# Viète's Formulas

(Redirected from Viete's Formulas)

This proof is about Viète's Formulas in the context of Polynomial Theory. For other uses, see Viète's Formula for Pi.

## Theorem

Let $P$ be a polynomial of degree $n$ with real or complex coefficients:

 $\ds \map P x$ $=$ $\ds \sum_{i \mathop = 0}^n a_i x^i$ $\ds$ $=$ $\ds a_n x^n + a_{n - 1} x^{n - 1} + \dotsb + a_1 x + a_0$

where $a_n \ne 0$.

Let $z_1, \ldots, z_k$ be real or complex roots of $P$, not assumed distinct.

Let $P$ be expressible in the form:

$\ds \map P x = a_n \prod_{k \mathop = 1}^n \paren {x - z_k}$

Then:

 $\ds \paren {-1}^k \dfrac {a_{n - k} } {a_n}$ $=$ $\ds e_k \paren {\set {z_1, \ldots, z_n} }$ that is, the elementary symmetric function on $\set {z_1, \ldots, z_n}$ $\ds$ $=$ $\ds \sum_{1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n} z_{i_1} \dotsm z_{i_k}$ for $k = 1, 2, \ldots, n$.

Listed explicitly:

 $\ds \paren {-1} \, \dfrac {a_{n-1} } {a_n}$ $=$ $\ds z_1 + z_2 + \cdots + z_n$ $\ds \paren {-1}^2 \dfrac {a_{n - 2} } {a_n}$ $=$ $\ds \paren {z_1 z_2 + \cdots + z_1 z_n} + \paren {z_2 z_3 + \cdots + z_2 z_n} + \cdots + \paren {z_{n - 1} z_n}$ $\ds$ $\vdots$ $\ds$ $\ds \paren {-1}^n \dfrac {a_0} {a_n}$ $=$ $\ds z_1 z_2 \cdots z_n$

## Proof

It is sufficient to consider the case $a_n = 1$:

$\ds \map P x = \prod_{k \mathop = 1}^n \paren {x - z_k}$

The proof proceeds by induction.

Let $\map {\Bbb P} n$ be the statement that the identity below holds for all sets $\set {z_1, \ldots, z_n}$.

 $\ds \prod_{j \mathop = 1}^n \paren {x - z_j}$ $=$ $\ds x^n + \sum_{j \mathop = 1}^n \paren {-1}^{n - j + 1} e_{n - j + 1} \paren {\set {z_1, \ldots, z_n} } \, x^{j - 1}$ $\ds$ $=$ $\ds x^n + \paren {-1} \, e_1 \paren {\set {z_1, \ldots, z_n} } \, x^{n - 1} + \paren {-1}^2 \, e_2 \paren {\set {z_1, \ldots, z_n} } \, x^{n - 2} + \cdots + \paren {-1}^n e_n \paren {\set {z_1, \ldots, z_n} }$

$\map {\Bbb P} 1$ holds because $\map {e_1} {\set {z_1} } = z_1$.

Induction Step $\map {\Bbb P} n$ implies $\map {\Bbb P} {n + 1}$:

Assume $\map {\Bbb P} n$ holds and $n \ge 1$.

Let for given values $\set {z_1, \ldots, z_n, z_{n + 1} }$:

$\ds \map Q x = \paren {x - z_{n + 1} } \prod_{k \mathop = 1}^n \paren {x - z_k}$

Expand the right side product above using induction hypothesis $\map {\Bbb P} n$.

Then $\map Q x$ equals $x^{n + 1}$ plus terms for $x^{j - 1}$, $1 \le j \le n + 1$.

If $j = 1$, then one term occurs for $x^{j - 1}$:

$\ds \paren {-x_{n + 1} } \, \paren {\paren {-1}^{n - 1 + 1} \map {e_{n - 1 + 1} } {\set {z_1, \ldots, z_n} } x^{1 - 1} } = \paren {-1}^{n + 1} \map {e_n} {\set {z_1, \ldots, z_n, z_{n + 1} } }$

If $2 \le j \le n + 1$, then two terms $T_1$ and $T_2$ occur for $x^{j - 1}$:

 $\ds T_1$ $=$ $\ds \paren x \paren {\paren {-1}^{n - j + 2} \, \map {e_{n - j + 2} } {\set {z_1, \ldots, z_n} } x^{j - 2} }$ $\ds T_2$ $=$ $\ds \paren {-z_{n + 1 } } \, \paren {\paren {-1}^{n - j + 1} \, \map {e_{n - j + 1} } {\set {z_1, \ldots, z_n} } x^{j - 1} }$

The coefficient $c$ of $x^{j - 1}$ for $2 \le j \le n + 1$ is:

 $\ds c$ $=$ $\ds \dfrac {T_1 + T_2} {x^{j - 1} }$ $\ds$ $=$ $\ds \paren {-1}^m \, \map {e_m} {\set {z_1, \ldots, z_n} } + \paren {-1}^m \, \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } z_{n + 1}$ where $m = n - j + 2$.

Use recursion identity to simplify the expression for $c$:

 $\ds \map {e_m} {\set {z_1, \ldots, z_n, z_{n + 1} } }$ $=$ $\ds z_{n + 1} \, \map {e_{m - 1} } {\set {z_1, \ldots, z_n} } + \map {e_m} {\set {z_1, \ldots, z_n} }$ $\ds \leadsto \ \$ $\ds c$ $=$ $\ds \paren {-1}^{n - j + 2} \, \map {e_{n - j + 2} } {\set {z_1, \ldots, z_n, z_{n + 1} } }$

Thus $\map {\Bbb P} {n + 1}$ holds and the induction is complete.

Set equal the two identities for $\map P x$:

$\displaystyle x^n + \sum_{k \mathop = 0}^{n - 1} a_k x^k = x^n + \paren {-1} \, \map {e_1} {\set {z_1, \ldots, z_n} } x^{n - 1} + \paren {-1}^2 \, \map {e_2} {\set {z_1, \ldots, z_n} } x^{n - 2} + \cdots + \paren {-1}^n \map {e_n} {\set {z_1, \ldots, z_n} }$

Linear independence of the powers $1, x, x^2, \ldots$ implies polynomial coefficients match left and right.

Hence the coefficient $a_k$ of $x^k$ on the left hand side matches $\paren {-1}^{n - k} \, \map {e_{n - k} } {\set {z_1, \ldots, z_n} }$ on the right hand side.

$\blacksquare$

## Examples

### Sum of Roots of Quadratic Equation

Let $P$ be the quadratic equation $a x^2 + b x + c = 0$.

Let $\alpha$ and $\beta$ be the roots of $P$.

Then:

$\alpha + \beta = -\dfrac b a$

### Product of Roots of Quadratic Equation

Let $P$ be the quadratic equation $a x^2 + b x + c = 0$.

Let $\alpha$ and $\beta$ be the roots of $P$.

Then:

$\alpha \beta = \dfrac c a$

### Coefficients of Cubic

Consider the cubic equation:

$x^3 + a_1 x^2 + a_2 x^1 + a_3 = 0$

Let its roots be denoted $x_1$, $x_2$ and $x_3$.

Then:

 $\ds x_1 + x_2 + x_3$ $=$ $\ds -a_1$ $\ds x_1 x_2 + x_2 x_3 + x_3 x_1$ $=$ $\ds a_2$ $\ds x_1 x_2 x_3$ $=$ $\ds -a_3$

### Coefficients of Quartic

Consider the quartic equation:

$x^4 + a_1 x^3 + a_2 x^2 + a_3 x + a_4 = 0$

Let its roots be denoted $x_1$, $x_2$, $x_3$ and $x_4$.

Then:

 $\ds x_1 + x_2 + x_3 + x_4$ $=$ $\ds -a_1$ $\ds x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_1 + x_1 x_3 + x_2 x_4$ $=$ $\ds a_2$ $\ds x_1 x_2 x_3 + x_2 x_3 x_4 + x_1 x_2 x_4 + x_1 x_3 x_4$ $=$ $\ds -a_3$ $\ds x_1 x_2 x_3 x_4$ $=$ $\ds a_4$

### Two Numbers whose Sum is $4$ and whose Product is $8$

Let $z_1$ and $z_2$ be two numbers whose sum is $4$ and whose product is $8$.

Then:

 $\ds z_1$ $=$ $\ds 2 + 2 i$ $\ds z_2$ $=$ $\ds 2 - 2 i$

### Cubic with all Equal Roots

The coefficient of $x$ in the expansion of cubic $\paren {x - 1}^3$ is $3$.

### Monic Polynomial Formulas

Let:

 $\ds \map P x$ $=$ $\ds x^N + \displaystyle \sum_{k \mathop = 0}^{N - 1} b_k x^k$ Monic polynomial of degree $N$.

Let $U$ be the set of $N$ roots of equation $\map P x = 0$.

Then:

 $\ds b_k$ $=$ $\ds \paren {-1}^{N - k} \, \map {e_{N - k} } U, \quad 0 \le k \le N - 1$ Definition of Elementary Symmetric Function $\map {e_m} U$

## Also known as

Viète's Formulas are also known (collectively) as Viète's Theorem or (the) Viète Theorem.

The Latin form of his name (Vieta) is also often seen.

## Source of Name

This entry was named for François Viète.