Vieta's formulas

Any general polynomial of degree n P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 (with the coefficients being real or complex numbers and ) has (not necessarily distinct) complex roots by the fundamental theorem of algebra. Vieta's formulas relate the polynomial coefficients to signed sums of products of the roots as follows: {{NumBlk||\begin{cases} r_1 + r_2 + \dots + r_{n-1} + r_n = -\dfrac{a_{n-1}}{a_n} \\[1ex] (r_1 r_2 + r_1 r_3 + \cdots + r_1 r_n) + (r_2r_3 + r_2r_4+\cdots + r_2r_n)+\cdots + r_{n-1}r_n = \dfrac{a_{n-2}}{a_{n}} \\[1ex] {} \quad \vdots \\[1ex] r_1 r_2 \cdots r_n = (-1)^n \dfrac{a_0}{a_n}. \end{cases}|}} Vieta's formulas can equivalently be written as \sum_{1\le i_1 for (the indices are sorted in increasing order to ensure each product of roots is used exactly once). The left-hand sides of Vieta's formulas are the elementary symmetric polynomials of the roots. Vieta's system can be solved by Newton's method through an explicit simple iterative formula, the Durand-Kerner method. ==Generalization to rings==

Vieta's formulas are frequently used with polynomials with coefficients in any integral domain . Then, the quotients a_i/a_n belong to the field of fractions of (and possibly are in itself if a_n happens to be invertible in ) and the roots r_i are taken in an algebraically closed extension. Typically, is the ring of the integers, the field of fractions is the field of the rational numbers and the algebraically closed field is the field of the complex numbers. Vieta's formulas are then useful because they provide relations between the roots without having to compute them. For polynomials over a commutative ring that is not an integral domain, Vieta's formulas are only valid when a_n is not a zero-divisor and P(x) factors as a_n(x-r_1)(x-r_2)\dots(x-r_n). For example, in the ring of the integers modulo 8, the quadratic polynomial P(x) = x^2-1 has four roots: 1, 3, 5, and 7. Vieta's formulas are not true if, say, r_1=1 and r_2=3, because P(x)\neq (x-1)(x-3). However, P(x) does factor as (x-1)(x-7) and also as (x-3)(x-5), and Vieta's formulas hold if we set either r_1=1 and r_2=7 or r_1=3 and r_2=5. ==Example==

Vieta's formulas applied to quadratic and cubic polynomials: The roots r_1, r_2 of the quadratic polynomial P(x) = ax^2 + bx + c satisfy r_1 + r_2 = -\frac{b}{a}, \quad r_1 r_2 = \frac{c}{a}. The first of these equations can be used to find the minimum (or maximum) of ; see . The roots r_1, r_2, r_3 of the cubic polynomial P(x) = ax^3 + bx^2 + cx + d satisfy r_1 + r_2 + r_3 = -\frac{b}{a}, \quad r_1 r_2 + r_1 r_3 + r_2 r_3 = \frac{c}{a}, \quad r_1 r_2 r_3 = -\frac{d}{a}. ==Proof==

Direct proof Vieta's formulas can be proved by considering the equality a_n x^n + a_{n-1}x^{n-1} +\cdots + a_1 x+ a_0 = a_n (x-r_1) (x-r_2) \cdots (x-r_n) (which is true since r_1, r_2, \dots, r_n are all the roots of this polynomial), expanding the products in the right-hand side, and equating the coefficients of each power of x between the two members of the equation. Formally, if one expands (x-r_1) (x-r_2) \cdots (x-r_n) and regroup the terms by their degree in , one gets :\sum_{k=0}^n (-1)^{n-k}x^k \left(\sum_{\stackrel{(\forall i)\; b_i\in\{0,1\}}{b_1+\cdots+b_n=n-k}} r_1^{b_1}\cdots r_n^{b_n}\right), where the inner sum is exactly the th elementary symmetric function As an example, consider the quadratic f(x) = a_2x^2 + a_1x + a_0 = a_2(x - r_1)(x - r_2) = a_2(x^2 - x(r_1 + r_2) + r_1 r_2). Comparing identical powers of x, we find a_2=a_2, a_1=-a_2 (r_1+r_2) and a_0 = a_2 (r_1r_2) , with which we can for example identify r_1+r_2 = - a_1/a_2 and r_1r_2 = a_0/a_2 , which are Vieta's formula's for n=2. Proof by mathematical induction Vieta's formulas can also be proven by induction as shown below. Inductive hypothesis: Let {P(x)} be polynomial of degree n, with complex roots {r_1},{r_2},{\dots},{r_n} and complex coefficients a_0,a_1,\dots,a_n where { a_n} \neq 0. Then the inductive hypothesis is that{P(x)} = {a_n}{x^n}+{{a_{n-1}}{x^{n-1}}}+{\cdots}+{{a_{1}}{x}}+{{a}_{0}} = {{a_n}{x^{n}}}-{a_n}{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ {{(-1)^{n}}{ (a_n)}{({r_1}{r_2}{\cdots}{r_n})}} Base case, n = 2 (quadratic): Let {a_2},{a_1} be coefficients of the quadratic and a_0 be the constant term. Similarly, let {r_1},{r_2} be the roots of the quadratic:{a_2 x^2}+{a_1 x} + a_0 = {a_2}{(x-r_1)(x-r_2)}Expand the right side using distributive property:{a_2 x^2}+{a_1 x} + a_0 = {a_2}{({x^2}-{r_1x}-{r_2x}+{r_1}{r_2})}Collect like terms:{a_2 x^2}+{a_1 x} + a_0 = {a_2}{({x^2}-{({r_1}+{r_2}){x}}+{r_1}{r_2})}Apply distributive property again:{a_2 x^2}+{a_1 x} + a_0 = {{a_2}{x^2}-{{a_2}({r_1}+{r_2}){x}}+{a_2}{({r_1}{r_2})}}The inductive hypothesis has now been proven true for n = 2. Induction step: Assuming the inductive hypothesis holds true for all n\geqslant 2, it must be true for all n+1 .{P(x)} = {a_{n+1}}{x^{n+1}}+{{a_{n}}{x^{n}}}+{\cdots}+{{a_{1}}{x}}+{{a}_{0}}By the factor theorem, {(x-r_{n+1})} can be factored out of P(x) leaving a 0 remainder. Note that the roots of the polynomial in the square brackets are r_1,r_2,\cdots,r_n:{P(x)} = {(x-r_{n+1})} {[{\frac{{a_ {n+ 1}}{x^ {n+1}}+{{a_{n}}{x^{n}}}+{\cdots}+{{a_{1}}{x}}+{{a}_{0}}}{x- r_{n +1}}}]}Factor out a_{n+1}, the leading coefficient P(x), from the polynomial in the square brackets:{P(x)} ={(a_{n+{1}})}{(x-r_{n+1})} {[{\frac{{x^ {n+1}}+ {\frac{{a_{n}} {x^{n}}}{(a_{n+{1}})}}+{\cdots}+{\frac {a_{1}}{(a_{n+{1}})} {x}}+ {{\frac{a_0}{{(a_{n+{1}})}}}}} {x- r_{n +1}}}]}For simplicity sake, allow the coefficients and constant of polynomial be denoted as \zeta:P(x) = {(a_ {n+1})}{(x-r_ {n+1})}{[{x^n}+{\zeta_{n-1}x^{n-1}}+{\cdots}+{\zeta_0}]}Using the inductive hypothesis, the polynomial in the square brackets can be rewritten as:P(x) = {(a_ {n+1})} {(x-r_ {n+1})} {[{{x^{n}}}-{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ {{(-1)^{n}}{({r_1}{r_2}{\cdots}{r_n})}}]}Using distributive property:P(x) = {(a_ {n+1})}{({x} {[{{x^{n}}}-{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ {{(-1)^{n}}{({r_1}{r_2}{\cdots}{r_n})}}]} {- r_ {n+1}} {[{{x^{n}}}-{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ {{(-1)^{n}}{({r_1}{r_2}{\cdots}{r_n})}}]} )}After expanding and collecting like terms:\begin{align} {P(x)} = {{a_{n+1}}{x^{n+1}}}-{a_{n+1}}{({r_1}+{r_2}+{\cdots}+{r_n}+{r_{n+1}}){x^{n}}}+{\cdots}+ {{(-1)^{n+1}}{({r_1}{r_2}{\cdots}{r_n}{r_{n+1}})}} \\ \end{align}The inductive hypothesis holds true for n+1, therefore it must be true \forall n \in \mathbb{N} Conclusion:{a_ n}{x^n}+{{a_{n-1}}{x^{n-1}}}+{\cdots}+{{a_{1}}{x}}+{{a}_{0}} = {{a_n}{x^{n}}}-{a_n}{({r_1}+{r_2}+{\cdots}+{r_n}){x^{n-1}}}+{\cdots}+ {{(-1)^{n}}{({r_1}{r_2}{\cdots}{r_n})}}By dividing both sides by a_{n}, it proves the Vieta's formulas true. == History ==

A method similar to Vieta's formula can be found in the work of the 12th century Islamic mathematician Sharaf al-Din al-Tusi. It is plausible that algebraic advancements made by other Islamic mathematicians such as Omar Khayyam, al-Tusi, and al-Kashi influenced 16th-century algebraists, with Vieta being the most prominent among them. The formulas were derived by the 16th-century French mathematician François Viète, for the case of positive roots. In the opinion of the 18th-century British mathematician Charles Hutton, as quoted by Funkhouser, the general principle (not restricted to positive real roots) was first understood by the 17th-century French mathematician Albert Girard: ...[Girard was] the first person who understood the general doctrine of the formation of the coefficients of the powers from the sum of the roots and their products. He was the first who discovered the rules for summing the powers of the roots of any equation. ==See also==