# 應用於最優化的牛頓法

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

## 方法描述

${\displaystyle f(x)}$ ${\displaystyle x=x_{n}}$ 處的二階泰勒展開式${\displaystyle f_{T}(x)}$ 為:

${\displaystyle f_{T}(x)=f_{T}(x_{n}+\Delta x)\approx f(x_{n})+f^{\prime }(x_{n})\Delta x+{\frac {1}{2}}f^{\prime \prime }(x_{n})\Delta x^{2}.}$

${\displaystyle 0={\frac {\text{d}}{{\text{d}}\Delta x}}(f(x_{n})+f^{\prime }(x_{n})\Delta x+{\frac {1}{2}}f^{\prime \prime }(x_{n})\Delta x^{2})=f^{\prime }(x_{n})+f^{\prime \prime }(x_{n})\Delta x.}$

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

## 幾何意義

${\displaystyle ax_{n}^{2}+bx_{n}+c=f(x_{n}),}$
${\displaystyle 2ax_{n}+b=f^{\prime }(x_{n}),}$
${\displaystyle 2a=f^{\prime \prime }(x_{n}).}$

${\displaystyle x_{n+1}=-{\frac {b}{2a}}.}$

## 高維問題求解

${\displaystyle x_{n+1}=x_{n}-[\mathbf {H} f(x_{n})]^{-1}\nabla f(x_{n}),n\geq 0.}$

${\displaystyle x_{n+1}=x_{n}-\gamma [\mathbf {H} f(x_{n})]^{-1}\nabla f(x_{n}).}$

${\displaystyle [\mathbf {H} f(x_{n})]\Delta x=-\nabla f(x_{n})}$