Mostly a mathematician.
When working at scale, it is often slow or impossible to form full-sized matrices. Instead, we make use of black-box functions to calculate matrix-vector products \(Ab\). There is thus a need for solvers which accept these functions instead of requiring matrix representations of our linear maps – and, in particular, for conjugate gradient solvers for the symmetric positive definite case.
Define \[\langle p, q \rangle = \int_{-1}^1 p(x)q(x)\,dx\] to be an inner product on \( \mathscr{P}_n(\mathbb{R}) \), the space of polynomials of degree at most \(n\) with real coefficients. Note that \(\varphi(p)=p(0)\) is a linear functional on \( \mathscr{P}_n(\mathbb{R}) \). Therefore, by the Riesz representation theorem, there must exist a unique polynomial \(q_n(x)\) such that \[\int_{-1}^1 p(x)q(x)\,dx = p(0)\] for all \(p\) with degree at most \(n\).
It has been said that the hottest thing a man can do is go to therapy. Well, I’m always down to do hot things, but I don’t really have time for therapy in the usual sense. It’s all about your emotions and your relationships and your inner life, and that takes forever to talk about. I’m an applied mathematician. I am efficient.
One of the fun things about being specifically an applied mathematician is knowing when my colleagues are dealing with a combinatorial explosion. Recently, a friend of mine asked me if there were an easy way to determine how many of the partitions of a set of sixteen fixed elements obeyed a certain set of conditions. Doing this with brute force was a non-starter: there were over \(10^{10}\) of them, and if we were able to check sixteen thousand partitions per second the calculation would still have taken more than a week.
This blog post is a lemma for this one. If you haven’t read that one yet, I would suggest you start there.
It is known that, for \(\vert r \vert<1\), \(\sum_{n=0}^\infty ar^n=\frac{a}{1-r}\). The usual proof goes something as follows: let \(S=\sum_{n=0}^\infty ar^n\). Then \(rS=\sum_{n=0}^\infty ar^{n+1}=\sum_{n=1}^\infty ar^n\). Subtracting, \(S-rS=\sum_{n=0}^\infty ar^n-\sum_{n=1}^\infty ar^n=a\), so \(S=\frac{a}{1-r}\) as required. It’s very elegant, but it seems a little bit magical.
Back in high school, I had a math teacher called Shirley J. Perrett. (She would introduce herself as “Mrs. \(Pe^2r^2t^2\), but not in that order.”) She had a textbook called A2 Core Maths for Edexcel, by Emanuel, Wood, and Crawshaw. And that textbook had a problem in chapter 15 that none of the students Mrs. Perrett had ever taught had solved without hints.
Vincent Baker’s Rock of Tahamaat is a fascinating little RPG. It also uses the word “concubine” way too much to be in what I consider good taste, and it’s missing a certain frisson of punk that would make my regular group enjoy it a little bit more. What follows is a reskin, not a hack: no game mechanics are changed, but the setting is different.
Let \(f\) be a function that you want to interpolate on the interval \([0, 1]\). Let \(k\) be some number of points. Say you know the values of \(f_i=f(i/k)\) for \(i=0, \dots, k-1\). If you knew that \(f\) were periodic with period \(1\), or in general if you knew that \(f\) decomposed into a sum of sinusoids with low frequencies, how might you leverage that knowledge to estimate \(f_{i+1/2}=f((i+1/2)/k)\)? (I’m assuming that the interpolation you have in mind exactly doubles the density of the grid. If you want to interpolate even more points, just repeat your procedure.)
When I was young, I consumed a piece of media – I’m being vague because I genuinely don’t remember any more than this – starring a dogmatic food critic who had published a book called How to Eat Food: A Comprehensive Guide to the Food You Should Like and the Food You Shouldn’t. Ever since, I’ve thought that this would be a fantastic title for something sufficiently tongue-in-cheek to pull it off. I don’t know anywhere near enough about food to write that, but I do drink quite a lot of bubble tea, and I’ve been told that my thoughts about it are more systematic than those of the average bear, so here goes. This is my comprehensive guide to the boba you should like and the boba you shouldn’t.