How to solve this non-linear 6 varriable equations using newton rapson method.x

How to solve this non-linear 6 varriable equations using newton rapson method. i want the values of a,b,c.d.e.f for what it will become 0.

eq1 = cosd(5*a)+cosd(5*b)+cosd(5*c)+cosd(5*d)+cosd(5*e)+cosd(5*f);

eq2 = cosd(7*a)+cosd(7*b)+cosd(7*c)+cosd(7*d)+cosd(7*e)+cosd(7*f);

eq3 = cosd(11*a)+cosd(11*b)+cosd(11*c)+cosd(11*d)+cosd(11*e)+cosd(11*f);

eq4 = cosd(13*a)+cosd(13*b)+cosd(13*c)+cosd(13*d)+cosd(13*e)+cosd(13*f);

eq5 = cosd(17*a)+cosd(17*b)+cosd(17*c)+cosd(17*d)+cosd(17*e)+cosd(17*f);

eq6 = cosd(19*a)+cosd(19*b)+cosd(19*c)+cosd(19*d)+cosd(19*e)+cosd(19*f);

[John D'Errico]

Just use fsolve. Never write code for something like Newton Raphson when existing professionally written code already exists. Unless of course, it is homework, where you were told to use Newton-Raphson.

Note that in any case, there will be infintely many solutions, and that no such solver can return all solutions.

[Debarshi Chakraborty]

clc

options=optimset('Display','iter');

x0 = [-5;-5;-5;-5;-5;-5];

options = optimset('Display','iter');

[x,fval] = fsolve(@myfun,x0,options);

x,fval

sir, can u please tell me how can add ( 0<x1<x2<x3<x4<x5<x6<1.5708) this for x on fsolve

[John D'Errico]

You cannot do so. (Nor did I ever suggest you could, since that was not part of your original question.) Fsolve does not allow a way to implement inequality constraints on the solution. However, I would point out that neither does Newton-Raphson, in any form, and that was what you were originally asking.

You could try reformulating the problem in a form that an optimizer like fmincon can solve thus a sum of squares of residuals.) It does allow inequality constraints, although it will not solve them in a form of strict inequalities. Only >= and <= constraints are allowed even there.

[Walter Roberson]

Do your regular solving. Then sort the results. The equations are symmetric, the order of the coefficients does not not change the outcome.

[Debarshi Chakraborty]

X0 = [-5;=5;=5;=5;=5;-5];

maxIter = 1000;

tolX = 10e-6;

% Computation

X = X0;

Xold = X0;

R = inv(j);

for i =1:maxIter

[f,j]=new(X);

X = (X- R*f);

err(:,i) = (X-Xold)

Xold = X;

if (err(:,i)<tolX)

break;

end

end

% How can i add ( 0<x1<x2<x3<x4<x5<x6<1.5708) this condition on this code

[Walter Roberson]

What do you want to have happen when the condition is violated, which is highly likely to happen? The algorithm you are using has no provision to "back up and try again".

[John D'Errico]

Walter is correct about the symmetry of your equations, in that any solution found by fsolve, if it is not sorted in increasing order can arbtrarily be sorted into increasing order. This will still be a solution.

[Debarshi Chakraborty]

I want the value of of X when its staisfied that condition ( 0<x1<x2<x3<x4<x5<x6<1.5708)

so after every iteratin i want to check if its follows that condition

[Walter Roberson]

What do you want to do if you determine that the condition is not met?

[Debarshi Chakraborty]

if this condition is not tru then i dont need that value,, i want the value of only if it's on that range

[Walter Roberson]

Ok so after (say) 2 iterations when the error is still quite high, and it generates an x that does not meet the conditions, then you want to... What, give up on the complete solving process?

You might perhaps be thinking that in that situation you want to loop back and try different initial values, but the problem with that is that it is likely that every initial starting point that is not exactly right already will end up violating the conditions.

Newton Raphson works by successive over-projection and recovery, so as long as you are using Newton Raphson, you will encounter current values that violate the conditions and yet are needed to recover to a better position.