The equation $x - Rx^{-1} = 0$ has $x \pm \sqrt{R}$ for its solution. Establish the Newton's iterative scheme, in simplified form, for this situation (to compute the root). There is no need to carry out the actual computation
So far I have
$x - \frac{R}{x}$
so $f'(x) = 1 + \frac{R}{x^2}$ But now what do I do? I'm a little confused.
Here is what I have:
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \implies x_n - \frac{x_n - \frac{R}{x_n}}{1 + \frac{R}{x^2_n}} \implies x_n - [ \frac{x^2_n - R}{x_n} * \frac{x^2_n}{x^2_n + R}]$
which $\implies x_n - [ x^2_n - R * \frac{x_n}{x^2_n + R} ]$
so... now I'm getting $\frac{x^3_n + x_n R - x^3_n + x_n R}{x^2_n + R}$
Finally I am getting: $\frac{2x_n R}{x^2_n + R}$
Steps are mostly in my OP thanks to @André Nicolas .
Computing the final result I got: $$x_{n+1}=\frac{2x_n R}{x^2_n + R},$$ which I further turned into $\frac{2}{1} \left( \frac{x_n R}{x^2_n + R} \right)$.