Viète's Formulas

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This proof is about Viète's Formulas in the context of Polynomial Theory. For other uses, see Viète's Formula for Pi.


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Theorem

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

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

where $a_n \ne 0$.

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

Let $P$ be expressible in the form:

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


Then:

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


Listed explicitly:

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


Proof

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

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


The proof proceeds by induction.

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

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


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

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

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 $x^{j - 1}$:

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


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

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


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

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


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

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

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


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

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

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.

$\blacksquare$


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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.


Also see


Source of Name

This entry was named for François Viète.


Sources

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