Category Archives: Georgia Tech

Tiontobl: A combinatorial game

As a sophomore at Georgia Tech, I took a class on Combinatorial Game Theory with two good friends, David Hollis (now at Reckless Abandon Labs, which he founded) and Michelle Delcourt (now working towards her PhD at UIUC). As a final project, we were supposed to analyze a game combinatorially. The three of us ended up creating a game, called Tiontobl, and we wrote a brief paper. We submitted it to the journal Integers, but we were asked to revise and expand part of the paper. At some point in time, we’ll finish revising the paper and submit it again (it’s harder now, since we’re split across the country – but it will happen).

Nonetheless, I was talking about it the other day, and I thought I should put the current paper out there.

The paper can be found here (tiontobl).

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Reading Math

First, a recent gem from MathStackExchange:

Task: Calculate $latex \displaystyle \sum_{i = 1}^{69} \sqrt{ \left( 1 + \frac{1}{i^2} + \frac{1}{(i+1)^2} \right) }$ as quickly as you can with pencil and paper only.

Yes, this is just another cute problem that turns out to have a very pleasant solution. Here’s how this one goes. (If you’re interested – try it out. There’s really only a few ways to proceed at first – so give it a whirl and any idea that has any promise will probably be the only idea with promise).


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Integration by Parts

I suddenly have college degrees to my name. In some sense, I think that I should feel different – but all I’ve really noticed is that I’ve much less to do. Fewer deadlines, anyway. So now I can blog again! Unfortunately, I won’t quite be able to blog as much as I might like, as I will be traveling quite a bit this summer. In a few days I’ll hit Croatia.

Georgia Tech is magnificent at helping its students through their first few tough classes. Although the average size of each of the four calculus classes is around 150 students, they are broken up into 30 person recitations with a TA (usually a good thing, but no promises). Some classes have optional ‘Peer Led Undergraduate Study’ programs, where TA-level students host additional hours to help students master exercises over the class material. There is free tutoring available in many of the freshmen dorms every on most, if not all, nights of the week. If that doesn’t work, there is also free tutoring available from the Office of Minority Education or the Department of Success Programs – the host of the so-called 1-1 Tutoring program (I was a tutor there for two years). One can schedule 1-1 appointments between 8 am and something like 9 pm, and you can choose your tutor. For the math classes, each professor and TA holds office hours, and there is a general TA lounge where most questions can be answered, regardless of whether one’s TA is there. Finally, there is also the dedicated ‘Math Lab,’ a place where 3-4 highly educated math students (usually math grad students, though there are a couple of math seniors) are available each hour between 10 am and 4 pm (something like that – I had Thursday from 1-2 pm, for example). It’s a good theory.

During Dead Week, the week before finals, I had a group of Calc I students during my Math Lab hour. They were asking about integration by parts – when in the world is it useful? At first, I had a hard time saying something that they accepted as valuable – it’s an engineering school, and the things I find interesting do not appeal to the general engineering population of Tech. I thought back during my years at Tech (as this was my last week as a student there, it put me in a very nostalgic mood), and I realized that I associate IBP most with my quantum mechanics classes with Dr. Kennedy. In general, the way to solve those questions was to find some sort of basis of eigenvectors, normalize everything, take more inner products than you want, integrate by parts until it becomes meaningful, and then exploit as much symmetry as possible. Needless to say, that didn’t satisfy their question.

There are the very obvious answers. One derives Taylor’s formula and error with integration by parts:

$latex \begin{array}{rl}
f(x) &= f(0) + \int_0^x f'(x-t) \,dt\\
&= f(0) + xf'(0) + \displaystyle \int_0^x tf”(x-t)\,dt\\
&= f(0) + xf'(0) + \frac{x^2}2f”(0) + \displaystyle \int_0^x \frac{t^2}2 f”'(x-t)\,dt
$ … and so on.

But in all honesty, Taylor’s theorem is rarely used to estimate values of a function by hand, and arguing that it is useful to know at least the bare bones of the theory behind one’s field is an uphill battle. This would prevent me from mentioning the derivation of the Euler-Maclaurin formula as well.

I appealed to aesthetics: Taylor’s Theorem says that $latex \displaystyle \sum_{n\ge0} x^n/n! = e^x$, but repeated integration by parts yields that $latex \displaystyle \int_0^\infty x^n e^{-x} dx=n!$. That’s sort of cool – and not as obvious as it might appear at first. Although I didn’t mention it then, we also have the pretty result that n integration by parts applied to $latex \displaystyle \int_0^1 \dfrac{ (-x\log x)^n}{n!} dx = (n+1)^{-(n+1)}$. Summing over n, and remembering the Taylor expansion for $latex e^x$, one gets that $latex \displaystyle \int_0^1 x^{-x} dx = \displaystyle \sum_{n=1}^\infty n^{-n}$.

Finally, I decided to appeal to that part of the student that wants only to do well on tests. Then for a differentiable function $latex f$ and its inverse $latex f^{-1}$, we have that:
$latex \displaystyle \int f(x)dx = xf(x) – \displaystyle \int xf'(x)dx = $
$latex = xf(x) – \displaystyle \int f^{-1}(f(x))f'(x)dx = xf(x) – \displaystyle \int f^{-1}(u)du$.
In other words, knowing the integral of $latex f$ gives the integral of $latex f^{-1}$ very cheaply, and this is why we use integration by parts to integrate things like $latex \ln x$, $latex \arctan x$, etc. Similarly, one gets the reduction formulas necessary to integrate $latex \sin^n (x)$ or $latex \cos^n (x)$. If one believes that being able to integrate things is useful, then these are useful.There is of course the other class of functions such as $latex \cos(x)\sin(x)$ or $latex e^x \sin(x)$, where one integrates by parts twice and solves for the integral. I still think that’s really cool – sort of like getting something for nothing.

And at the end of the day, they were satisfied. But this might be the crux of the problem that explains why so many Tech students, despite having so many resources for success, still fail – they have to trudge through a whole lost of ‘useless theory’ just to get to the ‘good stuff.’

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2401: Additional Examples for Test 3

In the past, I have talked about how good a supplemental source of information the Khan Academy is. Again, it is supplementary. But it seems to have lots of fully worked and fully explained examples of the concepts of chapters 17 and chapter 18 (sections 1 through 4) — the topics for your next exam. I have placed the relevant links below.

Double Integrals ( I, II, III, IV, V, VI)
Triple Integrals( I, II, III)
Line Integrals( I, II, III, IV)
Clever Line Integrals (I, II, III, IV, V, VI, VII, VIII, IX)

As always, if you have any questions let me know. I will be hosting a review session in the Math Lab at 5 – come prepared and with questions. I suspect we’ll be focusing on the iterated integrals of Chapter 17. Good luck!

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2401: Missing recitation

As I went to visit Brown on the 16th-18th, I had my friend Matt cover recitation. As we have started considering double and triple integrals, and iterated integrals in particular, I thought I could point out a very good site for brushing up on material. I think it can act as a wonderful supplement to the lecture and recitation material. The Khan Academy is an online information center built around the idea that video presentations and video lectures can give intuition without adding any pressure – you can rewatch anything you’ve missed, repeat important parts, etc. all without feeling like you’re wasting someone’s time. While most of the Khan material is aimed at primary and secondary school, they happen to have a multivariable calculus section (although it’s far insufficient to Tech’s course material – don’t think of this as a replacement, but instead as merely a supplementary way to build intuition).

Here are links to the Khan Academy website, and links to 5 lectures that I think are relevant to material I would have covered.

Khan website
Intro to iterated integrals

I encourage you to check them out before we meet again for recitation.

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Test Solution

Due to the amount of confusion and the large number of emails, I have written up the solution to Problem 1 from Test 2.

The Problem
Determine the path of steepest descent along the surface $latex z = 2 + x + 2y – x^2 – 3y^2 $ from the point $latex (0,0,2).$

There are a few things to note – the first thing we must do is find which direction points ‘downwards’ the most. So we note that for a function $latex f(x,y) = z, $ we know that $latex \nabla f $ points ‘upwards’ the most at all points where it isn’t zero. So at any point $latex P, $ we go in the direction $latex -\nabla f.$

The second thing to note is that we seek a path, not a direction. So let us take a curve that parametrizes our path: $latex {\bf C} (t) = x(t) \hat{i} + y(t) \hat{j}.$

So $latex -\nabla f = (2x -1)\hat{i} + (6y -2)\hat{j}.$
As the velocity of the curve points in the direction of the curve, our path satisfies:

$latex x'(t) = 2x(t) -1; x(0) = 0$
$latex y'(t) = 6y(t) – 2; y(0) = 0$

These are two ODEs that we can solve by separation of variables (something that is, in theory, taught in 1502 – for more details, look at chapter 9 in Salas, Hille, and Etgen). Let’s solve the y one:

$latex y’ = 6y – 2$
$latex \frac{dy}{dt} = 6y – 2$
$latex \frac{dy}{6y-2} = dt$
$latex ln(6y-2)(1/6) = t + k$ for a constant k
$latex 6y = e^{6t + k} + 2= Ae^{6t}$ for a constant A
$latex y = Ae^{6t} + 1/3$ for a new constant A
$latex y(0) = 0 \Rightarrow A = -1/3 $

Solving both yields:

$latex x = \frac{1}{2} -\frac{1}{2} e^{2t} $
$latex y = \frac{1}{3} – \frac{1}{3} e^{6t} $

Now let’s get rid of the t. Note that $latex (3y -1) = e^{6t}$ and $latex (2x -1) = e^{2t}$. Using these together, we can get rid of t by noting that $latex \dfrac{3y-1}{(2x – 1)^3} = 1.$ Rewriting, we get $latex 3y = (2x-1)^3 + 1.$

So the path is given by $latex 3y = (2x-1)^3 + 1$

Good luck on your next test!

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