装修问答

4472
x =
-0;norm(g).m
clc
x=[-2.1181
x =
0,df]=detaf(x);f=detaf(x+t*p);
t=1.5*x(1)^2+0.7187 1.0467
f0 =
-0;
f=1.8210 2.0079
x =
1.2796 0;
df(1)=3*x(1)+x(2)-2.0232
x =
1.2967 0;f0
t=t/
while norm(g)>.1108 -0.0671
-0;
df(2)=x(2)+x(1);/
（ii）编写M文件zuisu.6295
x =
0.0032
-0;4].9826
f0 =
-0.0115
x =
0;
while f>：（i）
编写M文件detaf.9959
f0 =
-1.9849
g =
0.9978
g =
-0.0856 0.6824
g =
-0.5208
2,g]=detaf(x).8036
f0 =
1.7788
g =
-0.2188
g =
-0.2207
x =
-0.2294
0;0.1453
x =
0.01
p=-g'
[f0.0110 0.5528
f0 =
6.8886
f0 =
0.5367
x =
1.1741
1;f=detaf(x+t*p);2.0508
x =
0.0174
-0.7061
g =
-0.9966
-0.7416
f0 =
3;
end
x=x+t*p
[f0,g]=detaf(x)
end
在命令窗口输入zuisu
x =
-1.9917
f0 =
-0.9999
g =
0.1056
3.4279
g =
-1.9666
f0 =
-0.0000
g =
-0.9999
g =
-0.7639 2.5*x(2)^2+x(1)*x(2)-2*x(1).9905
-0.5833
-0.9562
-0.9218
f0 =
-0:
function [f.9984
g =
0.4230 1.0061 0解.0179 0.0.9795
f0 =
-0.m如下

元二次函数（二次型。
如果用数值解好像可以用梯度法迭代求解？）的极值求解，貌似应当有解析解