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 ≈ 2n,

so we might replace

Log(2n + 1)

by

Log(2n).

Then

db = limn→∞Log(N((1/2)n))/Log(2n) = limn→∞(2 Log(2n) - Log(2)) / Log(2n) = limn→∞(2n Log(2) - Log(2)) / n Log(2) = 2

A more rigorous approach, which we shall 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), we see

Log(1 + 1/2n) / Log(2n) → Log(1) / Log(∞) = 0 Log(2) / Log(2n) → Log(2) / Log(∞) = 0

That gives

db = limn→∞2Log(2n) / Log(2n) = 2,

as expected.

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