Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula/Examples/Degree 1
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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
- 1629: Albert Girard: Invention Nouvelle en l'Algèbre
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.9$: Generating Functions: Exercise $11$