# Newton's identities

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In mathematics, Newton's identities, also known as the Newton–Girard formulae, give relations between two types of symmetric polynomials, namely between power sums and elementary symmetric polynomials. Evaluated at the roots of a monic polynomial P in one variable, they allow expressing the sums of the k-th powers of all roots of P (counted with their multiplicity) in terms of the coefficients of P, without actually finding those roots. These identities were found by Isaac Newton around 1666, apparently in ignorance of earlier work (1629) by Albert Girard. They have applications in many areas of mathematics, including Galois theory, invariant theory, group theory, combinatorics, as well as further applications outside mathematics, including general relativity.

## Mathematical statement

### Formulation in terms of symmetric polynomials

Let x1,…, xn be variables, denote for k ≥ 1 by pk(x1,…,xn) the k-th power sum:

$p_k(x_1,\ldots,x_n)=\sum\nolimits_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,$

and for k ≥ 0 denote by ek(x1,…,xn) the elementary symmetric polynomial that is the sum of all distinct products of k distinct variables, so in particular

\begin{align} e_0(x_1,\ldots,x_n) &= 1,\\ e_1(x_1,\ldots,x_n) &= x_1+x_2+\cdots+x_n,\\ e_2(x_1,\ldots,x_n) &= \textstyle\sum_{1 \leq in.\\ \end{align}

Then Newton's identities can be stated as

$ke_k(x_1,\ldots,x_n) = \sum_{i=1}^k(-1)^{i-1} e_{k-i} (x_1,\ldots,x_n) p_i(x_1,\ldots,x_n),$

valid for all n ≥ k ≥ 1. Concretely, one gets for the first few values of k:

\begin{align} e_1(x_1,\ldots,x_n) &= p_1(x_1,\ldots,x_n),\\ 2e_2(x_1,\ldots,x_n) &= e_1(x_1,\ldots,x_n)p_1(x_1,\ldots,x_n)-p_2(x_1,\ldots,x_n),\\ 3e_3(x_1,\ldots,x_n) &= e_2(x_1,\ldots,x_n)p_1(x_1,\ldots,x_n) - e_1(x_1,\ldots,x_n)p_2(x_1,\ldots,x_n) + p_3(x_1,\ldots,x_n).\\ \end{align}

The form and validity of these equations do not depend on the number n of variables (although the point where the left-hand side becomes 0 does, namely after the n-th identity), which makes it possible to state them as identities in the ring of symmetric functions. In that ring one has

\begin{align} e_1 &= p_1,\\ 2e_2 &= e_1p_1-p_2,\\ 3e_3 &= e_2p_1 - e_1p_2 + p_3,\\ 4e_4 &= e_3p_1 - e_2p_2 + e_1p_3 - p_4,\\ \end{align}

and so on; here the left-hand sides never become zero. These equations allow to recursively express the ei in terms of the pk; to be able to do the inverse, one may rewrite them as

\begin{align} p_1 &= e_1,\\ p_2 &= e_1p_1-2e_2,\\ p_3 &= e_1p_2 - e_2p_1 + 3e_3 ,\\ p_4 &= e_1p_3 - e_2p_2 + e_3p_1 - 4e_4, \\ & {}\ \ \vdots \end{align}

### Application to the roots of a polynomial

Now view the xi as parameters rather than as variables, and consider the monic polynomial in t with roots x1,…,xn:

$\prod_{i=1}^n \left( t - x_i \right) = \sum_{k=0}^n (-1)^{k} a_k t^{n-k},$

where the coefficients $a_k$ are given by the elementary symmetric polynomials in the roots: ak = ek(x1,…,xn). Now consider the power sums of the roots

$s_k = p_k(x_1,\ldots,x_n)= \sum_{i=1}^n x_i^k.$

Then according to Newton's identities these can be expressed recursively in terms of the coefficients of the polynomial using

\begin{align} s_1 &= a_1,\\ s_2 &= a_1 s_1 - 2 a_2,\\ s_3 &= a_1 s_2 - a_2 s_1 + 3 a_3,\\ s_4 &= a_1 s_3 - a_2 s_2 + a_3 s_1 - 4 a_4, \\ & {} \ \ \vdots \end{align}

### Application to the characteristic polynomial of a matrix

When the polynomial above is the characteristic polynomial of a matrix A (in particular when A is the companion matrix of the polynomial), the roots $x_i$ are the eigenvalues of the matrix, counted with their algebraic multiplicity. For any positive integer k, the matrix Ak has as eigenvalues the powers xik, and each eigenvalue $x_i$ of A contributes its multiplicity to that of the eigenvalue xik of Ak. Then the coefficients of the characteristic polynomial of Ak are given by the elementary symmetric polynomials in those powers xik. In particular, the sum of the xik, which is the k-th power sum sk of the roots of the characteristic polynomial of A, is given by its trace:

$s_k = \operatorname{tr} ( A^k )\,.$

The Newton identities now relate the traces of the powers Ak to the coefficients of the characteristic polynomial of A. Using them in reverse to express the elementary symmetric polynomials in terms of the power sums, they can be used to find the characteristic polynomial by computing only the powers Ak and their traces.

This computation requires computing the traces of matrix powers Ak and solving a triangular system of equations. Both can be done in complexity class NC (solving a triangular system can be done by divide-and-conquer). Therefore, characteristic polynomial of a matrix can be computed in NC. By the Cayley-Hamilton theorem, every matrix satisfies its characteristic polynomial, and a simple transformation allows to find the matrix inverse in NC.

Rearranging the computations into an efficient form leads to the Fadeev-Leverrier algorithm (1840), a fast parallel implementation of it is due to L. Csanky (1976). Its disadvantage is that it requires division by integers, so in general the field should have characteristic 0.

### Relation with Galois theory

For a given n, the elementary symmetric polynomials ek(x1,…,xn) for k = 1,…, n form an algebraic basis for the space of symmetric polynomials in x1,…. xn: every polynomial expression in the xi that is invariant under all permutations of those variables is given by a polynomial expression in those elementary symmetric polynomials, and this expression is unique up to equivalence of polynomial expressions. This is a general fact known as the fundamental theorem of symmetric polynomials, and Newton's identities provide explicit formulae in the case of power sum symmetric polynomials. Applied to the monic polynomial $\textstyle t^n+\sum_{k=1}^n (-1)^{k} a_k t^{n-k}$ with all coefficients ak considered as free parameters, this means that every symmetric polynomial expression S(x1,…,xn) in its roots can be expressed instead as a polynomial expression P(a1,…,an) in terms of its coefficients only, in other words without requiring knowledge of the roots. This fact also follows from general considerations in Galois theory (one views the ak as elements of a base field, the roots live in an extension field whose Galois group permutes them according to the full symmetric group, and the field fixed under all elements of the Galois group is the base field).

The Newton identities also permit expressing the elementary symmetric polynomials in terms of the power sum symmetric polynomials, showing that any symmetric polynomial can also be expressed in the power sums. In fact the first n power sums also form an algebraic basis for the space of symmetric polynomials.

## Related identities

There are a number of (families of) identities that, while they should be distinguished from Newton's identities, are very closely related to them.

### A variant using complete homogeneous symmetric polynomials

Denoting by hk the complete homogeneous symmetric polynomial that is the sum of all monomials of degree k, the power sum polynomials also satisfy identities similar to Newton's identities, but not involving any minus signs. Expressed as identities of in the ring of symmetric functions, they read

$kh_k=\sum_{i=1}^kh_{k-i}p_i,$

valid for all n ≥ k ≥ 1. Contrary to Newton's identities, the left-hand sides do not become zero for large k, and the right-hand sides contain ever more non-zero terms. For the first few values of k, one has

\begin{align} h_1 &= p_1,\\ 2h_2 &= h_1p_1+p_2,\\ 3h_3 &= h_2p_1 + h_1p_2 + p_3.\\ \end{align}

These relations can be justified by an argument analogous to the one by comparing coefficients in power series given above, based in this case on the generating function identity

$\sum_{k=0}^\infty h_k(X_1,\ldots,X_n)t^k = \prod_{i=1}^n\frac1{1-X_it}.$

Proofs of Newton's identities, like these given below, cannot be easily adapted to prove these variants of those identities.

### Expressing elementary symmetric polynomials in terms of power sums

As mentioned, Newton's identities can be used to recursively express elementary symmetric polynomials in terms of power sums. Doing so requires the introduction of integer denominators, so it can be done in the ring ΛQ of symmetric functions with rational coefficients:

\begin{alignat}2 e_1 &= p_1,\\ e_2 &= \textstyle\frac12p_1^2 - \frac12p_2 &&= \textstyle\frac12 ( p_1^2 - p_2 ),\\ e_3 &= \textstyle\frac16p_1^3 - \frac12p_1 p_2 + \frac13p_3 &&= \textstyle\frac{1}{6} ( p_1^3 - 3 p_1 p_2 + 2 p_3 ),\\ e_4 &= \textstyle\frac1{24}p_1^4 - \frac14p_1^2 p_2 + \frac18p_2^2 + \frac13p_1 p_3 - \frac14p_4 &&= \textstyle\frac1{24} ( p_1^4 - 6 p_1^2 p_2 + 3 p_2^2 + 8 p_1 p_3 - 6 p_4 ),\\ \end{alignat}

and so forth. Applied to a monic polynomial, these formulae express the coefficients in terms of the power sums of the roots: replace each ei by ai and each pk by sk.

### Expressing complete homogeneous symmetric polynomials in terms of power sums

The analogous relations involving complete homogeneous symmetric polynomials can be similarly developed, giving equations

\begin{alignat}2 h_1 &= p_1,\\ h_2 &= \textstyle\frac12p_1^2 + \frac12p_2 &&= \textstyle\frac12 ( p_1^2 + p_2 ),\\ h_3 &= \textstyle\frac16p_1^3 + \frac12p_1 p_2 + \frac13p_3 &&= \textstyle\frac{1}{6} ( p_1^3 + 3 p_1 p_2 + 2 p_3 ),\\ h_4 &= \textstyle\frac1{24}p_1^4 + \frac14p_1^2 p_2 + \frac18p_2^2 + \frac13p_1 p_3 + \frac14p_4 &&= \textstyle\frac1{24} ( p_1^4 + 6 p_1^2 p_2 + 3 p_2^2 + 8 p_1 p_3 + 6 p_4 ),\\ h_n &= \sum_{m_1+2m_2+\cdots+nm_n=n} \prod_{i=1}^n \frac{p_i^{m_i}}{m_i ! i^{m_i}}\\ \end{alignat}

and so forth, in which there are only plus signs. These expressions correspond exactly to the cycle index polynomials of the symmetric groups, if one interprets the power sums pi as indeterminates: the coefficient in the expression for hk of any monomial p1m1p2m2\dotsplml is equal to the fraction of all permutations of k that have m1 fixed points, m2 cycles of length 2, …, and ml cycles of length l. Explicitly, this coefficient can be written as $1/N$ where $N=\Pi_{i=1}^l(m_i!\,i^{m_i})$; this N is the number permutations commuting with any given permutation π of the given cycle type. The expressions for the elementary symmetric functions have coefficients with the same absolute value, but a sign equal to the sign of π, namely (−1)m2+m4+\dots.

It can be proved by：

$nf(m_1,...,m_n)=f(m_1-1,...,m_n)+...+f(m_1,...,m_n-1)$
$m_1\prod_{i=1}^n \frac{1}{i^{m_i} m_i!}+...+nm_n \prod_{i=1}^n \frac{1}{i^{m_i} m_i!}=n\prod_{i=1}^n \frac{1}{i^{m_i} m_i!}$

### Expressing power sums in terms of elementary symmetric polynomials

One may also use Newton's identities to express power sums in terms of symmetric polynomials, which does not introduce denominators:

\begin{align} p_1 &= e_1,\\ p_2 &= e_1^2 - 2 e_2,\\ p_3 &= e_1^3 - 3 e_2 e_1 + 3 e_3,\\ p_4 &= e_1^4 - 4 e_2 e_1^2 + 4 e_3 e_1 + 2 e_2^2 - 4 e_4,\\ p_5 &= e_1^5 - 5 e_2 e_1^3 + 5 e_3 e_1^2 + 5 e_2^2 e_1 - 5 e_4 e_1 - 5 e_3e_2 + 5 e_5,\\ p_6 &= e_1^6 - 6 e_2 e_1^4 + 6 e_3 e_1^3 + 9 e_2^2 e_1^2 - 6 e_4 e_1^2 - 12 e_3 e_2 e_1 + 6 e_5 e_1 - 2 e_2^3 + 3 e_3^2 + 6 e_4 e_2 - 6e_6,\\ p_m &= \sum_{r_i=0}^{\lfloor \frac{m}{i} \rfloor} \frac{m(r_1+r_2+\cdots+r_n -1)!}{r_1!r_2!\cdots r_n!} \prod_{i=1}^n (-e_i)^{r_i}\\ \end{align}

which can be proved by：

\begin{align}f(m,r_1,\cdots,r_n)&=f(m-1,r_1-1,\cdots,r_n)+\cdots+f(m-n,r_1,\cdots,r_n-1)\\ &=\frac{(m-1)(r_1+\cdots+r_n-2)!}{(r_1-1)!\cdots r_n!}+\cdots+\frac{(m-n)(r_1+\cdots+r_n-2)!}{r_1!\cdots(r_n-1)!}\\ &=\frac{[r_1(m-1)+\cdots+r_n(m-n)](r_1+\cdots+r_n-2)!}{r_1!\cdots r_n!}\\ &=\frac{[m(r_1+\cdots+r_n)-m](r_1+\cdots+r_n-2)!}{r_1!\cdots r_n!}\\ &=\frac{m(r_1+\cdots+r_n-1)!}{r_1!\cdots r_n!}\end{align}

### Expressing power sums in terms of complete homogeneous symmetric polynomials

Finally one may use the variant identities involving complete homogeneous symmetric polynomials similarly to express power sums in term of them:

\begin{align} p_1 &= + h_1,\\ p_2 &= - h_1^2 + 2 h_2,\\ p_3 &= + h_1^3 - 3 h_2 h_1 + 3 h_3,\\ p_4 &= - h_1^4 + 4 h_2 h_1^2 - 4 h_3 h_1 - 2 h_2^2 + 4 h_4,\\ p_5 &= + h_1^5 - 5 h_2 h_1^3 + 5 h_2^2 h_1 + 5 h_3 h_1^2 - 5 h_3h_2 - 5 h_4 h_1 + 5 h_5,\\ p_6 &= - h_1^6 + 6 h_2 h_1^4 - 9 h_2^2 h_1^2 - 6 h_3 h_1^3 + 2 h_2^3 + 12 h_3 h_2 h_1 + 6 h_4 h_1^2 - 3 h_3^2 - 6 h_4 h_2 - 6 h_1 h_5 + 6h_6,\\ \end{align}

and so on. Apart from the replacement of each ei by the corresponding hi, the only change with respect to the previous family of identities is in the signs of the terms, which in this case depend just on the number of factors present: the sign of the monomial $\Pi_{i=1}^l h_i^{m_i}$ is −(−1)m1+m2+m3+…. In particular the above description of the absolute value of the coefficients applies here as well.

### Expressions as determinants

One can obtain explicit formulas for the above expressions in the form of determinants, by considering the first n of Newton's identities (or it counterparts for the complete homogeneous polynomials) as linear equations in which the elementary symmetric functions are known and the power sums are unknowns (or vice versa), and apply Cramer's rule to find the solution for the final unknown. For instance taking Newton's identities in the form

\begin{align} e_1 &= 1p_1,\\ 2e_2 &= e_1p_1-1p_2,\\ 3e_3 &= e_2p_1 - e_1p_2 + 1p_3,\\ \vdots &= \vdots \\ ne_n &= e_{n-1}p_1 - e_{n-2} p_2 + \cdots +(-1)^ne_1p_{n-1}+(-1)^{n-1}p_n\\ \end{align}

we consider $p_1$, ${-p_2}$, $p_3$, ..., $(-1)^np_{n-1}$ and $p_n$ as unknowns, and solve for the final one, giving

$p_n=\frac{\begin{vmatrix}1 & 0 & \cdots && e_1 \\ e_1 & 1 & 0 & \cdots & 2e_2 \\ e_2 & e_1 & 1& & 3e_3 \\ \vdots&&\ddots&\ddots&\vdots \\ e_{n-1} & \cdots & e_2 & e_1 & ne_n \end{vmatrix}}{\begin{vmatrix}1 & 0 & \cdots & \\ e_1 & 1 & 0 & \cdots \\ e_2 & e_1 & 1& \\ \vdots&&\ddots&\ddots \\ e_{n-1} & \cdots & e_2 & e_1 & (-1)^{n-1} \end{vmatrix}} =\frac{\begin{vmatrix}1 & 0 & \cdots && e_1 \\ e_1 & 1 & 0 & \cdots & 2e_2 \\ e_2 & e_1 & 1& & 3e_3 \\ \vdots&&\ddots&\ddots&\vdots \\ e_{n-1} & \cdots & e_2 & e_1 & ne_n \end{vmatrix}}{(-1)^{n-1}} =\begin{vmatrix}e_1 & 1 & 0 & \cdots\\ 2e_2 & e_1 & 1 & 0 & \cdots\\ 3e_3 & e_2 & e_1 & 1 \\ \vdots &&& \ddots & \ddots \\ ne_n & e_{n-1} & \cdots & & e_1 \end{vmatrix}.$

Solving for $e_n$ instead of for $p_n$ is similar, as the analogous computations for the complete homogeneous symmetric polynomials; in each case the details are slightly messier than the final results, which are (Macdonald 1979, p. 20):

$e_n=\frac1{n!} \begin{vmatrix}p_1 & 1 & 0 & \cdots\\ p_2 & p_1 & 2 & 0 & \cdots \\ \vdots&& \ddots & \ddots \\ p_{n-1} & p_{n-2} & \cdots & p_1 & n-1 \\ p_n & p_{n-1} & \cdots & p_2 & p_1 \end{vmatrix}, \qquad p_n=(-1)^{n-1} \begin{vmatrix}h_1 & 1 & 0 & \cdots\\ 2h_2 & h_1 & 1 & 0 & \cdots\\ 3h_3 & h_2 & h_1 & 1 \\ \vdots &&& \ddots & \ddots \\ nh_n & h_{n-1} & \cdots & & h_1 \end{vmatrix}, \qquad h_n=\frac1{n!} \begin{vmatrix}p_1 & -1 & 0 & \cdots\\ p_2 & p_1 & -2 & 0 & \cdots \\ \vdots&& \ddots & \ddots \\ p_{n-1} & p_{n-2} & \cdots & p_1 & 1-n \\ p_n & p_{n-1} & \cdots & p_2 & p_1 \end{vmatrix}.$

Note that the use of determinants makes that the formula for $h_n$ has additional minus signs compared to the one for $e_n$, while the situation for the expanded form given earlier is opposite. As remarked in (Littlewood 1950, p. 84) one can alternatively obtain the formula for $h_n$ by taking the permanent of the matrix for $e_n$ instead of the determinant, and more generally an expression for any Schur polynomial can be obtained by taking the corresponding immanant of this matrix.

## Derivation of the identities

Each of Newton's identities can easily be checked by elementary algebra; however, their validity in general needs a proof. Here are some possible derivations

### From the special case n = k

One can obtain the k-th Newton identity in k variables by substitution into

$\prod_{i=1}^k (t - x_i) = \sum_{i=0}^k (-1)^{k-i} e_{k-i}(x_1,\ldots,x_k)t^i$

as follows. Substituting xj for t gives

$0= \sum_{i=0}^k (-1)^{k-i} e_{k-i}(x_1,\ldots,x_k){x_j}^i \quad\text{for }1\leq j\leq k$

Summing over all j gives

$0= (-1)^kke_k(x_1,\ldots,x_k)+\sum_{i=1}^k(-1)^{k-i} e_{k-i}(x_1,\ldots,x_k)p_i(x_1,\ldots,x_k),$

where the terms for i = 0 were taken out of the sum because p0 is (usually) not defined. This equation immediately gives the k-th Newton identity in k variables. Since this is an identity of symmetric polynomials (homogeneous) of degree k, its validity for any number of variables follows from its validity for k variables. Concretely, the identities in n < k variables can be deduced by setting k − n variables to zero. The k-th Newton identity in n > k variables contains more terms on both sides of the equation than the one in k variables, but its validity will be assured if the coefficients of any monomial match. Because no individual monomial involves more than k of the variables, the monomial will survive the substitution of zero for some set of n − k (other) variables, after which the equality of coefficients is one that arises in the k-th Newton identity in k (suitably chosen) variables.

### Comparing coefficients in series

Another derivation can be obtained by computations in the ring of formal power series R[[t]], where R is Z[x1,…, xn], the ring of polynomials in n variables x1,…, xn over the integers.

Starting again from the basic relation

$\prod_{i=1}^n (t - x_i) = \sum_{k=0}^n (-1)^{k} a_k t^{n-k}$

and "reversing the polynomials" by substituting 1/t for t and then multiplying both sides by tn to remove negative powers of t, gives

$\prod_{i=1}^n (1- x_it) = \sum_{k=0}^n (-1)^{k} a_k t^k.$

(the above computation should be performed in the field of fractions of R[[t]]; alternatively, the identity can be obtained simply by evaluating the product on the left side)

Swapping sides and expressing the ai as the elementary symmetric polynomials they stand for gives the identity

$\sum_{k=0}^n (-1)^{k} e_k(x_1,\ldots,x_n) t^k=\prod_{i=1}^n (1- x_it).$

One formally differentiates both sides with respect to t, and then (for convenience) multiplies by t, to obtain

\begin{align} \sum_{k=0}^n (-1)^{k}k e_k(x_1,\ldots,x_n) t^k &= t \sum_{i=1}^n \left((-x_i) \prod\nolimits_{j\neq i} (1- x_jt)\right)\\ &= -\left(\sum_{i=1}^n \frac{x_it}{1-x_it}\right) \prod\nolimits_{j=1}^n (1- x_jt)\\ &= -\left(\sum_{i=1}^n \sum_{j=1}^\infty(x_it)^j\right) \left(\sum_{\ell=0}^n (-1)^\ell e_\ell(x_1,\ldots,x_n) t^\ell\right)\\ &= \left(\sum_{j=1}^\infty p_j(x_1,\ldots,x_n)t^j\right) \left(\sum_{\ell=0}^n (-1)^{\ell-1} e_\ell(x_1,\ldots,x_n) t^\ell\right),\\ \end{align}

where the polynomial on the right hand side was first rewritten as a rational function in order to be able to factor out a product out of the summation, then the fraction in the summand was developed as a series in t, using the formula

$\frac{X}{1 - X} = X + X^2 + X^3 + X^4 + X^5 + \cdots$,

and finally the coefficient of each t j was collected, giving a power sum. (The series in t is a formal power series, but may alternatively be thought of as a series expansion for t sufficiently close to 0, for those more comfortable with that; in fact one is not interested in the function here, but only in the coefficients of the series.) Comparing coefficients of tk on both sides one obtains

$(-1)^{k}k e_k(x_1,\ldots,x_n) = \sum_{j=1}^k (-1)^{k-j-1} p_j(x_1,\ldots,x_n)e_{k-j}(x_1,\ldots,x_n),$

which gives the k-th Newton identity.

### As a telescopic sum of symmetric function identities

The following derivation, given essentially in (Mead, 1992), is formulated in the ring of symmetric functions for clarity (all identities are independent of the number of variables). Fix some k > 0, and define the symmetric function r(i) for 2 ≤ i ≤ k as the sum of all distinct monomials of degree k obtained by multiplying one variable raised to the power i with k − i distinct other variables (this is the monomial symmetric function mγ where γ is a hook shape (i,1,1,…1)). In particular r(k) = pk; for r(1) the description would amount to that of ek, but this case was excluded since here monomials no longer have any distinguished variable. All products pieki can be expressed in terms of the r(j) with the first and last case being somewhat special. One has

$p_ie_{k-i}=r(i)+r(i+1)\quad\text{for }1

since each product of terms on the left involving distinct variables contributes to r(i), while those where the variable from pi already occurs among the variables of the term from eki contributes to r(i + 1), and all terms on the right are so obtained exactly once. For i = k one multiplies by e0 = 1, giving trivially

$p_ke_0=p_k=r(k)\,$.

Finally the product p1ek−1 for i = 1 gives contributions to r(i + 1) = r(2) like for other values i < k, but the remaining contributions produce k times each monomial of ek, since any one of the variables may come from the factor p1; thus

$p_1e_{k-1}=ke_k+r(2)\,$.

The k-th Newton identity is now obtained by taking the alternating sum of these equations, in which all terms of the form r(i) cancel out.