多元函数的泰勒展开式公式推导

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多元函数的泰勒展开式公式推导

2023-07-31 02:42| 来源: 网络整理| 查看: 265

前言

最优化问题中总会遇到多元函数的泰勒展开公式,没有推导过总是感觉很抽象,参考如下链接,文章讲得很清晰。

多元函数的泰勒(Taylor)展开式

推导过程 一元函数在点 X k \mathcal{X}_{k} Xk​处的泰勒展开式为: f ( x ) = f ( x k ) + ( x − x k ) f ′ ( x k ) + 1 2 ! ( x − x k ) 2 f ′ ′ ( x k ) + o n f(x)=f\left(x_{k}\right)+\left(x-x_{k}\right) f^{\prime}\left(x_{k}\right)+\frac{1}{2 !}\left(x-x_{k}\right)^{2} f^{\prime \prime}\left(x_{k}\right)+o^{n} f(x)=f(xk​)+(x−xk​)f′(xk​)+2!1​(x−xk​)2f′′(xk​)+on二元函数在点 ( x k , y k ) \left(x_{k}, y_{k}\right) (xk​,yk​)处的泰勒展开式为: f ( x , y ) = f ( x k , y k ) + ( x − x k ) f x ′ ( x k , y k ) + ( y − y k ) f y ′ ( x k , y k ) + 1 2 ! ( x − x k ) 2 f x x ′ ′ ( x k , y k ) + 1 2 ! ( x − x k ) ( y − y k ) f x y ′ ′ ( x k , y k ) + 1 2 ! ( x − x k ) ( y − y k ) f y x ′ ′ ( x k , y k ) + 1 2 ! ( y − y k ) 2 f y y ′ ′ ( x k , y k ) + o n \begin{array}{c} f(x, y)=f\left(x_{k}, y_{k}\right)+\left(x-x_{k}\right) f_{x}^{\prime}\left(x_{k}, y_{k}\right)+\left(y-y_{k}\right) f_{y}^{\prime}\left(x_{k}, y_{k}\right) \\ +\frac{1}{2 !}\left(x-x_{k}\right)^{2} f_{x x}^{\prime \prime}\left(x_{k}, y_{k}\right)+\frac{1}{2 !}\left(x-x_{k}\right)\left(y-y_{k}\right) f_{x y}^{\prime \prime}\left(x_{k}, y_{k}\right) \\ +\frac{1}{2 !}\left(x-x_{k}\right)\left(y-y_{k}\right) f_{y x}^{\prime \prime}\left(x_{k}, y_{k}\right)+\frac{1}{2 !}\left(y-y_{k}\right)^{2} f_{y y}^{\prime \prime}\left(x_{k}, y_{k}\right) \\ +o^{n} \end{array} f(x,y)=f(xk​,yk​)+(x−xk​)fx′​(xk​,yk​)+(y−yk​)fy′​(xk​,yk​)+2!1​(x−xk​)2fxx′′​(xk​,yk​)+2!1​(x−xk​)(y−yk​)fxy′′​(xk​,yk​)+2!1​(x−xk​)(y−yk​)fyx′′​(xk​,yk​)+2!1​(y−yk​)2fyy′′​(xk​,yk​)+on​ 上公式化成矩阵形式如下,可以和后面的多元函数泰勒展开的矩阵形式相互对照理解。 f ( x , y ) = f ( x k , y k ) + [ f x ′ ( x k , y k ) f y ′ ( x k , y k ) ] [ x − x k y − y k ] f(x, y)=f\left(x_{k}, y_{k}\right)+\left[f_{x}^{\prime}\left(x_{k}, y_{k}\right) \quad f_{y}^{\prime}\left(x_{k}, y_{k}\right)\right]\left[\begin{array}{c} x-x_{k} \\ y-y_{k} \end{array}\right] f(x,y)=f(xk​,yk​)+[fx′​(xk​,yk​)fy′​(xk​,yk​)][x−xk​y−yk​​] + 1 2 ! [ x − x k y − y k ] [ f x x ′ ′ ( x k , y k ) f x y ′ ′ ( x k , y k ) f y x ′ ′ ( x k , y k ) f y y ′ ′ ( x k , y k ) ] [ x − x k y − y k ] +\frac{1}{2 !}\left[x-x_{k} \quad y-y_{k}\right]\left[\begin{array}{cc} f_{x x}^{\prime \prime}\left(x_{k}, y_{k}\right) & f_{x y}^{\prime \prime}\left(x_{k}, y_{k}\right) \\ f_{y x}^{\prime \prime}\left(x_{k}, y_{k}\right) & f_{y y}^{\prime \prime}\left(x_{k}, y_{k}\right) \end{array}\right]\left[\begin{array}{c} x-x_{k} \\ y-y_{k} \end{array}\right] +2!1​[x−xk​y−yk​][fxx′′​(xk​,yk​)fyx′′​(xk​,yk​)​fxy′′​(xk​,yk​)fyy′′​(xk​,yk​)​][x−xk​y−yk​​]多元函数在 ( x k 1 , x k 2 , … , x k n ) \left(x_{k}^{1}, x_{k}^{2}, \dots, x_{k}^{n}\right) (xk1​,xk2​,…,xkn​) 处的泰勒展开公式: f ( x 1 , x 2 , … , x n ) = f ( x k 1 , x k 2 , … , x k n ) + ∑ i = 1 n ( x i − x k i ) f x i ′ ( x k 1 , x k 2 , … , x k n ) + 1 2 ! ∑ i , j = 1 n ( x i − x k i ) ( x j − x k j ) f i j ′ ′ ( x k 1 , x k 2 , … , x k n ) + o n \begin{array}{c} f\left(x^{1}, x^{2}, \ldots, x^{n}\right)=f\left(x_{k}^{1}, x_{k}^{2}, \ldots, x_{k}^{n}\right)+\sum_{i=1}^{n}\left(x^{i}-x_{k}^{i}\right) f_{x^{i}}^{\prime}\left(x_{k}^{1}, x_{k}^{2}, \ldots, x_{k}^{n}\right) \\ +\frac{1}{2 !} \sum_{i, j=1}^{n}\left(x^{i}-x_{k}^{i}\right)\left(x^{j}-x_{k}^{j}\right) f_{i j}^{\prime \prime}\left(x_{k}^{1}, x_{k}^{2}, \ldots, x_{k}^{n}\right) \\ +o^{n} \end{array} f(x1,x2,…,xn)=f(xk1​,xk2​,…,xkn​)+∑i=1n​(xi−xki​)fxi′​(xk1​,xk2​,…,xkn​)+2!1​∑i,j=1n​(xi−xki​)(xj−xkj​)fij′′​(xk1​,xk2​,…,xkn​)+on​多元函数泰勒展开式写成矩阵的形式: f ( x ) = f ( x k ) + [ ∇ f ( x k ) ] T ( x − x k ) + 1 2 ! [ x − x k ] T H ( x k ) [ x − x k ] + o n f(\mathbf{x})=f\left(\mathbf{x}_{k}\right)+\left[\nabla f\left(\mathbf{x}_{k}\right)\right]^{T}\left(\mathbf{x}-\mathbf{x}_{k}\right)+\frac{1}{2 !}\left[\mathbf{x}-\mathbf{x}_{k}\right]^{T} H\left(\mathbf{x}_{k}\right)\left[\mathbf{x}-\mathbf{x}_{k}\right]+o^{n} f(x)=f(xk​)+[∇f(xk​)]T(x−xk​)+2!1​[x−xk​]TH(xk​)[x−xk​]+on H ( x k ) = [ ∂ 2 f ( x k ) ∂ x 1 2 ∂ 2 f ( x k ) ∂ x 1 ∂ x 2 ⋯ ∂ 2 f ( x k ) ∂ x 1 ∂ x n ∂ 2 f ( x k ) ∂ x 2 ∂ x 1 ∂ 2 f ( x k ) ∂ x 2 2 ⋯ ∂ 2 f ( x k ) ∂ x 2 ∂ x n ⋮ ⋮ ⋱ ⋮ ∂ 2 f ( x k ) ∂ x n ∂ x 1 ∂ 2 f ( x k ) ∂ x n ∂ x 2 ⋯ ∂ 2 f ( x k ) ∂ x n 2 ] H\left(\mathbf{x}_{k}\right)=\left[\begin{array}{cccc} \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{1}^{2}} & \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{1} \partial x_{2}} & \cdots & \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{1} \partial x_{n}} \\ \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{2} \partial x_{1}} & \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{2}^{2}} & \cdots & \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{2} \partial x_{n}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{n} \partial x_{1}} & \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{n} \partial x_{2}} & \cdots & \frac{\partial^{2} f\left(x_{k}\right)}{\partial x_{n}^{2}} \end{array}\right] H(xk​)=⎣⎢⎢⎢⎢⎢⎡​∂x12​∂2f(xk​)​∂x2​∂x1​∂2f(xk​)​⋮∂xn​∂x1​∂2f(xk​)​​∂x1​∂x2​∂2f(xk​)​∂x22​∂2f(xk​)​⋮∂xn​∂x2​∂2f(xk​)​​⋯⋯⋱⋯​∂x1​∂xn​∂2f(xk​)​∂x2​∂xn​∂2f(xk​)​⋮∂xn2​∂2f(xk​)​​⎦⎥⎥⎥⎥⎥⎤​当为二元时, ∇ f ( x k ) = [ f x ′ ( x k , y k ) f y ′ ( x k , y k ) ] \nabla f\left(x_{k}\right)=\left[\begin{array}{l} f_{x}^{\prime}\left(x_{k}, y_{k}\right) \\ f_{y}^{\prime}\left(x_{k}, y_{k}\right) \end{array}\right] ∇f(xk​)=[fx′​(xk​,yk​)fy′​(xk​,yk​)​]

[ x − x k ] = [ x − x k y − y k ] \left[x-x_{k}\right]=\left[\begin{array}{l} x-x_{k} \\ y-y_{k} \end{array}\right] [x−xk​]=[x−xk​y−yk​​] H ( x k ) = [ f x x ′ ′ ( x k , y k ) f x y ′ ′ ( x k , y k ) f y x ′ ′ ( x k , y k ) f y y ′ ′ ( x k , y k ) ] H\left(x_{k}\right)=\left[\begin{array}{ll} f_{x x}^{\prime \prime}\left(x_{k}, y_{k}\right) & f_{x y}^{\prime \prime}\left(x_{k}, y_{k}\right) \\ f_{y x}^{\prime \prime}\left(x_{k}, y_{k}\right) & f_{y y}^{\prime \prime}\left(x_{k}, y_{k}\right) \end{array}\right] H(xk​)=[fxx′′​(xk​,yk​)fyx′′​(xk​,yk​)​fxy′′​(xk​,yk​)fyy′′​(xk​,yk​)​]



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