ATRACTOR

We have \(B=\begin{pmatrix}-1 & 2\\ 1 & -1 \end{pmatrix}^{-1}=\begin{pmatrix}1 & 2\\ 1 & 1 \end{pmatrix}\). Diagonalizing the matrix \(B\) we get: \[B=\begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}1-\sqrt{2} & 0\\ 0 & 1+\sqrt{2} \end{pmatrix}\begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}^{-1}\]

Therefore, the successive powers of \(B\) are given by: \[\begin{array}{cll} B^{n} & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}1-\sqrt{2} & 0\\ 0 & 1+\sqrt{2} \end{pmatrix}^{n}\begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}^{-1}\\ & = & \begin{pmatrix}-\sqrt{2} & \sqrt{2}\\ 1 & 1 \end{pmatrix}\begin{pmatrix}\left(1-\sqrt{2}\right)^{n} & 0\\ 0 & \left(1+\sqrt{2}\right)^{n} \end{pmatrix}\begin{pmatrix}-\frac{1}{2\sqrt{2}} & \frac{1}{2}\\ \frac{1}{2\sqrt{2}} & \frac{1}{2} \end{pmatrix}\\ & = & \begin{pmatrix}\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2} & \frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{\sqrt{2}}\\ \frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2\sqrt{2}} & \frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2} \end{pmatrix} \end{array}\]

So we get: \[\begin{array}{cll} P_{n} & = & B^{n}P_{0}\\ & = & \begin{pmatrix}\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2} & \frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{\sqrt{2}}\\ \frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2\sqrt{2}} & \frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2} \end{pmatrix}\begin{pmatrix}x_{0}\\ y_{0} \end{pmatrix}\\ & = & \begin{pmatrix}\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}\\\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}y_{0} \end{pmatrix} \end{array}\]
\[x_{n}=\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}\] \[y_{n}=\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}y_{0}\]

Note that the sequences \(a_{n}=\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}\) and \(b_{n}=\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}\) can be recursively defined in the following way: \[\begin{array}{ccc} a_{n}=a_{n-2}+2a_{n-1}, & \; a_{1}=1, & \; a_{2}=3\\ b_{n}=b_{n-2}+2b_{n-1}, & \; b_{1}=1, & \; b_{2}=2 \end{array}\]

Thus, since \(x_{n}=a_{n}x_{0}+2b_{n}y_{0}\) and \(y_{n}=b_{n}x_{0}+a_{n}y_{0}\), the sequences \(x_{n}\) and \(y_{n}\) can be recursively defined in the following way: \[\begin{array}{cll} x_{n}=x_{n-2}+2x_{n-1}, & \; x_{1}=x_{0}+2y_{0}, & \; x_{2}=3x_{0}+4y_{0}\\ y_{n}=y_{n-2}+2y_{n-1}, & \; y_{1}=x_{0}+y_{0}, & \; y_{2}=2x_{0}+3y_{0} \end{array}\]

If \(x_{0}\) and \(y_{0}\) are positive integer numbers, we deduce by induction that \(x_{n}\) and \(y_{n}\) are also positive integer numbers for all \(n\in\mathbb{N}\).

On the other hand, if \(x_{0}\) and \(y_{0}\) are relatively prime, then so are \(x_{n}\) e \(y_{n}\) for all \(n\in\mathbb{N}\), which can also be proved by induction.

Indeed, we have: \[\begin{pmatrix}x_{n}\\ y_{n} \end{pmatrix}=\begin{pmatrix}1 & 2\\ 1 & 1 \end{pmatrix}\begin{pmatrix}x_{n-1}\\ y_{n-1} \end{pmatrix}=\begin{pmatrix}x_{n-1}+2y_{n-1}\\ x_{n-1}+y_{n-1} \end{pmatrix}\]
\[x_{n}=x_{n-1}+2y_{n-1}\] \[y_{n}=x_{n-1}+y_{n-1}\]

Assuming that \(x_{n-1}\) and \(y_{n-1}\) are relatively prime, that is, \(\mbox{gcd}\left(x_{n-1},\: y_{n-1}\right)=1\), we get: \[\begin{array}{cl} \mbox{gdc}\left(x_{n},\: y_{n}\right) & =\mbox{gcd}\left(x_{n-1}+2y_{n-1},\: x_{n-1}+y_{n-1}\right)\\ & =\mbox{gcd}\left(x_{n-1}+2y_{n-1}-\left(x_{n-1}+y_{n-1}\right),\: x_{n-1}+y_{n-1}\right)\\ & =\mbox{gcd}\left(y_{n-1},\: x_{n-1}+y_{n-1}\right)\\ & =\mbox{gcd}\left(y_{n-1},\: x_{n-1}+y_{n-1}-y_{n-1}\right)\\ & =\mbox{gcd}\left(y_{n-1},\: x_{n-1}\right)\\ & =1 \end{array},\] that is, \(x_{n}\) and \(y_{n}\) are relatively prime.

The sequence \(r_{n}=\frac{x_{n}}{y_{n}}\) can then be defined in two different ways:

- by its generic term: \[\begin{array}{cl} r_{n} & =\frac{x_{n}}{y_{n}}\\ & =\frac{\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}x_{0}+\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{\sqrt{2}}y_{0}}{\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2\sqrt{2}}x_{0}+\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}y_{0}}\end{array}\]

- recursively: \[\begin{array}{cl} r_{n} & =\frac{x_{n}}{y_{n}}\\ & =\frac{x_{n-1}+2y_{n-1}}{x_{n-1}+y_{n-1}}\\ & =\frac{\frac{x_{n-1}}{y_{n-1}}+2}{\frac{x_{n-1}}{y_{n-1}}+1}\\ & =\frac{r_{n-1}+2}{r_{n-1}+1}\\ & =1+\frac{1}{1+r_{n-1}} \end{array}\]

Remarks:

- For each pair of positive integer values of \(x_{0}\) and \(y_{0}\), we obtain a sequence of rationals converging to \(\sqrt{2}\).

- In the above case, where we had \(x_{0}=1\) and \(y_{0}=1\), the terms of the sequence \(r_{n}\) can be written in the following way: \[r_{0}=1\] \[r_{1}=1+\frac{1}{1+r_{0}}=1+\frac{1}{1+1}=1+\frac{1}{2}\] \[r_{2}=1+\frac{1}{1+r_{1}}=1+\frac{1}{1+1+\frac{1}{2}}=1+\frac{1}{2+\frac{1}{2}}\] \[r_{3}=1+\frac{1}{1+r_{2}}=1+\frac{1}{1+1+\frac{1}{2+\frac{1}{2}}}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}\] \[\sqrt{2}=\mbox{lim }r_{n}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+...}}}}}}\]

We say that \(\sqrt{2}\) is represented as a continued fraction. To know more about continued fractions, consult this page.

- There are sequences which converge to \(\sqrt{2}\) faster than the ones obtained by the method above, such as the following recursively defined sequence: \[s_{0}=1\] \[s_{n}=\frac{1}{2}\left(s_{n-1}+\frac{2}{s_{n-1}}\right),\; n>0\]

We have: \[s_{1}=\frac{3}{2}=1,5\] \[s_{2}=\frac{17}{12}=1,416666666666...\] \[s_{3}=\frac{577}{408}=1,414215686274...\] \[s_{4}=\frac{665857}{470832}=1,414213562374...\] where \(\sqrt{2}=1,414213562373...\)

Indeed, \(\left(s_{n}\right)_{n\in\mathbb{N}}\) is the subsequence of the sequence \[\begin{array}{cl} r_{n} & =\frac{\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}.1+\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{\sqrt{2}}.0}{\frac{-\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2\sqrt{2}}.1+\frac{\left(1-\sqrt{2}\right)^{n}+\left(1+\sqrt{2}\right)^{n}}{2}.0}\\ & =\frac{\left(1+\sqrt{2}\right)^{n}+\left(1-\sqrt{2}\right)^{n}}{\left(1+\sqrt{2}\right)^{n}-\left(1-\sqrt{2}\right)^{n}}\sqrt{2},\; n>0 \end{array}\] formed by the terms whose order is an integer power of \(2\), that is, \(s_{n}=r_{2^{n}}\), which can be seen using induction on \(n\).

For \(n=0\), we have \[s_{0}=r_{2^{0}}=r_{1}=1\]

Assuming now that we have \[s_{n-1}=r_{2^{n-1}}=\frac{\left(1+\sqrt{2}\right)^{2^{n-1}}+\left(1-\sqrt{2}\right)^{2^{n-1}}}{\left(1+\sqrt{2}\right)^{2^{n-1}}-\left(1-\sqrt{2}\right)^{2^{n-1}}}\sqrt{2}\] we get \[\begin{array}{cl} s_{n} & =\frac{1}{2}\left(s_{n-1}+\frac{2}{s_{n-1}}\right)\\ & =\frac{1}{2}\left(\frac{\left(1+\sqrt{2}\right)^{2^{n-1}}+\left(1-\sqrt{2}\right)^{2^{n-1}}}{\left(1+\sqrt{2}\right)^{2^{n-1}}-\left(1-\sqrt{2}\right)^{2^{n-1}}}\sqrt{2}+\frac{2}{\frac{\left(1+\sqrt{2}\right)^{2^{n-1}}+\left(1-\sqrt{2}\right)^{2^{n-1}}}{\left(1+\sqrt{2}\right)^{2^{n-1}}-\left(1-\sqrt{2}\right)^{2^{n-1}}}\sqrt{2}}\right)\\ & =\frac{1}{2}\left(\frac{\left(1+\sqrt{2}\right)^{2^{n-1}}+\left(1-\sqrt{2}\right)^{2^{n-1}}}{\left(1+\sqrt{2}\right)^{2^{n-1}}-\left(1-\sqrt{2}\right)^{2^{n-1}}}\sqrt{2}+\frac{\left(1+\sqrt{2}\right)^{2^{n-1}}-\left(1-\sqrt{2}\right)^{2^{n-1}}}{\left(1+\sqrt{2}\right)^{2^{n-1}}+\left(1-\sqrt{2}\right)^{2^{n-1}}}\sqrt{2}\right)\\ & =\frac{1}{2}\frac{\left(\left(1+\sqrt{2}\right)^{2^{n-1}}+\left(1-\sqrt{2}\right)^{2^{n-1}}\right)^{2}+\left(\left(1+\sqrt{2}\right)^{2^{n-1}}-\left(1-\sqrt{2}\right)^{2^{n-1}}\right)^{2}}{\left(\left(1+\sqrt{2}\right)^{2^{n-1}}-\left(1-\sqrt{2}\right)^{2^{n-1}}\right)\left(\left(1+\sqrt{2}\right)^{2^{n-1}}+\left(1-\sqrt{2}\right)^{2^{n-1}}\right)}\sqrt{2}\\ & =\frac{1}{2}\frac{2\left(\left(1+\sqrt{2}\right)^{2^{n-1}}\right)^{2}+2\left(\left(1-\sqrt{2}\right)^{2^{n-1}}\right)^{2}}{\left(\left(1+\sqrt{2}\right)^{2^{n-1}}\right)^{2}-\left(\left(1-\sqrt{2}\right)^{2^{n-1}}\right)^{2}}\sqrt{2}\\ & =\frac{\left(1+\sqrt{2}\right)^{2^{n}}+\left(1-\sqrt{2}\right)^{2^{n}}}{\left(1+\sqrt{2}\right)^{2^{n}}-\left(1-\sqrt{2}\right)^{2^{n}}}\sqrt{2}\\ & =r_{2^{n}} \end{array}\]

What happens in the case of the regular hexagon and the pentagon?

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