多项式插值
$$\det (A) = \left| {\begin{array}{*{20}{c}}
1&{{x_0}}&{x_0^2}& \cdots &{x_0^n}&{}\\
1&{{x_1}}&{x_1^2}& \cdots &{x_1^n}&{}\\
{}& \cdots & \cdots & \cdots & \cdots & \cdot \\
1&{{x_n}}&{x_n^2}& \cdots &{x_n^n}&{}
\end{array}} \right|{\rm{ }}! = {\rm{ 0}}$$
拉格朗日插值
1779年英国数学家爱德华·华林于最早发现拉格朗日插值法。不久后1783年,由莱昂哈德·欧拉再次发现。直到1795年,拉格朗日在其著作《师范学校数学基础教程》中发表了这个插值方法。
$${l_i}(x) = \frac{{\left( {x - {x_0}} \right) \cdots \left( {x - {x_{i - 1}}} \right)\left( {x - {x_{i + 1}}} \right) \cdots \left( {x - {x_n}} \right)}}{{\left( {{x_i} - {x_0}} \right) \cdots \left( {{x_i} - {x_{i - 1}}} \right)\left( {{x_i} - {x_{i + 1}}} \right) \cdots \left( {{x_i} - {x_n}} \right)}} = \prod\limits_{\begin{array}{*{20}{c}}
{j = 0}\\
{j \ne i}
\end{array}}^n {\frac{{x - {x_j}}}{{{x_i} - {x_j}}}} ,\quad (i = 0,1, \cdots ,n)$$
注:$\frac{{x - {x_j}}}{{{x_i} - {x_j}}}$就是求解$p(x)$的$n+1$个基函数。如果对推导过程感兴趣,参考链接里有,很简单的。
$${l_i}\left( {{x_j}} \right) = \left\{ {\begin{array}{*{20}{l}}
0&{j \ne i}\\
1&{j = i}
\end{array}} \right.$$
令
$${L_n}(x) = \sum\limits_{i = 0}^n {{y_i}} {l_i}(x) = \sum\limits_{i = 0}^n {{y_i}} \left( {\prod\limits_{\begin{array}{*{20}{c}}
{j = 0}\\
{j = i}
\end{array}}^n {\frac{{x - {x_j}}}{{{x_i} - {x_j}}}} } \right)$$
这就是$n$次的拉格朗日插值多项式,且$n+1$的$n$次拉格朗日插值多项式是惟一的。
function y=lagrange(x0,y0,x) %n个数据以数组x0,y0输入,m个插值点以数组x输入,输出数组y为m个插值。 n=length(x0);m=length(x); for i=m; z=x(i); s=0.0; for k=1:n p=1.0; for j=1:n; if j~=k p=p*(z-x0(j))/(x0(k)-x0(j)); end end s=p*y0(k)+s; end y(i)=s; end
牛顿插值
1. 差商
2. 牛顿插值公式
对比拉格朗日插值法:与拉格朗日插值法相比,牛顿插值法增加节点时,不需要重新计算,在某些情景下比拉格朗日插值法更好用,而且在计算余项时,牛顿插值的余项由于不需要导数,故$f(x)$是由离散点或者导数不存在时仍然适用,这是拉格朗日余项计算所不能比拟的。差商与导数的关系为:$$f\left[x_{0}, x_{1}, \cdots, x_{n}\right]=\frac{f^{(n)}(\xi)}{n !}$$其中,$\xi \in(\alpha, \beta), \alpha=\min _{0 \leq i \leq n}\left\{x_{i}\right\}, \beta=\max _{0 \leq i \leq n}\left\{x_{i}\right\}$
泰勒插值
埃尔米特插值
不少实际的插值问题不但要求在节点上的函数值相等,而且还要求对应的导数值也相等,甚至要求高阶导数也相等,满足这种要求的插值多项式就是埃尔米特插值多项式。
这意味着,拟合曲线与实际曲线不仅都过点$\left(x_{i}, y_{i}\right),\left(x_{i+1}, y_{i+1}\right)$,而且两点处还有相同的切线。
设在节点$a \leq x_{0}<x_{1}<\cdots<x_{n} \leq b$上,$y_{j}=f\left(x_{j}\right)$,$m_{j}=f^{\prime}\left(x_{j}\right) \quad(j=0,1, \cdots, n)$
即满足条件:$H\left(x_{j}\right)=y_{j}, \quad H^{\prime}\left(x_{j}\right)=m_{j}, \quad(j=0,1, \cdots, n)$
这里共有$2n+2$个插值条件,可唯一确定一个次数不超过$2n+1$的多项式$H_{2 n+1}(x)=H(x)$,其形式为:
$$H_{2 n+1}(x)=a_{0}+a_{1} x+\cdots+a_{2 n+1} x^{2 n+1}$$
采用拉格朗日插值多项式的基函数方法的思想:
先求出$2n+2$个插值基函数,每个基函数都是$2n+1$次多项式,且满足条件:
$$\begin{array}{l}{\alpha_{j}\left(x_{k}\right)=\delta_{j k}=\left\{\begin{array}{ll}{0,} & {j \neq k} \\ {1,} & {j=k}\end{array}, \quad \alpha_{j}^{\prime}\left(x_{k}\right)=0\right.} \\ {\beta_{j}\left(x_{k}\right)=0, \quad \beta_{j}^{\prime}\left(x_{k}\right)=\delta_{j k} \quad(j, k=0,1, \cdots, n)}\end{array}$$
用基函数表示插值多项式:
$$H_{2 n+1}(x)=\sum_{j=0}^{n}\left[y_{j} \alpha_{j}(x)+m_{j} \beta_{j}(x)\right]$$
由插值基函数所满足的条件,有$H_{2 n+1}\left(x_{k}\right)=y_{k}, \quad H_{2 n+1}^{\prime}\left(x_{k}\right)=m_{k}, \quad(k=0,1, \cdots, n)$
然后求出基函数$\alpha_{j}(x)$与$\beta_{j}(x)$
利用拉格朗日插值基函数$l_{j}(x)$,
令$\alpha_{j}(x)=(a x+b) l_{j}^{2}(x)$
得到
$$\begin{array}{l}{\alpha_{j}\left(x_{j}\right)=\left(a x_{j}+b\right) l_{j}^{2}\left(x_{j}\right)=1} \\ {\alpha_{j}^{\prime}\left(x_{j}\right)=l_{j}\left(x_{j}\right)\left[a l_{j}\left(x_{j}\right)+2\left(a x_{j}+b\right) l_{j}^{\prime}\left(x_{j}\right)\right]=0}\end{array}$$
整理得
$$\left\{\begin{array}{l}{a x_{j}+b=1} \\ {a+2 l_{j}^{\prime}\left(x_{j}\right)=0}\end{array}\right.$$
解出
$$a=-2 l_{j}^{\prime}\left(x_{j}\right), \quad b=1+2 x_{j} l_{j}^{\prime}\left(x_{j}\right)$$
由于
$$l_{j}(x)=\frac{\left(x-x_{0}\right) \cdots\left(x-x_{j-1}\right)\left(x-x_{j+1}\right) \cdots\left(x-x_{n}\right)}{\left(x_{j}-x_{0}\right) \cdots\left(x_{j}-x_{j-1}\right)\left(x_{j}-x_{j+1}\right) \cdots\left(x_{j}-x_{n}\right)}$$
两边取对数再求导,得
$$l_{j}^{\prime}\left(x_{j}\right)=\sum_{k=0 \atop k \neq j}^{n} \frac{1}{x_{j}-x_{k}}$$
于是,
$$\alpha_{j}(x)=\left(1-2\left(x-x_{j}\right) \sum_{k=0 \atop k \neq j}^{n} \frac{1}{x_{j}-x_{k}}\right) l_{j}^{2}(x)$$
同理,得到
$$\beta_{j}(x)=\left(x-x_{j}\right) l_{j}^{2}(x)$$
综上所述,埃尔米特插值多项式为:
$$\left\{ {\begin{array}{*{20}{l}}
{{H_{2n + 1}}(x) = \sum\limits_{j = 0}^n {\left[ {{y_j}{\alpha _j}(x) + {m_j}{\beta _j}(x)} \right]} }\\
{{y_j} = f\left( {{x_j}} \right),{m_j} = {f^\prime }\left( {{x_j}} \right)}\\
{{\alpha _j}(x) = \left( {1 - 2\left( {x - {x_j}} \right)\sum\limits_{k = 0}^n {\frac{1}{{{x_j} - {x_k}}}} } \right)l_j^2(x),k \ne j}\\
{{\beta _j}(x) = \left( {x - {x_j}} \right)l_j^2(x)}
\end{array}} \right..$$
仿照拉格朗日插值余项的证明方法,得到埃尔米特插值余项:
$$R(x)=f(x)-H_{2 n+1}(x)=\frac{f^{(2 n+2)}(\xi)}{(2 n+2) !} \omega_{n+1}^{2}(x)$$
$n$取1时,是一个非常重要的特例,即两点三次埃尔米特插值,比较典型和常用(4个基函数均为三次多项式,$n=1$)
插值基函数满足条件:
$$\begin{array}{l}{\alpha_{k}\left(x_{k}\right)=1, \quad \alpha_{k}\left(x_{k+1}\right)=0, \quad \alpha_{k}^{\prime}\left(x_{k}\right)=\alpha_{k}^{\prime}\left(x_{k+1}\right)=0} \\ {\alpha_{k+1}\left(x_{k}\right)=0, \quad \alpha_{k+1}\left(x_{k+1}\right)=1, \quad \alpha_{k+1}^{\prime}\left(x_{k}\right)=\alpha_{k+1}^{\prime}\left(x_{k+1}\right)=0} \\ {\beta_{k}\left(x_{k}\right)=\beta_{k}\left(x_{k+1}\right)=0, \beta_{k}^{\prime}\left(x_{k}\right)=1, \quad \beta_{k}^{\prime}\left(x_{k+1}\right)=0} \\ {\beta_{k+1}\left(x_{k}\right)=\beta_{k+1}\left(x_{k+1}\right)=0, \beta_{k+1}\left(x_{k}\right)=0, \quad \beta_{k+1}^{\prime}\left(x_{k+1}\right)=1}\end{array}$$
两点三次埃尔米特插值多项式为:
$$H_{3}(x)=y_{k} \alpha_{k}(x)+y_{k+1} \alpha_{k+1}(x)+m_{k} \beta_{k}(x)+m_{k+1} \beta_{k+1}(x)$$
余项为:
$$R_{3}(x)=\frac{1}{4 !} f^{(4)}(\xi)\left(x-x_{k}\right)^{2}\left(x-x_{k+1}\right)^{2}, \quad \xi \in\left(x_{k}, x_{k+1}\right)$$
function y-hermite(x0,y0,y1,x); %x0,y0为样本点数据,y1为导数指,m个插值点以数组x输入,输出数组y为m个插值 n=length(x0);m=length(x); for k=1:m; yy=0.0; for i=1:n h=1.0; a=0.0; for j=i:n if j~=i h=h*((x(k)-x0(j))/(x0(i)-x0(j)))^2; a=1/(x0(i)-x0(j))+a; end end yy=yy+h*((x0(i)-x0(k))*(2*a*y0(i)-y(i))+y0(i)); end y(k)=yy; end
分段插值
分段线性插值
$${l_i}(x) = \left\{ {\begin{array}{*{20}{l}}
{\frac{{x - {x_{i - 1}}}}{{{x_i} - {x_{i - 1}}}},}&{x \in \left[ {{x_{i - 1}},{x_i}} \right]{\rm{ }}i = 0}\\
{\frac{{x - {x_{i + 1}}}}{{{x_i} - {x_{i + 1}}}},}&{x \in \left[ {{x_i},{x_{i + 1}}} \right]{\rm{ }}i = n}\\
{0,}&{else}
\end{array}{\rm{ }}} \right.$$
MATLAB:
method:
nearest | 最近项插值 |
linear | 线性插值 |
spline | 逐段3次样条插值 |
cubic | 保凹凸性3次插值 |
n=11; m=61; x=-5:10/(m-1):5; y=1./(1+x.^2); z=0*x; x0=-5:10/(n-1):5; y0=1./(1+x0.^2); y1=interp1(x0, y0, x);%interp1(x0,y0,x)为Matlab中现成的分段线性插值程序 plot(x, z, 'r', x, y, 'k:', x, y1, 'r') gtext('Piece. –linear.'), gtext('y=1/(1+x^2)') title('Piecewise Linear')
分段三次埃尔米特插值
根据两点三次埃尔米特插值多项式,得到${I_h}(x)$在$\left[x_{k}, x_{k+1}\right]$的表达式为:
若在整个区间$[a, b]$上定义一组分段三次插值基函数$\alpha_{j}(x)$及$\beta_{j}(x)$,$(j=0,1, \cdots, n)$,则$I_{h}(x)$可表示为:
$${\alpha _j}(x) = \left\{ {\begin{array}{*{20}{c}}
{{{\left( {\frac{{x - {x_{j - 1}}}}{{{x_j} - {x_{j - 1}}}}} \right)}^2}\left( {1 + 2\frac{{x - {x_j}}}{{{x_{j - 1}} - {x_j}}}} \right){x_{j - 1}} \le x \le {x_j}}\\
\begin{array}{l}
{\left( {\frac{{x - {x_{j + 1}}}}{{{x_j} - {x_{j + 1}}}}} \right)^2}\left( {1 + 2\frac{{x - {x_j}}}{{{x_{j + 1}} - {x_j}}}} \right){x_j} \le x \le {x_{j + 1}}\\
{\rm{ }}0
\end{array}
\end{array}} \right.$$
$${\beta _j}(x) = \left\{ {\begin{array}{*{20}{c}}
{{{\left( {\frac{{x - {x_{j - 1}}}}{{{x_j} - {x_{j - 1}}}}} \right)}^2}\left( {x - {x_j}} \right){x_{j - 1}} \le x \le {x_j}}\\
\begin{array}{l}
{\left( {\frac{{x - {x_{j + 1}}}}{{{x_j} - {x_{j + 1}}}}} \right)^2}\left( {x - {x_j}} \right){x_j} \le x \le {x_{j + 1}}\\
{\rm{ }}0
\end{array}
\end{array}} \right.$$
MATLAB:来自官方文档 https://ww2.mathworks.cn/help/matlab/ref/pchip.html
spline
和 pchip
为两个不同函数生成的插值结果进行比较。x = -3:3; y = [-1 -1 -1 0 1 1 1]; xq1 = -3:.01:3; p = pchip(x,y,xq1); %分段三次埃尔米特插值 s = spline(x,y,xq1); %逐段3次样条插值plot(x,y,'o',xq1,p,'-',xq1,s,'-.') legend('Sample Points','pchip','spline','Location','SouthEast')
三次样条插值
分段插值的优点是误差小,稳定性高,但是曲线的光滑性不好,而许多实际问题需要插值曲线有较高阶的整体光滑性,理论上是可以使用高次埃尔米特插值或分段高次埃尔米特插值,但是必须知道每一个点的导数是不现实的。所以,样条插值诞生了。理论推导参考:https://www.cnblogs.com/duye/p/8671820.html1、y=interp1(${x_0}$,${y_0}$,$x$,'spline');
% spline改成linear,则变成线性插值
2、y=spline(${x_0}$,${y_0}$,$x$);
% 这个是根据己知的x,y数据,用样条函数插值出${x_i}$处的值。
% 由己知的x,y数据,求出它的样条函数表达式,不过该表达式不是用矩阵直接表示,要求点x`的值,要用函数
3、pp=csape(x,y,conds); y=ppval(pp,$x$);
% 各种边界条件的三次样条插值
conds是指边界条件,边界类型可为:
①'complete': 给定边界一阶导数,默认的边界条件
②'not-a-knot':非扭结条件,不用给边界值.
③'periodic': 周期性边界条件,不用给边界值.
④'second': 给定边界二阶导数.
⑤'variational': 自然样条(边界二阶导数为[0,0]。
clear clc x0=[0,3,5,7,9,11,12,13,14,15]; y0=[0,1.2,1.7,2.0,2.1,2.0,1.8,1.2,1.0,1.6]; t=0:0.05:15; y1=spline(x0,y0,t); dy1=(spline(x0,y0,0.0001)-spline(x0,y0,0))/0.0001%x=0处斜率 min1=min(spline(x0,y0,13:0.001:15))%13到15最小值 figure plot(x0,y0,'ro',t,y1);%画出曲线 title('三次样条插值');