II型插值条件下三次样条函数误差分析

Error Analysis of Cubic Spline Function Under Type II Interpolation

摘要: 样条函数是函数逼近理论一个非常活跃的分支，促使了研究人员需要深刻认识样条函数的本质及性质。本文介绍了基于Hermite两点三次公式的三转角插值算法。三转角以插值节点的一阶导数为未知量构建样条函数，在此基础上，研究插值节点均匀分布时，在第二类边界条件下，即II型插值条件下，当边界初值发生扰动时，对应的三次样条函数在插值节点的一阶导数值如何随第二边界初值的扰动而变化，基于Doolittle分解和Crout分解性质，推导出2个定理，即误差估计的表达式，这些定理为三次样条函数在二阶导数边界初值变化时的误差分析提供了可行的方法。

Abstract: Spline function is a very active branch of function approximation theory, which makes researchers need to deeply understand the essence and properties of spline function. This paper introduces the three-angle interpolation algorithm based on Hermite two-point cubic formula. The three-angle spline function is constructed with the first derivative of the interpolating node as an unknown quantity. On this basis, when interpolating nodes are evenly distributed, under the second type of boundary condition, that is, under the type II interpolation condition, when the initial value of the boundary is disturbed, the corresponding cubic spline function in the interpolating node’s first derivative value changes with the disturbance of the initial value of the second boundary. Based on the properties of Doolittle decomposition and Crout decomposition, two theorems, namely the expression of error estimation, are derived. These theorems provide a feasible method for error analysis of cubic spline function when the initial value of the second derivative boundary changes.

1. 引言

三次样条函数误差估计是三次样条函数理论和应用研究的重要内容。自从Hall [1]等学者给出三次样条函数误差估计表达式，针对不同误差阶的最优估计参数设置及误差界等问题，不少学者对其进行了改进和完善。这些研究优化了三次样条函数在插值区间内具有普适意义的误差界估计。

对于三次样条函数在插值结点处的误差估计，保明堂[2]推导了自然边界条件下插值结点处的一阶和二阶导数误差估计。常庚哲[3]对该结果[2]进行了改进。金坚明推导了I型插值条件下插值结点处的误差结果[4]，并推广至双三次B样条插值函数和II型插值条件。

本文推导了II型插值条件下三次样条在插值结点处误差估计的等式表达，这有别于已有文献提供的不等式结果。从这个意义上讲，本文结果比现有文献更精确。

2. 定义和符号

定义[5]：设在区间$\left[a,b\right]$ 上，给定$n+1$ 个插值节点$a=x_0<x_1<\cdots <x_{n-1}<x_n=b$ ，在节点处的函数值$y_k=f\left(x_k\right)\left(k=0,1,\cdots ,n\right)$ ，若插值函数$s\left(x\right)$ ，满足条件：

1) $s\left(x_k\right)=f\left(x_k\right)=y_k,k=0,1,\cdots ,n$ ；

2) 在每个小区间$\left[x_i,x_{i+1}\right],\left(i=0,1,\cdots ,n-1\right)$ 上$s\left(x\right)$ 是三次多式$s_i\left(x\right)$ ；

3) $s\left(x\right)$ 在$\left[a,b\right]$ 上二阶连续可微，即$s\left(x\right)\in C^2\left[a,b\right]$ 。则称$s\left(x\right)$ 为$f\left(x\right)$ 的三次样条插值函数。

三转角插值算法[6] [7]是一种以插值节点的一阶导数为未知量，来构造$s\left(x\right)$ 的。设插值区间$\left[a,b\right]$ 上的分割记为$\Delta$ ：$a=x_0<x_1<\cdots <x_{n-1}<x_n=b$ ，插值结点处函数值为$y_k=f\left(x_k\right)\left(k=0,1,\cdots ,n\right)$ ，$m_k$ 为三次样条函数$s\left(x\right)\in C^2\left[a,b\right]$ 在插值结点$x_k\left(k=0,1,\cdots ,n\right)$ 处的一阶导数。对于$k=1,2,\cdots ,n-1$ ，记

$h_k=x_k-x_{k-1},{\mu }_k=\frac{h_k}{h_k+h_{k+1}},{\lambda }_k=\frac{h_{k+1}}{h_k+h_{k+1}},d_k=3\left({\mu }_k\frac{y_{k+1}-y_k}{h_{k+1}}+{\lambda }_k\frac{y_k-y_{k-1}}{h_k}\right)$ . 1)

根据两点三次Hermite插值公式[8]，得到三次样条函数$s\left(x\right)$ 在区间$\left[x_{k-1},x_k\right]\left(k=1,\cdots ,n\right)$ 的表达式：

$\begin{array}{l}{s}_k\left(x\right)=\frac{y_{k-1}}{h_k^3}{\left(x-x_k\right)}^2\left[h_k+2\left(x-x_{k-1}\right)\right]+\frac{y_k}{h_k^3}{\left(x-x_{k-1}\right)}^2\left[h_k+2\left(x_k-x\right)\right]\\ +\frac{m_{k-1}}{h_k^2}\left(x-x_{k-1}\right){\left(x-x_k\right)}^2+\frac{m_k}{h_k^2}\left(x-x_k\right){\left(x-x_{k-1}\right)}^2\end{array}$ . (2)

其中$m_k$ 满足如下方程组。式(3)中有$n+1$ 个未知数，但只有$n-1$ 个方程，需要补充2个边界条件才能对其求解。

${\lambda }_k{m}_{k-1}+2m_k+{\mu }_k{m}_{k+1}=d_k,\left(k=1,2,\cdots ,n\right)$ . (3)

考虑II型插值条件作为边界条件，即给定插值区间$\left[a,b\right]$ 端点处二阶导数$s^{″}\left(a\right)=s^{″}\left(x_0\right)=M_0={y^{″}}_0$ 和$s^{″}\left(b\right)=s^{″}\left(x_n\right)=M_n={y^{″}}_n$ ，则方程组(3)转化为如下三对角方程组。

$\left[\begin{array}{ccccccc}2& 1& 0& \cdots & 0& 0& 0\\ {\lambda }_1& 2& {\mu }_1& \ddots & 0& 0& 0\\ 0& {\lambda }_2& 2& \ddots & 0& 0& 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& \ddots & 2& {\mu }_{n-2}& 0\\ 0& 0& 0& \ddots & {\lambda }_{n-1}& 2& {\mu }_{n-1}\\ 0& 0& 0& \cdots & 0& 1& 2\end{array}\right]\left[\begin{array}{c}{m}_{\text{0}}\\ m_1\\ m_2\\ ⋮\\ m_{n-2}\\ m_{n-1}\\ m_n\end{array}\right]=\left[\begin{array}{c}{d}_{\text{0}}\\ d_{\text{1}}\\ d_2\\ ⋮\\ d_{n-2}\\ d_{n-1}\\ d_n\end{array}\right]$ . (4)

其中$d_0=3\frac{y_1-y_0}{h_1}-\frac{h_1}{2}{y^{″}}_0$ ，$d_n=3\frac{y_n-y_{n-1}}{h_n}+\frac{h_n}{2}{y^{″}}_n$ 。

本文的主要工作在于推导当II型插值条件存在误差，即$M_0$ 存在误差${{\epsilon }^{″}}_0$ 和$M_n$ 存在误差${{\epsilon }^{″}}_n$ 时，三次样条函数$s\left(x\right)$ 在插值结点处的一阶导数$m_k$ 误差如何用${{\epsilon }^{″}}_0$ 与${{\epsilon }^{″}}_n$ 精确表达。

3. 几个引理

本文在推导过程中主要利用了矩阵的Doolittle分解和Crout分解及其性质。为此，先引入三对角矩阵LU分解及其性质的几个引理。设

$A=\left[\begin{array}{llllll}{b}_1\hfill & c_1\hfill & 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill \\ a_2\hfill & b_2\hfill & c_2\hfill & \ddots \hfill & \hfill & ⋮\hfill \\ 0\hfill & a_3\hfill & b_3\hfill & \ddots \hfill & 0\hfill & ⋮\hfill \\ ⋮\hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ ⋮\hfill & \hfill & 0\hfill & \ddots \hfill & b_{n-1}\hfill & c_{n-1}\hfill \\ 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill & a_n\hfill & b_n\hfill \end{array}\right]$ . (5)

定义${\alpha }_i\left(i=1,2,\cdots ,n\right)$ 如下：

${\alpha }_i\text{=}\{\begin{array}{l}{b}_{1}i=1\hfill \\ b_i-\frac{a_i{c}_{i-1}}{{\alpha }_{i-1}}i=2,3,\cdots ,n\hfill \end{array}$ . (6)

引理1 (Doolittle分解) [9]：式(5)矩阵$A$ 在${\alpha }_i\ne 0\left(i=1,2,\cdots ,n\right)$ 时，其LU分解结果可表示为式(7)。其中${\beta }_i=c_i\left(i=1,2,\cdots ,n-1\right),{\gamma }_i=\frac{a_i}{{\alpha }_{i-1}}\left(i=2,3,\cdots ,n\right)$ 。式(7)中${\alpha }_i$ 根据式(6)计算得到。

$A=LU,L=\left[\begin{array}{llllll}1\hfill & 0\hfill & 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill \\ {\gamma }_2\hfill & 1\hfill & 0\hfill & \ddots \hfill & \hfill & ⋮\hfill \\ 0\hfill & {\gamma }_3\hfill & 1\hfill & \ddots \hfill & 0\hfill & ⋮\hfill \\ ⋮\hfill & 0\hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ ⋮\hfill & \hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill & {\gamma }_n\hfill & 1\hfill \end{array}\right],U=\left[\begin{array}{llllll}{\alpha }_1\hfill & {\beta }_1\hfill & 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill \\ 0\hfill & {\alpha }_2\hfill & {\beta }_2\hfill & \ddots \hfill & \hfill & ⋮\hfill \\ 0\hfill & 0\hfill & {\alpha }_3\hfill & \ddots \hfill & 0\hfill & ⋮\hfill \\ ⋮\hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ ⋮\hfill & \hfill & 0\hfill & \ddots \hfill & \ddots \hfill & {\beta }_{n-1}\hfill \\ 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill & 0\hfill & {\alpha }_n\hfill \end{array}\right]$ . (7)

证明：用矩阵$U$ 的第$i$ 行乘以矩阵$L$ 的第$i+1$ 列$\left(i=1,2,\cdots ,n-1\right)$ ，可得${\beta }_i=c_i\left(i=1,2,\cdots ,n-1\right)$ 。

用矩阵$U$ 的第$i$ 行乘以矩阵$L$ 的第$i-1$ 列$\left(i=2,\cdots ,n\right)$ ，可得${\gamma }_i=\frac{a_i}{{\alpha }_{i-1}}\left(i=2,3,\cdots ,n\right)$ 。

用矩阵$U$ 的第$i$ 行乘以矩阵$L$ 的第$i$ 列$\left(i=1,2,\cdots ,n\right)$ ，即可得${\alpha }_1=b_1$ 和${\alpha }_i=b_i-{\beta }_{i-1}{\gamma }_i\left(i=2,\cdots ,n-1,n\right)$ 。把${\beta }_i$ 和${\gamma }_i$ 代入，则所有结论成立。

引理2 (Crout分解)：设A为形如式(5)的三对角矩阵，且满足式(8)中的${\theta }_i\ne 0\left(i=1,2,\cdots ,n\right)$ ，则A可以分解为$A=U_1{L}_1$ ，$U_1$ 和$L_1$ 如式(17)所示，$U_1$ 和$L_1$ 的下标主要用于区别式(7)中的矩阵。

$U_1=\left[\begin{array}{llllll}1\hfill & {\beta }_1\hfill & 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill \\ 0\hfill & 1\hfill & {\beta }_2\hfill & \ddots \hfill & \hfill & ⋮\hfill \\ 0\hfill & 0\hfill & 1\hfill & \ddots \hfill & 0\hfill & ⋮\hfill \\ ⋮\hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ ⋮\hfill & \hfill & 0\hfill & \ddots \hfill & \ddots \hfill & {\beta }_{n-1}\hfill \\ 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill & 0\hfill & 1\hfill \end{array}\right],L_1=\left[\begin{array}{llllll}{\theta }_1\hfill & 0\hfill & 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill \\ {\gamma }_2\hfill & {\theta }_2\hfill & 0\hfill & \ddots \hfill & \hfill & ⋮\hfill \\ 0\hfill & {\gamma }_3\hfill & {\theta }_3\hfill & \ddots \hfill & 0\hfill & ⋮\hfill \\ ⋮\hfill & 0\hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ ⋮\hfill & \hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill & {\gamma }_n\hfill & {\theta }_n\hfill \end{array}\right]$ . (8)

且满足${\gamma }_i=a_i\left(i=2,3,\cdots ,n\right)$ ，${\theta }_n=b_n$ 。${\beta }_i=\frac{c_i}{{\theta }_{i+1}},{\theta }_i=b_i-\frac{c_i}{{\theta }_{i+1}}{a}_{i+1}\left(i=1,2,\cdots ,n-1\right)$ 。

证明：用矩阵$U_1$ 的第$i+1$ 行乘以矩阵$L_1$ 的第$i$ 列$\left(i=1,2,\cdots ,n-1\right)$ ，可得${\gamma }_i=a_i\left(i=2,3,\cdots ,n\right)$ 。

用矩阵$U_1$ 的第$i$ 行乘以矩阵$L_1$ 的第$i+1$ 列$\left(i=1,2,\cdots ,n-1\right)$ ，可得${\beta }_i=\frac{c_i}{{\theta }_{i+1}}\left(i=1,2,\cdots ,n-1\right)$ 。

用矩阵$U_1$ 的第$i$ 行乘以矩阵$L_1$ 的第$i$ 列$\left(i=1,2,\cdots ,n\right)$ ，即可得${\theta }_n=b_n$ 和${\theta }_i=b_i-{\beta }_i{\gamma }_{i+1}\left(i=1,2,\cdots ,n-1\right)$ 。把${\beta }_i$ 和${\gamma }_i$ 代入，则所有结论成立。

引理3 [10]：若式(7)中$L$ 非奇异，设$Q={\left(q_{ij}\right)}_{1\le i,j\le n}=L^{-1}$ ，则$Q$ 中第$j$ 列$\left(j=1,2,\cdots ,n\right)$ 元素按式(9)计算得到。

$q_{k,j}=0\left(k<j\right),q_{k,k}=q_{j,j}=1\left(k=j\right),q_{k,j}={\left(-1\right)}^{k-j}{\displaystyle \prod _{l=j}^{k-1}{\gamma }_{l+1}}\left(k>j\right)$ . (9)

4. 主要结论

对于式(4)，引入

$A_c=\left[\begin{array}{ccccccc}2& 1& 0& \cdots & 0& 0& 0\\ {\lambda }_{\text{1}}& 2& {\mu }_1& \ddots & 0& 0& 0\\ 0& {\lambda }_{\text{2}}& 2& \ddots & 0& 0& 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& 0& 0& \ddots & 2& {\mu }_{n-\text{2}}& 0\\ 0& 0& 0& \ddots & {\lambda }_{n-\text{1}}& 2& {\mu }_{n-\text{1}}\\ 0& 0& 0& \cdots & 0& 1& 2\end{array}\right]$ ，$M_c=\left[\begin{array}{c}{m}_{\text{0}}\\ m_1\\ m_2\\ ⋮\\ m_{n-2}\\ m_{n-1}\\ m_n\end{array}\right]$ ，$G=\left[\begin{array}{c}{d}_{\text{0}}\\ d_{\text{1}}\\ d_2\\ ⋮\\ d_{n-2}\\ d_{n-1}\\ d_n\end{array}\right]$ . (10)

则式(4)可记为$A_c{M}_c=G$ 。这里引入$A_c$ 是因为$A_c$ 为$n+1$ 阶方阵，以区别于式(5)中的n阶方阵$A$ 。$A_c$ 的下标c代表三次样条(cubic spline)。在推导有关结论之前，本文先证明如下引理。

引理4 [11]：对于式(4)所示三次样条函数系数矩阵$A_c$ ，若对应插值区间$\left[a,b\right]$ 为等间距均匀分割，则$A_c$ 形如式(7)分解得到矩阵$U$ 中${\alpha }_i\left(i=0,1,\cdots ,n\right)$ 由下式(11)计算得到，其中$x_1=1+\frac{\sqrt{3}}{2}$ ，$x_2=1-\frac{\sqrt{3}}{2}$ 。

${\alpha }_i=\{\begin{array}{l}\begin{array}{l}2i=0\\ \frac{7}{4}i=1\end{array}\hfill \\ \begin{array}{l}\frac{\left(3+2\sqrt{3}\right)x_1^i-\left(3-2\sqrt{3}\right)x_2^i}{\left(3+2\sqrt{3}\right)x_1^{i-1}-\left(3-2\sqrt{3}\right)x_2^{i-1}}i=2,3,\cdots ,n-1\\ \frac{1}{2}\frac{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}i=n\end{array}\hfill \end{array}$ . (11)

证明：据均匀分割及式(1)，有${\lambda }_i={\mu }_i=\frac{1}{2}\left(i=1,2,\cdots ,n-1\right)$ 。由式(4)或(10)，有${\lambda }_n={\mu }_0=1$ 。再由式(5)和式(6)以及引理1，有${\alpha }_0=2,{\alpha }_1=2-\frac{{\lambda }_1{\mu }_0}{{\alpha }_0}=\frac{7}{4}$ 。进一步，对于$i=2,3,\cdots ,n-1$ ，有${\alpha }_i=2-\frac{{\lambda }_i{\mu }_{i-1}}{{\alpha }_{i-1}}=\frac{8{\alpha }_{i-1}-1}{4{\alpha }_{i-1}}$ 。接下来利用不动点方法求解序列${\alpha }_i\left(i=2,3,\cdots ,n-1\right)$ 的一般表达式。先解得方程$x=\frac{8x-1}{4x}$ 的根为$x_{1,2}=1\pm \frac{\sqrt{3}}{2}$ 。进而对于$i=2,3,\cdots ,n-1$ ，有$\frac{{\alpha }_i-x_1}{{\alpha }_i-x_2}=\frac{{\alpha }_1-x_1}{{\alpha }_1-x_2}\times {\left(\frac{x_2}{x_1}\right)}^{i-1}$ 。再由${\alpha }_1=\frac{7}{4}$ ，有${\alpha }_i=\frac{\left(3+2\sqrt{3}\right)x_1^i-\left(3-2\sqrt{3}\right)x_2^i}{\left(3+2\sqrt{3}\right)x_1^{i-1}-\left(3-2\sqrt{3}\right)x_2^{i-1}}$ $\left(i=2,3,\cdots ,n-1\right)$ 。而对于${\alpha }_n$ ，由${\lambda }_n=1,{\mu }_{n-1}=\frac{1}{2}$ 及${\alpha }_n=2-\frac{{\lambda }_n{\mu }_{n-1}}{{\alpha }_{n-1}}=2-\frac{1}{2{\alpha }_{n-1}}$ ，代入得到${\alpha }_n=\frac{1}{2}\frac{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}$ 。

接下来推导插值区间为等间距均匀分割时，在II型插值条件下，当边界初值误差给定时，式(4)对应的三次样条函数中$m_k\left(k=0,1,\cdots ,n\right)$ 的误差计算公式。

定理1 [11]：对于插值区间为均匀分割的II型插值条件下三次样条函数，其系数矩阵满足式(4)，当右侧边界初值$M_n$ 误差为${{\epsilon }^{″}}_n$ ，即${M^{\prime }}_n=M_n+{{\epsilon }^{″}}_n$ ，左侧边界初值$M_0$ 为任意常数，则三次样条插值结点处一阶导数$m_i\left(i=0,1,\cdots ,n\right)$ 的误差${\epsilon }_i$ 按式(12)计算得到，其中$x_{1,2}$ 同引理4。

$\begin{array}{l}\left|{\epsilon }_n\right|=h_n\frac{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}\left|{{\epsilon }^{″}}_n\right|\\ \left|{\epsilon }_i\right|=\{\begin{array}{l}{\left(\frac{1}{2}\right)}^{n-i}\frac{4\sqrt{3}\left|{\epsilon }_n\right|}{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}i=0,1\\ {\left(\frac{1}{2}\right)}^{n-i}\frac{\left(3+2\sqrt{3}\right)x_1^{i-1}-\left(3-2\sqrt{3}\right)x_2^{i-1}}{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}\left|{\epsilon }_n\right|i=2,3,\cdots ,n-1\end{array}\end{array}$ . (12)

证明：根据引理1，$A_c$ 的Doolittle LU分解$A_c=L_d{U}_d$ 如式(13)，其中${\beta }_i={\mu }_i\left(i=0,1,\cdots ,n-1\right)$ ， ${\gamma }_i=\frac{{\lambda }_i}{{\alpha }_{i-1}}\left(i=1,2,\cdots ,n\right)$ 。$L_d$ 和$U_d$ 的下标d代表Doolittle。由引理4，显然有${\alpha }_i\ne 0\left(i=0,1,\cdots ,n\right)$ ，所以$A_c$ 满足引理1的分解条件。

$A_c=L_d{U}_d,L_d=\left[\begin{array}{llllll}1\hfill & 0\hfill & 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill \\ {\gamma }_1\hfill & 1\hfill & 0\hfill & \ddots \hfill & \hfill & ⋮\hfill \\ 0\hfill & {\gamma }_2\hfill & 1\hfill & \ddots \hfill & 0\hfill & ⋮\hfill \\ ⋮\hfill & 0\hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ ⋮\hfill & \hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill & {\gamma }_n\hfill & 1\hfill \end{array}\right],U_d=\left[\begin{array}{llllll}{\alpha }_0\hfill & {\beta }_0\hfill & 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill \\ 0\hfill & {\alpha }_1\hfill & {\beta }_1\hfill & \ddots \hfill & \hfill & ⋮\hfill \\ 0\hfill & 0\hfill & {\alpha }_2\hfill & \ddots \hfill & 0\hfill & ⋮\hfill \\ ⋮\hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & \ddots \hfill & 0\hfill \\ ⋮\hfill & \hfill & 0\hfill & \ddots \hfill & \ddots \hfill & {\beta }_{n-1}\hfill \\ 0\hfill & \cdots \hfill & \cdots \hfill & 0\hfill & 0\hfill & {\alpha }_n\hfill \end{array}\right]$ . (13)

由$A_c{M}_c=G$ 和$A_c=L_d{U}_d$ ，有$L_d{U}_d{M}_c=G$ 。显然$L_d$ 非奇异，从而有

$U_d{M}_c=L_d^{-1}G$ . (14)

根据引理(3)，可计算得到$L_d^{-1}$ 。再把$U_d$ 和$L_d^{-1}$ 代入式(14)，解出$M_c$ 中各元素。

$\{\begin{array}{l}{m}_n=\frac{{\left(-1\right)}^n{\displaystyle \prod }_{l=\text{0}}^{n-1}{\gamma }_{l+1}}{{\alpha }_n}{d}_{\text{0}}+\frac{{\left(-1\right)}^{n-\text{1}}{\displaystyle \prod }_{l=\text{1}}^{n-\text{1}}{\gamma }_{l+1}}{{\alpha }_n}{d}_1+\cdots +\frac{\left(-{\gamma }_n\right)}{{\alpha }_n}{d}_{n-\text{1}}+\frac{1}{{\alpha }_n}{d}_n\\ m_i=\frac{{\left(-1\right)}^i{\displaystyle \prod }_{l=0}^{i-1}{\gamma }_{l+1}}{{\alpha }_i}{d}_{\text{0}}+\frac{{\left(-1\right)}^{i-1}{\displaystyle \prod }_{l=1}^{i-\text{1}}{\gamma }_{l+1}}{{\alpha }_i}{d}_1+\cdots +\frac{\left(-{\gamma }_i\right)}{{\alpha }_i}{d}_{i-1}+\frac{1}{{\alpha }_i}{d}_i-\frac{{\beta }_i}{{\alpha }_i}{m}_{i+\text{1}}\\ \left(i=1,2,\cdots ,n-1\right)\\ m_{\text{0}}=\frac{1}{{\alpha }_{\text{0}}}{d}_{\text{0}}-\frac{{\beta }_{\text{0}}}{{\alpha }_0}{m}_{\text{1}}\end{array}$ . (15)

由于$d_n=3\frac{y_n-y_{n-1}}{h_n}+\frac{h_n}{2}{y^{″}}_n$ ，其中${y^{″}}_n=M_n$ 。当$M_n$ 误差为${{\epsilon }^{″}}_n$ 时，即${M^{\prime }}_n=M_n+{{\epsilon }^{″}}_n$ ，则$d_n$ 的计算将受$M_n$ 的误差影响，记

$d_n^{\text{'}}=3\frac{y_n-y_{n-1}}{h_n}+\frac{h_n}{2}\left(M_n+{{\epsilon }^{″}}_n\right)=d_n+\frac{h_n}{2}{{\epsilon }^{″}}_n$ .

下面计算当$M_n$ 存在误差${{\epsilon }^{″}}_n$ 时，式(4)中$m_i\left(i=0,1,\cdots ,n\right)$ 产生的误差${\epsilon }_i\left(i=0,1,\cdots ,n\right)$ 。由式(13)和引理1，有${\beta }_i={\mu }_i\left(i=0,1,\cdots ,n-1\right)$ ，结合式(15)，有

$\{\begin{array}{l}{\epsilon }_n={m^{\prime }}_n-m_n=\frac{h_n}{2{\alpha }_n}{{\epsilon }^{″}}_n\\ {\epsilon }_i={m^{\prime }}_i-m_i={\left(-1\right)}^{n-i}{\displaystyle \prod _{l=i}^{n-1}\frac{{\beta }_l}{{\alpha }_l}{\epsilon }_n}i=1,2,\cdots ,n-1\\ {\epsilon }_0={m^{\prime }}_0-m_0={\left(-1\right)}^n{\displaystyle \prod _{l=0}^{n-1}\frac{{\beta }_l}{{\alpha }_l}{\epsilon }_n}\end{array}$ . (16)

下面计算式(16)中的${\displaystyle \prod _{l=i}^{n-1}\frac{{\beta }_l}{{\alpha }_l}}\left(i=0,1,\cdots ,n-1\right)$ 。由等间距均匀分割特性和式(1)，有${\lambda }_i={\mu }_i=\frac{1}{2}\left(i=1,2,\cdots ,n-1\right)$ 。特别地，由式(4)得到${\lambda }_n={\mu }_0=1$ 。根据引理4，对于${\alpha }_l\left(l=0,1,\cdots ,n\right)$ ，有${\alpha }_0=2,{\alpha }_1=2-\frac{{\lambda }_1{\mu }_0}{{\alpha }_0}=\frac{7}{4}$ ，${\alpha }_i=2-\frac{{\lambda }_i{\mu }_{i-1}}{{\alpha }_{i-1}}\left(i=2,3,\cdots ,n\right)$ 。同时由于${\beta }_i={\mu }_i\left(i=0,1,\cdots ,n-1\right)$ ，进而有

${\displaystyle \prod _{l=i}^{n-1}\frac{{\beta }_l}{{\alpha }_l}}\text{=}\{\begin{array}{l}\frac{{\beta }_0}{{\alpha }_0}\frac{{\beta }_1}{{\alpha }_1}\frac{{\beta }_2}{{\alpha }_2}\cdots \frac{{\beta }_{n-1}}{{\alpha }_{n-1}}=\left(\frac{{\beta }_0}{{\alpha }_0}\frac{{\beta }_1}{{\alpha }_1}\right)\left(\frac{{\beta }_2\cdots {\beta }_{n-1}}{{\alpha }_2\cdots {\alpha }_{n-1}}\right)=\frac{4\sqrt{3}{\left(\frac{1}{2}\right)}^n}{\left(3+2\sqrt{3}\right)x_1{}^{n-1}-\left(3-2\sqrt{3}\right)x_2{}^{n-1}}i=0\hfill \\ \frac{{\beta }_1}{{\alpha }_1}\frac{{\beta }_2}{{\alpha }_2}\cdots \frac{{\beta }_{n-1}}{{\alpha }_{n-1}}=\left(\frac{{\beta }_1}{{\alpha }_1}\right)\left(\frac{{\beta }_2\cdots {\beta }_{n-1}}{{\alpha }_2\cdots {\alpha }_{n-1}}\right)=\frac{4\sqrt{3}{\left(\frac{1}{2}\right)}^{n-1}}{\left(3+2\sqrt{3}\right)x_1{}^{n-1}-\left(3-2\sqrt{3}\right)x_2{}^{n-1}}i=1\hfill \\ \frac{{\beta }_i}{{\alpha }_i}\frac{{\beta }_{i+1}}{{\alpha }_{i+1}}\cdots \frac{{\beta }_{n-1}}{{\alpha }_{n-1}}=\left(\frac{{\beta }_i\cdots {\beta }_{n-1}}{{\alpha }_i\cdots {\alpha }_{n-1}}\right)={\left(\frac{1}{2}\right)}^{n-i}\frac{\left(3+2\sqrt{3}\right)x_1{}^{i-1}-\left(3-2\sqrt{3}\right)x_2{}^{i-1}}{\left(3+2\sqrt{3}\right)x_1{}^{n-1}-\left(3-2\sqrt{3}\right)x_2{}^{n-1}}i=2,3,\cdots ,n-1\hfill \end{array}$

在式(16)两边取绝对值，得到$\left|{\epsilon }_i\right|$ 如下。

$\begin{array}{l}\left|{\epsilon }_i\right|=\left|{\left(-1\right)}^{n-i}{\displaystyle \prod _{l=i}^{n-1}\frac{{\beta }_l}{{\alpha }_l}{\epsilon }_n}\right|=\{\begin{array}{l}\frac{4\sqrt{3}{\left(\frac{1}{2}\right)}^{n-i}}{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}\left|{\epsilon }_n\right|i=0,1\hfill \\ {\left(\frac{1}{2}\right)}^{n-i}\frac{\left(3+2\sqrt{3}\right)x_1^{i-1}-\left(3-2\sqrt{3}\right)x_2^{i-1}}{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}\left|{\epsilon }_n\right|i=2,3,\cdots ,n-1\hfill \end{array}\\ \left|{\epsilon }_n\right|=\left|\frac{h_n}{2{\alpha }_n}{{\epsilon }^{″}}_n\right|=h_n\frac{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}\left|{{\epsilon }^{″}}_n\right|\end{array}$

定理2：对于均匀分割II型插值条件下三次样条函数式(4)，当左侧边界初值$M_0$ 误差为${{\epsilon }^{″}}_n$ ，即${M^{\prime }}_0=M_0+{{\epsilon }^{″}}_0$ ，右侧边界初值$M_n$ 为任意常数，则三次样条插值结点$m_i\left(i=0,1,\cdots ,n\right)$ 处误差${\epsilon }_i$ 按式(17)计算得到，其中$x_{1,2}$ 同引理4。

$\begin{array}{l}\left|{\epsilon }_n\right|={\left(\frac{1}{2}\right)}^n\frac{4\sqrt{3}h\left|{{\epsilon }^{″}}_0\right|}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}\\ \left|{\epsilon }_i\right|={\left(\frac{1}{2}\right)}^i\frac{\left(3+2\sqrt{3}\right)x_1^{n-i-1}-\left(3-2\sqrt{3}\right)x_2^{n-i-1}}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}h\left|{{\epsilon }^{″}}_0\right|i=0,1,2,3,\cdots ,n-1\end{array}$ . (17)

证明：$U_1{L}_{1}M=D$ ，$L_{1}M=U_1^{-1}D$ ，具体公式如下：

$\left[\begin{array}{cccccc}1& {\beta }_0& 0& \cdots & \cdots & 0\\ 0& 1& {\beta }_1& \ddots & & ⋮\\ 0& 0& 1& \ddots & 0& ⋮\\ ⋮& \ddots & \ddots & \ddots & \ddots & 0\\ ⋮& & 0& \ddots & \ddots & {\beta }_{n-1}\\ 0& \cdots & \cdots & 0& 0& 1\end{array}\right]\left[\begin{array}{cccccc}{\theta }_0& 0& 0& \cdots & \cdots & 0\\ {\gamma }_1& {\theta }_1& 0& \ddots & & ⋮\\ 0& {\gamma }_2& {\theta }_2& \ddots & 0& ⋮\\ ⋮& 0& \ddots & \ddots & \ddots & 0\\ ⋮& & \ddots & \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& {\gamma }_n& {\theta }_n\end{array}\right]\left[\begin{array}{c}{m}_0\\ m_1\\ m_2\\ ⋮\\ ⋮\\ m_n\end{array}\right]=\left[\begin{array}{c}{d}_0\\ d_1\\ d_2\\ ⋮\\ ⋮\\ d_n\end{array}\right]$

$\left[\begin{array}{cccccc}{\theta }_0& 0& 0& \cdots & \cdots & 0\\ {\gamma }_1& {\theta }_1& 0& \ddots & & ⋮\\ 0& {\gamma }_2& {\theta }_2& \ddots & 0& ⋮\\ ⋮& 0& \ddots & \ddots & \ddots & 0\\ ⋮& & \ddots & \ddots & \ddots & 0\\ 0& \cdots & \cdots & 0& {\gamma }_n& {\theta }_n\end{array}\right]\left[\begin{array}{c}{m}_0\\ m_1\\ m_2\\ ⋮\\ ⋮\\ m_n\end{array}\right]=\left[\begin{array}{cccc}\begin{array}{cc}\begin{array}{l}{\begin{array}{cc}1& p\end{array}}_{01}\\ \begin{array}{cc}0& 1\end{array}\end{array}& \begin{array}{l}{p}_{02}\\ p_{12}\end{array}\end{array}& \begin{array}{l}\cdots \\ \ddots \end{array}& \begin{array}{l}\cdots \\ \end{array}& \begin{array}{l}{p}_{0n}\\ p_{1n}\end{array}\\ \begin{array}{cc}\begin{array}{cc}0& 0\end{array}& 1\end{array}& \ddots & \ddots & p_{2n}\\ \begin{array}{ccc}⋮& \ddots & \ddots \end{array}& \ddots & \ddots & ⋮\\ \begin{array}{ccc}\begin{array}{l}⋮\\ 0\end{array}& \begin{array}{l}\\ \cdots \end{array}& \begin{array}{l}\ddots \\ \cdots \end{array}\end{array}& \begin{array}{l}\ddots \\ 0\end{array}& \begin{array}{l}\ddots \\ 0\end{array}& \begin{array}{l}⋮\\ 1\end{array}\end{array}\right]\left[\begin{array}{c}{d}_0\\ d_1\\ d_2\\ ⋮\\ ⋮\\ d_n\end{array}\right]$

其中$p_{kj}={\left(-1\right)}^{j-k}{\displaystyle \prod _{l=k}^{j-1}{\beta }_l}\left(j>k\right)$ 。

由此可解出$m_i$ 。

$\{\begin{array}{l}{m}_0=\frac{d_0}{{\theta }_0}+\cdots +\frac{a_{0n}}{{\theta }_0}{d}_n\hfill \\ m_1=\frac{d_1}{{\theta }_1}+\cdots +\frac{a_{1n}}{{\theta }_1}{d}_n-\frac{{\gamma }_1{m}_0}{{\theta }_1}\hfill \\ \cdots \hfill \\ \begin{array}{l}{m}_i=\frac{d_i}{{\theta }_i}+\cdots +\frac{a_{in}}{{\theta }_i}{d}_n-\frac{{\gamma }_i{m}_{i-1}}{{\theta }_i}\hfill \\ \cdots \hfill \\ m_n=\frac{d_n}{{\theta }_n}-\frac{{\gamma }_n{m}_{n-1}}{{\theta }_n}\hfill \end{array}\hfill \end{array}$ . (18)

$\{\begin{array}{l}{d}_0=3\frac{y_1-y_0}{h}-\frac{h}{2}{M}_0\hfill \\ d_n=3\frac{y_n-y_{n-1}}{h}+\frac{h}{2}{M}_n\hfill \end{array}$ . (19)

由(19)式可知，$M_0$ 发生偏移只与$d_0$ 有关。而$d_0$ 只与$m_0$ 的第一项有关，于是${M^{\prime }}_0=M_0+{{\epsilon }^{″}}_0$ ，${d^{\prime }}_0=d_0-\frac{h}{2}{{\epsilon }^{″}}_0$ ，将其代入式(18)中，可得如下方程组：

$\{\begin{array}{l}\begin{array}{c}{\epsilon }_0=-\frac{1}{{\theta }_0}\frac{h{{\epsilon }^{″}}_0}{2}\\ {\epsilon }_1=\frac{{\gamma }_1}{{\theta }_1}\frac{1}{{\theta }_0}\frac{h{{\epsilon }^{″}}_0}{2}\\ \cdots \end{array}\hfill \\ {\epsilon }_i={\left(-1\right)}^{i+1}{\displaystyle \prod _{l=1}^i\frac{{\gamma }_l}{{\theta }_l}\frac{1}{{\theta }_0}\frac{h{{\epsilon }^{″}}_0}{2}}\hfill \\ \cdots \hfill \\ {\epsilon }_n={\left(-1\right)}^{n+1}{\displaystyle \prod _{l=1}^n\frac{{\gamma }_l}{{\theta }_l}\frac{1}{{\theta }_0}\frac{h{{\epsilon }^{″}}_0}{2}}\hfill \end{array}$

${\epsilon }_i$ 表示当左端点发生偏移时，内部第插值节点所对应的一阶导数偏移量。其中$b_i=2\left(i=0,1,\cdots ,n\right)$ ，$a_i=\frac{1}{2}\left(i=1,\cdots ,n-1\right),a_n=1$ ，$c_i=\frac{1}{2}\left(i=1,\cdots ,n-1\right),c_0=1$ 。

从而可得${\gamma }_i=a_i=\frac{1}{2}\left(i=1,\cdots ,n-1\right),{\gamma }_n=a_n=1$ ，${\theta }_n=b_n=2$ ，${\theta }_{n-1}=b_{n-1}-\frac{c_{n-1}}{{\theta }_n}{a}_n=2-\frac{1}{2}\times \frac{1}{2}=\frac{7}{4}$ ，${\theta }_i=b_i-\frac{c_i}{{\theta }_{i+1}}{a}_{i+1}=2-\frac{1}{4{\theta }_{i+1}}$ ${\theta }_0=b_0-\frac{c_0}{{\theta }_1}{a}_1=2-\frac{1}{2{\theta }_1}$

${\theta }_i=\{\begin{array}{l}\begin{array}{l}2i=n\\ \frac{7}{4}i=n-1\end{array}\hfill \\ \begin{array}{l}\frac{\left(3+2\sqrt{3}\right)x_1^{n-i}-\left(3-2\sqrt{3}\right)x_2^i}{\left(3+2\sqrt{3}\right)x_1^{n-i-1}-\left(3-2\sqrt{3}\right)x_2^{n-i-1}}i=1,2,3,\cdots ,n-2\\ \frac{1}{2}\frac{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}{\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}}i=0\end{array}\hfill \end{array}$

现在来计算$\left|{\epsilon }_0\right|$

${\epsilon }_0=\frac{-\left(\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}\right)h{{\epsilon }^{″}}_0}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}$

$\left|{\epsilon }_0\right|=\frac{\left(\left(3+2\sqrt{3}\right)x_1^{n-1}-\left(3-2\sqrt{3}\right)x_2^{n-1}\right)h\left|{{\epsilon }^{″}}_0\right|}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}$

现在来计算$\frac{1}{{\theta }_0}{\displaystyle \prod _{l=1}^i\frac{{\gamma }_l}{{\theta }_l}\left(i=1,2,\cdots ,n-1\right)}$ 。

$\frac{1}{{\theta }_0}{{\displaystyle \prod _{l=1}^i\frac{{\gamma }_l}{{\theta }_l}=\left(\frac{1}{2}\right)}}^i\frac{1}{{\theta }_0{\theta }_1\cdots {\theta }_i}$

${\theta }_0{\theta }_1\cdots {\theta }_i=\frac{1}{2}\frac{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}{\left(3+2\sqrt{3}\right)x_1^{n-i-1}-\left(3-2\sqrt{3}\right)x_2^{n-i-1}}$

$\frac{1}{{\theta }_0}{\displaystyle \prod _{l=1}^i\frac{{\gamma }_l}{{\theta }_l}=}{\left(\frac{1}{2}\right)}^{i-1}\frac{\left(3+2\sqrt{3}\right)x_1^{n-i-1}-\left(3-2\sqrt{3}\right)x_2^{n-i-1}}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}$

从而可得

${\epsilon }_i={\left(-1\right)}^{i+1}{\left(\frac{1}{2}\right)}^i\frac{\left(3+2\sqrt{3}\right)x_1^{n-i-1}-\left(3-2\sqrt{3}\right)x_2^{n-i-1}}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}h{{\epsilon }^{″}}_0$

$\left|{\epsilon }_i\right|={\left(\frac{1}{2}\right)}^i\frac{\left(3+2\sqrt{3}\right)x_1^{n-i-1}-\left(3-2\sqrt{3}\right)x_2^{n-i-1}}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}h\left|{{\epsilon }^{″}}_0\right|$

综上所述，$\left|{\epsilon }_i\right|\left(i=0,1,2,\cdots ,n-1\right)$ 的表达式

$\left|{\epsilon }_i\right|={\left(\frac{1}{2}\right)}^i\frac{\left(3+2\sqrt{3}\right)x_1^{n-i-1}-\left(3-2\sqrt{3}\right)x_2^{n-i-1}}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}h\left|{{\epsilon }^{″}}_0\right|$

现在来计算$\frac{1}{{\theta }_0}{\displaystyle \prod _{l=1}^n\frac{{\gamma }_l}{{\theta }_l}}$

$\frac{1}{{\theta }_0}{\displaystyle \prod _{l=1}^n\frac{{\gamma }_l}{{\theta }_l}}=\frac{1}{{\theta }_0}\frac{{\gamma }_n}{{\theta }_n}{\displaystyle \prod _{l=1}^{n-1}\frac{{\gamma }_l}{{\theta }_l}}={\left(\frac{1}{2}\right)}^n\frac{1}{{\theta }_0{\theta }_1\cdots {\theta }_{n-1}}={\left(\frac{1}{2}\right)}^{n-1}\frac{\left(3+2\sqrt{3}\right)x_1^0-\left(3-2\sqrt{3}\right)x_2^0}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}$

从而可得

$\begin{array}{l}{\epsilon }_n={\left(-1\right)}^{n+1}{\left(\frac{1}{2}\right)}^n\frac{4\sqrt{3}h{{\epsilon }^{″}}_0}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}\\ \left|{\epsilon }_n\right|={\left(\frac{1}{2}\right)}^n\frac{4\sqrt{3}h\left|{{\epsilon }^{″}}_0\right|}{{\left(3+2\sqrt{3}\right)}^2{x}_1^{n-2}-{\left(3-2\sqrt{3}\right)}^2{x}_2^{n-2}}\end{array}$

至此定理2得证。

5. 数值算例

上一节研究了三转角插值算法，在第二类边界条件下，在插值节点均匀分布的情况下，端点处的二阶导数值发生偏移时，对三次样条函数本身的影响，本章针对上一章推导和证明的定理，设计符合条件的数值算例，用以验证第三章所推导定理的正确性和有效性。

例题1：对于函数$y={\text{e}}^x,x\in \left[0,5\right]$ ，插值节点为$\left[0,1,2,3,4,5\right]$ ，并对其用三转角插值算法在第二类边界条件下构造三次样条插值，其中两端点处的二阶导数值分布设为：假设$M_0=0,M_n=0,{{\epsilon }^{″}}_n=1$ ，其中$M_0$ 表示左端点的二阶导数值，$M_n$ 表示右端点的二阶导数值，$m_i$ 表示节点处的一阶导数，$\left|\epsilon \right|$ 表示节点处一阶导数的扰动，如表1所示。

解：根据式定理1我们求出$\left|{\epsilon }_i\right|$ 。

Table 1. Perturbation values of the second derivative of the right endpoint

表1. 右端点二阶导数扰动时$\left|\epsilon \right|$

$M_0$	0	0	$\left|m-m^{\prime }\right|$	$\left|\epsilon \right|$
$m_0$	1.36940091157271	1.36860346340684	0.000797448165869241	0.000797448165869219
$m_1$	2.41604366223172	2.41763855856346	0.00159489633173848	0.00159489633173844
$m_2$	8.13359273629236	8.12801059913128	0.00558213716108469	0.00558213716108453
$m_3$	17.1513506767847	17.1720843290973	0.0207336523125967	0.0207336523125997
$m_4$	64.8882863592096	64.8109338871203	0.0773524720893164	0.0773524720893142
$m_5$	108.278370424544	108.567046660588	0.288676236044637	0.288676236044657
$M_5$	0	1

例题2：对于函数$y={\text{e}}^x,x\in \left[0,5\right]$ ，插值节点为$\left[0,1,2,3,4,5\right]$ ，并对其用三转角插值算法在第二类边界条件下构造三次样条插值，其中两端点处的二阶导数值分布设为：假设$M_0=0,M_n=0,{{\epsilon }^{″}}_0=1$ ，其中$M_0$ 表示左端点的二阶导数值，$M_n$ 表示右端点的二阶导数值，$m_i$ 表示节点处的一阶导数，$\left|\epsilon \right|$ 表示节点处一阶导数的扰动，如表2所示。

解：根据定理2我们求出$\left|{\epsilon }_i\right|$ 。

Table 2. Perturbation values of the second derivative of the left endpoint

表2. 左端点二阶导数扰动时$\left|\epsilon \right|$

$M_0$	0	1	$\left|m-m^{\prime }\right|$	$\left|\epsilon \right|$
$m_0$	1.36940091157271	1.08072467552805	0.288676236044657	0.288676236044657
$m_1$	2.41604366223172	2.49339613432104	0.0773524720893137	0.0773524720893142
$m_2$	8.13359273629236	8.11285908397976	0.0207336523126003	0.0207336523125997
$m_3$	17.1513506767847	17.1569328139458	0.00558213716108469	0.00558213716108453
$m_4$	64.8882863592096	64.8866914628778	0.00159489633173848	0.00159489633173844
$m_5$	108.278370424544	108.279167872710	0.000797448165869241	0.000797448165869219
$M_5$	0	0