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We know that for $x_{1} \neq x_{2}$, $$ \sum_{\substack{\mbox{}\\ i_{1},\ i_{2}\ \in\ \mathbb{N}_{\ \geq\ 0} \\[1mm] i_{1}\ +\ i_{2}\ =\ n}} x_{1}^{i_{1}}\,x_{2}^{i_{2}}\ =\ \frac{x_{1}^{n + 1} - x_{2}^{n + 1}}{x_{1} - x_{2}} $$ See, e.g., a previous question.

Q: Can we extend this to higher orders ?, i.e., can we have a nice formula for $$ \sum_{\substack{\mbox{}\\ i_{1},\ i_{2},\ \ldots,\ i_{k}\ \in\ \mathbb{N}_{\,\geq\ 0}\\[1mm] i_{1}\ +\ i_{2}\ +\ \cdots\ +\ i_{k}\ =\ n}} \rule{4mm}{0pt} \prod_{s\ =\ 1}^{k}x_{s}^{i_{s}}\,\,\, {\large ?}. $$

What I have tried: Is this correct?

  • Using generating function \begin{align} \operatorname{G}\left(x\right) & = \prod_{s\ =\ 1}^{k}\frac{1}{1 - x_{s}x}, \\ \mbox{we have}\quad \frac{1}{1 - x_{s}x} & = \sum_{m\ =\ 0}^{\infty}x_{s}^{m}\,x^{m} \\ \mbox{and thus}\quad \operatorname{G}\left(x\right) & = \sum_{m\ =\ 0}^{\infty}x^{m} \left(\sum_{\substack{\mbox{}\\ i_{1},\ i_{2},\ \ldots,\ i_{k}\ \in\ \mathbb{N}_{\,\geq\ 0} \\[1mm] i_{1}\ +\ i_{2}\ +\ \cdots\ +\ i_{k}\ =\ m}} \rule{5mm}{0pt} \prod_{s\ =\ 1}^{k}x_{s}^{i_{s}}\right) \end{align} Therefore, the coefficient of $x^{n}$ in $\operatorname{G}\left(x\right)$ would be the answer.
  • Assuming all $x_{i}$'s are distinct, use partial fraction decomposition: $$ \operatorname{G}\left(x\right) = \sum_{s\ =\ 1}^{k}\frac{A_{s}}{1 - x_{s}x} $$
  • Multiply both sides by $\prod_{s\ =\ 1}^{k}\, \left(1 - x_{s}x\right)$: \begin{align} 1 & = \sum_{s}A_{s} \prod_{\substack{\mbox{}\\ \ell\ \neq\ s}} \left(1 - x_{\ell}x\right) \\[2mm] \mbox{Let}\ x = {1 \over x_{s}}:\quad 1 & = A_{s}\prod_{\substack{\mbox{}\\ \ell\ \neq\ s}}\left(1 - \frac{x_{\ell}}{x_{s}}\right) \\[2mm] \mbox{and thus}\quad A_{s} & = \frac{1}{\prod_{\ell\ \neq\ s} (1 - \frac{x_\ell}{x_s})} \end{align}
  • Hence, the coefficient of $x^{n}$ in $\operatorname{G}\left(x\right)$ is: $$ \sum_{s}A_{s}\,x_{s}^{n}\ =\ \sum_{s}\frac{x_{s}^{n}} {\prod_{\ell\ \neq\ s}\, \left(1 - x_{\ell}/x_{s}\right)} $$
Felix Marin
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Vezen BU
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2 Answers2

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To rephrase @Calvin Lin's answer in a more elegant, compact and symmetric form, $$ \sum\limits_{i_0+\dots+i_k=n} x_0^{i_0} \dots x_k^{i_k} = \sum\limits_{i=0}^k x_i^{n+k} \prod\limits_{j \neq i} \frac{1}{x_i-x_j} $$

In particular, for $n=k=2$ we have

$$ a^2+b^2+c^2+ab+bc+ca = \frac{a^4}{(a-b)(a-c)}+\frac{b^4}{(b-a)(b-c)}+\frac{c^4}{(c-a)(c-b)} $$

This actually seems to match the expression that you got in the question statement.

Alternatively, you can also formulate it in the Vandermonde's matrix terms:

$$ \begin{vmatrix} 1 & x_0 & x_0^2 & \dots & x_0^{k} \\ 1 & x_1 & x_1^2 & \dots & x_1^{k} \\ 1 & x_2 & x_2^2 & \dots & x_2^{k} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_k & x_k^2 & \dots & x_k^{k} \end{vmatrix} \sum\limits_{i_0+\dots+i_k=n} x_0^{i_0} \dots x_k^{i_k} = \begin{vmatrix} 1 & x_0 & x_0^2 & \dots & x_0^{n+k} \\ 1 & x_1 & x_1^2 & \dots & x_1^{n+k} \\ 1 & x_2 & x_2^2 & \dots & x_2^{n+k} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_k & x_k^2 & \dots & x_k^{n+k} \end{vmatrix} $$ I couldn't find a direct interpretation to it, but from Cramer's rule it means that in the system

$$ \begin{bmatrix} 1 & x_0 & x_0^2 & \dots & x_0^{k} \\ 1 & x_1 & x_1^2 & \dots & x_1^{k} \\ 1 & x_2 & x_2^2 & \dots & x_2^{k} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & x_k & x_k^2 & \dots & x_k^{k} \end{bmatrix} \begin{bmatrix}a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_k\end{bmatrix} = \begin{bmatrix}x_0^{n+k} \\ x_1^{n+k} \\ x_2^{n+k} \\ \vdots \\ x_k^{n+k}\end{bmatrix}, $$

$a_k$ is exactly the expression that you're looking for. Maybe there is also a decent formula for $a_i$?

Oleksandr Kulkov
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3

(Not a complete solution. The proof is left as an exercise.)

Let $S_k$ denote the permutation group on $k$ elements.
For a permutation $ \sigma \in S_k$, $(-1)^\sigma$ denotes it's parity.

Claim: $ \displaystyle \prod_{1\leq i<j \leq k} (x_i - x_j) = \sum_{\sigma \in S_k} \left[ (-1)^\sigma \prod_i x_i^{k - \sigma(i) }\right] $.

As an explicit example, for $k = 3$,
$$ \begin{array} {l l l } & & (a-b)(a-c)(b-c) \\ & = & a^{3-1}b^{3-2}c^{3-3} - a^{3-1}b^{3-3}c^{3-2} - a^{3-2}b^{3-1}c^{3-3} + a^{3-2}b^{3-3}c^{3-1} + a^{3-3}b^{3-1}c^{3-2} - a^{3-3} b^{3-2}c^{3-1} \\ & = & a^2b-a^2c+b^2c - b^2a + c^2a-c^2 b. \end{array}$$

Let OP's expression be denoted as $ \displaystyle f(n, k) = \sum_{i_1, i_2, \ldots, i_k \in\mathbb{N}_{\geq0} \\ i_1+i_2+\cdots+i_k=n} \prod_{s = 1}^{k} x_s^{i_s}.$

Claim: $ f(n, k) \displaystyle \prod_{1\leq i<j \leq k} (x_i - x_j) = \sum_{\sigma \in S_k, \sigma(j) = 1} \left[ (-1)^\sigma x_j ^{n+k-1} \prod_{i\neq j} x_i^{k - \sigma(i) } \right] $.

As an explicit example, for $n=2, k = 3$,
$$ \begin{array} { l l } & & (a^2+b^2+c^2+ab+bc+ca)(a-b)(a-c)(b-c) \\ & = & a^{2+3-1}b^{3-2}c^{3-3} - a^{2+3-1}b^{3-3}c^{3-2} - a^{3-2}b^{3-1}c^{2+3-3} + a^{3-2}b^{3-3}c^{2+3-1} + a^{3-3}b^{2+3-1}c^{3-2} - a^{3-3} b^{3-2}c^{2+3-1} \\ & = & a^4b-a^4c+b^4c - b^4a + c^4a-c^4 b. \end{array} $$

Corollary: This gives us a "nice formula" that is the generalization of OP's observation for $k = 2$. Even though $f(n,k)$ has ${n+k-1 \choose n } $terms, it can be expressed as a fraction with $k!$ terms in the numerator and $k!$ terms in the denominator that match term wise:

$$ \displaystyle f(n, k) = \frac{\displaystyle \sum_{\sigma \in S_k, \sigma(j) = 1} \left[ (-1)^\sigma x_j ^{n+k-1} \prod_{i\neq j} x_i^{k - \sigma(i) } \right]}{ \displaystyle \sum_{\sigma \in S_k} \left[ (-1)^\sigma \prod_i x_i^{k - \sigma(i) }\right] } $$

As an explicit example, for $n = 2, k = 3$,

$$ (a^2+b^2+c^2+ab+bc+ca) = \frac{ a^4b-a^4c+b^4c - b^4a + c^4a-c^4 b}{ a^2b-a^2c+b^2c - b^2a + c^2a-c^2 b }. $$

To the uninitiated, it may seem surprising that these polynomials divide out nicely.
If we apply Remainder Factor Theorem, it is clear why the numerator has a factor of $x_i - x_j$, hence is a multiple of the denominator.

Calvin Lin
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