Jacobian and Newton Rhapson Matrix running infintely

I have this code that I am trying to use to find what the values of theta 4, 5, 6, and length 3 are by estimating using the loops below, I based the code off of an example piece of code, but when it runs it doesn't stop, I thought there could be an issue with my while loop, but I haven't been able to see it

syms the4 the5 the6 L3
f1 = 1.8*cos(the6)+2.69*cos(the5)-1.36*cos(the4)-2.1023;
f2 = 1.8*sin(the6)+2.69*sin(the5)-1.36*sin(the4)-3.2373;
f3 = 0.6365+L3*cos(the4);
f4 = L3*sin(the4)-0.6907; %setting up eqs

J = [diff(f1,the4), diff(f1,the5), diff(f1, the6), diff(f1, L3)
    diff(f2,the4), diff(f2,the5), diff(f2, the6), diff(f2, L3)
    diff(f3,the4), diff(f3,the5), diff(f3, the6), diff(f3, L3)
    diff(f4,the4), diff(f4,the5), diff(f4, the6), diff(f4, L3)]; %finding jacobian
F = [f1;f2;f3;f4];
X = [the4,the5,the6,L3]'; %setting up the X, also variables to be solved
X_new = X-J^-1*F; %Newtown Rhapson equation

the40 = (3*pi)/4; %initial guesses
the50 = pi/3;
the60 = pi/2;
L30 = 1;

epsilon = 1e-10; %used to stop code from running infinitely, gets us close enough to answer
iter_count = 0;
iter_max = 100;

Fvalue = subs(F,{the4,the5,the6,L3},{the40,the50,the60,L30}); %replaces the old x with new x in F
while(iter_count<=iter_max)
    iter_count = iter_count+1;
    X_new_value = subs(X_new, [the4,the5,the6,L3],[the40,the50,the60,L30]) %continues to replace x values, closer to approximate
    the40 = X_new_value(1);
    the50 = X_new_value(2);
    the60 = X_new_value(3);
    L30 = X_new_value(4);
    Fvalue = subs(F,[the4,the5,the6,L3],[the40,the50,the60,L30]);
    Fvalue = double(Fvalue);
end
if iter_count>iter_max
    disp('No solutions are found');
else
    disp(['theta_4 = ',num2str(double(theta_40))])
    disp(['theta_5 = ',num2str(double(theta_50))])
    disp(['theta_6 = ',num2str(double(theta_60))])
    disp(['L3 = ' ,num2str(double(L30))])
end
   

[KALYAN ACHARJYA]

Sir, I don't think it is running infintely, but it takes time. please check code is taking high time at iter=4

iter_count =
     1
Elapsed time is 0.052071 seconds.
iter_count =
     2
Elapsed time is 0.150448 seconds.
iter_count =
     3
Elapsed time is 2.922127 seconds.
iter_count =
     4
     .......

Sure, the iteration will be terminated at iter_max. Better to focus on improving the code or change in imlementation, so that it can takes less time.

Answer 1

[Alan Stevens]

Runs much quicker without symbolics (the Jacobian is easy to obtain by hand). Like so

F = @(X) [1.8*cos(X(3))+2.69*cos(X(2))-1.36*cos(X(1))-2.1023;
          1.8*sin(X(3))+2.69*sin(X(2))-1.36*sin(X(1))-3.2373;
          0.6365+X(4)*cos(X(1));
          X(4)*sin(X(1))-0.6907]; %setting up eqs
                         
J = @(X) [1.36*sin(X(1)), -2.69*sin(X(2)), -1.8*sin(X(3)), 0;
         -1.36*cos(X(1)), 2.69*cos(X(2)), 1.8*cos(X(3)), 0;
         -X(4)*sin(X(1)), 0, 0, cos(X(1));
          X(4)*cos(X(1)), 0, 0, sin(X(1))];  % Jacobian
                             
the40 = (3*pi)/4; %initial guesses
the50 = pi/3;
the60 = pi/2;
L30 = 1;
X = [the40; the50; the60; L30];

epsilon = 1e-10; %used to stop code from running infinitely, gets us close enough to answer
iter_count = 0;
iter_max = 100;
error = 1;

while iter_count<=iter_max && error>epsilon
    iter_count = iter_count+1;
    X0 = X;
    X = X0 - J(X0)\F(X0);  
    error = max(abs(X-X0));
end
if iter_count>iter_max
    disp('No solutions are found');
else
    disp(['theta_4 = ',num2str(X(1))])
    disp(['theta_5 = ',num2str(X(2))])
    disp(['theta_6 = ',num2str(X(3))])
    disp(['L3 = ' ,num2str(X(4))])
end