Viète'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 $P_n$ be a polynomial of degree $n$ with real or complex coefficients:

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

where $a_n \ne 0$.

Let $z_1, \ldots, z_n$ be the roots of $P_n$ (be they real or complex), not assumed distinct.

Then:

 $\ds \forall k \in \set {0, 1, \ldots, n}: \,$ $\ds \paren {-1}^k \dfrac {a_{n - k} } {a_n}$ $=$ $\ds \map {e_k} {\set {z_1, \ldots, z_n} }$ $\ds$ $=$ $\ds \sum_{\substack {1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n \\ 1 \mathop \le k \mathop \le n} } z_{i_1} \dotsm z_{i_k}$

where $\map {e_k} {\set {z_1, \ldots, z_n} }$ denotes the elementary symmetric function of degree $k$ on $\set {z_1, \ldots, z_n}$.

Listed explicitly:

 $\ds \dfrac {a_n} {a_n}$ $=$ $\ds \map {e_0} {\set {z_1, \ldots, z_n} }$ $\ds = 1$ $\ds -\dfrac {a_{n - 1} } {a_n}$ $=$ $\ds \map {e_1} {\set {z_1, \ldots, z_n} }$ $\ds = z_1 + z_2 + \cdots + z_n$ $\ds \dfrac {a_{n - 2} } {a_n}$ $=$ $\ds \map {e_2} {\set {z_1, \ldots, z_n} }$ $\ds = \paren {z_1 z_2 + \cdots + z_1 z_n} + \paren {z_2 z_3 + \cdots + z_2 z_n} + \cdots + \paren {z_{n - 1} z_n}$ $\ds$ $\vdots$ $\ds$ $\ds \paren {-1}^n \dfrac {a_0} {a_n}$ $=$ $\ds \map {e_n} {\set {z_1, \ldots, z_n} }$ $\ds = z_1 z_2 \cdots z_n$

Proof

First we note that from the Polynomial Factor Theorem:

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

To shorten the exposition, let us define:

$E_{\tuple {r, s} } := \map {e_r} {\set {z_1, \ldots, z_s} }$

for $r, s \in \Z_{\ge 1}$.

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

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

The proof then proceeds by induction as follows.

Let $\map {\Bbb P} n$ be the statement:

 $\ds \forall \set {z_1, z_2, \ldots, z_n}: \,$ $\ds \prod_{j \mathop = 1}^n \paren {x - z_j}$ $=$ $\ds \sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - j}$ $\ds$ $=$ $\ds E_{\tuple {0, n} } x^n - E_{\tuple {1, n} } x^{n - 1} + E_{\tuple {2, n} } x^{n - 2} - \cdots + \paren {-1}^n E_{\tuple {n, n} }$

Basis for the Induction

We have:

 $\ds \forall \set {z_1}: \,$ $\ds \prod_{j \mathop = 1}^1 \paren {x - z_j}$ $=$ $\ds x - z_1$ $\ds$ $=$ $\ds E_{\tuple {0, 1} } x^1 - E_{\tuple {1, 1} } x^0$ Definition of Elementary Symmetric Function: $E_{\tuple {0, 1} } = \map {e_0} {\set {z_1} } = 1$, $E_{\tuple {1, 1} } = \map {e_1} {\set {z_1} } = z_1$ $\ds$ $=$ $\ds \sum_{j \mathop = 0}^1 \paren {-1}^j E_{\tuple {j, 1} } x^{1 - j}$ Definition of Summation

Hence $\map {\Bbb P} 1$ holds.

This is our basis for the induction.

Induction Hypothesis

Now it needs to be shown that if $\map {\Bbb P} n$ is true then it logically follows that $\map {\Bbb P} {n + 1}$ is true.

This is the induction hypothesis:

 $\ds \forall \set {z_1, z_2, \ldots, z_n}: \,$ $\ds \prod_{j \mathop = 1}^n \paren {x - z_j}$ $=$ $\ds \sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - j}$ $\ds$ $=$ $\ds E_{\tuple {0, n} } x^n - E_{\tuple {1, n} } x^{n - 1} + E_{\tuple {2, n} } x^{n - 2} - \cdots + \paren {-1}^n E_{\tuple {n, n} }$

from which it is to be shown that:

 $\ds \forall \set {z_1, z_2, \ldots, z_{n + 1} }: \,$ $\ds \prod_{j \mathop = 1}^{n + 1} \paren {x - z_j}$ $=$ $\ds \sum_{j \mathop = 0}^{n + 1} \paren {-1}^j E_{\tuple {j, n + 1} } x^{n + 1 - j}$ $\ds$ $=$ $\ds E_{\tuple {0, n + 1} } x^{n + 1} - E_{\tuple {1, n + 1} } x^n + E_{\tuple {2, n + 1} } x^{n - 1} - \cdots + \paren {-1}^{n + 1} E_{\tuple {n + 1, n + 1} }$

Induction Step

This is our induction step:

Let it be assumed that $\map {\Bbb P} n$ holds for some $n \ge 1$.

Let $\map {Q_{n + 1} } x$ be an arbitrary polynomial of degree $n + 1$ with real or complex coefficients:

 $\ds \map {Q_{n + 1} } x$ $=$ $\ds \sum_{i \mathop = 0}^{n + 1} q_i x^i$ $\ds$ $=$ $\ds q_{n + 1} x^{n + 1} + q_n x^n + \dotsb + q_1 x + q_0$

We can divide all the coefficients by $q_{n + 1}$ to force $\map {Q_{n + 1} } x$ to be monic.

Hence after renaming those coefficients:

 $\ds \map {Q_{n + 1} } x$ $=$ $\ds \sum_{i \mathop = 0}^{n + 1} q_i x^i$ $\ds$ $=$ $\ds x^{n + 1} + q_n x^n + \dotsb + q_1 x + q_0$

and so, from the Polynomial Factor Theorem:

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

where $\set {z_1, \ldots, z_n, z_{n + 1} }$ are the roots of $\map {Q_{n + 1} } x$.

Hence:

 $\ds \map {Q_{n + 1} } x$ $=$ $\ds \paren {x - z_{n + 1} } \prod_{j \mathop = 1}^n \paren {x - z_j}$ Definition of Continued Product $\ds$ $=$ $\ds \paren {x - z_{n + 1} } \paren {\sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - j} }$ Induction Hypothesis $\ds$ $=$ $\ds x \paren {\sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - j} } - z_{n + 1} \paren {\sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - j} }$ Distributive Laws of Arithmetic $\ds$ $=$ $\ds \sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n + 1 - j} - z_{n + 1} \paren {\sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - j} }$ simplifying $\ds$ $=$ $\ds \sum_{j \mathop = 0}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - \paren {j - 1} } - z_{n + 1} \paren {\sum_{j \mathop = 1}^{n + 1} \paren {-1}^{j - 1} E_{\tuple {j - 1, n} } x^{n - \paren {j - 1} } }$ Translation of Index Variable of Summation $\ds$ $=$ $\ds E_{\tuple {0, n} } x^{n + 1} + \sum_{j \mathop = 1}^n \paren {-1}^j E_{\tuple {j, n} } x^{n - \paren {j - 1} } - z_{n + 1} \paren {\sum_{j \mathop = 1}^n \paren {-1}^{j - 1} E_{\tuple {j - 1, n} } x^{n - \paren {j - 1} } + \paren {-1}^n E_{\tuple {n, n} } x^0}$ separation out of terms in $E_{\tuple {0, n} }$ and $E_{\tuple {n, n} }$ $\ds$ $=$ $\ds E_{\tuple {0, n} } x^{n + 1} + \sum_{j \mathop = 1}^n \paren {-1}^j \paren {E_{\tuple {j, n} } + z_{n + 1} E_{\tuple {j - 1, n} } } x^{n + 1 - j} - z_{n + 1} \paren {-1}^n E_{\tuple {n, n} }$ simplifying and merging terms in $x^{n - \paren {j - 1} } = x^{n + 1 - j}$ $\ds$ $=$ $\ds E_{\tuple {0, n} } x^{n + 1} + \sum_{j \mathop = 1}^n \paren {-1}^j E_{\tuple {j, n + 1} } x^{n + 1 - j} + z_{n + 1} \paren {-1}^{n + 1} E_{\tuple {n, n} }$ Recursion Property of Elementary Symmetric Function $\ds$ $=$ $\ds \sum_{j \mathop = 0}^{n + 1} \paren {-1}^j E_{\tuple {j, n + 1} } x^{n + 1 - j}$ subsuming the indices for $j = 0$ and $j = n + 1$, using Example: $z_{n + 1} E_{\tuple {n, n} } = E_{\tuple {n + 1, n + 1} }$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$

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:

$\ds \map P x = x^N + \sum_{k \mathop = 0}^{N - 1} b_k x^k$

be a monic polynomial of degree $N$.

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

Then:

$b_k = \paren {-1}^{N - k} \map {e_{N - k} } U, \quad 0 \le k \le N - 1$

where $\map {e_m} U$ denotes an elementary symmetric function.

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.