aperçu

a brief survey or sketch

an immediate impression
Take 4 at a stable blog
I don’t understand how a book like The Dutch House can be so captivating. There are no overarching villain, nor fantastical world building or gimmicks. The sole driving force lies in the ability of Ann Patchett to deliver a soulful story stemming from the Cinderellaesque expulsion of two siblings, Maeve and Danny, from their family “home,” the Dutch House.
The two trudged into the future, with Danny obtaining a medical degree but ultimately eschewing it by becoming a real estate investor while Maeve became a successful CFO figure in a frozen vegetable company, while never letting go of the past, returning to the edges of the Dutch House time after time; partly out of habit, and partly to reminisce. It was in those quiet times where the author really shined and captured my attention, and drove me to keep on reading. Turns out, I have been and will always be a sucker for good prose.
I know I am facing a lot of personal issues this quarantine, and the following quote resonated:
There are a few times in life when you leap up and the past that you’d been standing on falls away behind you, and the future you mean to land on is not yet in place, and for a moment you’re suspended, knowing nothing and no one, not even yourself.
I am at a crossroads in my life with finishing up my degree. I’ve just been through a tough breakup. I will land on solid ground though. I need to focus on not what I have lost, but rather what I can achieve:
I’d never been in the position of getting my head around what I’d been given. I only understood what I’d lost.
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 BaskinRobbins 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:
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:
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.
I finished this series relatively quickly, probably in the span of a month total for three books. Looking back, the best book was probably the first two. There was just an air of mystery surrounding the nature of the invading aliens. Who are they? Why are they coming? What kinds of technology do they have? These questions really drives the first novel into a satisfying conclusion.
In the second and third books, where time skips anywhere from one to a few dozen years, a bleak picture of the universe is painted by Liu (the author). To no surprise, the universe of the novel is populated with lifeforms who mistrust each other and seek to destroy one another. Everything is explained quite thoroughly, but sometimes a bit too much. I wish he left some deduction for the readers to make ourselves rather than spoon feeding all the details.
There are a few more criticisms I have of the second and third books:
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 saddlepoint 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 highceiling platformers like Supermeatboy or the ageold 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.
This is a post regarding the paper Efficient Preconditioning for the $p$Version Finite Element Method in Two Dimensions. In Lemma 3.3, the basis required is the bubble (or interior) polynomials, the set of edge functions orthogonal to the interiors, and linear (of bilinear) functions.
The edge functions orthogonality is explicitly stated as $\hat a(u, v) = 0$ for all $u \in \Gamma_i$ (edge space) and $v \in \mathcal{I}$ interior space. A very natural question is why the vertex functions does not need to be orthogonal to the interior functions. The fun fact is that it secretly is.
Note that the paper is for the $H^1$ seminorm, hence $\hat a(u, v) = \int_T \nabla u \nabla v \, dx$. Now let $u$ be a hat (or bilinear) function, and let $v \in \mathcal{I}$. Then we have that
\begin{align*}
\int_T \nabla u \nabla v = \int_{\partial T} u \partial_n v – \int_T \Delta u v = 0.
\end{align*}
The first term is 0 due to the bubble functions vanishing on the boundary, and that $\Delta u = 0$ because it is linear.