Vieta's formulas

File:Francois Viete.jpeg

François Viète

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 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 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

Vieta's formulas can equivalently be written as $$\sum _{1\le i_1<i_2<\cdots <i_k\le n}\left(\prod _{j=1}^k{r}_{i_j}\right)=(-1{)}^k\frac{a_{n-k}}{a_n}$$

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 rings

Vieta's formulas are frequently used with polynomials with coefficients in any integral domain Template:Mvar. Then, the quotients $a_i/a_n$ belong to the field of fractions of Template:Mvar (and possibly are in Template:Mvar itself if $a_n$ happens to be invertible in Template:Mvar) and the roots $r_i$ 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 $a_n$ 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 $P(x)\ne (x-1)(x-3)$. 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 $P(x)=a{x}^2+bx+c$ satisfy $$r_1+r_2=-\frac{b}{a},r_1{r}_2=\frac{c}{a}.$$

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

The roots $r_1,r_2,r_3$ of the cubic polynomial $P(x)=a{x}^3+b{x}^2+cx+d$ satisfy $$r_1+r_2+r_3=-\frac{b}{a},r_1{r}_2+r_1{r}_3+r_2{r}_3=\frac{c}{a},r_1{r}_2{r}_3=-\frac{d}{a}.$$

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 Template:Tmath, one gets

$\sum _{k=0}^n(-1{)}^{n-k}{x}^k(\sum _{\stackrel{(\mathrm{\forall }i)b_i\in \{0,1\}}{b_1+\cdots +b_n=n-k}}{r}_1^{b_1}\cdots r_n^{b_n}),$

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

As an example, consider the quadratic $$f(x)=a_2{x}^2+a_{1}x+a_0=a_2(x-r_1)(x-r_2)=a_2(x^2-x(r_1+r_2)+r_1{r}_2).$$

Comparing identical powers of $x$, we find $a_2=a_2$, $a_1=-a_2(r_1+r_2)$ and $a_0=a_2(r_1{r}_2)$, with which we can for example identify $r_1+r_2=-a_1/a_2$ and $r_1{r}_2=a_0/a_2$, 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 $a_n\ne 0$. 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:$$a_2{x}^2+a_{1}x+a_0=a_2(x^2-r_{1}x-r_{2}x+r_1{r}_2)$$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 $n⩾2$, 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:$$\begin{array}{r}P(x)=a_{n+1}{x}^{n+1}-a_{n+1}(r_1+r_2+\cdots +r_n+r_{n+1})x^n+\cdots +(-1{)}^{n+1}(r_1{r}_2\cdots r_n{r}_{n+1})\\ \end{array}$$The inductive hypothesis holds true for $n+1$, therefore it must be true $\mathrm{\forall }n\in \mathbb{\mathbb{N}}$

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

• Content (algebra)
• Descartes' rule of signs
• Newton's identities
• Gauss–Lucas theorem
• Properties of polynomial roots
• Rational root theorem
• Symmetric polynomial and elementary symmetric polynomial

References

• Template:Springer