Vieta's Formulas

Revision as of 13:56, 24 July 2006 by 4everwise (talk | contribs)


Let $P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0$, where the coefficient of $\displaystyle x^{i}$ is $\displaystyle {a}_i$. As a consequence of the Fundamental Theorem of Algebra, we can also write $P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)$, where $\displaystyle {r}_i$ are the roots of $\displaystyle P(x)$. We thus have that

$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).$

Expanding out the right hand side gives us

$a_nx^n - a_n(r_1+r_2+\cdots+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 + \cdots + r_{n-1}r_n)x^{n-2} + \cdots + (-1)^na_n r_1r_2\cdots r_n.$

We can see that the coefficient of $\displaystyle  x^k$ will be the $\displaystyle k$th symmetric sum. The $k$th symmetric sum is just the sum of the roots taken $k$ at a time. For example, the 4th symmetric sum is $\displaystyle r_1r_2r_3r_4 + r_1r_2r_3r_5+\cdots+r_{n-3}r_{n-2}r_{n-1}r_n.$ Notice that every possible combination of four roots shows up in this sum.

We now have two different expressions for $\displaystyle P(x)$. These must be equal. However, the only way for two polynomials to be equal for all values of $\displaystyle x$ is for each of their corresponding coefficients to be equal. So, starting with the coefficient of $\displaystyle  x^n$, we see that

$\displaystyle a_n = a_n$
$a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)$
$a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)$
$a_0 = (-1)^n a_n r_1r_2\cdots r_n$

More commonly, these are written with the roots on one side and the $\displaystyle a_i$ on the other (this can be arrived at by dividing both sides of all the equations by $\displaystyle a_n$).

If we denote $\displaystyle \sigma_k$ as the $\displaystyle k$th symmetric sum, then we can write those formulas more compactly as $\displaystyle \sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}$, for $\displaystyle 1\le k\le {n}$.

See also

Related Links

Mathworld's Article

Invalid username
Login to AoPS