Timeline for Binomial ID $\sum_{k=m}^p(-1)^{k+m}\binom{k}{m}\binom{n+p+1}{n+k+1}=\binom{n+p-m}{n}$
Current License: CC BY-SA 3.0
5 events
when toggle format | what | by | license | comment | |
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Dec 5, 2016 at 14:25 | comment | added | Noam D. Elkies | I suppose you can also use a (formal) differential equation to prove this. | |
Dec 5, 2016 at 12:38 | vote | accept | Matt Majic | ||
Dec 5, 2016 at 5:54 | comment | added | Fedor Petrov | @Noam but why formal series $(1+t)^x=\sum \binom{x}{n} t^n$ satisfy this equality? It may be justified by some abstract nonsense like 'this is known for reals or for positive integers, but the coefficients are polynomials in $x,y$, thus it holds formally too.' This argument may be reduced to bit less abstract: 'both parts of CV are polynomials in $x,y$ of degree at most $\ell$, thus it suffices to check that their values agree for integers $x,y\geqslant 0, x+y\leqslant \ell$ (this triangle is an interpolating set for polynomials of degree at most $\ell$) , where identity is just obvious.' | |
Dec 5, 2016 at 4:32 | comment | added | Noam D. Elkies | Right, with the same generatingfunctionological explanation: $(1+t)^x (1+t)^y = (1+t)^{x+y}$. | |
Dec 5, 2016 at 4:15 | history | answered | Fedor Petrov | CC BY-SA 3.0 |