Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula

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Theorem

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

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

Let $m \in \Z_{>0}$ be a (strictly) positive integer.


Let:

\(\ds h_m\) \(=\) \(\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} \paren {\prod_{k \mathop = 1}^m x_{j_k} }\)
\(\ds \) \(=\) \(\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m}\)

That is, $h_m$ is the product of all $m$-tuples of elements of $U$ taken $m$ at a time.


For $r \in \Z_{> 0}$, let:

$S_r = \ds \sum_{j \mathop = a}^b {x_j}^r$


Let $S_m$ be expressed in the form:

$S_m = \ds \sum_{k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} A_m {h_1}^{k_1} {h_2}^{k_2} \cdots {h_m}^{k_m}$

for $k_1, k_2, \ldots, k_m \ge 0$.


Then :

$A_m = \paren {-1}^{k_1 + k_2 + \cdots + k_m - 1} \dfrac {m \paren {k_1 + k_2 + \cdots + k_m - 1}! } {k_1! \, k_2! \, \cdots k_m!}$


Proof

Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.

\(\ds \sum_{m \mathop \ge 1} \dfrac {S_m z^m} m\) \(=\) \(\ds \map \ln {\map G z}\) Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2
\(\ds \) \(=\) \(\ds \map \ln {1 + h_1 z + h_2 z^2 + \cdots}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop \ge 1} \dfrac {\paren {-1}^{k - 1} } k \paren {h_1 z + h_2 z^2 + \cdots}^k\) Power Series Expansion for Logarithm of 1 + x
\(\ds \leadsto \ \ \) \(\ds S_m\) \(=\) \(\ds m \sum_{k \mathop \ge 1} \dfrac {\paren {-1}^{k - 1} } k \paren {\sum_{j \mathop = 1} h_j z^j}^k\)


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Examples

Degree $1$

$S_1 = h_1$


Degree $2$

$S_2 = 2 h_2 - {h_1}^2$


Degree $3$

$S_3 = 3 h_3 - 3 h_1 h_2 + {h_1}^3$


Degree $4$

$S_4 = 4 h_4 - 4 h_1 h_3 - 2 {h_2}^2 + 4 {h_1}^2 h_1 - {h_1}^4$


Sources

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