Generating Function for Elementary Symmetric Function

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 $\left\{ {x_a, x_{a + 1}, \ldots, x_b}\right\}$.

Let $m \in \Z_{>0}$ be a (strictly) positive integer.


Let $a_m$ be an elementary symmetric function:

$a_m = \displaystyle \sum_{a \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le b} x_{j_1} x_{j_2} \cdots x_{j_m}$


Let $G \left({z}\right)$ be a generating function for the sequence $\left\langle{a_m}\right\rangle$.


Then:

$G \left({z}\right) = \displaystyle \prod_{k \mathop = a}^b \left({1 + x_k z}\right)$


Proof

We have by definition of generating function that:

$G \left({z}\right) = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$

We have that:

$a_0 = 1$



Suppose $a = b$.

Let $G_a \left({z}\right)$ be the generating function for $\left\langle{a_m}\right\rangle$ under this condition.

Then:

$a \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le a$

can be fulfilled by only one set $\left\{ {j_1, j_2, \ldots, j_m}\right\}$, that is:

$j_1 = a$


Thus in this case:

\(\displaystyle a_m\) \(=\) \(\displaystyle x_a \delta_{m 1}\) where $\delta_{m 1}$ is the Kronecker delta.
\(\displaystyle \leadsto \ \ \) \(\displaystyle G_a \left({z}\right)\) \(=\) \(\displaystyle \sum_{n \mathop \ge 0} x_a \delta_{n 1} z^n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle G_a \left({z}\right)\) \(=\) \(\displaystyle 1 + x_a z\)

Then by Product of Generating Functions, it follows that:

$G \left({z}\right) = \left({1 + x_a z}\right) \left({1 + x_{a + 1} z}\right) \cdots \left({1 + x_b z}\right)$



Sources