Viète's Formulas
This proof is about sums and products of coefficients of polynomials. For other uses, see Viète's Formula for Pi.
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Contents
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 $\set {z_1, \ldots, z_n}$.
| \(\displaystyle \prod_{j \mathop = 1}^n \paren {x - z_j}\) | \(=\) | \(\displaystyle x^n + \sum_{j \mathop = 1}^n \paren {-1}^{n - j + 1} e_{n - j + 1} \paren {\set {z_1, \ldots, z_n} } \, x^{j - 1}\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle x^n + \paren {-1} \, e_1 \paren {\set {z_1, \ldots, z_n} } \, x^{n - 1} + \paren {-1}^2 \, e_2 \paren {\set {z_1, \ldots, z_n} } \, x^{n - 2} + \cdots + \paren {-1}^n e_n \paren {\set {z_1, \ldots, z_n} }\) |
$\map {\Bbb P} 1$ 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.
$\blacksquare$
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Examples
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:
| \(\displaystyle x_1 + x_2 + x_3\) | \(=\) | \(\displaystyle -a_1\) | |||||||||||
| \(\displaystyle x_1 x_2 + x_2 x_3 + x_3 x_1\) | \(=\) | \(\displaystyle a_2\) | |||||||||||
| \(\displaystyle x_1 x_2 x_3\) | \(=\) | \(\displaystyle -a_3\) |
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:
| \(\displaystyle x_1 + x_2 + x_3 + x_4\) | \(=\) | \(\displaystyle -a_1\) | |||||||||||
| \(\displaystyle x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_1 + x_1 x_3 + x_2 x_4\) | \(=\) | \(\displaystyle a_2\) | |||||||||||
| \(\displaystyle x_1 x_2 x_3 + x_2 x_3 x_4 + x_1 x_2 x_4 + x_1 x_3 x_4\) | \(=\) | \(\displaystyle -a_3\) | |||||||||||
| \(\displaystyle x_1 x_2 x_3 x_4\) | \(=\) | \(\displaystyle a_4\) |
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:
| \(\displaystyle z_1\) | \(=\) | \(\displaystyle 2 + 2 i\) | |||||||||||
| \(\displaystyle z_2\) | \(=\) | \(\displaystyle 2 - 2 i\) |
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
- Viète's Formula for Pi (an unrelated result)
Source of Name
This entry was named for François Viète.
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $103$
- Viète theorem. Michiel Hazewinkel (originator),Encyclopedia of Mathematics. URL: https://www.encyclopediaofmath.org/index.php?title=Viète_theorem: Theorem for a field.
- Weisstein, Eric W. "Vieta's Formulas." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/VietasFormulas.html: Theorem for distinct roots.
