Question

紧急求助!一个matlab编程问题

有公式:y=a(exp(bx)),已知x=4,6,8,12,20,26,36,56,95,240,360,541,2147,12000;对应的y=0.25,0.22,0.2 ,0.18,0.16,0.14,0.12,0.1,0.08,0.06,0.04,0.02,0,0.现要用matlab的非线性最小二乘法求a,b要具体的matlab程序.并画图
本人急用此程序,希望matlab高手指教,万分感谢!!!
这回显示的的结果是??? Undefined function or variable 'nlinfit'.因为刚开始用不怎么会，麻烦了

Answer 1

【补充】也许你的matlab版本低，这样也行，结果一样。
clc;clear
x=[4,6,8,12,20,26,36,56,95,240,360,541,2147];%,12000];
y=[0.25,0.22,0.2 ,0.18,0.16,0.14,0.12,0.1,0.08,0.06,0.04,0.02,0];%,0.];
funy1=inline('a(1)*(exp(a(2)*x))','a','x')
a=nlinfit(x,y,funy1,[0.2 -0.01])
funy2=inline(' b(1)*exp(b(2)*x)+b(3)*exp(b(4)*x)','b','x')
b=nlinfit(x,y,funy2,[0.2 -0.01 0.2 -0.01])
xx=0:x(end);
y1=a(1)*exp(a(2)*xx);
y2=b(1)*exp(b(2)*xx)+b(3)*(exp(b(4)*xx));
plot(x,y,'o',xx,[y1;y2])

【原来的】提供的方程不太好，推荐一个 f(x) = a*exp(b*x) + c*exp(d*x)，效果更好。看看对比。

clc;clear
x=[4,6,8,12,20,26,36,56,95,240,360,541,2147];%,12000];
y=[0.25,0.22,0.2 ,0.18,0.16,0.14,0.12,0.1,0.08,0.06,0.04,0.02,0];%,0.];
funy1=@(a,x) a(1)*(exp(a(2)*x))
a=nlinfit(x,y,funy1,[0.2 -0.01])
funy2=@(b,x) b(1)*exp(b(2)*x)+b(3)*exp(b(4)*x)
b=nlinfit(x,y,funy2,[0.2 -0.01 0.2 -0.01])
xx=0:x(end);
y1=a(1)*exp(a(2)*xx);
y2=b(1)*exp(b(2)*xx)+b(3)*(exp(b(4)*xx));
plot(x,y,'o',xx,[y1;y2])

结果：
a =

0.2272 -0.0137

b =

0.1645 -0.0638 0.1135 -0.0030