I have a question about using Newton-Raphson to refine a guess of the reciprocal square root function. The reciprocal square root of $a$ is the number $x$ which satisfies the following equation:
$$x^{-2} = a$$
So we are looking for the root of the following equation:
$$x^{-2} - a = 0$$
Applying the Newton-Raphson method then leads to the following:
$$x_{n+1} = x_n - {f(x_n) \over f'(x_n)} = x_n - {x_n^{-2} - a \over -2x_n^{-3}} = x_n(1.5 - 0.5ax_n^2)$$
Before looking up the above standard solution, I tried to come up with my own equation:
$$x^2 = {1 \over a}$$
In this case, we are looking for the root of a different equation:
$$x^2 - {1 \over a} = 0$$
And the Newton-Raphson method gives us:
$$x_{n+1} = x_n - {f(x_n) \over f'(x_n)} = x_n - {x_n^2 - {1 \over a} \over 2x_n} = 0.5(x_n + {1 \over ax_n})$$
Is there anything wrong with this alternative approach, and why would I choose one over the other?