## Wednesday, February 25, 2015

### The 6 Stages of Math Writing

Here's some news! I've decided to devote all of March to the basics of distributional calculus. In undergrad, I had a professor that taught distributional calculus from a purely theoretical standpoint and refused to match this with intuition, so this will be an adventure in explaining math for me as well.

It goes so well with the blog title---we're the Analyisisters! Let's throw some analysis at everyone!

In the meantime, sit tight while I pretend to be Seinfeld.

How would you write out the solution to this problem at each stage of university life?

Let $A$, $B$ be matrices in $\mathbb{R}^n$. If $AB = I$, then $A^kB^k=I$ for all $k \in \mathbb{N}$.

LEVEL 1: FROSH

\begin{align}
B^kA^k &= B\ldots BBAA\ldots A\\
&= B\ldots BIA \ldots A\\
&= B \ldots BA \ldots A\\
&= B \ldots BIA \ldots A = I
\end{align}

Yes, I know math homework is supposed to be written in complete sentences, but, why bother? I'm the chosen one who will be able to understand what this means 12 years later.

I mean, come on. I understand it right now. It's really easy.

LEVEL 2: SOPHOMORE

$BA = I$. Then $B^{k+1}A^{k+1} = B^kBAA^k=B^kIA^k=B^kA^k=I$.

Oh, you were serious about that sentence thing? And the sentences have to end with periods? Are you sure? Okay.

Hey, are you going to take points off if I don't put it in a sentence? Why are you doing that? I didn't know that was going to happen.

We know that $BA=I$. Suppose $B^kA^k=I$. Therefore,
\begin{align}
B^{k+1}A^{k+1}&=B^kBAA^{k}\\
&=B^kIA^k.
\end{align}Therefore, using the properties of the identity, $B^{k+1}A^{k+1}=B^kA^k = I$. Therefore, this proves our statement.

They'll never guess my favorite connecting word.

This can be solved using induction. We are given that $BA = I$, providing the base case, so we suppose that $B^kA^k = I$ to show that $B^{k+1}A^{k+1}= I$. We then find that
\begin{align}
B^{k+1}A^{k+1}&= B^kBAA^k\\
&= B^kA^k = I,
\end{align}as desired.

Wow, can you believe how I wrote as a frosh? Who even thinks that's okay? I guess that it shows that I know how important math writing is. That.

We proceed inductively with the given base case $BA=I$. Suppose $B^kA^k = I$ towards demonstrating $B^{k+1}A^{k+1}$ to be the identity as well. Using the definition of integer exponents and both given/inductive hypotheses, we conclude
$$B^{k+1}A^{k+1}=B^kBAA^k=B^kA^k=I;$$that is, the conditions of induction are satisfied and the original statement follows. This fact can be used to show equivalency of left and right inverses (i.e., $AB = I$ iff $BA = I$ for square $A,\ B$ of concordant dimensions).

Varying sentence structures, excessively clear logic, weird punctuation marks, parenthetical statements. Look! Revel in my competence! Feel the 2 hours I spent formatting the answer until it was textbook perfect!

Do you want to see my personalized LaTeX class with a multi-page macro set designed specifically for this field? I made it while my friends were at the bar.

LEVEL 6: PROFESSOR/VERY CONFIDENT GRAD STUDENT

Given $BA=I$ as the inductive hypothesis, observe that
$$B^{k+1}A^{k+1}=B^kBAA^k=B^kA^k=I$$implies the above.

There's no way I'm spending more than 5 minutes on this trivial problem. Why are you even showing it to me? I have several papers to review and two classes to prepare for. This it pointless.

Stay tuned next week, where we do absolutely nothing funny and go down the rabbit hole of formal math! (I need to update my macro set.)