# The Classic Greek Ladder and Newton's Method

Author(s):
Robert J. Wisner (New Mexico State University)

### Extending the Connection

What about Newton's Method and the $\sqrt{3}$ Greek Ladder? From [7], the standard version of this ladder begins with rung $\langle1\quad1\rangle$, each rung is followed by $\langle a+b\quad3a+b\rangle$, and each rung is "reduced." It begins $\begin{array}{ccc} 1 & 1 & \frac{1}{1}=1.00000\\ 1 & 2 & \frac{2}{1}=2.00000\\ 3 & 5 & \frac{5}{3}\approx1.66667\\ 4 & 7 & \frac{7}{4}=1.75000 \end{array} \qquad \begin{array}{ccc} 11 & 19 & \frac{19}{11}\approx1.72727\\ 15 & 26 & \frac{26}{15}\approx1.73333\\ 41 & 71 & \frac{71}{41}\approx1.73171\\ 56 & 97 & \frac{97}{56}\approx1.73214 \end{array}$ Here is what happens when Newton's Method is applied and iterated. $\begin{array}{c} x_{1}=\frac{1}{1}-\frac{\left( \frac{1}{1}\right) ^{2}-3}{2\times\frac{1}{1}}=\frac{2}{1}\quad\text{from Rung 2}\\ x_{2}=\frac{2}{1}-\frac{\left( \frac{2}{1}\right) ^{2}-3}{2\times2}=\frac{7}{4}\quad\text{from Rung 4}\\ x_{3}=\frac{7}{4}-\frac{\left( \frac{7}{4}\right) ^{2}-3}{2\times\frac{7}{4}}=\frac{97}{56}\quad\text{from Rung 8} \end{array}$ So the doubling pattern comports with the case for $\sqrt{3}$ to this point. Can that pattern be proved? Let us see.

The above version of the $\sqrt{3}$ ladder is standard, but finding a general term for the sequences $\{a_{n}\}$ and $\{b_{n}\}$ from the rungs $\langle a_{n}\quad b_{n}\rangle$ as presented is not immediate, since the rungs have been reduced when possible. Here is the $\sqrt{3}$ Greek Ladder without rung reductions: $\begin{array}{cc} 1 & 1\\ 2 & 4\\ 6 & 10\\ 16 & 28 \end{array} \qquad \begin{array}{cc} 44 & 76\\ 120 & 208\\ 328 & 568\\ 896 & 1552 \end{array}$ And now, a recursive definition of the rungs $\langle a_{n}\quad b_{n}\rangle$ of this equivalent version of the $\sqrt{3}$ Greek Ladder is suggested and given by \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1} +2a_{n-2};\\ b_{1} & =1\text{, }b_{2}=4\text{, and for }n>2\text{, }b_{n}=2b_{n-1}+2b_{n-2} \end{align*} (It is left to the reader to show that this is equivalent to the standard definition given earlier for the $\sqrt{3}$ ladder.) Using the same method of [4] again, the $n$-th rung $\langle a_{n}\quad b_{n}\rangle$ is $\left\langle \frac{\left( 1+\sqrt{3}\right) ^{n}-\left( 1-\sqrt{3}\right)^{n}}{2\sqrt{3}}\qquad\frac{\left( 1+\sqrt{3}\right) ^{n}+\left( 1-\sqrt{3}\right) ^{n}}{2}\right\rangle$

So let us see what happens when $\frac{b_{n}}{a_{n}}$ is chosen as an estimate in Newton's Method. The corresponding approximation for $\sqrt{3}$ deriving from the $n$-th rung simplifies to $\frac{b_{n}}{a_{n}} = \frac{\sqrt{3}\left( \sqrt{3}+1\right) ^{n}+\sqrt{3}\left( 1-\sqrt{3}\right) ^{n}}{\left( \sqrt{3}+1\right) ^{n}-\left( 1-\sqrt{3}\right) ^{n}}$ When this fraction is subjected to Newton's Method for $\sqrt{3}$, you can check that the result is $\frac{\sqrt{3}\left[ \left( 1+\sqrt{3}\right) ^{2n}+\left( 1-\sqrt{3}\right) ^{2n}\right] }{\left(1+\sqrt{3}\right) ^{2n}-\left(1-\sqrt{3}\right) ^{2n}}$ which is the reduced form of the fraction $\frac{b_{2n}}{a_{2n}}$ from the $2n$-th rung of the ladder, so the doubling is established for the $\sqrt{3}$ ladder.

Since the doubling pattern holds for the ladders of $\sqrt{2}$ and $\sqrt{3}$, it is natural to ask if it holds for the $\sqrt{k}$ Greek Ladder, in which each rung $\langle a\quad b\rangle$ is followed by the rung $\langle a+b\quad ka+b\rangle$ That ladder begins $\begin{array}{ccc} 1 & 1 & \frac{1}{1}\\ 2 & k+1 & \frac{k+1}{2}\\ k+3 & 3k+1 & \frac{3k+1}{k+3}\\ 4k+4 & k^{2}+6k+1 & \frac{k^{2}+6k+1}{4k+4}\\ k^{2}+10k+5 & 5k^{2}+10k+1 & \frac{5k^{2}+10k+1}{k^{2}+10k+5}\end{array}$ giving rise to the recursive definition \begin{align*} a_{1} & =1\text{, }a_{2}=2\text{, and for }n>2\text{, }a_{n}=2a_{n-1}+\left( k-1\right) a_{n-2};\\ b_{1} & =1\text{, }b_{2}=k+1\text{, and for }n>2\text{, }b_{n} =2b_{n-1}+\left( k-1\right) b_{n-2} \end{align*} It is again left to the reader to show equivalence in definitions. Working from [4] again leads to $a_{n}=\frac{\left( 1+\sqrt{k}\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}{2\sqrt{k}}\text{ and }b_{n}=\frac{\left( 1+\sqrt{k}\right)^{n}+\left( 1-\sqrt{k}\right)^{n}}{2}$ So the corresponding Greek Ladder approximation to $\sqrt{k}$ is $\frac{b_{n}}{a_{n}}=\frac{\sqrt{k}\left( 1-\sqrt{k}\right)^{n}+\sqrt{k}\left( \sqrt{k}+1\right)^{n}}{\left( \sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}$ and the Newton formula yields \begin{align*} & \frac{\sqrt{k}\left( 1-\sqrt{k}\right) ^{n}+\sqrt{k}\left( \sqrt{k}+1\right)^{n}}{\left( \sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}-\frac{\left( \frac{\sqrt{k}\left( 1-\sqrt{k}\right)^{n}+\sqrt {k}\left( \sqrt{k}+1\right)^{n}}{\left( \sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}\right)^{2}-k}{2\times\frac{\sqrt{k}\left(1-\sqrt{k}\right)^{n}+\sqrt{k}\left( \sqrt{k}+1\right)^{n}}{\left(\sqrt{k}+1\right)^{n}-\left( 1-\sqrt{k}\right)^{n}}}\\ & =\frac{\sqrt{k}\left( \sqrt{k}+1\right)^{2n}+\sqrt{k}\left( 1-\sqrt{k}\right)^{2n}}{\left( \sqrt{k}+1\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}} \end{align*} But is this the fraction from the $2n$-th rung of the ladder? That rung is $\left\langle \frac{\left( 1+\sqrt{k}\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}}{2\sqrt{k}}\qquad\frac{\left( 1+\sqrt{k}\right)^{2n}+\left(1-\sqrt{k}\right)^{2n}}{2}\right\rangle$ which gives the approximation $\frac{\frac{\left( 1+\sqrt{k}\right)^{2n}+\left( 1-\sqrt{k}\right)^{2n}}{2}}{\frac{\left( 1+\sqrt{k}\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}}{2\sqrt{k}}}=\frac{\sqrt{k}\left( \sqrt{k}+1\right) ^{2n}+\sqrt{k}\left( 1-\sqrt{k}\right)^{2n}}{\left( \sqrt{k}+1\right)^{2n}-\left( 1-\sqrt{k}\right)^{2n}}$ as it should.