利用MATLAB求解这样一个非线性自回归模型，y=ax^2+bx+c ,求系数a,b,c的值，感激不尽 急！！！matlab指数拟合，模型已知，但是不知道怎么求系数
\u975e\u7ebf\u6027\u6700\u5c0f\u4e8c\u4e58\u6cd5matlab\u6765\u6c42y=a/(b*x+c)\u53ef\u4ee5\u7528\u7ebf\u6027\u6700\u5c0f\u4e8c\u4e58\u6cd5\uff0c\u4e3a\u5565\u8981\u7528\u975e\u7ebf\u6027\u6700\u5c0f\u4e8c\u4e58\u6cd5\uff1f\u5148\u6c42\u5012\u6570y*=(b/a)x+c/a,\u7b97\u51fa\u622a\u8ddd\u548c\u659c\u7387\u540e\u6c42\u51faa\uff0cb\uff0cc\u5c31\u884c\u4e86\u3002\u6700\u5c0f\u4e8c\u4e58\u6cd5\u7684\u793a\u4f8b\u89c1http://baike.baidu.com/view/139822.htm
\u9898\u4e3b\u7684\u6307\u6570\u62df\u5408\u6a21\u578by=a*P^b*GDP^c*T^d*W^e\uff0c\u53ef\u4ee5\u7528matlab\u7684lsqcurvefit\u975e\u7ebf\u6027\u6700\u5c0f\u4e8c\u4e58\u6cd5\u51fd\u6570\u6c42\u89e3\u5176\u62df\u5408\u7cfb\u6570a\u3001b\u3001c\u3001d\u3001e\u3002\u7531\u4e8e\u6570\u636e\u6837\u672c\u6709\u70b9\u504f\u5dee\uff08\u79bb\u6563\uff09\uff0c\u4e25\u91cd\u4f9d\u8d56a\u3001b\u3001c\u3001d\u3001e\u7684\u521d\u503c\uff0c\u7ecf\u591a\u6b21\u62df\u5408\u5f97\u5230\u76f8\u5bf9\u8f83\u597d\u7684\u503c\uff0ca=-1.603e-27\uff0cb=-2.6004\uff0cc=2.0349\uff0cd=0.74572\uff0ce=12.7048\u6c42\u89e3\u4e3b\u8981\u4ee3\u7801\uff1ax=[P' GDP' T' W']; %\u6570\u636ek0=[...] %\u521d\u503c[k,resnorm]= lsqcurvefit(@(k,x)nh_fun(k,x),k0,x,yg);nh_fun(k,x) %\u6307\u6570\u62df\u5408\u6a21\u578b\u51fd\u6570
y=[105 96 102 101];
fun=inline('a(1)*x^2+a(2)*x+a(3)','a','x');
a=lsqcurvefit(fun,[1, 1, 1],x,y);
c=a(3),b=a(2),a=a(1)
c =
-2.2001e+003
>> b=a(2)
b =
46.5663
>> a=a(1)
a =
-0.2353
多项式拟合
x=[100 105 96 102];
y=[105 96 102 101];
t=polyfit(x,y,2);
a=t(1)
b=t(2)
c=(3)
>> x=[100 105 96 102]';
y=[105 96 102 101]';
X=[ones(size(x)) x x.^2];
t=regress(y,X);
a=t(3)
b=t(2)
c=t(1)
a*x.^2+b*x+c
----------------------------------
结果如下：
a =
-0.2353
b =
46.5665
c =
-2.2001e+003
ans =
104.0099
95.7066
102.1834
102.1001
>>
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