Fractals/Iterations in the complex plane/qpolynomials

Complex quadratic polynomial

=intro=
 * " The dynamics of polynomials is much better understood than the dynamics of general rational maps" due to the Bottcher’s theorem
 * infinity is always is superattracting fixed point for polynomials.
 * A polynomial of degree n has at most n real zeros and n−1 turning points.

=Forms =

z^2+c
Complex quadratic polynomial of the form :

$$ f_c(z) = z^2 + c \,$$

belongs to the class of the functions :


 * $$ z^n + c \,$$

notation
"... typical notational convention is to parameterize critically preperiodic polynomials $$z \to z^2 + c$$ by an angle $$\theta$$ of an external ray landing at the critical value rather than by c. In the event that more than one ray lands at the critical value, there may be multiple parameters referring to the same polynomial. As a simpler example, the polynomial:


 * $$f_{1/6} = f_i$$
 * for $$\theta = 1/4 $$ c= -0.228155493653962 +1.115142508039937 i
 * $$ f_{9/56} = f_{11/56} = f_{15/56} $$

How to compute iteration
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 point
A critical point of $$f_c\,$$ is a point $$ z_{cr} \,$$ in the dynamical plane such that the derivative vanishes :


 * $$f_c'(z_{cr}) = 0. \,$$

Since


 * $$f_c'(z) = \frac{d}{dz}f_c(z) = 2z $$

implies


 * $$ z_{cr} = 0\,$$

One can see that :
 * the only (finite) critical point of $$f_c \,$$ is the point $$ z_{cr} = 0\,$$
 * critical point is the same for all c parameters

$$z_{cr}$$ is an initial point for Mandelbrot set iteration.

Dynamic plane
period 1 ( = fixed) points :

$$S_1(f_c) = \frac{1}{2} \pm \frac{\sqrt{1 - 4c}}{2}$$

period 2 points :

$$S_2(f_c) = -\frac{1}{2} \pm \frac{\sqrt{-3 - 4c}}{2}$$

1/2
First compute muliplier m of the fixed points using internal angle p/q and Maxima CAS:

$$\frac{p}{q} = \frac{1}{2} $$

(%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*z
Complex quadratic polynomial of the form :

$$ f_m(z) = z^2 + m z \,$$

which has an indifferent fixed point with multiplier

$$\lambda = m = e^{2 \pi t i} \,$$

at the origin

belongs to the class of the functions :


 * $$ z^n + m z \,$$

How to compute iteration
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 point
A critical point of $$f_c\,$$ is a point $$ z_{cr} \,$$ in the dynamical plane such that the derivative vanishes :


 * $$f_m'(z_{cr}) = 0. \,$$

Since


 * $$f_m'(z) = \frac{d}{dz}f_m(z) = 2z + m $$

implies


 * $$ z_{cr} = -\frac{m}{2}\,$$

One can see that :
 * critical point is related with m value and have to be computed for every m parameters

$$z_{cr}$$ is an initial point for Mandelbrot set iteration.

period 1 components
(%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)

1/1
First compute parameter of the function : p:1$ q:1$ m:exp(2*%pi*%i*p/q);

The parameter is:

$$m = e^{2\pi i} = 1 $$

then the function is:

$$f_m(z) = z^2 + z$$

it gives the same Julia set ( cauliflower ) as function :

$$f_c(z) = z^2 + \frac{1}{4}$$

Compute fixed points :

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

Find it's stability index = abs(multiplier) of the fixed point :

(%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

Critical point :

$$z_{cr} = -\frac{m}{2} = - \frac{1}{2}$$

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));

1/2
First compute m = muliplier of the fixed points = parameter of the function f  using internal angle p/q and Maxima CAS:

$$\frac{p}{q} = \frac{1}{2} $$

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

so function f is :

$$f_m = z^2 +mz = z^2 -z$$

How to compute iteration  $$z_{n+1} = f_m(z_n) $$ ? (%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 :

$$ z_f : \{ z : f_m(z) = z \} $$

(%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 $$z_{cr}$$ : (%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 :

$$-2z^3$$

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
 * coefficient of above term a = -2

Attracing vectore satisfy :

$$nav^n = -1$$

so here :

$$-4v^2 = -1$$

$$v^2 = \frac{1}{4}$$

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 :
 * $$V_{a1} = \overrightarrow{v_{a1}0}$$ from $$z=-\frac{1}{2}$$ to the origin
 * $$V_{a2} = \overrightarrow{v_{a2}0}$$ from $$z=\frac{1}{2}$$ to the origin

Critical point z=1/2 lie on attracting vector $$V_{a1} $$. Thus critical orbits tend straight to the origin under the iteration

Repelling vectors satisfy :

$$nav^n = 1$$

so here :

$$-4v^2 = 1$$

$$v^2 = - \frac{1}{4}$$

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]

1/3
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

1/7
How to speed up computations ?

Approximate $$f^7$$ by :

$$f_a^7(z) = (245.4962434402444i-234.5808769813032)*z^8 + z $$

How to compute $$f_a^7$$ : (%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)
Description
 * at Mu-Ency
 * at wikibook Pictures of Julia and Mandelbrot Sets

Critical points
critical points :
 * z = 1/2
 * z = ∞

period 1 components
(%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

Dynamic plane
"Note that each member of the family of quadratic polynomials

$$\{ g_{\lambda} : z \to z(1- \lambda z) \}_{\lambda \in C\setminus\{0\}} $$

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 $$g_{\lambda}$$ is given by

$$\frac{1}{2\lambda}$$

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

$$z \to z^2 +\frac{1}{4} $$

via Mobius transformations

$$h_{\lambda}(z) = - \lambda z +\frac{1}{2}$$

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

HD(J(z^2 +1/4)) ≈ 1.0812 "

z-z^2
Description

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

$$\frac{p}{q} = \frac{1}{2} $$

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

so function f is :

$$f_m = z(1+mz) = z-z^2 $$

How to compute iteration  $$z_{n+1} = f_m(z_n) $$ ?

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 :

$$ z_f : \{ z : f_m(z) = z \} $$

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

Stability of the fixed points :

(%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 $$z_{cr}$$ :

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

=References=