Timeline for Solving partial difference equation
Current License: CC BY-SA 2.5
4 events
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| Oct 4, 2010 at 4:22 | comment | added | Mike Spivey | But David's comment implies that you can decouple your recurrence relation into two separate ones: One for the $n \equiv k \bmod{2}$ case and one for the $n \equiv k+1 \bmod{2}$ case. If you do that, rearranging indices and plowing through some algebra, the first case turns into the recurrence $B^n_k = (2k-n+1)B^{n-1}_k + (2n-2k+1)B^{n-1}_{k-1}$, $B^0_0 = 1$, and the second case turns into the recurrence $C^n_k = (2k-n+2)C^{n-1}_k + (2n-2k)C^{n-1}_{k-1}$, $C^0_0 = 1$. Maybe these aren't much better, but they are at least "first order" in both $n$ and $k$ now. | |
| Aug 1, 2010 at 12:50 | comment | added | Ross Tang | Furthermore, for the Eulerian numbers, the order of difference equation for both the variables k and n are 1. But for the equation I asked, it is first order in n, but 2nd order in k. So I think they differ quite a lot. | |
| Jul 27, 2010 at 13:47 | comment | added | Ross Tang |
Thank you for your answer. I read the page you mentioned in wikipedia. There is indeed closed form to Eulerian numbers: $A(n,m)=\sum_{k=0}^{m}(-1)^k \binom{n+1}{k} (m+1-k)^n.$ I don't know if I read it wrongly or not. Thanks!
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| Jul 27, 2010 at 12:47 | history | answered | David E Speyer | CC BY-SA 2.5 |