ATRACTOR

Given an initial \(n\)-sided polygon with vertices wih abscissa \(x_r\), \(r\in\{0,1,\ldots,n-1\}\), if we link the points that divide the sides in two segments, being the length of the first the double of the second, we get a new \(n\)-sided polygon whose vertices have abscissa \[x'_r-x_r\;=\;\frac{2}{3}(x_{r+1}-x_r)\] that is, \[x'_r\;=\;\frac{x_r+2x_{r+1}}{3}\] where \(x_n=x_0\). Consider the Fourier representation of abscissa \(x_r\), given by \[x_r\;=\;\sum_{j=0}^{\lfloor n/2\rfloor} \left(P_j\cos\frac{2jr\pi}{n}+Q_j\sin\frac{2jr\pi}{n}\right) \] where \(\lfloor n/2\rfloor=m\) denotes the floor of \(n/2\). Writing vector \((P_j,Q_j)\) in polar coordinates \((C_j\cos\theta_j,C_j\sin\theta_j)\), we have \[x_r\;=\;\sum_{j=0}^m C_j\cos\left(\frac{2jr\pi}{n}-\theta_j\right) \] Then \[\begin{array}{ll}x'_r & =\;\frac{x_r+2x_{r+1}}{3}\;=\\ & =\;\frac{1}{3}\sum_{j=0}^m C_j\left(\cos\left(\frac{2jr\pi}{n}-\theta_j\right)+ 2\cos\left(\frac{2jr\pi}{n}-\theta_j+ \frac{2j\pi}{n}\right)\right)\;=\\ & =\;\sum_{j=0}^m C_j d_j \cos\left(\frac{2jr\pi}{n}-\theta_j+\psi_j\right) \end{array}\] where \[d_j\;=\;\frac{1}{3}\sqrt{(1+2\cos\frac{2j\pi}{n})^2+ (2\sin\frac{2j\pi}{n})^2}\;=\; \frac{1}{3}\sqrt{5+4\cos\frac{2j\pi}{n}} \] and \[\psi_j\;=\;\arctan\frac{2\sin\frac{2j\pi}{n}}{1+2\cos\frac{2j\pi}{n}} \] More generally, \[x_r^{(k)}\;=\;\frac{x_r^{(k-1)}+2x_{r+1}^{(k-1)}}{3}\;=\; X+\sum_{j=1}^m C_j d_j^k \cos\left(\frac{2jr\pi}{n}-\theta_j+k\psi_j\right) \] When \(k\) tends to infinite, all summands in this sum tend to 0 faster than the first one so we may forget them and consider the following approximation: \[x_r^{(k)}\;\approx\; X+C_1 d_1^k \cos\left(\frac{2r\pi}{n}-\theta_1+k\psi_1\right) \] for big values of \(k\). Making \(C=C_{1}d_1^k\) and \(\theta=\theta_1-k\psi_1\) we have \[x_r^{(k)}\;\approx\;C \cos\left(\frac{2r\pi}{n}-\theta\right) \] that is, \[x_r^{(k)}\;\approx\;X+P\cos\frac{2r\pi}{n}+Q\sin\frac{2r\pi}{n} \] where \(P=C\cos\theta\) and \(Q=C\sin\theta\). Similarly, \[y_r^{(k)}\;\approx\;Y+R\cos\frac{2r\pi}{n}+S\sin\frac{2r\pi}{n} \] and, assuming the points in euclidean space, \[z_r^{(k)}\;\approx\;Z+T\cos\frac{2r\pi}{n}+U\sin\frac{2r\pi}{n} \] Hence, in a way similar to the bisection case, the points \(P_{r}^{(k)}\) approach the vertices of a polygon obtained by applying a linear function to a regular \(n\)-sided polygon centered at the origin followed by a translation. This polygon is also inscribed in a ellipse centered at \((X,Y,Z)\) and the relations of parallelism of the sides of the regular polygon are kept in the corresponding sides of the new polygon.

Note also that \[x_r^{(k+3)}\;\approx\;X+C_1 d_1^{k+3} \cos\left(\frac{2r\pi}{n}-\theta_1+(k+3)\psi_1\right) \] \[x_r^{(k+3)}-X\;\approx\;C_1 d_1^k d_1^3 \cos\left(\frac{2r\pi}{n}-\theta_1+k\psi_1+3\psi_1\right) \] \[x_r^{(k+3)}-X\;\approx\;d_1^3 C_1 d_1^k \cos\left(\frac{2(r+2)\pi}{n}-\theta_1+k\psi_1+3\psi_1- \frac{4\pi}{n}\right) \] But, since \[x_{r+2}^{(k)}-X\;\approx\;C_1 d_1^k \cos\left(\frac{2(r+2)\pi}{n}-\theta_1+k\psi_1\right) \] we have \[x_r^{(k+3)}-X\;\approx\;d_1^3 (x_{r+2}^{(k)}-X) \] provided that the approximation \(3\psi_1\approx\frac{4\pi}{n}\) is valid, that is, \(\psi_1\approx\frac{4\pi}{3n}\). In fact, we have \[\psi_1\;=\;\arctan\frac{2\sin\frac{2\pi}{n}}{1+2\cos\frac{2\pi}{n}} \;=\;\frac{4\pi}{3n}+\frac{1}{8}\left(\frac{2\pi}{n}\right)^3+ \frac{1}{972}\left(\frac{2\pi}{n}\right)^5+\ldots \] so that, for large values of \(n\), we have \(\psi_1\approx\frac{4\pi}{3n}\). Hence, the points obtained by applying the given process three times are, approximately, the same points obtained by an homothety of center in the centroid of that polygon and ratio \(d_{1}^{3}\). The greater the values of \(k\) and \(n\) are, the better the approximation is). However, we should bear in mind that, because of the approximation, there is always a difference between the shape of a polygon and that of the polygon that is obtained by applying the given process three times, and that this difference, unlike what happened with bisection, does not tend to 0 when \(k\) tends to infinity (that is, after a large number of iterations).

What will happen if, more generally, we divide each side of a polygon into two segments with lengths \(p\) and \(1-p\) times greater than the length of that side, for (0 < p < 1\) ?