2. B. Box-Counting Dimension

Box-Counting Dimension of the Product of a Sierpinski Gasket and a Line Segment

The straightforward approach may appear to run into some trouble. Specifically, how can we simplify

Log(3ⁿ + 2ⁿ)

when the Log does not behave nicely with respect to sums?

The trick is to turn the sum into a product:

3ⁿ + 2ⁿ = 3ⁿ(1 + (2/3)ⁿ)

With this, the calculation is fairly straightforward.

db	= limn->infinity Log(N((1/2)n)) / Log(1/((1/2)n))
	= limn->infinityLog(3n + 2n) / Log(2n)
	= limn->infinityLog(3n(1 + (2/3)n)) / Log(2n)
	= limn->infinity(Log(3n) + Log((1 + (2/3)n)) / Log(2n)
	= limn->infinityLog(3n) / Log(2n)
	because limn->infinityLog((1 + (2/3)n) = Log(1) = 0
	= limn->infinity(nLog(3)) / (nLog(2))
	= Log(3)/Log(2)

Note this is the larger of the box-counting dimension of the line segment (d_b = 1) and the Gasket (d_b = Log(3)/Log(2)).

Return to Box-Counting Dimension of the Product of a Sierpinski Gasket and a Line Segment.