Polynomial expansion of a product, and Stirling numbers?

Solution 1:

These $S_n$s are called the Elementary Symmetric Polynomials and they are very well studied. Almost any question you have about them has probably been answered somewhere. Unfortunately, I don't know of a "formula" for computing $S_j$ that's better than the definition

$$S_j(x_1, \ldots, x_n) = \sum_{J \subseteq [n], |J| = j} \prod_{j \in J} x_j$$

but since it seems like you're interested in natural numbers and stirling coefficients, I do know of a closed form that might be of interest to you:

$$S_j(1,2,3,\ldots,n) = s(n,j)$$

where $s(n,j)$ are the signed stirling numbers of the first kind. See here or here for instance.

Again, much is known about the stirling numbers, so if you can phrase your question in that language, I'm sure somebody has an answer!

For instance, since you're looking for a connection to falling factorials, it's "well known" that

$$ x^{\underline{n}} = \sum_{j=0}^n s(n,k) x^k $$

so the stirling numbers act as a kind of "change of basis" from the usual polynomials $x^j$ to the falling-power polynomials $x^{\underline{j}}$.

I hope this helps ^_^