A shape of dimension 1.26
Tonight's idea
Double the size of a line segment and it is made of 2 copies of the original. Double a square: 4 copies. Double a cube: 8. Triple them and you get 3, 9 and 27. Every time, the number of copies \(N\) is the scale factor \(s\) raised to the dimension \(d\):
\[N = s^d.\]
Turn that round to measure dimension. Take logarithms, in any base:
\(d = \dfrac{\log N}{\log s}\), the similarity dimension, for copies that do not overlap (touching is fine).
Now try it on the Koch curve. Start with a segment, replace its middle third with a two-sided bump, and repeat on every new segment for ever.
The finished curve is 4 copies of itself, each a third of the size. Scale by 3, get 4 copies, so \(3^d = 4\) and
\[d = \frac{\log 4}{\log 3} \approx 1.262.\]
Not 1 and not 2. That fits the picture: the curve is infinitely long, yet covers no area. Shapes like this are fractals.
Watch one: the Sierpinski triangle
Take a solid triangle. Join the midpoints of its sides and remove the middle triangle. Three corner triangles are left, each half the size. Do the same to each of them, for ever.
What remains is made of 3 copies of itself at half size. Scale by 2, get 3 copies:
\[d = \frac{\log 3}{\log 2} \approx 1.585.\]
Each step keeps \(\tfrac34\) of the area, so the area shrinks to 0, which fits a dimension below 2.
Step through the first six stages of the Koch snowflake, with each new set of bumps in orange and the perimeter growing by \(\tfrac43\) a step: mathdetours.com/demos/koch.html.
Your turn
- Scale a cube by 4. How many copies of the original fit? Check that the formula gives \(d = 3\).
- The Cantor set: remove the middle third of a segment, then the middle third of each piece left, for ever. How many copies of itself is it made of, at what scale? Find its dimension to 3 decimal places.
- The Sierpinski carpet: cut a square into 9 equal squares, remove the centre one, and repeat in each of the 8 left. Find its dimension. Then find the dimension of a curve in which every segment is replaced by 8 segments, each a quarter as long. Give that one exactly.
- After \(n\) steps the Koch curve, built on a segment of length 1, has \(4^n\) pieces of length \(r = 3^{-n}\). So a ruler of length \(r\) measures its length as \(L = \left(\tfrac43\right)^n\). Use \(3^d = 4\) to show that \(L = r^{1 - d}\). What happens to \(L\) as the ruler gets shorter?
Stretch: dust of dimension 1
Cut a square into 16 equal squares and keep only the four corner ones. Repeat in each, for ever. Find the dimension of what is left. It contains no curve at all, only scattered points. What does that tell you about what this dimension measures?
Answers
- \(4^3 = 64\) copies. \(d = \dfrac{\log 64}{\log 4} = 3\), since \(64 = 4^3\).
- Two copies, each a third of the size. \(d = \dfrac{\log 2}{\log 3} \approx 0.631\). More than a point, less than a line.
- Carpet: 8 copies at a third of the size, so \(d = \dfrac{\log 8}{\log 3} \approx 1.893\). Curve: 8 copies at a quarter of the size, so \(d = \dfrac{\log 8}{\log 4} = \dfrac{3 \log 2}{2 \log 2} = \dfrac32\) exactly.
- \(r^{1 - d} = \left(3^{-n}\right)^{1 - d} = 3^{-n} \times \left(3^{d}\right)^{n} = 3^{-n} \times 4^n = \left(\tfrac43\right)^n = L\). Since \(1 - d \approx -0.262\) is negative, \(L\) grows without limit as \(r\) shrinks. The shorter the ruler, the longer the curve. The same thing happens when you measure a rugged coastline.
Stretch. Four copies, each a quarter of the width: \(d = \dfrac{\log 4}{\log 4} = 1\). The dust has the same dimension as a line without being one. Similarity dimension measures how the amount of stuff grows when you scale up, not whether the pieces join together.