牛顿迭代法

设 $f(x)$ 在 $\mathbb R$ 有二阶连续导数，且 $f'(x) \neq 0$ 则
$\forall x_0, x \in \mathbb R,$ 若 $f(x_0) = 0$ 则
$f(x_0) = f(x) + f'(x) \Delta x + \dfrac {1} {2} f''(\varepsilon) \Delta x ^2 = 0$
$\Rightarrow$
$x_0 - x = \Delta x = - \dfrac {1} { f'(x)} \left [ f(x) + \dfrac {1} {2} f''(\varepsilon) \Delta x ^2 \right ]$
$= - \dfrac { f(x) } { f'(x)} - \dfrac {1} {2} \dfrac { f''(\varepsilon)} { f'(x)} \Delta x ^2$
$\Rightarrow$
$x_0 = x - \dfrac { f(x) } { f'(x)} - \dfrac {1} {2} \dfrac { f''(\varepsilon)} { f'(x)} \Delta x ^2$
由于 $\lim \limits_{x \to x_0} \left [- \dfrac {1} {2} \dfrac { f''(\varepsilon)} { f'(x)} \Delta x ^2 \right ] = 0$
因此 $\lim \limits_{x \to x_0} \left [ x - \dfrac { f(x) } { f'(x)} \right ] = x_0$
令 $F(x) = x - \dfrac { f(x) } { f'(x)}$，则 $\lim \limits_{x \to x_0} F(x) = x_0$
由此得到迭代公式：
$x_{k + 1} = F(x_k) = x_k - \dfrac { f(x_k) } { f'(x_k)}$