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Finite Newton recurrences for the slice zeros

Statement

Let $α_{1}, \dots, α_{d} \in C$ , let $p_{n} : = α_{1}^{n} + \dots + α_{d}^{n}$ , and let $e_{n}$ be the $n$ th elementary symmetric polynomial in the $α_{j}$ . Then, with $e_{0} : = 1$ ,

$n e_{n} = \sum_{j = 1}^{n} (- 1)^{j - 1} e_{n - j} p_{j} (1 \leq n \leq d) .$

Consequently each $e_{n}$ is a polynomial with rational coefficients in $p_{1}, \dots, p_{n}$ . In particular, whenever the $p_{j}$ vary holomorphically, so do the $e_{n}$ .

Facts & Assumptions

Given: Complex numbers $α_{1}, \dots, α_{d}$ and the associated power sums $p_{n}$ and elementary symmetric functions $e_{n}$ .

[L1]

The power sums attached to slice zeros vary holomorphically with the parameter (The power sums of the slice zeros vary holomorphically).

[A1]

Put $Q (ζ) : = \prod_{j = 1}^{d} (ζ - α_{j}) = ζ^{d} + c_{1} ζ^{d - 1} + \dots + c_{d},$ so $c_{n} = (- 1)^{n} e_{n}$ .

Proof

technique · direct

1.1A1algebra

By logarithmic differentiation of the polynomial in [A1], $\frac{Q^{'} (ζ)}{Q (ζ)} = \sum_{j = 1}^{d} \frac{1}{ζ - α_{j}} .$ Multiplying by $ζ$ and expanding each summand for large $ζ$ gives the formal Laurent identity $ζ \frac{Q^{'} (ζ)}{Q (ζ)} = d + \sum_{n \geq 1} p_{n} ζ^{- n} .$

2.1step 1.1A1algebra

Multiplying the identity from step 1.1 by $Q (ζ) = \sum_{n = 0}^{d} c_{n} ζ^{d - n}$ with $c_{0} = 1$ and comparing the coefficient of $ζ^{d - n}$ for $1 \leq n \leq d$ yields $n c_{n} + \sum_{j = 1}^{n} c_{n - j} p_{j} = 0.$ Substituting $c_{n} = (- 1)^{n} e_{n}$ gives $n e_{n} = \sum_{j = 1}^{n} (- 1)^{j - 1} e_{n - j} p_{j} .$ Since $n \neq 0$ in $C$ , this determines $e_{n}$ recursively as a polynomial in $p_{1}, \dots, p_{n}$ .

3.1step 2.1L1algebra∎

The displayed recursion uses only addition, multiplication, and division by the nonzero scalar $n$ . Therefore if the power sums $p_{j}$ vary holomorphically, then so do the $e_{n}$ ; the slice-zero case mentioned in [L1] is exactly such a holomorphic family.

Depends on

Used by

Weierstrass preparation theorem Theorem

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jiří Lebl, Tasty Bits of Several Complex Variables, Exercise 6.2.1 (standard reference, not scraped)
Jaap Korevaar and Jan Wiegerinck, Several Complex Variables, Section 4.4 (standard reference, not scraped)