Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula/Examples/Degree 4

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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:

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


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 $k_1, k_2, k_3, k_4 \in \Z_{\ge 0}$ such that:

$k_1 + 2 k_2 + 3 k_3 + 4 k_4 = 4$

Thus $\tuple {k_1, k_2, k_3, k_4}$ can be:

$\tuple {4, 0, 0, 0}$
$\tuple {2, 1, 0, 0}$
$\tuple {0, 2, 0, 0}$
$\tuple {1, 0, 1, 0}$
$\tuple {0, 0, 0, 1}$


Hence:

\(\ds S_4\) \(=\) \(\ds \paren {-1}^{4 + 0 + 0 + 0 - 1} \dfrac {4 \paren {4 + 0 + 0 + 0 - 1}! } {4! \, 0! \, 0! \, 0!} {h_1}^4 {h_2}^0 {h_3}^0 {h_4}^0\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{2 + 1 + 0 + 0 - 1} \dfrac {4 \paren {2 + 1 + 0 + 0 - 1}! } {2! \, 1! \, 0! \, 0!} {h_1}^2 {h_2}^1 {h_3}^0 {h_4}^0\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{0 + 2 + 0 + 0 - 1} \dfrac {4 \paren {0 + 2 + 0 + 0 - 1}! } {0! \, 2! \, 0! \, 0!} {h_1}^0 {h_2}^2 {h_3}^0 {h_4}^0\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{1 + 0 + 1 + 0 - 1} \dfrac {4 \paren {1 + 0 + 1 + 0 - 1}! } {1! \, 0! \, 1! \, 0!} {h_1}^1 {h_2}^0 {h_3}^1 {h_4}^0\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-1}^{0 + 0 + 0 + 1 - 1} \dfrac {4 \paren {0 + 0 + 0 + 1 - 1}! } {0! \, 0! \, 0! \, 1!} {h_1}^0 {h_2}^0 {h_3}^0 {h_4}^1\)
\(\ds \) \(=\) \(\ds -\dfrac {4 \times 6 {h_1}^4} {24} + \dfrac {4 \times 2 {h_1}^2 {h_2}^1} 2 - \dfrac {4 \times 1 {h_2}^2} 2 - \dfrac {4 \times 1 {h_1}^1 {h_3}^1} 1 + \dfrac {4 \times 1 {h_4}^1} 1\)
\(\ds \) \(=\) \(\ds 4 h_4 - 4 h_1 h_3 - 2 {h_2}^2 + 4 {h_1}^2 h_1 - {h_1}^4\)

$\blacksquare$


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