Excel quad iteration M-set iterator Movie maker 75.

Excel quad iteration

M-set iterator

Movie maker 75

The Fractal Geometryof the Mandelbrot Set

How the computer has revolutionized mathematics


You need to know:

How to count


You need to know:


How to add

How to count

You need to know:

Many people know thepretty pictures...

but few know the evenprettier mathematics.

Oh, that's nothing but the 3/4 bulb ....

...hanging off the period 16 M-set.....

...lying in the 1/7 antenna...

...attached to the 1/3 bulb...

...hanging off the 3/7 bulb...

...on the northwest side of the main cardioid.

Oh, that's nothing but the 3/4 bulb, hanging off the period 16 M-set, lying in the 1/7 antenna of the 1/3 bulb attached to the 3/7 bulb on the northwest side of the main cardioid.

Start with a function:

x + constant2

Start with a function:

x + constant2

and a seed:

x0

Then iterate:

x = x + constant1 02

Then iterate:



Then iterate:



x = x + constant3 2

2

Then iterate:



x = x + constant3 2

2

x = x + constant4 3

2

Then iterate:



x = x + constant3 2

2

x = x + constant4 3

2

Orbit of x0

etc.

Goal: understand the fate of orbits.

Example: x + 1 Seed 02

x = 00x = 1x = 2

x = 3

x = 4

x = 5

x = 6


x = 00x = 11x =2

x = 3

x = 4

x = 5

x = 6


x = 00x = 11x = 22

x = 3

x = 4

x = 5

x = 6


x = 00x = 11x = 22

x = 53

x = 4

x = 5

x = 6


x = 00x = 11x = 22

x = 53

x = 264

x = 5

x = 6


x = 00x = 11x = 22

x = 53

x = 264

x = big5

x = 6


x = 00x = 11x = 22

x = 53

x = 264

x = big5

x = BIGGER6


x = 00x = 11x = 22

x = 53

x = 264

x = big5

x = BIGGER6

“Orbit tends to infinity”


x = 00x = 1x = 2

x = 3

x = 4

x = 5

x = 6


x = 00x = 01x = 2

x = 3

x = 4

x = 5

x = 6


x = 00x = 01x = 02

x = 3

x = 4

x = 5

x = 6


x = 00x = 01x = 02

x = 03

x = 4

x = 5

x = 6


x = 00x = 01x = 02

x = 03

x = 04

x = 05

x = 06

“A fixed point”

Example: x - 1 Seed 02

x = 00x = 1x = 2

x = 3

x = 4

x = 5

x = 6


x = 00x = -11x = 2

x = 3

x = 4

x = 5

x = 6


x = 00x = -11x = 02

x = 3

x = 4

x = 5

x = 6


x = 00x = -11x = 02

x = -13

x = 4

x = 5

x = 6


x = 00x = -11x = 02

x = -13

x = 04

x = 5

x = 6


x = 00x = -11x = 02

x = -13

x = 04

x = -15

x = 06

“A two- cycle”

Example: x - 1.1 Seed 02

x = 00x = 1x = 2

x = 3

x = 4

x = 5

x = 6


x = 00x = -1.11x = 2

x = 3

x = 4

x = 5

x = 6


x = 00x = -1.11x = 0.112

x = 3

x = 4

x = 5

x = 6


x = 00x = -1.11x = 0.112

x = 3

x = 4

x = 5

x = 6

time for the computer!

Excel + OrbDgm

Observation:

For some real values of c, the orbit of 0 goes to infinity, but for other values, the orbit of 0 does not escape.

Complex Iteration

Iterate z + c2

complexnumbers

Example: z + i Seed 02

z = 00z = 1z = 2

z = 3

z = 4

z = 5

z = 6


z = 00z = i1z = 2

z = 3

z = 4

z = 5

z = 6


z = 00z = i1z = -1 + i2

z = 3

z = 4

z = 5

z = 6


z = 00z = i1z = -1 + i2

z = -i 3

z = 4

z = 5

z = 6


z = 00z = i1z = -1 + i2

z = -i 3

z = -1 + i 4

z = 5

z = 6


z = 00z = i1z = -1 + i2

z = -i 3

z = -1 + i 4

z = -i 5

z = 6


z = 00z = i1z = -1 + i2

z = -i 3

z = -1 + i 4

z = -i 5

z = -1 + i 6

2-cycle


1-1

i

-i


-i

-1 1

i


1-1

i

-i


-i

-1 1

i


1-1

i

-i


-i

-1 1

i

Example: z + 2i Seed 02

z = 00z = 1z = 2

z = 3

z = 4

z = 5

z = 6

Example: z + 2i Seed 02

z = 00z = 2i1z = -4 + 2i 2

z = 12 - 14i3

z = -52 + 336i 4

z = big 5

z = BIGGER 6

Off toinfinity

Same observation

Sometimes orbit of 0 goes to infinity, other times it does not.

The Mandelbrot Set:

All c-values for which the orbit of 0 does NOT go to infinity.

Algorithm for computing M

Start with a grid of complex numbers


Each grid point is a complex c-value.


Compute the orbitof 0 for each c. Ifthe orbit of 0 escapes,color that grid point.

red = fastest escape



orange = slower



yellowgreenblueviolet


Compute the orbitof 0 for each c. Ifthe orbit of 0 does not escape, leave that grid pointblack.

The eventual orbit of 0


3-cycle


4-cycle


5-cycle


2-cycle


fixed point


goes to infinity


gone to infinity

How understand the periods of the bulbs?

junction point

three spokes attached

Period 3 bulb

junction point

three spokes attached

Period 4 bulb

Period 5 bulb

Period 7 bulb

Period 13 bulb

Filled Julia Set:

Filled Julia Set:

Fix a c-value. The filled Julia set is all of the complex seeds whose

orbits do NOT go to infinity.

Example: z2

Seed: In Julia set?

0

Example: z2

Seed: In Julia set?

0 Yes

Example: z2

Seed: In Julia set?

0 Yes

1

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1 Yes

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1 Yes

i

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1 Yes

i Yes

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1 Yes

i Yes

2i

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1 Yes

i Yes

2i No

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1 Yes

i Yes

2i No

5

Example: z2

Seed: In Julia set?

0 Yes

1 Yes

-1 Yes

i Yes

2i No

5 No way

Filled Julia Set for z 2

All seeds on and inside the unit circle.

i

1-1

Other filled Julia sets

Choose c from some componentof the Mandelbrot set, then use the

same algorithm as before:colored points escape to ∞ and soare not in the filled Julia set;

black points form the filled Julia set.

M-set computer

If c is in the Mandelbrot set, then the filled Julia set is always a connected set.

Other filled Julia sets

But if c is not in the Mandelbrot set, then the filled Julia set is totally disconnected.

Amazingly, the orbit of 0 knows it all:

Theorem: For z2 + c:

If the orbit of 0 goes to infinity, the Julia set is a Cantor set (totally disconnected, “fractal dust”), and c is not in the Mandelbrot set.

But if the orbit of 0 does not go to infinity,the Julia set is connected (just one piece),and c is in the Mandelbrot set.

M-set movie maker --- frame # 200

Animations:

In and out of M

arrangementof the bulbs

Saddle node

Period doubling

Period 4 bifurcation

How do we understand the arrangement of the bulbs?

How do we understand the arrangement of the bulbs?

Assign a fraction p/q to eachbulb hanging off the main cardioid;

q = period of the bulb.

?/3 bulb

principal spoke

shortest spoke

1/3 bulb

1/3 bulb

1/3

??? bulb

1/3

1/4 bulb

1/3

1/4 bulb

1/3

1/4

??? bulb

1/3

1/4

2/5 bulb

1/3

1/4

2/5 bulb

1/3

1/42/5

??? bulb

1/3

1/42/5

3/7 bulb

1/3

1/42/5

3/7 bulb

1/3

1/42/5

3/7

3/7 bulb

1/3

1/43/7

2/5

??? bulb

1/3

1/43/7

2/5

1/2 bulb

1/3

1/43/7

1/2

2/5

??? bulb

1/3

1/43/7

1/2

2/5

2/3 bulb

1/3

1/43/7

1/2

2/3

2/5

How to count

1/4

How to count

1/3

1/4

How to count

1/3

1/42/5

How to count

1/3

1/42/5

3/7

How to count

1/3

1/42/5

3/7

1/2

How to count

1/3

1/42/5

3/7

1/2

2/3

How to count

1/3

1/42/5

3/7

1/2

2/3

The bulbs are arranged in the exactorder of the rational numbers.

How to count

1/3

1/42/5

3/7

1/2

2/3

The bulbs are arranged in the exactorder of the rational numbers.

1/101

32,123/96,787

How to count

Animations:

Mandelbulbs

Spiralling fingers

How to add

How to add

1/2

How to add

1/2

1/3

How to add

1/2

1/3

2/5

How to add

1/2

1/3

2/5

3/7

+ =

1/2 + 1/3 = 2/5

+ =

1/2 + 2/5 = 3/7

Undergrads who add fractions this way will be subject to a minimum of five years in jail where

they must do at least 500 integrals per day.

Only PhDs in mathematics are allowed to add fractions this way.

221/2

0/1

Here’s an interesting sequence:

221/2

0/1

Watch the denominators

1/3

221/2

0/1


1/3

2/5

221/2

0/1


1/3

2/5

3/8

221/2

0/1


1/3

2/5

3/85/13

221/2

0/1

What’s next?

1/3

2/5

3/85/13

221/2

0/1

What’s next?

1/3

2/5

3/85/13

8/21

221/2

0/1

The Fibonacci sequence

1/3

2/5

3/85/13

8/2113/34

The Farey Tree

€

0

1

€

1

1

The Farey Tree

€

0

1

€

1

1

How get the fraction in betweenwith the smallest denominator?

The Farey Tree

€

0

1

€

1

1

€

1

2

Farey addition

How get the fraction in betweenwith the smallest denominator?

The Farey Tree

€

0

1

€

1

1

€

1

2

€

1

3

€

2

3

The Farey Tree

€

0

1

€

1

1

€

1

2

€

1

3

€

2

3

€

2

5

€

1

4

€

3

5

€

3

4

The Farey Tree

€

0

1

€

1

1

€

1

2

€

1

3

€

2

3

€

2

5

€

1

4

€

3

5

€

3

4

€

3

8

€

5

13

....

essentially the golden number

Another sequence (denominatorsonly)

1

2


1

2

3


1

2

3

4


1

2

3

4

5


1

2

3

4

5

6


1

2

3

4

5

67

sequence

1

2

3

4

5

67

Devaney

The Dynamical Systems and Technology Project at Boston University

website: math.bu.edu/DYSYS:

Have fun!

Mandelbrot set explorer;Applets for investigating M-set;Applets for other complex functions;Chaos games, orbit diagrams, etc.

Farey.qt

Farey tree

D-sequence

Continued fraction expansion

Far from rationals

Other topics

Website


Let’s rewrite the sequence:

1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34, ..... as a continued fraction:


12

= 12

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....


13

= 12 + 1

1

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....


25

= 12 + 1

1 + 11

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....


38

= 12 + 1

1 + 11 1

1+

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....


= 12 + 1

1 + 11 1

1+

11

+

513

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....


= 12 + 1

1 + 11 1

1+

11

+

821

11

+

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....


= 12 + 1

1 + 11 1

1+

11

+

1334

11

+

11

+

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....


= 12 + 1

1 + 11 1

1+

11

+

1334

11

+

11

+

essentially the1/golden number

the sequence: 1/2, 1/3, 2/5, 3/8, 5/13, 8/21, 13/34,.....

We understand what happens for

= 1a + 1

b + 1c 1

d+

1e

+

1f

+

1g

+

where all entries in the sequence a, b, c, d,.... are bounded above. But if that sequence grows too quickly, we’re in trouble!!!

etc.€

θ

The real way to prove all this:

Need to measure: the size of bulbs the length of spokes the size of the “ears.”

There is an external Riemann map : C - D C - Mtaking the exterior of the unit disk to the exterior of the Mandelbrot set.

€

Φ

€

Φ

€

Φ

€

Φ takes straight rays in C - D to the “external rays” in C - M

01/2

1/3

2/3 €

γ0

€

γ1/3

€

γ2/3€

γ1/2

external ray of angle 1/3

€

1

3→2

3→1

3→

1

7→2

7→4

7→1

7→

1

5→2

5→4

5→3

5→1

5→

Suppose p/q is periodic of period k under doubling mod 1:

period 2

period 3

period 4

€

1

3→2

3→1

3→

1

7→2

7→4

7→1

7→

1

5→2

5→4

5→3

5→1

5→

Suppose p/q is periodic of period k under doubling mod 1:

period 2

period 3

period 4

Then the external ray of angle p/qlands at the “root point” of a period k bulb in the Mandelbrot set.

€

Φ0

1/3

2/3

€

γ00 is fixed under angle doubling, so lands at the cusp of the main cardioid.

€

γ0

€

Φ0

1/3

2/3

€

γ1/3

€

γ2/3

€

γ0

€

γ1/3

€

γ2/31/3 and 2/3 have period 2 under doubling, so and land at the root of the period 2 bulb.

2

€

Φ0

1/3

2/3

€

γ1/3

€

γ2/3

€

γ0

€

γ1/3

€

γ2/3And if lies between 1/3 and 2/3,then lies between and .

2

€

θ

€

γθ

€

θ

€

γθ

€

Φ0

1/3

2/3

€

γ1/3

€

γ2/3

€

γ0

So the size of the period 2 bulb is, by definition, the length of the set of rays

between the root point rays, i.e., 2/3-1/3=1/3.

2

€

Φ0

1/3

2/3

€

γ1/3

€

γ2/3

€

γ0

1/15 and 2/15 have period 4, andare smaller than 1/7....

1/72/7

3/7

4/7

5/7

6/7

€

γ1/7

€

γ2/7

€

γ3/7

€

γ4 /7

€

γ5/7

€

γ6/7

2

3

3

1/15

2/15

€

Φ0

1/3

2/3

€

γ1/3

€

γ2/3

€

γ0

1/15 and 2/15 have period 4, andare smaller than 1/7....

1/72/7

3/7

4/7

5/7

6/7

€

γ1/7

€

γ2/7

€

γ3/7

€

γ4 /7

€

γ5/7

€

γ6/7

2

3

3

1/15

2/15

€

γ1/15€

γ2/15

€

Φ0

1/3

2/3

€

γ1/3

€

γ2/3

€

γ0

1/72/7

3/7

4/7

5/7

6/7

€

γ1/7

€

γ2/7

€

γ3/7

€

γ4 /7

€

γ5/7

€

γ6/7

2

3

3

1/15

2/15

€

γ1/15€

γ2/15

3/15 and 4/15 have period 4, andare between 1/7 and 2/7....

€

Φ0

1/3

2/3

€

γ1/3

€

γ2/3

€

γ0


1/72/7

3/7

4/7

5/7

6/7

€

γ1/7

€

γ2/7

€

γ3/7

€

γ4 /7

€

γ5/7

€

γ6/7

2

3

3

1/15

2/15

€

γ1/15€

γ2/15

1/72/7


1/72/7


3/154/15

So what do we know about M?

All rational external rays land at a single point in M.


All rational external rays land at a single point in M.

Rays that are periodic under doubling land at root points of a bulb.

Non-periodic rational raysland at Misiurewicz points(how we measure lengthof antennas).


“Highly irrational” rays also land at unique points, and we understand what goes on here.

“Highly irrational" = “far”from rationals, i.e.,

€

θ −pq>c

qk

So what do we NOT know about M?

But we don't know if irrationals that are “close” to rationals land.

So we won't understandquadratic functions untilwe figure this out.

MLC Conjecture:

The boundary of the M-setis “locally connected” ---if so, all rays land and we are in heaven!. But if not......

The Dynamical Systems and Technology Project at Boston University

website: math.bu.edu/DYSYS

Have fun!

A number is far from the rationals if:

€

θ

€

|θ − p /q |

€

>


€

θ

€

|θ − p /q |

€

>

€

c /qk


€

θ

€

|θ − p /q |

€

>

€

c /qk

This happens if the “continued fraction expansion” of has only bounded terms.

€

θ

Excel quad iteration M-set iterator Movie maker 75.

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Transcript of Excel quad iteration M-set iterator Movie maker 75.