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Mathematical economics is old enough to be respectable, but not all economists respect it. It has powerful supporters and impressive testimonials, yet many capable economists deny that mathematics, except as a shorthand or expository device, can be applied to economic reasoning. There have even been rumors that mathematics is used in economics (and in other social sciences) either for the deliberate purpose of mystification or to confer dignity upon common places as French was once used in diplomatic communications.

In J. R. Newman (ed.) The World of Mathematics, New York: Simon and Schuster, 1956.

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# The Classic Greek Ladder and Newton's Method

### 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.

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Wisner, Robert J., "The Classic Greek Ladder and Newton's Method," Loci (August 2009), DOI: 10.4169/loci003330