the koch snowflake curve - boise state universitywright/courses/m598/agana... · 2014-12-12 ·...

The Koch Snowflake Curve

Monica J. Agana

Boise State University

Fall 2014

Monica J. Agana (Boise State University) Fall 2014 1 / 16

Motivation

To introduce the audience to the Koch snowflake construction, and discussseveral of its main properties.

Monica J. Agana (Boise State University) Fall 2014 2 / 16

Main Properties

Infinite Perimeter

Finite Area

Continuous Everywhere

Nowhere Differentiable

Monica J. Agana (Boise State University) Fall 2014 3 / 16

Main Properties

Infinite Perimeter

Finite Area

Continuous Everywhere

Nowhere Differentiable

Monica J. Agana (Boise State University) Fall 2014 3 / 16

Main Properties

Infinite Perimeter

Finite Area

Continuous Everywhere

Nowhere Differentiable

Monica J. Agana (Boise State University) Fall 2014 3 / 16

Main Properties

Infinite Perimeter

Finite Area

Continuous Everywhere

Nowhere Differentiable

Monica J. Agana (Boise State University) Fall 2014 3 / 16

Koch Curve

Start with a line segment of any length.

Monica J. Agana (Boise State University) Fall 2014 4 / 16

Koch Curve

Monica J. Agana (Boise State University) Fall 2014 5 / 16

Koch Snowflake

Monica J. Agana (Boise State University) Fall 2014 6 / 16

Historical Background

(1900’s) Niels Fabian Helge von Koch, a Swedish mathematician [2].

(1904) Koch snowflake: Sur une courbe continue sans tangente,obtenue par une construction geometrique elementaire” [3].

(1906) Koch curve: Une methode geometrique elementaire pourl’etude de certaines questions de la theorie des courbes plane [2].

Continuous everywhere but nowhere differentiable functions [2].

Monica J. Agana (Boise State University) Fall 2014 7 / 16

Historical Background

(1900’s) Niels Fabian Helge von Koch, a Swedish mathematician [2].

(1904) Koch snowflake: Sur une courbe continue sans tangente,obtenue par une construction geometrique elementaire” [3].

(1906) Koch curve: Une methode geometrique elementaire pourl’etude de certaines questions de la theorie des courbes plane [2].

Continuous everywhere but nowhere differentiable functions [2].

Monica J. Agana (Boise State University) Fall 2014 7 / 16

Historical Background

(1900’s) Niels Fabian Helge von Koch, a Swedish mathematician [2].

(1904) Koch snowflake: Sur une courbe continue sans tangente,obtenue par une construction geometrique elementaire” [3].

(1906) Koch curve: Une methode geometrique elementaire pourl’etude de certaines questions de la theorie des courbes plane [2].

Continuous everywhere but nowhere differentiable functions [2].

Monica J. Agana (Boise State University) Fall 2014 7 / 16

Historical Background

(1900’s) Niels Fabian Helge von Koch, a Swedish mathematician [2].

(1904) Koch snowflake: Sur une courbe continue sans tangente,obtenue par une construction geometrique elementaire” [3].

(1906) Koch curve: Une methode geometrique elementaire pourl’etude de certaines questions de la theorie des courbes plane [2].

Continuous everywhere but nowhere differentiable functions [2].

Monica J. Agana (Boise State University) Fall 2014 7 / 16

Infinite length

Sketch of Proof: [1]Enough to show that one side of the equilateral triangle has infinite length.

At stage 0 we begin with a line segment with length, L representingone side.

At stage 1, apply the first iteration. We should have 4 segments eachof length L

3 . Then the total length of the sides is 43 · L.

At stage 2, apply second iteration. We should have 16 segments eachof length 1

3 ·L3 = L

9 . Total length is 43 ·

43 = (43)

2.

Monica J. Agana (Boise State University) Fall 2014 8 / 16

Infinite length

Sketch of Proof: [1]Enough to show that one side of the equilateral triangle has infinite length.

At stage 0 we begin with a line segment with length, L representingone side.

At stage 1, apply the first iteration. We should have 4 segments eachof length L

3 . Then the total length of the sides is 43 · L.

At stage 2, apply second iteration. We should have 16 segments eachof length 1

3 ·L3 = L

9 . Total length is 43 ·

43 = (43)

2.

Monica J. Agana (Boise State University) Fall 2014 8 / 16

Infinite length

Sketch of Proof: [1]Enough to show that one side of the equilateral triangle has infinite length.

At stage 0 we begin with a line segment with length, L representingone side.

At stage 1, apply the first iteration. We should have 4 segments eachof length L

3 . Then the total length of the sides is 43 · L.

At stage 2, apply second iteration. We should have 16 segments eachof length 1

3 ·L3 = L

9 . Total length is 43 ·

43 = (43)

2.

Monica J. Agana (Boise State University) Fall 2014 8 / 16

Repeat process up to n stages. So the total length, Ltotal is (43)n.

Final curve:

limn→∞(4

3)n =∞,

Monica J. Agana (Boise State University) Fall 2014 9 / 16

Repeat process up to n stages. So the total length, Ltotal is (43)n.

Final curve:

limn→∞(4

3)n =∞,

Monica J. Agana (Boise State University) Fall 2014 9 / 16

Finite Area

Find area of equilateral triangle, and build the formula

Aequilateral =

√3s2

4,

where s is the (finite) length of one side of this equilateral triangle.

Using this formula, we build the area for the Koch snowflake as

Akoch =

√3s2

4+ 3 ·

√3

4(s

3)2 + 12 ·

√3

4(s

9)2 + · · ·

Use algebra to modify and simplify the series:

Akoch =8

5·Aequilateral

Monica J. Agana (Boise State University) Fall 2014 10 / 16

Finite Area

Find area of equilateral triangle, and build the formula

Aequilateral =

√3s2

4,

where s is the (finite) length of one side of this equilateral triangle.

Using this formula, we build the area for the Koch snowflake as

Akoch =

√3s2

4+ 3 ·

√3

4(s

3)2 + 12 ·

√3

4(s

9)2 + · · ·

Use algebra to modify and simplify the series:

Akoch =8

5·Aequilateral

Monica J. Agana (Boise State University) Fall 2014 10 / 16

Finite Area

Find area of equilateral triangle, and build the formula

Aequilateral =

√3s2

4,

where s is the (finite) length of one side of this equilateral triangle.

Using this formula, we build the area for the Koch snowflake as

Akoch =

√3s2

4+ 3 ·

√3

4(s

3)2 + 12 ·

√3

4(s

9)2 + · · ·

Use algebra to modify and simplify the series:

Akoch =8

5·Aequilateral

Monica J. Agana (Boise State University) Fall 2014 10 / 16

Finite Area

Find area of equilateral triangle, and build the formula

Aequilateral =

√3s2

4,

where s is the (finite) length of one side of this equilateral triangle.

Using this formula, we build the area for the Koch snowflake as

Akoch =

√3s2

4+ 3 ·

√3

4(s

3)2 + 12 ·

√3

4(s

9)2 + · · ·

Use algebra to modify and simplify the series:

Akoch =8

5·Aequilateral

Monica J. Agana (Boise State University) Fall 2014 10 / 16

Continuous everywhere but nowhere differentiable

Nowhere differentiable - no tangent line at any point.

The Koch Curve: A Geometric Proof by Sıme Ungar [4].

Monica J. Agana (Boise State University) Fall 2014 11 / 16

Continuous everywhere but nowhere differentiable

Nowhere differentiable - no tangent line at any point.

The Koch Curve: A Geometric Proof by Sıme Ungar [4].

Monica J. Agana (Boise State University) Fall 2014 11 / 16

Continuous everywhere but nowhere differentiable

Suffices to assume [0, 1] is the interval under consideration [under alinear transformation, a finite interval can be put into (0, 1),preserving the continuity of f .

Define a continuous mapping f : [0, 1]→ R2 by

f := limn→∞fn.

and let K be the Koch curve.

Monica J. Agana (Boise State University) Fall 2014 12 / 16

Continuous everywhere but nowhere differentiable

Suffices to assume [0, 1] is the interval under consideration [under alinear transformation, a finite interval can be put into (0, 1),preserving the continuity of f .

Define a continuous mapping f : [0, 1]→ R2 by

f := limn→∞fn.

and let K be the Koch curve.

Monica J. Agana (Boise State University) Fall 2014 12 / 16

Continuous everywhere but nowhere differentiable

Consider the sequence of piecewise linear functions [0, 1]→ R2.

Monica J. Agana (Boise State University) Fall 2014 13 / 16

Continuous everywhere but nowhere differentiable

Consider the sequence of piecewise linear functions [0, 1]→ R2.

Monica J. Agana (Boise State University) Fall 2014 13 / 16

Continuity everywhere

To obtain K, limit of the fn’s w.r.t. the Hausdorff metric:C([0, 1],R2), the set of all continuous maps from [0, 1] into the planewith,

‖g‖ := supt∈[0,1]‖f(t)‖

Uses this to show K is a Jordan arc (i.e. f is an injection).

Monica J. Agana (Boise State University) Fall 2014 14 / 16

Continuity everywhere

To obtain K, limit of the fn’s w.r.t. the Hausdorff metric:C([0, 1],R2), the set of all continuous maps from [0, 1] into the planewith,

‖g‖ := supt∈[0,1]‖f(t)‖

Uses this to show K is a Jordan arc (i.e. f is an injection).

Monica J. Agana (Boise State University) Fall 2014 14 / 16

Nowhere differentiable

Idea behind it:

Iterate each side up to a point, and it becomes nondifferentiable. Youcan do this at all points.

For a proof of this refer to [2] or [4].

Monica J. Agana (Boise State University) Fall 2014 15 / 16

Nowhere differentiable

Idea behind it:

Iterate each side up to a point, and it becomes nondifferentiable. Youcan do this at all points.

For a proof of this refer to [2] or [4].

Monica J. Agana (Boise State University) Fall 2014 15 / 16

Applications of Series. N.p., n.d. Web. 27 Nov. 2014.

H. von Koch, Une methode geometrique elementaire pour l’etude decertaines questions de la theorie des curves plane, Acta Math. 30.(1906) 145-174.

“Niels Fabian Helge Von Koch.” Koch Biography. N.p., n.d. Web. 27Nov. 2014.

Ungar, Sime. “The Koch Curve: A Geometric Proof.” The AmericanMathematical Monthly 114.1 (2007): 61-66. JSTOR. Web. 11 Dec.2014.

Monica J. Agana (Boise State University) Fall 2014 16 / 16