Timeline for hypergeometric function $_2F_1(-n;-r;1;2)$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 14, 2012 at 16:59 | comment | added | Campello | What do you mean comparable? $r=O(n)$ ? Also it is not clear for me how the $i=n+r−\sqrt{n^2+r^2}$ comes out... | |
| Oct 11, 2012 at 23:17 | comment | added | Noam D. Elkies | The summands are all positive, and logarithmically concave down, so it's just a matter of locating the peak and estimating its width. If I did it right, when $r,n$ are large and comparable, the largest summand is around $i = n + r - \sqrt{n^2+r^2}$. Stirling's formula for $N!$ will let you estimate this maximal coefficient, and then a Gaussian-integral approximation around that peak will tell you the multiple of $\sqrt n$ to use for the peak width. The resulting formula won't be pretty but it should at least be correct. | |
| Oct 11, 2012 at 21:48 | comment | added | Campello | Thank you Noam. The mistake was corrected. Gerald, the approximation $2^r n^r/r!$ (volume of ball in the $l_1$ norm) is good for fixed $r$ and (very) large $n$. However, when $r$ also increases (for example, if r = O(n)), what should be the behavior? | |
| Oct 11, 2012 at 21:28 | history | edited | Campello | CC BY-SA 3.0 |
There was a mistake on the set \mbox{#} \lbrace x \in \mathbb{Z}^n : |x_1| + \cdots + |x_n| \leq r \rbrace . It now counts the number of points within the l_1 norm, indeed.
|
| Oct 10, 2012 at 0:24 | comment | added | Ira Gessel | These numbers are sometimes called Delannoy numbers; see mathworld.wolfram.com/DelannoyNumber.html. | |
| Oct 9, 2012 at 15:25 | comment | added | Gerald Edgar | For fixed $r$, it is a polynomial in $n$ of degree $r$, so take leading term: ${}_2F_1(-n,-r;1;2) \sim {2^r n^r}/{r!}$ as $n \to \infty$. | |
| Oct 9, 2012 at 14:26 | comment | added | Noam D. Elkies | The formula counts integers in the $l_2$ norm, not $l_1$. Presumably $l_1$ is what you meant, i.e. $|x_1|+\cdots+|x_n| \leq r$, not $|x_1|^2+\cdots+|x_n|^2 \leq r$. | |
| Oct 9, 2012 at 14:12 | history | asked | Campello | CC BY-SA 3.0 |