I applied to eight colleges: Caltech, Case Western, Columbia, Harvard, MIT, Northeastern, Princeton, and Stanford. Four offered me an interview. One was different in that the interview wasn’t really optional but an essential part of the application.

Although the interviews all had a general script of questions in common (such as tell me a little bit about yourself), the dynamic of each interview was subtly different.

MIT was my first experience back in December in a Starbucks at the Whitman mall. My interviewer was a mathematics professor well into his years, although his personality didn’t show it. He still embodied what I considered “the MIT spirit.” I had a little trouble getting going as I didn’t know where to start. Usually my general outline starts with talking about math and my conflict between math or engineering; I then move on to something like music and tennis or some of the more personal projects that I’ve been working on. Somewhere in there I talk about growing up in Astoria, Queens and my childhood and my turbulent family dynamics. It seems fairly strewn but my monologues are fairly consistent from interview to interview.

MIT’s interview was also the only time I brought my laptop to show my interviewer some of the projects I hyperlinked above. He seemed genuinely impressed but not blown away. If anything, I’m sure I proved to him that I had a passion for something and I’d gotten far with it. This was also the only time I had an extremely academic discussion with my interviewer about his research, which was symplectic geometry. Although I had trouble understanding much of what he said, I was fairly intrigued and very interested. It was fun to see him try to explain it to me, anyway.

Next was much later, in the middle of January. Princeton was downtown in the village at a Panera. The man speaking with me was middle-aged, old but not nearly as old as my previous interviewer. His presence gave off a sense that he had a firm grasp on how the world works, and that he was well-read. I also feel like this was my weakest interview, which was strange because he said that I made his day at the end.

The other three happened in the span of a single week. Harvard was at a different Panera in town and Columbia was at another Starbucks twenty minutes away. Not sure why, but Paneras and Starbucks seem like staple locales for college interviews.

I set up Stanford’s interview to be this Sunday, the day after Columbia’s at a cute coffee shop in the city. The reason I arranged for Sunday instead of Saturday was because I go to Columbia’s SHP in the morning, which runs past noon which was the time my interviewer wanted. (Psst: if you’re not a senior yet but live ~1 hour from Columbia and like science or math, apply for SHP. It’s free and awesome.) Much to my confusion, I unlocked my phone after class on Saturday to see three text messages and two missed calls. I had apparently stood up my Stanford interviewer for a half hour and counting.

I didn’t know if I had genuinely made a mistake or had a brain fart or it was something on his end, but of course I apologized profusely. I felt really bad for not responding to my interviewer for a half hour while he was slowly accruing salt at a table for two in a far-off LPQ. Eventually, the conflict was resolved. I had indeed scheduled the interview for Sunday and it was “his bad.” Whew. A sigh of relief. In immediate retrospect I thought that I could have called the interview off as I had some right to do so. After all, it wasn’t my fault. And he had backtracked his words and immediately shifted his tone of voice from annoyance to apologetic. Before, he said, I believe I cannot give Stanford a constructive interview of you any longer. It seems that I have put more effort into making this happen than you have. Now, he says, I’ll see you Sunday at noon, man, okay? 

In a not-so-immediate retrospect, calling off the interview would have been a very bad idea. I had a great interview- probably the best out of all that I’d had so far. The man I thought was slightly tense and stuck-up was in fact extremely laid-back and easygoing. He sipped his cup of tea+milk as he heard me talk about engineering, what my dad did for a living, and how I’d spent two months at Stanford’s summer school this year.

So what’s the point of writing all this down? Not much. But here are a few suggestions/some advice: the interview will very, very rarely make or break you. It’s not the end of the world if you have a bad interview, nor should you be complacent because you had an extremely strong interview. It doesn’t hold too much weight with regards to the rest of your application. In fact, the interview will probably reflect your application fairly closely; if you’re a strong candidate, a strong interview will confirm that. If you’re leaning towards the reject pile, a subpar interview also serves as confirmation. Interviews matter most when you find yourself in the middle, i.e. admissions purgatory. And even then interviews don’t really matter. So just take a deep breath and be yourself, and don’t be afraid to let your passions show, whether that be mathematics or field hockey or video games.




Integration by parts for multiple functions (n>2)

Let me preface this by saying that the things I’ll get into aren’t even remotely useful from a practical standpoint. This is basically for shits and giggles. They might be useful if you don’t have a calculator/computer and want to rip your hair out, however.

If you take/took BC Calculus or Calc II you’ve probably heard of the integration by parts technique. You can look at it as an inverse product rule that lets you take the integral of a product of two functions f(x)=u and g(x)=v. An example would be

\displaystyle \int x^2\sin{x}\ dx.

In the above case you would apply IBP twice to decrement the power of x^2 to a constant. If you continue on I’m just going to assume that you’ve seen this already and know how to do these kinds of problems.

What if we wanted to find the integral of a product of three functions, or four, or n? Sometimes such an integral might not be expressed in terms of elementary functions, but there’s probably an approximation out there using a computer. Sometimes the solution exists, however.

For two functions, the solution to integration by parts is

\displaystyle \int u\ dv = uv-\int v\ du.

This is derived by taking the integral on both sides of the product rule. So to find the n-function integration by parts formula, it makes sense to first derive the general n-function product rule.

Let’s start small with the product of three functions, uvw. Taking (uvw)' is equivalent to (uv)'w+(uv)w' by treating uv like a single function. Now we use the product rule again on uv to get

\displaystyle (uvw)'=u'vw+uv'w+uvw'.

This seems to make sense; for a product rule of three functions, there are three terms with each function having one and only one derivative for each term. Let’s see if this trend continues for four functions f_1 f_2 f_3 f_4. We can consider f_1 f_2 f_3 as an entire function and apply the product rule to get (f_1 f_2 f_3 f_4)' = (f_1 f_2 f_3)'f_4+(f_1 f_2 f_3)f_4'. But we already found what (f_1 f_2 f_3)' is! We just used different letters. So we now know that

\displaystyle (f_1 f_2 f_3 f_4)'=f_1'f_2f_3f_4+f_1f_2'f_3f_4+f_1f_2f_3'f_4+f_1f_2f_3f_4'.

Our pattern is now confirmed! We can continue like this on towards infinity, since if we do it this way we have a formula for an (n-1)-function product rule that we can apply towards the n-function product rule. The only small challenge is formulating this pattern concisely. Using a combination of pi and sigma notation, we can state the following proposition.

Proposition 1 (General product rule). Let f_1, f_2, \cdots, f_n be n different differentiable functions of x. Then

\displaystyle \frac{d}{dx}\prod_{i=1}^{n}f_i = \sum_{i=1}^{n}\frac{f_i'}{f_i}\prod_{j=1}^{n}f_j.

Now that we have the general product rule, we just have to take the integral of both sides with respect to x to get the general integration by parts formula. Doing so, we get

\displaystyle \int\prod_{i=1}^{n}f_i = \int\sum_{i=1}^{n}\frac{f_i'}{f_i}\prod_{j=1}^{n}f_j\ dx.

Although this is technically a valid formulation of the general IBP formula, I claim that integral signs are interchangeable, that is,

\displaystyle \int\sum_{i=1}^{n}f_i\ dx = \sum_{i=1}^{n}\int f_i\ dx.

This is due to the sum rule of integration: \int(f+g)\ dx=\int f\ dx + \int g\ dx.

Proposition 2 (General integration by parts). Let f_1, f_2, \cdots, f_n be n different continuous and differentiable functions of x. Then integration by parts is given by

\displaystyle \prod_{i=1}^{n}f_i = \sum_{i=1}^{n}\int\frac{f_i'}{f_i}\prod_{j=1}^{n}f_j\ dx.

This isn’t in a ‘useable’ form per se yet since we need to expand that sum and product notation to actually apply it in an integration, but this is as clean as the theorem is going to get. To see IBP in its familiar form, we make the substitutions u_1=f_1(x), u_2=f_2(x),\cdots,u_n=f_n(x) to get

\displaystyle \prod_{i=1}^{n}u_i = \sum_{i=1}^{n}\int\frac{du_i}{u_i}\prod_{j=1}^{n}u_j.

Let’s see what this form looks like when n=3.

\displaystyle \prod_{i=1}^{3}u_i = \sum_{i=1}^{3}\int\frac{du_i}{u_i}\prod_{j=1}^{3}u_j

\displaystyle u_1 u_2 u_3 = \int\frac{du_1}{u_1}(u_1u_2u_3)+\int\frac{du_2}{u_2}(u_1u_2u_3)+\int\frac{du_3}{u_3}(u_1u_2u_3)

\displaystyle \int u_1 u_2\ du_3 = u_1u_2u_3 -\int u_2u_3\ du_1 - \int u_1u_3\ du_2.

This last form is finally something we can work with when trying to find, say, the fiendish integral

\displaystyle \int xe^x\cos{x}\ dx.

I’ll leave the work up to you if you so choose, but using the n=3 IBP formula I got a solution of

\displaystyle \int xe^x\cos{x}\ dx = \frac{e^x(x\cos{x}+(x-1)\sin{x})}{2}.

That’s all for now.