多元函式的泰勒(Taylor)展開式

阿新 • • 發佈：2019-01-03

實際優化問題的目標函式往往比較複雜。為了使問題簡化，通常將目標函式在某點附近展開為泰勒(Taylor)多項式來逼近原函式。

一元函式在點 $x_{k}$ 處的泰勒展開式為：

$f (x) = f (x_{k}) + (x - x_{k}) f^{'} (x_{k}) + \frac{1}{2!} (x - x_{k})^{2} f^{″} (x_{k}) + o^{n}$
二元函式在點 $(x_{k}, y_{k})$ 處的泰勒展開式為：

$f (x, y) = f (x_{k}, y_{k}) + (x - x_{k}) f_{x}^{'} (x_{k}, y_{k}) + (y - y_{k}) f_{y}^{'} (x_{k}, y_{k}) + \frac{1}{2!} (x - x_{k})^{2} f_{x x}^{″} (x_{k}, y_{k}) + \frac{1}{2!} (x - x_{k}) (y - y_{k}) f_{x y}^{″} (x_{k}, y_{k}) + \frac{1}{2!} (x - x_{k}) (y - y_{k}) f_{y x}^{″} (x_{k}, y_{k}) + \frac{1}{2!} (y - y_{k})^{2} f_{y y}^{″} (x_{k}, y_{k}) + o^{n}$
多元函式(n)在點 $x_{k}$ 處的泰勒展開式為：

$f (x^{1}, x^{2}, \dots, x^{n}) = f (x_{k}^{1}, x_{k}^{2}, \dots, x_{k}^{n}) + \sum_{i = 1}^{n} (x^{i} - x_{k}^{i}) f_{x^{i}}^{'} (x_{k}^{1}, x_{k}^{2}, \dots, x_{k}^{n}) + \frac{1}{2!} \sum_{i, j = 1}^{n} (x^{i} - x_{k}^{i}) (x^{j} - x_{k}^{j}) f_{i j}^{″} (x_{k}^{1}, x_{k}^{2}, \dots, x_{k}^{n}) + o^{n}$