Fractals and measuring a piece of paper

Question

Traditionally, I was taught that a piece of paper was 2d, so I decided to measure it, as if it was a square, since for a square, the area is just the length of one side squared. So for an 8.5 by 11 inch sheet of paper, the area would be 93.5 in^2. What I'm confused about right now is how 11 raised to the (log base 11 of (8.5)) plus one is 1.89, which is less than 2d, while 8.5 raised to the (log base 8.5 of (11)) plus one is greater than 2. I feel as though there is a correct way to do this, and I'm doing something wrong here. Thanks !

Answer 1

Perhaps you're using a different model of fractal dimension, but the one I'm familiar with has to do with scaling, not size. If you scale the piece of paper by a factor of $n$, you scale the, say, measure of the paper by a factor of $n^2$ since $A'=(nb)(nh)=n^2A$. In general, we say an object has dimension $D$ if scaling the object by a factor of $n$ results in scaling the (what we're calling) measure by a factor of $n^D$. Thus, in the case of the paper, we get $D=2$.

Answer 2

One way of characterizing the dimension of an object is the similarity dimension. Roughly speaking, we ask "If we scale our set down by some fixed factor, how many copies of the scaled set are needed to reconstruct the original set?" Note that this does assume that a set can be made out of scaled copies of itself, i.e. that the set is self-similar. We also probably need to assume some separation properties, but that is maybe beyond the scope of this answer---again, we are speaking very roughly.

For example:

1. A line segment of length 1. If we scale the segment by a factor of $C<1$, we get a segment of length $C$. In general, we will need $N = \frac{1}{C}$ copies of the scaled segment to reconstruct the original. As a particular case, suppose that $C = \frac{1}{4}$. Then, after scaling, it would take four copies of a segment of length $\frac{1}{4}$ to get the original interval back again. In this case, note that $$ N = \frac{1}{C}.$$
2. A square with side length 1. If we scale the square down by a factor of $C<1$, we end up with a square of side length $C$. Such a square has an area of $C^2$, and the original square had an area of 1, so it will generally take $N = \frac{1}{C^2}$ copies of the scaled square to rebuild the original. Again, as a particular case, note that it will take $16$ squares of side length $\frac{1}{4}$ to recover the original square. Hence $$ N = \frac{1}{C^2}. $$
3. A cube of side length 1. In this case, it takes $C^3$ scaled cubes to get the original cube (e.g. consider a Rubik's cube, which is made up of 27 cubes, each of which has $\frac{1}{3}$ the side length of the entire Rubik's cube). Here $$ N = \frac{1}{C^3}. $$

While this isn't entirely rigorous, we might note the following pattern: \begin{array}{llll} \textbf{Shape:} & \textbf{Dimension:} & \textbf{Copies vs Scaling:} & \textbf{Exponent:} \\ \text{segment} & \text{one} & N = \frac{1}{C} & 1 \\ \text{square} & \text{two} & N = \frac{1}{C^2} & 2 \\ \text{cube} & \text{three} & N = \frac{1}{C^3} & 3 \\ \end{array} What we might notice is that an $s$-dimensional object can be reconstructed $\frac{1}{C^s}$ copies of itself, each of which has been scaled by a factor of $C$. Thus the dimension of an object is the solution $s$ to the equation $$ N = \frac{1}{C^s}. $$ Solving, we get $$ N = \frac{1}{C^s} \implies \log(N) = -s\log(C) \implies s = -\frac{\log(N)}{\log(C)}.$$ Thus the dimension is obtained by comparing the logarithms of the number of copies and the scaling ratio. Note that $s$ will be positive, since $N$ is (typically) an integer (and will be greater than 1 in any event), and $C$ was assumed to be less than 1 (so that $\log(C) < 0$).

In the case of an $8.5\times 11$ inch piece of paper, if you scale it down by a factor of $C = \frac{1}{2}$, you end up with a piece of paper that is $4.25\times 5.5$ inches. We can tape/staple/glue $N=4$ such pieces of paper together in order to recover the original piece of paper, thus $$\dim(\text{piece o' paper}) = -\frac{\log(4)}{\log(\frac{1}{2})} = -\frac{\log(2^2)}{\log(2^{-1})} = -\frac{2\log(2)}{-\log(2))} = 2.$$ This implies that a piece of paper is two-dimensional, which is the expected result.

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themathandlanguagetutor's answer gives a very slightly (and subtly) different notion of dimension, called the Minkowski dimension. Instead of comparing the number of copies of a set needed to rebuild the set to the scaling ratio, the Minkowski dimension compares the measure (length, or area, or volume, etc.) of a set to the scaling ratio. In nice cases, these two ideas agree, which is gratifying!