TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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Newton's identities: $k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i}$

Statement

Put $e_{0} = 1$ and $e_{k} = 0$ for $k > n$ . For every $k \geq 1$ ,

$k e_{k} = \sum_{i = 1}^{k} (- 1)^{i - 1} e_{k - i} p_{i} .$

In particular, for $k \leq n$ this recursively relates $e_{k}$ to $p_{1}, \dots, p_{k}$ , while for $k > n$ it gives

$p_{k} - e_{1} p_{k - 1} + \dots + (- 1)^{n} e_{n} p_{k - n} = 0.$

No division is used, so the identities hold over every commutative ring.

Facts & Assumptions

Given: A commutative ring $R$ and variables $x_{1}, \dots, x_{n}$ .

[L1]

The power sum is $p_{i} = \sum_{j} x_{j}^{i}$ , and $H (t) = \sum_{r \geq 0} h_{r} t^{r}$ (Power sums $p_{k}$ and complete homogeneous symmetric polynomials $h_{k}$ ).

[L2]

The formal-series identity is $E (- t) H (t) = 1$ , where $E (- t) = \prod_{j} (1 - x_{j} t)$ (The generating-series identity $E (- t) H (t) = 1$ ).

[L3]

The formal derivative of a polynomial $\sum_{r} a_{r} t^{r}$ is $\sum_{r \geq 1} r a_{r} t^{r - 1}$ (The formal derivative of a polynomial).

[L4]

Over a commutative ring, formal differentiation of polynomials is additive and satisfies $(f g)^{'} = f^{'} g + f g^{'}$ (Linearity, power rule, Leibniz rule and the degree bound for the formal derivative).

Proof

technique · direct

1.1L2L3L4algebra

Write $G (t) : = E (- t) = \prod_{j = 1}^{n} (1 - x_{j} t)$ , which by [L2] is a polynomial in $t$ of degree at most $n$ over $R [x_{1}, \dots, x_{n}]$ , so [L3] and [L4] apply to it. Iterating the Leibniz rule of [L4] over the $n$ factors, and using $(1 - x_{j} t)^{'} = - x_{j}$ from [L3], gives $G^{'} (t) = - \sum_{j} x_{j} \prod_{ℓ \neq j} (1 - x_{ℓ} t)$ .

2.1step 1.1L1L2algebra

Multiply by the power series $H (t) = G (t)^{- 1}$ from [L2]. Then $- G^{'} (t) H (t) = \sum_{j} x_{j} / (1 - x_{j} t) = \sum_{i \geq 1} p_{i} t^{i - 1}$ , where the last equality is coefficientwise geometric expansion.

3.1step 2.1L2L3algebra

Multiply step 2.1 by $G (t)$ and use $G (t) H (t) = 1$ from [L2]; no derivative of the infinite series $H$ is taken. This gives $- G^{'} (t) = G (t) \sum_{i \geq 1} p_{i} t^{i - 1}$ . Since $G (t) = \sum_{i = 0}^{n} (- 1)^{i} e_{i} t^{i}$ , [L3] evaluates the left side as $\sum_{k \geq 1} (- 1)^{k - 1} k e_{k} t^{k - 1}$ . Comparing the coefficient of $t^{k - 1}$ yields $(- 1)^{k - 1} k e_{k} = \sum_{i = 1}^{k} (- 1)^{k - i} e_{k - i} p_{i}$ .

4.1step 3.1algebra∎

Multiplying the identity in step 3.1 by $(- 1)^{k - 1}$ proves the displayed Newton identity. When $k > n$ , the term $k e_{k}$ is zero and reindexing gives the stated recurrence for $p_{k}$ .

Depends on

Used by

Dependency tree · two levels

9 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

K. Conrad, Symmetric Polynomials, Section 3 (standard reference, not scraped)
D. Grinberg, An Introduction to Algebraic Combinatorics, Chapter 7, Section 7.1 (standard reference, not scraped)

Newton's identities: kek=∑i=1k(−1)i−1ek−ipikek​=∑i=1k​(−1)i−1ek−i​pi​