插值节点具有高阶导数信息的多项式插值算法

Polynomial Interpolation Algorithm with Higher-Order Derivative Information for Interpolation Nodes

摘要: 目前大部分对插值的研究都只研究函数的多项式插值，少数文章给出了一阶或二阶导数的多项式插值，本文给出了对函数及其高阶导数的多项式插值的理论算法，构造高阶导数的多项式插值听起来是复杂的，一般的算法也是庞大的，本文采取与经典研究方法不同的思路，运用行列式函数求导法则直接待定插值多项式，并给出证明，最后通过K值法和罗尔定理计算了Lagrange型插值余项，便于误差的分析，至此，要求在各插值节点上插值多项式函数与被插函数的函数值相等，且任意阶导数值也相等的问题在理论上解决了。

Abstract: At present, most of the researches on interpolation only study the polynomial interpolation of func-tion. A few papers give the polynomial interpolation of the first or second derivative. This paper gives the theoretical algorithm of polynomial interpolation of function and its higher derivative, the construction of polynomial interpolation of high order derivative sounds complicated, and the gen-eral algorithm is huge. This paper takes a different idea from the classical research methods. Finally, the residual terms of Lagrange-type interpolation are calculated by K-value method and Rohr’s theorem, which is convenient for error analysis. So far, the problems that the interpolation polyno-mial function and the interpolated function are equal on each interpolation node, and the derivative values of any order are equal, have been solved theoretically.

1. 引言

插值理论在工程近似计算领域发挥着重要作用，而多项式函数由于其计算和表达上的简单性，在数值近似理论以及工程计算等方面都有着广泛应用。在实际应用过程中，相当一部分多项式插值问题要求在插值节点处，插值多项式不仅与被插值函数的值相等，而且与它的导数值也相等。对于插值节点有一、二阶导数信息的Hermite插值算法已经有学者进行了相关研究 [1]，并给出了较为良好的算法设计。本文提供了在一定条件下，插值节点具有高阶导数信息的多项式插值一般算法，并利用行列式函数高阶导给出证明，最后计算了Lagrange型插值余项。

2. 预备知识

引理设函数 $f_{ij}\left(x\right)\left(i,j=1,2,\cdots ,n\right)$ 在区间I上可导，则行列式函数

$f\left(x\right)=\left|\begin{array}{cccc}{f}_{11}\left(x\right)& f_{12}\left(x\right)& \cdots & f_{1n}\left(x\right)\\ f_{21}\left(x\right)& f_{22}\left(x\right)& \cdots & f_{2n}\left(x\right)\\ ⋮& ⋮& & ⋮\\ f_{n1}\left(x\right)& f_{n2}\left(x\right)& \cdots & f_{nn}\left(x\right)\end{array}\right|$

也在区间I上可导，且有

$f^{\prime }\left(x\right)={\displaystyle \underset{i=1}{\overset{n}{\sum }}\begin{array}{cccc}{f}_{11}\left(x\right)& f_{12}\left(x\right)& \cdots & f_{1n}\left(x\right)\\ ⋮& ⋮& & ⋮\\ {f^{\prime }}_{i1}\left(x\right)& {f^{\prime }}_{i2}\left(x\right)& \cdots & {f^{\prime }}_{in}\left(x\right)\\ ⋮& ⋮& & ⋮\\ f_{n1}\left(x\right)& f_{n2}\left(x\right)& \cdots & f_{nn}\left(x\right)\end{array}}$ .

引理的证明在参考文献 [2] 中已有学者给出，此处不作证明。

3. 主要结论及证明

定理1 [3] 设函数f为数集 $ℝ$ 上的实函数，且对于 $ℝ$ 上的各点 $x_1,x_2,\cdots ,x_p\left(x_1<x_2<\cdots <x_p\right)$ ，有

$f^{\left(k\right)}\left(x_i\right)=r_{ik}\left(i=1,2,\cdots ,p;k=0,1,\cdots ,n_i\right)$ ，记 $n={\displaystyle \underset{i=1}{\overset{p}{\sum }}\left(n_i+1\right)}-1$ ， $x_n\left(x\right)=\left(1,x,\cdots ,x^n\right)$ ，若

$\left|\begin{array}{c}{x}_n\left(x_1\right)\\ ⋮\\ x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮\\ x_n\left(x_p\right)\\ ⋮\\ x_n^{\left(n_p\right)}\left(x_p\right)\end{array}\right|\ne 0$ ，则满足 $\begin{array}{cc}p\left(x\right)& x_n\left(x\right)\\ f\left(x_1\right)& x_n\left(x_1\right)\\ ⋮& ⋮\\ f^{\left(n_1\right)}\left(x_1\right)& x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮& ⋮\\ f\left(x_p\right)& x_n\left(x_p\right)\\ ⋮& ⋮\\ f^{\left(n_p\right)}\left(x_p\right)& x_n^{\left(n_p\right)}\left(x_p\right)\end{array}=0$

的多项式 $p\left(x\right)$ 即为f的插值多项式，即 $p^{\left(k\right)}\left(x_i\right)=r_{ik}\left(i=1,2,\cdots ,p;k=0,1,\cdots ,n_i\right)$ 。

证明只需证明对于 $\forall x_j\in \left\{x_1,x_2,\cdots ,x_p\right\},\forall l\in \left\{0,1,\cdots ,n_j\right\}$ ，有 $p^{\left(l\right)}\left(x_j\right)=f^{\left(l\right)}\left(x_j\right)$ ，再由 $x_j$ 和l的任意性即可证出结论。

对于等式

$\begin{array}{cc}p\left(x\right)& x_n\left(x\right)\\ f\left(x_1\right)& x_n\left(x_1\right)\\ ⋮& ⋮\\ f^{\left(n_1\right)}\left(x_1\right)& x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮& ⋮\\ f\left(x_p\right)& x_n\left(x_p\right)\\ ⋮& ⋮\\ f^{\left(n_p\right)}\left(x_p\right)& x_n^{\left(n_p\right)}\left(x_p\right)\end{array}=0$

观察发现，等式左端的行列式函数除首行外各元素均为实数，因此，利用引理对该式求一、二阶导可得

$\begin{array}{cc}{p}^{\prime }\left(x\right)& {x^{\prime }}_n\left(x\right)\\ f\left(x_1\right)& x_n\left(x_1\right)\\ ⋮& ⋮\\ f^{\left(n_1\right)}\left(x_1\right)& x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮& ⋮\\ f\left(x_p\right)& x_n\left(x_p\right)\\ ⋮& ⋮\\ f^{\left(n_p\right)}\left(x_p\right)& x_n^{\left(n_p\right)}\left(x_p\right)\end{array}=0,\begin{array}{cc}{p}^{″}\left(x\right)& {x^{″}}_n\left(x\right)\\ f\left(x_1\right)& x_n\left(x_1\right)\\ ⋮& ⋮\\ f^{\left(n_1\right)}\left(x_1\right)& x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮& ⋮\\ f\left(x_p\right)& x_n\left(x_p\right)\\ ⋮& ⋮\\ f^{\left(n_p\right)}\left(x_p\right)& x_n^{\left(n_p\right)}\left(x_p\right)\end{array}=0$

以此类推，对该式求l阶导可得

$\begin{array}{cc}{p}^{\left(l\right)}\left(x\right)& x_n^{\left(l\right)}\left(x\right)\\ f\left(x_1\right)& x_n\left(x_1\right)\\ ⋮& ⋮\\ f^{\left(n_1\right)}\left(x_1\right)& x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮& ⋮\\ f\left(x_p\right)& x_n\left(x_p\right)\\ ⋮& ⋮\\ f^{\left(n_p\right)}\left(x_p\right)& x_n^{\left(n_p\right)}\left(x_p\right)\end{array}=0$

将 $x=x_j$ 代入上式，得到

$\begin{array}{cc}{p}^{\left(l\right)}\left(x_j\right)& x_n^{\left(l\right)}\left(x_j\right)\\ f\left(x_1\right)& x_n\left(x_1\right)\\ ⋮& ⋮\\ f^{\left(n_1\right)}\left(x_1\right)& x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮& ⋮\\ f\left(x_p\right)& x_n\left(x_p\right)\\ ⋮& ⋮\\ f^{\left(n_p\right)}\left(x_p\right)& x_n^{\left(n_p\right)}\left(x_p\right)\end{array}=0$

对等式左端行列式函数进一步变形，将 $f^{\left(l\right)}\left(x_j\right)$ 所在行，即 $\left(f^{\left(l\right)}\left(x_j\right),x_n^{\left(l\right)}\left(x_j\right)\right)$ 乘负一加到首行，则有

$\begin{array}{cc}{p}^{\left(l\right)}\left(x_j\right)-f^{\left(l\right)}\left(x_j\right)& 0\\ f\left(x_1\right)& x_n\left(x_1\right)\\ ⋮& ⋮\\ f^{\left(n_1\right)}\left(x_1\right)& x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮& ⋮\\ f\left(x_p\right)& x_n\left(x_p\right)\\ ⋮& ⋮\\ f^{\left(n_p\right)}\left(x_p\right)& x_n^{\left(n_p\right)}\left(x_p\right)\end{array}=0$

把行列式函数按首行展开，则

$\left(p^{\left(l\right)}\left(x_j\right)-f^{\left(l\right)}\left(x_j\right)\right)\left|\begin{array}{c}{x}_n\left(x_1\right)\\ ⋮\\ x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮\\ x_n\left(x_p\right)\\ ⋮\\ x_n^{\left(n_p\right)}\left(x_p\right)\end{array}\right|=0$

而其中，

$\left|\begin{array}{c}{x}_n\left(x_1\right)\\ ⋮\\ x_n^{\left(n_1\right)}\left(x_1\right)\\ ⋮\\ x_n\left(x_p\right)\\ ⋮\\ x_n^{\left(n_p\right)}\left(x_p\right)\end{array}\right|\ne 0$ ，故 $p^{\left(l\right)}\left(x_j\right)-f^{\left(l\right)}\left(x_j\right)=0$

由此，定理得证。

定理2设f为数集 $ℝ$ 上的实函数，对于 $ℝ$ 上的各点 $x_1,x_2,\cdots ,x_p\left(x_1<x_2<\cdots <x_p\right)$ ，有

$f^{\left(k\right)}\left(x_i\right)=r_{ik}\left(i=1,2,\cdots ,p;k=0,1,\cdots ,n_i\right)$ ，且f在 $ℝ$ 上 $n+1$ 阶可导，其中， $n={\displaystyle \underset{i=1}{\overset{p}{\sum }}\left(n_i+1\right)}-1$ ，对于它的插

值 [4] 多项式 $p\left(x\right)$ ，存在 $\xi \in ℝ$ ，满足

$f\left(x\right)=p\left(x\right)+\frac{f^{\left(n+1\right)}\left(\xi \right)}{\left(n+1\right)!}{\displaystyle \underset{i=1}{\overset{p}{\prod }}{\left(x-x_i\right)}^{n_i+1}}$

证明 [5] 令

$g\left(t\right)=f\left(t\right)-\left(p\left(t\right)+k{\displaystyle \underset{i=1}{\overset{p}{\prod }}{\left(t-x_i\right)}^{n_i+1}}\right)$

由于 $p^{\left(l\right)}\left(x_j\right)=f^{\left(l\right)}\left(x_j\right)$ ，所以显然有 $g^{\left(k\right)}\left(x_i\right)=0\left(i=1,2,\cdots ,p;k=0,1,\cdots ,n_i\right)$ 且 $g\left(x\right)=0$ 。

对 $g\left(x\right)=g\left(x_i\right)=0\left(i=1,2,\cdots ,p\right)$ 应用Roll定理，则 $g^{\prime }\left(x\right)$ 增加了n个零点，结合 $g^{\prime }\left(x_i\right)=0$ ，再次应

用Roll定理，以此类推，最终得到 $g^{\left(m\right)}\left(x\right)$ 一共存在 $n-m+2$ 个零点，其中， $m=\underset{1\le i\le n}{\max }{n}_i$ 。

由于 $n-m+2>0$ ，所以可以对 $g^{\left(m\right)}\left(x\right)$ 继续应用 $n-m+1$ 次Roll定理，于是存在 $\xi \in ℝ$ ，满足 $g^{\left(n+1\right)}\left(\xi \right)=0$ ，即

$f^{\left(n+1\right)}\left(\xi \right)-k\left(n+1\right)!=0$

于是解得 $k=\frac{f^{\left(n+1\right)}\left(\xi \right)}{\left(n+1\right)!}$ ，代入可得

$f\left(x\right)=p\left(x\right)+\frac{f^{\left(n+1\right)}\left(\xi \right)}{\left(n+1\right)!}{\displaystyle \underset{i=1}{\overset{p}{\prod }}{\left(x-x_i\right)}^{n_i+1}}$

4. 总结

本文解决了满足各插值节点上与被插函数的函数值相等的插值多项式的理论算法，并给出了拉格朗日型余项，可用于估计误差。

NOTES

^*通讯作者。