2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.