value in each iteration

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)?

[Mario Malic]

What is the purpose of this question on which you got the answer?

https://ch.mathworks.com/matlabcentral/answers/595906-stopping-criterion-in-gradient-descent-algorithm?s_tid=srchtitle

[saja mk]

i need to clculate

this ,

[Walter Roberson]

That f needs a vector of length 2 for its parameter, not a scalar i.

[saja mk]

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]

@Walter Roberson

so how can i implement it ?

[Walter Roberson]

You cannot. Your gradient descent formula is only valid for functions of one variable.

Answer 1

[saja mk]

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

[Walter Roberson]

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