Identity, Maths Club of IISER Kolkata

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.