Polynomial expansion of a product, and Stirling numbers?

Question

I've recently came across a problem that demanded to find the value of all the $x_i$ such that $P=x_1 x_2 x_3 \cdots x_n$ is Maximal, given $S=x_1+x_2+\ldots+x_n=a$ where $a$ is some natural number.

The first step is to prove that the maximal value of $P$ is achieved when all the $x_i$s are equal, which is easy to see via the arithmetico-geometric inequality, but i went ahead and tried to prove it in another way.

Let's start with the case when $n = 2$, let $K= \frac{x_1 + x_2}{2}$ so we get $x_1=K+b$ and $x_2=K-b$ for some $b$, so $P=K^2 - b^2$. In order for $P$ to be maximal, $b$ needs to be $0$, so $x_1=x_2$.

Let's try to generalize this. We have $K= \frac{x_1 + x_2 + \ldots + x_n}{n}$, and let $b_n$ be a sequence s.t $x_i=K+b_i$, and $b_1+b_2+ \ldots +b_n=0$.

So we get $P=(K+b_1)(K+b_2)\cdots(K+b_n)$

I tried to expand this, for $n=4$ for example, you get

$$ P= K^4 + K^3(x_1+x_2+x_3+x_4) + K^2 (x_1 x2 + x_1 x_3 + x_1 x_4 + \ldots) + K (x_1 x_2 x_3 + x_1 x_2 x_4 + \ldots) + x_1 x_2 x_3 x_4 $$

You can probably see the pattern, let's let $S_j$ be the sum of the products of the elements of all possible sets containing $j$ elements of all the $x_n$, we can see that

$$P=K^n+K^{n-1} S_1 + K^{n-2} S_2 + \ldots + K^{n-j} S_j+ \ldots+ KS_{n-1} + S_n$$

What is this $S_n$, and is there a formula for it? I know this has a connection with Stirling numbers and falling factorial, for $b_j=j-1$

Answer 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 $x^{\underline{j}}$.

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I hope this helps ^_^