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→∞ Log(N((1/2)n)) / Log(1/((1/2)n))
= limn→∞Log(3n + 2n) / Log(2n)
= limn→∞Log(3n(1 + (2/3)n)) / Log(2n)
= limn→∞(Log(3n) + Log((1 + (2/3)n)) / Log(2n)
= limn→∞Log(3n) / Log(2n)
because limn→∞Log((1 + (2/3)n) = Log(1) = 0
= limn→∞(n⋅Log(3)) / (n⋅Log(2))
= Log(3)/Log(2)
Note this is the larger of the box-counting dimension of the line segment (db = 1) and the Gasket (db = Log(3)/Log(2)).

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