Viète's Formulas

Formulas

For a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ with roots $z_1 , z_2, \dots, z_n$ (not necessarily distinct),

$\displaystyle \sum_{1 \le i_1 < \cdots < i_k \le n} z_{i_1} \cdots z_{i_k} = (-1)^k \frac{a_{n-k}}{a_n} $

for $k = 1, 2, \dots, n$. Listed explicitly,

• $z_1 + z_2 + \cdots + z_n = -a_{n-1} / a_n$;
• $z_1 z_2 + \cdots + z_1 z_n + z_2 z_3 + \cdots + z_2 z_n \cdots + z_{n-1} z_n = a_{n-2} / a_n$;
• $\cdots$
• $z_1 z_2 \cdots z_n = (-1)^n a_0 / a_n$.

Proof

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

Setting $P(x) = a_n (x - z_1) \cdots (x- z_n)$, which is permitted by the fundamental theorem of algebra, it follows from a corollary to the Products of Sums formula that $P(x)$ foils as a sum of powers $x^k$ with coefficients as sums of all products of elements of subsets of $\{ z_1, \dots, z_n \}$ of complementary size $n-k$, hence equating pairwise with the original coefficients of $P(x)$ (divided through by $a_n$) obtains Viète's Formulas.

$\blacksquare$

Source of Name

This entry was named for François Viète.