Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula/Examples/Degree 3
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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:
- $S_3 = 3 h_3 - 3 h_1 h_2 + {h_1}^3$
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:
- $k_1 + 2 k_2 + \cdots + m k_m = m$
We need to find all sets of $k_1, k_2, k_3 \in \Z_{\ge 0}$ such that:
- $k_1 + 2 k_2 + 3 k_3 = 3$
Thus $\tuple{k_1, k_2, k_3}$ can be:
- $\tuple{3, 0, 0}$
- $\tuple{1, 1, 0}$
- $\tuple{0, 0, 1}$
Hence:
\(\ds S_3\) | \(=\) | \(\ds \paren{-1}^{3 + 0 + 0 - 1} \dfrac {3 \paren{3 + 0 + 0 - 1}! } {3! \, 0! \, 0!} {h_1}^3 {h_2}^0 {h_3}^0\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \paren{-1}^{1 + 1 + 0 - 1} \dfrac {3 \paren{1 + 1 + 0 - 1}! } {1! \, 1! \, 0!} {h_1}^1 {h_2}^1 {h_3}^0\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \paren{-1}^{0 + 0 + 1 - 1} \dfrac {3 \paren{0 + 0 + 1 - 1}! } {0! \, 0! \, 1!} {h_1}^0 {h_2}^0 {h_3}^1\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 \times 2 {h_1}^3} 6 - \dfrac {3 {h_1}^1 {h_2}^1} 1 + \dfrac {3 {h_3}^1} 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 h_3 - 3 h_1 h_2 + {h_1}^3\) |
$\blacksquare$
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 $11$