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Given $n$ variables there are $n \choose k$ different terms that are the product of $k$ different variables. For example, in the case that $n = 3$, the $k$-products of the variables $x_1, x_2, x_3$, are

  • $k = 1$: $x_1$, $x_2$, $x_3$;
  • $k = 2$: $x_1x_2$, $x_1x_3$, $x_2x_3$; and
  • $k = 3$: $x_1x_2x_3$.

Can we write the sum of all the $k$-products concisely using sum and product notation? If so, how? For $n = 3$, for instance, we want an expression using $\Sigma$-$\Pi$-notation, in terms of $k$ (and generally in terms of $n$ as well), for $s_{3,k}$, the sum of the $k$-products:

  • $s_{3,1} = x_1 + x_2 + x_3$;
  • $s_{3,2} = x_1x_2 + x_1x_3 + x_2x_3$; and
  • $s_{3,3} = x_1x_2x_3$;

This is basically just the multinomial theorem, with the coefficients and exponents neglected, and broken into several separate sums. But it's not clear to me whether it's possible in the conventional notation to express the "choice" needed to form the sums separately.

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1 Answer 1

One possibility to express the $k$-th symmetric polynomial is the following expression: $$s_{n,k}=\sum_{A\subseteq\{1,\ldots,n\} \wedge |A|=k}\ \ \prod_{i\in A} x_i$$

As a sidenote, an "implicit" definition of these sums could look like this: $$\prod_{i=1}^n (x+x_i) = \sum_{k=0}^n s_{n,k}x^{n-k}$$ (with the silent assumption that $s_{n,0}=1$) Borrowing the generating functions notation: $$s_{n,k}=[x^{n-k}] \prod_{i=1}^n (x+x_i)$$

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