【数值分析干货】第五章 插值与逼近

插值条件

$$\phi (x_i)=f(x_i)=y_i,i=0,1,...,n$$

唯一性定理

给定{(xi,yi)∣i=0,1,...,n}\left \{ (x_i,y_i) |i=0,1,...,n \right \}{(xi​,yi​)∣i=0,1,...,n},则满足插值条件的$n$次多项式$p_n(x)$唯一.

$n$次拉格朗日插值多项式

$$L_n(x)=\sum _{k=0}^n{l}_k(x)y_k$$

$$R_n(x)=\frac{f^{(n+1)}({\xi }_x)}{(n+1)!}{\omega }_{n+1}(x)$$

$${\xi }_x\in (a,b)且与x有关$$

$n$次拉格朗日基函数

形式
$$l_k(x)=\prod _{\begin{array}{c}j=0\\ j\ne k\end{array}}^n\frac{x-x_j}{x_k-x_j}$$
性质
$$取f(x)=x^k,k=0,1,...\Rightarrow \sum _{i=0}^n{l}_i(x)x_i^k=x^k$$

$$特别地，当k=0时,\sum _{i=0}^n{l}_i(x)=1$$

Newton插值多项式

差商

形式
$$f[x_0,x_1]=\frac{f(x_1)-f(x_0)}{x_1-x_0}$$

$$...$$

$$f[x_0,x_1,...,x_n]=\frac{f[x_1,x_2,...,x_n]-f[x_0,x_1,...,x_{n-1}]}{x_n-x_0}$$
性质
1.线性组合
2.任意排列
3.$n⩾k$时，$f[x_0,x_1,...,x_{n-1},x]$是$n-k$次多项式；$n<k$时，$f[x_0,x_1,...,x_{n-1},x]=0$
4.$f[x_0,x_1,...,x_k]=\frac{f^{(k)}(\xi )}{k!}$

Newton插值多项式

形式
$$f(x)=N_n(x)+R_n(x)$$

$$N_n(x)=f(x_0)+(x-x_0)f[x_0,x_1]+...+(x-x_0)(x-x_1)...(x-x_{n-1})f[x_0,x_1,...,x_n]$$

$$R_n(x)=(x-x_0)(x-x_1)...(x-x_n)f[x_0,x_1,...,x_n,x]$$

$$f[x_0,x_1,...,x_n,x]=\frac{f^{(n)}({\xi }_x)}{(n+1)!}$$

递推
$$N_{k+1}(x)=N_k(x)+{\omega }_{k+1}(x)f[x_0,x_1,...,x_{k+1}]$$

Hermite插值

待续。。。

正则多项式

待续。。。

最小二乘法

给定数据集{(xi,yi)∣i=0,1,...,m}\left \{ (x_i,y_i) |i=0,1,...,m \right \}{(xi​,yi​)∣i=0,1,...,m},经验函数$y={\sum }_{j=0}^n{\phi }_j(x)a_j$,则${\phi }_j=({\phi }_j(x_0),{\phi }_j(x_1),...,{\phi }_j(x_m){)}^T$，$f=(y_0,y_1,...,y_m{)}^T$，$({\phi }_j,{\phi }_k)={\sum }_{i=0}^m\rho (x_i){\phi }_j(x_i){\phi }_k(x_i)$，$(f,{\phi }_j)={\sum }_{i=0}^m\rho (x_i){\phi }_j(x_i)y_i$
正则方程组为
$$\left[\begin{array}{cccc}({\phi }_0,{\phi }_0)& ({\phi }_0,{\phi }_1)& ...& ({\phi }_0,{\phi }_n)\\ ({\phi }_1,{\phi }_0)& ...& ...& ...\\ ...& ...& ...& ...\\ ({\phi }_n,{\phi }_0)& ...& ...& ({\phi }_n,{\phi }_n)\end{array}\right]\left[\begin{array}{c}{a}_0\\ a_1\\ ...\\ a_n\end{array}\right]=\left[\begin{array}{c}(f,{\phi }_0)\\ (f,{\phi }_1)\\ ...\\ (f,{\phi }_n)\end{array}\right]$$
拟合曲线
$${\phi }^{\ast }(x)=p_n^{\ast }(x)=\sum _{i=0}^n{\phi }_i^{\ast }(x)a_i^{\ast }$$
均方误差
$${\Vert {\delta }^{\ast }\Vert }_2={\left[\sum _{i=0}^m{\rho }_i{\left({\phi }^{\ast }(x_i)-y_i\right)}^2\right]}^{\frac{1}{2}}$$