A belated Happy Basanta Panchami to all! Once this blog took off, our interest in discussing new ideas grew exponentially ... or is it really exponential ? Let's see. A few days ago, Gourav Banerjee shared this recursion relation (along with the solution after we got tired of trying) with us. \[x_{n+1} = 2x_n(1-x_n).\] Although this is non-linear and looks intimidating, the solution turns out to be quite elegant. The trick is to rewrite the relation as follows. \[ 1 - 2x_{n+1} = 1 - 4x_n + 4x_{n}^{2} = (1-2x_n)^2. \] This already looks much nicer! Taking the logarithm of both sides, \[\log (1-2x_{n+1}) = 2 \log (1-2x_n). \] Set \(b_n = \log(1-2x_n)\) ... whoa! This is a simple geometric progression , with \(b_{n+1} = 2b_n\). Thus, \(b_n = 2^nb_0\) for some initial \(b_0\), which yields \[1-2x_n = (1-2x_0)^{2^n}, \qquad x_n = \frac{1}{2} \left[1-(1-2x_0)^{2^n}\right]. \] Looks easy, right? This is an example of the more general logistic map , \[x_{n+1} = rx_n(1-x_...
This blog is for sharing short and interesting problems and random thoughts, with some mathematically informal discussion.