Timeline for Asymptotics of a recursion
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 1, 2016 at 11:13 | comment | added | Jan-Christoph Schlage-Puchta | Anyway, Majer's answer is better. The final result is the same, as the derivation of Stirling involves Euler-Maclaurin, but this way you save yourself a lot of work. | |
| Jan 1, 2016 at 11:09 | comment | added | Jan-Christoph Schlage-Puchta | If $f$ is a smooth function, then $\int f(t)$ is the first approximation to $\sum f(n)$. Euler-Maclaurin gives a sequence of correction terms, involving higher and higher derivatives of $f$. When applied to a sum of length $N$, using $k$-th derivatives usually gives an error of size $k!^C N^{-k}$, so taking $k$ too big does not pay off, but most of the time one or two terms suffice. | |
| Dec 31, 2015 at 16:48 | comment | added | Koen Van Tuijl | Do you mean by applying the Eucler-Maclaurin summation that I should replace the sum of logarithms by an integral over a logarithm? | |
| Dec 31, 2015 at 14:59 | history | answered | Jan-Christoph Schlage-Puchta | CC BY-SA 3.0 |