intro

• " The dynamics of polynomials is much better understood than the dynamics of general rational maps" due to the Bottcher’s theorem[2]
• 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.[3]

Forms

z^2+c

Complex quadratic polynomial of the form :

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

belongs to the class of the functions :

${\displaystyle z^{n}+c\,}$

notation

"... typical notational convention is to parameterize critically preperiodic polynomials ${\displaystyle z\to z^{2}+c}$  by an angle ${\displaystyle \theta }$  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:

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

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 ${\displaystyle f_{c}\,}$  is a point ${\displaystyle z_{cr}\,}$  in the dynamical plane such that the derivative vanishes :

${\displaystyle f_{c}'(z_{cr})=0.\,}$

Since

${\displaystyle f_{c}'(z)={\frac {d}{dz}}f_{c}(z)=2z}$

implies

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

One can see that :

• the only (finite) critical point of ${\displaystyle f_{c}\,}$  is the point ${\displaystyle z_{cr}=0\,}$
• critical point is the same for all c parameters

${\displaystyle z_{cr}}$  is an initial point for Mandelbrot set iteration.[4]

Dynamic plane

period 1 ( = fixed) points :[5]

${\displaystyle S_{1}(f_{c})={\frac {1}{2}}\pm {\frac {\sqrt {1-4c}}{2}}}$


period 2 points :

${\displaystyle S_{2}(f_{c})=-{\frac {1}{2}}\pm {\frac {\sqrt {-3-4c}}{2}}}$


1/2

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

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

(%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))
q:1$m:exp(2*%pi*%i*p/q);  The parameter is: ${\displaystyle m=e^{2\pi i}=1}$ then the function is: ${\displaystyle f_{m}(z)=z^{2}+z}$ it gives the same Julia set ( cauliflower [10] ) as function : ${\displaystyle f_{c}(z)=z^{2}+{\frac {1}{4}}}$  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  ${\displaystyle z_{cr}=-{\frac {m}{2}}=-{\frac {1}{2}}}$ 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: ${\displaystyle {\frac {p}{q}}={\frac {1}{2}}}$ (%i1) p:1$
(%i2) q:2$(%i3) m:exp(2*%pi*%i*p/q); (%o3) - 1  so function f is : ${\displaystyle f_{m}=z^{2}+mz=z^{2}-z}$ How to compute iteration ${\displaystyle z_{n+1}=f_{m}(z_{n})}$ ? (%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 : ${\displaystyle z_{f}:\{z:f_{m}(z)=z\}}$ (%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 ${\displaystyle z_{cr}}$ : (%i14) zcr:solve(d=0); (%o14) [z = 1/2] (%i15) multiplicities; (%o15) [1]  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 : ${\displaystyle -2z^{3}}$ 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 : ${\displaystyle nav^{n}=-1}$ so here : ${\displaystyle -4v^{2}=-1}$ ${\displaystyle v^{2}={\frac {1}{4}}}$ 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 :

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

Critical point z=1/2 lie on attracting vector ${\displaystyle V_{a1}}$ . Thus critical orbits tend straight to the origin under the iteration[11]

Repelling vectors satisfy :

${\displaystyle nav^{n}=1}$

so here :

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

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

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 ${\displaystyle f^{7}}$  by :

${\displaystyle f_{a}^{7}(z)=(245.4962434402444i-234.5808769813032)*z^{8}+z}$

How to compute ${\displaystyle f_{a}^{7}}$  :

(%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

Critical points

critical points :

• z = 1/2
• z = ∞

Parameter plane

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

z(1-mz)

Dynamic plane

"Note that each member of the family of quadratic polynomials

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


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

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


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

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


via Mobius transformations

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


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

HD(J(z^2 +1/4)) ≈ 1.0812


"[13]

z(1+ mz)

dynamic plane

z-z^2

Description [14]

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

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

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


so function f is :

${\displaystyle f_{m}=z(1+mz)=z-z^{2}}$

How to compute iteration ${\displaystyle z_{n+1}=f_{m}(z_{n})}$  ?

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 :

${\displaystyle z_{f}:\{z:f_{m}(z)=z\}}$

(%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 ${\displaystyle z_{cr}}$  :

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