多元函数的泰勒(Taylor)展开式

最新推荐文章于 2025-04-16 10:22:04 发布

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最新推荐文章于 2025-04-16 10:22:04 发布

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实际优化问题目标函数往往比较复杂，简化问题通常将目标函数在某点附近展开为Taylor多项式来逼近原函数：
$\cdot$ 一元函数在点 $x_k$ 处的泰勒展开为：

f (x) = f (x k) + (x - x k) f' (x k) + 1 2 ! (x - x k) 2 f'' (x k) + o n

$f(x)=f(x_k)+(x-x_k)f'(x_k)+\frac{1}{2!}(x-x_k)^2f''(x_k)+o^n$
$\cdot$ 二元函数在点 $(x_k,y_k)$ 处的泰勒展开为：

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) 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) f'' y y (x k, y k) + o n

$f(x,y)=f(x_k,y_k)+(x-x_k)f'_x(x_k,y_k)+(y-y_k)f'_y(x_k,y_k)+\frac{1}{2!}(x-x_k)f''_{xx}(x_k,y_k)+\frac{1}{2!}(x-x_k)(y-y_k)f''_{xy}(x_k,y_k)+\frac{1}{2!}(x-x_k)(y-y_k)f''_{yx}(x_k,y_k)+\frac{1}{2!}(y-y_k)f''_{yy}(x_k,y_k)+o^n$
⋅多元函数在点 $x_k$ 处的泰勒展开为：

f (x 1, x 2, \dots, x n) = f (x 1 k, x 2 k, \dots, x n k) + \sum i = 1 n (x i - x i k) f' x i (x 1, x 2, \dots, x n) + 1 2 ! \sum i, j = 1 n (x i - x i k) (x j - x j k) f'' i j (x 1, x 2, \dots, x n)

$f(x^1,x^2,\cdots,x^n)=f(x^1_k,x^2_k,\cdots,x^n_k)+\sum_{i=1}^{n}(x^i-x^i_k)f'_{x^i}(x^1,x^2,\cdots,x^n)+\frac{1}{2!}\sum_{i,j=1}^n(x^i-x^i_k)(x^j-x^j_k)f''_{ij}(x^1,x^2,\cdots,x^n)$
Taylor展开矩阵形式表达为：

f (x) = f (x k) + [\nabla f (x k)] T (x - x k) + 1 2 ! [x - x k] T H (x k) [x - x k] + o n

$f(x)=f(x_k)+[\nabla f(x_k)]^T(x-x_k)+\frac {1}{2!}[x-x_k]^TH(x_k)[x-x_k]+o^n$
其中：

H (x k) = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ \partial 2 f ( x k ) \partial x 2 1 \partial 2 f ( x k ) \partial x 2 \partial x 1 ⋮ \partial 2 f ( x k ) \partial x n \partial x 1 \partial 2 f ( x k ) \partial x 1 \partial x 2 \dots \partial 2 f ( x k ) \partial x 2 2 \dots ⋱ \partial 2 f ( x k ) \partial x n \partial x 2 \dots \partial 2 f ( x k ) \partial x 1 \partial x n \partial 2 f ( x k ) \partial x 2 \partial x n ⋮ \partial 2 f ( x k ) \partial x 2 n ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

$H(x_k)=\begin{bmatrix} \frac{\partial^2f(x_k)}{\partial x^2_1} &\frac{\partial^2f(x_k)}{\partial x_1\partial x_2} \cdots & \frac{\partial^2f(x_k)}{\partial x_1\partial x_n}\\\frac{\partial^2f(x_k)}{\partial x_2\partial x_1} &\frac{\partial^2f(x_k)}{\partial x^2_2} \cdots & \frac{\partial^2f(x_k)}{\partial x_2\partial x_n}\\ \vdots & \ddots & \vdots \\ \frac{\partial^2f(x_k)}{\partial x_n\partial x_1} &\frac{\partial^2f(x_k)}{\partial x_n\partial x_2} \cdots & \frac{\partial^2f(x_k)}{\partial x^2_n} \end{bmatrix}$