插值
$a_0,a_1,\cdots,a_n$ $a_0+a_1x+\cdots+a_nx^n$ $a,b,c$ $a\sin(bx+c)$
数学模型 $f(x)$ $f(x)$
好
定义 1 . $\{x_i\}_{i=0}^n$ $n+1$ $f(x)$ $f(x_i)$ $\Phi$ $\Phi$ $g(x)$
\[g(x_i)=f(x_i), i=0,1,\cdots,n
\]
$g(x)$ $f(x)$ 插值函数 $f(x)$ 原函数 $\{x_i\}$ 插值节点
$P_n=span\{1,x,x^2,\cdots,x^n\}$ $S_n=span\{1,\cos x,\sin x,\cos 2x, \sin 2x,\cdots,\cos nx,\sin nx\}$
插值函数的存在唯一性
$g(x)$
$\Phi=span\{\phi_0(x),\cdots,\phi_m(x)\}$ $m+1$
\[g(x)=a_0\phi_0(x)+\cdots+a_m\phi_m(x)
\]
$a_0,\cdots,a_m$ $g(x)$
$g(x)$
\[g(x_i)=a_0\phi_0(x_i)+\cdots+a_m\phi_m(x_i)=f(x_i), i=0,1,\cdots,n
\]
\[\begin{pmatrix}
\phi_0(x_0) & \phi_1(x_0) & \cdots & \phi_m(x_0) \\
\phi_0(x_1) & \phi_1(x_1) & \cdots & \phi_m(x_1) \\
\vdots & \vdots & & \vdots \\
\phi_0(x_n) & \phi_1(x_n) & \cdots & \phi_m(x_n) \\
\end{pmatrix}
\begin{pmatrix}
a_0 \\ a_1 \\ \vdots \\ a_m
\end{pmatrix}
=\begin{pmatrix}
f(x_0) \\ f(x_1) \\ \vdots \\ f(x_n)
\end{pmatrix}
\]
$a_0, \cdots, a_m$
定理 1 . $m=n$
\[\left|\begin{matrix}
\phi_0(x_0) & \phi_1(x_0) & \cdots & \phi_n(x_0) \\
\phi_0(x_1) & \phi_1(x_1) & \cdots & \phi_n(x_1) \\
\vdots & \vdots & & \vdots \\
\phi_0(x_n) & \phi_1(x_n) & \cdots & \phi_n(x_n) \\
\end{matrix}\right|\neq 0
\]
$g(x)$
插值多项式
$f(x)$ $n$ $f(a)=0$ $f(x)=g(x)(x-a)$ $g(x)$ $n-1$ $f(a)=0$ $f'(a)=0$ $f(x)=g(x)(x-a)^2$ $g(x)$ $n-2$
在以后的学习中，一直使用多项式来近似
$\{x_i\}_{i=0}^n$ $n$ $P_n=span\{1,x, x^2,\cdots, x_n\}$
\[\left|\begin{matrix}
1 & x_0 & \cdots & x_0^n \\
1 & x_1 & \cdots & x_1^n \\
\vdots & \vdots & & \vdots \\
1 & x_n & \cdots & x_n^n
\end{matrix}\right|=\prod_{i,j=0, \\ j>i}^n(x_j-x_i)\neq 0
\]
定理 2 . $n+1$ $n$
待定系数法得到插值多项式
$g(x)=a_0+a_1x+\cdots+a_n x^n$
\[\begin{cases}
a_0+a_1 x_0+\cdots+a_n x_0^n & = f(x_0) \\
a_0+a_1 x_1+\cdots+a_n x_1^n & = f(x_1) \\
\cdots \\
a_0+a_1 x_n+\cdots+a_n x_n^n & = f(x_n) \\
\end{cases}
\]
$a_0, \cdots, a_n$
例 1 . $(1, 3)$ $(2, 2)$ $(3, 4)$
解 .
$g(x)=a_0+a_1 x+a_2 x^2$
\[\begin{cases}
a_0+ a_1 + a_2 &=3 \\
a_0+2a_1 + 4 a_1&=2 \\
a_0+3a_1 + 9 a_2&=4 \\
\end{cases}
\]
解 .
\[g(x)=a_0+a_1(x-2)+a_2(x-2)^2
\]
\[\begin{cases}
a_0-a_1+a_2&=3 \\
a_0&=2 \\
a_0+a_1+a_2&=4
\end{cases}
\]
解 .
\[g(x)=a_0+a_1(x-2)+a_2(x-1)(x-3)
\]
\[\begin{cases}
a_0-a_1&=3 \\
a_0-a_2&=2 \\
a_0+a_1&=4
\end{cases}
\]
方法的不足
病态矩阵 $n$ $>10$
Lagrange插值
\[g(x)=f(x_0)l_0(x)+f(x_1)l_1(x)+\cdots+f(x_n)l_n(x)
\]
$l_i(x), i=0,\cdots,n$ $n$
\[f(x_0)=g(x_0)=f(x_0)l_0(x_0)+f(x_1)l_1(x_0)+\cdots+f(x_n)l_n(x_0)
\]
\[l_0(x_0)=1, l_i(x_0)=0, i=1,2,\cdots,n
\]
\[f(x_1)=g(x_1)=f(x_0)l_0(x_1)+f(x_1)l_1(x_1)+\cdots+f(x_n)l_n(x_1)
\]
\[l_0(x_1)=0, l_1(x_1)=1, l_i(x_1)=0, i=2,3,\cdots,n
\]
$i=0,1,\cdots,n$
\[\begin{aligned}
f(x_i)=g(x_i)=f(x_0)l_0(x_i)+\cdots+f(x_i)l_i(x_i)+\cdots+f(x_n)l_n(x_i)
\end{aligned}
\]
\[l_i(x_i)=1, l_j(x_i)=0, j=0,1,\cdots,j-1,j+1,\cdots,n
\]
$l_0(x)$ $l_1(x)$ $\cdots$ $l_n(x)$ $x_0$ $\cdots$ $x_1$ $\cdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $x_n$ $\cdots$
$l_i(x)$ $l_i(x_j)= \delta_{ij}=\begin{cases} 1, & i=j \\ 0, & i\neq j \end{cases}$
$l_i(x)$ $x_0$ $x_1$ $\cdots$ $x_{i-1}$ $x_{i+1}$ $\cdots$ $x_n$
\[l_i(x)=a_i(x-x_0)\cdots(x-x_{i-1})(x-x_{i+1})\cdots(x-x_n)
\]
$a_i$ $l_i(x_i)=1$
\[a_i(x_i-x_0)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n)=1
\]
\[a_i=\frac1{(x_i-x_0)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n)}
\]
\[l_i(x)=\frac{(x-x_0)\cdots(x-x_{i-1})(x-x_{i+1})\cdots(x-x_n)}{(x_i-x_0)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n)}
\]
定义 2 .
\[l_i(x)=\frac{(x-x_0)\cdots(x-x_{i-1})(x-x_{i+1})\cdots(x-x_n)}{(x_i-x_0)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n)}
\]
Lagrange基函数
\[f(x_0)l_0(x)+f(x_1)l_1(x)+\cdots+f(x_n)l_n(x)
\]
Lagrange插值 $L_n(x)$
例 2 .
解 . $x_0$ $x_1$
\[L_1(x)=f(x_0)\frac{x-x_1}{x_0-x_1}+f(x_1)\frac{x-x_0}{x_1-x_0}
\]
例 3 . $(x_0,f(x_0))$ $(x_1,f(x_1))$ $(x_2, f(x_2))$
例 4 . $(x_0, f(x_0))$
解 .
\[\begin{aligned}
l_0(x)=\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} \\
l_1(x)=\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} \\
l_2(x)=\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} \\
\end{aligned}
\]
\[\begin{aligned}
L_2(x)=&f(x_0)\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}
+f(x_1)\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} \\
&+f(x_2)\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} \\
\end{aligned}
\]
$(x_0, f(x_0))$
例 5 . $(1, 3)$ $(2, 2)$ $(3, 4)$
解 . $1,2,3$
\[\begin{aligned}
l_0(x)=&\frac{(x-2)(x-3)}{(1-2)(1-3)} \\
l_1(x)=&\frac{(x-1)(x-3)}{(2-1)(2-3)} \\
l_2(x)=&\frac{(x-1)(x-2)}{(3-1)(3-2)} \\
\end{aligned}
\]
\[g(x)=3\frac{(x-2)(x-3)}{(1-2)(1-3)}+2\frac{(x-1)(x-3)}{(2-1)(2-3)}
+4\frac{(x-1)(x-2)}{(3-1)(3-2)}
\]
$L_n(x)$ $n$ $L_n(x)$
$\{l_i(x)\}_{i=0}^n$ $\{x_{0},\cdots,x_{k-1},x_{k+1},\cdots,x_{n}\}$ $n-1$ $P_{k}(x)$
\[L_n(x)=\frac{(x-x_j)P_{j}(x)-(x-x_i)P_{j}(x)}{x_i-x_j}
\]
Lagrange插值的伪代码
fx=0.0 ;
for i=0 ,n:
li=1.0 ;
for j=0 ,i-1 :
li=li*(t-x[j])/(x[i]-x[j]);
end for
for j=i+1 ,n:
li=li*(t-x[j])(x[i]-x[j])
end for
fx = fx+y[i]*li;
end for
// fx 即为 插值函数 在 t 处的值
$f(x), x\in[a,b]$ $L_N(x)$ $L_N(x)$ $[a,b]$
\[E=\max_{x\in[a,b]} |f(x)-L_n(x)|
\]
$E$ $[a,b]$ $E$
\[E\approx\max\{|f(y_j)-L_N(y_j)|,j=0,1,2,\cdots,500\}
\]
\[f(x)=\dfrac{1}{1+x^2}, \quad x\in[-5,5]
\]
$N=5,10,20,40$
$x_i=-5+\dfrac{10}{N}i$ $i=0,1,\cdots,N$ $x_i=-5\cos(\dfrac{2i+1}{2N+2}\pi)$ $i=0,1,\cdots,N$
$L_N(x)$
\[E_N=\max\{|f(y_j)-L_N(y_j)|,j=0,1,\cdots,500\}
\]
$ y_j=-5+10/500*j , j=0,1,\cdots,500 $
$E_N$
第1组节点误差：
N=5 , X.XXXXXXXXXXXXXXXX E xxx
N=10 , X.XXXXXXXXXXXXXXXX E xxx
N=20 , X.XXXXXXXXXXXXXXXX E xxx
N=40 , X.XXXXXXXXXXXXXXXX E xxx
第2组节点误差：
N=5 , X.XXXXXXXXXXXXXXXX E xxx
N=10 , X.XXXXXXXXXXXXXXXX E xxx
N=20 , X.XXXXXXXXXXXXXXXX E xxx
N=40 , X.XXXXXXXXXXXXXXXX E xxx
误差分析
\[R_n(x_i)=f(x_i)-L_n(x_i)=0, i=0,1,\cdots,n
\]
\[R_n(x)=K_n(x)(x-x_0)(x-x_1)\cdots(x-x_n)
\]
$K_n(x)$
$K_n(x)$
$f(x)$ $a$
\[\phi(t)=f(t)-L_n(t)-K_n(a)(t-x_0)\cdots(t-x_n)
\]
$t$
\[\phi(x_i)=f(x_i)-L_n(x_i)=0, \quad i=0,1,\cdots,n
\]
\[\phi(a)=f(a)-L_n(a)-R_n(a)=0
\]
$\{x_i\}_{i=0}^n$ $a$ $\phi(t)$
$\phi(t)$ $\{x_0,\cdots,x_n,a\}$ $[b,c]$
$\phi(t)$ $n+2$ $\phi(t)$ $n+1$ $\phi'(t)$ $(b,c)$ $\phi'(t)$ $(b,c)$ $n$ $\phi''(t)$ $\phi(t)$ $n+1$ $\xi\in(b,c)$ $\phi^{(n+1)}(\xi)=0$ $\phi(t)$ $\phi(t)$ $f(t)$ $f(x)$ $m$ $\phi(t)$ $m$
$f(x)$ $n+1$ $\xi$
\[0=\phi^{(n+1)}(\xi)=f^{(n+1)}(\xi)-K_n(a)(n+1)!
\]
\[K_n(a)=\frac{f^{(n+1)}(\xi)}{(n+1)!}
\]
$a$ 误差表达式
\[R_n(x)=f(x)-L_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}\prod_{i=0}^n(x-x_i)
\]
注 $f(x)$
例 6 . $\sin\frac{\pi}6=\frac12$ $\sin\frac{\pi}4=\frac{\sqrt 2}2$ $\sin\frac{\pi}3=\frac{\sqrt 3}2$ $\sin50^\circ$
解 . $x=50^\circ=\frac{5\pi}{18}$ $x_0=\frac{\pi}6$ $x_1=\frac{\pi}4$ $x_2=\frac{\pi}3$
$x_0$ $x_1$
\[L_1(x)=\frac12\frac{x-\frac{\pi}4}{\frac{\pi}6-\frac{\pi}4}
+\frac{\sqrt2}2\frac{x-\frac{\pi}6}{\frac{\pi}4-\frac{\pi}6}
\]
\[R_1(x)=\frac{\sin^{(2)}(\xi)}{2!}(x-\frac{\pi}6)(x-\frac{\pi}4),\quad \xi\in(\frac{\pi}6,\frac{\pi}3)
\]
$\sin50^\circ\approx L_1(\frac{5\pi}{18})\dot=0.77614$
\[\sin^{(2)}(\xi)=-\sin(\xi)\in(-\frac{\sqrt3}2,-\frac12), \quad \xi\in(\frac{\pi}6,\frac{\pi}3)
\]
\[R_1(\frac{5\pi}{18})=-\sin(\xi)\frac12\frac{\pi}{18}\frac{4\pi}{18}
\in(-0.01319,-0.00762)
\]
$\sin50^\circ=0.7660444\cdots$ $-0.01001$
$x_1$ $x_2$ $\tilde R_1$
\[\sin50^\circ\approx0.76008, \quad \tilde R_1\in(0.00538, 0.00660)
\]
$0.00596$ $x_0$ $x_1$ $x_2$
\[\begin{aligned}
L_2(x)=&\frac12\frac{(x-\frac{\pi}4)(x-\frac{\pi}3)}{(\frac{\pi}6-\frac{\pi}4)(\frac{\pi}6-\frac{\pi}3)}
+\frac{\sqrt2}2\frac{(x-\frac{\pi}6)(x-\frac{\pi}3)}{(\frac{\pi}4-\frac{\pi}6)(\frac{\pi}4-\frac{\pi}3)} \\
&+\frac{\sqrt3}2\frac{(x-\frac{\pi}6)(x-\frac{\pi}4)}{(\frac{\pi}3-\frac{\pi}6)(\frac{\pi}3-\frac{\pi}4)}
\end{aligned}
\]
\[R_2(x)=\frac{\sin^{(3)}(\xi)}{3!}(x-\frac{\pi}6)(x-\frac{\pi}4)(x-\frac{\pi}3), \xi\in(\frac{\pi}6,\frac{\pi}3)
\]
\[L_2(\frac{5\pi}{18})\approx0.76543, \quad R_2(\frac{5\pi}{18})\in(0.00044,0.00077)
\]
$0.00061$
$x$ $\{x_i\}_{i=0}^n$ 外插 内插的误差要小于外插的误差 高阶插值的误差要小于低阶插值的误差
事后误差估计
$f^{(n+1)}(\xi)$
$f(x)$ $\xi$ $f^{(n+1)}(\xi)$ $f(x)$
事后误差估计
$\{x_0,x_1,\cdots,x_n\}$ $L_n(x)$
\[f(x)-L_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)\cdots(x-x_n)
\]
$\{x_1,x_2,\cdots,x_{n+1}\}$ $\tilde L_n(x)$
\[f(x)-\tilde L_n(x)=\frac{f^{(n+1)}(\eta)}{(n+1)!}(x-x_1)\cdots(x-x_{n+1})
\]
假设 $f^{(n+1)}(\eta)\approx f^{(n+1)}(\xi)$
\[\frac{f(x)-L_n(x)}{f(x)-\tilde L_n(x)}\approx\frac{x-x_0}{x-x_{n+1}}
\]
\[f(x)\approx\frac{x-x_{n+1}}{x_0-x_{n+1}}L_n(x)+\frac{x-x_0}{x_{n+1}-x_0}\tilde L_n(x)
\]
$L_n(x)$
\[f(x)-L_n(x)\approx\frac{x-x_0}{x_{n+1}-x_0}\left(\tilde L_n(x)-L_n(x)\right)
\]
事后误差估计
注意
$f^{(n+1)}(\eta)\approx f^{(n+1)}(\xi)$
$x$ $\tilde x$ $r$
$|r|<r_0$
\[|x-\tilde x|<r_0 \Rightarrow \tilde x-r_0<x<\tilde x+r_0
\]
$x$ $|r|>r_1$
\[|x-\tilde x|>r_1 \Rightarrow x>\tilde x+r_1 \,\mbox{or}\, x<\tilde x- r_1
\]
$x$ $\pm\infty$ $\tilde x$
注