用MATLAB计算—最小二乘法及均方误差

用fit函数或者nlinfit,nlinfit是最小平方估计比如第一个>> y= [19。0 32。3 49。0 73。3 97。8]y =19。0000 32。3000 49。0000 73。3000 97。8000>> x=[19 25 31 38 44]x =19 25 31 38 44>> fit(x',y','a+b*x^2')Warning:Start point not provided,choosing random start point。> In fit>handlewarn at 715In fit at 315ans =General model:ans(x) = a+b*x^2Coefficients (with 95% confidence bounds):a = 0。9726 (0。7579,1。187)b = 0。05004 (0。04986,0。05021)用nlinfit建立m文件mymodel。mfunction y=mymodel(beta,x)y=beta(1)+beta(2)*x。^2;console下输入：nlinfit(x,y,@mymodel,[0 0])ans =0。9726 0。0500第二个我不知道函数模型,没法拟合 x=[0 5 10 15 20 25 30 35 40 45 50 55];y=[0 1。27 2。16 2。86 3。44 3。87 4。15 4。37 4。51 4。58 4。62 4。64];建立m文件mymodel。mfunction y=mymodel(beta,x)y=beta(1)*exp(-beta(2)。/x);console下输入：nlinfit(x,y,@mymodel,[1 1])ans =5。5031 8。7873