Category: Blarghing

On the very long drive down to Florida, I listened to the Crazy Rich Asians audiobook. The whole time, I just couldn’t understand how the movie made such a big splash when the book just felt was so unrelatable on so many levels, with a prose that is simply… adequate. I honestly thought “shit, how did this book get turned into a film. I don’t want this book to reflect Asian-Americans.”

The main plot between Rachel and Nick was, in most ways, a standard feel-good romcom variations, but subplots involving Astrid and Eddie were jarring. Were we suppose to sympathize with a multimillionaire heiress  who suspects her husband was cheating on him? What about a conceited “family” man who’s main purpose in the plot is to further magnify the fact that money doesn’t buy happiness? Like the people in the story, it just seems to me that there’s too much fluff between the covers, and not enough substance

… which is probably why the movie was such a big hit. The math is simple: glamour + good actors = pretty decent film.

I’ve learned a new word to describe an incredibly specific event:

Mondegreen: mishearing or misinterpretation of a phrase as a result of near-homophony, in a way that gives it a new meaning

Surprisingly, the Wikipedia page never mentions the Taylor Swift song where people all think it’s “Starbucks lovers” instead of “list of ex lovers.”

And here I was for a few weeks thinking it’s a love song about selling out and corporate greed.

Otherwise known as a SMURFETTE, and unwitting namesake for a principle describing how in many popular culture “a group of male buddies will be accented by a lone female, stereotypically defined.”

Honestly, what a great name for an absolutely terrible trend in media. Maybe Joe Biden can learn something about this.

The recent Sunday crossword puzzle had a clue relating to the source of bay leaves, whose answer was LAUREL. I was curious afterwards and stumbled upon two horrifying uses:

1. They’re used in entomology as an “active ingredient in killing jars.” Apparently, the tender leaves releases vapors which slowly kills small insects. It turns out, the essential oil concentration is the active ingredients and not some intrinsic mechanic property, which really begs the question of why people fall for those MLMs shelling oils.

   Speaking of…

2. Maybe burning a bay leaf does have noticeable effects on the human body as this website suggests, but it seems there’s a non-zero population in the world who like to think that burning this dried plant leaf can also bring fame and fortune. Maybe it’s because it’s the color green? Maybe because bay leaf sounds somewhat like bay life, and San Francisco is part of Silicon Valley? Maybe it’s just humans wanting to believe in the supernatural? Or just an attractive person able to convert superstition into a successful YouTube career?

On the issue of using them in cooking, I firmly believe they are overhyped and doesn’t really contribute much.

The year started off like any other year, with the exception that there were significantly more ophthalmology based puns, until a little virus blossomed into a pandemic. The resulting quarantine is messy: toilet paper became a commodity worth it’s volume in gold, Zoom overtook Skype as the de facto way to FaceTime people, and the Baskin-Robbins logo is associated with Joe Exotic.

Another tragedy is my haircut. I’ve never really liked how my coiffure looked after an appointment, and I always say “it’ll look better after a few days.” This was a lie. The truth is, my hair didn’t get better. It was moreso I settled. It was (and still is) basically an unhappy relationship.

The barber always asked me how I wanted it cut, and I always replied “It was four weeks ago since my last haircut” then they trim off four weeks worth of hair and happily take my twenty dollars plus tip. The problem is, I usually never liked what my hair looked like four weeks ago, nor that haircut from eight weeks ago ad infinitum. Just a reminder, my hair looked like:

Please ignore the Transition glasses.

Within the last few months, I actually became more comfortable with my hair. And now, with this stupid quarantine, I’m going to stroll into the barbershop looking like the Geico caveman

and telling them “It’s been … two months since my last haircut. Please save me” and they happily take my $20 with tip.

This nice problem was in the analysis section of Putnam and Beyond: prove

\begin{align*}
\lim_{n\to \infty} n^2 \int_0^{1/n} x^{x+1} \, dx = 1/2.
\end{align*}

The solution is quite nice, and simply relies on the fact that $\lim_{n\to 0^+} x^x = 1$, hence for $n$ large enough, we can approximate the integral with $\int_0^{1/n} x\, dx$ instead.
There’s an easy generalization of this problem:
\begin{align*}
\lim_{n\to \infty} n^{k+1} \int_0^{1/n} x^{x+k} \, dx = 1/(k + 1).
\end{align*}

Generalizing this fact, we don’t even need the composite exponential as the proof just need a $f(x)$ to be a function such that $\lim_{x\to 0^+} f(x) = 1$ with an integral bound approaching $0$.

My adviser told me about this meshing of a cube (or any hexahedral) into 6 different tetrahedrons which is easy to draw. For the sake of exposition, we will consider the cube $(-1,1)^3$ The procedure is as follows:

1. Draw a diagonal from $(-1,-1, -1)$ to $(1,1,1)$.
2. Now, project the diagonal to each of the 6 faces of the cube, which will result in a mesh of the cube.

While the procedure is simple enough, the individual tetrahedrons were a bit difficult to visualize. To help with that, I’ve made a small Mathematica script that one can play with:

pts = {{-1, -1, -1}, {1, -1, -1}, {1, 1, -1}, {-1, 1, -1}, {-1, -1, 
    1}, {1, -1, 1}, {1, 1, 1}, {-1, 1, 1}};
ti = {{1, 2, 6, 7}, {1, 2, 3, 7}, {1, 6, 7, 5}, {4, 1, 8, 7}, {1, 5, 
   8, 7}, {1, 4, 3, 7}}; Graphics3D[{Opacity@0.4, 
  Table[Tetrahedron[pts[[i]]], {i, ti}]}]

From that, we can easily see that mesh now.

So what does the “conforming” part of the title mean? Of course, there is an easier way to tile the cube using only 5 tetrahedrons, but if you put together multiple cubes, one have to be careful of how you orient them. Using the above meshing, as long as the cubes are not too distorted and can easily create the tetrahedral mesh by drawing the diagonal in the same direction.

For example, below we have a eight hexahedral elements laid in a cube, but there are three slab, three columns, and two cubes (with one significantly smaller). This whole thing was needed so that I can construct something as anisotropic as the mesh below without resorting to fancy software.

One can solve Poisson’s problem $-\Delta u = f$ in $d$ dimensions with homogeneous Dirichlet boundary conditions using a mixed formulation as explained below:

Let $\sigma = \nabla u$, then for a sufficiently smooth function $\tau$, by Green’s theorem
\begin{align*}
(\sigma, \tau) &= (\nabla u, \tau) \\
&= -(u, \textrm{div } \tau).
\end{align*}
Again, choosing $v$ a function sufficiently smooth, we have
\begin{align*}
f = -\textrm{div } \sigma \implies (f, v) = (-\textrm{div } \sigma, v).
\end{align*}
This gives the saddle-point problem: find $(u, \sigma) \in V \times M$ such that
\begin{align*}
(\sigma, \tau) + (u, \textrm{div } \tau) &= 0\\
(\textrm{div } \sigma, v) &= -(f, v)
\end{align*}
hold for all $(\tau, v) \in V \times M$. Note that we don’t have to take a derivative of $u$, hence it’s natural to try $M = L^2$, but what about the space $V$?

One very easy choice to guess is $V = [H^1(\Omega)]^d$ as we want the divergence to be all defined, but unfortunately this doesn’t work as the gradient of the solution to Poisson’s problem can easily not be in $[H^1(\Omega)]^d$.

In order to illustrate this, consider $u =\left(r^{2/3}-r^{5/3}\right) \sin \left(\frac{2 \theta }{3}\right)$ on the domain of the unit circle with bottom left quarter taken out. It’s not hard to see that $u = 0$ on the boundary of the domain, and we can easily find the $f$ such that it satisfies Poisson’s equation. Now, we can either calculate the gradient exactly or argue as follows.

First, recall how to take a gradient in polar coordinates. Note that $\partial_r u \approx r^{-1/3}$ plus higher order terms and that $\frac{1}{r}\partial_\theta u \approx r^{-1/3}$ plus higher order terms also. Now, one can easily calculate the $H^1$ seminorm to see that the derivative is unbounded as we’re integrating over $[0,1]$ with $(r^{-4/3})^2r$ terms (the extra $r$ comes from the change of basis from polar integration).

The above is an example of why the space $H(\textrm{div})$ is needed.

Once again, Dunkey has proven himself to be a modern day Donald Draper… in some sense. I bought Celeste almost strictly due to how fun it seemed. It truly is a great game with tight controls and extremely interesting level design.

The first point is starting to get quite standard now, but I want to reiterate on the latter point. It seems many high-ceiling platformers like Supermeatboy or the age-old N game constantly rely on precision. Celeste throws that away with an emphasis on when/where/how you use the air dash. That air dash, that one extra mechanic, really is the crux. Honestly, it reminded me of Ori’s dash, but the level design here is more like a puzzle.

Of course, the game can get a bit annoying. There are places where precise timing is the only way through the level (or so it seems to me?). Flag number 9 is particularly annoying. One can turn on the assist mode, but it really breaks the game by making it very easy. Another quirk is that it doesn’t save automatically when quitting from the Switch; this caused me to have to beat certain levels twice as I was switching between games.

All in all, quite a fun game with a ton of content. Worth it for just $20.