Newton-Girard Formulas/Examples/Order 4

Theorem
Let $a, b \in \Z$ be integers such that $b \ge a$.

Let $U$ be a set of $n = b - a + 1$ numbers $\left\{ {x_a, x_{a + 1}, \ldots, x_b}\right\}$.

Then:
 * $\displaystyle \sum_{a \mathop \le j_1 \mathop < j_2 \mathop < j_3 \mathop < j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4} = \dfrac { {S_1}^4} {24} - \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 - \dfrac {S_4} 4$

where:
 * $\displaystyle S_r := \sum_{k \mathop = a}^b {x_k}^r$.

Proof
From Newton-Girard Formulas:


 * $\displaystyle \sum_{a \mathop \le j_1 \mathop < \mathop \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\left({-S_2}\right)^{k_2} } {2^{k_2} k_2 !} \dfrac { {S_3}^{k_3} } {3^{k_3} k_3 !} \cdots \dfrac {\left({\left({-1}\right)^{m + 1} S_m}\right)^{k_m} } {m^{k_m} k_m !}$

where:
 * $S_j = \displaystyle \sum_{k \mathop = a}^b {x_k}^j$ for $j \in \Z_{\ge 0}$.

Setting $m = 4$:

We need to find all sets of $k_1, k_2, k_3, k_4 \in \Z_{\ge 0}$ such that:


 * $k_1 + 2 k_2 + 3 k_3 + 4 k_4 = 4$

Thus $\left({k_1, k_2, k_3, k_4}\right)$ can be:


 * $\left({4, 0, 0, 0}\right)$
 * $\left({2, 1, 0, 0}\right)$
 * $\left({0, 2, 0, 0}\right)$
 * $\left({1, 0, 1, 0}\right)$
 * $\left({0, 0, 0, 1}\right)$

Hence: