n}(z)\to z_{0}}$  for every z ∈ D

### Julia set

Let B be a finite Blaschke product of degree d > 1. Julia set ${\displaystyle J_{B}}$  , the set on which the iterates ${\displaystyle B^{n}}$  fail to be normal on any neighbourhood is either the unit circle ${\displaystyle S^{1}}$  or a Cantor subset[15][16]

• if B is elliptic, J(B) = ∂D,
• if B is hyperbolic, J(B) is a Cantor subset of D,
• if B is parabolic and z0 ∈ ∂D is the Denjoy-Wolff point of B,
• J(B) = ∂D if B(z0) = 0
• J(B) is a Cantor subset of ∂D if ${\displaystyle B''(z0)\neq 0.}$

## McMullen maps

singularly perturbed maps, also called McMullen maps[17]

 ${\displaystyle R_{\lambda }(z)=z^{m}+\lambda /z^{d}}$


## degree 2

Function: ${\displaystyle f(z)={\frac {z^{2}}{z^{2}-1}}}$


maxima

Maxima 5.41.0 http://maxima.sourceforge.net
using Lisp GNU Common Lisp (GCL) GCL 2.6.12
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) display2d:false;

(%o1) false
(%i2) f:z^2/(z^2-1);

(%o2) z^2/(z^2-1)
(%i3) dz:diff(f,z,1);

(%o3) (2*z)/(z^2-1)-(2*z^3)/(z^2-1)^2
(%i4) s:solve(f=z);

(%o4) [z = -(sqrt(5)-1)/2,z = (sqrt(5)+1)/2,z = 0]
(%i5) s:map('float,s);

(%o5) [z = -0.6180339887498949,z = 1.618033988749895,z = 0.0]
(%i6)


So fixed points ${\displaystyle z:f(z)=z}$ :

• z = -0.6180339887498949
• z = 1.618033988749895
• z = 0.0

### The Bubble bath Julia set

${\displaystyle f(z)={\frac {1-z^{2}}{z^{2}}}}$

The derivative wrt z is

${\displaystyle f'(z)={\frac {-2}{z^{3}}}}$

The Julia set for f is called the Bubble Bath Julia set.[18] It is called the Bubble Bath for its visual similarity to a tub of bubbles.

Function f is defined for all z in the Riemann sphere ${\displaystyle {\hat {\mathbf {C} }}}$  = it is defined on the whole Riemann sphere

${\displaystyle f:{\hat {\mathbf {C} }}\to {\hat {\mathbf {C} }}}$

• The Fatou set of f is the basin of attraction of the 3-cycle consisting of the points 0, −1, and infinity. It is the only one attracting cycle and it is superattracting
• The Julia set J(f) is the set of points whose orbits are not attracted to the above 3-cycle
• the only critical points of f are:
• z = 0 because it is the pole of order 3 of d(z), the zero of 1/d(z)
• z = infinity because it is the zero of function d(z)

Maxima CAS code :

kill(all);
remvalue(all);
display2d:false;

define(f(z), (1 -z^2)/(z^2));

(%o3) f(z):=(1-z^2)/z^2
define( d(z), ratsimp(diff(f(z),z,1)));

(%o13) d(z):=-2/z^3
(%i14) limit(d(z),z,infinity);

(%o14) 0
(%i15) limit(d(z),z,0);

(%o15) infinity


(%i2) f(-1);
(%o2)                                  0
(%i3) limit(d(z),z,0);
(%o3)                             limit  d(z)
z -> 0
(%i4) limit(f(z),z,0);
(%o4)                                 inf
(%i5) limit(f(z),z,inf);
(%o5)                                 - 1


Satbility of periodic cycle:

kill(all);
display2d:false;
ratprint : false; /* remove "rat :replaced " */

define(f(z), (1 -z^2)/(z^2));

F(z0):= block(
[z],
if is(z0 = 0) then 	z: limit(f(z),z,0)
elseif is(z0 = infinity) then z: limit(f(z),z,infinity)
elseif is(z0 = inf) then z: limit(f(z),z,inf)
else z:f(z0),

return(z)
)$define( dz(z), ratsimp(diff(f(z),z,1))); Dz(z0) := block( [m,z], if is(z0 = 0) then m: limit(dz(z),z,0) elseif is(z0 = infinity) then m: limit(dz(z),z,infinity) elseif is(z0 = inf) then m: limit(dz(z),z,inf) else m:dz(z0), return(m) )$

GiveStability(z0, p):=block(
[z,d],

/* initial values */
d : 1,
z : z0,

for i:1 thru p step 1 do (

d : Dz(z)*d,
z: F(z)
/*print("i = ", 0, "  d =",d, "  z = ", z)*/
),

return (cabs(d))
)$GiveStability(-1,3);  See also : ### degree 2 by Michael Becker #### 1 period 4 basin Julia set for f(z)=(z2+a) over (z2+b) a=-0.2+0.7i , b=0.917 ## degree 3 The components of the map ${\displaystyle f(z)=z-(z^{3}-1)/3z^{2}}$ contain the attracting points that are the solutions to ${\displaystyle z^{3}=1}$ . This is because the map is the one to use for finding solutions to the equation ${\displaystyle z^{3}=1}$ by Newton–Raphson formula. The solutions must naturally be attracting fixed points. ### degree 3 by Michael Becker #### 1 period 2 basin Julia set f(z)=1 over z3+z*(-3-3*I) ${\displaystyle f(z)={\frac {1}{z^{3}+z*(-3-3*I)}}}$ 2 critical points : { -0.4550898605622273*I -1.098684113467809, 0.4550898605622273*I+1.098684113467809}; Both critical points tend to the periodic cycle. There is only one attractive period cycle : period 2 cycle = {0, infinity}. Whole plane ( sphere) is a basin of attraction of period 2 cycle ( which is divided into 2 components ). Julia set is a boundary. #### 2 period 2 basins Function  ${\displaystyle f(z)={\frac {1}{z^{3}+a*z+b}}}$  where • a = 2.099609375 • b = 0.349609375 Derivative:  d(z):=-(3*z^2+2.099609375)/(z^3+2.099609375*z+0.349609375)^2  Critical points:  [-0.8365822085525526*%i,0.8365822085525526*%i]  One can check it also using Wolfram Alpha  solve (3*z^2+2.099609375)/(z^3+2.099609375*z+0.349609375)^2=0  the result:  z = ± (5/16)* i* sqrt(43/6))  These are 2 finite critical points. Infinity is a critical point too, as the 1st derivative's denominator degree is strictly greater than the numerator's. In numerical computations one can use the critical value (an image of critical point)  ${\displaystyle f(\infty )=0}$  There are two period 2 cycle: • { +0.4101296722285255 +0.5079485669960778*I , +0.4101296722285255 -0.5079485669960778*I }; • { +1.6890328811664648 +0.0000000000000000*I , +0.1147519899962205 +0.0000000000000000*I }; Both finite critical points fall into first cycle. Infinity ( or it's image zero) falls into the second cycle ( on the horizontal axis) Infinity is not a fixed point remvalue(all); display2d:false; define(f(z), 1/(z^3+ 2.099609375*z + 0.349609375)); (%i5)limit(f(z),z,infinity); (%o5) 0 (%i6) limit(f(z),z,0); (%o6) 2.860335195530726  ## degree 5 ### by L. Javier Hernandez Paricio the rational map ${\displaystyle h(z)={\frac {1+4z^{5}}{5z^{4}}}}$ has six fixed points: • ∞ ( repelling) • −0,809017 − 0,587785i • −0,809017 +0,587785i • 0,309017 − 0,951057i • 0,309017 + 0,951057i • 1 the basin of an end point associated to a fixed point (6= ∞) of f is the same that the attraction basin of the Newton-Raphson numerical method when it is applied to find the roots of the equation ${\displaystyle z^{5}-1=0}$ [19] ## degree 6 Julia set of rational function f(z)=z^2(3 − z^4 ) over 2.png The Julia set of the degree 6 function f :[20] ${\displaystyle f(z)=z^{2}{\frac {3-z^{4}}{2}}}$ There are 3 superattracting fixed points at : • z = 0 • z = 1 • z = ∞ All other critical points are in the backward orbit of 1. How to compute iteration : z:x+y*%i; z1:z^2*(3-z^4)/2; realpart(z1); ((x^2−y^2)*(−y^4+6*x^2*y^2−x^4+3)−2*x*y*(4*x*y^3−4*x^3*y))/2 imagpart(z1); (2*x*y*(−y^4+6*x^2*y^2−x^4+3)+(x^2−y^2)*(4*x*y^3−4*x^3*y))/2  Find fixed points using Maxima CAS : z1:z^2*(3-z^4)/2; s:solve(z1=z); s:float(s);  result : [z=−1.446857247913871,z=.7412709105660023,z=−1.357611535209976*%i−.1472068313260655,z=1.357611535209976*%i−.1472068313260655,z=1.0,z=0.0]  check multiplicities of the roots : multiplicities; [1,1,1,1,1,1]   z1:z^2*(3-z^4)/2; s:solve(z1=z)$
s:map(rhs,s)$f:z1; k:diff(f,z,1); define(d(z),k); m:map(d,s)$
m:map(abs,m)\$
s:float(s);
m:float(m);


Result : there are 6 fixed point 2 of them are supperattracting ( m=0 ), rest are repelling ( m>1 ):

 [−1.446857247913871,.7412709105660023,−1.357611535209976*%i−.1472068313260655,1.357611535209976*%i−.1472068313260655,1.0,0.0]
[14.68114348748323,1.552374536603988,10.66447061028112,10.66447061028112,0.0,0.0]


Critical points :

[%i,−1.0,−1.0*%i,1.0,0.0]


## degree 9 by Michael Becker

Julia set for f(z)=z2 over (z9-z+0.025)