Standard equations

Equation for periodic point for period ${\displaystyle p\,}$ is :

${\displaystyle \ F_{p}(z,c)=0\,}$

so in Maxima :

for i:1 thru 4 step 1 do disp(i,expand(Fn(i,z,c)=0));
1
z^2-z+c=0
2
z^4+2*c*z^2-z+c^2+c=0
3
z^8+4*c*z^6+6*c^2*z^4+2*c*z^4+4*c^3*z^2+4*c^2*z^2-z+c^4+2*c^3+c^2+c=0
4
z^16+8*c*z^14+28*c^2*z^12+4*c*z^12+56*c^3*z^10+24*c^2*z^10+70*c^4*z^8+60*c^3*z^8+6*c^2*z^8+2*c*z^8+56*c^5*z^6+80*c^4*z^6+
24*c^3*z^6+8*c^2*z^6+28*c^6*z^4+60*c^5*z^4+36*c^4*z^4+16*c^3*z^4+4*c^2*z^4+8*c^7*z^2+24*c^6*z^2+24*c^5*z^2+16*c^4*z^2+
8*c^3*z^2- z+c^8+4*c^7+6*c^6+6*c^5+5*c^4+2*c^3+c^2+c=0

Degree of standard equations is ${\displaystyle 2^{p}\,}$

These equations give periodic points for period p and its divisors. It is because these polynomials are product of lower period polynomials ${\displaystyle G_{p}\,}$:

for i:1 thru 4 step 1 do disp(i,factor(Fz(i,z,c)));
1
z^2-z+c
2
(z^2-z+c)*(z^2+z+c+1)
3
(z^2-z+c)*(z^6+z^5+3*c*z^4+z^4+2*c*z^3+z^3+3*c^2*z^2+3*c*z^2+z^2+c^2*z+2*c*z+z+c^3+2*c^2+c+1)
4
(z^2-z+c)*(z^2+z+c+1)(z^12+6*c*z^10+z^9+15*c^2*z^8+3*c*z^8+4*c*z^7+20*c^3*z^6+12*c^2*z^6+z^6+6*c^2*z^5+2*c*z^5+15*c^4*z^4+18*c^3*z^4+3*
c^2*z^4+4*c*z^4+4*c^3*z^3+4*c^2*z^3+z^3+6*c^5*z^2+12*c^4*z^2+6*c^3*z^2+5*c^2*z^2+c*z^2+c^4*z+2*c^3*z+c^2*z+2*c*z+c^6+
3*c^5+3*c^4+3*c^3+2*c^2+1)

so

${\displaystyle \ F_{1}(z,c)=z^{2}-z+c=G_{1}\,}$

${\displaystyle \ F_{2}(z,c)=(z^{2}-z+c)*(z^{2}+z+c+1)=G_{1}*G_{2}\,}$

${\displaystyle \ F_{3}(z,c)=(z^{2}-z+c)*(z^{6}+z^{5}+3*c*z^{4}+z^{4}+2*c*z^{3}+z^{3}+3*c^{2}*z^{2}+3*c*z^{2}+z^{2}+c^{2}*z+2*c*z+z+c^{3}+2*c^{2}+c+1)=G_{1}*G_{3}\,}$

${\displaystyle \ F_{4}(z,c)=G_{1}*G_{2}*G_{4}\,}$

${\displaystyle \ F_{5}(z,c)=G_{1}*G_{5}\,}$

Standard equations can be reduced to equations ${\displaystyle G_{p}\,}$ giving periodic points for period p without its divisors.

See also : prime factors in wikipedia

Reduced equations

Reduced equation is

${\displaystyle G_{p}(z,c)=0\,}$

so in Maxima gives :

for i:1 thru 5 step 1 do disp(i,expand(G[i,z,c]=0));
1
z^2-z+c=0
2
z^2+z+c+1=0
3
z^6+z^5+3*c*z^4+z^4+2*c*z^3+z^3+3*c^2*z^2+3*c*z^2+z^2+c^2*z+2*c*z+z+c^3+2*c^2+c+1=0
4
z^12+6*c*z^10+z^9+15*c^2*z^8+3*c*z^8+4*c*z^7+20*c^3*z^6+12*c^2*z^6+z^6+6*c^2*z^5+2*c*z^5+15*c^4*z^4+18*c^3*z^4+
3*c^2*z^4+4*c*z^4+4*c^3*z^3+4*c^2*z^3+z^3+6*c^5*z^2+12*c^4*z^2+
6*c^3*z^2+5*c^2*z^2+c*z^2+c^4*z+2*c^3*z+c^2*z+2*c*z+c^6+3*c^5+3*c^4+3*c^3+2*c^2+1=0

Fixed points ( period = 1 )

Equation defining fixed points :

 z^2-z+c = 0

in Maxima :

Define c value

(%i1) c:%i;
(%o1) %i

Compute fixed points

(%i2) p:float(rectform(solve([z*z+c=z],[z])));
(%o2) [z=0.62481053384383*%i-0.30024259022012,z=1.30024259022012-0.62481053384383*%i]

Find which is repelling :

abs(2*rhs(p[1]));

Show that sum of fixed points is 1

(%i10) p:solve([z*z+c=z], [z]);
(%o10) [z=-(sqrt(1-4*%i)-1)/2,z=(sqrt(1-4*%i)+1)/2]
(%i14) s:radcan(rhs(p[1]+p[2]));
(%o14) 1

Draw points :

(%i15) xx:makelist (realpart(rhs(p[i])), i, 1, length(p));
(%o15) [-0.30024259022012,1.30024259022012]
(%i16) yy:makelist (imagpart(rhs(p[i])), i, 1, length(p));
(%o16) [0.62481053384383,-0.62481053384383]
(%i18) plot2d([discrete,xx,yy], [style, points]);

One can add point z=1/2 to the lists:

(%i31) xx:cons(1/2,xx);
(%o31) [1/2,-0.30024259022012,1.30024259022012]
(%i34) yy:cons(0,yy);
(%o34) [0,0.62481053384383,-0.62481053384383]
(%i35) plot2d([discrete,xx,yy], [style, points]);

In C :

complex double GiveFixed(complex double c){ /*

Equation defining fixed points : z^2-z+c = 0 z*2+c = z z^2-z+c = 0

coefficients of standard form ax^2+ bx + c

	a = 1 , b = -1 , c = c

The discriminant d is

d=b^2- 4ac d = 1 - 4c

alfa =  (1-sqrt(d))/2 

• /

complex double d = 1-4*c; complex double z = (1-csqrt(d))/2.0; return z;

}

</source>

double complex GiveAlfaFixedPoint(double complex c)
{
  // Equation defining fixed points : z^2-z+c = 0
  // a = 1, b = -1 c = c
  double dx, dy; //The discriminant is the  d=b^2- 4ac = dx+dy*i
  double r; // r(d)=sqrt(dx^2 + dy^2)
  double sx, sy; // s = sqrt(d) = sqrt((r+dx)/2)+-sqrt((r-dx)/2)*i = sx + sy*i
  double ax, ay; // alfa = ax+ay*I
 
 // d=1-4c = dx+dy*i
 dx = 1 - 4*creal(c);
 dy = -4 * cimag(c);
 // r(d)=sqrt(dx^2 + dy^2)
 r = sqrt(dx*dx + dy*dy);
 //sqrt(d) = s =sx +sy*i
 sx = sqrt((r+dx)/2);
 sy = sqrt((r-dx)/2);
 // alfa = ax +ay*i = (1-sqrt(d))/2 = (1-sx + sy*i)/2
 ax = 0.5 - sx/2.0;
 ay =  sy/2.0;

return ax+ay*I;
}

period 2 points

Using non-reduced equation in Maxima CAS :

(%i2) solve([(z*z+c)^2+c=z], [z]);
(%o2) [z=-(sqrt(-4*c-3)+1)/2,z=(sqrt(-4*c-3)-1)/2,z=-(sqrt(1-4*c)-1)/2,z=(sqrt(1-4*c)+1)/2]

Show that z1+z2 = -1

(%i4) s:radcan(rhs(p[1]+p[2]));
(%o4) -1

Show that z2+z3=1

(%i6) s:radcan(rhs(p[3]+p[4]));
(%o6) 1

Reduced equation :

 ${\displaystyle z^{2}+z+c+1=0}$

so standard coefficient names of single-variable quadratic polynomial^[6]

${\displaystyle ax^{2}+bx+c}$

are :

 a = 1
 b = 1
 c = c+1

One can use the quadratic formula^[7] to find the exact solution ( roots of quadratic equation):^[8]

${\displaystyle z_{1,2}={\frac {-b\pm {\sqrt {d}}}{2a}}={\frac {-1\pm {\sqrt {d}}}{2}}}$

where :

${\displaystyle d=b^{2}-4ac=1-4(c+1)=-4c-3}$

Another formula^[9] can be used to solve square root of complex number "

${\displaystyle s={\sqrt {d}}={\sqrt {r}}{\frac {d+r}{|d+r|}}}$

where :

${\displaystyle r=|d|}$

so

Period 3 points

In Maxima CAS

kill(all);
remvalue(z);

f(z,c):=z*z+c; /* Complex quadratic map */
 finverseplus(z,c):=sqrt(z-c);
 finverseminus(z,c):=-sqrt(z-c);

/* */
fn(p, z, c) :=
  if p=0 then z
  elseif p=1 then f(z,c)
  else f(fn(p-1, z, c),c);

/*Standard polynomial F_p \, which roots are periodic z-points of period p and its divisors */
F(p, z, c) := fn(p, z, c) - z ;

/* Function for computing reduced polynomial G_p\, which roots are periodic z-points of period p without its divisors*/
G[p,z,c]:=
block(
[f:divisors(p),
t:1], /* t is temporary variable = product of Gn for (divisors of p) other than p */
f:delete(p,f), /* delete p from list of divisors */
if p=1
then return(F(p,z,c)),
for i in f do 
 t:t*G[i,z,c],
g: F(p,z,c)/t,
return(ratsimp(g))
)$

 GiveRoots(g):=
 block(
 [cc:bfallroots(expand(%i*g)=0)],
 cc:map(rhs,cc),/* remove string "c=" */
 cc:map('float,cc),
 return(cc)
  )$

compile(all);

k:G[3,z,c]$
c:1;
s:GiveRoots(ev(k))$
s;

algorithm

For given ( and fixed = constant values):

• period p
• parameter c
• initial aproximation ${\displaystyle z_{0}}$ = initial value of Newton iteration

find one periodic point ${\displaystyle z}$ of complex quadratic polynomial ${\displaystyle f}$:

  ${\displaystyle z:f^{p}(z)=z}$

where

  ${\displaystyle f(z)=z^{2}+c}$

Note that periodic point ${\displaystyle z}$ is only one of p points of periodic cycle.

To find periodic point one have to solve equation ( = find a single zero of a function F):

${\displaystyle f^{p}(z)-z=0}$

Lets call left side of this equation (arbitrary name) function ${\displaystyle F}$:

${\displaystyle F_{p}(z)=f^{p}(z)-z}$

Derivative of function ${\displaystyle F}$ with respect to variable z :

 ${\displaystyle F'_{p}(z)=f^{p'}(z)-1}$

It can be checked using Maxima CAS:

(%i1) f:z*z+c;
(%o1)                                                                                                            z^2  + c
(%i2) F:f-z;
(%o2)                                                                                                          z^2  - z + c
(%i3) diff(F,z,1);
(%o3)                                                                                                           2 z - 1

Newton iteration is :

${\displaystyle z_{n+1}=z_{n}-{\frac {F_{p}(z_{n})}{F'_{p}(z_{n})}}=z_{n}-{\frac {f_{p}(z_{n})-z_{n}}{f'_{p}(z_{n})-1}}=z_{n}-N_{p}(z_{n})}$

gives sequence of ${\displaystyle z_{n}}$ points:

${\displaystyle z_{0}}$

${\displaystyle z_{1}=z_{0}-N_{p}(z_{0})}$

${\displaystyle z_{2}=z_{1}-N_{p}(z_{1})}$

${\displaystyle ...}$

${\displaystyle z_{n+1}=z_{n}-N_{p}(z_{n})}$

where:

• n is a number of Newton iteration
• p is a period ( fixed positive integer)
• ${\displaystyle z_{0}}$ is an initial aproximation of periodic point
• ${\displaystyle z_{n}}$ is a final aproximation of periodic point
• ${\displaystyle f_{p}(z_{n})}$ is computed using quadratic iteration with ${\displaystyle z_{n}}$ as a starting point
• ${\displaystyle f'_{p}(z_{n})}$ is a value of first derivative . It computed using iteration
• ${\displaystyle N_{p}}$ is called Newton function. Sometimes different and shorter notation is used : ${\displaystyle N_{p}(z_{n})=z_{n}-{\frac {F_{p}(z_{n})}{F'_{p}(z_{n})}}}$

stoping criteria

• ${\displaystyle \epsilon _{succes}}$ : predefined accuracy threshold ( succes = converged )^[10]
• ${\displaystyle n_{max}}$ : maximal number of iterations ( fail to converge)

${\displaystyle \epsilon =|z_{n+1}-z_{n}|}$

Usually one starts with :

 ${\displaystyle \epsilon _{succes}=10^{-16}}$

for long double floating point number format

 ${\displaystyle n_{max}=10^{d}}$

precision

Precision is proportional to the degree of polynomial F, which is proportional to the period p

 ${\displaystyle d=d(p)=2^{p-1}}$

so

 ${\displaystyle period\to degree\to accuracy\to numberformat}$

IEEE 754 - 2008	Common name	C++ data type	Base ${\displaystyle b}$	Precision ${\displaystyle r}$	Machine epsilon = ${\displaystyle b^{-(r-1)}}$
binary16	half precision	short	2	11	2−10 ≈ 9.77e-04
binary32	single precision	float	2	24	2−23 ≈ 1.19e-07
binary64	double precision	double	2	53	2−52 ≈ 2.22e-16
	extended precision	_float80[11]	2	64	2−63 ≈ 1.08e-19
binary128	quad(ruple) precision	_float128[11]	2	113	2−112 ≈ 1.93e-34

where :

• precision = number of fraction's binary digits
• accuracy < machine epsilon

 precision =  -log2(accuracy)
 

Relation between accuracy and degree d:^[12]

• accuracy = 1/d
• accuracy = 1/(2 d log d)

Compare:

• precision for computing centers with Maxima CAS
• precision for zoom

number of iterations

code

• cpp from mandel

/* 
gcc -std=c99 -Wall -Wextra -pedantic -O3 m.c -lm
./a.out
*/

#include <complex.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

const double pi = 3.141592653589793;

static inline double cabs2(complex double z) {
  return (creal(z) * creal(z) + cimag(z) * cimag(z));
}

/* 
newton function 
 N(z) = z - (fp(z)-z)/fp'(z)) 
 used for computing periodic point 
 of complex quadratic polynomial
 f(z) = z*z +c

*/

complex double N( complex double c, complex double zn , int pMax, double er2){

  
complex double z = zn;
complex double d = 1.0; /* d = first derivative with respect to z */
int p;

for (p=0; p < pMax; p++){

   
   d = 2*z*d; /* first derivative with respect to z */
   z = z*z +c ; /* complex quadratic polynomial */
   if (cabs(z) >er2) break;
    
}

      z = zn - (z - zn)/(d - 1) ;

return z;
}

/* 
compute periodic point 
of complex quadratic polynomial
using Newton iteration = numerical method

*/

complex double GivePeriodic(complex double c, complex double z0, int period, double eps2, double er2){

complex double z = z0;
complex double zPrev = z0; // prevoiuus value of z
int n ; // iteration
const int nMax = 64;

for (n=0; n<nMax; n++) {
     
    z = N( c, z, period, er2);
    if (cabs2(z - zPrev)< eps2) break;
    
    zPrev = z; }

return z;
}

int main() {

 
 const double ER2  = 2.0 * 2.0; // ER*ER
 const double EPS2 = 1e-100 * 1e-100; // EPS*EPS

  int period = 3;
  complex double c = -0.12565651+0.65720*I ; // https://commons.wikimedia.org/wiki/File:Fatou_componenets_3.png
  complex double z0 = 0.0 ; // initial aproximation for Newton method; usuallly crotical point
 
  complex double zp; // periodic point
  
 
  
  
  
  
  zp = GivePeriodic( c , z0, period, EPS2, ER2); //
  
  
  
 
  
  
  printf (" c = %f ; %f \n",  creal(c), cimag(c)); 
  printf (" period = %d  \n",  period); 
  printf (" z0 = %f ; %f \n",  creal(z0), cimag(z0));
  printf (" zp = %.16f ; %.16f \n",  creal(zp), cimag(zp));

  return 0;
}

/*

batch file for Maxima CAS

find periodic point of iterated function

solve : 
c0: -0.965 + 0.085*%i$ //     period of critical orbit = 2
2						1 ( alfa ) repelling 						2						1 ( beta) repelling
[0.08997933216560844 %i - 0.972330689471866, 	0.0385332450804691 %i - 0.6029437025417168, 	(- 0.0899793321656081 %i) - 0.02766931052813359, 	1.602943702541716 - 0.03853324508046944 %i] // periodic z points
[1.952970301777068, 				1.208347498712429, 				0.1882750218384916, attr					3.206813573935709] // stability ( 1)
[0.3676955262170045, 				1.460103677644584, 				0.3676955262170032, attr					10.28365329797831] // stability (2)

Newton : 
z0: 0 				 zn : - 0.08997933216560859 %i - 0.02766931052813337
z0: -0.86875-0.03125*%i 	 zn: 0.08997933216560858*%i-0.9723306894718666

(%o21) (-0.08997933216560859*%i)-0.02766931052813337
(%i22) zn2:Periodic(period,c0,(-0.86875)-0.03125*%i,ER,eps)
(%o22) 0.08997933216560858*%i-0.9723306894718666

*/

kill(all); /* if you want to delete all variables and functions. Careful: this does not completely reset the environment to its initial state. */
remvalue(z);
reset(); /*  Resets many (global) system variables */
ratprint : false; /* remove "rat :replaced  " */
display2d:false$ /* display in one line */

mode_declare(ER, real)$
ER: 1e100$
mode_declare(eps,real)$
eps:1e-100$
mode_declare( c0, complex)$
c0: -0.965 + 0.085*%i$ /*     period of critical orbit = 2*/
mode_declare(z0, complex)$
z0:0$
mode_declare(period, integer)$
period:2$

/* ----------- functions  ------------------------ */

/*
 https://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/qpolynomials
complex quadratic polynomial
*/

f(z,c):=z*z+c$

/* iterated map */

fn(p, z, c) :=
  if p=0 then z
  elseif p=1 then f(z,c)
  else f(fn(p-1, z, c),c);
  
  
Fn(p,z,c):=  fn(p, z, c) - z$

  
  
/* find periodic point using solve = symbolic method */  
S(p,c):=
(
s: solve(Fn(p,z,c)=0,z),
s: map( rhs, s),
s: float(rectform(s)))$

/* newton function : N(z) = z - (fp(z)-z)/(f'(z)-1) */

N(zn, c, iMax, _ER):=block(

[z, d], /* d = first derivative with respect to z */
/* mode_declare([z,zn, c,d], complex), */

d  : 1+0*%i,
z  : zn,

catch( for i:1 thru iMax step 1 do (

 
 /*  print("i =",i,", z=",z, ", d = ",d),   */

  
  d : float(rectform(2*z*d)), /* first derivative with respect to z */
  z : float(rectform(z^2+c)), /* complex quadratic polynomial */
  
  
  if ( cabs(z)> _ER)  /* z is escaping */
     then  throw(z)

  )), /* catch(for i */
  
 
   if (d!=0) /* if not : divide by 0 error */ 
     then  z:float(rectform( zn - (z - zn)/(d-1))),
      
   
 return(z) /*  */

)$

/* compute periodic point 
  of complex quadratic polynomial
using Newton iteration = numerical method  */
Periodic(p, c, z0, _ER, _eps) :=block
(
 [z, n, temp], /* local variables */

 if (p<1) 
   then print("to small p ")
   else (

 	n:0,
 	z: z0,  /*  */
 	temp:0,
 	/* print("n = ", n, "zn = ", z ),   */
 
 
        catch(
 	  while (1) do (
 
 		z : N(z, c, p, _ER),   
 		n:n+1,
 		
   	       /*  print("n = ", n, "zn = ", z ),  */
   		
   		
   		if(cabs(temp-z)< _eps) then  throw(z),
   		temp:z,
 
 		if(n>64) then throw(z) ))), /* catch(while) */
 
 	

 
 return(z) 
 
  
  
  )$
  
  
 compile(all)$

/* ------------------ run ---------------------------------------- */

zn1: Periodic(period, c0, z0, ER, eps);   
zn2 : Periodic(period, c0,-0.868750000000000-0.031250000000000*%i, ER, eps);

(%i21) zn1:Periodic(period,c0,z0,ER,eps)
(%o21) (-0.08997933216560859*%i)-0.02766931052813337
(%i22) zn2:Periodic(period,c0,(-0.86875)-0.03125*%i,ER,eps)
(%o22) 0.08997933216560858*%i-0.9723306894718666

test

"Viete tests^[14] on the roots found:

• if p is any polynomial of degree d and normalized (i.e. the leading coefficient is 1), then the sum of all roots of p must be equal to the negative of the coefficient of degree d − 1.
• Moreover, the product of the roots must be equal to (−1)^d times the constant coefficient. If one root vanishes, then the product of the remaining d − 1 roots is equal to (−1)^(d−1) times the linear term. "

${\displaystyle z_{1}+z_{2}+\dots +z_{d-1}+z_{d}=-{\dfrac {a_{d-1}}{a_{d}}}}$

${\displaystyle z_{1}z_{2}\dots z_{d}=(-1)^{d}{\dfrac {a_{0}}{a_{d}}}}$

"The sum of all roots should be 0 = the coefficient a of the degree d − 1 term of our polynomial. "^[15]

${\displaystyle a_{d-1}=0}$

${\displaystyle a_{d}=1}$

The constant term is a0. Its value depends on the period (degree).

/*

batch file for Maxima CAS

find all periodic point of iterated function
using solve

*/

kill(all);         /* if you want to delete all variables and functions. Careful: this does not completely reset the environment to its initial state. */
remvalue(all);
reset();           /*  Resets many (global) system variables */
ratprint : false;  /* remove "rat :replaced  " */
display2d:false$   /* display in one line */

mode_declare(c0, complex)$
c0: -0.965 + 0.085*%i$ /*     period of critical orbit = 2*/
mode_declare(z, complex)$

mode_declare(period, integer)$
period:2  $

/* ----------- functions  ------------------------ */

/*
 https://en.wikibooks.org/wiki/Fractals/Iterations_in_the_complex_plane/qpolynomials
complex quadratic polynomial
*/

f(z,c):=z*z+c$

/* iterated map */

fn(p, z, c) :=
  if p=0 then z
  elseif p=1 then f(z,c)
  else f(fn(p-1, z, c),c);
  
  
Fn(p,z,c):=  fn(p, z, c) - z$

  
  
/* find periodic point using solve = symbolic method */  
S(p,c):=
(
s: solve(Fn(p,z,c)=0,z),
s: map( rhs, s),
s: float(rectform(s)))$

/* gives a list of coefficients */
GiveCoefficients(P,x):=block
( [a, degree],
  P:expand(P), /* if not expanded */
  degree:hipow(P,x),
  a:makelist(coeff(P,x,degree-i),i,0,degree));

GiveConstCoefficient(period, z, c):=(
 [l:[]],
 l :  GiveCoefficients( Fn(period,z,c), z),
 l[length(l)]
)$

/*

https://en.wikipedia.org/wiki/Vieta%27s_formulas
sum of all roots , here it should be zero

*/

VietaSum(l):=(
abs(sum(l[i],i, 1, length(l)))

)$

/*

it should be = (-1)^d* a0 = constant coefficient of Fn

*/

VietaProduct(l):=(
rectform(product(l[i],i, 1, length(l)))
)$

compile(all);

s: S(period, c0)$

vs: VietaSum(s);

vp : VietaProduct(s);

a0: GiveConstCoefficient( period, z,c0);

 z^4+2*c*z^2-z+c^2+c = 0

${\displaystyle d=2^{p-1}=2^{2-1}=2^{2}=4}$

[0.08997933216560844*%i-0.972330689471866, 0.0385332450804691*%i-0.6029437025417168, (-0.0899793321656081*%i)-0.02766931052813359, 1.602943702541716-0.03853324508046944*%i]

 ${\displaystyle sum=6.938893903907228e-18\approx 0}$

  ${\displaystyle product=(-0.07904999999999943*\%i)-0.04100000000000011\approx a_{0}=c^{2}+c\approx (-0.07905*\%i)-0.04100000000000004}$