matlab中用0.618法求minf(x)=e^(-x)+x^2在区间(0,1)上的极小值,精确到0.03.

clc
clear all;
elp=0.03;
tao=0.618;
N=fix(log(elp)/log(tao))+1;
k=1;
a(k)=0;
b(k)=1;
r(1)=b(k)-tao*(b(k)-a(k));
u(1)=a(k)+tao*(b(k)-a(k));
for i=1:N
%f1=(r(k)).^2-r(k);
f1 = exp(-r(k))+r(k).^2;
f2 = exp(-u(k))+u(k).^2;
%f2=(u(k)).^2-u(k);
if(f1>f2)
a(k+1)=r(k);
b(k+1)=b(k);
r(k+1)=u(k);
u(k+1)=a(k+1)+tao*(b(k+1)-a(k+1));
x=(b(k+1)+a(k+1))/2;
else
a(k+1)=a(k);
b(k+1)=u(k);
u(k+1)=r(k);
r(k+1)=b(k+1)-tao*(b(k+1)-a(k+1));
x=(b(k+1)+a(k+1))/2;
end
k=k+1;
end
x
y=exp(-x)+x.^2;
y
结果：
x = 0.3501
y = 0.8272
验证：
>> x1=0:0.001:1;
>> y1=exp(-x1)+x1.^2;
>> min(y1)
ans =
0.8272