Recursion Property of Elementary Symmetric Function/Proof 1

Theorem

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

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

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

Proof

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:

\(\ds \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } }\)

\(=\)

\(\ds \sum_A z_{p_1} \cdots z_{p_m}\)

\(\ds \)

\(=\)

\(\ds \sum_B z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}\)

\(\ds \)

\(=\)

\(\ds z_{n + 1} \sum_D z_{p_1} \cdots z_{p_m} + \sum_C z_{p_1} \cdots z_{p_m}\)

\(\ds \)

\(=\)

\(\ds z_{n + 1} \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} }\)

$\blacksquare$