Array indices must be positive integers or logical values

clc
f1= (-(3/5)*X(1))+((1/4)*X(2))+((1/4)*cos(X(3)))+1.43;
f2= ((1/4)*X(1))-((3/5)*X(2))-((1/4)*sin(X(3)))-1.24;
f3= ((1/4)*sin(X(1)))-((1/4)*exp(-X(2)))-((3/5)*X(3))-1.17;
F=  [f1;f2;f3];
%J= jacobian([f1,f2,f3],[X(1),X(2),X(3)]);
J = [(-3/5) (1/4) (-1/4)*sin(X(3));
    1/4 -3/5 (-1/4)*cos(X(3));
    (1/4)*cos(X(1)) (1/4)*exp(-X(2)) -3/5];
x0=0; %initial guess
y0=0; 
z0=0; 
X = [x0;y0;z0];
TC=10^-8;
i = 0;
i_max = 100;
while(error>TC)
    
    i=i+1;
    X0=X;
    X = X0 - J(X0)\F(X0);
    error = max(abs(X-X0));
end
if i>i_max
    disp('No solutions are found');
else
    disp(['x = ',num2str(X(1))])
    disp(['y = ',num2str(X(2))])
    disp(['z = ',num2str(X(3))])
    disp(error)
end

I get error

" Array indices must be positive integers or logical values. "

Can someone help me where is the problem?

also can you help me with J= jacobian([f1,f2,f3],[X(1),X(2),X(3)])? when I run this with upper code it give me error of incorrect arrangement

[Jan]

Please post the complete error message. Then the readers do not have to guess, in which line the error occurs.

We cannot run your code due to the missing inputs, so it would be a good idea to post all details you have in the command line available already.

"it give me error of incorrect arrangement" - see above: Post a copy of the message instead of a rough paraphrasation.

Answer 1

[Jan]

I guess boldly, that this line is failing:

X = X0 - J(X0)\F(X0);

J and F are defined as vectors, not as functions.

I've removed the pile of parentheses from the equation and inserted a check of i<i_max in the loop condition:

F = @(X) [-(3/5)*X(1) + X(2) / 4 + cos(X(3)) / 4 + 1.43; ...
          X(1) / 4 - (3/5)*X(2) - sin(X(3)) / 4 - 1.24; ...
          sin(X(1)) / 4 - exp(-X(2)) / 4 - (3/5)*X(3) - 1.17];
J = @(X) [-3/5, 1/4, -sin(X(3) / 4);
          1/4, -3/5, -cos(X(3) / 4);
          cos(X(1)) / 4, exp(-X(2)) / 4, -3/5];

Err   = Inf;
X     = [0;0;0];
TC    = 1e-8;
i     = 0;
i_max = 100;
while Err > TC && i < i_max
    i  = i+1;
    X0 = X;
    X  = X0 - J(X0) \ F(X0);
    Err = max(abs(X - X0));
end
if i == i_max
    disp('No solutions are found');
else
    disp(X)
    disp(Err)
end

    1.3984
   -1.8385
   -4.1593

   5.2661e-09

[Syed Abdul Rafay]

Is there are way to use jacobian command in this code without writing the jacobian matrix by myself and taking its inverse and adding that in the loop?

[Syed Abdul Rafay]

Also thanks a lot. It made my day.