ATRACTOR

In the case of the regular hexagon and the pentagon, the conclusions we would reach would be similar to those in the case of the square. In particular, we would have:

- a sequence of rational numbers converging to \(\frac{1+\sqrt{5}}{2}\), given by: \[r_{n}=\frac{r_{n-1}+1}{r_{n-1}}=1+\frac{1}{r_{n-1}},\; r_{0}=1\]

- a sequence of rational numbers converging to \(\sqrt{3}\), given by: \[t_{n}=\frac{t_{n-1}+3}{t_{n-1}+1}=1+\frac{2}{1+t_{n-1}},\; t_{0}=1\]

So, we get: \[r_{1}=1+\frac{1}{r_{0}}=1+\frac{1}{1}\] \[r_{2}=1+\frac{1}{r_{1}}=1+\frac{1}{1+\frac{1}{1}}\] \[r_{3}=1+\frac{1}{r_{2}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{1}}}\]
\[\frac{1+\sqrt{5}}{2}=\mbox{lim }r_{n}=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}}}}=[1,1,1,1,1,1,1,...]\]
\[\begin{array}{cl} t_{n} & =1+\frac{2}{1+t_{n-1}}\\ & =1+\frac{2}{1+1+\frac{2}{1+t_{n-2}}}\\ & =1+\frac{2}{2+\frac{2}{1+t_{n-2}}}\\ & =1+\frac{1}{1+\frac{1}{1+t_{n-2}}},\: n>1 \end{array}\] \[t_{0}=1\] \[t_{1}=1+\frac{2}{1+1}=1+\frac{2}{2}=1+\frac{1}{1}\] \[t_{2}=1+\frac{1}{1+\frac{1}{1+t_{0}}}=1+\frac{1}{1+\frac{1}{1+1}}=1+\frac{1}{1+\frac{1}{2}}\] \[t_{3}=1+\frac{1}{1+\frac{1}{1+t_{1}}}=1+\frac{1}{1+\frac{1}{1+1+\frac{1}{1}}}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1}}}\] \[t_{4}=1+\frac{1}{1+\frac{1}{1+t_{2}}}=1+\frac{1}{1+\frac{1}{1+1+\frac{1}{1+\frac{1}{2}}}}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2}}}}\] \[t_{5}=1+\frac{1}{1+\frac{1}{1+t_{3}}}=1+\frac{1}{1+\frac{1}{1+1+\frac{1}{1+\frac{1}{2+\frac{1}{1}}}}}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1}}}}}\] \[t_{6}=1+\frac{1}{1+\frac{1}{1+t_{4}}}=1+\frac{1}{1+\frac{1}{1+1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2}}}}}}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2}}}}}}\]
\[\sqrt{3}=\mbox{lim }t_{n}=\mbox{lim }t_{2n-1}-\mbox{lim }t_{2n}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}}}=[1,1,2,1,2,1,2,...]\]

Remark: if we look at the second proof of the incommensurability between the diagonal and the side of the pentagon, we have another sequence of rational numbers converging to \(\frac{1+\sqrt{5}}{2}\), given by: \[s_{n}=\frac{2s_{n-1}+1}{s_{n-1}+1}=1+\frac{1}{1+\frac{1}{s_{n-1}}},\; s_{0}=1\]

Using induction, we see that \(s_{n}\) is the subsequence of \(r_{n}\) formed by the even order terms, that is, \(s_{n}=r_{2n}\). This is not unexpected, since the successive pentagons obtained in the second proof are congruent with the even order pentagons obtained in the first proof. Check this!