2 ms·
In analytic number theory we usually only care about growth rates in a very coarse sense- up to scaling by some constant, and asymptotically. Because the log of
by ravi-delia 3y ago
In analytic number theory we usually only care about growth rates in a very coarse sense- up to scaling by some constant, and asymptotically. Because the log of any base is precisely the same up to constants, it doesn't actually matter. If you look at the expression, to the right of the equals sign is a big O- that's the same big O as the one you might be familiar with in complexity.
That being said, in math more generally when we need a concrete log the natural log is pretty much always the way to go- I haven't seen ln in a little while.
- danielam 3y ago> Because the log of any base is precisely the same up to constants i.e., \log_{b_1} n = \frac{1}{\log_{b_2} b_1} \log_{b_2} n where \frac{1}{\log_{b_2} b_1} is the constant fixed for a given choice of `b_1` and `b_2`, i.e., the bases.
- NooneAtAll3 3y agoexcept that in that formula _exact_ "k-b" is used - coarseness is hidden in O(1/sqrt(k)) that follows