# FormsEdit

## z^2+cEdit

Complex quadratic polynomial of the form :

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

belongs to the class of the functions :

$z^n + c \,$

### 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 $f_c\,$ 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.[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:

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

(%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 result is m = 1 so f(z) = z^2 + z 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: $\frac{p}{q} = \frac{1}{2}$ (%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 • cooefficient 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[7]

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]


#### 1/7Edit

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)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

#### z-z^2Edit

Description [9]

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]
<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 $z_{cr}$ :

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