一般的最小二乘逼近(曲线拟合的最小二乘法)
最小二乘法的一般提法是:对给定的一组数据
(i=0,1,...,m),要求在函数类
中找一个函数
,使误差平方和
![\left \| \delta \right \|_{2}^{2}=\sum_{i=0}^{m}\delta _{i}^{2}=\sum_{i=0}^{m}\left [ S^{*}(x_{i})-y_{i} \right ]^{2}=_{S(x)\in \varphi }^{min}\sum_{i=0}^{m}\left [S(x_{i})-y_{i} \right ]^{2}](/s/com/codecogs/latex/G.http/gif.latex?%5Cleft%20%5C%7C%20%5Cdelta%20%5Cright%20%5C%7C_%7B2%7D%5E%7B2%7D=%5Csum_%7Bi=0%7D%5E%7Bm%7D%5Cdelta%20_%7Bi%7D%5E%7B2%7D=%5Csum_%7Bi=0%7D%5E%7Bm%7D%5Cleft%20%5B%20S%5E%7B*%7D%28x_%7Bi%7D%29-y_%7Bi%7D%20%5Cright%20%5D%5E%7B2%7D=_%7BS%28x%29%5Cin%20%5Cvarphi%20%7D%5E%7Bmin%7D%5Csum_%7Bi=0%7D%5E%7Bm%7D%5Cleft%20%5BS%28x_%7Bi%7D%29-y_%7Bi%7D%20%5Cright%20%5D%5E%7B2%7D)
其中
(n<m)
带权的最小二乘法:
![\left \| \delta\right \|_{2}^{2}=\sum_{i=0}^{m}\omega (x_{i})\left [ S(x_{i})-f(x_{i}) \right ]^{2}](/s/com/codecogs/latex/G.http/gif.latex?%5Cleft%20%5C%7C%20%5Cdelta%5Cright%20%5C%7C_%7B2%7D%5E%7B2%7D=%5Csum_%7Bi=0%7D%5E%7Bm%7D%5Comega%20%28x_%7Bi%7D%29%5Cleft%20%5B%20S%28x_%7Bi%7D%29-f%28x_%7Bi%7D%29%20%5Cright%20%5D%5E%7B2%7D)
其中ω(x)≥0是[a,b]上的权函数。
用最小二乘法求曲线拟合的问题,就是在S(x)中求一函数
,使
取的最小。它转化为求多元函数
![I(a_{0},a_{1},...a_{n})=\sum_{i=0}^{m}\omega (x_{i}) \sum_{j=0}^{n}\left [ a_{j}\varphi_{j}(x_{i})-f(x_{i}) \right ]^{2}](/s/com/codecogs/latex/G.http/gif.latex?I%28a_%7B0%7D,a_%7B1%7D,...a_%7Bn%7D%29=%5Csum_%7Bi=0%7D%5E%7Bm%7D%5Comega%20%28x_%7Bi%7D%29%20%5Csum_%7Bj=0%7D%5E%7Bn%7D%5Cleft%20%5B%20a_%7Bj%7D%5Cvarphi_%7Bj%7D%28x_%7Bi%7D%29-f%28x_%7Bi%7D%29%20%5Cright%20%5D%5E%7B2%7D)
的极小点
问题。
由求多元函数极值的必要条件,有
![\frac{\partial I}{\partial a_{k}}=2\sum_{i=0}^{m}\omega(x_{i})\left [ \sum_{j=0}^{n} a_{j}\varphi _{j}(x_{i})-f(x_{i}) \right ] \varphi_{k}(x_{i})=0](/s/com/codecogs/latex/G.http/gif.latex?%5Cfrac%7B%5Cpartial%20I%7D%7B%5Cpartial%20a_%7Bk%7D%7D=2%5Csum_%7Bi=0%7D%5E%7Bm%7D%5Comega%28x_%7Bi%7D%29%5Cleft%20%5B%20%5Csum_%7Bj=0%7D%5E%7Bn%7D%20a_%7Bj%7D%5Cvarphi%20_%7Bj%7D%28x_%7Bi%7D%29-f%28x_%7Bi%7D%29%20%5Cright%20%5D%20%5Cvarphi_%7Bk%7D%28x_%7Bi%7D%29=0)
(k=0,1,...,n)
若记


(k=0,1,...,n)
则上式可写为:

(k=0,1,...,n);
这个方程称为法方程,矩阵形式
G a=d.
其中
,
,

由于
线性无关,故
,方程存在唯一解
(k=0,1,...,n),
从而得到函数f(x)的最小二乘解为
