Timeline for How to prove the combinatorial equality? [closed]
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 12, 2018 at 23:32 | history | closed |
Peter Heinig Benjamin Steinberg Dietrich Burde David Handelman Chris Godsil |
Not suitable for this site | |
| Mar 12, 2018 at 23:24 | comment | added | Alexander Burstein | This is really a question for Math Stack Exchange. | |
| Mar 12, 2018 at 18:32 | answer | added | LSpice | timeline score: 1 | |
| Mar 12, 2018 at 17:58 | comment | added | Martin Sleziak | Simply putting the formula into Approach0 returns some posts on math.SE which seem relevant - at least at the first glance - this one seems the closest: Two identities with binomial coefficients. | |
| Mar 12, 2018 at 17:49 | comment | added | მამუკა ჯიბლაძე | Oh I see sorry - I was checking on Mathematica and it seems to have unusual convention with some $\binom pq$ with negative $q$ nonzero... | |
| Mar 12, 2018 at 17:34 | comment | added | LSpice | Well, sure enough, that's just what Jaynes says. Google Books link. | |
| Mar 12, 2018 at 17:32 | comment | added | ilya.gromov | @LSpice, in that case it would be Vandermonde convolution. It's not what I'm looking for. The exprassion I wrote I found in "Probability Theory. The Logic of Science" of Jaynes on page 153. | |
| Mar 12, 2018 at 17:32 | comment | added | LSpice | Oh, but then we have to change the RHS to $\binom N n$, and ensure that $R \le n$, so maybe that's too many changes to be what the OP intended. In its modified form, this is Vandermonde's convolution. | |
| Mar 12, 2018 at 17:28 | comment | added | LSpice | @მამუკაჯიბლაძე, right, sorry, I should have said to replace $\sum_{R = 0}^N$ by $\sum_{r = 0}^R$, in which case I think that it is correct (and the apparent dependence of the LHS on $R$ is illusory). | |
| Mar 12, 2018 at 17:17 | review | Close votes | |||
| Mar 12, 2018 at 23:32 | |||||
| Mar 12, 2018 at 17:13 | comment | added | მამუკა ჯიბლაძე | @LSpice With that extra sum the equality does not hold (neither with sum up to $n$) | |
| Mar 12, 2018 at 17:11 | comment | added | LSpice | Probably you want an extra $\sum_{r = 0}^R$ in there. | |
| Mar 12, 2018 at 17:10 | comment | added | მამუკა ჯიბლაძე | your lhs depends on $r$ while the rhs does not | |
| Mar 12, 2018 at 17:09 | comment | added | Peter Heinig | Dear @ilya.gromov: the formula you gave is not a sentence, hence not even meaningful. I am confident you will get kind help at math.stackexchange.com. | |
| Mar 12, 2018 at 16:52 | review | First posts | |||
| Mar 12, 2018 at 16:58 | |||||
| Mar 12, 2018 at 16:50 | history | asked | ilya.gromov | CC BY-SA 3.0 |