Viète's formulas

:"For Viète's formula for computing π, see that article."In mathematics, more specifically in algebra, Viète's formulas, named after François Viète, are formulas which relate the coefficients of a polynomial to signed sums and products of its roots.

The formulas

Any general polynomial of degree $nge\; 1$

:$p(X)=a\_nX^n\; +\; a\_\{n-1\}X^\{n-1\}\; +cdots\; +\; a\_1\; X+\; a\_0$

(with the coefficients being real or complex numbers and $a\_n\; e\; 0$) is known by the fundamental theorem of algebra to have $n$ (not necessarily distinct) complex roots $x\_1,\; x\_2,\; dots,\; x\_n.$

Viète's formulas relate the polynomial's coefficients $lbrace\; a\_k\; brace$ to signed sums and products of its roots $lbrace\; x\_i\; brace$ as follows:

:

Equivalently stated, the $(n-k)$'th coefficient $a\_\{n-k\}$ is related to a signed sum of all possible subproducts of roots, taken $k$-at-a-time:

:

for each $k=1,\; 2,\; dots,\; n$ (where we wrote the indices $\{i\_k\}$ in increasing order to ensure each subproduct of roots is used exactly once).

Generalization to rings

Viète's formulas hold more generally for polynomials with coefficients in any integral domain, as long as the leading coefficient "a"_"n" is invertible (so that the divisions make sense) and the polynomial has $n$ distinct roots in that ring. The condition of being an integral domain is needed to assure that a polynomial of degree "n" cannot have more than "n" roots, and that is if has "n" roots then it is determined (up to a scalar) by those roots. However, if one replaces the assumption that "x"₁, …, "x"_"n" are "the roots" of the polynomial by the simpler requirement that one has the relation

:$a\_nX^n\; +\; a\_\{n-1\}X^\{n-1\}\; +cdots+\; a\_1\; X\; +\; a\_0\; =\; a\_n(X-x\_1)(X-x\_2)cdots(X-x\_n),$

then Viète's formulas even hold in any commutative ring, and they merely express the way the coefficients on the left hand side are formed when expanding the product on the left. Note that the given relation has to be required even in the case of an integral domain, if one wishes to allow some of the roots to coincide, and therefore needs to specify what multiplicity should be associated to each root.

Example

Viète's formulas applied to quadratic and cubic polynomial:

For the second degree polynomial (quadratic) $p(X)=aX^2\; +\; bX\; +\; c$, roots $x\_1,\; x\_2$ of the equation $p(X)=0$ satisfy:$x\_1\; +\; x\_2\; =\; -\; frac\{b\}\{a\},\; quad\; x\_1\; x\_2\; =\; frac\{c\}\{a\}$

The first of these equations can be used to find the minimum (or maximum) of "p". See second order polynomial.

For the cubic polynomial $p(X)=aX^3\; +\; bX^2\; +\; cX\; +\; d$, roots $x\_1,\; x\_2,\; x\_3$ of the equation $p(X)=0$ satisfy:$x\_1\; +\; x\_2\; +\; x\_3\; =\; -\; frac\{b\}\{a\},\; quad\; (x\_1\; x\_2\; +\; x\_1\; x\_3\; +\; x\_2\; x\_3)\; =\; frac\{c\}\{a\},\; quad\; x\_1\; x\_2\; x\_3\; =\; -\; frac\{d\}\{a\}$

Proof

Viète's formulas can be proven by expanding the equality

: $a\_nX^n\; +\; a\_\{n-1\}X^\{n-1\}\; +cdots\; +\; a\_1\; X+\; a\_0\; =\; a\_n(X-x\_1)(X-x\_2)cdots\; (X-x\_n)$

(which is true since $x\_1,\; x\_2,\; dots,\; x\_n$ are all the roots of this polynomial), multiplying through the factors on the right-hand side, and identifying the coefficients of each power of $X.$

ee also

* Newton's identities
* Elementary symmetric polynomial
* Symmetric polynomial
* Properties of polynomial roots

References

*cite book
last = Vinberg
first = E. B.
title = A course in algebra
publisher = American Mathematical Society, Providence, R.I
date = 2003
pages =
isbn = 0821834134

*cite book
last = Djukić
first = Dušan, et al.
coauthors =
title = The IMO compendium: a collection of problems suggested for the International Mathematical Olympiads, 1959-2004
publisher = Springer, New York, NY
date = 2006
pages =
isbn = 0387242996