Post on 14-Jan-2016
description
RememberingBenoit
Mandelbrot
20 November 1924 – 14 October 2010
First Citizenof
Science
(1924 – 2010)
Fatherof
Fractal Geometry
(1924 – 2010)
Theoryof
Roughness
(1924 – 2010)
The FractalGeometryof Nature
1977
1982
1985
The year when I metBenoit Mandelbrot
andRichard F. Voss
December 6, 1982
Leo Kadanoff
University of Utah
Mandelbrot Set 1980
1986
The mathematics behindthe Mandelbrot Set
University of California at Santa Cruz, October 1987
1988
Publishing all the algorithms
known at that time
How Mountains turn into Clouds …
A Masterpiece by Richard F. Voss
A completely synthetic mathematical
construction of mountains and clouds
1991...
1991...MaletskyPerciante
Yunker
PeitgenJürgensSaupe
1992
Mandelbrot Set:
The most complex object mathematics has ever seen
Iteration
Iteration of rational functions
Theory of Julia & Fatou~1918
€
Choose z0 in the complex plane.
Then iterate, which means compute
zn+1 = f (zn ) for n = 0,1,2,3,...
€
f (z) =p(z)
q(z), where p(z) and q(z) are polynomials
€
Example : f (z) =2z3 +1
3z2
I studied thatin the fall of 1982
at the University of Utah
Newton's Method for x3-1
Julia Sets
€
Given a rational function f (z),
collect all starting points z for which the
iteration does not go to infinity
J = z | z→ f (z) → f ( f (z)) → ...{ → ∞}
"The iteration does not escape to infinity"
"The Prisoner Set"
€
The Set of Complex Numbers C
z = a+ bi, i = −1
€
Addition
z = a+ bi, w = c + di
z + w = (a+ c) + (b+ d)i
€
Multiplication
z = a+ bi, w = c + di
z • w = (a+ bi) + (c + di) = (ac −bd) + (ad + bc)i
a
b z
€
The Set of Complex Numbers C
z = a+ bi, i = −1
€
Division? Find inverse to z = a+ bi :
1
z=
1
a+ bi=
1
a+ bi•a−bi
a−bi=a−bi
a2 + b2
a
b z
1/z
€
The Set of Complex Numbers C
z = a+ bi, i = −1
€
Modulus
z = a+ bi
| z | = a2 + b2
a
b z
€
The Quadratic Family
f (z) = z2 + c, c ∈ C
€
z0,
z1 = z02 + c,
z2 = z12 + c = z0
2 + c( )2
+ c = z04 + 2cz0
2 + c 2 + c
z3 = z22 + c = z0
4 + 2cz02 + c 2 + c( )
2+ c = ...
Julia Set
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
i.e. choose c and then
collect all starting points for which the iteration
does not go to infinity (Prisoner Set)
Theorem of Julia & Fatou
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
is
- either one piece (connected)
- or an infinite dust (Cantor Set)
Theorem of Julia & Fatou
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
is connected if and only if
c → c 2 + c → (c 2 + c)2 + c → ... → ∞
connected not connected
dust
connected not connected
(super) infinite dust
Cantor Set
Two simple Julia Sets
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
€
c = 0 :
z→ z2 → z4 → z8 → ...
Two simple Julia Sets
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
1
€
z <1⇒ z2 = z2
= z • z < z
€
z >1⇒ z2 = z2
= z • z > z
€
c = 0 :
z→ z2 → z4 → z8 → ...
Two simple Julia Sets
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
1
€
c = 0 :
z→ z2 → z4 → z8 → ...
€
Is it connected? Need to check :
c → c 2 + c → (c 2 + c)2 + c → ...
€
c = 0, compute
c → c 2 + c → (c 2 + c)2 + c → ...
Two simple Julia Sets
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
+2
€
−2 ≤ z ≤ 2 ⇔ z ≤ 2
⇒ z2 − 2 ≤ 4 − 2 = 2
-2€
c = −2 :
z→ z2 − 2 → (z2 − 2)2 − 2 → ...
Two simple Julia Sets
€
Jc = z | z→ z2 + c → (z2 + c)2 + c → ...{ → ∞}
+2-2€
c = −2 :
z→ z2 − 2 → (z2 − 2)2 − 2 → ...
€
Is it connected? Check for c = −2 :
c → c 2 + c → (c 2 + c)2 + 2 → ...
The Mandelbrot Set
€
M = c | Jc is connected{ }
⇔
M = c | c → c 2 + c → (c 2 + c)2 + c → ...→ ∞{ }
€
{−2,0}∈ M
The Mandelbrot Set
€
M = c | c → c 2 + c → (c 2 + c)2 + c → ...→ ∞{ }
sequence becomes unbounded"escapes"
sequence remains bounded"imprisoned"
Making a picture:(b/w)
1980
Computer (Pixel) Graphics
C64: 1982 16 colors
Macintosh: 1984 b/w--------------------------RGB 256x256x256only in few research labsUniversity of Utah
1/4-2
1
-1
The Mandelbrot Set
€
M = c | c → c 2 + c → (c 2 + c)2 + c → ...→ ∞{ }
all sequences become unbounded"escape"
some sequences remain bounded"imprisoned"
2
Making a picture:b/w
€
c 2 + c − c ≤ c 2 + c + c ⇒
€
c 2 + c ≥ c 2 − c
€
=c 2
€
When c > 2 then c → c 2 + 2 → c 2 + 2( )2
+ c → ... escapes
€
Whe need the Triangle Inequality :
a +b ≤ a + b
€
Whe will show :
c > 2⇒ c 2 + c > c
€
⇒ c 2 + c ≥ c 2 − c = c c −1( )
€
⇒ c 2 + c > c
The Mandelbrot Set
€
M = c | c → c 2 + c → (c 2 + c)2 + c → ...→ ∞{ }
"imprisoned"
2
"escapes"takes 5 steps to land
outside circle
"escapes"takes 13 steps to land
outside circle
Making a picture:(color)
1982/83Salt Lake City
Around the Mandelbrot Set
Powers of Ten
Similarity between
Julia Sets
and the
Mandelbrot Set
1/(period)2
Mandelbrot Set 1990 (Peitgen/Jürgens/Saupe)
Electrostatic Potential(key for mathematical understanding)
Flying the Mandelbrot Set
Interview Bremen1986
We will always remember