Viète's Formulas
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 $\set {z_1, \ldots, z_n}$.
\(\ds \prod_{j \mathop = 1}^n \paren {x - z_j}\) | \(=\) | \(\ds x^n + \sum_{j \mathop = 1}^n \paren {-1}^{n - j + 1} e_{n - j + 1} \paren {\set {z_1, \ldots, z_n} } x^{j - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds x^n + \paren {-1} \, e_1 \paren {\set {z_1, \ldots, z_n} } x^{n - 1} + \paren {-1}^2 \map {e_2} {\set {z_1, \ldots, z_n} } x^{n - 2} + \cdots + \paren {-1}^n \map {e_n} {\set {z_1, \ldots, z_n} }\) |
$\map {\Bbb P} 1$ 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
- 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. https://mathworld.wolfram.com/VietasFormulas.html: Theorem for distinct roots.