# Newton-Raphson for reciprocal square root

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?

• You would choose the first if you implement it on a computer because multiplication is faster than division. – Daniel Fischer Dec 15 '13 at 13:41

The slowness of division vs. multiplication on early mainframes motivated the division-free Newton iteration for solving $$x^{-2} = a$$, as noted in the Question:
$$x_{n+1} = x_n(1.5 - 0.5 * a * x_n^2)$$
Note that if $$\sqrt{a}$$ is really needed, rather than $$1/\sqrt{a}$$, we can get it with one additional multiplication:
$$\sqrt{a} = a \cdot (1/\sqrt{a})$$