# Sum of power functions over a simplex

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 $(a_1,\cdots,a_d) \in {\mathbb R}^d$.

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: $$Numerator = x_1 x_2 \sum\limits_{(l_1,l_2) \in {\mathbb N}_+^2} {\mathcal A}_{l_1,l_2} x_1^{l_1} x_2^{l_2}$$ where $${\mathcal A}_{l_1,l_2} = (-1)^{l_1+l_2} \left[ \binom{a_1+n}{l_1} \binom{a_2+m+1}{l_2} +\binom{a_1+n}{l_1} (\binom{a_2+m+1}{l_2+1} - \binom{a_2+n+1}{l_2+1}) -\binom{a_2+m+1}{l_2} (\binom{a_1+m+1}{l_1+1} - \binom{a_1+n}{l_1+1}) \right] \tag{I}$$ The denominator is of the form: $$Denominator = x_1 x_2 \sum\limits_{(l_1,l_2) \in \{0,1\}^2 , l_1+l_2 \ge 1} (-1)^{l_1+l_2+1} x_1^{l_1} x_2^{l_2}$$ 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: $${\mathcal A}_{l_1,l_2} = {\mathcal B}_{l_1,l_2-1} 1_{l_2\ge 1} + {\mathcal B}_{l_1-1,l_2}1_{l_1\ge 1} - {\mathcal B}_{l_1-1,l_2-1}1_{l_1\ge 1} 1_{l_2 \ge 1}$$ The solution to this recursion relation reads: $${\mathcal B}_{l_1,l_2} = \sum\limits_{j=0}^{l_2} \sum\limits_{j_1=j}^{l_2} (-1)^j \binom{j_1}{j} {\mathcal A}_{l_1+1+j,l_2-j_1} \tag{II}$$ Now, the sum in question reads: $${\mathcal B}_{b_1,b_2} (-1)^{b_1+b_2}$$ 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}