前言
本篇为插值法专栏第四篇内容讲述,此章主要讲述 Hermite(埃尔米特)插值法及matlab代码,其中也给出详细的例子让大家更好的理解Hermite插值法提示
之前已经介绍牛顿插值法和三次样条插值,如果没看过前两篇的可以点击以下链接阅读
- 数值分析(一)牛顿插值法
- 数值分析(二)三次样条插值法
- 数值分析(二续) 三次样条插值二类边界完整matlab代码
一、Hermite插值
在许多实际应用中,不仅要求函数值相等,而且要求若干阶导数也相等。
1. Hermite定理
定义:满足函数值相等且导数也相等的插值方法 f ( x ) ≈ φ ( x ) f(x) \approx \varphi (x) f(x)≈φ(x), φ ( x i ) = f ( x i ) ( i = 0 , 1 , … , n ) \varphi ({x_i}) = f({x_i}) (i=0,1,\ldots ,n) φ(xi)=f(xi)(i=0,1,…,n) φ ′ ( x i ) = f ′ ( x i ) \varphi '({x_i}) = f'({x_i}) φ′(xi)=f′(xi) φ ( 2 ) ( x i ) = f ( 2 ) ( x i ) \varphi ^{(2)} ({x_i}) = f ^{(2)} ({x_i}) φ(2)(xi)=f(2)(xi) ⋮ \vdots ⋮ φ ( m ) ( x i ) = f ( m ) ( x i ) \varphi ^{(m)} ({x_i}) = f ^{(m)} ({x_i}) φ(m)(xi)=f(m)(xi)
定理:设 f ( x ) ∈ C n [ a , b ] , x 0 , . . . , x n f(x) \in {C^n}[a,b],{x_0},...,{x_n} f(x)∈Cn[a,b],x0,...,xn 为 [ a , b ] [a,b] [a,b] 上的互异节点,则 f [ x 0 , . . . , x n ] f[{x_0},... ,{x_n}] f[x0,...,xn] 是其变量的连续函数
2. 重节点差商
注意:
差商知识不清楚的话可以看之前这篇 数值分析(一)牛顿插值法 中有对差商详细的讲解
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f[{x_0},{x_0}] = \mathop {\lim }\limits_{{x_1} \to {x_0}} f[{x_0},{x_1}] = \frac{{f({x_1}) - f({x_0})}}{{{x_1} - {x_0}}} = f'({x_0})
f[x0,x0]=x1→x0limf[x0,x1]=x1−x0f(x1)−f(x0)=f′(x0)
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f[{x_0},{x_0},{x_0}] = \mathop{\lim }\limits_{{x_1} \to {x_0}\\{{x_2} \to {x_0}}}f[{x_0},{x_1},{x_2}] = \frac{1}{{2!}}f''({x_0})
f[x0,x0,x0]=x1→x0x2→x0limf[x0,x1,x2]=2!1f′′(x0)
一般
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f[{x_0},{x_0},\cdots,{x_0}] = \mathop{\lim }\limits_{{x_i} \to {x_0}}f[{x_0},{x_1},\cdots,{x_n}] = \frac{1}{{n!}}f^{(n)}({x_0})
f[x0,x0,⋯,x0]=xi→x0limf[x0,x1,⋯,xn]=n!1f(n)(x0)
3. 重节点Newton插值
在Newton插值公式中,令 x i → x 0 , i = 1 , ⋯ , n x_i \to x_0, i = 1, \cdots, n xi→x0,i=1,⋯,n,则 N n ( x ) = f ( x 0 ) + f [ x 0 , x 1 ] ( x − x 0 ) + ⋯ + f [ x 0 , x 1 , ⋯ , x n ] ∏ i = 1 n − 1 ( x − x i ) = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + ⋯ + f ( n ) ( x 0 ) n ! ( x − x 0 ) n N_n(x) = f(x_0)+f[x_0, x_1](x - x_0)+ \cdots + f[x_0, x_1, \cdots, x_n]\prod_{i=1 }^{n-1} (x-x_i) \\ = f(x_0) + f'(x_0)(x-x_0)+ \cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n Nn(x)=f(x0)+f[x0,x1](x−x0)+⋯+f[x0,x1,⋯,xn]i=1∏n−1(x−xi)=f(x0)+f′(x0)(x−x0)+⋯+n!f(n)(x0)(x−x0)n
担心有些同学看不懂推导,解释一下如何转化为第二个等式(即,Taylor 插值多项式):
将1.1中讲述的重节点差商 带入Newton插值公式中: lim x i → x 0 f [ x 0 , x 1 , ⋯ , x n ] = 1 n ! f ( n ) ( x 0 ) \mathop{\lim }\limits_{{x_i} \to {x_0}}f[{x_0},{x_1},\cdots,{x_n}]=\frac{1}{{n!}}f^{(n)}({x_0}) xi→x0limf[x0,x1,⋯,xn]=n!1f(n)(x0)所有的 x i → x 0 x_i \to x_0 xi→x0,所以 ∏ i = 1 n − 1 ( x − x i ) \prod_{i=1 }^{n-1} (x-x_i) ∏i=1n−1(x−xi) 就变成了 ( x − x 0 ) n (x-x_0)^n (x−x0)n
插值余项 R n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x − x 0 ) n + 1 R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} Rn(x)=(n+1)!f(n+1)(ξ)(x−x0)n+1
4. Hermite 插值公式
一般来说,给定 m + 1 m+1 m+1 个插值条件,就可以构造出一个 m m m 次Hermite 插值多项式,接下来介绍两个典型的Hermite插值:三点三次 Hermite插值 和 两点三次 Hermite 插值
4.1 三点三次 Hermite插值
插值节点:
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x_0,x_1, x_2
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插值条件:
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P(x_i) = f(x_i),i=0,1,2,P'(x_1) = f'(x_1)
P(xi)=f(xi),i=0,1,2,P′(x1)=f′(x1)
设:
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P(x) = f(x_0) + f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)+A(x-x_0)(x-x_1)(x-x_2)
P(x)=f(x0)+f[x0,x1](x−x0)+f[x0,x1,x2](x−x0)(x−x1)+A(x−x0)(x−x1)(x−x2)
将
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P'(x_1) = f'(x_1)
P′(x1)=f′(x1)代入可得
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A = \frac{f'(x_1)-f[x_0,x_1] - f[x_0,x_1,x_2](x_1-x_0)}{(x_1-x_0)(x_1-x_2)}
A=(x1−x0)(x1−x2)f′(x1)−f[x0,x1]−f[x0,x1,x2](x1−x0)
由于
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R(x)的零点,且
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x1是二重零点,故可设,余项公式:
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R(x) = f(x) - P(x) = k(x)(x-x_0)(x-x_1)^2(x-x_2)
R(x)=f(x)−P(x)=k(x)(x−x0)(x−x1)2(x−x2)与 Lagrange 插值余项公式的推导过程类似,可得
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R(x) = \frac{f^{(4)}(\xi_x)}{4!}(x-x_0)(x-x_1)^2(x-x_2)
R(x)=4!f(4)(ξx)(x−x0)(x−x1)2(x−x2)
其中
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4.2 两点三次 Hermite插值
插值节点:
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插值条件:
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P(x_i) = f(x_i)=y_i,P'(x_i) = f'(x_i)=m_i,i=0,1
P(xi)=f(xi)=yi,P′(xi)=f′(xi)=mi,i=0,1
模仿Lagrange 多项式的思想,设
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H_3(x) = y_0\alpha_0(x)+y_1\alpha_1(x)+m_0\beta_0(x)+m_1\beta_1(x)
H3(x)=y0α0(x)+y1α1(x)+m0β0(x)+m1β1(x)其中
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\alpha_0(x), \alpha_1(x), \beta_0(x), \beta_1(x)
α0(x),α1(x),β0(x),β1(x)均为3次多项式,且满足
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\alpha_j(x_i) = \delta_{ij}, \alpha'_j(x_i)=0,\\ \beta_j(x_i)=0, \beta'_j(x_i) = \delta_{ji} \\ (i, j =0, 1)
αj(xi)=δij,αj′(xi)=0,βj(xi)=0,βj′(xi)=δji(i,j=0,1)
上述中的 δ i j \delta_{ij} δij表达式在Lagrange多项式中提到过,怕有些同学忘了这里帮忙回顾一下
l k ( x i ) = δ k i = { 1 ( i = k ) 0 ( i ≠ k ) l_k(x_i) = \delta_{ki} = \left \{\begin{matrix}1&(i=k) \\ 0&(i \ne k)\end{matrix}\right. lk(xi)=δki={10(i=k)(i=k)
将插值条件代入立即可得 α 0 ( x ) \alpha_0(x) α0(x)求解过程
Step 1:根据前面可以得到: α 0 ( x 0 ) = 1 , α 0 ′ ( x 0 ) = 0 , α 0 ( x 1 ) = 0 , α 0 ′ ( x 1 ) = 0 \alpha_0(x_0) = 1, \alpha'_0(x_0)=0, \alpha_0(x_1) = 0, \alpha'_0(x_1)=0 α0(x0)=1,α0′(x0)=0,α0(x1)=0,α0′(x1)=0
Step 2:由上面提到过
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\alpha_0(x), \alpha_1(x), \beta_0(x), \beta_1(x)
α0(x),α1(x),β0(x),β1(x)都是3次多项式,
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\alpha_0(x_1) = 0, \alpha'_0(x_1)=0
α0(x1)=0,α0′(x1)=0,根据以上条件可以构造函数为:
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\alpha_0(x) = (ax + b)(\frac{x-x_1}{x_0-x_1})^2
α0(x)=(ax+b)(x0−x1x−x1)2
Step 3:将
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α0(x0)=1,α0′(x0)=0 代入上式,可以求解得到
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a = -\frac{2}{x_0-x_1}, b = \frac{3x_0-x_1}{x_0-x_1} = 1+ \frac{2x_0}{x_0-x_1}
a=−x0−x12,b=x0−x13x0−x1=1+x0−x12x0
Step 4:将求得出来的
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α0(x) 中,即得:
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\alpha_0(x) = (1+2\frac{x-x_0}{x_1-x_0})(\frac{x-x_1}{x_0-x_1})^2
α0(x)=(1+2x1−x0x−x0)(x0−x1x−x1)2
注意
:我在推公式的过程感觉是2前面是个减号(很容易出现自己推导和给出的不一致),我仔细一看发现是这么一回事,那么我来把省略的一些步骤给他详细写一下
a x + b = − 2 x x 0 − x 1 + 1 + 2 x 0 x 0 − x 1 = 1 − 2 x − x 0 x 0 − x 1 = 1 + 2 x − x 0 x 1 − x 0 ax+b = -\frac{2x}{x_0-x_1}+1+ \frac{2x_0}{x_0-x_1} = 1 - 2\frac{x-x_0}{x_0-x_1} = 1 + 2\frac{x-x_0}{x_1-x_0} ax+b=−x0−x12x+1+x0−x12x0=1−2x0−x1x−x0=1+2x1−x0x−x0才发现将-
放到了分母
同理: α 1 ( x ) = ( 1 + 2 x − x 1 x 0 − x 1 ) ( x − x 0 x 1 − x 0 ) 2 \alpha_1(x) = (1+2\frac{x-x_1}{x_0-x_1})(\frac{x-x_0}{x_1-x_0})^2 α1(x)=(1+2x0−x1x−x1)(x1−x0x−x0)2
类似得可以得到: β 0 ( x ) = ( x − x 0 ) ( x − x 1 x 0 − x 1 ) 2 \beta_0(x) = (x-x_0)(\frac{x-x_1}{x_0-x_1})^2 β0(x)=(x−x0)(x0−x1x−x1)2 β 1 ( x ) = ( x − x 1 ) ( x − x 0 x 1 − x 0 ) 2 \beta_1(x) = (x-x_1)(\frac{x-x_0}{x_1-x_0})^2 β1(x)=(x−x1)(x1−x0x−x0)2
满足插值条件
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P(x_0) = f(x_0)=y_0,P'(x_0) = f'(x_0)=m_0 \\ P(x_1) = f(x_1)=y_1,P'(x_1) = f'(x_1)=m_1
P(x0)=f(x0)=y0,P′(x0)=f′(x0)=m0P(x1)=f(x1)=y1,P′(x1)=f′(x1)=m1 的三次Hermite插值多项式为
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H_3(x) = y_0(1+2\frac{x-x_0}{x_1-x_0})(\frac{x-x_1}{x_0-x_1})^2+y_1 (1+2\frac{x-x_1}{x_0-x_1})(\frac{x-x_0}{x_1-x_0})^2 \\ +m_0(x-x_0)(\frac{x-x_1}{x_0-x_1})^2+m_1(x-x_1)(\frac{x-x_0}{x_1-x_0})^2
H3(x)=y0(1+2x1−x0x−x0)(x0−x1x−x1)2+y1(1+2x0−x1x−x1)(x1−x0x−x0)2+m0(x−x0)(x0−x1x−x1)2+m1(x−x1)(x1−x0x−x0)2
余项 R 3 ( x ) = f ( 4 ) ( ξ x ) 4 ! ( x − x 0 ) 2 ( x − x 1 ) 2 R_3(x) = \frac{f^{(4)}(\xi_x)}{4!}(x-x_0)^2(x-x_1)^2 R3(x)=4!f(4)(ξx)(x−x0)2(x−x1)2其中 ξ x ∈ ( x 0 , x 1 ) \xi_x \in (x_0, x_1) ξx∈(x0,x1)
4.3 2 n + 1 2n+1 2n+1次Hermite插值多项式
由上面 4.2 中给出了
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\alpha_0(x), \alpha_1(x), \beta_0(x), \beta_1(x)
α0(x),α1(x),β0(x),β1(x) 的表达式,那么依次类推推导出
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2n+1 次Hermite插值多项式,且条件为
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H(x_i) = f(x_i),H'(x_i) = f'(x_i),(i=0,1,\cdots,n)
H(xi)=f(xi),H′(xi)=f′(xi),(i=0,1,⋯,n)
求解:
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\alpha_j(x) = [1-2(x-x_j)\sum_{\begin{matrix} k = 0\\ k\ne j \end{matrix}}^{n}\frac{1}{x_j-x_k}]l_j^2(x)
αj(x)=[1−2(x−xj)k=0k=j∑nxj−xk1]lj2(x)
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\beta_j(x) = (x-x_j)l_j^2(x)
βj(x)=(x−xj)lj2(x)
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H_{2n+1}(x) = \sum_{j=0}^{n}[1-2(x-x_j)\sum_{\begin{matrix} k = 0\\ k\ne j \end{matrix}}^{n}\frac{1}{x_j-x_k}]l_j^2(x)f(x_j)+\sum_{j=0}^{n}(x-x_j)l_j^2(x)f'(x_j)
H2n+1(x)=j=0∑n[1−2(x−xj)k=0k=j∑nxj−xk1]lj2(x)f(xj)+j=0∑n(x−xj)lj2(x)f′(xj)
上述中的 l j ( x ) l_j(x) lj(x) 在 Lagrange 多项式中提到过不太清楚得同学可以去回顾一下
二、Hermite插值算法及matlab代码
Matlab 版本号 2022b
1. 2 n + 1 2n+1 2n+1次Hermite插值matlab代码实现
function y=Hermitezi(X,Y,Y1,x)
% 2n+1次Hermite插值函数
% X为已知数据点的x坐标
% Y为已知数据点的y坐标
% Y1为数据点y值导数
% x0为插值点的x坐标
if(length(X) == length(Y))
if(length(X) == length(Y1))
n=length(X);
else
disp('y和y的导数维数不相等');
renturn;
end
else
disp('x和y的维数不相等');
return;
end
%以上为输入判断和确定“n”的值
m=length(x);
for t=1:m
z=x(t);s=0.0;
for i=1:n
h=1.0;
a=0.0;
for j=1:n
if(j~= i)
h=h*(z-X(j))^2/((X(i)-X(j))^2);%求得值为(li(x))^2
a=a+1/(X(i)-X(j)); %求得ai(x)表达式之中的累加部分
end
end
s=s+h*((X(i)-z)*(2*a*Y(i)-Y1(i))+Y(i));
end
y(t)=s;
end
end
2. 例题
主函数:
x=[0,1];
y=[1,2];
y1=[0.5,0.5];
y=Hermitezi(x,y,y1,0.724)
结果:
三、总结
这篇blog我推迟了1年才开始写,本应该很早就把这篇写完,由于比较忙遇到一些事情,现在重新得拾起开始撰写分享给大家,也非常感谢大家对我的关注,今年的计划也是将这个插值法专栏更新完,之后会再转换到另一个专栏的撰写。下方有专栏链接,可以订阅一下防止找不到。文章来源:https://www.toymoban.com/news/detail-497897.html
四、插值法专栏
专栏链接:插值法专栏,如果对你有帮助的话可以点个赞,点个订阅,我将完善此专栏文章来源地址https://www.toymoban.com/news/detail-497897.html
- 数值分析(一) 牛顿插值法及matlab代码
- 数值分析(二) 三次样条插值法matlab程序
- 数值分析(二续) 三次样条插值二类边界完整matlab代码
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