Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1

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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.


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

Then:

\(\ds \map G z\) \(=\) \(\ds \prod_{k \mathop = a}^b \dfrac 1 {1 - x_k z}\)
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \cdots \paren {1 - x_b z} }\)


Proof

For each $k \in \set {a, a + 1, \ldots, b}$, the product of $x_k$ taken $m$ at a time is simply ${x_k}^m$.

Thus for $n = 1$ we have:

$h_m = {x_k}^m$

Let the generating function for such a $\sequence {h_m}$ be $\map {G_k} z$.

From Generating Function for Sequence of Powers of Constant:

$\map {G_k} z = \dfrac 1 {1 - x_k z}$


By Product of Summations, we have:

$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \prod_{k \mathop = a}^b \sum_{j \mathop = 1}^m x_j$


Hence:

\(\ds \map G z\) \(=\) \(\ds \sum_{k \mathop \ge 0} h_k z^k\) Definition of Generating Function
\(\ds \) \(=\) \(\ds \prod_{k \mathop = a}^b \dfrac 1 {1 - x_k z}\) Product of Generating Functions: General Rule
\(\ds \) \(=\) \(\ds \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \dotsm \paren {1 - x_b z} }\)

$\blacksquare$


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