How do I find the error of nth iteration in Newton's Raphson's method without knowing the exact root In our calculus class, we were introduced to the numerical approximation of root by Newton Raphson method. The question was to calculate the root of a function up to nth decimal places.
Assuming that the function is nice and our initial value does lead to convergence. Our teacher terminated the algorithm when the two successive iterations had the same first n digits and told us that the the approximation was correct up to the nth digit!
I feel that the termination step is valid if $f(x_n)$ and $f(x_{n+1})$ has different signs but my teacher disagrees. How do I sort this out??
Futhermore how do I find the error in the  nth iteration without knowing the exact root?

This is a complement to Pp.'s answer.
Newton's method converges, and with the rate given in Pp.'s answer, only if certain assumptions are met. In simple numerical examples these assumptions are usually not tested in advance. Instead, one chooses a reasonable starting point $x_0$ and will then soon find out whether the process converges to the intended root.
Simple sufficient conditions are the following: You have made out an interval $I:=[a,b]$ such that $f(a)f(b)<0$, and that $f'(x)\ne0$, $f''(x)\ne0$ for all $x\in I$. Depending on the signs of $f$, $f'$, $f''$ at the endpoints of $I$ you should chose either $x_0:=a$ or $x_0:=b$ and then can be sure that the $x_n$ converge to the unique root $\xi\in I$. E.g., if $f(b)>0$, $f'(x)>0$ and $f''(x)>0$ $\>(x\in I)$ you should choose $x_0:=b$. Note that in this case $x_n>\xi$ for all $n$ (draw a figure!), so that the lower estimate $a$ of the root is never improved.
