# 韦达定理

$F(x)=\sum_{m=1}^{n}a_mx^m + a_0 = 0$

$\begin{cases} \sum_{k=1}^{n}x_k = -\frac{a_{n-1}}{a_n} \\ \sum_{m=1}^{n-1}\prod_{k=m+1}^{n}x_mx_k = \frac{a_{n-2}}{a_n} \\ \vdots \\\Pi_{k=1}^{n}x_k = (-1)^n \frac{a_0}{a_n}. \end{cases}$

$x_1\,$$x_2\,$是一元二次方程$ax^2+bx+c=0\,$的两根，那么
$x_1+x_2=-\frac{b}{a}$$x_1x_2=\frac{c}{a}$

$\alpha + \beta =-\frac{b}{a}$$\alpha \beta =\frac{c}{a}$
$\alpha\,$$\beta\,$必定是一元二次方程$ax^2+bx+c=0\,$的两个根。

## 定理特例的证明

$x_1\,$$x_2\,$一元二次方程$ax^2+bx+c=0\,$ (a≠0)的两个解，且不妨令$x_1 \ge x_2$

$x_1=\frac{-b + \sqrt {b^2-4ac}}{2a}$$x_2=\frac{-b - \sqrt {b^2-4ac}}{2a}$

$x_1+x_2=\frac{-b + \sqrt {b^2-4ac} + \left (-b \right) - \sqrt {b^2-4ac}}{2a} =-\frac{b}{a}$

$x_1x_2=\frac{ \left (-b + \sqrt {b^2-4ac} \right) \left (-b - \sqrt {b^2-4ac} \right)}{\left (2a \right)^2} =\frac{c}{a}$

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$.........(1)

$(x-x_1)(x-x_2) = 0$,

∴对照式子(1)与(2)，得 $x_1+x_2=-\frac{b}{a}$$x_1x_2=\frac{c}{a}$

## 定理的证明

$x_1\,,x_2\,,\cdots \,,x_n$一元n次方程$\sum_{m=1}^{n}a_mx^m + a_0 = 0,(n\ge 1, a_n\ne0)$的n个解。

$\begin{cases} a_{n-1}=a_n\sum_{k=1}^{n}x_k \\ a_{n-2}=a_n\sum_{m=1}^{n-1}(\Pi_{k=m}^{m+1}(-x_k)) \\ a_{n-3}=a_n\sum_{m=1}^{n-2}(\Pi_{k=m}^{m+2}(-x_k)) \\ \vdots \\ a_0=a_n\Pi_{k=1}^{n}(-x_k). \end{cases}$

$\begin{cases} \sum_{k=1}^{n}x_k = (-1)^1\,\frac{a_{n-1}}{a_n} \\ \sum_{m=1}^{n-1}(\Pi_{k=m}^{m+1}(-x_k)) = (-1)^2\,\frac{a_{n-2}}{a_n} \\ \sum_{m=1}^{n-2}(\Pi_{k=m}^{m+2}(-x_k)) = (-1)^3\,\frac{a_{n-3}}{a_n} \\ \vdots \\ \Pi_{k=1}^{n}(-x_k) = (-1)^n\,\frac{a_0}{a_n} . \end{cases}$

## 其它有关定理

$\sum_{k=1}^n \frac{1}{x_0-x_k}=\frac{F'(x_0)}{F(x_0)}$

$F(x)= (x-x_1)(x-x_2)...(x-x_n)$

$F'(x)=(x-x_2)...(x-x_n)+(x-x_1)...(x-x_n)+...+(x-x_1)(x-x_2)...(x-x_{n-1})$

$\frac{F'(x)}{F(x)}=\frac{1}{x-x_1}+\frac{1}{x-x_2}+...+\frac{1}{x-x_n}$