Recursion Property of Elementary Symmetric Function

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Theorem

Let $\set {z_1, z_2, \ldots, z_{n + 1} }$ be a set of $n + 1$ numbers, duplicate values permitted.


Then for $1 \le m \le n$:

$\map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } } = z_{n + 1} \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} }$

where $\map {e_m} {\set {z_1, \ldots, z_n} }$ denotes the elementary symmetric function of degree $m$ on $\set {z_1, \ldots, z_n}$.


Proof 1

Case $m = 1$ holds because $e_0$ is $1$ and $e_1$ is the sum of the elements.


Assume $2 \le m \le n$.

Define four sets:

$A = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n + 1}$
$B = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_{m - 1} \le n, p_m = n + 1}$
$C = \set {\set {p_1, \ldots, p_m} : 1 \le p_1 < \cdots < p_m \le n}$
$D = \set {\set {p_1, \ldots, p_{m - 1} } : 1 \le p_1 < \cdots < p_{m - 1} \le n}$


Then $A = B \cup C$ and $B \cap C = \O$ implies:


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$\ds \sum_A z_{p_1} \cdots z_{p_m} = \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}$

Simplify:

$\ds \sum_B z_{p_1} \cdots z_{p_m} = z_{n + 1} \sum_D z_{p_1} \cdots z_{p_{m - 1} }$


By definition of elementary symmetric function:

\(\ds \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } }\) \(=\) \(\ds \sum_A z_{p_1} \cdots z_{p_m}\)
\(\ds \sum_D z_{p_1} \cdots z_{p_{m - 1} }\) \(=\) \(\ds \map {e_{m - 1} } {\set {z_1, \ldots, z_n} }\)
\(\ds \sum_C z_{p_1} \cdots z_{p_m}\) \(=\) \(\ds \map {e_m} {\set {z_1, \ldots, z_n} }\)


Assemble the preceding equations:

\(\ds \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } }\) \(=\) \(\ds \sum_A z_{p_1} \cdots z_{p_m}\)
\(\ds \) \(=\) \(\ds \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}\)
\(\ds \) \(=\) \(\ds z_{n + 1} \sum_D z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}\)
\(\ds \) \(=\) \(\ds z_{n + 1} \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} }\)

$\blacksquare$


Proof 2

Recall the definition of elementary symmetric function:

\(\ds \map {e_m} {\set {z_1, z_2, \ldots, z_n} }\) \(=\) \(\ds \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le n} \paren {\prod_{i \mathop = 1}^m z_{j_i} }\)
\(\ds \) \(=\) \(\ds \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le n} z_{j_1} z_{j_2} \cdots z_{j_m}\)


Consider the summands of $\map {e_m} {\set {z_1, z_2, \ldots, z_n, z_{n + 1} } }$:

$z_{j_1} z_{j_2} \cdots z_{j_m}$

where $1 \le j_1 < j_2 < \cdots j_m \le n + 1$.


They consist of $2$ types:

Type $(1)$: such that $j_m < n + 1$
Type $(2)$: such that $j_m = n + 1$.


We have that:

the summands of Type $(1)$ are exactly the summands of $\map {e_m} {\set {z_1, z_2, \ldots, z_n} }$
the summands of Type $(2)$ consist of the summands of $\map {e_{m - 1} } {\set {z_1, z_2, \ldots, z_n} }$ multiplied by $z_{n + 1}$.


Hence the result.

$\blacksquare$


Examples

Example: $x_{n + 1} \map {e_n} {\set {x_1, x_2, \ldots, x_n} } = \map {e_{n + 1} } {\set {x_1, x_2, \ldots, x_n, x_{n + 1} } }$

$x_{n + 1} \map {e_n} {\set {x_1, x_2, \ldots, x_n} } = \map {e_{n + 1} } {\set {x_1, x_2, \ldots, x_n, x_{n + 1} } }$
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