最小二乘拟合曲线

发布于:2025-05-15 ⋅ 阅读:(20) ⋅ 点赞:(0)

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= (φ0φ0)(φ1φ0)(φnφ0)(φ0φ1)(φ1φ1)(φnφ1)(φ0φn)(φ1φn)(φnφn)

a = ( a 0 a 1 ⋮ a n ) \boldsymbol{a}=\begin{pmatrix} a_0\\a_1\\\vdots\\a_n \end{pmatrix} a= a0a1an

d = ( d 0 d 1 ⋮ d n ) \boldsymbol{d}=\begin{pmatrix} d_0\\d_1\\\vdots\\d_n \end{pmatrix} d= d0d1dn

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=0mω(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=0nω(xi)f(xi)φj(xi)

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

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=4n=1 φ 0 ( x ) = 1 , φ 1 ( x ) = x \varphi_0(x)=1,\varphi_1(x)=x φ0(x)=1φ1(x)=x


( 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=04ωi=8=(φ1,φ0)=i=04ωixi=22=i=04ωixi2=74=i=04ωifi=47=i=04ωixifi=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)