Timeline for A difficult determinant
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2018 at 16:30 | answer | added | Marcel | timeline score: 5 | |
| Apr 13, 2018 at 3:24 | comment | added | Zach Teitler | With $x_i = N+a_i-i$, $y_j = N+b_j-j$, you are asking for $\det((x_i+y_j)!)$. If you multiply the $i$th row by $(x_i!)^{-1}$ and the $j$th column by $(y_j!)^{-1}$, then up to a factor $\prod_i x_i! \prod_j y_j!$, you just need $\det(\binom{x_i+y_j}{x_i})$. The Lindström-Gessel-Viennot lemma gives a combinatorial interpretation of the latter determinant in terms of lattice paths between the set of points $\{(-x_i,0) \mid i=1,\dotsc,N\}$ and the set of points $\{(0,y_j) \mid j=1,\dotsc,N\}$. | |
| Apr 12, 2018 at 23:24 | comment | added | Marcel | @user64494 As I said in the edit, all I did was remove some superfluous denominators. Nothing substantial has changed. | |
| Apr 12, 2018 at 20:08 | history | edited | Marcel | CC BY-SA 3.0 |
added 315 characters in body
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| Apr 12, 2018 at 19:58 | comment | added | user64494 | Your $D(\vec{a},\vec{b})=\det\left((2N+a_i+b_j-i-j)!\right)=? $ was changed without indicating it in your edit. My pevious comment was about the previous version of$D(\vec{a},\vec{b})=\det\left((2N+a_i+b_j-i-j)!/(\left( N-i \right) !\, \left( N+a_{{i}}-i \right) !)\right)$. | |
| Apr 12, 2018 at 19:43 | history | edited | Marcel | CC BY-SA 3.0 |
edited body
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| Apr 12, 2018 at 19:32 | history | edited | Marcel | CC BY-SA 3.0 |
Removed some superfluous factors
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| Apr 12, 2018 at 16:43 | comment | added | user64494 | Math experiment done with Maple brings nothing of the kind for $N=3, N=4.$ The executed code on demand. | |
| Apr 12, 2018 at 16:41 | comment | added | Per Alexandersson | Does it factor? I mean, a CAS can factor some expressions.. | |
| Apr 12, 2018 at 14:57 | history | asked | Marcel | CC BY-SA 3.0 |