Roots and Chaos
Davar Khoshnevisan
Department of Mathematics
University of Utah
http://www.math.utah.edu/~davar
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
1 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
2. Randomness in statistical sciences [sampling]
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
2. Randomness in statistical sciences [sampling]
3. Randomness in biological sciences [population genetics]
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
2. Randomness in statistical sciences [sampling]
3. Randomness in biological sciences [population genetics]
4. Randomness in the financial sciences [stock market, option
pricing,. . . ]
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
2. Randomness in statistical sciences [sampling]
3. Randomness in biological sciences [population genetics]
4. Randomness in the financial sciences [stock market, option
pricing,. . . ]
5. What is random? [E.g., “is 4.167 random?”]
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
2. Randomness in statistical sciences [sampling]
3. Randomness in biological sciences [population genetics]
4. Randomness in the financial sciences [stock market, option
pricing,. . . ]
5. What is random? [E.g., “is 4.167 random?”]
6. At least two ways of reaching randomness:
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
2. Randomness in statistical sciences [sampling]
3. Randomness in biological sciences [population genetics]
4. Randomness in the financial sciences [stock market, option
pricing,. . . ]
5. What is random? [E.g., “is 4.167 random?”]
6. At least two ways of reaching randomness:
6.1 (Human-made) models for understanding complicated problems;
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
What is “random”?
1. Randomness in physical sciences [quantum mechanics, statistical
physics, astrophysics, . . . ]
2. Randomness in statistical sciences [sampling]
3. Randomness in biological sciences [population genetics]
4. Randomness in the financial sciences [stock market, option
pricing,. . . ]
5. What is random? [E.g., “is 4.167 random?”]
6. At least two ways of reaching randomness:
6.1 (Human-made) models for understanding complicated problems;
6.2 Intrinsic randomness [today’s main topic].
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
2 / 18
Randomness in modeling
1. What happens during a coin toss? Can we predict the outcome?
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
3 / 18
Randomness in modeling
1. What happens during a coin toss? Can we predict the outcome?
2. A standard method: Find a mathematical model for the outcome
[Pr(Heads) = Pr(Tails) = 1/2].
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
3 / 18
Randomness in modeling
1. What happens during a coin toss? Can we predict the outcome?
2. A standard method: Find a mathematical model for the outcome
[Pr(Heads) = Pr(Tails) = 1/2].
3. Probability theory studies the model [law of large numbers, central
limit theorem, etc.].
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
3 / 18
Randomness in modeling
1. What happens during a coin toss? Can we predict the outcome?
2. A standard method: Find a mathematical model for the outcome
[Pr(Heads) = Pr(Tails) = 1/2].
3. Probability theory studies the model [law of large numbers, central
limit theorem, etc.].
4. Is the model any good?
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
3 / 18
Intrinsic Randomness
1. Today’s topic: Sometimes complex systems act as if they were truly
governed by random underlying rules.
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
4 / 18
Intrinsic Randomness
1. Today’s topic: Sometimes complex systems act as if they were truly
governed by random underlying rules.
2. Complex 6= complicated [as we shall see].
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
4 / 18
Intrinsic Randomness
1. Today’s topic: Sometimes complex systems act as if they were truly
governed by random underlying rules.
2. Complex 6= complicated [as we shall see].
3. Might draw conclusions about the existence of random patterns in
various disciplines.
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
4 / 18
A
Basic Example
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
If area of square = 4, then side = 2 (2 × 2 = 4)
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
5 / 18
A
Basic Example
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
I
If area of square = 4, then side = 2 (2 × 2 = 4)
√
I.e., 4 = 2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
5 / 18
A
Basic Example
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
I
I
If area of square = 4, then side = 2 (2 × 2 = 4)
√
I.e., 4 = 2
√
√
9 = 3, 16 = 4, . . .
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
5 / 18
A
Basic Example
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
I
I
I
If area of square = 4, then side = 2 (2 × 2 = 4)
√
I.e., 4 = 2
√
√
9 = 3, 16 = 4, . . .
√
2=?
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
5 / 18
A
Basic Example
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
If area of square = 4, then side = 2 (2 × 2 = 4)
√
I.e., 4 = 2
√
√
9 = 3, 16 = 4, . . .
√
2=?
I
By a calculator:
I
I
I
D. Khoshnevisan (Univ of Utah)
√
2 ≈ 1.414213562373095
Roots & chaos
June 16, 2015
5 / 18
A
Basic Example
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
I
I
I
I
I
If area of square = 4, then side = 2 (2 × 2 = 4)
√
I.e., 4 = 2
√
√
9 = 3, 16 = 4, . . .
√
2=?
√
2 ≈ 1.414213562373095
√
How is this done? First, a few facts about 2
D. Khoshnevisan (Univ of Utah)
By a calculator:
Roots & chaos
June 16, 2015
5 / 18
On
√
2
≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
√
I
2 is an irrational number (ascribed to Hippasus; around 5 BC)
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
6 / 18
On
√
2
≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
√
I
I
2 is an irrational number (ascribed to Hippasus; around 5 BC)
√
Open Question: How many zeros in the decimal expansion of 2?
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
6 / 18
On
√
2
≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
√
I
I
I
2 is an irrational number (ascribed to Hippasus; around 5 BC)
√
Open Question: How many zeros in the decimal expansion of 2?
Conjecture: [Émile Borel, Comptes Rendus Acad Sci Paris 230, 1950,
pp. 591–593] √
all digits 0–9 are equidistributed in the decimal
expansion of 2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
6 / 18
Mathematical
Experimentation
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
√
What does 2 “look like”? Here is a simple first attempt:
a
a2
√
1
√1
⇒1< 2<2
2
2
2
4
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
7 / 18
Mathematical
Experimentation
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
I
√
What does 2 “look like”? Here is a simple first attempt:
a
a2
√
1
√1
⇒1< 2<2
2
2
2
4
√
So 2 ≈ 1.??. Therefore:
a
a2
1.1 1.21
1.2 1.44
√
1.3 1.69 ⇒ 1.4 < 2 < 1.5
1.4 1.96
√
2
2
1.5 2.25
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
7 / 18
Mathematical
Experimentation
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
. . . So
a
1.41
√
2
1.42
√
2 ≈ 1.4??. Therefore:
a2
√
1.9881
⇒ 1.41 < 2 < 1.42
2
2.0164
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
8 / 18
Mathematical
Experimentation
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
I
. . . So
a
1.41
√
2
1.42
√
2 ≈ 1.4??. Therefore:
a2
√
1.9881
⇒ 1.41 < 2 < 1.42
2
2.0164
√
. . . So 2 ≈ 1.41??. Therefore:
a
a2
1.411 1.990921
1.412 1.993744
√
1.413 1.996569 ⇒ 1.414 < 2 < 1.415 etc. . . .
1.414
1.999396
√
2
2
1.415 2.002225
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
8 / 18
The Babylonian Clay Tablet
(circa 1600–1800 BC; Yale Collection 7289)
√
10
24
51
2 ≈ 1 + 60
+ 60
1.41421
2 + 603 = |
{z } 296296 · · ·
X
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
9 / 18
The Babylonian Way (likely)
Newton’s
Method for
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
Want to solve x 2 = 2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
10 / 18
The Babylonian Way (likely)
Newton’s
Method for
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
Want to solve x 2 = 2
I
Equivalently,
x
2
= x1 ; i.e., x =
D. Khoshnevisan (Univ of Utah)
x
2
+
1
x
Roots & chaos
June 16, 2015
10 / 18
The Babylonian Way (likely)
Newton’s
Method for
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
Want to solve x 2 = 2
I
Equivalently,
I
Try an initial guess [or “seed”] x0 [say x0 := 1]
x
2
= x1 ; i.e., x =
D. Khoshnevisan (Univ of Utah)
x
2
+
1
x
Roots & chaos
June 16, 2015
10 / 18
The Babylonian Way (likely)
Newton’s
Method for
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
Want to solve x 2 = 2
I
Equivalently,
I
Try an initial guess [or “seed”] x0 [say x0 := 1]
I
Next guess: x1 :=
x
2
= x1 ; i.e., x =
D. Khoshnevisan (Univ of Utah)
x0
2
+
1
x0
x
2
+
1
x
[say x1 :=
Roots & chaos
3
2
= 1.5]
June 16, 2015
10 / 18
The Babylonian Way (likely)
Newton’s
Method for
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
Want to solve x 2 = 2
I
Equivalently,
I
Try an initial guess [or “seed”] x0 [say x0 := 1]
I
Next guess: x1 :=
I
Next guess: x2 :=
x
2
= x1 ; i.e., x =
D. Khoshnevisan (Univ of Utah)
x0
2
x1
2
+
+
1
x0
1
x1
x
2
+
1
x
[say x1 :=
[say x2 :=
Roots & chaos
3
2 = 1.5]
17
1.41 6̄]
12 = |{z}
June 16, 2015
10 / 18
The Babylonian Way (likely)
Newton’s
Method for
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
Want to solve x 2 = 2
I
Equivalently,
I
Try an initial guess [or “seed”] x0 [say x0 := 1]
I
Next guess: x1 :=
I
I
x
2
= x1 ; i.e., x =
+
Next guess: x2 :=
x0
2
x1
2
Next guess: x3 :=
x2
2
D. Khoshnevisan (Univ of Utah)
x
2
+
1
x
[say x1 :=
+
1
x0
1
x1
+
1
x2
[say x3 :=
[say x2 :=
Roots & chaos
3
2 = 1.5]
17
1.41 6̄]
12 = |{z}
577
408
≈ 1.41421
| {z } 568627451]
June 16, 2015
10 / 18
The Babylonian Way (likely)
Newton’s
Method for
√
2 ≈ 1.41421356237309504880168872420969807856967187537694807317667973799 · · ·
I
Want to solve x 2 = 2
I
Equivalently,
I
Try an initial guess [or “seed”] x0 [say x0 := 1]
I
Next guess: x1 :=
I
x
2
= x1 ; i.e., x =
+
Next guess: x2 :=
x0
2
x1
2
I
Next guess: x3 :=
I
Next guess: x4 :=
D. Khoshnevisan (Univ of Utah)
x
2
+
1
x
[say x1 :=
+
1
x0
1
x1
x2
2
+
1
x2
[say x3 :=
577
408
x3
2
+
1
x3
[say x4 :=
665857
470832
[say x2 :=
Roots & chaos
3
2 = 1.5]
17
1.41 6̄]
12 = |{z}
≈ 1.41421
| {z } 568627451]
≈ 1.41421356237
|
{z
} 469]
June 16, 2015
10 / 18
An Amusing Aside
I
x=
√
2 is the only positive solution to x 2 = 2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
(x − 1)(x + 1) = 1
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
(x − 1)(x + 1) = 1
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
1
(x − 1)(x + 1) = 1 equivalently, x − 1 = 1+x
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
1
(x − 1)(x + 1) = 1 equivalently, x − 1 = 1+x
1
Equivalently, x = 1 + 1+x
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
1
(x − 1)(x + 1) = 1 equivalently, x − 1 = 1+x
1
Equivalently, x = 1 + 1+x
This is a “self-referential” description of x:
x
= 1+
D. Khoshnevisan (Univ of Utah)
1
1+x
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
1
(x − 1)(x + 1) = 1 equivalently, x − 1 = 1+x
1
Equivalently, x = 1 + 1+x
This is a “self-referential” description of x:
x
= 1+
D. Khoshnevisan (Univ of Utah)
1
1+x
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
1
(x − 1)(x + 1) = 1 equivalently, x − 1 = 1+x
1
Equivalently, x = 1 + 1+x
This is a “self-referential” description of x:
x
= 1+
D. Khoshnevisan (Univ of Utah)
1
1
=1+
1
1+x
2 + 1+x
Roots & chaos
June 16, 2015
11 / 18
An Amusing Aside
I
I
I
I
I
√
x = 2 is the only positive solution to x 2 = 2
⇔ x 2 − 1 = 1, which when factored yields . . .
1
(x − 1)(x + 1) = 1 equivalently, x − 1 = 1+x
1
Equivalently, x = 1 + 1+x
This is a “self-referential” description of x:
x
= 1+
1
1
1
=1+
=1+
.
1
1+x
2 + 1+x
2 + 2+1 1
1+x
I
Theorem. The preceding can be carried out ad infinitum; that is,
√
1
2=1+
1
2+
1
2+
2 + ···
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
11 / 18
Extensions
I
Can apply our method to
D. Khoshnevisan (Univ of Utah)
√
3 ≈ 1.732050807568877 . . . :
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
D. Khoshnevisan (Univ of Utah)
⇔
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
Roots & chaos
x=
x
2
+
3
2x
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
I
x2 = 3
x0 = 2
D. Khoshnevisan (Univ of Utah)
⇔
[1 <
√
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
3 < 2]
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
⇔
√
I
I
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
x0 = 2
[1 < 3 < 2]
x1 = x20 + 2x30 = 47 = |{z}
1.7 5
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
⇔
√
I
I
I
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
x0 = 2
[1 < 3 < 2]
x1 = x20 + 2x30 = 47 = |{z}
1.7 5
x1
3
97
x2 = 2 + 2x1 = 56 ≈ 1.732
| {z } 142857142857
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
⇔
√
I
I
I
I
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
x0 = 2
[1 < 3 < 2]
x1 = x20 + 2x30 = 47 = |{z}
1.7 5
x1
3
97
x2 = 2 + 2x1 = 56 ≈ 1.732
| {z } 142857142857
x3 =
x2
2
+
3
2x2
D. Khoshnevisan (Univ of Utah)
=
37634
21728
≈ 1.7320508
| {z } 10014728 . . .
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
⇔
√
I
I
I
I
I
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
x0 = 2
[1 < 3 < 2]
x1 = x20 + 2x30 = 47 = |{z}
1.7 5
x1
3
97
x2 = 2 + 2x1 = 56 ≈ 1.732
| {z } 142857142857
x3 =
x2
2
+
3
2x2
=
37634
21728
≈ 1.7320508
| {z } 10014728 . . .
Aside. For a continued-fraction expansion, start with
x2 = 3 ⇔ x2 − 1 = 2 ⇔
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
⇔
√
I
I
I
I
I
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
x0 = 2
[1 < 3 < 2]
x1 = x20 + 2x30 = 47 = |{z}
1.7 5
x1
3
97
x2 = 2 + 2x1 = 56 ≈ 1.732
| {z } 142857142857
x3 =
x2
2
+
3
2x2
=
37634
21728
≈ 1.7320508
| {z } 10014728 . . .
Aside. For a continued-fraction expansion, start with
x2 = 3 ⇔ x2 − 1 = 2 ⇔
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
⇔
√
I
I
I
I
I
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
x0 = 2
[1 < 3 < 2]
x1 = x20 + 2x30 = 47 = |{z}
1.7 5
x1
3
97
x2 = 2 + 2x1 = 56 ≈ 1.732
| {z } 142857142857
x3 =
x2
2
+
3
2x2
=
37634
21728
≈ 1.7320508
| {z } 10014728 . . .
Aside. For a continued-fraction expansion, start with
x 2 = 3 ⇔ x 2 − 1 = 2 ⇔ (x − 1)(x + 1) = 2 ⇔
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
12 / 18
Extensions
I
Can apply our method to
I
x2 = 3
⇔
√
I
I
I
I
I
x
2
√
=
3 ≈ 1.732050807568877 . . . :
3
2x
⇔
x=
x
2
+
3
2x
x0 = 2
[1 < 3 < 2]
x1 = x20 + 2x30 = 47 = |{z}
1.7 5
x1
3
97
x2 = 2 + 2x1 = 56 ≈ 1.732
| {z } 142857142857
x3 =
x2
2
+
3
2x2
=
37634
21728
≈ 1.7320508
| {z } 10014728 . . .
Aside. For a continued-fraction expansion, start with
x 2 = 3 ⇔ x 2 − 1 = 2 ⇔ (x − 1)(x + 1) = 2 ⇔ x = 1 +
I
This yields
√
2
3=1+
2
2+
2+
D. Khoshnevisan (Univ of Utah)
2
1+x .
Roots & chaos
2
2 + ···
June 16, 2015
12 / 18
Related Exercises
I
The method can be used to compute
D. Khoshnevisan (Univ of Utah)
Roots & chaos
√
5, etc. . . .
June 16, 2015
13 / 18
Related Exercises
I
I
√
The method can be used to compute 5, etc. . . .
√
E.g., x = 5 can be written as
4
...
x 2 − 1 = 4 ⇔ (x − 1)(x + 1) = 4 ⇔ x = 1 + 1+x
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
13 / 18
Related Exercises
I
I
I
√
The method can be used to compute 5, etc. . . .
√
E.g., x = 5 can be written as
4
...
x 2 − 1 = 4 ⇔ (x − 1)(x + 1) = 4 ⇔ x = 1 + 1+x
This yields
√
4
5=1+
4
2+
4
2+
2 + ···
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
13 / 18
Related Exercises
I
√
The method can be used to compute 5, etc. . . .
√
E.g., x = 5 can be written as
4
...
x 2 − 1 = 4 ⇔ (x − 1)(x + 1) = 4 ⇔ x = 1 + 1+x
This yields
√
4
5=1+
4
2+
4
2+
2 + ···
I
Or x 2 − 4 = 1 ⇔ (x − 2)(x + 2) = 1 ⇔ x = 2 +
I
I
D. Khoshnevisan (Univ of Utah)
Roots & chaos
1
2+x .
June 16, 2015
13 / 18
Related Exercises
I
I
I
I
I
√
The method can be used to compute 5, etc. . . .
√
E.g., x = 5 can be written as
4
...
x 2 − 1 = 4 ⇔ (x − 1)(x + 1) = 4 ⇔ x = 1 + 1+x
This yields
√
4
5=1+
4
2+
4
2+
2 + ···
Or x 2 − 4 = 1 ⇔ (x − 2)(x + 2) = 1 ⇔ x = 2 +
This yields
√
1
5=2+
1
4+
1
4+
4 + ···
D. Khoshnevisan (Univ of Utah)
Roots & chaos
1
2+x .
June 16, 2015
13 / 18
A silly extension . . . or is it?
I
Now try: x 2 = −1
D. Khoshnevisan (Univ of Utah)
⇔
x
2
1
= − 2x
Roots & chaos
⇔
x=
x
2
−
June 16, 2015
1
2x
14 / 18
A silly extension . . . or is it?
I
Now try: x 2 = −1
I
x0 = 0.1
D. Khoshnevisan (Univ of Utah)
⇔
x
2
1
= − 2x
Roots & chaos
⇔
x=
x
2
−
June 16, 2015
1
2x
14 / 18
A silly extension . . . or is it?
I
Now try: x 2 = −1
I
x0 = 0.1
I
x1 =
0.1
2
−
1
2×0.1
D. Khoshnevisan (Univ of Utah)
⇔
x
2
1
= − 2x
⇔
x=
x
2
−
1
2x
= −4.95
Roots & chaos
June 16, 2015
14 / 18
A silly extension . . . or is it?
I
Now try: x 2 = −1
I
x0 = 0.1
I
x1 =
I
x2 =
⇔
x
2
1
= − 2x
⇔
x=
x
2
−
1
2x
0.1
1
2 − 2×0.1 = −4.95
1
−4.95
− 2×(−4.95)
≈ −2.3740
2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
14 / 18
A silly extension . . . or is it?
x
2
1
= − 2x
x
2
1
2x
I
Now try: x 2 = −1
I
x0 = 0.1
I
x1 =
I
x2 =
I
x3 ≈ −0.9764, x4 ≈ 0.0239, x5 ≈ −20.9027, x6 ≈ −10.4274, . . .
⇔
⇔
x=
−
0.1
1
2 − 2×0.1 = −4.95
1
−4.95
− 2×(−4.95)
≈ −2.3740
2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
14 / 18
A silly extension . . . or is it?
x
2
1
= − 2x
x
2
1
2x
I
Now try: x 2 = −1
I
x0 = 0.1
I
x1 =
I
x2 =
I
x3 ≈ −0.9764, x4 ≈ 0.0239, x5 ≈ −20.9027, x6 ≈ −10.4274, . . .
I
x7 ≈ −5.1658, x8 = −2.4861, x9 = −1.0419, . . .
⇔
⇔
x=
−
0.1
1
2 − 2×0.1 = −4.95
1
−4.95
− 2×(−4.95)
≈ −2.3740
2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
14 / 18
A silly extension . . . or is it?
x
2
1
= − 2x
x
2
1
2x
I
Now try: x 2 = −1
I
x0 = 0.1
I
x1 =
I
x2 =
I
x3 ≈ −0.9764, x4 ≈ 0.0239, x5 ≈ −20.9027, x6 ≈ −10.4274, . . .
I
x7 ≈ −5.1658, x8 = −2.4861, x9 = −1.0419, . . .
I
Do see a pattern?
⇔
⇔
x=
−
0.1
1
2 − 2×0.1 = −4.95
1
−4.95
− 2×(−4.95)
≈ −2.3740
2
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
14 / 18
A2 silly extension . . . or is it?
x = −1 . . . x =
x
2
−
1
2x
. . . 5000 iterations
1600
1400
1200
1000
800
600
400
200
0
−200
−400
0
500
1000
D. Khoshnevisan (Univ of Utah)
1500
2000
2500
3000
3500
Roots & chaos
4000
4500
5000
June 16, 2015
15 / 18
A2 silly extension . . . or is it?
x = −1 . . . x =
x
2
−
1
2x
. . . 5000 iterations . . . “heavy tails” . . . n versus
2
π
arctan(xn )
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
0
500
1000
D. Khoshnevisan (Univ of Utah)
1500
2000
2500
3000
3500
Roots & chaos
4000
4500
5000
June 16, 2015
16 / 18
A2 silly extension . . . or is it?
x = −1 . . . x =
x
2
−
1
2x
. . . 5000 iterations . . . “heavy tails” . . . n versus
2
π
arctan(xn )
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
0
500
1000
D. Khoshnevisan (Univ of Utah)
1500
2000
2500
3000
3500
Roots & chaos
4000
4500
5000
June 16, 2015
17 / 18
A2 silly extension . . . or is it?
x = −1 . . . x =
x
2
−
1
2x
. . . 50,000 iterations . . . “heavy tails” . . . n versus
2
π
arctan(xn )
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
4
x 10
D. Khoshnevisan (Univ of Utah)
Roots & chaos
June 16, 2015
18 / 18