Examples of specializations of elementary symmetric polynomials

Question

Let $\mathcal{S}_{x}=\{x_{1,},x_{2},\ldots x_{n}\}$ be a set of $n$ indeterminates. The $h^{th}$elementary symmetric polynomial is the sum of all monomials with $h$ factors \begin{eqnarray*} e_{h}(\mathcal{S}_{x}) & = & \sum_{1\leqslant i_{1}<i_{2}<\ldots<i_{h}\leqslant n}x_{i_{1}}x_{i_{2}}\ldots x_{i_{h-1}}x_{i_{h}} \end{eqnarray*} which, from a generating function standpoint, can be built up as the coefficients of the $h^{th}$ power of the following linear factorization \begin{eqnarray*} \prod_{i=1}^{n}(1+x_{i}z) & = & (1+x_{1}z)(1+x_{2}z)(1+x_{3}z)\ldots(1+x_{n}z)\\ & = & \sum_{h=0}^{n}e_{h}(\mathcal{S}_{x})z^{h} \end{eqnarray*}

Some usual specializations of the set $\mathcal{S}_{x}$ lead to known families of numbers and multiplicative identities: binomial coefficients for $x_{i}=1_{i}$, to $q$-binomial coefficients for $x_{i}=q^{i}$ and Stirling numbers of the first kind for $x_{i}=i$;

(i) For $\mathcal{S}_{1}=\{1_{1},1_{2},1_{3},\ldots,1_{n}\}$ \begin{eqnarray*} (1+z)^{n} & = & (1+1_{1}z)(1+1_{2}z)(1+1_{3}z)\ldots(1+1_{n}z)\\ & = & \sum_{h=0}^{n}{n \choose h}z^{h} \end{eqnarray*} binomial coefficients arise $e_{h}(\mathcal{S}_{1})={n \choose h}$

(ii) For $\mathcal{S}_{q^{i}}=\{q,q^{2},q^{3}\ldots,q^{n-1},q^{n}\}$ \begin{eqnarray*} \prod_{i=1}^{n}(1+q^{i}z) & = & (1+q^{1}z)(1+q^{2}z)(1+q^{3}z)\ldots(1+q^{(n-1)}z)\\ & = & \sum_{h=0}^{n}{n \choose h}_{q}q^{{h+1 \choose 2}}z^{h} \end{eqnarray*} we get the $q$-binomial coefficients (or Gaussian coefficients) $e_{h}(\mathcal{S}_{q^{i}})={n \choose h}_{q}q^{{h+1 \choose 2}}$

(iii) And for $\mathcal{S}_{i}=\{1,2,3,\ldots n-1\}$ \begin{eqnarray*} \prod_{i=1}^{n-1}(1+iz) & = & (1+1z)(1+2z)(1+3z)\ldots(1+(n-1)z)\\ & = & \sum_{h=0}^{n}\left[\begin{array}{c} n\\ n-h \end{array}\right]z^{h} \end{eqnarray*} Stirling numbers of the first kind arise $e_{h}(\mathcal{S}_{i})=\left[\begin{array}{c} n\\ n-h \end{array}\right]$

In this context, are there other specializations of the set $\mathcal{S}_{x}=\{x_{1,},x_{2},\ldots x_{n}\}$ which lead to other families of numbers or identities?

Answer 1

Perhaps also worth mentioning: $$ \prod_{n\geqslant1}(1+n^{-s}z)=1+\sum_{h\geqslant1}\zeta(\underbrace{s,...,s}_h)z^h, $$ where the multiple zeta values of the repeating $s$ are given (e. g. using Newton identities) as \begin{align*} \zeta(s,s)&=\frac12(\zeta(s)^2-\zeta(2s)),\\ \zeta(s,s,s)&=\frac16(\zeta(s)^3-3\zeta(s)\zeta(2s)+2\zeta(3s)),\\ \zeta(s,s,s,s)&=\frac1{24}(\zeta(s)^4-6\zeta(s)^2\zeta(2s)+3\zeta(2s)^2+8\zeta(s)\zeta(3s)-6\zeta(4s)),\\ &... \end{align*} Added later

A better version: $$ \prod_{n\geqslant1}(1+n^{-s}z)=\exp\left(z\zeta(s)-\frac12z^2\zeta(2s)+\frac13z^3\zeta(3s)-...\right) $$

Answer 2

The specialization $\mathcal{S} = \{1, -1, \frac12, -\frac12, \frac13, -\frac13, \ldots \}$, in the limit $n\to \infty$, produces the Weierstrass factorization $$ \frac{\sin \pi z}{\pi z} = \prod_{n\geq 1}\left(1+\frac1{n}z\right)\left(1-\frac1{n}z\right).$$ The Taylor coefficients of this function are $$ e_h(\mathcal{S}) = \frac{(-1)^k}{(2k+1)!} \pi^{2k}$$ if $h = 2k$ is even, and $e_h(\mathcal{S}) = 0$ if $h$ is odd.

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Equivalently, the specialization $\mathcal{S} = \{-1, -\frac14, -\frac19, \ldots \}$ gives the Weierstarss factorization of $\frac{\sin \pi z^{1/2}}{\pi z^{1/2}}$ which has Taylor series coefficients $$ e_h(\mathcal{S}) = \frac{(-1)^h}{(2h+1)!} \pi^{2h} .$$

Answer 3

Under the specialization $\mathcal{S} = \{1, \frac12, \frac13, \ldots, \frac1n \}$, the corresponding products to not converge to a limit as $n\to\infty$ since the harmonic series diverges. However, in the limit $n\to \infty$ these products may be understood in relation to the gamma function $\Gamma(z)$, or rather the reciprocal gamma function $1/\Gamma(z)$. Namely,

$$\prod_{k=1}^n \left( 1 + \frac1k z \right) \to \frac{1}{z\Gamma(z)}\prod_{k=1}^n \left(1+\frac1k\right)^z = \frac{1}{z\Gamma(z)}(1+n)^z$$ where the ''covergence'' means that as $n\to\infty$, the ratio of the two sides goes to 1.

The exponential factor $(1+n)^z$ has Taylor coefficients which get large as $n\to\infty$:

$$(1+n)^z = \sum_{k\geq 0}\frac{\log(1+n)^k}{k!} z^k.$$

The reciprocal gamma factor, on the other hand, has Taylor coefficients which seem to approach 0 in size. The coefficients satisfy a recursive relation which implies they are rational algebraic combinations of the constant $\gamma \approx 0.577$ and the Riemann zeta values $\zeta(j)$ at positive integers. The first few terms are $$\frac1{z\Gamma(z)} = 1 + \gamma z + \left(\frac{\gamma^2}{2} - \frac{\pi^2}{12}\right)z^2 + \cdots .$$