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Content Iteration of Complex Function
source/reference: Iteration of Complex FunctionIteration of Functionfⁿ is called the nth iterate of f Let f(z)=z+1 f²(z)=f(f(z))=f(z+1)=(z+1)+1=z+2 f³(z)=f(f²(z))=f(z+2)=(z+2)+1=z+3 … fⁿ(z)=f(fⁿ⁻¹(z))=f(z+(n1))=(z+(n1))+1=z+n Let f(z)=3z f²(z)=f(f(z))=f(3z)=3(3z)=3²z f³(z)=f(f²(z))=f(3²z)=3(3²z)=3³z … fⁿ(z)=f(fⁿ⁻¹(z))=f(3ⁿ⁻¹z)=3(3ⁿ⁻¹z)=3ⁿz Iteration of Quadratic PolynomialsThe general quadratic polynomial is of the form p(z)=az²+bz+d, where constants a, b, d∊ℂ. However, for each triple of constants (a,b,d), there is exactly one constant c such that polynomial f(z)=z²+c behave the same as p(z)=az²+bz+d under iteration. Given p(z)=az²+bz+d, where constants a, b, and d∊ℂ. Define functons φ(z)=az+b 2 and f(z)=z²+c where c=ad+b 2(b 2)² . ⇒f(φ(z))=(φ(z))²+c=(az+b 2)²+ad+b 2(b 2)² =a²z²+abz+ad+b 2=a(az²+bz+d)+b 2 ⇒f(φ(z))=a(p(z))+b 2=φ(p(z)) ⇒φ⁻¹(f(φ(z)))=p(z) ∴p(z)=φ⁻¹(f(φ(z))) for all z rewrite as p=φ⁻¹∘f∘φ where ∘ means composed with ⇒p∘p=(φ⁻¹∘f∘φ)∘(φ⁻¹∘f∘φ)=φ⁻¹∘f∘φ∘φ⁻¹∘f∘φ=φ⁻¹∘f∘f∘φ ⇒p²=φ⁻¹∘f²∘φ ⇒p³=φ⁻¹∘f³∘φ … ⇒pⁿ=φ⁻¹∘fⁿ∘φ It thus suffices to study the iteration of quadratic polynomials of the form f(z)=z²+c. The Set of IterationThe Julia set, named after the French mathematician Gaston Julia (18931978), of f(z)=z²+c is the set of all z∊ℂ for which the behavior of the iterates is "chaotic" in a neighborhood. the Fatou set, named after the French mathematician {oerre Fatpi (18781929), is the set of all z∊ℂ for which the iterates behave "normally" in a neighborhood. The iterates of f behave normally near z if nearby point remain nearby under iteration. The iterates of behave chaotically at z if in any small neighborhood of z the behavior of the iterates depends sensitively on the initial point. For example, f(z)=z²+c. Let c=0, f(z)=z², then fⁿ(z)=z2ⁿ Let z=re𝑖θ, then fⁿ(z)=r2ⁿe𝑖2ⁿθ If z<1, then fⁿ(z)=z2ⁿ→0 as n→∞, so fⁿ(z)→0 as n→∞. If z>1, then fⁿ(z)=z2ⁿ→∞ as n→∞, so fⁿ(z)→∞ as n→∞. If z=1, then z=re𝑖θ=e𝑖θ, so fⁿ(z)=e𝑖2ⁿθ, thus fⁿ(z)=1 for all n In any little disk around a point z with z=1, there are points w with w>1, for which fⁿ(w)→∞, and other points w with w<1, for which fⁿ(w)→0. The unit circle {z:z=1} is thus the locus of chaotic behavior, whereas {z:z>1} with iterates attracted to ∞ and {z:z<1} with iterates attracted to 0 form the locus of normal behavior. Julia set J(f)={z:z=1}, the unit circle. and Fatou set F(f)={z:z>1}∪{z:z<1} The Basin of Attraction to ∞Let f(z)=z²+c. Let A(∞)={z:fⁿ(z)→∞} "basin of attraction to ∞" By theorem. The set A(∞) is open, connected and unbounded. It is contained in the Fatou set of f. The Julia set of f coincides with the boundary of A(∞) , which is a closed and bounded subset of ℂ. Where
Involution or conjugationLet f(z)=z²2. f²(z)=f(z²2)=(z²2)²2. But it is hard to calculate and understand the iterates, fⁿ(z). Conjugate f with φ(w)=w+1
w,φ:{w:w>1}→ℂ\[2,2].
f maps [2,2] to[2,2] and ℂ\[2,2] to ℂ\[2,2]. Thus look at φ⁻¹∘f∘φ. Let f(z)=z²2. φ(w)=w+1 w, ⇒f(φ(w))=(φ(w))²2=(w+1 w)²2=w²+2w w+1 w²2=w²+1 w²=φ(w²) ⇒φ⁻¹(f(φ(w)))=w² Let g(w)=w² ⇒φ⁻¹(f(φ(w)))=g(w) ⇒ f(φ(w))=φ(g(w)) Let φ(w)=z⇒w=φ⁻¹(z) ⇒ f(z)=φ(g(φ⁻¹(z))) Thus, on ℂ\[2,2], the function f(z)=z²2 behaves like g(w)=w² behaves on the exterior of the closed unit disk. Since the iterates gⁿ(w) tend to ∞ for w>1. therefore fⁿ(z)→∞ as n→∞ for all z∊ℂ\[2,2]. Thus A(∞)=ℂ\[2,2], and thus J(f)=[2,2], the closed interval from 2 to 2 on the real axis. Both the Julia sets of f(z)=z² and f(z)=z²2 are the only smooth Julia set examples amongst all f(z)=z²+c. Julia SetsFor f(z)=z²+c, where c∊ℂ,the Julia set of f is the boundary of A(∞), where A(∞) is the "basin of attraction to infinity", i.e. A(∞)={z∊ℂ:fⁿ(z)→∞ as n→∞}. Let K(f) be the filledin Julia set of f for which z∊ℂ and {fⁿ(z)} stays bounded, i.e. K(f)={z∊ℂ : {fⁿ(z)} is bounded}. For examples f(z)=z² (c=0): Then K(f)={z:z≤1} and J(f)={z:z=1} f(z)=z²2 (c=2): Then K(f)=[2,2] and J(f)=[2,2] For f(z)=z²1 (c=1) Let f(z)=z²1 z=0: f(0)=1, f²(0)=f(1)=0, f³(0)=f(0)=1, f⁴(0)=0, …. The orbit is thus 0, 1, 0, 1, 0, 1, …. This is called a periodic orbit, and 0 is a periodic point of period 2. clearly, 0∊K(f). z=1: 1, 0, 1, 0, 1, 0, 1, …. So 1 isnot itself a periodic point (the orbit never returen to 1), but it's the next best thing;it runs into a periodic orbit. It thus is called 1 a preperiodic point. Again, 1∊K(f). z=1+√5 2:(1+√5 2)² 1=1+2√5+5 41=2+2√5 4=1+√5 2=z. The point 1+√5 2is a fixed point of f and thus belongs to K(f) as well. Other z z=2: 2, 3, 8, 63, …, →∞. Thus 2∊A(∞). z=: 2, 3, 8, 63, …, →∞. Thus 2∊A(∞). First, if z lies on the real axis and is smaller than (1+√5 More generally, if z∊ℂ and z>1+√5 And more generally, if z∊ℂ and fⁿ(z)>1+√5 A similar condition holds for general quadratic polynomials f(z)=z²+c, By theorem, Let f(z)=z²+c, and let R= 1+√(1+4c) Computer Algorithm finding the FilledIn Julia SetsComputer algorithm:
Prettier PicturesThe pictures of the filledin Julia sets produced so far are only black and white. Fro colorful pictures, choose colors c₀, c₁, c₂, …, cmaxiter for A(∞) as well as a color for the filledin Julia set. If z>R, color the corresponding pixel with color c₀. Otherwise, if f(z)>R, color the corresponding pixel with color c₁. Otherwise, if f(f(z))>R, color the corresponding pixel with color c₂. …. Otherwise, i.e. if fⁿ(z)>R for all n≤maxiter, color the corresponding pixel with corresponding color cₙ for the filledin Julia set. The larger constant maxiter, the more detail of the pictures, but the longer the calculation will take. For small windows (i.e. zooms into the Julia set), a higher precision is required to avoid roundoff errors. The Mandelbrot SetThe Julia set of f(z)=z²+c is either "in one piece" or "totally dusty" By definition. The Mandelbrot set M is the set of all parameters c∊ℂ for which the Julia set J(f) of f(z)=z²+c is connected in one piece. That is M={c∊ℂ: J(z²+c) is connected} Note: The Mandelbrot set is a subset of the parameter space (the space of all possible cvalues), whereas Julia sets are sets of zvalues. Examples 0∊M since J(z²)={z"z=1} 2∊M since J(z²2)=[2,2] ¼∊M according to picture of J(z²+¼) 1∉M according to picture of J(z²+1) By theorem. Let f(z)=z²+c. Then J(f) is connected if and only if z=0 does not belong to A(∞), that is if and only if the orbit {fⁿ(0)} remains bounded under iteration. In fact, it is possible to show the following: By theorem. A complex number c belongs to M if and only if fⁿ(0)≤2 for all n≥1 (where f(z)=z²+c). Computer Algorithm to Plot the Mandelbrot SetComputer algorithm
A Prettier PictureAgain, you can use different colors for those parameters c∊ℂ for which 0 escapes to infinity under iteration, depending on how quickly the escape happens"
Zooming into the Mandelbrot set and coloring parameters by escape time yields beautiful pictures. Properties of the Mandelbrot SetProperties of the Mandelbrot Set
Misiurewicz PointsMany points in the boundary of M are cocalled Misiurewicz points:
Big Open ConjectureHere is one of the big outstanding conjectures in the field of complex dynamics: By Conjecture. The Mandelbrot set is locally connected, that is, for every c∊M and every open set V with c∊V, there exists an open set U such that c∊U⊂V and U∩M is connected. ©sideway ID: 190300019 Last Updated: 19/3/2019 Revision: 0 Latest Updated Links

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