Using Newton's Method for approximating \(\sqrt{3}\) (the relevant function being \(y=x^{2}-3\), whence \(\frac{dy}{dx}=2x),\) the first estimate will be chosen as \(a_{0}=\)\(\frac{1}{1}\) so as to make the comparisons fair. Recall that the method for approximating \(\sqrt{3}\) consists of following each estimate \(a_{n}\) by \[a_{n+1}=a_{n}-\frac{a_{n}^{2}-3}{2a_{n}}.\] So the calculations begin as follows.
\[\frac{1}{1}-\frac{\left(\frac{1}{1}\right)^{2}-3}{2\left(\frac{1}{1}\right)}=\frac{2}{1}=2.000000\]
\[\frac{2}{1}-\frac{\left(\frac{2}{1}\right)^{2}-3}{2\left(\frac{2}{1}\right)}=\frac{7}{4}=1.750000\]
\[\frac{7}{4}-\frac{\left(\frac{7}{4}\right)^{2}-3}{2\left(\frac{7}{4}\right)}=\frac{97}{56}\approx 1.732143\]
\[\frac{97}{56}-\frac{\left(\frac{97}{56}\right)^{2}-3}{2\left(\frac{97}{56}\right)}=\frac{18817}{10864}\approx 1.732051\]
Notice that the first three of these steps are the 2nd, 4th, and 8th rungs, respectively, of the Greek ladder on page 3, and you can check that if the Greek ladder were extended to sixteen rungs, the result would be the rung \(\langle 10864\quad 18817 \rangle\).
Such is the "doubling" pattern that was proved in [6]. In any case, the Greek ladder gives six-place accuracy with but a three-digit denominator \((780)\), as do the continued fractions, while Newton's Method requires the five-place denominator \(10864.\) Moreover, the classical Greek ladder and the continued fraction calculations on pages 3 and 4 use only simple arithmetic, while Newton's Method requires calculating derivatives.