最小二乘拟合曲线
最小二乘拟合曲线
给定逼近函数列 { φ k } k = 0 n \left\{\varphi_k \right\}_{k=0}^{n} {φk}k=0n,和样本点列 ( x i , y i ) , i = 0 , 1 , 2 , 3 , . . . , m (x_i,y_i),i=0,1,2,3,...,m (xi,yi),i=0,1,2,3,...,m,参数向量方程组如下给出:
G a = d \boldsymbol{Ga=d} Ga=d
其中:
G = ( ( φ 0 , φ 0 ) ( φ 0 , φ 1 ) ⋯ ( φ 0 , φ n ) ( φ 1 , φ 0 ) ( φ 1 , φ 1 ) ⋯ ( φ 1 , φ n ) ⋮ ⋮ ⋯ ⋮ ( φ n , φ 0 ) ( φ n , φ 1 ) ⋯ ( φ n , φ n ) ) \boldsymbol{G}=\begin{pmatrix} (\varphi_0,\varphi_0) &(\varphi_0,\varphi_1) &\cdots &(\varphi_0,\varphi_n) \\ (\varphi_1,\varphi_0) &(\varphi_1,\varphi_1) &\cdots &(\varphi_1,\varphi_n) \\ \vdots &\vdots &\cdots & \vdots\\ (\varphi_n,\varphi_0) &(\varphi_n,\varphi_1) &\cdots &(\varphi_n,\varphi_n) \\ \end{pmatrix} G= (φ0,φ0)(φ1,φ0)⋮(φn,φ0)(φ0,φ1)(φ1,φ1)⋮(φn,φ1)⋯⋯⋯⋯(φ0,φn)(φ1,φn)⋮(φn,φn)
a = ( a 0 a 1 ⋮ a n ) \boldsymbol{a}=\begin{pmatrix} a_0\\a_1\\\vdots\\a_n \end{pmatrix} a= a0a1⋮an
d = ( d 0 d 1 ⋮ d n ) \boldsymbol{d}=\begin{pmatrix} d_0\\d_1\\\vdots\\d_n \end{pmatrix} d= d0d1⋮dn
Remark
-
( φ j , φ k ) = ∑ i = 0 m ω ( x i ) φ j ( x i ) φ k ( x i ) (\varphi_j,\varphi_k)=\sum_{i=0}^{m}\omega(x_i)\varphi_j(x_i)\varphi_k(x_i) (φj,φk)=i=0∑mω(xi)φj(xi)φk(xi)
-
d k = ( f , φ k ) = ∑ i = 0 n ω ( x i ) f ( x i ) φ j ( x i ) d_k=(f,\varphi_k)=\sum_{i=0}^{n}\omega(x_i)f(x_i)\varphi_j(x_i) dk=(f,φk)=i=0∑nω(xi)f(xi)φj(xi)
例
已知一组实验数据如表,求它的拟合曲线。
x i x_i xi | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|
f ( x i ) f(x_i) f(xi) | 4 | 4.5 | 6 | 8 | 8.5 |
ω ( x i ) \omega (x_i) ω(xi) | 2 | 1 | 3 | 1 | 1 |
解
令 S 1 ( x ) = a 0 + a 1 x S_1(x)=a_0 + a_1x S1(x)=a0+a1x,这里 m = 4 , n = 1 m = 4,n = 1 m=4,n=1, φ 0 ( x ) = 1 , φ 1 ( x ) = x \varphi_0(x)=1,\varphi_1(x)=x φ0(x)=1,φ1(x)=x
故
( φ 0 , φ 0 ) = ∑ i = 0 4 ω i = 8 ( φ 0 , φ 1 ) = ( φ 1 , φ 0 ) = ∑ i = 0 4 ω i x i = 22 ( φ 1 , φ 1 ) = ∑ i = 0 4 ω i x i 2 = 74 ( φ 0 , f ) = ∑ i = 0 4 ω i f i = 47 ( φ 1 , f ) = ∑ i = 0 4 ω i x i f i = 145.5 \begin{align*} (\varphi_0,\varphi_0)&=\sum_{i = 0}^{4}\omega_i = 8\\ (\varphi_0,\varphi_1)&=(\varphi_1,\varphi_0)=\sum_{i = 0}^{4}\omega_ix_i = 22\\ (\varphi_1,\varphi_1)&=\sum_{i = 0}^{4}\omega_ix_i^2 = 74\\ (\varphi_0,f)&=\sum_{i = 0}^{4}\omega_if_i = 47\\ (\varphi_1,f)&=\sum_{i = 0}^{4}\omega_ix_if_i = 145.5 \end{align*} (φ0,φ0)(φ0,φ1)(φ1,φ1)(φ0,f)(φ1,f)=i=0∑4ωi=8=(φ1,φ0)=i=0∑4ωixi=22=i=0∑4ωixi2=74=i=0∑4ωifi=47=i=0∑4ωixifi=145.5
带入法方程得到:
( a 0 a 1 ) = ( 2.5648 1.2037 ) \begin{pmatrix} a_0\\ a_1 \end{pmatrix}=\begin{pmatrix} 2.5648\\ 1.2037 \end{pmatrix} (a0a1)=(2.56481.2037)