Box-Counting Dimension of a Filled-in Triangle

First, recall a formula from algebra

1 + 2 + 3 + ... + k = k(k+1)/2

and so

1 + 2 + 3 + ... + 2ⁿ = 2ⁿ(2ⁿ + 1)/2

From this we see

Log(N((1/2)ⁿ)) = Log(2ⁿ(2ⁿ + 1)/2) = Log(2ⁿ) + Log(2ⁿ + 1) - Log(2)

For large n,

2ⁿ + 1 is very close to 2ⁿ,

so we might replace

Log(2ⁿ + 1)

by

Log(2ⁿ).

A more rigorous approach, which we will need for later calculations, is to note

2ⁿ + 1 = 2ⁿ(1 + 1/2ⁿ),

and so

Log(2n + 1)

= Log(2n(1 + 1/2n))

= Log(2n) + Log(1 + 1/2n).

Putting all this together, we have

Log(N((1/2)n))

= Log(2n) + Log(2n) + Log(1 + 1/2n) - Log(2)

= 2Log(2n) + Log(1 + 1/2n) - Log(2),

and so

Log(N((1/2)n)) / Log(1/(1/2)n)

= Log(N((1/2)n)) / Log(2n)

= (2Log(2n) / Log(2n)) + (Log(1 + 1/2n) / Log(2n)) - (Log(2) / Log(2n))

As n gets larger (so (1/2)ⁿ gets smaller), the second and third terms go to 0, and the Log(2ⁿ) cancels out of the first, so we are left with d_b = 2, as expected.

Return to Box-Counting Dimension of a Filled-in Triangle.