Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula/Examples/Degree 1

From ProofWiki
Jump to navigation Jump to search

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

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:

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


Then:

$S_1 = h_1$


Proof

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

$S_m = \ds \sum_{k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} \paren {-1}^{k_1 + k_2 + \cdots + k_m - 1} \dfrac {m \paren {k_1 + k_2 + \cdots + k_m - 1}! } {k_1! \, k_2! \, \cdots k_m!} {h_1}^{k_1} {h_2}^{k_2} \cdots {h_m}^{k_m}$

where:

$k_1 + 2 k_2 + \cdots + m k_m = m$

When $m = 1$ we have only one set of $k_j$:

$k_1 = 1$

Thus:

\(\ds S_1\) \(=\) \(\ds \paren {-1}^{1 - 1} \dfrac {1 \paren {1 - 1}! } 1 {h_1}^1\)
\(\ds \) \(=\) \(\ds h_1\)

$\blacksquare$


Sources