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Let $P$ and $Q$ be polynomials over $\mathbb C$, and $n\in\mathbb N$ be a positive integer. I'm interested in the root sums of the form $$ \sum_{P(x)=0}\frac{Q(x)}{P'(x)^n},$$ where the sum runs over all roots of $P$, which we assume to be distinct (simple roots). Since $P'$ is a polynomial, we sum a rational function over the roots of polynomial, which using the elementary symmetric polynomials gives a rational function in the coefficients of $P$ and $Q$.

Motivation: These sums occur in calculations of partition functions in physics.

Goal: Express this as a function of $P$, $Q$ and $n$.

Intermediate goal: Consider the polynomial $Q(x)=1$. As discussed in Math.stackexchange 4955603, I believe this gives $$ \sum_{P(x)=0}\frac{1}{P'(x)^n}=\frac{R(P)}{D(P)^{\lfloor \frac n2 \rfloor}}$$ Here, $D(P)$ denotes the discriminant of the polynomial $P$, $\lfloor \cdot \rfloor$ is the floor function, and $R(P)$ is another polynomial in the coefficients of $P$. For odd $n$ there is a cancellation of a common factor of numerator and denominator.

How does this extend to the cases where $Q$ is an arbitrary polynomial? Are these root sums known in the algebra literature?

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  • $\begingroup$ I'm having a hard time making sense of this post. In the first paragraph, is $R$ the same as $Q$? I'm not sure why you would expect this to be a symmetric polynomial in the coefficients of $P$ when $Q$, which has no connection to $P,$ appears in the numerator... $\endgroup$ Commented Sep 23, 2024 at 21:37
  • $\begingroup$ Apologies, I've been sloppy with changing notation from the other post. Does it make sense now? $\endgroup$ Commented Sep 23, 2024 at 22:49

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I don't know if these sums have a name, but you can compute them using resultants. Let $P$ be a degree-$d$ polynomial over $\mathbb{C}$ with not-necessarily-simple roots $\alpha_1, \ldots, \alpha_d \in \mathbb{C}$, and let $p_d \in \mathbb{C}$ be the leading coefficient of $P$. Suppose we have two polynomials $Q$ and $R$ over $\mathbb{C}$, with the added promise that $R$ and $P$ share no common roots (in your setting, we have $R = P'$ and $P$ is assumed to have no repeated roots). Writing the sum $\sum_{i=1}^d \frac{Q(\alpha_i)}{R(\alpha_i)}$ over a common denominator, we see that we need to compute

  1. the numerator $\sum_{i=1}^d Q(\alpha_i) \prod_{j \neq i} R(\alpha_j)$, and
  2. the denominator $\prod_{i=1}^d R(\alpha_i)$.

Both of these terms appear in the expansion of the polynomial $G(y) := \prod_{i=1}^d (Q(\alpha_i) + y \cdot R(\alpha_i))$, where $y$ is a new variable, so it suffices to compute the coefficients of $G(y)$. The resultant of $P(x)$ and $Q(x) + y \cdot R(x)$ with respect to the variable $x$ is precisely $$ \mathrm{res}_x(P(x), Q(x) + y \cdot R(x)) = p_d^e \prod_{i=1}^d (Q(\alpha_i) + y \cdot R(\alpha_i)) = p_d^e \cdot G(y), $$ where $e = \max(\deg(Q), \deg(R))$. This resultant can be computed by, e.g., finding the determinant of the Sylvester matrix of $P(x)$ and $Q(x) + y \cdot R(x)$.

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    $\begingroup$ Nice! It might be worth mentioning that any question similar to this one can be answered using resultants. That's because the transition matrix from monomial symmetric functions to the elementary symmetric functions, involving the inverse Kostka matrix, is related to the coefficients of the resultant. See, e.g., Sec 7 of arxiv.org/abs/2407.04429 $\endgroup$ Commented Oct 10, 2024 at 14:18
  • $\begingroup$ Amazing, thank you! I had the resultants in mind but couldn't make the connection. $\endgroup$ Commented Oct 11, 2024 at 13:46

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