曲线拟合_线性最小二乘法及其MATLAB程序

1 曲线拟合的线性最小二乘法及其MATLAB 程序

例7.2.1 给出一组数据点),(i i y x 列入表7–2中，试用线性最小二乘法求拟合曲线，并用（7.2），（7.3）和（7.4）式估计其误差，作出拟合曲线.

表7–2 例7.2.1的一组数据),(y x

解 （1）在MATLAB 工作窗口输入程序

>> x=[-2.5 -1.7 -1.1 -0.8 0 0.1 1.5 2.7 3.6];

y=[-192.9 -85.50 -36.15 -26.52 -9.10 -8.43 -13.12 6.50 68.04];

plot(x,y,'r*'),

legend('实验数据(xi,yi)')

xlabel('x'), ylabel('y'),

title('例7.2.1的数据点(xi,yi)的散点图')

运行后屏幕显示数据的散点图（略）.

（3）编写下列MA TLAB 程序计算)(x f 在),(i i y x 处的函数值，即输入程序

>> syms a1 a2 a3 a4

x=[-2.5 -1.7 -1.1 -0.8 0 0.1 1.5 2.7 3.6];

fi=a1.*x.^3+ a2.*x.^2+ a3.*x+ a4

运行后屏幕显示关于a 1,a 2, a 3和a 4的线性方程组

fi =[ -125/8*a1+25/4*a2-5/2*a3+a4,

-4913/1000*a1+289/100*a2-17/10*a3+a4,

-1331/1000*a1+121/100*a2-11/10*a3+a4,

-64/125*a1+16/25*a2-4/5*a3+a4,

a4, 1/1000*a1+1/100*a2+1/10*a3+a4,

27/8*a1+9/4*a2+3/2*a3+a4, 19683/1000*a1+729/100*a2+27/10*a3+a4, 5832/125*a1+324/25*a2+18/5*a3+a4]

编写构造误差平方和的MATLAB 程序

>> y=[-192.9 -85.50 -36.15 -26.52 -9.10 -8.43 -13.12 6.50

68.04];

fi=[-125/8*a1+25/4*a2-5/2*a3+a4,

-4913/1000*a1+289/100*a2-17/10*a3+a4,

-1331/1000*a1+121/100*a2-11/10*a3+a4,

-64/125*a1+16/25*a2-4/5*a3+a4, a4, 1/1000*a1+1/100*a2+1/10*a3+a4,

27/8*a1+9/4*a2+3/2*a3+a4,

19683/1000*a1+729/100*a2+27/10*a3+a4,

5832/125*a1+324/25*a2+18/5*a3+a4];

fy=fi-y; fy2=fy.^2; J=sum(fy.^2)

运行后屏幕显示误差平方和如下

J=

(-125/8*a1+25/4*a2-5/2*a3+a4+1929/10)^2+(-4913/1000*a1+2

89/100*a2-17/10*a3+a4+171/2)^2+(-1331/1000*a1+121/100*a2-11/10*a3+a4+723/20)^2+(-64/125*a1+16/25*a2-4/5*a3+a4+663/25)^2+(a4+91/10)^2+(1/1000*a1+1/100*a2+1/10*a3+a4+843/100)^2+(27/8*a1+9/4*a 2+3/2*a3+a4+328/25)^2+(19683/1000*a1+729/100*a2+27/10*a3+a4-13/

2)^2+(5832/125*a1+324/25*a2+18/5*a3+a4-1701/25)^2

为求4321,,,a a a a 使J 达到最小，只需利用极值的必要条件0=∂∂k

a J )4,3,2,1(=k ，

得到关于4321,,,a a a a 的线性方程组，这可以由下面的MA TLAB 程序完成，即输入程序

>> syms a1 a2 a3 a4

J=(-125/8*a1+25/4*a2-5/2*a3+a4+1929/10)^2+(-4913/1000*a1+

289/100*a2-17/10*a3+a4...+171/2)^2+(-1331/1000*a1+121/100*a2-11/10*a3+a4+723/20)^2+(-64/125*a1+16/25*a2-4/5*a3+a4+663/25)^2+(a 4+91/10)^2+(1/1000*a1+1/100*a2+1/10*a3+a4+843/100)^2+(27/8*a1+9/4*a2+3/2*a3+a4+328/25)^2+(19683/1000*a1+729/100*a2+27/10*a3+a4-13/2)^2+(5832/125*a1+324/25*a2+18/5*a3+a4-1701/25)^2;

Ja1=diff(J,a1); Ja2=diff(J,a2); Ja3=diff(J,a3); Ja4=diff(J,a4);

Ja11=simple(Ja1), Ja21=simple(Ja2), Ja31=simple(Ja3), Ja41=simple(Ja4),

运行后屏幕显示J 分别对a 1, a 2 ,a 3 ,a 4的偏导数如下

Ja11=

56918107/10000*a1+32097579/25000*a2+1377283/2500*a3+

23667/250*a4-8442429/625

Ja21 =

32097579/25000*a1+1377283/2500*a2+23667/250*a3+67*a4

+767319/625

Ja31 =

1377283/2500*a1+23667/250*a2+67*a3+18/5*a4-232638/125

Ja41 =

23667/250*a1+67*a2+18/5*a3+18*a4+14859/25

解线性方程组Ja 11 =0，Ja 21 =0，Ja 31 =0，Ja 41 =0，输入下列程序

>>A=[56918107/10000, 32097579/25000, 1377283/2500, 23667/250; 32097579/25000, 1377283/2500, 23667/250, 67; 1377283/2500, 23667/250, 67, 18/5; 23667/250, 67, 18/5, 18];

B=[8442429/625, -767319/625, 232638/125, -14859/25];

C=B/A, f=poly2sym(C)

运行后屏幕显示拟合函数f 及其系数C 如下

C = 5.0911 -14.1905 6.4102 -8.2574

f=716503695845759/140737488355328*x^3

-7988544102557579/562949953421312*x^2

+1804307491277693/281474976710656*x

-4648521160813215/562949953421312

故所求的拟合曲线为

8.25746.410214.19055.0911)(23-+-=x x x x f .

（4）编写下面的MA TLAB 程序估计其误差，并作出拟合曲线和数据的图形.输入程序

>> xi=[-2.5 -1.7 -1.1 -0.8 0 0.1 1.5 2.7 3.6];

y=[-192.9 -85.50 -36.15 -26.52 -9.10 -8.43 -13.12 6.50 68.04];

n=length(xi);

f=5.0911.*xi.^3-14.1905.*xi.^2+6.4102.*xi -8.2574;

x=-2.5:0.01: 3.6;

F=5.0911.*x.^3-14.1905.*x.^2+6.4102.*x -8.2574;

fy=abs(f-y); fy2=fy.^2; Ew=max(fy),

E1=sum(fy)/n, E2=sqrt((sum(fy2))/n)

plot(xi,y,'r*'), hold on, plot(x,F,'b-'), hold off

legend('数据点(xi,yi)','拟合曲线y=f(x)'),

xlabel('x'), ylabel('y'),

title('例7.2.1的数据点(xi,yi)和拟合曲线y=f(x)的图形')

运行后屏幕显示数据),(i i y x 与拟合函数f 的最大误差E w ，平均误差E 1和均方根误差E 2及其数据点),(i i y x 和拟合曲线y =f (x )的图形（略）.

Ew = E1 = E2 =

3.105 4 0.903 4 1.240 9