Box-counting dimension

Suppose you want to determine dimension of a set S. Put this set on

evenly-spaced grid of size ε, and count how many boxes of size ε are

needed to cover the set.

Box-counting dimension (also called Minkowski-Bouligand) isdefined as

D = limε→0

lnN(ε)

ln(1/ε)

Note that when ε is small, we have

D ≈ lnN(ε)

ln(1/ε)

henceD ln(1/ε) ≈ lnN(ε)

ln(1/εD) ≈ lnN(ε)

N(ε) ≈ 1

εD

Example: linear segment

ε

Suppose length of a linear segment is L. If the grid size is ε, weneed N(ε) = L/ε boxes to cover the entire segment. If L = 1, andε = 1/4, we need 4.

D = limε→0

lnN(ε)

ln(1/ε)

D = limε→0

ln Lε

ln(1/ε)= lim

ε→0

lnL− ln ε

− ln ε= 1

Linear segment is one-dimensional!

Example: solid square

ε

Consider L×L square. If the grid size is ε, we need N(ε) = L2/ε2

boxes to cover the entire segment. If L = 1, and ε = 1/4, we need16.

D = limε→0

lnN(ε)

ln(1/ε)

D = limε→0

ln L2

ε2

ln(1/ε)= lim

ε→0

lnL2 − 2 ln ε

− ln ε= 2

Solid square is two-dimensional!

Example: Koch curve

Not all geometric object have integer dimensions!

There exist a class of objects, called fractals, for whichdimension is fractional.

Consider a curve proposed by Swedish mathematician NielsFabian Helge von Koch in 1904 in his paper Sur une courbecontinue sans tangente, obtenue par une constructiongeometrique elementaire (“On a continuous curve withouttangents constructible from elementary geometry”).

This was one of the first fractals proposed.

It is constructed from a linear segment in recursively definedstages.

Koch curve: construction

Figure from S. Lynch, Dynamical Systems with Application using Maple Birkhauser 2001

Koch curve on a grid

Figures on next four slides from S. Lynch, Dynamical Systems with Application using Maple Birkhauser 2001

Koch curve on a grid

Using l as ε we have

lnN(l) ≈ D · (− ln(l))

Thus slope of the line of best fitof lnN(l) versus − ln(l)yields the dimension.Here slope ≈ 1.2246, thus

D ≈ 1.2246.

Exact dimension (by other methods)

D = ln 4/ ln 3 ≈ 1.2619.

Henon attractor on a grid

Henon attractor on a grid

Fo orbit with 5000 pointsthe slope ≈ 1.0562,thus box-countingdimension is

D ≈ 1.0562

Henon attractor is more than a line,but only slightly so.Exact dimension is unknown.

Sierpinski traingle

Another fractal example: Sierpinski triangle (gasket), proposed byPolish mathematician Wac law Franciszek Sierpinski in 1915.

Dimension = ln 3/ ln 2 ≈ 1.585.In 3D, analogous construction is known as tetrix (Sierpinskipyramid). Surprisingly, it has integer dimension(D = ln 4/ ln 2 = 2)

For the purpose of this course, we will adopt the followingdefinition.

Definition

Fractal is a geometrical object which has non-integer box-countingdimension.

One of the most striking properties of fractals is self-similarity.

Informally, this means that a part of the fractal, when magnified, iscongruent to the whole fractal. Sometimes, especially in the caseof fractal-like structures occurring in nature, one needs to replace“congruent” by “very strongly resembling”.

We have seen in in the case of Henon attractor. Koch curve isself-similar too.

Self-similarity of Koch curve

Figure from H-O Peitgen et al., Chaos and Fractals, Springer 2004

Self-similarity of coast line

Figure from H-O Peitgen et al., Chaos and Fractals, Springer 2004

Cantor set

An important example of fractal is Cantor set, introduced byGerman mathematician Georg Cantor in 1883.It is constructed from [0, 1] by removing middle third. We thenrepeat removal of middle third from thus obtained segments, adinfinitum.

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What is the total length Ln of all segments at stage n? At stage nwe have 2n segments of length 1

3n each. Thus

Ln =1

3n· 2n =

(2

3

)n

.

limn→∞

Ln = 0

Cantor set has total length 0, but it is not empty! We say it is aset of measure zero.

Dimension of Cantor set

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At stage 0, one needs 3 boxes of size 13 to cover the set.

At stage 1, one needs 6 boxes of size 132

to cover the set.At stage 2, one needs 12 boxes of size 1

33to cover the set.

At stage n, one needs 3 · 2n boxes of size 13n+1 to cover the set.

D = limε→0

lnN(ε)

ln(1/ε)= lim

n→∞

ln(3 · 2n)ln(3n+1)

= limn→∞

ln 3 + n ln 2

ln 3 + n ln 3

D =ln 2

ln 3≈ 0.63093

Cantor set is thus “denser” than a point, but “less dense” than aline.

Another example: Koch square

Figure from S. Lynch, Dynamical Systems with Application using Maple Birkhauser 2001