Timeline for Is this similar to a known combinatorial identity?
Current License: CC BY-SA 3.0
13 events
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| Jul 15, 2013 at 20:20 | vote | accept | CommunityBot | moved from User.Id=5117 by developer User.Id=36770 | |
| Jul 15, 2013 at 20:02 | answer | added | user5117 | timeline score: 3 | |
| Jul 12, 2013 at 2:05 | answer | added | Zack Wolske | timeline score: 8 | |
| Jul 8, 2013 at 14:59 | history | edited | user5117 | CC BY-SA 3.0 |
fixed some formatting broken by migration
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| Jun 6, 2013 at 19:59 | comment | added | user5117 | David: thanks for the extra info. Maybe I was too timid in what I was trying to do. (Or else I just didn't see the appropriate package...) I'll look into it more! | |
| Jun 6, 2013 at 19:45 | comment | added | David Lehavi | Hi Artie, it's been a long while since I played with it. But if memory serves (and also see the statement in 2.6 in A=B), the main difference is that "Zeilberger's algorithm .... is guaranteed to work every time", whereas - as far as I understand - Gosper's algorithm would always work only if you can reformulate the equality as some hyper-goemetric identity. | |
| Jun 5, 2013 at 20:48 | comment | added | user5117 | Dear David: I'm not so familiar with the Wilf--Zeilberger machinery, but it does seem applicable here. For the record, the Mathematica package gosper.m available from the site you link is capable of simplifying parts of these expressions, but so far it does not see capable of digesting the whole thing... | |
| Jun 5, 2013 at 20:46 | comment | added | user5117 | Dear Zack: I think the discrepancy in our answers comes from what the binomial coefficient (-1 choose 0) means; sorry if I was being a bit sloppy here. Anyway, for my purposes I want to define this to be 1, even if that is nonstandard. I think with that convention, the formula I wrote really does work out. (I checked it by computer for enough cases to satisfy myself, although others might be more exacting.) Anyway, thanks for your input! | |
| Jun 5, 2013 at 1:23 | comment | added | David Lehavi | Doesn't the Wilf-Zeilberger technique work for this ? See math.rutgers.edu/~zeilberg/programsAB.html for Maple the packages, and math.upenn.edu/~wilf/AeqB.html for the book. | |
| Jun 4, 2013 at 23:00 | comment | added | Zack Wolske | For example, if I haven't made any mistakes, then taking $n=i$ makes all the binomial coefficients of the final term on the left $0$, and taking $i=2$ makes the first summation empty and leaves the others with just one term each, $2x^{i−2}$ and $(i-2)x^{i−1}/2$. So when $n=i=x=2$, the left side is $2$. If the upper index and binomial coefficient in the final term are changed to be in line with the other terms, then everything cancels, and $(x−2)$ divides the left side. | |
| Jun 4, 2013 at 22:41 | comment | added | Zack Wolske | Is the upper index $i$ and the binomial choosing $i-j$ in the last term on the left correct? It seems that taking $x=2$, the second term cancels with the $-2$ part of the $(i-2-l)$ part of the third term, and leaves something remarkably close to the final term, but not equal to it. | |
| Jun 4, 2013 at 21:58 | history | edited | user5117 | CC BY-SA 3.0 |
spelling fix
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| Jun 4, 2013 at 21:07 | history | asked | user5117 | CC BY-SA 3.0 |