Vieta's formulas

In mathematics, Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. They are named after François Viète (1540-1603), more commonly referred to by the Latinised form of his name, "Franciscus Vieta."

Basic formulasEdit

Any general polynomial of degree n <math display="block">P(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0</math> (with the coefficients being real or complex numbers and Template:Math) has Template:Math (not necessarily distinct) complex roots Template:Math by the fundamental theorem of algebra. Vieta's formulas relate the polynomial coefficients to signed sums of products of the roots Template:Math as follows: Template:NumBlk{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}</math>|Template:EquationRef}}

Vieta's formulas can equivalently be written as <math display="block">\sum_{1\le i_1 < i_2 < \cdots < i_k\le n} \left(\prod_{j = 1}^k r_{i_j}\right)=(-1)^k\frac{a_{n-k}}{a_n}</math>

for Template:Math (the indices Template:Math are sorted in increasing order to ensure each product of Template:Math roots is used exactly once).

The left-hand sides of Vieta's formulas are the elementary symmetric polynomials of the roots.

Vieta's system Template:EquationNote can be solved by Newton's method through an explicit simple iterative formula, the Durand-Kerner method.

Generalization to ringsEdit

Vieta's formulas are frequently used with polynomials with coefficients in any integral domain Template:Mvar. Then, the quotients <math>a_i/a_n</math> belong to the field of fractions of Template:Mvar (and possibly are in Template:Mvar itself if <math>a_n</math> happens to be invertible in Template:Mvar) and the roots <math>r_i</math> are taken in an algebraically closed extension. Typically, Template:Mvar 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 <math>a_n</math> is not a zero-divisor and <math>P(x)</math> factors as <math>a_n(x-r_1)(x-r_2)\dots(x-r_n)</math>. For example, in the ring of the integers modulo 8, the quadratic polynomial <math>P(x) = x^2-1</math> has four roots: 1, 3, 5, and 7. Vieta's formulas are not true if, say, <math>r_1=1</math> and <math>r_2=3</math>, because <math>P(x)\neq (x-1)(x-3)</math>. However, <math>P(x)</math> does factor as <math>(x-1)(x-7)</math> and also as <math>(x-3)(x-5)</math>, and Vieta's formulas hold if we set either <math>r_1=1</math> and <math>r_2=7</math> or <math>r_1=3</math> and <math>r_2=5</math>.

ExampleEdit

Vieta's formulas applied to quadratic and cubic polynomials:

The roots <math>r_1, r_2</math> of the quadratic polynomial <math>P(x) = ax^2 + bx + c</math> satisfy <math display="block"> r_1 + r_2 = -\frac{b}{a}, \quad r_1 r_2 = \frac{c}{a}.</math>

The first of these equations can be used to find the minimum (or maximum) of Template:Math; see Template:Slink.

The roots <math>r_1, r_2, r_3</math> of the cubic polynomial <math>P(x) = ax^3 + bx^2 + cx + d</math> satisfy <math display="block"> 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}.</math>

ProofEdit

Direct proofEdit

Vieta's formulas can be proved by considering the equality <math display="block">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)</math> (which is true since <math>r_1, r_2, \dots, r_n</math> are all the roots of this polynomial), expanding the products in the right-hand side, and equating the coefficients of each power of <math>x</math> between the two members of the equation.

Formally, if one expands <math>(x-r_1) (x-r_2) \cdots (x-r_n)</math> and regroup the terms by their degree in Template:Tmath, one gets

<math>\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),</math>

where the inner sum is exactly the Template:Tmathth elementary symmetric function

As an example, consider the quadratic <math display=block>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).</math>

Comparing identical powers of <math>x</math>, we find <math>a_2=a_2</math>, <math>a_1=-a_2 (r_1+r_2) </math> and <math> a_0 = a_2 (r_1r_2) </math>, with which we can for example identify <math> r_1+r_2 = - a_1/a_2 </math> and <math> r_1r_2 = a_0/a_2 </math>, which are Vieta's formula's for <math>n=2</math>.

Proof by mathematical inductionEdit

Vieta's formulas can also be proven by induction as shown below.

Inductive hypothesis:

Let <math>{P(x)}</math> be polynomial of degree <math>n</math>, with complex roots <math>{r_1},{r_2},{\dots},{r_n}</math> and complex coefficients <math>a_0,a_1,\dots,a_n</math> where <math>{ a_n} \neq 0</math>. Then the inductive hypothesis is that<math display="block">{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})}}</math>

Base case, <math>n = 2 </math> (quadratic):

Let <math>{a_2},{a_1}</math> be coefficients of the quadratic and <math>a_0 </math>be the constant term. Similarly, let <math>{r_1},{r_2}</math> be the roots of the quadratic:<math display="block">{a_2 x^2}+{a_1 x} + a_0 = {a_2}{(x-r_1)(x-r_2)}</math>Expand the right side using distributive property:<math display="block">{a_2 x^2}+{a_1 x} + a_0 = {a_2}{({x^2}-{r_1x}-{r_2x}+{r_1}{r_2})}</math>Collect like terms:<math display="block">{a_2 x^2}+{a_1 x} + a_0 = {a_2}{({x^2}-{({r_1}+{r_2}){x}}+{r_1}{r_2})}</math>Apply distributive property again:<math display="block">{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})}}</math>The inductive hypothesis has now been proven true for <math>n = 2</math>.

Induction step:

Assuming the inductive hypothesis holds true for all <math>n\geqslant 2</math>, it must be true for all <math>n+1 </math>.<math display="block">{P(x)} = {a_{n+1}}{x^{n+1}}+{{a_{n}}{x^{n}}}+{\cdots}+{{a_{1}}{x}}+{{a}_{0}}</math>By the factor theorem, <math>{(x-r_{n+1})}</math> can be factored out of <math>P(x) </math> leaving a 0 remainder. Note that the roots of the polynomial in the square brackets are <math>r_1,r_2,\cdots,r_n</math>:<math display="block">{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}}}]}</math>Factor out <math>a_{n+1}</math>, the leading coefficient <math>P(x)</math>, from the polynomial in the square brackets:<math display="block">{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}}}]}</math>For simplicity sake, allow the coefficients and constant of polynomial be denoted as <math>\zeta</math>:<math display="block">P(x) = {(a_ {n+1})}{(x-r_ {n+1})}{[{x^n}+{\zeta_{n-1}x^{n-1}}+{\cdots}+{\zeta_0}]}</math>Using the inductive hypothesis, the polynomial in the square brackets can be rewritten as:<math display="block">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})}}]}</math>Using distributive property:<math display="block">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})}}]} )}</math>After expanding and collecting like terms:<math display="block">\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}</math>The inductive hypothesis holds true for <math>n+1</math>, therefore it must be true <math>\forall n \in \mathbb{N}</math>

Conclusion:<math display="block">{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})}}</math>By dividing both sides by <math>a_{n}</math>, it proves the Vieta's formulas true.

HistoryEdit

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 mathematician such as Omar Khayyam, al-tusi, and al-Kashi influenced 16th-century algebraists, with Vieta being the most prominent among them.<ref>Template:Cite journal</ref><ref>{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>

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,<ref>Template:Harv</ref> 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.

NotesEdit

Template:Reflist