For this section, we need, for reference, to repeat and extend the classic ladder, this time with rungs numbered in bold: \[ \begin{array}{ccccccccccccccc} \mathbf{1} & 1 & 1 & & \mathbf{5} & 29 & 41 & & \mathbf{9} & 985 & 1393 & & \mathbf{13} & 33461 & 47321\\ \mathbf{2} & 2 & 3 & & \mathbf{6} & 70 & 99 & & \mathbf{10} & 2378 & 3363 & & \mathbf{14} & 80782 & 114243\\ \mathbf{3} & 5 & 7 & & \mathbf{7} & 169 & 239 & & \mathbf{11} & 5741 & 8119 & & \mathbf{15} & 195025 & 275807\\ \mathbf{4} & 12 & 17 & & \mathbf{8} & 408 & 577 & & \mathbf{12} & 13860 & 19601 & & \mathbf{16} & 470832 & 665857

\end{array} \]

A standard topic in calculus is Newton's Method for approximating roots, so it is natural to ask how the classic Greek Ladder compares with Newton's Method for approximating \( \sqrt{2} \). To that end, we let \(s\left( x\right) =x^{2}-2\), so then \(s^{\prime}\left( x\right) =2x\), and the Newton scheme is to start with an estimate \(x_{0}\) and compute for each \(x_{n}\) the next estimate \(x_{n+1}\), where \[ x_{n+1}=x_{n}-\frac{s\left( x_{n}\right) }{s^{\prime}\left( x_{n}\right) } \] The classic Greek Ladder begins with the first estimate of \(\frac{1}{1}\), so

for comparison, we choose the corresponding Newton estimate as \(x_{0}=\frac {1}{1}\), and this is what ensues: \[ \begin{array}{lll} x_{1}=\frac{1}{1}-\frac{\left( \frac{1}{1}\right) ^{2}-2}{2\times\frac{1}{1}}=\frac{3}{2}\quad\text{from Rung 2} & & x_{3}=\frac{17}{12}-\frac{\left( \frac{17}{12}\right) ^{2}-2}{2\times\frac{17}{12}}=\frac{577}{408}\quad\text{from Rung 8}\\x_{2}=\frac{3}{2}-\frac{\left( \frac{3}{2}\right) ^{2}-2}{2\times\frac{3}{2}}=\frac{17}{12}\quad\text{from Rung 4} & & x_{4}=\frac{577}{408} -\frac{\left( \frac{577}{408}\right) ^{2}-2}{2\times\frac{577}{408}} =\frac{665857}{470832}\quad\text{from Rung 16}\end{array}\] So the \(n\)-th Newton iterate equals the \(2^{n}\)-th Greek Ladder approximation for \(n=1\) to \(n=4\). The next Newton iterates comport with the approximations from rung \(32\) of the ladder, which is \(\langle627013566048\quad 886731088897\rangle\); the \(64\)-th rung; and on up to rungs too wide for a page here. (These computations were obtained by the software *Scientific WorkPlace*, and if you ask it to "Evaluate" an expression involving fractions, it gives fractional output. "Evaluate Numerically" gives decimal estimates as output.) In any case, it seems that more than a sufficiency of such calculations have now been made so as to provoke the statement of this**Theorem.** *The classical* \(\sqrt{2}\) *Greek Ladder has the property that its rungs numbered* \(2^{n}\) *yield fractions equal to the* \(n^{th}\) *Newton's Method iterates* \(x_{n}\) *approximating* \(\sqrt{2}\), *with* \(x_{0}=\frac{1}{1}\).*Proof.* From the recursive definition of the columns of the classical Greek Ladder, which is \[ \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1} +a_{n-2};\\ b_{1} & =1\text{, }b_{2}=3\text{, and for }n>2\text{, }b_{n}=2b_{n-1}+b_{n-2}\end{align*} \]

we fix on the sequences \(\left\{ a_{i}\right\}\) and \(\left\{b_{i}\right\}\). But first, it needs to be shown that this recursive definition is equivalent to the original. To this end, look at the rungs beginning with \(\langle a_{n-2}\quad b_{n-2}\rangle\) and use the original system progressing therefrom: \[ \begin{array}{cc} a_{n-2} & b_{n-2}\\ a_{n-1}=a_{n-2}+b_{n-2} & b_{n-1}=2a_{n-2}+b_{n-2}\\ a_{n}=3a_{n-2}+2b_{n-2} & b_{n}=4a_{n-2}+3b_{n-2} \end{array} \] We can see from this that \[a_{n}=3a_{n-2}+2b_{n-2}=2\left( a_{n-2} +b_{n-2}\right) +a_{n-2}=2a_{n-1}+a_{n-2}\] which is the relationship in the recursive definition of \(a_{n}\). A similar analysis gives \(b_{n}\) in terms of \(b_{n-1}\) and \(b_{n-2}\).

The splendid paper by Hendel [4] gives various methods for deriving general terms of such recursive sequences. The section of that paper titled "Number Theory" seems most direct for the purposes at hand and gives \[ a_{n}=\frac{\left( 1+\sqrt{2}\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n} }{2\sqrt{2}}\text{ and }b_{n}=\frac{\left( 1+\sqrt{2}\right) ^{n}+\left(1-\sqrt{2}\right) ^{n}}{2}. \] So the \(n\)-th rung of the classical Greek Ladder is \[ {\left\langle \frac{\left( 1+\sqrt{2}\right) ^{n}-\left( 1-\sqrt{2}\right)^{n}}{2\sqrt{2}}\qquad\frac{\left( 1+\sqrt{2}\right) ^{n}+\left( 1-\sqrt{2}\right) ^{n}}{2}\right\rangle}.\] The corresponding approximation for \(\sqrt{2}\) deriving from the \(n\)-th rung is thus \[ \frac{\frac{\left( 1+\sqrt{2}\right) ^{n}+\left( 1-\sqrt{2}\right) ^{n}}{2}}{\frac{\left( 1+\sqrt{2}\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n}}{2\sqrt{2}}}=\frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left(1-\sqrt{2}\right) ^{n}}{\left( \sqrt{2}+1\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n}} \] and we want to see what happens when this fraction from the \(n\)-th rung is subjected to Newton's Method. Here it is: \[\begin{align*} & \frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left( 1-\sqrt {2}\right) ^{n}}{\left( \sqrt{2}+1\right) ^{n}-\left( 1-\sqrt{2}\right)^{n}}-\frac{\left( \frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left( 1-\sqrt{2}\right) ^{n}}{\left( \sqrt{2}+1\right)^{n}-\left(1-\sqrt{2}\right) ^{n}}\right) ^{2}-2}{2\times\frac{\sqrt{2}\left( \sqrt{2}+1\right) ^{n}+\sqrt{2}\left( 1-\sqrt{2}\right) ^{n}}{\left( \sqrt{2}+1\right) ^{n}-\left( 1-\sqrt{2}\right) ^{n}}}\\ & =\frac{\sqrt{2}\left[ \left( 1+\sqrt{2}\right) ^{2n}+\left( 1-\sqrt{2}\right) ^{2n}\right] }{\left( 1+\sqrt{2}\right) ^{2n}-\left(1-\sqrt{2}\right) ^{2n}}\end{align*}\] which comes from the \(2n\)-th rung, \[ \left\langle \frac{\left( 1+\sqrt{2}\right) ^{2n}-\left( 1-\sqrt{2}\right)^{2n}}{2\sqrt{2}}\qquad\frac{\left( 1+\sqrt{2}\right) ^{2n}+\left(1-\sqrt{2}\right) ^{2n}}{2}\right\rangle,\] whence the doubling pattern of the theorem is established.

It is implicit in this proof that the doubling takes place for *any* rung, so the process can begin by starting not just with \(\langle1\quad1\rangle\) but with any rung whatsoever.