多元函数泰勒展开式

实际优化问题的目标函数往往比较复杂。为了使问题简化，通常将目标函数在某点附近展开为泰勒(Taylor)多项式来逼近原函数

一元函数\(f(x)\)在\(x_k\)处的泰勒展开式为

\[f(x)=f(x_k)+(x-x_k)f^\prime(x_k)+\frac{1}{2!}(x-x_k)^2f^{\prime\prime}(x_k)+O^n \]

二元函数\(f(x,y)\)在\((x_k,y_k)\)处泰勒展开式为

\[f(x,y)=f(x_k,y_k)+(x-x_k)f^\prime_x(x_k,y_k)+(y-y_k)f^\prime_y(x_k,y_k)+ \\\\ \frac{1}{2!}(x-x_k)^2f^{\prime\prime}_{xx}(x_k,y_k)+\frac{1}{2!}(y-y_k)^2f^{\prime\prime}_{yy}(x_k,y_k)+ \\\\ \frac{1}{2!}(x-x_k)(y-y_k)f^{\prime\prime}_{xy}(x_k,y_k)+\frac{1}{2!}(x-x_k)(y-y_k)f^{\prime\prime}_{yx}(x_k,y_k)+O^n \]

三元函数\(f(x^1,x^2,x^3)\)在\(\vec{x_k}=(x^1_k,x^2_k,x^3_k)\)处泰勒展开式为，（二阶项展开共有9项）

\[f(x^1,x^2,x^3)=f(x^1_k,x^2_k,x^3_k)+\sum_{i=1}^{3}(x^i-x^i_k)f^\prime_{x^i}(x^1_k,x^2_k,x^3_k)+\\\\ \frac{1}{2!}\sum_{i=1}^{3}\sum_{j=1}^{3}(x^i-x^i_k)(x^j-x^j_k)f^{\prime\prime}_{ij}(x^1_k,x^2_k,x^3_k)+O^n \]

多元函数\(f(x^1,x^2,...,x^n)\)在\(\vec{x_k}=(x^1_k,x^2_k,...,x^n_k)\)处泰勒展开式为

\[f(x^1,x^2,...,x^n)=f(x^1_k,x^2_k,...,x^n_k)+\sum_{i=1}^{n}(x^i-x^i_k)f^\prime_{x^i}(x^1_k,x^2_k,...,x^n_k)+\\\\ \frac{1}{2!}\sum_{i=1}^{n}\sum_{j=1}^{n}(x^i-x^i_k)(x^j-x^j_k)f^{\prime\prime}_{ij}(x^1_k,x^2_k,...,x^n_k)+O^n \]

写成矩阵形式

其中，\(H(x_k)\)为Hessian矩阵

参考：https://blog.csdn.net/red_stone1/article/details/70260070