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 x_{n} and after the iteration you are going to get x_{n+1}. Now we can write

f(a)=f(x_{n}) + (a-x_{n})f'(x_{n}) + (a-x_{n})^{2}*f''(x_{n})/2 + ...

Since f(a) is zero, we get dividing throughout by f'(x_{n}):

-f(x_{n})/f'(x_{n}) = (a-x_{n})+(a-x_{n})^{2} f''(x_{n})/f'(x_{n})

The term on the left is just x_{n+1} - x_{n}. So we get

x_{n+1}-x_{n} = (a-x_{n})+(a-x_{n})^{2} f''(x_{n})/f'(x_{n})
...(1)

Okay thats the main formula and it gives us two interesting results. The first is obtained by cancelling x_{n} from both sides. We get:

x_{n+1} - a = (a-x_{n})^{2} f''(x_{n})/f'(x_{n})

If we write the error at the n^{th} step as e_{n} then we have

e_{n+1} = e_{n}^{2}*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

x_{n+1} - x_{n} = (a-x_{n} ) = e_{n}

So an estimate for the error is just the difference between the points at consecutive iterations!