newton's method on $\frac{1}{x}$

Question

I was trying to use newton's method to approximate $\frac{1}{x}$, by finding the solution to $\frac{1}{x}=0$,I got $$x_{k+1}=x_k-\frac{\frac{1}{x_k}-n} {-\frac{1}{x_k^2}} = 2x_k-nx_k^2$$ I was trying to find where it is convergent to the root, $0$ and $1$ both don't work, and I did not manage to find any other candidate.
can any of you help me find where it converges?
I'm pretty sure this isn't a very efficient way to approximate it, are there any other better ways?

Answer 1

You have $$x_{k+1}-\frac{1}{n} = -n(x_k^2-2\frac{1}{n}x_k + \frac{1}{n^2}) =-n(x_k-\frac{1}{n})^2$$ So, for all $k\in \mathbb{N}$ $$x_k-\frac{1}{n} = (-1)^k n^k (x_0-\frac{1}{n})^{2k}$$

The sequence converges if and only if the initial value of the sequence $x_0$ satisfies the following condition: $$\sqrt{n}|x_0-\frac{1}{n}|<1$$ or $$\frac{1}{n} - \frac{1}{\sqrt{n}}< x_0 < \frac{1}{n} + \frac{1}{\sqrt{n}}$$