Timeline for Asymptotic behaviour of sequence
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2014 at 9:27 | comment | added | alexbailey | Yes I thought it would be OK, I was just a little concerned when you pulled the $n$ out of the integral, but I guess it all cancels out in the end? Many thanks again. | |
| Jun 15, 2014 at 19:18 | comment | added | Kirill | @alexbailey ${m\choose n}$ is by definition $0$ when $0\leq m<n$ (the expression for it in terms of gamma functions makes this very clear). It doesn't change anything because the function still reaches its maximum on the domain. In practical terms, if you knew where the maximum was, you could sum only the few terms around it and it would be enough. | |
| Jun 15, 2014 at 19:14 | comment | added | alexbailey | Just one more question, I was being a bit sloppy when I said the sum is from $0$ to $n$ as the terms when $k^2<n-k$ are not really defined, I guess it should really be something like $k$ from $0$ to $\left\lfloor{\frac{1}{2}\left(1+\sqrt{1+4n}\right)}\right\rfloor$. Does this change anything with the analytic approach? | |
| Jun 15, 2014 at 19:05 | comment | added | Kirill | @alexbailey I don't know about that :) The way I learned this, I'd say: Yay for mathematical physics. There is a really nice book by de Bruijn that explains this type of thing. | |
| Jun 15, 2014 at 19:01 | comment | added | alexbailey | This is beautiful. Yay for analytic number theory! | |
| Jun 15, 2014 at 19:00 | vote | accept | alexbailey | ||
| Jun 14, 2014 at 6:40 | history | answered | Kirill | CC BY-SA 3.0 |