Talk:Chu-Vandermonde Identity

Where should this go?

The Chu-Vandermonde Identity implies this relationship and vice-versa.

Help please. --Robkahn131 (talk) 04:27, 9 December 2022 (UTC)


 * Haven't a clue. --prime mover (talk) 06:49, 9 December 2022 (UTC)


 * I took the liberty of removing the source citation (link was broken, btw) as it did not contain the contents of this page. --prime mover (talk) 06:52, 9 December 2022 (UTC)

As you have noted below, you need more than just Chu-Vandermonde Identity, so instead I would try to give the identity a separate name. For example, it is very similar to the Binomial Theorem. So I suggest calling it like "Variant of Binomial Theorem", "Modified Binomial Theorem", "Rising Factorial Binomial Theorem" or something analogous. I also recommend transcluding it to the Binomial Theorem.--Julius (talk) 11:42, 9 December 2022 (UTC)


 * Thank you!! --Robkahn131 (talk) 11:56, 9 December 2022 (UTC)


 * Now I've had a chance to think about this, the logic does not really make sense. We know that the Chu-Vandermonde Identity is true, we don't need to hypothesise it. --prime mover (talk) 18:08, 9 December 2022 (UTC)

Theorem
Let $r, s \in \R, n \in \Z_{\ge 0}$.

The Chu-Vandermonde Identity:
 * $\ds \sum_{k \mathop = 0}^n \binom r k \binom s {n - k} = \binom {r + s} n $

Implies the following identity:
 * $\ds \leadsto \sum_{k \mathop = 0}^n \dbinom n k r^{\overline k} s^{\overline {n-k} } = \paren {r + s}^{\overline n}$


 * A quick teachable moment: Help:Editing/House Style. Bewilders me how this is such a common mistake. So many do it. --prime mover (talk) 06:49, 9 December 2022 (UTC)

Proof
From Rising Factorial as Factorial by Binomial Coefficient, we have:

Therefore: