Why do I get this error: A(I): index out of bounds; value 2 out of bound 1 (line 13)

function root=newtonmethod(f,fprime,x0,maxiter,tol)

%input:
% f string that names the function f(x).
% fprime string that names the derivative f’(x).
% x0 the initial point
% tol is the termination tolerances
% maxiter the maximum number of iteration

x(1) = x0;
n = 2;

while abs(f(x(n))) > tol & n < maxiter

      m = n+1;

      x(m) = x(n) - f(x(n))/fprime(x(n));    

      if x(n) - f(x(n))/fprime(x(n)) < 0
        root = x(n);
        break;
      end

end

root = x(n)

Answer 1

[Daniel kiracofe]

Because the very first time the while statement's condition is checked, n=2, and x is a vector that has 1 element. You are asking to evaluate the 2nd element of an array that has only 1 element. You probably want to start with n=1 instead of n=2.

Answer 2

[Roger Stafford]

Besides the error that Daniel gives you, another error is that ’n’ never changes within your while-loop, so the (corrected) loop would never stop unless you just happened to start with the right value.