Fractals/Iterations in the complex plane/qpolynomials

Complex quadratic polynomialwikipedia : Complex quadratic polynomial - definition intro " The dynamics of polynomials is much better understood than the dynamics of general rational maps" due to the Bottcher’s theoremON THE NOTIONS OF MATING by CARSTEN LUNDE PETERSEN AND DANIEL MEYER 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.theknowledgeburrow : what-is-the-difference-between-a-polynomial-and-rational-function Forms z^2+c Complex quadratic polynomial of the form : belongs to the class of the functions : notation "... typical notational convention is to parameterize critically preperiodic polynomials by an angle 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: for c= -0.228155493653962 +1.115142508039937 i 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 thumbnail|right|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.Java program by Dieter Röß showing result of changing initial point of Mandelbrot iterations Dynamic plane period 1 ( = fixed) points :Algebraic Geometry of Discrete Dynamics The case of one variable V.Dolotin and A.Morozov period 2 points : 1/2 thumbnail|right|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*z Complex quadratic polynomial of the form : which has an indifferent fixed pointwikipedia : fixed point with multiplierwikipedia : multiplier at the originwikipedia : originMichael Yampolsky, Saeed Zakeri : Mating Siegel quadratic polynomials. belongs to the class of the functions : 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 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. thumbnail|right|Parameter plane lambda parameter plane 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) Dynamic plane 1/1 thumbnail|right|Julia set for f(z) = z^2+z or f(z) = z^2 + 1/4 or f(z)= z-z^2 thumbnail|right|Domains for Fatou coordinate right|thumb| 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 parameter is: then the function is: it gives the same Julia set ( cauliflower images of Cauliflower_Julia_set ) as function : 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 : 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: (%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 than 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 coefficient 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 iterationMark McClure in stackexchange questions : what-is-the-shape-of-parabolic-critical-orbit 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] 1/3 thumbnail|right|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 1/7 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) Description at Mu-Ency lambda map at Mu-Ency at wikibook Pictures of Julia and Mandelbrot Sets 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 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 "REAL ANALYTICITY OF HAUSDORFF DIMENSION OF DISCONNECTED JULIA SETS OF CUBIC PARABOLIC POLYNOMIALS Hasina Akter z(1+ mz) dynamic plane z-z^2 Description S Lapan : On the existence of attracting domains for maps tangent to the identity. Ph.D. Thesis, University of Michigan 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] 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 : (%i16) zcr:solve(d=0); (%o16) [z=1/2] References