proof of Newton-Girard formula for symmetric polynomials


The following is a proof of Newton-Girard formula using formal power series. Let z be an indeterminateMathworldPlanetmath and f⁢(z) be the polynomialMathworldPlanetmathPlanetmath

1-E1⁢z+…+(-1)n⁢En⁢zn.

Take log and differentiate both sides of the equation

f⁢(z)=∏i=1n(1-xi⁢z).

We obtain

f′⁢(z)/f⁢(z)=∑i=1n-xi1-xi⁢z, (1)

where f′⁢(z) is the derivative of f⁢(z)

f′⁢(z)=-E1+2⁢E2⁢z-…+(-1)n⁢n⁢En⁢zn-1.

The right hand side of (1) is equal to

-∑i=1n∑k=0∞xik+1⁢zk=-∑k=0∞Sk+1⁢zk.

By equating coefficients of

f′⁢(z)=-f⁢(z)⁢(S1+S2⁢z+S3⁢z2+…)

we get the Newton-Girard formula.

Title proof of Newton-Girard formula for symmetric polynomials
Canonical name ProofOfNewtonGirardFormulaForSymmetricPolynomials
Date of creation 2013-03-22 15:34:37
Last modified on 2013-03-22 15:34:37
Owner kshum (5987)
Last modified by kshum (5987)
Numerical id 4
Author kshum (5987)
Entry type Proof
Classification msc 11C08