Paul Michael Kielstra

When I have computer problems, as I often do, my usual response is somewhere on the spectrum from frustration to anger. This usually works pretty well as a motivator. A few months ago, I called a friend and told him that I was chasing down a subtle bug and it was difficult to keep up my motivation. “Michael,” he immediately told me, “You have to name it.” So I did. I picked a name more or less at random, and immediately I was emotionally ready to go again. Fred was not going to keep me down. I was going to get him if it were the last thing I did. I had a correct code by the end of the next day.

NSF ACCESS is a fantastic high-performance computing program. If you’re affiliated with a university or other institution, you can upload a CV, write a paragraph describing your project, and get approved for several thousand CPU-hours within just a few days. If you don’t have existing grants it’s totally free. It even works if you’re a grad student, although then you’ll need your advisor to provide a form letter saying that you’re legit. But they even have suggested language for that.

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.