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