1
$\begingroup$

Let $d \ge 1$ be a positive integer. Let $n,m$ be another positive integers subject to $m\ge n+d$. Let $\vec{x} := (x_1,x_2,\cdots,x_d)$ be real numbers such that all of them cannot be equal to unity at the same time. By successively summing over each dimension , using the geometric sum formula and then simplifying the result I showed that: \begin{eqnarray} S_n^m(\vec{x}) := \sum\limits_{n \le i_1 < i_2 < \cdots < i_d \le m} \prod\limits_{l=1}^d {x_l^{i_l}} &=& \sum\limits_{j=-1}^{d-1} (-1)^{j+1} \frac{(\prod\limits_{l=1}^{d-j-1} x_l^{n+l-1}) (\prod\limits_{l=d-j}^d x_l^{m+1})}{\prod\limits_{l=0}^j(1- \prod\limits_{\eta=0}^l x_{d-j+\eta}) \cdot \prod\limits_{l=0}^{d-j-2}(1- \prod\limits_{\eta=0}^l x_{d-j-1-\eta})}\\ &=& \left(\prod\limits_{l=1}^d x_l^{n-1}\right)\frac{\det(-\prod\limits_{l=i}^d x_l + \prod\limits_{l=i}^d x_l^{j+1+(m-n+1-d)\delta_{j,d}})_{i,j=1}^d}{\det(-1+\prod\limits_{l=i}^d x_l^j)_{i,j=1}^d} \end{eqnarray} Now, the sum in question is defined for every $\vec{x} \in {\mathbb R}^d$ including $\vec{x} = (1,1,\cdots,1)$. However the right hand side is not determined at $\vec{x} = (1,1,\cdots,1)$ . The question is how do I compute all partial derivatives of the sum at $(1,1,\cdots,1)$. In other words I need : \begin{eqnarray} \left. \left(\prod\limits_{l=1}^d \frac{1}{b_l!}\frac{d^{b_l}}{d x_l^{b_l}} x_l^{a_l} \right) S^m_n(\vec{x}) \right|_{\vec{x} = (1,1,\cdots,1)} = ? \end{eqnarray} for $(b_1,\cdots,b_d) \in {\mathbb N}^d_+$ and some $(a_1,\cdots,a_d) \in {\mathbb R}^d$.

$\endgroup$
1
$\begingroup$

Let us answer this question for $d=2$. We have: \begin{eqnarray} &&x_1 x_2 (x_1+x_2-x_1 x_2) S^m_n(1-x_1,1-x_2) = \\ &&x_1(1-x_1)^{a_1+n} (1-x_2)^{a_2+n+1} - (x_1+x_2-x_1 x_2)(1-x_1)^{a_1+n} (1-x_2)^{a_2+m+1} + x_2(1-x_1)^{a_1+m+1} (1-x_2)^{a_2+m+1} \end{eqnarray} We see that the sum $S^m_n$ is a rational function of two variables $x_1$ and $x_2$. Clearly the numerator is of the form: \begin{equation} Numerator = x_1 x_2 \sum\limits_{(l_1,l_2) \in {\mathbb N}_+^2} {\mathcal A}_{l_1,l_2} x_1^{l_1} x_2^{l_2} \end{equation} where \begin{equation} {\mathcal A}_{l_1,l_2} = (-1)^{l_1+l_2} \left[ \binom{a_1+n}{l_1} \binom{a_2+m+1}{l_2} +\binom{a_1+n}{l_1} (\binom{a_2+m+1}{l_2+1} - \binom{a_2+n+1}{l_2+1}) -\binom{a_2+m+1}{l_2} (\binom{a_1+m+1}{l_1+1} - \binom{a_1+n}{l_1+1}) \right] \tag{I} \end{equation} The denominator is of the form: \begin{equation} Denominator = x_1 x_2 \sum\limits_{(l_1,l_2) \in \{0,1\}^2 , l_1+l_2 \ge 1} (-1)^{l_1+l_2+1} x_1^{l_1} x_2^{l_2} \end{equation} Dividing the numerator by the denominator we get another polynomial in two variables $\sum\limits_{l_1,l_2}{\mathcal B}_{l_1,l_2} x_1^{l_1} x_2^{l_2}$. The coefficients of the later poly6nomial satisfy following recursion relations: \begin{equation} {\mathcal A}_{l_1,l_2} = {\mathcal B}_{l_1,l_2-1} 1_{l_2\ge 1} + {\mathcal B}_{l_1-1,l_2}1_{l_1\ge 1} - {\mathcal B}_{l_1-1,l_2-1}1_{l_1\ge 1} 1_{l_2 \ge 1} \end{equation} The solution to this recursion relation reads: \begin{equation} {\mathcal B}_{l_1,l_2} = \sum\limits_{j=0}^{l_2} \sum\limits_{j_1=j}^{l_2} (-1)^j \binom{j_1}{j} {\mathcal A}_{l_1+1+j,l_2-j_1} \tag{II} \end{equation} Now, the sum in question reads: \begin{equation} {\mathcal B}_{b_1,b_2} (-1)^{b_1+b_2} \end{equation} Now instering (I) into (II) and simplifying the result we get: \begin{eqnarray} &&\sum\limits_{n \le i_1 < i_2 \le m} \binom{i_1+a_1}{b_1} \binom{i_2+a_2}{b_2} = \\ && \sum\limits_{j=0}^{b_2} (-1)^{j+1} \left. \left[\binom{a_1+n}{1+b_1+j} \binom{a_2+\xi-j}{1+b_2-j} - \binom{1+a_1+\xi}{2+b_1+j} \binom{a_2+m-j}{b_2-j}\right] \right|_{\xi=n}^{\xi=m} \end{eqnarray}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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