steepest descent using secant method

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Hacer Busra KELES 2021-5-4

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https://ww2.mathworks.cn/matlabcentral/answers/821410-steepest-descent-using-secant-method

评论： SALAH ALRABEEI 2021-6-21

I am trying the sovee function the steepest descent algrot using secant method but ı guess my code is wrong how can fixed;

syms X1 x2;
f = x1.^4/8 + x2.^2/4 -2*x1-4*x1*x2;
x1(1) = 2;
x2(1) = 2;
df_dx1 = diff(f, X);
df_dx2 = diff(f, Y);
g0 = [subs(df_dx1,[X,Y], [x1(1),x2(1)]) subs(df_dx2, [X1,X2], [x1(1),x2(1)])];
d = -(g0);
for i = 1:5
I = [x1(i),x1(i)];
syms t
g = subs(f, [X,Y], [x1(i)+d(1)*t,x2(i)+t*d(2)]);
dg_dt = diff(g,t);
    t = solve(dg_dt, t);
x1(i+1) = I(1)+t*d(1);
x2(i+1) = I(2)+t*d(2);
g_o = [subs(df_dx,[X,Y], [x1(i),y(i)]) subs(df_dy, [X1,X2], [x(i),x2(i)])];
    i = i+1;
    g_1 = [subs(df_dx1,[X1,X2], [x1(i),X2(i)]) subs(df_dx2, [X1,X2], [x1(i),x2(i)])];
    d=-(g_1);
end
Iter = 1:i;
X_coordinate = x1';
Y_coordinate = x2';
Iterations = Iter';
T = table(Iterations,X_coordinate,Y_coordinate);
fcontour(f, 'Fill', 'On');
hold on;
plot(x1,x2,'*-r');
fprintf('Initial Objective Function Value: %d\n\n',subs(f,[X1,X2], [x1(1),x2(1)]));
fprintf('Number of Iterations for Convergence: %d\n\n', i);
fprintf('Point of Minima: [%d,%d]\n\n', x1(i), x2(i));
fprintf('Objective Function Minimum Value after Optimization: %d\n\n', subs(f,[X1,X2], [x1(i),x2(i)]));
disp(T)

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SALAH ALRABEEI 2021-6-21

your function depends on X1 and X2 variables, while you differentiated wrt X and Y, HOW COME!

SALAH ALRABEEI 2021-6-21

You have to double chceck your code first, your function has X1 and X2 variables, so whenever you use diff or subs, you have to keep them. here you have also y and x2 as both number of variable.

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