多元函数的泰勒展开式公式推导

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

f(x)=f(xk)+(x−xk)f′(xk)+12!(x−xk)2f′′(xk)+onf(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(xk,yk)\left(x_{k}, y_{k}\right)(xk​,yk​)处的泰勒展开式为：

f(x,y)=f(xk,yk)+(x−xk)fx′(xk,yk)+(y−yk)fy′(xk,yk)+12!(x−xk)2fxx′′(xk,yk)+12!(x−xk)(y−yk)fxy′′(xk,yk)+12!(x−xk)(y−yk)fyx′′(xk,yk)+12!(y−yk)2fyy′′(xk,yk)+on\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(xk,yk)+[fx′(xk,yk)fy′(xk,yk)][x−xky−yk]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​​]

+12![x−xky−yk][fxx′′(xk,yk)fxy′′(xk,yk)fyx′′(xk,yk)fyy′′(xk,yk)][x−xky−yk]+\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(xk1,xk2,…,xkn)\left(x_{k}^{1}, x_{k}^{2}, \dots, x_{k}^{n}\right)(xk1​,xk2​,…,xkn​)处的泰勒展开公式：

f(x1,x2,…,xn)=f(xk1,xk2,…,xkn)+∑i=1n(xi−xki)fxi′(xk1,xk2,…,xkn)+12!∑i,j=1n(xi−xki)(xj−xkj)fij′′(xk1,xk2,…,xkn)+on\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(xk)+[∇f(xk)]T(x−xk)+12![x−xk]TH(xk)[x−xk]+onf(\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(xk)=[∂2f(xk)∂x12∂2f(xk)∂x1∂x2⋯∂2f(xk)∂x1∂xn∂2f(xk)∂x2∂x1∂2f(xk)∂x22⋯∂2f(xk)∂x2∂xn⋮⋮⋱⋮∂2f(xk)∂xn∂x1∂2f(xk)∂xn∂x2⋯∂2f(xk)∂xn2]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(xk)=[fx′(xk,yk)fy′(xk,yk)]\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−xk]=[x−xky−yk]\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(xk)=[fxx′′(xk,yk)fxy′′(xk,yk)fyx′′(xk,yk)fyy′′(xk,yk)]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​)​]