11.1 Fractals

Hold a fern leaf up to the light. The whole frond is made of leaflets, and each leaflet is a shrunken version of the entire frond. Lean in closer and each leaflet is itself built from tinier sub-leaflets that repeat the very same outline. The design keeps reappearing as you zoom in, smaller and smaller, apparently without end.

A shape that hides smaller replicas of itself like this is called a fractal, and the underlying property is self-similarity.

The Key Idea A fractal is a shape that is self-similar: magnify any part of it and you meet the same (or nearly the same) pattern you saw in the whole, repeated again and again at ever-smaller scales. Nature is crowded with them, a river branching into streams and rivulets, broccoli florets, frost on a window, coastlines, mountain ridges, forks of lightning, and the everyday fern.

Mathematicians cooked up fractals of their own, built by applying one simple rule over and over. Four of the most striking are the Sierpinski Carpet, the Sierpinski Triangle, the Koch Snowflake and the Cantor Set. Let us construct them one at a time.

The Sierpinski Carpet

The Polish mathematician Wacław Sierpiński introduced this fractal. Its recipe is delightfully short:

  1. Stage 0: begin with one solid square.
  2. Slice the square into a 3\times 3 grid of 9 little squares, and delete the middle one. (Eight squares survive and one square hole opens up.)
  3. Apply the same cut to each of the 8 surviving squares.
  4. Carry on without stopping.
Stage 0 Stage 1 Stage 2

Two things shift at every stage: the surviving squares become smaller and more plentiful, while fresh holes keep opening. Write R_n for the count of remaining (filled) squares at stage n, and H_n for the count of holes.

Counting the survivors. Each surviving square at stage n breaks into 8 survivors at stage n+1. Hence

R_{n+1} = 8\,R_n.

Starting from R_0 = 1:

R_0 = 1,\quad R_1 = 8,\quad R_2 = 8\times 8 = 8^2,\quad R_3 = 8^3,\ \dots \quad\Rightarrow\quad \boxed{R_n = 8^n.}

Counting the holes. Every surviving square cuts one fresh hole at the next stage, while all earlier holes stay put. So the new holes that appear at stage n+1 number R_n, giving

H_{n+1} = H_n + R_n.

Spelling this out:

H_0 = 0,\quad H_1 = 1,\quad H_2 = 1 + 8,\quad H_3 = 1 + 8 + 8^2,\ \dots

so H_n = 1 + 8 + 8^2 + \dots + 8^{n-1}, the running tally of every square that ever lived before stage n. Numerically the holes go 0,\ 1,\ 9,\ 73,\ 585,\ \dots

Shrinking area Let the starting square have area 1 square unit. At each stage we retain 8 of the 9 equal sub-squares, so the surviving area is multiplied by \tfrac{8}{9} every time. After n stages the leftover area is

\left(\tfrac{8}{9}\right)^n.

This slides towards 0 as n climbs, the true Sierpinski Carpet has zero area, and yet it is anything but empty!

The Sierpinski Triangle (Gasket)

Sierpiński cooked up a second fractal by the same trick, but starting from a triangle:

  1. Stage 0: begin with an equilateral triangle.
  2. Join the midpoints of its three sides. This carves it into 4 matching smaller equilateral triangles; delete the middle one.
  3. Repeat on each of the 3 surviving triangles. Keep going.
Stage 0 Stage 1 Stage 2

Math Talk: why exactly 4 equal triangles? Joining the midpoints of an equilateral triangle produces a small central triangle plus three corner triangles. Each corner triangle has two sides that are halves of the original sides, and its angles are untouched (the join lines run parallel to the opposite sides, the Midpoint Theorem). Therefore each of the four pieces is equilateral with half the side length of the parent. Four congruent equilateral triangles that tile the whole exactly, that is the reason the recipe works.

For the Sierpinski Triangle, at stage n the count of surviving triangles is T_n = 3^n (each one splits into 3), and the surviving area is \left(\tfrac{3}{4}\right)^n (we keep 3 of every 4 equal sub-triangles). This too dwindles to 0.

Figure it Out: Sierpinski fractals

  1. Sketch the opening stages (at least to Stage 2) of the sequence that builds the Sierpinski Triangle.
  2. Find the number of holes and the surviving triangles at each stage of the Sierpinski Triangle.
  3. Find the area of the region that remains at the nth stage for both Sierpinski fractals (carpet and triangle). Take the starting square/triangle to have area 1 sq. unit.
  1. Stage 0 is a single filled equilateral triangle; Stage 1 keeps the three corner triangles with the centre gone (1 hole); Stage 2 repeats the rule inside each of those three (see the figure above).
  2. Surviving triangles: each one becomes 3, so T_n = 3^n: that is 1, 3, 9, 27, \dots for n = 0, 1, 2, 3. New holes open in the centre of each surviving triangle, so 3^{\,n-1} new holes form at stage n, and the total is H_n = 1 + 3 + 3^2 + \dots + 3^{\,n-1}: that is 0, 1, 4, 13, \dots for n = 0, 1, 2, 3.
  3. Sierpinski Carpet: keep \tfrac{8}{9} of the area each stage, so remaining area = \left(\dfrac{8}{9}\right)^n. Sierpinski Triangle: keep \tfrac{3}{4} each stage, so remaining area = \left(\dfrac{3}{4}\right)^n. Both head towards 0 as n \to \infty.

The Koch Snowflake

Named for the Swedish mathematician Helge von Koch, who set it down in 1904. Here we add to the shape rather than carving from it:

  1. Stage 0: begin with an equilateral triangle.
  2. On each side, split it into 3 equal parts, build a small equilateral triangle on the middle part, then rub out that middle part. Each straight side turns into a four-segment “bump”.
  3. Apply the same step to every side of the new shape. Keep going.
Stage 0 Stage 1 (bumps) Stage 2 (snowflake)

(The Stage 1 and Stage 2 sketches are only schematic, the genuine Koch curve is best drawn slowly by hand on grid paper. The point to hold on to is that every straight side becomes a bumpier four-piece path.)

Counting the sides. Each side breaks into 4 new sides at every stage, and Stage 0 has 3 sides. So

\text{sides at stage } n = 3\times 4^n: \qquad 3,\ 12,\ 48,\ 192,\ 768,\ \dots

The perimeter grows without limit. Each stage swaps every side of length \ell for four pieces of length \tfrac{\ell}{3}, total \tfrac{4}{3}\ell. So every stage multiplies the whole perimeter by \tfrac{4}{3}. Starting from a triangle of side length 1 (perimeter 3):

\text{perimeter at stage } n = 3\times\left(\tfrac{4}{3}\right)^n: \qquad 3,\ 4,\ \tfrac{16}{3},\ \tfrac{64}{9},\ \dots

Finite area, infinite perimeter Here is the jaw-dropping part. The Koch Snowflake sits comfortably inside a small circle, so it pens in a finite area. Yet its perimeter is multiplied by \tfrac{4}{3} > 1 at every stage, so it swells without bound, the boundary is infinitely long! A curve of unlimited length fencing a finite patch of plane: that is the peculiar magic of fractals.

Figure it Out: Koch Snowflake

  1. Sketch the opening stages (at least to Stage 2) of the sequence that builds the Koch Snowflake.
  2. Find the number of sides at the nth stage.
  3. Find the perimeter of the shape at the nth stage. Take the starting triangle to have side length 2 units.
  1. Begin with an equilateral triangle (Stage 0). For Stage 1 plant a small triangular bump on the middle third of each of the 3 sides, giving a six-pointed star outline with 12 sides. For Stage 2 do the same to all 12 sides.
  2. Each side becomes 4 sides per stage, and Stage 0 has 3 sides, so the count is 3\times 4^n: namely 3, 12, 48, 192, \dots
  3. With side length 2, the starting perimeter is 3\times 2 = 6, and every stage multiplies by \tfrac{4}{3}. So the perimeter at stage n is 6\times\left(\dfrac{4}{3}\right)^n, i.e. 6,\ 8,\ \tfrac{32}{3},\ \tfrac{128}{9},\dots It rises without limit.

The Cantor Set: a fractal on a line

Enrichment: a fractal in one dimension Fractals need not be flowery. The simplest of all lives on a single line segment, and the German mathematician Georg Cantor described it in 1883.

  1. Stage 0: start with the segment from 0 to 1.
  2. Erase the open middle third \left(\tfrac{1}{3}, \tfrac{2}{3}\right), leaving two segments.
  3. Erase the middle third of each remaining segment. Keep going forever.
S0 S1 S2

At stage n there are 2^n segments, each of length \left(\tfrac{1}{3}\right)^n, so the total length left is \left(\tfrac{2}{3}\right)^n, which marches to 0. The total length vanishes, yet uncountably many points (such as 0, 1 and \tfrac{1}{4}) never get erased. A “dust” of points with zero total length: the line-segment version of the carpet’s vanishing area.

Fractals in Art

Did You Know: Fractals across cultures Fractals are not the private property of modern mathematics; people have built them into art for a thousand years.

  • The Kandariya Mahadev Temple at Khajuraho, Madhya Pradesh (finished around 1025 CE) climbs as one towering form assembled from smaller copies of that same form, each of which carries still smaller copies, a fractal in stone. Comparable self-similar designs grace temple towers at Madurai, Hampi, Rameswaram and Varanasi.
  • In West Africa, the wedding blankets of the Fulani people of Nigeria are woven with diamonds nested inside diamonds inside diamonds, a fractal in cloth.
  • The Dutch printmaker M. C. Escher played with fractals in works such as “Smaller and Smaller,” in which a single lizard shrinks copy by copy towards the centre.