Fractals/Iterations in the complex plane/qpolynomials

< Fractals

FormsEdit

z^2+cEdit

Complex quadratic polynomial of the form :

belongs to the class of the functions :

How to compute iterationEdit

In Maxima CAS :

(%i28) z:zx+zy*%i;
(%o28) %i*zy+zx
(%i37) c:cx+cy*%i;
(%o37) %i*cy+cx
(%i38) realpart(z^2+c);
(%o38) -zy^2+zx^2+cx
(%i39) imagpart(z^2+c);
(%o39) 2*zx*zy+cy

Critical pointEdit

Critical orbits for various parabolic parameters on boundary of Main component of Mandelbrot set

A critical point of is a point in the dynamical plane such that the derivative vanishes :

Since

implies

One can see that :

  • the only (finite) critical point of is the point
  • critical point is the same for all c parameters

is an initial point for Mandelbrot set iteration.[2]

Dynamic planeEdit

1/2Edit

Julia set for parabolic parameter on the end fo 1/2 internal ray of main component of Mandelbrot set

First compute muliplier m of the fixed points using internal angle p/q and Maxima CAS:

(%i1) p:1$
(%i2) q:2$
(%i3) m:exp(2*%pi*%i*p/q);
(%o3)                                 - 1

Now compute parameter c of the function :

(%i1) GiveC(t,r):=
(
 [w,c],
 /* point of  unit circle   w:l(internalAngle,internalRadius); */
 w:r*%e^(%i*t*2*%pi),  /* point of circle */
 c:w/2-w*w/4, /* point on boundary of period 1 component of Mandelbrot set */
 float(rectform(c))    
)$

(%i3) c:GiveC(1/2,1);
(%o3) −0.75


Find fixed points z

(%i4) z1:z^2+c;
(%o4) z^2−0.75
(%i2) f:z^2+c;
(%o2)                              z^2  - 0.75
(%i3) d:diff(f,z,1);
(%o3)                                 2 z
(%i6) s:solve(z1=z);
(%o6)                              [z = 3/2, z = -1/2]
(%i7) s:map(rhs,s);
(%o7)                             [z = 3/2, z = -1/2]
(%i8) z:s[1];
(%o8)                                  3/2
(%i9) abs(float(rectform(ev(d))));
(%o9)                                 3.0
(%i10) z:s[2];
(%o10)                                - 1/2
(%i11) abs(float(rectform(ev(d))));
(%o11)                                1.0

So z=-1/2 is a parabolic fixed point.

z^2 + m*zEdit

Complex quadratic polynomial of the form :

which has an indifferent fixed point[3] with multiplier[4]

at the origin[5][6]

belongs to the class of the functions :

How to compute iterationEdit

In Maxima CAS :

(%i1) z:zx+zy*%i;
(%o1) %i*zy+zx
(%i2) m:mx+my*%i;
(%o2) %i*my+mx
(%i3) z1:z^2+m*z;
(%o3) (%i*zy+zx)^2+(%i*my+mx)*(%i*zy+zx)
(%i4) realpart(z1);
(%o4) -zy^2-my*zy+zx^2+mx*zx
(%i5) imagpart(z1);
(%o5) 2*zx*zy+mx*zy+my*zx

Critical pointEdit

A critical point of is a point in the dynamical plane such that the derivative vanishes :

Since

implies

One can see that :

  • critical point is related with m value and have to be computed for every m parameters

is an initial point for Mandelbrot set iteration.

Parameter plane lambda

parameter planeEdit

period 1 componentsEdit

(%i1) e1:z^2+m*z=z;
(%o1) z^2+m*z=z
(%i2) e2:2*z+m=w;
(%o2) 2*z+m=w
(%i3)  s:eliminate ([e1,e2], [z]);
(%o3) [-(m-w)*(w+m-2)]
(%i4) s:solve([s[1]], [m]);
(%o4) [m=2-w,m=w]

It means that there are 2 components of period 1 :

  • one with radius=1 and center=0 ( m=w )
  • second with radius=1 and center= -2 ( m=2-w)


How to compute boundary points of first component :

(%i1) m:exp(2*%pi*%i*p/q);
(%o1) %e^((2*%i*%pi*p)/q)
(%i2) realpart(m);
(%o2) cos((2*%pi*p)/q)
(%i3) imagpart(m);
(%o3) sin((2*%pi*p)/q)

Dynamic planeEdit

1/1Edit

Julia set for f(z) = z^2+z or f(z) = z^2 + 1/4 or f(z)= z-z^2
Domains for Fatou coordinate
Orbits of some points inside Julia set are shown ( white points)

First compute parameter of the function :

p:1$
q:1$
m:exp(2*%pi*%i*p/q);

The result is m = 1 so f(z) = z^2 + z. Julia set is a cauliflower.

Compute fixed points :

(%i3) solve(z=z^2+z);
(%o3) [z=0]
(%i4) multiplicities;
(%o4) [2]


Find it's stability index ( abs(multiplier)) :

(%i1) f:z^2+z;
(%o1) z^2+z
(%i2) d:diff(f,z,1);
(%o2) 2*z+1
(%i7) z:0;
(%o7) 0
(%i8) abs(float(rectform(ev(d))));
(%o8) 1.0


Iteration :

f(z):= z^2+z;

fn(n, z) :=
  if n=0 then z
  elseif n=1 then f(z)
  else f(fn(n-1, z));

1/3Edit

Critical orbit for f(z)=z^2 + mz where p over q=1 over 3

First compute parameter of the function :

/* Maxima CAS session */
(%i1) p:1;
      q:3;
      m:exp(2*%pi*%i*p/q);
(%o1) 1
(%o2) 3
(%o3) (sqrt(3)*%i)/2-1/2
(%i9) float(rectform(m));
(%o9) 0.86602540378444*%i-0.5


Then find fixed points :

/* Maxima CAS session */
(%i10) f:z^2+m*z;
(%o10) z^2+((sqrt(3)*%i)/2-1/2)*z
(%i11) z1:f;
(%o11) z^2+((sqrt(3)*%i)/2-1/2)*z
(%i12) solve(z1=z);
(%o12) [z=-(sqrt(3)*%i-3)/2,z=0]
(%i13) multiplicities;
(%o13) [1,1]

Compute multiplier of the fixed point :

(%i23) d:diff(f,z,1);
(%o23) 2*z+(sqrt(3)*%i)/2-1/2

Check stability of fixed points :

(%i12) s:solve(z1=z);
(%o12) [z=-(sqrt(3)*%i-3)/2,z=0]
(%i20) s:map(rectform,s);
(%o20) [3/2-(sqrt(3)*%i)/2,0]
(%i21) s:map('float,s);
(%o21) [1.5-0.86602540378444*%i,0.0]
(%i24) z:s[1];
(%o24) 1.5-0.86602540378444*%i;
(%i31) abs(float(rectform(ev(d))));
(%o31) 2.645751311064591

It means that fixed point z=1.5-0.86602540378444*%i is repelling.

Second point z=0 is parabolic :

(%i33) z:s[2];
(%o33) 0.0
(%i34) abs(float(rectform(ev(d))));
(%o34) 1.0

Find critical point :

(%i1) solve(2*z+(sqrt(3)*%i)/2-1/2);
(%o1) [z=-(sqrt(3)*%i-1)/4]
(%i2) s:solve(2*z+(sqrt(3)*%i)/2-1/2);
(%o2) [z=-(sqrt(3)*%i-1)/4]
(%i3) s:map(rhs,s);
(%o3) [-(sqrt(3)*%i-1)/4]
(%i4) s:map(rectform,s);
(%o4) [1/4-(sqrt(3)*%i)/4]
(%i5) s:map('float,s);
(%o5) [0.25-0.43301270189222*%i]
(%i6) abs(s[1]);
(%o6) 0.5

1/2Edit

First compute m = muliplier of the fixed points = parameter of the function f using internal angle p/q and Maxima CAS:

(%i1) p:1$
(%i2) q:2$
(%i3) m:exp(2*%pi*%i*p/q);
(%o3)                                 - 1

so function f is :


How to compute iteration  ?

(%i29) z1;
(%o29)                              z^2  - z
(%i30) z:zx+zy*%i;
(%o30)                            %i zy + zx
(%i32) realpart(ev(z1));
(%o32)                         - zy^2  + zx^2  - zx
(%i33) imagpart(ev(z1));
(%o33)                           2 zx zy - zy

Then find fixed points of f :

(%i4) z1:z^2+m*z;
(%o4)                               z^2  - z
(%i5) zf:solve(z1=z);                                                  
(%o5)                           [z = 0, z = 2]
(%i6) multiplicities;
(%o6)                               [1, 1]

Stability of the fixed points :

(%i7) f:z1;
(%o7)                               z^2  - z
(%i8) d:diff(f,z,1);
(%o8)                               2 z - 1
(%i9) z:zf[1];   
(%o9)                                z = 0
(%i10) abs(ev(d));
(%o10)                         abs(2 z - 1) = 1
(%i11) z:zf[2];
(%o11)                               z = 2
(%i12) abs(ev(d));
(%o12)                         abs(2 z - 1) = 3
(%i13) 

So fixed point :

  • z=0 is parabolic ( stability index = 1)
  • z=2 is repelling ( stability indexs = 3 , greater then 1 )


Find critical point  :

(%i14) zcr:solve(d=0);
(%o14)                              [z = 1/2]
(%i15) multiplicities;
(%o15)                                [1]

Attracting vectors

Because q=2, thus we examine 2-th iteration of f :

(%i16) z1;
(%o16)                              z^2  - z
(%i17) z2:z1^2-z1;
(%o17)                        (z^2  - z)^2  - z^2  + z
(%i18) taylor(z2,z,0,20);
taylor: z = 2 cannot be a variable.
 -- an error. To debug this try: debugmode(true);
(%i19) remvalue(z);
(%o19)                                [z]
(%i20) z;
(%o20)                                 z
(%i21) taylor(z2,z,0,20);
(%o21)/T/                    z - 2 z^3  + z^4  + . . .

Next term after z is a :

so here :

  • degree of above term is k=3
  • number of attracting directions ( and petals) is n= k-1 = 2 ( also n = e*q)
  • the parabolic degeneracy e = n/q = 1
  • cooefficient of above term a = -2

Attracing vectore satisfy :

so here :

One can solve it in Maxima CAS :

(%i22)  s:solve(z^2=1/4);
(%o22)                         [z = - 1/2, z =1/2]
(%i23) s:map(rhs,s);
(%o23)                             [-1/2, 1/2]
(%i24) carg_t(z):=
 block(
 [t],
 t:carg(z)/(2*%pi),  /* now in turns */
 if t<0 then t:t+1, /* map from (-1/2,1/2] to [0, 1) */
 return(t)
)$
(%i25)  s:map(carg_t,s);
(%o25)                              [1/2, 0]

So attracting vectors are :

  • from to the origin
  • from to the origin


Critical point z=1/2 lie on attracting vector . Thus critical orbits tend straight to the origin under the iteration[7]

Repelling vectors satisfy :


so here :

One can solve it in Maxima CAS :

(%i26) s:solve(z^2=-1/4);
(%o26)                        [z = - %i/2, z = %i/2]
(%i27) s:map(rhs,s);
(%o27)                            [- %i/2, %i/2 ]
(%i28) s:map(carg_t,s);
(%o28)                              [3/4, 1/4]

1/7Edit

How to speed up computations ? Approximate by :

How to compute  :

(%i1) z:x+y*%i;
(%o1) %i*y+x
(%i2) z7:(245.4962434402444*%i-234.5808769813032)*z^8 + z;
(%o2) (245.4962434402444*%i-234.5808769813032)*(%i*y+x)^8+%i*y+x
(%i3) realpart(z7);
(%o3) -234.5808769813032*(y^8-28*x^2*y^6+70*x^4*y^4-28*x^6*y^2+x^8)-245.4962434402444*(-8*x*y^7+56*x^3*y^5-56*x^5*y^3+8*x^7*y)+x
(%i4) imagpart(z7);
(%o4) 245.4962434402444*(y^8-28*x^2*y^6+70*x^4*y^4-28*x^6*y^2+x^8)-234.5808769813032*(-8*x*y^7+56*x^3*y^5-56*x^5*y^3+8*x^7*y)+y

m*z*(1-z)Edit

Description

Critical pointsEdit

critical points :

  • z = 1/2
  • z = ∞

Parameter planeEdit

period 1 componentsEdit

(%i1) e1:m*z*(1-z)=z;
(%o1) m*(1-z)*z=z
(%i2) d:diff(m*z*(1-z),z,1);
(%o2) m*(1-z)-m*z
(%i3) e2:d=w;
(%o3) m*(1-z)-m*z=w
(%i4) s:eliminate ([e1,e2], [z]);
(%o4) [m*(m-w)*(w+m-2)]
(%i5) s:solve([s[1]], [m]);
(%o5) [m=2-w,m=w,m=0]

It means that there are 2 period 1 components :

  • discs of radius 1 and centre in 0
  • disc of radius 1 and centre = 2

z(1-mz)Edit

Dynamic planeEdit

"Note that each member of the family of quadratic polynomials


is parabolic since for each λ ∈ C \ {0}, the polynomial gλ has a parabolic fixed point 0 with miltiplicity 2 and the only finite critical point of is given by

 

which is contained in the basin of 0. The study of this family is too trivial since all its members are conjugate to


via Mobius transformations

 

and therefore all their Julia sets J(gλ) have the same Hausdorff dimension as

HD(J(z^2 +1/4)) ≈ 1.0812 

"[9]

z(1+ mz)Edit

dynamic planeEdit

z-z^2Edit

Description [10]


First compute m = muliplier of the fixed points = parameter of the function f using internal angle p/q and Maxima CAS:

(%i1) p:1$
(%i2) q:2$
(%i3) m:exp(2*%pi*%i*p/q);
(%o3)                                 - 1

so function f is :


How to compute iteration  ?

Find it using Maxima CAS :

(%i1) z:x+y*%i;
(%o1) %i*y+x
(%i2) z1:z-z^2;
(%o2) −(%i*y+x)^2+%i*y+x
(%i3) realpart(z1);
(%o3) y^2−x^2+x
(%i4) imagpart(z1);
(%o4) y−2*x*y

Then find fixed points of f :

(%i6) remvalue(z);
(%o6) [z]
(%i7) zf:solve(z-z^2=z);
(%o7) [z=0]
(%i9) multiplicities;
(%o9) [2]
<pre>


Stability of the fixed points :

<pre>
(%i11) f:z-z^2;
(%o11) z−z^2
(%i12) d:diff(f,z,1);
(%o12) 1−2*z
(%i13) zf:solve(z-z^2=z);
(%o13) [z=0]
(%i14) z:zf[1];
(%o14) z=0
(%i15) abs(ev(d));
(%o15) abs(2*z−1)=1

It means that fixed point z=0 is a parabolic point ( stability indeks = 1 ).


Find critical point  :

(%i16) zcr:solve(d=0);
(%o16) [z=1/2]

ReferencesEdit