Vieta's formulas

Short description: Relating coefficients and roots of a polynomial

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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 (more commonly referred to by the Latinised form of his name, "Franciscus Vieta").

Basic formulas

Any general polynomial of degree n $P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0}$ (with the coefficients being real or complex numbers and $a n \neq 0$ ) has $n$ (not necessarily distinct) complex roots $r 1, r 2, ..., r n$ by the fundamental theorem of algebra. Vieta's formulas relate the polynomial coefficients to signed sums of products of the roots $r 1, r 2, ..., r n$ as follows:

${\begin{cases} r_{1} + r_{2} + \dots + r_{n - 1} + r_{n} = - \frac{a_{n - 1}}{a_{n}} \\ (r_{1} r_{2} + r_{1} r_{3} + \dots + r_{1} r_{n}) + (r_{2} r_{3} + r_{2} r_{4} + \dots + r_{2} r_{n}) + \dots + r_{n - 1} r_{n} = \frac{a_{n - 2}}{a_{n}} \\ ⋮ \\ r_{1} r_{2} \dots r_{n} = (- 1)^{n} \frac{a_{0}}{a_{n}} . \end{cases}$

([[#equation_*|*]])

Vieta's formulas can equivalently be written as $\sum_{1 \leq i_{1} < i_{2} < \dots < i_{k} \leq n} (\prod_{j = 1}^{k} r_{i_{j}}) = (- 1)^{k} \frac{a_{n - k}}{a_{n}}$ for $k = 1, 2, ..., n$ (the indices $i k$ are sorted in increasing order to ensure each product of $k$ 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 $R$ . Then, the quotients $a_{i} / a_{n}$ belong to the field of fractions of $R$ (and possibly are in $R$ itself if $a_{n}$ happens to be invertible in $R$ ) and the roots $r_{i}$ are taken in an algebraically closed extension. Typically, $R$ 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 $P (x) = a x^{2} + b x + c$ 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 $P$ ; see Quadratic equation § Vieta's formulas.

The roots $r_{1}, r_{2}, r_{3}$ of the cubic polynomial $P (x) = a x^{3} + b x^{2} + c x + d$ 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

Vieta's formulas can be proved by expanding the equality $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} = a_{n} (x - r_{1}) (x - r_{2}) \dots (x - r_{n})$ (which is true since $r_{1}, r_{2}, \dots, r_{n}$ are all the roots of this polynomial), multiplying the factors on the right-hand side, and identifying the coefficients of each power of $x .$

Formally, if one expands $(x - r_{1}) (x - r_{2}) \dots (x - r_{n}),$ the terms are precisely $(- 1)^{n - k} r_{1}^{b_{1}} \dots r_{n}^{b_{n}} x^{k},$ where $b_{i}$ is either 0 or 1, accordingly as whether $r_{i}$ is included in the product or not, and k is the number of $r_{i}$ that are included, so the total number of factors in the product is n (counting $x^{k}$ with multiplicity k) – as there are n binary choices (include $r_{i}$ or x), there are $2^{n}$ terms – geometrically, these can be understood as the vertices of a hypercube. Grouping these terms by degree yields the elementary symmetric polynomials in $r_{i}$ – for x^k, all distinct k-fold products of $r_{i} .$

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 $n = 2$ .

Alternate Proof (Mathematical Induction)

Vieta's formulas also formulas can be proven by induction.

Inductive Hypothesis:

Let $P (x)$ be a n degree polynomial, with real or complex roots $r_{1}, r_{2}, \dots, r_{n}$ . $P (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} = a_{n} x^{n} - a_{n} (r_{1} + r_{2} + \dots + r_{n}) x^{n - 1} + \dots + (- 1)^{n} (a_{n}) (r_{1} r_{2} \dots r_{n}), a_{n} \neq 0$ 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 $n + 1$ . $P (x) = a_{n + 1} x^{n + 1} + a_{n} x^{n} + \dots + a_{1} x + a_{0}$ 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 $r_{1}, r_{2}, \dots, r_{n}$ : $P (x) = (x - r_{n + 1}) [\frac{a_{n + 1} x^{n + 1} + a_{n} x^{n} + \dots + a_{1} x + a_{0}}{x - r_{n + 1}}]$ Factor out $a_{n + 1}$ , the leading coefficient $P (x)$ , from the polynomial in the square brackets:Failed to parse (syntax error): {\displaystyle {P(x)} ={(a_{n+{1}})}{(x-r_{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 $ζ$ : $P (x) = (a_{n + 1}) (x - r_{n + 1}) [x^{n} + ζ_{n - 1} x^{n - 1} + \dots + ζ_{0}]$ Using the inductive hypothesis, the polynomial in the square brackets can be rewritten as: $P (x) = (a_{n + 1}) (x - r_{n + 1}) [x^{n} - (r_{1} + r_{2} + \dots + r_{n}) x^{n - 1} + \dots + (- 1)^{n} (r_{1} r_{2} \dots r_{n})]$ Using distributive property: $P (x) = (a_{n + 1}) (x [x^{n} - (r_{1} + r_{2} + \dots + r_{n}) x^{n - 1} + \dots + (- 1)^{n} (r_{1} r_{2} \dots r_{n})] - r_{n + 1} [x^{n} - (r_{1} + r_{2} + \dots + r_{n}) x^{n - 1} + \dots + (- 1)^{n} (r_{1} r_{2} \dots r_{n})])$ After expanding and collecting like terms: $\begin{aligned} P (x) = a_{n + 1} x^{n + 1} - a_{n + 1} (r_{1} + r_{2} + \dots + r_{n} + r_{n + 1}) x^{n} + \dots + (- 1)^{n + 1} (r_{1} r_{2} \dots r_{n} r_{n + 1}) \end{aligned}$ The inductive hypothesis holds true for $n + 1$ , therefore it must be true $\forall n \in ℕ$

Conclusion: $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} = a_{n} x^{n} - a_{n} (r_{1} + r_{2} + \dots + r_{n}) x^{n - 1} + \dots + (- 1)^{n} (r_{1} r_{2} \dots r_{n})$ By dividing both sides both sides by $a_{n}$ , it proves the Vieta's formulas true.

History

As reflected in the name, the formulas were discovered 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,^[1] 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.

References

↑ (Funkhouser 1930)

Hazewinkel, Michiel, ed. (2001), "Viète theorem", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/v096630
Funkhouser, H. Gray (1930), "A short account of the history of symmetric functions of roots of equations", American Mathematical Monthly (Mathematical Association of America) 37 (7): 357–365, doi:10.2307/2299273
Vinberg, E. B. (2003), A course in algebra, American Mathematical Society, Providence, R.I, ISBN 0-8218-3413-4
Djukić, Dušan (2006), The IMO compendium: a collection of problems suggested for the International Mathematical Olympiads, 1959–2004, Springer, New York, NY, ISBN 0-387-24299-6

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[1] (Funkhouser 1930)

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