Dec 21, 2017

Dancing in the dark

It is not particularly gloomy, as I see a light. And I know I have to keep dancing, which is not terribly a bad idea, as the show must go on anyway. But, nevertheless, it is not easy to negotiate between serving and asking or, perhaps, between loving and wanting.

I understand it is not a dichotomy but rather a spectrum, on which there seems to be a sweet spot to figure out. Still, I cannot help wondering maybe I am dancing on the wrong stage; the matter is more complex, requiring different perspectives. Well, there are lots to learn in living with someone, and I am happy to dance in the dark.

Aug 10, 2017

At thirty eight

I usually pay very little attention to my age and sometimes have trouble exactly telling it off the top of my head. Recently, however, I had a series of conversations that burned into my mind the number 38. Nevertheless, the immediate reaction was “so what?” I did not think of what to do with it or how to interpret it.

Although I still hesitate to make any general statement of getting older, I can at least compare my own thirty eight and twenties. The biggest difference is that I am better at identifying and working for things important for me. In my twenties, I really could not tell whether I wanted to cherish someone or be cherished by everyone, or whether I wanted to recognize something great or be recognized as great. Even after bumping into what mattered to me, I did not know how to take risks and dedicate myself to them.

Now, I am at thirty eight and lucky enough to have found a few things I know important and love to serve.

Aug 4, 2017

Unqualified to be helpful

In the early afternoon, one of the colleagues in the department asked me for some help on machine learning. I gave some advice and promised to get back to him with more information. Since I had some clues, I was confident of further help and naturally felt good.

A few hours later, a woman who I recently had got acquainted with called me and said she had to cancel the lunch next day. There was a quiver in her voice. While she tried to tell me the reason, I did not quite get it but learned only the fact that something terrible had happened to her. She hung up and I was left texting “Please let me know anything I can help you with. Please.” I knew that I was not close enough to be asked for any help in her emergency. The text message looked pointless and made me feel useless.

I must be prepared, think, and take actions in order to be helpful for those who I try to help.

May 26, 2017

A conference poster of BTABM

Here is a conference poster of the ABM project that I have been working on with Paul Mitchell.

(linked to the original file)

Bacillus thuringiensis (Bt) corn is a transgenic seed variety, used for controlling pest damage, whose adoption involves many stakeholders. We think that it is appropriate to model the situation as a complex system. Like many others, I like Python and so use Mesa as a platform. We try to get it done during the summer!

May 15, 2017

The irrationality of √2

When we start to learn analysis or in particular real numbers, a proof for the irrationality of $\sqrt{2}$ typically goes like this.

Assume, for contradiction, there exists $p, q \in \mathbb{Z}$ satisfying $$ \left( \frac{p}{q} \right)^2 = 2 $$ where $p$ and $q$ are relatively prime and $q \neq 0$. Then, rearranging, we have $p^2 = 2q^2$ and so $p$ is even or $p = 2r$ where $r\in \mathbb{Z}$. But, substituting it above, we also have $q^2 = 2r^2$ and $q$ is even as well. This is a contradiction to being relatively prime.

In the lecture on Field automorphism (at 6:35), Professor Macauley gives another proof based on the following proposition. If $\phi$ is an automorphism of $\mathbb{Q}(\sqrt{2})$ ($\mathbb{Q}$ adjoining $\sqrt{2}$), then, $$ \phi(q) = q, \quad \forall q\in \mathbb{Q} .$$ One of such automorphisms is $\phi(a+b\sqrt{2}) = a-b\sqrt{2}$. But, now, $\phi(\sqrt{2}) = -\sqrt{2} \neq \sqrt{2}$. Thus, $\sqrt{2}$ is not rational. Constructing the extension field $\mathbb{Q}(\sqrt{2})$, it is tautological but nonetheless cute.

May 4, 2017

RR[x]/(x²+1) isomorphic to CC

It is a very basic problem in ring theory, but I keep forgetting the key ideas involved. The trick is to identify the following evaluation homomorphism: $$ \varphi_i: \mathbb{R}[x] \to \mathbb{C},\quad \varphi_i(p(x)) = p(i) .$$ To me, this tells a lot. First, it maps all the real coefficients of $\mathbb{R}[x]$ into $\mathbb{C}$, reducing $\mathbb{R}^n$ to $\mathbb{R}^2$. Second, it maps $x$ into $i$ by essentially evaluating each polynomial $p(x)\in\mathbb{R}[x]$ at $i\in\mathbb{C}$. Although $i \notin \mathbb{R}$, it seems okay because the codomain $\mathbb{C}$ is an extension field of $\mathbb{R}$. Finally, it is surjective because for any $a+bi \in \mathbb{C}$ there exists $a+bx \in \mathbb{R}[x]$ such that $\varphi(a+bx) = a+bi$. The kernel captures the very motive for going to this trouble, i.e. no roots of $x^2+1$ in $\mathbb{R}$. $$ \ker(\varphi) = \{p(x)\in\mathbb{R}[x] \mid p(i)=0\} = \{(x^2+1)q(x) \mid q(x)\in\mathbb{R}[x]\} = \langle x^2+1 \rangle .$$ Thus, by the first isomorphism theorem, $\mathbb{R}[x]/\langle x^2+1 \rangle \cong \mathbb{C}$.

Apr 27, 2017

Being a unit in ZZn

$$ \text{$\overline{a}$ is invertible in $\mathbb{Z}_n$ iff $a$ and $n$ are relatively prime.}$$ The following proof from Pinter (p.228) is nice. Use a notation: $\overline{x} = x \pmod n$. Then, \begin{align*} & \text{$a$ and $n$ are relatively prime} \\ \Leftrightarrow\quad& \text{gcd}(a,n) = 1 \\ \Leftrightarrow\quad& sa + tn = 1 \quad (\exists s,t \in \mathbb{Z})\\ \Leftrightarrow\quad& 1 - sa = tn \\ \Leftrightarrow\quad& \overline{sa} = \overline{1} \\ \Leftrightarrow\quad& \overline{s}\:\overline{a} = \overline{1} \\ \end{align*}

Mar 9, 2017

Normality of the Commutator Subgroup

The problem is:
Given a group $G$, show that a subgroup $N = \{xyx^{-1}y^{-1} \mid x,y \in G \}$ is normal.
Since it was difficult for me, I searched online and found a proof on Stack Exchange. As many others, I think it is slick.

Let $g \in G$ and $n,n' \in N$. Then, $$ gng^{-1} = gng^{-1}(n^{-1}n) = (gng^{-1}n^{-1})n = n'n \in N .$$

Feb 24, 2017

Fiber and Preimage

In the algebra class, I encountered a concept called fiber, which sounded similar to preimage, and I didn't get the difference. According to Wikipedia, given $f: X \to Y$, the preimage of a set $B \subseteq Y$ under $f$ is a subset of $X$ defined by $f^{-1}(B)=\{\,x\in X\,|\,f(x)\in B\}$. If $B$ is a singleton set, i.e. $B = \{y\}$, the preimage $f^{−1}(y)$ is also called the fiber of $y$.