Trouble interpreting a generating function equation

Question

Given this generating function equation:

$\prod^n_{i=1}\frac{x_iz}{e^{zx_i}-1} = \sum^\infty_{j=0}B_j^{(n)}(x_1, x_2, ...x_n)\frac{z^j}{j!}$

I am having trouble breaking down the pieces of this. I have looked at generating functions, Bernoulli polynomials, and symmetric functions that resemble the above, but each time I think I understand and can formulate one side from the other, something is just a bit off.

My biggest struggle is trying to formalize $B_j^{(n)}$, as I can tell it's a symmetric function by the identity but can't find a way to prove this. I'm trying to start by computing the first few iterations of $B_j^{(n)}$, and expanding from there but haven't had any luck.

Something is just not clicking, and if anyone could break the equation down for me or help me make sense of this it would be much appreciated!

Answer 1

Since $\frac{z}{e^z-1}=\sum_{j=0}^\infty B_j\frac{z^j}{j}$ is a generating function of the Bernoulli numbers we can represent $B_j^{(n)}(x_1,x_2,\ldots,x_n)$ as multivariate polynomial with coefficients in terms of these numbers.

We obtain \begin{align*} \sum_{j=0}^\infty&\color{blue}{B_j^{(n)}(x_1,x_2,\ldots,x_n)}\frac{z^j}{j!}\\ &=\prod_{i=1}^n\frac{x_iz}{e^{x_iz}-1}\\ &=\prod_{i=1}^n\sum_{j=0}^\infty B_j\frac{(x_iz)^j}{j!}\\ &=\sum_{k=0}^\infty\sum_{{j_1+j_2+\cdots+j_n}\atop{j_1,j_2,\ldots,j_n\geq 0}} B_{j_1}B_{j_2}\cdots B_{j_n}\frac{(x_1z)^{j_1}(x_2z)^{j_2}\cdots(x_nz)^{j_n}}{j_1!j_2!\cdots j_n!}z^k\\ &=\sum_{k=0}^\infty\color{blue}{\sum_{{j_1+j_2+\cdots+j_n}\atop{j_1,j_2,\ldots,j_n\geq 0}} \binom{k}{j_1,j_2,\ldots,j_n}B_{j_1}B_{j_2}\cdots B_{j_n}x_1^{j_1}x_2^{j_2}\cdots x_n^{j_n}}\frac{z^k}{k!} \end{align*}