Matlab code for non linear equation using newton Method

2021 12 19
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更新时间：2021 12 19
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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

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Torsten
Torsten 2021-12-19
编辑：Torsten 2021-12-19
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
Nameer Ahmad 2021-12-19
编辑：Torsten 2021-12-19
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
Torsten 2021-12-19
编辑：Torsten 2021-12-19
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
Nameer Ahmad 2021-12-19
Tell me whch norm should i use to get 11 iterations

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Torsten
Torsten
2021-12-19

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