Matlab code for non linear equation using newton Method

clc;
clear all;

% Function f1 and f2
func = @(V,S) [V*sin(S) - 0.5; V^2 - V*cos(S)];

% Jacobian
jac = @(V,S) [sin(S), V*cos(S); 2*V - cos(S), V*sin(S)];

maxit=50; % Maximum iteration
tol=10^-6; % Tolerance
% Roots
V = zeros(maxit);
S = zeros(maxit);
% Initial guess
V(1) = 1; S(1) = 0;

% Newton Raphshon method of solving
for i = 1: maxit+1
    F = func(V(i),S(i));
    J = jac(V(i),S(i));
    Dx = -J\F;
    V(i+1) = V(i) + Dx(1);
    S(i+1) = S(i) + Dx(2);
    F = func(V(i+1),S(i+1));
    if norm (F)<tol
        break
    end
   
    disp(V(i+1));
    disp(S(i+1));
end

% Output
fprintf('No. of iterations: %d\n',i);
fprintf('Roots: V = %f and S = %f rad \n',V(i+1),S(i+1));

i Want to stop this at 11 ietrations help me to get 11 iteration ansr with same tolernce

[Torsten]

Everything's fine with your code.

You need 10 iterations to reach out, and this is the correct answer.

Maybe 11 iterations are required if a different norm is used to evaluate F.

[Nameer Ahmad]

x=[1;0]; 
f = @(v,q) [v*sin(q)-0.5;((v^2)-v*cos(q))]; 
df = @(v,q) [sin(q) v*cos(q);((2*v)-cos(q)) v*sin(q)]; 
tolerance = [1*10^-6;1*10^-6];
for i = 1:20 
    disp(x); 
    error = abs(f(x(1),x(2))); 
    x=x-(inv(df(x(1),x(2)))*f(x(1),x(2))); 
    %disp(error); 
    if error<tolerance 
        break 
    end 
end 
disp(x);

With this code it ends of 11 iteration and manualy also 11 plzz help

[Torsten]

But you evaluate the error before the iteration here. This does not make much sense.

Furthermore, you use a different error norm (namely componentwise absolute value), but this does not play a role in this case.

And you should not use "inv". Better use "\" as in the first code in order to calculate Dx.

[Nameer Ahmad]

Tell me whch norm should i use to get 11 iterations