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Including Complex Dynamics in Complex Analysis Courses
Mandelbrot and Julia sets
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Benefits: Introduces “modern” topics in mathematics into the course
Illustrates different ways to use complex analysis “tools”
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Topics: Mandelbrot set (complex square roots) Attracting cycles (Schwarz Lemma) Understanding chaos (rational maps) Exponential dynamics (geometry of exp(z))
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Begin by recalling iteration
Start with a function: 2 x constant and a seed: x0
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Then iterate: x = x + constant x = x + constant x = x + constant
2 x = x constant 1 2 x = x constant 2 1 Orbit of x 2 x = x constant 3 2 2 x = x constant 4 3 etc. Goal: understand the fate of orbits.
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Example: x + 1 Seed 0 x = 0 x = 1 x = 2 x = 5 x = 26 x = big
x = 1 1 x = 2 2 x = 5 “Orbit tends to infinity” 3 x = 26 4 x = big 5 x = BIGGER 6
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Example: x + 0 Seed 0 x = 0 x = 0 x = 0 x = 0 x = 0 x = 0 x = 0
2 Example: x Seed 0 x = 0 x = 0 1 x = 0 2 “A fixed point” x = 0 3 x = 0 4 x = 0 5 x = 0 6
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Example: x - 1 Seed 0 x = 0 x = -1 x = 0 x = -1 x = 0 x = -1 x = 0
2 Example: x Seed 0 x = 0 x = -1 1 x = 0 2 x = -1 “A two- cycle” 3 x = 0 4 x = -1 5 x = 0 6
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Example: x - 1.1 Seed 0 x = 0 x = -1.1 x = 0.11 x = x = x = x =
2 Example: x Seed 0 x = 0 x = 1 x = 2 x = 3 time for the computer! x = 4 x = 5 x = 6
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Observation Sometimes orbit of 0 goes to
infinity, other times it does not.
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Big Question: Answer: How do we understand the chaotic
behavior that occurs? Answer: We move to the complex plane and use tools from complex analysis
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Complex Iteration 2 Iterate z + c complex numbers
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Example: z + i Seed 0 z = 0 z = i z = -1 + i z = -i z = -1 + i z = -i
2 Example: z + i Seed 0 z = 0 z = i 1 z = i 2 z = -i 3 z = i 4 z = -i 5 2-cycle z = i 6
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Example: z + 2i Seed 0 z = 0 z = 2i z = -4 + 2i z = 12 - 14i
z = 2i 1 z = i 2 Off to infinity z = i 3 z = i 4 z = big 5 z = BIGGER 6
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Same observation Sometimes orbit of 0 goes to
infinity, other times it does not.
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The Mandelbrot Set: All c-values for which orbit
of 0 does NOT go to infinity. Why the heck do we care about the orbit of 0?
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Why use the orbit of 0? Answer: 0 is a “critical point” So spend some time talking about behavior of complex functions near critical points (not one-to-one) and near other points (always one-to-one)
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The Filled Julia Set: The Julia Set:
Fix a c-value. The filled Julia set is all of the complex seeds whose orbits do NOT go to infinity. The Julia Set: The boundary of the filled Julia set
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Example: z Seed: In Filled Julia set? Yes 1 Yes -1 Yes i Yes 2i No 5
Yes 1 Yes -1 Yes i Yes 2i No 5 I doubt it...
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The filled Julia set of z2 is the unit disk
A little complex analysis: |z| < 1: orbit of z |z| > 1: orbit of z infinity |z| = 1: orbit of z remains on the unit circle
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The filled Julia set of z2 is the unit disk
The Julia set is the unit circle
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
3 nearby seeds Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Chaos occurs on the Julia set
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Nearby orbits on the Julia set have vastly different behavior
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i.e., the map just doubles angles
On the Julia set the map is ei e2i i.e., the map just doubles angles
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On the Julia set the map is ei e2i
1/5 Rational angles on the Julia set are eventually periodic.
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Rational angles on the Julia set are eventually periodic.
1/5 2/5 Rational angles on the Julia set are eventually periodic.
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Rational angles on the Julia set are eventually periodic.
1/5 2/5 4/5 Rational angles on the Julia set are eventually periodic.
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Rational angles on the Julia set are eventually periodic.
1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.
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Rational angles on the Julia set are eventually periodic.
1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.
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Rational angles on the Julia set are eventually periodic.
1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.
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Rational angles on the Julia set are eventually periodic.
1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.
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Rational angles on the Julia set are eventually periodic.
1/5 2/5 3/5 4/5 Rational angles on the Julia set are eventually periodic.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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But irrational angles on the Julia set are very different.
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Irrational angles have orbits that are dense on the circle.
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So why do we use the orbit of 0 to plot the Mandelbrot set?
The orbit of the critical point knows “everything!”
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So why do we use the orbit of 0 to plot the Mandelbrot set?
The orbit of the critical point knows “everything!” To see this, we need the geometry of complex square roots
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The square root of a circle (s.c.c.)
that does not surround the origin.... i 1
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is a pair of s.c.c.’s that do not
The square root of a circle (s.c.c.) that does not surround the origin.... i z z2 1 is a pair of s.c.c.’s that do not surround the origin
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The square root of a circle (s.c.c.) that touches the origin....
1
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The square root of a circle (s.c.c.) that touches the origin....
z z2 1 is a figure 8 that passes through the origin
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The square root of a circle (s.c.c.) that surrounds the origin....
1
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The square root of a circle (s.c.c.) that surrounds the origin....
z z2 1 is one s.c.c. that now surrounds the origin
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To find the preimages of a s.c.c.
under z2 + c, we solve: so that:
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To find the preimages of a s.c.c.
under z2 + c, we solve: so that: So to find the preimage of a s.c.c., we translate the curve by -c, then take the complex square root....
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So now suppose the orbit of 0
escapes to infinity c
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This large circle is mapped outside itself c
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This large circle is mapped outside itself c
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so all blue points escape c
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so all blue points escape To find the preimage of this curve, subtract c and take the square root... c
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so all blue points escape To find the preimage of this curve, subtract c and take the square root... but subtracting c means the curve still encircles 0 c
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so all blue points escape so this is the preimage c
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so all blue points escape so these points also escape c
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so all blue points escape so these points also escape c
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so all blue points escape so these points also escape as do these points c
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so all blue points escape so these points also escape as do these points c
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so all blue points escape so these points also escape as do these points c
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8 so all blue points escape so these points also escape
as do these points c 8
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Eventually 0 is in a preimage of the
See what’s happening? Eventually 0 is in a preimage of the circle, so we get a figure 8 preimage. And then..... c = .4975
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So the filled Julia set consists of infinitely
many distinct components if 0 escapes (in fact it is a Cantor set). But when 0 does not escape, we never get a figure 8, so the filled Julia set is a connected set. That’s why the critical point “knows it all.”
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Attracting cycles Understanding chaos Dynamics of the exponential
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If F is complex analytic and has an
attracting cycle, then there must be a critical point inside its immediate basin of attraction. So F(z) = z2 + c can have at most one attracting cycle, and the orbit of 0 must find it.
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Reason: suppose F has an attracting fixed point p:
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Then there is a neighborhood U of p mapped 1-1 inside itself.
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Then there is a neighborhood U of p mapped 1-1 inside itself.
So we have an inverse map F-1: F(U) U with a repelling fixed point at p. F-1 p U
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Since there are no critical points in the basin, we can continue
F-1 p U
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Since there are no critical points in
the basin, and on and on ..... F-1 p U
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Eventually get an open disk V that is
mapped 1-1 to itself by F-1 F-1 ... p U V
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Eventually get an open disk V that is
mapped 1-1 to itself by F-1 F-1 p V
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But there is a repelling fixed point in V, so this can’t happen by the
Schwarz Lemma F-1 p V
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So this begins to describe the structure of the Mandelbrot set:
each “bulb” consists of parameters for which there is an attracting cycle of some given period
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The eventual orbit of 0 Eventual orbit
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The eventual orbit of 0 Eventual orbit
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The eventual orbit of 0 3-cycle
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The eventual orbit of 0 3-cycle
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The eventual orbit of 0 3-cycle
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The eventual orbit of 0 3-cycle
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The eventual orbit of 0
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The eventual orbit of 0
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The eventual orbit of 0 4-cycle
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The eventual orbit of 0 4-cycle
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The eventual orbit of 0 4-cycle
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The eventual orbit of 0 4-cycle
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The eventual orbit of 0 4-cycle
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The eventual orbit of 0
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The eventual orbit of 0
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The eventual orbit of 0 5-cycle
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The eventual orbit of 0 5-cycle
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The eventual orbit of 0 5-cycle
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The eventual orbit of 0 5-cycle
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The eventual orbit of 0 5-cycle
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The eventual orbit of 0 5-cycle
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The eventual orbit of 0 5-cycle
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The eventual orbit of 0 2-cycle
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The eventual orbit of 0 2-cycle
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The eventual orbit of 0 2-cycle
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The eventual orbit of 0 fixed point
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The eventual orbit of 0 fixed point
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The eventual orbit of 0 fixed point
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How understand the periods of the bulbs?
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How understand the periods of the bulbs?
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junction point three spokes attached
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junction point three spokes attached Period 3 bulb
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Period 4 bulb
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Period 5 bulb
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Period 7 bulb
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Period 13 bulb
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Complex exponential dynamics
Let E (z) = exp(z) where > 0. Differences: no critical points, but one “asymptotic value” at z = 0.
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Complex exponential dynamics
Let E (z) = exp(z) where > 0. Differences: no critical points, but one “asymptotic value” at z = 0. H is wrapped infinitely often around 0, which plays the role of the critical value H
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Complex exponential dynamics
Let E (z) = exp(z) where > 0. Differences: no critical points, but one “asymptotic value” at z = 0. The Julia set is now the closure of the set of escaping points, not the boundary of this set
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When 0 < < 1/e, the Julia set is
a “Cantor bouquet,” i.e., infinitely many curves extending to infinity in the right half plane.
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0 < < 1/e
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attracting fixed point q
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q p repelling fixed point
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q x0 p
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So where is J?
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So where is J?
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So where is J? Green points lie in the basin of q, so not in the Julia set
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So where is J? Green points lie in the basin of q, so not in the Julia set
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So where is J? Green points lie in the basin of q, so not in the Julia set
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So where is J? Green points lie in the basin of q, so not in the Julia set
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So where is J? Green points lie in the basin of q, so not in the Julia set
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The Julia set is a collection of curves (hairs) in the right half plane, each with an endpoint
and a stem. hairs endpoints stems
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A “Cantor bouquet” q p
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Colored points escape to and so are in the Julia set.
q p
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When = 1/e, E undergoes a “saddle node bifurcation”
but much more happens...
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When = 1/e, E undergoes a “saddle node bifurcation”
The two fixed points on the real line disappear and the orbit of the asymptotic value now escapes....
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And, because of this, the Julia instantaneously
becomes the entire complex plane!
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Replay
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How understand the chaotic behavior?
z2 z i
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so use that information
How understand the chaotic behavior? We understand what’s going on here, so use that information to understand what’s going on here.
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How understand the chaotic behavior?
These angles are just rotated by z2
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How understand the chaotic behavior?
These angles are just rotated by z2 So we can find a “conjugacy” that creates angles here that map in the same way
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Example: z2 - 2 z2 z2 - 2 How to map the exterior of the
circle to the exterior of [-2,2]? -2 2 z2 z2 - 2
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Consider H(z) = z + 1/z -2 2 z2 z2 - 2
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Consider H(z) = z + 1/z H takes rays to hyperbolas -2 2 z2 z2 - 2
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Consider H(z) = z + 1/z Well, usually that’s the case -2 2 z2 z2 - 2
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Consider H(z) = z + 1/z z2 z2 - 2 And H takes the circle to [-2, 2] -2
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Consider H(z) = z + 1/z z2 z2 - 2 And H takes the circle to [-2, 2]
since eit + e-it = 2 cos(t) -2 2 z2 z2 - 2
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On the circle we have: H H On [-2,2] we have: What is this function???
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On the circle we have: H H
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On the circle we have: H H
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On the circle we have: H H So this function is w w2 - 2
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So the orbits of F(z) = z2 on the unit circle (which
we completely understand are taken to orbits of G(z) = z by the “conjugacy” H(z) = z + 1/z. z F(z) F2(z) F3(z) F4(z) ... H H H H H w G(w) G2(w) G2(w) G4(z) ... And that’s how we understand the chaotic behavior on the Julia sets.
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