Hi, can anyone tell me what is the problem of this? It says "Operands to the || and && operators must be convertible to logical scalar values."

function [ x,niter ] = NewtonRaphson( x0,tol,nmax )

%DATA
N = 20;
niter = 0;
error = 1;
x0 = zeros(N-1,1);
dx = zeros(N-1,1);

while ((error > tol) && (niter < nmax))

      niter = niter + 1;
      x = x0 - ((F_springs(x0))./(J_springs(x0)));
      x0 = x;
      error = max(abs(x-F_springs(x0)));

end

for i = 2:N-2

      dx(i) = x(i)-x(i-1);

end

end

Answer 1

[Walter Roberson]

x0 = zeros(N-1,1);

so x0 is a vector.

x = x0 - ((F_springs(x0))./(J_springs(x0)));
x0 = x;

with x0 being a vector, x will be a vector; and then you assign it to x0 so x0 will stay a vector.

error = max(abs(x-F_springs(x0)));

with x and x0 being vectors, error is going to be a vector.

while ((error > tol) && (niter < nmax))

error is a vector, so error > tol is a vector. You now have a vector on the left side of && . But the && operator is strictly for scalars, never for vectors. For vectors you would need to use the & operator, as in

    while ((error > tol) & (niter < nmax))

As niter<nmax is a scalar, the effect of that would be to calculate (niter < nmax) and to "and" it with each element of (error > tol), giving a vector of results.

The definition of if or while with a vector is to consider the condition to be true only of all of the items in the vector are non-zero. This would be equivalent to

      while all((error > tol) & (niter < nmax))

which would stop as soon as any location in error was not greater than tol.

You should re-think why you want x0 to be a vector. If you have a good reason for it being a vector, then think about whether you truly want the entire vector to be involved in the calculations each time.

[Muhamad Mohd-Adly]

Hi Sir,

Thank you very much for your help. Im quite clear now.