Newton iteration method i need some help here.
My function is $f(x) =x^{3}$ . I was asked to find the number of iterations that are needed to reach the precission $10^{-5}$ if $x_{0} = 0.9$
I was wondering if there is a general formula to find the number of iteration, hence, the formula i know is  
$$x_{n+1}  = x_{n} - \frac{f(x_{n})}{f'(x_{n})}.$$ 
I know that it is easy to find the number of iteration by this formula, but what if the number of iteration for reaching my precission is at 40 ? , do i need to calculate all this iterations ? or there is a general formula ? 
So please if someone knows please help.
BTW, this is for Newton
Thank you
 A: When using $f(x)=x^3$ the recursion becomes $x_{n+1}=x_n-\frac{x_n^3}{3 x_n^2}=\frac{2}{3}x_n$ and hence can explicitly be solved as
$$ x_n = \left(\frac{2}{3}\right)^n x_0.$$
Now you just have to plug this into the inequality $x_n\leq 10^{-5}$, take the logarithm and solve for $n$, which gives
$$n\ln\frac{2}{3}\leq\ln\frac{10^{-5}}{0.9}\quad\Rightarrow\quad n\geq\frac{\ln (10^{-5}/0.9)}{\ln(2/3)}\approx 28.2$$
Note that $\leq$ turns to $\geq$ as you divide through a negative logarithm.
This means, the 29th iteration is the first one to be inside the $10^{-5}$-neighborhood.
A: Ingeneral the error in Newton's method satisfies:
$$e_{n+1}=-\frac{f''(\theta_n)}{2f'(x_n)}e_n^2$$
where $\theta_n$ is certain number between $x_n$ and the root.
Put $e_0$ equal to the length of the interval where your root is. Assume you can bound $$|-\frac{f''(\theta_n)}{2f'(x_n)}|<M.$$
Then $$|e_n|\le M|e_{n-1}|^2\le M^{1+2}|e_{n-2}|^4\le ...\le M^{1+2+4+...+2^n}|e_0|^{2^n}=|e_0M|^{2^{n}}M^2$$.
If we want $d$ digits we can put $|e_0M|^{2^{n}}M^2<10^{-d}$ and solve for $n$.
We get ${2^{n}}>\frac{\ln(\frac{10^{-d}}{M^2})}{\ln|e_0M|}$. So
$$n>\log_2\left(\frac{\ln(\frac{10^{-d}}{M^2})}{\ln|e_0M|}\right).$$
