ATRACTOR

If, given an initial polygon with \(n=2m\) sides whose vertices have abscissa \(x_r\) with \(r\in\{0,1,\ldots,n-1\}\), we consider the diagonals defined by opposite vertices and link the points that divide those diagonals in two segment lines of length \(p\) and \(1-p\) times the length of the diagonal, with \(0 < p < 1\), we obtain a new polygon with \(n\) sides whose vertices have abscissa \[x'_r\;=\;(1-p)x_r+px_{m+r} \] where \(x_{n+r}=x_r\) for every \(r\in\{0,1,\ldots,m-1\}\). Let us consider the Fourier representation of abscissa \(x_r\) given by \[x_r\;=\;\sum_{j=0}^m \left(P_j\cos\frac{2jr\pi}{n}+Q_j\sin\frac{2jr\pi}{n}\right) \] By writing vector \((P_j,Q_j)\) in polar coordinates \((C_j\cos\theta_j,C_j\sin\theta_j)\), we get \[x_r\;=\;\sum_{j=0}^m C_j\cos\left(\frac{2jr\pi}{n}-\theta_j\right) \] Then \[\begin{array}{ll}x'_r & =\;(1-p)x_r+px_{m+r}\;=\\ & =\;\sum_{j=0}^m C_j \left((1-p)\cos\left(\frac{2jr\pi}{n}-\theta_j\right)+ p\cos\left(\frac{2jr\pi}{n}-\theta_j+ \frac{2jm\pi}{n}\right)\right)\;=\\ & =\;\sum_{j=0}^m C_j \left((1-p)\cos\left(\frac{2jr\pi}{n}-\theta_j\right)+ p\cos\left(\frac{2jr\pi}{n}-\theta_j+j\pi\right)\right)\;=\\ & =\;\sum_{j=0}^m C_j (1-p+(-1)^j p) \cos\left(\frac{2jr\pi}{n}-\theta_j\right)\;=\\ & =\;\sum_{j=0}^m C_j d_j \cos\left(\frac{2jr\pi}{n}-\theta_j\right) \end{array}\] where \(d_j=1\) if \(j\) is even and \(d_j=1-2p\) otherwise. More generally, we have \[x^{(k)}_r\;=\;(1-p)x_r^{(k-1)}+px_{m+r}^{(k-1)}\;=\; \sum_{j=0}^m C_j d_j^k \cos\left(\frac{2jr\pi}{n}-\theta_j\right) \] As \(k\) tends to infinity, the summands of odd index tend to 0, so that we may forget them and take the following approximation: \[x^{(k)}_r\;\approx\;\sum_{i=0}^{\lfloor m/2\rfloor} C_{2i} \cos\left(\frac{4ir\pi}{n}-\theta_{2i}\right) \] for big values of \(k\). Note that the sum above does not depend on \(k\), that is, the vertices of the polygon sequence converge to certain fixed points. These points are the vertices of a polygon with \(m\) sides whose abscissa are given by \[x^\star_r\;=\;\sum_{j=0}^{\lfloor m/2\rfloor} C_j^\star \cos\left(\frac{4jr\pi}{n}-\theta_j^\star\right) \] where \(C_j^\star=C_{2j}\) and \(\theta_j^\star=\theta_{2j}\). Note that \[\begin{array}{ll}\frac{x_r+x_{r+m}}{2}\\ & =\;\frac{1}{2}\sum_{j=0}^m C_j \left(\cos\left(\frac{2jr\pi}{n}-\theta_j\right)+ \cos\left(\frac{2jr\pi}{n}-\theta_j+ \frac{2jm\pi}{n}\right)\right)\;=\\ & =\;\sum_{j=0}^m \frac{C_j}{2} \left(\cos\left(\frac{2jr\pi}{n}-\theta_j\right)+ \cos\left(\frac{2jr\pi}{n}-\theta_j+j\pi\right)\right)\;=\\ & =\;\sum_{j=0}^m C_j \frac{1+(-1)^j}{2} \cos\left(\frac{2jr\pi}{n}-\theta_j\right)\;=\\ & =\;\sum_{i=0}{\lfloor m/2\rfloor} C_{2i} \cos\left(\frac{4ir\pi}{n}-\theta_{2i}\right)\;=\;x_r^\star \end{array}\] Hence, the vertices of this \(m\)-sided polygon are precisely the midpoints of the diagonals considered in this method.

Note that the rate of convergence to this polygon increases with the approximation of \(1-2p\) to 0, that is, when the parameter \(p\) approaches 1/2 (in particular, for \(p=1/2\), the polygon is obtained already in the first iteration).