4. Hermite插值

理论和应用中提出的某些插值问题，要求插值函数 $p(x)$ 具有一定的光滑度，即在插值节点处满足一定的导数条件，这类插值问题称为Hermite插值问题。题目大多以三次Hermite插值为主。三次Hermite插值需要四个条件，二次Hermite插值需要三个条件，分类如下：

4.1 待定系数法求Hermite插值

Hermite插值我们一定会知道给定点的函数值，和某些点的导数值。由多项式插值得到的插值函数是唯一的，我们可以先用已知点构造拉格朗日或牛顿插值，得到一部分插值多项式，再根据导数值往插值多项式加入待定系数项，并代入导数值求解。这种方法比后两种快很多。

[!example]-
已知
$$\begin{array}{cccccccc}x& ||& 0& |& 1& |& 2& |\\ =& =& =& =& =& =& =& =\\ f(x)& ||& 5& |& 3& |& 9& |\\ -& -& -& -& -& -& -& -\\ f^{\prime }(x)& ||& -4& |& 0& |& & |\end{array}$$
用已知的五个条件求四次Hermite插值多项式。
解：先用三个点的函数值求拉格朗日插值多项式
$$\begin{array}{rl}L(x)=& 5\frac{(x-1)(x-2)}{(0-1)(0-2)}+3\frac{(x-0)(x-2)}{(1-0)(1-2)}+9\frac{(x-0)(x-1)}{(2-0)(2-1)}\\ & \\ =& 4x^2-6x+5\end{array}$$
令
$$\begin{array}{rl}H(x)=& L(x)+(ax+b)(x-0)(x-1)(x-2)\\ & \\ H^{\prime }(x)=& 8x-6+4a{x}^3+(3b-9a)x^2+(4a-6b)x+2b\\ & \\ H^{\prime }(0)=& -6+2b=-4\\ & \\ H^{\prime }(1)=& 2+4a+3b-9a+4a-6b+2b=2-a-b=0\end{array}$$
$$\therefore a=1,b=1$$
$$H(x)=x^4-2x^3+3x^2-4x+5$$

4.2 两点三次Hermite插值

确定三次插值多项式 $H_3(x)$ 使满足：
$$\begin{array}{r}{H}_3(x_0)=f(x_0)=y_0,H_3(x_1)=f(x_1)=y_1\\ \\ H_3^{\prime }(x_0)=f^{\prime }(x_0)=y_0^{\prime },H_3^{\prime }(x_1)=f^{\prime }(x_1)=y_1^{\prime }\end{array}$$
方法：基函数法，基函数为三次多项式。
步骤：

1. 设 $H_3(x)$ 和 $H_3^{\prime }(x)$ 的表达式
   $$H_3(X)={\varphi }_0(x)y_0+{\varphi }_1(x)y_1+{\psi }_0(x)y_1^{\prime }+{\psi }_1(x)y_1^{\prime }\text{\hspace{1em}(1)}$$
   $$H_3^{\prime }(X)={\varphi }_0^{\prime }(x)y_0+{\varphi }_1^{\prime }(x)y_1+{\psi }_0^{\prime }(x)y_1^{\prime }+{\psi }_1^{\prime }(x)y_1^{\prime }\text{\hspace{1em}(2)}$$
2. 根据表达式列表
   $$\begin{array}{cccc}将x_0代入(1)& 将x_1代入(1)& 将x_0代入(2)& 将x_1代入(2)\\ & & & \\ {\varphi }_0(x_0)=1& {\varphi }_0(x_1)=0& {\varphi }_0^{\prime }(x_0)=0& {\varphi }_0^{\prime }(x_1)=0\\ & & & \\ {\varphi }_1(x_0)=0& {\varphi }_1(x_1)=1& {\varphi }_1^{\prime }(x_0)=0& {\varphi }_1^{\prime }(x_1)=0\\ & & & \\ {\psi }_0(x_0)=0& {\psi }_0(x_1)=0& {\psi }_0^{\prime }(x_0)=1& {\psi }_0^{\prime }(x_1)=0\\ & & & \\ {\psi }_1(x_0)=0& {\psi }_1(x_1)=0& {\psi }_1^{\prime }(x_0)=0& {\psi }_1^{\prime }(x_1)=1\\ & & & \end{array}$$
3. 求基函数
   对 ${\varphi }_0$ 来说，有一个二重零点 $x_1$ ，再设一个零点，有
   $${\varphi }_0(x)=(x-x_1{)}^2(ax+b),(代入{\varphi }_0(x_0)=1,{\varphi }_0^{\prime }(x_0)=0求a,b)$$
   对 ${\varphi }_1$ 来说，有一个二重零点 $x_0$ ，再设一个零点，有
   $${\varphi }_1(x)=(x-x_0{)}^2(ax+b),(代入{\varphi }_1(x_1)=1,{\varphi }_1^{\prime }(x_1)=0求a,b)$$
   对 ${\psi }_0$ 来说，有一个一重零点 $x_0$ ，有一个二重零点 $x_1$ 有
   $${\psi }_0(x)=c(x-x_0)(x-x_1{)}^2,(代入{\psi }_0^{\prime }(x_0)=1求出c)$$
   对 ${\psi }_1$ 来说，有一个一重零点 $x_1$ ，有一个二重零点 $x_0$ 有
   $${\psi }_1(x)=c(x-x_1)(x-x_0{)}^2,(代入{\psi }_1^{\prime }(x_1)=1求出c)$$
   还需要知道基函数的导数形式以方便求解，若
   $$\varphi (x)=(x-x_i{)}^2(ax+b)=1$$
   $$\begin{array}{rl}{\varphi }^{\prime }(x)=& 2(x-x_i)(ax+b)+a(x-x_i{)}^2=0\end{array}$$
   解得
   $$\begin{array}{rl}a=& -\frac{2}{(x-x_i{)}^3}\\ & \\ b=& \frac{(3x-x_i)}{(x-x_i{)}^3}\end{array}$$
   若
   $${\psi }_i(x)=c(x-x_i)(x-x_j{)}^2$$
   $${\psi }_i^{\prime }(x)=c(x-x_j{)}^2+2c(x-x_j)(x-x_i)=1$$
   解得
   $$c=\frac{1}{(x-x_j{)}^2+2(x-x_i)(x-x_j)}{|}_{x=x_i}=\frac{1}{(x_i-x_j{)}^2}$$
   将基函数代入式（1）得到插值多项式。

[!example]-
设 $f(x)=x^4+2x^3+5$ 。对节点 $x_0=-3$ ， $x_1=-1$ ，求在区间 $[-3,-1]$ 上的三次Hermite插值多项式及其余项。
解：由题可得
$$y_0=32,y_1=4,y_0^{\prime }=-54,y_1^{\prime }=2$$
所以插值函数
$$H_3(x)=32{\varphi }_0(x)+4{\varphi }_1(x)-54{\psi }_0(x)+2{\psi }_1(x)$$
接下来求基函数
$$\begin{array}{l}a=-2/(-3+1{)}^3=1/4\\ \\ b=(-3\cdot 3+1)/(-3+1{)}^3=1\end{array}\}\to {\varphi }_0(x)=(x+1{)}^2(\frac{1}{4}x+1)$$
$$\begin{array}{l}a=-2/(-1+3{)}^3=-1/4\\ \\ b=(3\cdot (-1)+3)/(-1+3{)}^3=0\end{array}\}\to {\varphi }_1(x)=-\frac{1}{4}x(x+3{)}^2$$
$$c=\frac{1}{(-3+1{)}^2}=\frac{1}{4}\to {\psi }_0(x)=\frac{1}{4}(x+3)(x+1{)}^2$$
$$c=\frac{1}{(-1+3{)}^2}=\frac{1}{4}\to {\psi }_1(x)=\frac{1}{4}(x+1)(x+3{)}^2$$
代入插值多项式得
$$H_3(x)=-6x^3-22x^2-24x-4$$
由插值余项定理
$$\begin{array}{rl}|R_3(x)|=& |f(x)-H_3(x)|=|\frac{f^{(4)}(\xi )}{4!}(x+3{)}^2(x+1{)}^2|\\ & \\ =& |\frac{4!}{4!}(x+3{)}^2(x+1{)}^2|\end{array}$$

说明：

• 余项仿照拉格朗日余项， $x_0$ 点用了两次，所以平方， $x_1$ 点同理
• 类型 $1$ 中系数 $c$ 是相同的

4.3 三点三次Hermite插值

确定三次插值多项式 $H_3(x)$ 使满足：
$$\begin{array}{rl}{H}_3(x_0)=f(x_0)=y_0,H_3(x_1)=& f(x_1)=y_1,H_3(x_2)=f(x_2)=y_2\\ & \\ H_3^{\prime }(x_0)=& f^{\prime }(x_0)=y_0^{\prime }\end{array}$$
方法：基函数法，基函数为三次多项式。
步骤：

1. 设 $H_3(x)$ 和 $H_3^{\prime }(x)$ 的表达式
   $$H_3(X)={\varphi }_0(x)y_0+{\varphi }_1(x)y_1+{\varphi }_2(x)y_2+{\psi }_0(x)y_0^{\prime }\text{\hspace{1em}(1)}$$
   $$H_3^{\prime }(X)={\varphi }_0^{\prime }(x)y_0+{\varphi }_1^{\prime }(x)y_1+{\varphi }_2^{\prime }(x)y_2+{\psi }_0^{\prime }(x)y_0^{\prime }\text{\hspace{1em}(2)}$$
2. 根据表达式列表
   $$\begin{array}{cccc}将x_0代入(1)& 将x_1代入(1)& 将x_2代入(1)& 将x_0代入(2)\\ & & & \\ {\varphi }_0(x_0)=1& {\varphi }_0(x_1)=0& {\varphi }_0(x_2)=0& {\varphi }_0^{\prime }(x_0)=0\\ & & & \\ {\varphi }_1(x_0)=0& {\varphi }_1(x_1)=1& {\varphi }_1(x_2)=0& {\varphi }_1^{\prime }(x_0)=0\\ & & & \\ {\varphi }_2(x_0)=0& {\varphi }_2(x_1)=0& {\varphi }_2(x_2)=1& {\varphi }_2^{\prime }(x_0)=0\\ & & & \\ {\psi }_0(x_0)=0& {\psi }_0(x_1)=0& {\psi }_0(x_2)=0& {\psi }_0^{\prime }(x_0)=1\\ & & & \end{array}$$
3. 求基函数
   对 ${\varphi }_0$ 来说，有两个一重零点 $x_1,x_2$ ，再设一个零点，有
   $${\varphi }_0(x)=(x-x_1)(x-x_2)(ax+b),(代入{\varphi }_0(x_0)=1,{\varphi }_0^{\prime }(x_0)=0求a,b)$$
   对 ${\varphi }_1$ 来说，有两个一重零点 $x_0,x_2$ ，再设一个零点，有
   $${\varphi }_1(x)=(x-x_0)(x-x_2)(ax+b),(代入{\varphi }_1(x_1)=1,{\varphi }_1^{\prime }(x_1)=0求a,b)$$
   对 ${\varphi }_2$ 来说，有两个一重零点 $x_0,x_1$ ，再设一个零点，有
   $${\varphi }_2(x)=(x-x_0)(x-x_1)(ax+b),(代入{\varphi }_2(x_2)=1,{\varphi }_2^{\prime }(x_2)=0求a,b)$$
   对 ${\psi }_0$ 来说，有三个一重零点 $x_0,x_1,x_2$ ，有
   $${\psi }_0=c(x-x_0)(x-x_1)(x-x_2),(代入{\psi }_0^{\prime }(x_0)=1求出c)$$
   还需要知道基函数的导数形式以方便求解，若
   $$\varphi (x)=(x-x_i)(x-x_j)(ax+b)=1$$
   $${\varphi }^{\prime }(x)=(x-x_i)(ax+b)+(x-x_j)(ax+b)+a(x-x_i)(x-x_j)=0$$
   解得
   $$\begin{array}{rl}a=& \frac{2x-x_i-x_j}{(x-x_i{)}^2(x-x_j{)}^2}\\ & \\ b=& \frac{3x^2-2x(x_i+x_j)+x_i{x}_j}{(x-x_i{)}^2(x-x_j{)}^2}\end{array}$$
   若
   $$\psi (x)=c(x-x_i)(x-x_j)(x-x_k)$$
   $$\begin{array}{rl}{\psi }^{\prime }(x)=& [c(x-x_i)(x-x_j)+c(x-x_i)(x-x_k)+c(x-x_j)(x-x_k)]{|}_{x=x_i}\\ & \\ =& c(x_i-x_j)(x_i-x_k)=1\end{array}$$
   $$c=\frac{1}{(x_i-x_j)(x_i-x_k)}$$

[!example]-
已知
$$f(1)=5,f(2)=21,f(3)=53,f^{\prime }(1)=10$$
求三次Hermite插值多项式
解：设插值函数
$$H_3(x)=5\cdot {\varphi }_0(x)+21\cdot {\varphi }_1(x)+53\cdot {\varphi }_2(x)+10\cdot {\psi }_0(x)$$
求基函数
$$\begin{array}{l}a=-3/4\\ \\ b=-1/4\end{array}\}\to {\varphi }_0(x)=(x-2)(x-3)(-\frac{3}{4}x-\frac{1}{4})$$
$$\begin{array}{l}a=0\\ \\ b=12\end{array}\}\to {\varphi }_1(x)=12(x-1)(x-3)$$
$$\begin{array}{l}a=3/4\\ \\ b=65/4\end{array}\}\to {\varphi }_2(x)=(x-2)(x-3)(\frac{3}{4}x+\frac{65}{4})$$
$$c=\frac{1}{(1-2)(1-3)}=\frac{1}{2}\to {\psi }_0(x)=\frac{1}{2}(x-1)(x-2)(x-3)$$
所以
$$H_3(x)=41x^3+902x^2-5037x+5886$$

4.4 n+1个点2n+1次Hermite插值多项式

就是所有节点的函数值和导数值都用上了
$$H_{2n+1}(x)=\sum _{k=0}^n(1-2l_k^{\prime }(x_k)(x-x_k))l_k^2(x)f_k+\sum _{k=0}^n(x-x_k)l_k^2(x)f_k^{\prime }$$
余项公式
$$R_{2n+1}(x)=f(x)-H_{2n+1}(x)=\frac{f^{(2n+2)}(\xi )}{(2n+2)!}(\prod (x){)}^2,\prod (x)=\prod _{j=0}^n(x-x_j)$$
如果给定定义域，则右边积分可以得到一个 $2n+1$ 次的数值积分公式。
积分中值定理可得出积分误差
$$R[f]=\frac{f^{(2n+2)}(\eta )}{(2n+2)!}{\int }_a^b(\prod (x){)}^2\mathrm{d}x$$