Summation of Products of n Numbers taken m at a time with Repetitions/Recurrence Formula

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds h_m\) \(=\) \(\ds \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} \paren {\prod_{k \mathop = 1}^m x_{j_k} }\)
\(\ds \) \(=\) \(\ds \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m}\)


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:

$\ds S_r = \sum_{k \mathop = a}^b {x_k}^r$


A recurrence relation for $h_n$ can be given as:

\(\ds h_n\) \(=\) \(\ds \sum_{k \mathop = 1}^n \dfrac {S_k h_{n - k} } n\)
\(\ds \) \(=\) \(\ds \dfrac 1 n \paren {S_1 h_{n - 1} + S_2 h_{n - 2} + \cdots S_n h_0}\)

for $n \ge 1$.


Proof

\(\ds \map \ln {\map G z}\) \(=\) \(\ds \sum_{k \mathop \ge 1} \dfrac {S_k z^k} k\) Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2
\(\ds \leadsto \ \ \) \(\ds \map {\dfrac \d {\d z} } {\map \ln {\map G z} }\) \(=\) \(\ds \map {\dfrac \d {\d z} } {\sum_{k \mathop \ge 1} \dfrac {S_k z^k} k}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 {\map G z} \map {\dfrac \d {\d z} } {\map G z}\) \(=\) \(\ds \sum_{k \mathop \ge 1} \map {\dfrac \d {\d z} } {\dfrac {S_k z^k} k}\) Derivative of Logarithm Function, Chain Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \dfrac 1 {\sum_{m \mathop \ge 0} h_m z^m} \sum_{m \mathop \ge 0} \paren {m + 1} h_{m + 1} z^m\) \(=\) \(\ds \sum_{k \mathop \ge 1} S_k z^{k - 1}\) Derivative of Generating Function, Power Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \sum_{m \mathop \ge 0} \paren {m + 1} h_{m + 1} z^m\) \(=\) \(\ds \sum_{k \mathop \ge 1} S_k z^{k - 1} \sum_{m \mathop \ge 0} h_m z^m\)


This needs considerable tedious hard slog to complete it.
In particular: Follows from Product of Generating Functions
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.



Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Recurrence_Formula&oldid=565653"