Calculate $x=\frac{1}{a}$ indirectly For a given constant like $a \ne 0$, we are asked to calculate $\frac{1}{a}$ indirectly. The question also asks to determine the values of $x_0$ for which the sequence converges.
Note:
We have learned fixed-point iteration, False position, Newton, Aitken, Steffenson methods. I think that for this question, it's enough to provide a function with root $\frac{1}{a}$. So, is $g(x)=\frac{\frac{1}{a}+x}{2}$ good? If yes, how should I answer the second part of the question? I mean, How should I determine a bound for $x_0$ such that this bound, guarantees the sequence's convergence?  
Note 2: The first part of my question is similar to this question.
Note 3: For example, if you're using fixed-point iteration method, you should provide $g(x)$ before seeing $a$, and guarantee that it works for every arbitrary $a$ that you will be given.
Note 4: You can use whatever you want, But you should avoid the direct division $\frac{1}{a}$. Other divisions can be used.
 A: If $x_n$ is some approximation to the inverse, then $r_n=1-ax_n$ is small. As $ax_n=1-r_n$, so is $$ax_n(1+r_n)=1-r_n^2$$ by binomial theorems. As $r_n^2$ is a smaller error, $$x_{n+1}=x_n(1+r_n)=x_n(2-ax_n)=g(x_n)$$ is a better approximation.
You can get this method as Newton method for $$f(x)=\frac1x-a$$ or as per Arthur  using the geometric sequence with the identity $$\sum_{k=0}^\infty x^k=\prod_{n=0}^\infty(1+x^{2^n}).$$ The only condition is that the initial error needs to be smaller than $1$ in absolute value which can be achieved by finding the dyadic power $x_0=2^m$ so that $2^ma\in [\frac23,\frac43]$ or $[0.4,1.6]$. This you can do by extracting the floating point exponent using frexp or similar.
A: I think that you should change that $g(x)$ in order to respect the "spirit" of the problem. In any case, if you want to analyze convergence, then you have this result: if $g \in C^1\big([a,b]\big)$, $g\big([a,b]\big)\subset [a,b]$ and $g'(x)\leq K<1$ for all $x \in [a,b]$, then for every $x_0 \in [a,b]$ the recurrence
$$x_{n+1}=g(x_n)$$
converges to the only fixed point of $g$ on $[a,b]$.
A: Partial solution: For $-1<x<1$, we have $$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$$Insert $x=1-a$, and you have an approximation valid for $0<a<2$ that you can get as accurate as you'd like by just adding more terms. Of course, if $-2<a<0$, you can calculate $\frac1{-a}$ instead the same way, and just change the sign afterwards.
A: Solving $f(x)=0$ where
$$
f(x)=\frac1x-a
$$
using Newton's Method gives the iteration
$$
\begin{align}
x_{n+1}
&=x_n-\frac{f(x_n)}{f'(x_n)}\\
&=x_n-\frac{\frac1{x_n}-a}{-\frac1{x_n^2}}\\
&=x_n+x_n-ax_n^2
\end{align}
$$
that is,
$$
\bbox[5px,border:2px solid #C0A000]{x_{n+1}=2x_n-ax_n^2}
$$
Since
$$
\begin{align}
x_{n+1}-\frac1a
&=2x_n-ax_n^2-\frac1a\\
&=-a\left(x_n-\frac1a\right)^2
\end{align}
$$
the iteration is a contraction if $|ax_n-1|\lt1$.

A good choice is $x_0=10^k$ so that $0.1\lt ax_0\le1$.

Once it gets going, each iteration should about double the number of digits of accuracy.

Example
Find $\frac1{32}$: Let $a=32$ and $x_0=0.01$:
$$
\begin{array}{r|l}
n&x_n\\\hline
0&0.01\\
1&0.0168\\
2&0.02456832\\
3&0.0298213648760832\\
4&0.031184688053846759344\\
5&0.031249863499190069638\\
6&0.031249999999403760924\\
7&0.031250000000000000000\\
8&0.031250000000000000000
\end{array}
$$
