A symmetric function related to sums of square roots

Question

Let $x_1,x_2,\dots,x_n$ be indeterminates (say over $\mathbb{Q}$). For every sequence $\epsilon=(\epsilon_1, \dots,\epsilon_n)\in\{-1,1\}^n$ define $$ y_\epsilon = \sum_i \epsilon_i \sqrt{x_i}. $$ Let $f_{n,k}$ denote the $k$th elementary symmetric function of all the $y_\epsilon$'s. Thus $f_{n,2j+1}=0$ and $f_{n,2j}$ is a symmetric function of degree $j$ in the $x_i$'s. What can be said about this symmetric function? For instance, $f_{n,2}=-2^{n-1}(x_1+\cdots+x_n)$ and $f_{4,4}=12h_1^2+16h_2$, where $h_\lambda$ is a complete symmetric function.

In particular, does $f_{n,k}$ have an interesting expansion in terms of any of the ``classical'' bases for symmetric functions (monomial, power sum, elementary, complete, forgotten, Schur)? Is there some way to normalize $f_{n,k}$ so that it makes sense to let $n\to \infty$?

Answer 1

(Not an answer, just some thoughts I find useful.) There is a nice identity $$\sum_{(\varepsilon_1,\dots,\varepsilon_n)\in\{-1,1\}^n}\cos\left(\sum_{i=1}^n\varepsilon_iz_i\right)=2^n\prod_{i=1}^n\cos(z_i)$$ one can obtain by induction on $n$. More generally, if one defines $R_a=\{z\in\mathbb C:z^a=1\}$ and $$f_a(x)=\sum_{n\ge0}\frac{x^{an}}{(an)!}=\frac1a\sum_{r\in R_a}\exp(rx),$$ then $$\sum_{(\varepsilon_1,\dots,\varepsilon_n)\in R_a^n} f_a\left(t\cdot\sum_{i=1}^n\varepsilon_iz_i\right)=a^n\prod_{i=1}^nf_a(t\cdot z_i).$$ Expanding both sides as formal power series in $t$, it follows that $$\frac1{m!}\sum_{(\varepsilon_1,\dots,\varepsilon_n)\in R_a^n} \left(\sum_{i=1}^n\varepsilon_iz_i\right)^m = a^n\sum_{(\lambda_1,\dots,\lambda_n)\in P_m}\prod_{i=1}^n\frac{z_i^{\lambda_i}}{\lambda_i!},$$ where $P_m=\{(\lambda_1,\dots,\lambda_n)\in(a\mathbb Z_{\ge0})^n:\sum\lambda_i=m\}$, which is obviously empty for $a\nmid m$. So, if $z_i=\sqrt[a]{x_i}$, one gets the expansion of $ak$-th power sum of all $y_\varepsilon$ in terms of monomial symmetric polynomials. Unfortunately, Newton formulas seem to produce such a mess, so some amount of combinatorial work is required to get the coefficients of the required expansions. I'll add it later