4 ms·
The "area" that you want to minimize is the number of digits, d, times the base, b. A = d * b d is roughly equal to the log of the number represented, N,
by jetrink 2y ago
The "area" that you want to minimize is the number of digits, d, times the base, b.
A = d * b
d is roughly equal to the log of the number represented, N, base b.
d ~= ln(N)/ln(b)
Substituting,
A ~= b * ln(N) / ln(b)
Take the derivative of the area with respect to b and find where the derivative is zero to find the minimum. Using the quotient rule,
dA/db = ln(N) * (ln(b)*1 - b/b) / ln(b)^2
0 = ln(N) * (ln(b) - 1) / ln(b)^2
0 = ln(b) - 1
ln(b) = 1
b = e
I hope I got that right. Doing math on the internet is always dangerous.
- klyrs 2y ago> Doing math on the internet is always dangerous. Only if you see correctness as a thing to be avoided. In my experience, being Wrong On The Internet is the fastest way to get something proofread.
- l- 2y agoAlthough the cost function has a multiplication of a base times the floor of the log of the value with respect to that base plus one, area is a misleading analogy to describe the applicability as any geometric dimensional value has to taken with respect to a basis. For a visual, (directional) linear scaling is more in line so to say.