简介

埃尔米特插值不仅要求插值多项式 $P_n(x)$ 与被近似函数 $f(x)$ 的若干函数值重合，而且要求若干阶导数也重合。

INFO

$N+\phantom{0}$$\phantom{0}1$ 个条件可以确定 $N$ 次埃米尔特插值多项式。

泰勒插值

要求 $P_n(x)$ 与 $f(x)$ 在 $x_0$ 处函数值重合，且 $n$ 阶内导数均重合。

具体来说，需要满足以下 $n+\phantom{0}$$\phantom{0}1$ 个条件：

$$\{\begin{array}{l}{P}_n(x_0)=f(x_0)\\ P_n^{\prime }(x_0)=f^{\prime }(x_0)\\ P_n^{″}(x_0)=f^{″}(x_0)\\ \cdots \\ P_n^{(n)}(x_0)=f^{(n)}(x_0)\end{array}$$

设 $P_n(x)=\phantom{0}$$\phantom{0}{a}_0+\phantom{0}$$\phantom{0}{a}_1(x-\phantom{0}$$\phantom{0}{x}_0)+\phantom{0}$$\phantom{0}\cdots +\phantom{0}$$\phantom{0}{a}_n(x-\phantom{0}$$\phantom{0}{x}_0{)}^n$，代入以上，解得 $a_k=\phantom{0}$$\phantom{0}{\displaystyle \frac{f^{(k)}(x_0)}{k!}}$。即

$$P_n(x)=f(x_0)+\sum _{k=1}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k$$

余项

若 $f^{(n+1)}$ 在 $[x_0,\phantom{0}$$\phantom{0}x]$ 或 $[x,\phantom{0}$$\phantom{0}{x}_0]$ 上存在且连续，则泰勒插值产生的余项为

$$R_n(x)=f(x)-P_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$$

其中 $\xi$ 位于 $x_0$ 和 $x$ 之间（不包括）。

证

构造辅助函数

$$\phi (t)=R_n(t)-\frac{R_n(x)}{(x-x_0{)}^{n+1}}(t-x_0{)}^{n+1}$$

则 $\phi (t)=\phantom{0}$$\phantom{0}0$ 在 $t=\phantom{0}$$\phantom{0}{x}_0$ 有一个 $n+\phantom{0}$$\phantom{0}1$ 重根，在 $t=\phantom{0}$$\phantom{0}x$ 有一个一重根，共 $n+\phantom{0}$$\phantom{0}2$ 个零点。

应用 $n+\phantom{0}$$\phantom{0}1$ 次罗尔（Rolle）定理得，存在 $\xi$ 位于 $x_0$ 和 $x$ 之间（不包括），使得

$${\phi }^{(n+1)}(\xi )=0$$$$R_n^{(n+1)}(\xi )-\frac{R_n(x)}{(x_0-x{)}^{n+1}}\cdot (n+1)!=0$$$$R_n(x)=\frac{f^{(n+1)}(\xi )}{(n+1)!}(x-x_0{)}^{n+1}$$

两点三次埃米尔特插值

要求三次多项式 $H_3(x)$ 满足以下 $4$ 个条件：

$$\{\begin{array}{l}{H}_3(x_0)=f(x_0)\\ H_3(x_1)=f(x_1)\\ H_3^{\prime }(x_0)=f^{\prime }(x_0)\\ H_3^{\prime }(x_1)=f^{\prime }(x_1)\end{array}$$

设 $H_3(x)=\phantom{0}$$\phantom{0}f(x_0){\alpha }_0(x)+\phantom{0}$$\phantom{0}f(x_1){\alpha }_1(x)+\phantom{0}$$\phantom{0}{f}^{\prime }(x_0){\beta }_0(x)+\phantom{0}$$\phantom{0}{f}^{\prime }(x_1){\beta }_1(x)$，代入以上，得到

${\alpha }_0(x_0)=\phantom{0}$$\phantom{0}1$	${\alpha }_0(x_1)=0$	${\alpha }_0^{\prime }(x_0)=0$	${\alpha }_0^{\prime }(x_1)=0$
${\alpha }_1(x_0)=0$	${\alpha }_1(x_1)=\phantom{0}$$\phantom{0}1$	${\alpha }_1^{\prime }(x_0)=0$	${\alpha }_1^{\prime }(x_1)=0$
${\beta }_0(x_0)=0$	${\beta }_0(x_1)=0$	${\beta }_0^{\prime }(x_0)=\phantom{0}$$\phantom{0}1$	${\beta }_0^{\prime }(x_1)=0$
${\beta }_1(x_0)=0$	${\beta }_1(x_1)=0$	${\beta }_1^{\prime }(x_0)=0$	${\beta }_1^{\prime }(x_1)=\phantom{0}$$\phantom{0}1$

由于 ${\alpha }_0(x_1)=\phantom{0}$$\phantom{0}{\alpha }_0^{\prime }(x_1)=\phantom{0}$$\phantom{0}0$，故设 ${\alpha }_0(x)=\phantom{0}$$\phantom{0}(x-\phantom{0}$$\phantom{0}{x}_1{)}^2(ax+\phantom{0}$$\phantom{0}b)$。再代入 ${\alpha }_0(x_0)=\phantom{0}$$\phantom{0}1,\phantom{0}$$\phantom{0}{\alpha }_0(x_0)=\phantom{0}$$\phantom{0}0$ 有

$$\{\begin{array}{l}{\alpha }_0(x_0)=(x_0-x_1{)}^2(a{x}_0+b)=1\\ {\alpha }_0^{\prime }(x_0)=2(x_0-x_1)(a{x}_0+b)+a(x_0-x_1{)}^2=0\end{array}$$

解得

$$\{\begin{array}{l}a=-{\displaystyle \frac{2}{(x_0-x_1{)}^3}}\\ b={\displaystyle \frac{1}{(x_0-x_1{)}^2}}+{\displaystyle \frac{2x_0}{(x_0-x_1{)}^3}}\end{array}$$

故

$$\begin{array}{rl}{\alpha }_0(x)& =(x-x_1{)}^2(ax+b)\\ & =(x-x_1{)}^2[-{\displaystyle \frac{2}{(x_0-x_1{)}^3}}x+{\displaystyle \frac{1}{(x_0-x_1{)}^2}}+{\displaystyle \frac{2x_0}{(x_0-x_1{)}^3}}]\\ & ={\left(\frac{x-x_1}{x_0-x_1}\right)}^2(1+\frac{2x_0}{x_0-x_1}-\frac{2x}{x_0-x_1})\\ & ={\left(\frac{x-x_1}{x_0-x_1}\right)}^2(1+2\frac{x-x_0}{x_1-x_0})\\ & ={\ell }_0^2(x)[1+2{\ell }_1(x)]\end{array}$$

类似地，有

$$\begin{array}{c}{\alpha }_1(x)={\left(\frac{x-x_0}{x_1-x_0}\right)}^2(1+2\frac{x-x_1}{x_0-x_1})={\ell }_1^2(x)[1+2{\ell }_0(x)]\\ {\beta }_0(x)=(x-x_0){\left(\frac{x-x_1}{x_0-x_1}\right)}^2=(x-x_0){\ell }_0^2(x)\\ {\beta }_1(x)=(x-x_1){\left(\frac{x-x_0}{x_1-x_0}\right)}^2=(x-x_1){\ell }_1^2(x)\end{array}$$

即

$$\begin{array}{rl}{H}_3(x)& =f(x_0)\cdot {\ell }_0^2(x)[1+2{\ell }_1(x)]+f(x_1)\cdot {\ell }_1^2(x)[1+2{\ell }_0(x)]\\ & +f^{\prime }(x_0)\cdot (x-x_0){\ell }_0^2(x)+f^{\prime }(x_1)\cdot (x-x_1){\ell }_1^2(x)\end{array}$$

余项

若 $f^{(4)}(x)$ 在 $[x_0,\phantom{0}$$\phantom{0}{x}_1]$ 上连续，则两点三次埃米尔特插值产生的余项为

$$R_3(x)=f(x)-H_3(x)=\frac{f^{(4)}(\xi )}{4!}(x-x_0{)}^2(x-x_1{)}^2,\xi \in (x_0,x_1)$$

证

任取 $\hat{x}\in (x_0,\phantom{0}$$\phantom{0}{x}_1)$，构造辅助函数

$$\phi (t)=R_3(t)-\frac{R_3(\hat{x})}{(\hat{x}-x_0{)}^2(\hat{x}-x_1{)}^2}(t-x_0{)}^2(t-x_1{)}^2$$

则 $\phi (t)=\phantom{0}$$\phantom{0}0$ 有一个一重根 $\hat{x}$ 和两个二重根 $x_0,\phantom{0}$$\phantom{0}{x}_1$，共有 $5$ 个零点。

应用 $4$ 次罗尔（Rolle）定理即可得证。