Viète's Formulas

Theorem
Let


 * $\displaystyle \map P x = a_n x^n + a_{n - 1} x^{n - 1} + \dotsb + a_1 x + a_0$

be a polynomial of degree $n$ with coefficients real or complex numbers.

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

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

Then:

Equivalently:

Listed explicitly:

Proof
It suffices 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 ${\Bbb P} \paren { n }$ be the statement that the identity below holds for all sets $\set {z_1,\ldots,z_n}$.

Basis for the Induction:

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

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

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

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


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

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

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} e_{n-1+1} \paren { \set {z_1,\ldots,z_n } } x^{1-1} } =

\paren { -1 }^{n+1} e_{n} \paren { \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 = \paren { x } \paren { \paren {-1}^{n-j+2} e_{n-j+2} \paren { \set {z_1,\ldots,z_n} } x^{j-2}  },

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

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

Use recursion identity:


 * $\displaystyle e_m \paren { \set {z_1,\ldots,z_n,z_{n+1} } } = z_{n+1} e_{m-1} \paren { \set {z_1,\ldots,z_n} } + e_m \paren { \set {z_1,\ldots,z_n} }

$

to simplify the expression for $c$:


 * $\displaystyle c = \paren {-1}^{n-j+2} e_{n-j+2} \paren { \set {z_1,\ldots,z_n,z_{n+1} } }$

Then ${\Bbb P} \paren { n + 1 }$ 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 - e_1 \paren { \set {z_1,\ldots,z_n}  } x^{n-1} + e_2 \paren { \set {z_1,\ldots,z_n}  } x^{n-2} + \cdots +(-1)^n e_n \paren { \set {z_1,\ldots,z_n}  }$

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

Then coefficient $a_{k}$ of $x^k$ on the left matches $\paren {-1}^{n-k} e_{n-k} \paren { \set {z_1,\ldots,z_n} }$ on the right.

Also known as
Vieta's formulas, Vieta's theorem, Viète theorem, Viète's theorem. Singular form Viète's Formula is unrelated (computes $\pi$).

Also see
Definition:Elementary Symmetric Function

Elementary Symmetric Function/Examples/Monic Polynomial

Principle of Mathematical Induction