Identity, Maths Club of IISER Kolkata

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 ,  \[x_{n+1} = rx_n(1-x_n).\] It's

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

One of the most famous elementary problems in probability is Bertrand's Paradox . This stems from the seemingly simple question: what is the probability that the length of a random chord in a circle exceeds the length of the side of its inscribed equilateral triangle? To put the lengths into perspective, the side of an equilateral triangle inscribed in a circle of radius $r$ has length $\sqrt{3}r$. The YouTube channel Numberphile, in collaboration with Grant Sanderson from 3blue1brown, released an excellent video discussing this very problem. Don't forget to watch the extra footage ! I won't dwell on the mathematical details here; the video does a great job of explaining the problem with gorgeous visuals. Instead, an enthusiastic programmer's first instinct screams "let's code it"; and python is more than up to the task. Putting the usual caveats of generating (pseudo)randomness using computers aside, the numpy.random  module is a good place

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

Welcome to the official blog of Identity, the Maths Club of IISER Kolkata! Here, you'll find short posts by our members, small problems with neat solutions, and discussion around mathematics. Head on over to our main website for our articles, talks and recorded lectures, events, and more. PS. This site supports $\LaTeX$ (strictly speaking, MathJax ) expressions.