Newton’s Identities

Summary: §165–§166: the recurrence that converts the elementary symmetric coefficients of a polynomial (or a “polynomial of infinite degree”, i.e. a power series) into the power sums of its reciprocal roots. The pivotal computational engine of Chapter 10.

Sources: chapter10

Last updated: 2026-04-30

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Setup

Suppose

$1+Az+B{z}^2+C{z}^3+D{z}^4+\cdots =(1+\alpha z)(1+\beta z)(1+\gamma z)(1+\delta z)\cdots$

— with finitely or infinitely many factors. Multiplying out and matching powers of $z$ identifies $A,B,C,\dots$ as the elementary symmetric polynomials in $\alpha ,\beta ,\gamma ,\dots$:

$A=\alpha +\beta +\gamma +\delta +\cdots ,$

$B=\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta +\cdots ,$

$C=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta +\cdots ,$

$D=\alpha \beta \gamma \delta +\cdots ,\text{etc.}$

(source: chapter10, §165). Euler describes these as “products taken one at a time, two at a time, three at a time, …”

Power sums

Define

$P=\alpha +\beta +\gamma +\delta +\cdots ,Q={\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2+\cdots ,$

$R={\alpha }^3+{\beta }^3+{\gamma }^3+{\delta }^3+\cdots ,S={\alpha }^4+\cdots ,T={\alpha }^5+\cdots ,V={\alpha }^6+\cdots .$

These are the power sums of the roots $\alpha ,\beta ,\gamma ,\dots$ (which, in the parametrization above, are minus the reciprocals of the roots of the polynomial $1+Az+B{z}^2+\cdots$).

The recurrence

$\boxed{\begin{array}{rl}P& =A,\\ Q& =AP-2B,\\ R& =AQ-BP+3C,\\ S& =AR-BQ+CP-4D,\\ T& =AS-BR+CQ-DP+5E,\\ V& =AT-BS+CR-DQ+EP-6F,\\ & ⋮\end{array}}$

(source: chapter10, §166). Each line uses all previously-known $A,B,C,\dots$ and $P,Q,R,\dots$, so the sequence $P,Q,R,S,\dots$ can be computed mechanically once the elementary symmetrics $A,B,C,\dots$ are known.

The general pattern: the $n$-th line is

$P_n=A{P}_{n-1}-B{P}_{n-2}+C{P}_{n-3}-D{P}_{n-4}+\cdots +(-1{)}^{n+1}n(\text{coefficient of }{z}^n).$

Euler comments: “The truth of these formulas is intuitively clear, but a rigorous proof will be given in the differential calculus” (source: chapter10, §166).

Why it works

The squared sum identity $(\alpha +\beta +\gamma +\cdots {)}^2={\alpha }^2+{\beta }^2+\cdots +2(\alpha \beta +\alpha \gamma +\cdots )$ is the simplest case: $P^2=Q+2B$, hence $Q=P^2-2B=AP-2B$.

The general case follows from the same accounting: when one expands $P\cdot P_{n-1}$, one obtains the genuine $n$-th power sum $P_n$ together with all the “off-diagonal” products, which are exactly $B{P}_{n-2}$, $C{P}_{n-3}$, etc., with alternating signs from the elementary symmetric structure. A modern proof uses logarithmic differentiation: take $\log$ of the product side, differentiate, and read off the coefficient identity.

Use in Chapter 10

Euler’s strategy: take a transcendental function whose infinite product expansion is known (so the elementary symmetrics $A,B,C,\dots$ of its zeros are read off the power-series coefficients) and apply the recurrence to obtain the power sums of the reciprocals of the zeros.

The first application: let

$\frac{e^x-e^{-x}}{2x}=1+\frac{x^2}{1\cdot 2\cdot 3}+\frac{x^4}{1\cdot 2\cdot 3\cdot 4\cdot 5}+\cdots ={\prod }_{k=1}^{\infty }\left(1+\frac{x^2}{k^2{\pi }^2}\right).$

Substitute $x^2={\pi }^{2}z$. The roots in the new variable are $z=-1,-1/4,-1/9,\dots$, so the parametrization $1+Az+B{z}^2+\cdots =\prod (1+{\alpha }_{k}z)$ holds with ${\alpha }_k=1/k^2$. Hence:

• $A={\pi }^2/6$, $B={\pi }^4/120$, $C={\pi }^6/5040$, $D={\pi }^8/362880$, $\dots$ (read off from the power series).
• $P=\sum 1/k^2=A={\pi }^2/6$. → the Basel sum.
• $Q=\sum 1/k^4=AP-2B={\pi }^4/90$.
• $R=\sum 1/k^6={\pi }^6/945$.
• $S=\sum 1/k^8={\pi }^8/9450$, etc.

See zeta-at-even-integers for the table extended through $\zeta (26)$.

The same recurrence drives every subsequent computation in the chapter — from $\sum 1/(2k+1{)}^{2n}$ via the cosine-infinite-product (§169) to the character sums of §171–§180 and the csc partial fractions of §181–§183.

Modern footnote

These are the Newton–Girard formulas, also called Newton’s identities: a basic tool in the theory of symmetric functions and the bridge between elementary symmetric polynomials $e_n$ and power sum polynomials $p_n$. The variant Euler uses, with signs alternating because the parametrization is $\prod (1+\alpha z)$ rather than $\prod (z-\alpha )$, is one of two standard forms.

In modern notation, with $E(z)=\prod (1+{\alpha }_{k}z)$ and $\log E(z)={\sum }_k\log (1+{\alpha }_{k}z)={\sum }_k{\sum }_{n\ge 1}(-1{)}^{n+1}{\alpha }_k^n{z}^n/n$:

$\frac{E^{\prime }(z)}{E(z)}={\sum }_{n\ge 1}(-1{)}^{n+1}{P}_n{z}^{n-1}.$

Multiplying out $E^{\prime }(z)=E(z)\cdot E^{\prime }(z)/E(z)$ gives the Newton recurrence directly. Euler’s identity-by-identity verification and the modern logarithmic-derivative proof are the same calculation, just written differently.

Related pages

• basel-problem
• zeta-at-even-integers
• odd-and-alternating-zeta-decomposition
• circular-arc-series
• cotangent-partial-fraction
• exponential-infinite-product
• sine-infinite-product
• cosine-infinite-product
• chapter-10-on-the-use-of-the-discovered-factors-to-sum-infinite-series