value in each iteration

2020 9 19
1 个回答
更新时间：2020 9 19
3 次查看（30 天）

0 个投票

f=@(x) 0.5*(x(1)-1).^2+10*(x(2)+2).^2-2
miter=10000
and i have loop
for i=1:miter
then,who i can to calculate f(i)-f(i-1)?

6 个评论
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saja mk
saja mk 2020-9-19
i need to clculate
this ,
Walter Roberson
Walter Roberson 2020-9-19
That f needs a vector of length 2 for its parameter, not a scalar i.
saja mk
saja mk 2020-9-19
i al ready get the thev optimal by Gradient Descent Algorith
now ,i just need to change the tol as the equation in the previous comment
saja mk
saja mk 2020-9-19
@Walter Roberson
so how can i implement it ?
Walter Roberson
Walter Roberson 2020-9-19
You cannot. Your gradient descent formula is only valid for functions of one variable.

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回答（1 个）

saja mk
saja mk 2020-9-19

0 个投票

but i get the optimal in GDA
x0=[2 2]';
% tol=1e-6;
miter=10000;
% dxmin=1e-6;
alpha=0.01;
% gnorm=inf;
x=x0;
niter=1;
% dx=inf;
for i=1:miter
g=[(x(1)-1); 20*(x(2)+2)];
f=@(x) 0.5*(x(1)-1).^2+10*(x(2)+2).^2-2 ;
xnew = x - alpha*g;
fnew=f(xnew);
if abs(xnew-x)<=0.00001
break
end
% niter=niter+1;
x=xnew;
end
xopt=x;
niter=niter+1;
after run :
xopt =
1.0010
-2.0000

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Walter Roberson
Walter Roberson 2020-9-19
this does not use the epsilon formula from https://www.mathworks.com/matlabcentral/answers/596404-value-in-each-iteration#comment_1015369

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