# 函数插值法

## Hermite插值

$\forall i\in[0,n]: H(x_i)=y_i,H'(x_i)=y'_i\tag{5.1}$

### 基函数构造法

$h_i(x_j) = 0, j\not =i$ $h_i(x_j) = 1, j=i$

$h_i'(x_j) \equiv 0$

$H'_i(x_j) = 0, j\not =i$ $H'_i(x_j) = 1, j=i$

$H_i(x_j) \equiv 0$

\left\{ \begin{aligned} & h_i(x) = [a+b(x-x_i)]l_i^2(x) \\ & a = 1 \\ & b = -2l_i'(x_i) \end{aligned} \right. \left\{ \begin{aligned} & H_i(x) = cl_i^2(x) \\ & c = 1 \end{aligned} \right.

$l_i(x) = \frac{\prod_{j=0, j\not ={i}}^n(x-x_j)}{\prod_{j=0, j\not ={i}}^n(x_i-x_j)}$

$y_i[1-2l'_i(x_i)(x-x_i)]l_i^2(x)+y_i'(x-x_i)l_i^2(x)$

$H(x) = \sum_{i=0}^n \{y_i[1-2l'_i(x_i)(x-x_i)]l_i^2(x)+y_i'(x-x_i)l_i^2(x)\}$

Hermite插值法拥有误差表达式

$R_n(x) = \frac{f^{(2n+2)}(\xi)}{(2n+2)!}\omega_{n+1}^2(x)$

$R_n(x) = \frac{f^{(n+m+2)}(\xi)}{(n+m+2)!}\omega_{n+1}(x)\prod^{(m-1)}_{i=0}(x-x_i)$

### 降阶法

$N(x), s.t.\forall i\in[0,n], N(x_i) = y_i$

$H(x)-N(x) = P(x)\prod_{i=0}^n(x-x_i)$

Copyright AmachiInori 2017-2021. All Right Reserved.