Viète's Formulas

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Theorem

Let

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

be a polynomial of degree $n$ over a ring $R$.

Suppose $P$ can be expressed as:

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

where $z_1, \ldots, z_k \in R$.

That is, $z_1, \ldots, z_k$ are roots of $P$, not guaranteed to be unique in general.

Then:

$\displaystyle a_{n - k} = \paren {-1}^k a_n \sum_{1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n} z_{i_1} \dotsm z_{i_k}$

for $k = 1, 2, \ldots, n$.


Listed explicitly, supposing $a_n$ is invertible:

$z_1 + z_2 + \cdots + z_n = -a_{n-1} / a_n$
$z_1 z_2 + \cdots + z_1 z_n + z_2 z_3 + \cdots + z_2 z_n + \cdots + z_{n - 1} z_n = a_{n - 2} / a_n$
$\cdots$
$z_1 z_2 \cdots z_n = \paren {-1}^n a_0 / a_n$


Proof

Note that the indexing $1 \le i_1 \le \cdots i_k \le n$ represents all possible subsets of $\set {1, 2, \dotsc, n}$ of size $k$ up to order.

It follows from Product of Sums: Corollary that $\map P x$ foils as a sum of powers $x^k$ with coefficients as sums of all products of elements of subsets of $\set {z_1, \dotsc, z_n}$ of complementary size $n - k$, hence equating pairwise with the original coefficients of $\map P x$ obtains Viète's Formulas.



$\blacksquare$


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:

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


Source of Name

This entry was named for François Viète.


Sources