Generating Function for Elementary Symmetric Function

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Theorem

Let $U$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.

Define:

\(\displaystyle \map {e_m} {U}\) \(=\) \(\displaystyle \begin{cases} 1 & m = 0\\ \displaystyle \sum_{1 \mathop \le j_1 \mathop < \mathop \cdots \mathop < j_m \mathop \le n} x_{j_1} x_{j_2} \cdots x_{j_m} & 1 \leq m \leq n \\ 0 & m \gt n \\ \end{cases}\) elementary symmetric function
\(\displaystyle a_m\) \(=\) \(\displaystyle \map {e_m} U\) for $m=0,1,2,\ldots$
\(\displaystyle \map G z\) \(=\) \(\displaystyle \displaystyle \sum_{m=0}^\infty a_m z^m\) generating function for $\set {a_m}_{m=0}^\infty$

Then:

\(\displaystyle \map G z\) \(=\) \(\displaystyle \displaystyle \prod_{k \mathop = 1}^n \paren {1 + x_k z}\)


Explanation

Generating function discovery methods can find a formula for $\map G z$.

Let $n=1$.

Then $U$ is a singleton set $\set {x_1}$.

Expand the formal series:

\(\displaystyle \map G z\) \(=\) \(\displaystyle \map {e_0} U + \map {e_1} z + \sum_{m=2}^\infty 0 z^m\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle 1 + x_1 z\) Because $\map {e_0} U = 1$ and $\map {e_1} U = x_a$

Product of Generating Functions and experience with elementary symmetric functions suggests:

$\displaystyle \map G z = \paren {1 + x_1 z} \paren {1 + x_{2} z} \cdots \paren {1 + x_n z}$

Knuth (1997) section 1.2.9 discusses the technique and the issue of a valid proof.


Proof 1

The summation for $\map G z$ is a finite sum $m = 0, 1, \ldots, n$, which settles convergence issues.

Begin with Viete's Formulas:

$\displaystyle \prod_{k \mathop = a}^b \paren {x - x_k} = x^n + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } U \, x^m$

Change variables $x = -1 / z$:

\(\displaystyle \prod_{k \mathop = 1}^n \paren {-\frac 1 z - x_k}\) \(=\) \(\displaystyle \paren {-z}^{-n} + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } U \, \paren {-z}^{-m}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \prod_{k \mathop = 1}^n \paren {1 + x_k z}\) \(=\) \(\displaystyle z^n + \sum_{m \mathop = 0}^{n - 1} \map {e_{n - m} } U \, \paren z^{n - m}\) all signs cancel
\(\displaystyle \leadsto \ \ \) \(\displaystyle \prod_{k \mathop = 1}^n \paren {1 + x_k z}\) \(=\) \(\displaystyle \sum_{m \mathop = 0}^n \map {e_m} U \, z^m\)

$\blacksquare$


Proof 2

Apply mathematical induction on $n$.

Let $\map P n$ be the statement that

\(\displaystyle \map G z\) \(\equiv\) \(\displaystyle \displaystyle \sum_{m \mathop = 0}^{n+1} \map {e_m} { \set {x_1,\ldots,x_n} } z^{m}\)
\(\displaystyle \) \(=\) \(\displaystyle \displaystyle \prod_{k \mathop = 1}^n \paren {1+x_kz}\)

Basis for the induction:

Set $U = \set {x_1}$ for $n=1$.

Expand the formal series:

\(\displaystyle \map G z\) \(=\) \(\displaystyle \map {e_0} U + \map {e_1} {U} z + \displaystyle\sum_{m \mathop = 2}^\infty \map {e_m} {U} z^m\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \map {e_0} U + \map {e_1} {U} z\) The summation has all zero terms.
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle 1 + x_1 z\)

Then $\map P 1$ holds.

Induction step:

Assume $\map P n$ holds. Let's prove $\map P {n+1}$ holds.

The induction step uses a recursion relation:

\(\text {(1)}: \quad\) \(\displaystyle e_m \paren { \set {x_1,\ldots,x_n,x_{n+1} } }\) \(=\) \(\displaystyle x_{n+1} e_{m-1} \paren { \set {x_1,\ldots,x_n} } + e_m \paren { \set {x_1,\ldots,x_n} }\) Elementary Symmetric Function/Examples/Recursion

Let $\map G z$ be defined by statement $\map P n$.

Let $\map {G^*} z$ be defined by statement $\map P {n+1}$.

Then:

\(\displaystyle \map {G^*} z\) \(=\) \(\displaystyle \displaystyle \prod_{k \mathop = 1}^{n+1} \paren {1+x_kz}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \map G z \paren { 1 + x_{n+1} z }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \map G z + x_{n+1} z \map G z\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \sum_{m \mathop = 0}^n \map {e_m} {U} z^m + \sum_{m \mathop = 1}^{n+1} x_{n+1} \map {e_{m-1} } {U} z^{m}\) use hypothesis $\map P n$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \map {e_0} {U} + \sum_{m \mathop = 1}^{n+1} \paren { \map {e_m} {U} + x_{n+1} \map {e_{m-1} } {U} } z^{m}\) because $\map {e_{n+1} } {U} = 0$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \map {e_0} {U} + \sum_{m \mathop = 1}^{n+1} \map {e_m} {\set {x_1,\ldots,x_n,x_{n+1} } } z^{m}\) by recursion relation (1)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \sum_{m \mathop = 0}^{n+1} \map {e_m} {\set {x_1,\ldots,x_n,x_{n+1} } } z^{m}\) because $\map {e_0} {X} = 1$ for all sets $X$

Then $\map P {n+1}$ holds, completing the induction.

$\blacksquare$


Proof 3

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 $n=1$.

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

Then:

$1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le 1$

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

$j_1 = 1$


Thus in this case:

\(\displaystyle a_m\) \(=\) \(\displaystyle x_1 \delta_{m 1}\) where $\delta_{m 1}$ is the Kronecker delta.
\(\displaystyle \leadsto \ \ \) \(\displaystyle G \left({z}\right)\) \(=\) \(\displaystyle \sum_{n \mathop \ge 0} x_1 \delta_{n 1} z^n\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle G \left({z}\right)\) \(=\) \(\displaystyle 1 + x_1 z\)

Then by Product of Generating Functions, it follows that:

$G \left({z}\right) = \left({1 + x_1 z}\right) \left({1 + x_{2} z}\right) \cdots \left({1 + x_n z}\right)$



Sources

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