Viète's Formulas

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

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:

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:

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 complete induction as follows.

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

Basis for the Induction
We have:

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:

from which it is to be shown that:

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:

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:

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:

Then $\map {Q_{n + 1} } x$ equals $x^{n + 1}$ plus terms for $x^j$ where $0 \le j \le n$.

If $j = 0$, then one term occurs for $x^j$:


 * $\ds \paren {-x_{n + 1} } \, \paren {\paren {-1}^j E_{\tuple {j, n} } x^{1 - 1} } = \paren {-1}^{n + 1} E_{\tuple {n, n + 1} }$

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

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

Use Recursion Property of Elementary Symmetric Function to simplify the expression for $c$:

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

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


 * $\ds \sum_{k \mathop = 0}^n a_k x^k = x^n + \paren {-1} E_{\tuple {1, n} } x^{n - 1} + \paren {-1}^2 E_{\tuple {2, n} } x^{n - 2} + \cdots + \paren {-1}^n E_{\tuple {n, n} }$

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

Hence the coefficient $a_k$ of $x^k$ on the matches $\paren {-1}^{n - k} E_{\tuple {n - k, n} }$ on the.

Also see

 * Definition:Elementary Symmetric Function


 * Elementary Symmetric Function/Examples/Monic Polynomial