Identity, Maths Club of IISER Kolkata

Gourav Banerjee, a 21MS student, goes to the main canteen of IISER Kolkata for dinner at some arbitrarily scheduled time between 8 and 9 pm. He frequently meets an anonymous, beautiful girl in the mess and begins to wonder whether the girl is stalking him or if their meeting is just a coincidence. So he tries to compute the probability of meeting that girl in the mess during dinner time given the following constraints: Both Gourav and the girl go to mess for having dinner at some random time between 8 - 9 pm. Because of the Queue at the mess, both stay in the mess for minimum of 30 min. What do you think? Solution Let $x$ denote the time when Gourav enters the mess and let y denote the time when girl enters the mess. Here we take origin to be the 8 pm mark and a distance of 1 unit represents 1 hour on both $x$ and $y$ axis so all possible coordinates within the unit square $ABCD$ represents an event where Gourav and the girl both visit the canteen. Now the favourable coordinates which ...

Here's a fun problem from a course on Curves and Surfaces. Consider the following curve, called the twisted cubic. \[ \gamma\colon \mathbb{R} \to \mathbb{R}^3,\qquad \gamma(t) = (t, t^2, t^3). \] Show that no four distinct points on this curve can lie on the same plane. Method 1 - The scalar triple product The scalar triple product  is a good tool to have at one's disposal (thanks to Satbhav Voleti for sharing this solution); given three vectors $\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}$, their scalar triple product $\boldsymbol{u}\cdot(\boldsymbol{v}\times \boldsymbol{w})$ represents the volume of the parallelepiped whose legs are these vectors. Now if these three vectors all lie in the same plane, the corresponding parallelepiped is completely flat with zero volume, thus the scalar triple product is also zero. Coming back to our problem, suppose that the four distinct points on the twisted cubic $P_i = (t_i, t_i^2, t_i^3)$, $i = 0, 1, 2, 3$, all lie on the same plane. ...

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 post mainly concerns the age old question of classifying all right-angled triangles with integer sides. In other words, we wish to find all integer solutions for the equation \[ X^2 + Y^2 = Z^2. \tag{1} \] An observant reader may have noticed that if $(X, Y, Z)$ is a solution, then so is $(kX, kY, kZ)$ for any integer $k$, and vice versa. This means that we can discard all common factors of $X, Y, Z$, and focus on solving $(1)$, with the added condition $\gcd(X, Y, Z) = 1$. Such solutions are called primitive solutions . You may remember a parameterized expression for these solutions, but do you know how to derive it? This is what we will be discussing here. There are multiple ways of arriving at the parametrization, but the one we'll use traverses the poetic bridge between numbers and geometry (the two divine deities of mathematics) with such ease, that it definitely makes it the best among them all. Rational hunt There is another simplification we c...

Here's a curious polynomial identity I stumbled upon: for $r, n \in \mathbb{N}$ and $r > n$, \[ f(n, r) = r^n - \binom{r}{1}(r - 1)^n + \binom{r}{2}(r - 2)^n + \dots + (-1)^{r - 1}r = 0. \] This can be proved by brute force expansion, but there is a really nice combinatorial argument in connection with the following problem. How many surjective functions of the form $f\colon \{1, 2, \dots, n\} \to \{1, 2, \dots, r\}$ are there? Consider such a function $f$; for each $i \in \{1, 2, \dots, n\}$, we have $r$ choices for $f(i)$, giving us at most $r^n$ functions. However, $f$ must be surjective, so we must get rid of those functions which fail to hit some $j \in \{1, 2, \dots, r\}$ in the codomain. There are $(r - 1)^n$ functions whose codomain misses a particular $j$, and $\binom{r}{1}$ ways to choose this missing element, hence we take away their product. This isn't the end, since now we've removed those functions which miss at least two points ,...

A couple of days ago, Rohan Didmishe shared this problem with us: show that the function defined by \[ f\colon \mathbb{R} \to \mathbb{R}, \qquad f(x) = \begin{cases} x + x^2\sin(1 / x), &\text{ if }x \neq 0, \\ 0, &\text{ if } x = 0. \end{cases} \] is not monotonic (increasing or decreasing) in any interval $(-\delta, \delta)$ around zero. Graphing this function (say, using Desmos ) shows that it oscllates rapidly, curving up and down with increasing frequency the closer its gets to zero. This is due to the $x^2\sin(1 / x)$ term; the $x$ added in front 'tilts' the curve upwards. The first thing to look at is the derivative of $f$. Using $\lim_{x \to 0} x\sin(1 / x) = 0$ and the chain rule, we can compute \[ f'(x) = \begin{cases} 1 + 2x\sin(1 / x) - \cos(1 / x), &\text{ if }x \neq 0, \\ 1, &\text{ if } x = 0. \end{cases} \] Specifcally, $f'(0) = 1$ which seems to tell us that $f$ is increasing at $0$ ... or doe...