Viète's Formulas

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

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

Then:

Listed explicitly:

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


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

The proof proceeds by induction.

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

Basis for the Induction:

$\map {\Bbb P} 1$ 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 $\set {z_1, \ldots, z_n, z_{n + 1} }$:


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

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 $x^{j - 1}$:


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

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

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

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

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

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


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

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 matches $\paren {-1}^{n - k} \, \map {e_{n - k} } {\set {z_1, \ldots, z_n} }$ on the.

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.

Also see

 * Viète's Formula for Pi (an unrelated result)


 * Definition:Elementary Symmetric Function


 * Elementary Symmetric Function/Examples/Monic Polynomial