Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

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

share|improve this answer
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 $k$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.