Optimal way to pick the starting $x$ when computing the square root with the Babylonian method I was reading the example given in the Wikipedia article for the Babylonian method of computing the square root, and I wondered why did they set the starting $x_0$ to 600:

To calculate $\sqrt S$, where S = 125348, to six significant figures, use the rough estimation method above to get:
$x_0 = 6 * 10^2$ = 600.000

Did they pick 6 because the number $S$ has 6 digits, and multiplied by $10^2$ to get a number that has half as many digits as $S$?
Update: another reason I'm asking is that I've seen this algorithm that calculates the square root efficiently, but which lacks explanatory comments. I am fairly confident that what it does is compute the square root as if the input was a perfect square of a power of two, but I'm not sure if that's the optimal way in a wider mathematical sense.

In that section, it says that the number $6$ is for numbers in the form $\sqrt{S} = \sqrt{a} \cdot 10^n$, where $10 < a < 100$.
To minimise the absolute error (say $5$ is $3$ more than $2$), we take the arithmetic mean $\frac{\sqrt{10} + \sqrt{100}}{2} \approx 6.58$, and to minimise the relative error (say $5$ is $2.5$ times that of $2$), we take the geometric mean $\sqrt[2]{\sqrt{10} \cdot \sqrt{100}} \approx 5.62$. Therefore, as a rough estimate to $1$ significant figure, $6$ is the value that minimises both the arithmetic and the geometric mean.
