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 indeterminate and f(z) be the
polynomial
1-E1z+…+(-1)nEnzn. |
Take log and differentiate both sides of the equation
f(z)=n∏i=1(1-xiz). |
We obtain
f′(z)/f(z)=n∑i=1-xi1-xiz, | (1) |
where f′(z) is the derivative of f(z)
f′(z)=-E1+2E2z-…+(-1)nnEnzn-1. |
The right hand side of (1) is equal to
-n∑i=1∞∑k=0xk+1izk=-∞∑k=0Sk+1zk. |
By equating coefficients of
f′(z)=-f(z)(S1+S2z+S3z2+…) |
we get the Newton-Girard formula.
Title | proof of Newton-Girard formula for symmetric polynomials |
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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 |