# 牛顿法

## 方法说明

${\displaystyle 0=(x-x_{0})\cdot f'(x_{0})+f(x_{0})}$

${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}}$

${\displaystyle f'(x)\neq 0}$; 对于所有${\displaystyle x\in I}$，其中${\displaystyle I}$为区间[αr, α + r]，且${\displaystyle x_{0}}$在区间其中${\displaystyle I}$内，即 ${\displaystyle r\geqslant \left|a-x_{0}\right|}$ 的;

${\displaystyle x_{0}}$足够接近根 α

## 其它例子

### 第一个例子

${\displaystyle {\begin{matrix}x_{1}&=&x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}&=&0.5-{\frac {\cos(0.5)-0.5^{3}}{-\sin(0.5)-3\times 0.5^{2}}}&=&1.112141637097\\x_{2}&=&x_{1}-{\frac {f(x_{1})}{f'(x_{1})}}&=&\vdots &=&{\underline {0.}}909672693736\\x_{3}&=&\vdots &=&\vdots &=&{\underline {0.86}}7263818209\\x_{4}&=&\vdots &=&\vdots &=&{\underline {0.86547}}7135298\\x_{5}&=&\vdots &=&\vdots &=&{\underline {0.8654740331}}11\\x_{6}&=&\vdots &=&\vdots &=&{\underline {0.865474033102}}\end{matrix}}}$

### 第二个例子

${\displaystyle a}$${\displaystyle m}$次方根。

${\displaystyle x^{m}-a=0}$

${\displaystyle f(x)=x^{m}-a}$${\displaystyle f'(x)=mx^{m-1}}$

${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}}$

${\displaystyle x_{n+1}=x_{n}-{\frac {x_{n}^{m}-a}{mx_{n}^{m-1}}}}$

${\displaystyle x_{n+1}=x_{n}-{\frac {x_{n}}{m}}(1-ax_{n}^{-m})}$

（或 ${\displaystyle x_{n+1}=x_{n}-{\frac {1}{m}}\left(x_{n}-a{\frac {x_{n}}{x_{n}^{m}}}\right)}$

## 应用

### 求解最值问题

${\displaystyle x_{n+1}=x_{n}-{\frac {f^{\prime }(x_{n})}{f^{\prime \prime }(x_{n})}}.}$