A small point about the rate of convergence and the error. Lets say a is your true root, and you are at step n and your current guess is xn and after the iteration you are going to get xn+1. Now we can write
f(a)=f(xn) + (a-xn)f'(xn) + (a-xn)2*f''(xn)/2 + ...
Since f(a) is zero, we get dividing throughout by f'(xn):
-f(xn)/f'(xn) = (a-xn)+(a-xn)2 f''(xn)/f'(xn)
The term on the left is just xn+1 - xn. So we get
xn+1-xn = (a-xn)+(a-xn)2 f''(xn)/f'(xn) ...(1)
Okay thats the main formula and it gives us two interesting results. The first is obtained by cancelling xn from both sides. We get:
xn+1 - a = (a-xn)2 f''(xn)/f'(xn)
If we write the error at the nth step as en then we have
en+1 = en2*M
Where M is a bound for f''/f' on the interval of interest. This says that the Newton Raphson method is quadratically convergent.
The second interesting result is obtained by neglecting the quadratic term in formula (1) above. We get
xn+1 - xn = (a-xn ) = en
So an estimate for the error is just the difference between the points at consecutive iterations!