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Suppose $p$, $k$ and $s$ are integers with $s,k \le p$. Consider the following polynomial in $x$ and $y$, $$ \sum_{\ell=0}^k \binom{s}{\ell} \binom{p-s}{k-\ell} x^\ell y^{p-\ell}$$ Does this expression look familiar to anyone? Is there closed form?

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It is the sum of $pth$ degree homogeneous terms of $y^{p-k}(1 +x)^s(1+y)^{p-s}$

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Thanks. I appreciate it. I now think it is also possible to express it in terms of Jacobi polynomials. Something like $(x-y)^k P_k^{(\alpha,\beta)}(z)$ where $\alpha = s-k$, $\beta = p-k-s$ and $z = (x+y)/(x-y)$. – passerby51 Mar 5 '11 at 21:44
In fact, I am interested in the sum $\sum_{k=1}^p (-1)^{k-1} ( \cdots )$ where the dots are filed with my original expression. Any idea of a closed form for this? – passerby51 Mar 5 '11 at 21:49

We can transform

$$\sum_{\ell=0}^k \binom{s}{\ell} \binom{p-s}{k-\ell} x^\ell y^{p-\ell}$$

into a hypergeometric form as follows: we factor out $y^p$ like so

$$y^p\sum_{\ell=0}^k \binom{s}{\ell} \binom{p-s}{k-\ell} \left(\frac{x}{y}\right)^\ell$$

and then we can easily transform the first binomial coefficient to a Pochhammer symbol:

$$y^p\sum_{\ell=0}^k \frac{(-1)^\ell (-s)_\ell}{\ell!} \binom{p-s}{k-\ell} \left(\frac{x}{y}\right)^\ell$$

The second one requires a bit more work; using this identity, we have

$$y^p\binom{p-s}{k}\sum_{\ell=0}^k \frac{(-s)_\ell}{\ell!} (-1)^\ell \frac{(k-\ell+1)_\ell}{(p-s-k+1)_\ell}\left(\frac{x}{y}\right)^\ell$$

and then use this identity to yield

$$y^p\binom{p-s}{k}\sum_{\ell=0}^k \frac{(-s)_\ell (-k)_\ell}{(p-s-k+1)_\ell} \frac1{\ell!}\left(\frac{x}{y}\right)^\ell$$

from which we find that we have a finite ${}_2 F_1$ hypergeometric sum; more specifically we have

$$y^p\binom{p-s}{k} {}_2 F_1\left({{-k}\atop{}}{{}\atop{p-k-s+1}}{{-s}\atop{}}\mid \frac{x}{y}\right)$$

With either the hypergeometric expression or the sum before it, we find that the sum has $\min(k,s)$ terms.

From the hypergeometric expression, further transformations might be possible. Alternatively, the expression itself can directly be used for numerical evaluation, since hypergeometric functions satisfy a three-term recurrence; start with the expressions corresponding to $k=0$ and $k=1$ and recurse from there to numerically evaluate the expression for a given $k$.

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Thank you very much for writing this up. I didn't check back for a long time. So sorry for the very late reply. – passerby51 May 17 '12 at 16:06
No problem. It has been a long time, indeed... :) – J. M. May 17 '12 at 16:24

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