Viète's Formulas

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This proof is about sums and products of coefficients of polynomials. For other uses, see Viète's Formula for Pi.


Theorem

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

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

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:

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


Then:

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


Listed explicitly:

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


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 $\set {z_1, \ldots, z_n}$.

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


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 $x^{j - 1}$:

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


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

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


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

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

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

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:

\(\displaystyle z_1\) \(=\) \(\displaystyle 2 + 2 i\)
\(\displaystyle z_2\) \(=\) \(\displaystyle 2 - 2 i\)


Cubic with all Equal Roots

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


Monic Polynomial Formulas

Let:

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

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

Then:

\(\displaystyle b_k\) \(=\) \(\displaystyle \paren {-1}^{N - k} \, \map {e_{N - k} } U, \quad 0 \le k \le N - 1\) 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.


Also see


Source of Name

This entry was named for François Viète.


Sources