A Disquisition on the Square Root of Three - Newton's Method

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 2^nd, 4^th, and 8^th 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.