newton: convergence when calculating $x^2-2=0$ Find $x$ for which $x^2-2=0$ using the newton algorithm and $x_0 = 1.4$.
Then you get $x_{k+1} = x_k + \frac{1}{x_k} - \frac{x_k}{2}$.
How to show that you need 100 steps for 100 digits precision?
So I need to show for which $N$ it is $|x_N-\sqrt{2}| \leq 10^{-100}$ and therefore I am asked to show
$$|x_k - \sqrt{2}|\leq\delta<0.1 \implies |x_{k+1}-\sqrt{2}|\leq \frac{\delta^2}{2}$$
Then it obviously follows that you need 100 steps, but I don't manage to show this..

Try to show that $|x_{k+1}-\sqrt2|\leqslant\frac12(x_k-\sqrt2)^2$ hence $|x_k-\sqrt2|\leqslant2\delta_k$ implies that $|x_{k+1}-\sqrt2|\leqslant2\delta_{k+1}$ with $\delta_{k+1}=\delta_k^2$. 
Then note that $\delta_0\lt10^{-2}$ hence $\delta_k\leqslant10^{-2^{k+1}}$ for every $k$, in particular $2\delta_6\lt2\cdot10^{-128}$ ensures (slightly more than) $100$ accurate digits.
