# Category Archives: Teaching

## Math 420: Second Week Homework

Firstly, we have three administrative notes.

1. I’ve posted the second homework set. You can find it here.
2. I’ve also written solutions to the first homework set. You can find those here.
3. After feedback from the first week, I’m setting stable office hours. My office hours will be from 1-3pm on Monday and 2:30-3:30pm on Tuesday (immediately following our class). [Or we can set up an appointment].

I’ll see you on Tuesday, when we will continue to talk about the Euclidean Algorithm and greatest common divisors.

Posted in Brown University, Math 420, Mathematics, Teaching | Tagged , , , | Leave a comment

## Math 420: First Week Homework and References

Firstly, there are three administrative notes.

1. I’ve posted the first homework set. This is due on Thursday, and you can find it here.
2. I haven’t set official office hour times yet. But I will have office hours on Monday from noon to 2pm on Monday, 1 Feb 2016, in my office at the Science Library.
3. If you haven’t yet, I encourage you to read the syllabus.

We mentioned several good and interesting “number theoretic” problems in class today. I’d like to remind you of some of them, and link you to some additional places for information.

### Pythagorean Theorem

We’ve found all primitive Pythagorean triples in integers, which is a very nice theorem for an hour. But I also mentioned some of the history of the Pythagorean Theorem and the significance of numbers and number theory to the Greeks.

I told the class a story about how the Pythagorean student who revealed that there were irrational numbers was stoned. This is apocryphal. In fact, there is little exact record, but his name was Hippasus and it is more likely that he was drowned for releasing this information.

For this and other reasons, the Pythagorean school of thought split into two sects, one from Pythagoras and one from Hippasus.

### Goldbach’s Conjecture

Is it the case that every even integer is the sum of two primes? We think so. But we do not know.

I mentioned the Ternary Goldbach Conjecture, also known as the Weak Goldbach Conjecture, which says that every odd integer greater than $5$ is the sum of three odd primes. This was proved very recently. If you’re interested in what a mathematical paper looks like, you can give this paper a look. [Do not expect to be able to understand the paper — but it is interesting what sorts of tools can be used towards number theory]

### Fermat’s Last Theorem

Are there nontrivial integer solutions to $X^n + Y^n = Z^n$ where $n \geq 3$?

This is one of the most storied and studied problems in mathematics. I think this has to do with how simple the statement looks. Further, we managed to fully classify all solutions when $n = 2$ in one class period. It doesn’t seem like it should be too hard to extend that to other exponents, does it?

If time and interest permits, we will return to this topic at the end of the course. There is no way that we could present a proof, or even fully motivate the proof. But we might be able to say a few words about how progress towards the theorem spurred and created mathematics, and maybe we can give a hint of the breadth of the ideas used to finally produce a proof.

### Twin Prime Conjecture

Are there infinitely many primes $p$ such that $p+2$ is also prime? We think so, but we don’t know. Two years ago, we had absolutely no idea at all. Then Yitang Zhang had a brilliant idea (and not much later a graduate student named James Maynard had another brilliant idea) which allowed some sort of progress.

This culminated with the Polymath8 Project Bounded Gaps Between Primes. Math can be a social sport, and the polymath projects are massively collaborative online and open projects towards math problems. They’re still a bit new, and a bit experimental. But Polymath8 is certainly extremely successful.

What is known is that there exists at least one even number $H \leq 246$ such that $p$ and $p + H$ is prime infinitely often. In fact, James Maynard showed that you can make more complicated ensembles of prime distances.

The ideas that led to this result can likely be sharpened to give better results, but actually proving that there are infinitely many twin primes is almost certainly going to require a brand new idea and methodology.

The best related result comes from Chinese mathematician Chen Jingrun, who proved that every sufficiently large even integer can be written either as a sum of two primes, or as a sum of a prime and a number with exactly two prime factors. Although this seems very close, it is also likely that this idea cannot be sharpened further.

### Writing Numbers as Sums of Squares, Cubes, and So On

Can every integer be written as the sum of three squares? What about four squares? More generally, is there a number $n$ so that every integer can be written as a sum of at most $n$ squares?

Similarly, is there a number $n$ so that every integer can be written as a sum of at most $n$ cubes? What about fourth powers?

These problems are all associated to something called Waring’s Problem, about which much is known and much is unknown.

We also asked which primes can be written as a sum of two squares. Although we might have a hard time finding those primes right now, you might try to show that if $p$ is a prime that can be written as a sum of two squares, then either $p$ is $2$, or $p = 4z + 1$ for some integer $z$. The reasoning is very similar to some of the reasoning done in class today.

### Max’s Conjecture

For primitive Pythagorean triples $(a,b,c)$ with $a^2 + b^2 = c^2$, we showed that we can restrict out attention to cases where $a$ is odd, $b$ is even, and $c$ is odd. Max conjectured that those $c$ on the right are always of the form $4k + 1$ for some $k$, or equivalently $c$ is always an integer that leaves remainder $1$ after being divided by $4$.

We didn’t return to this in class, but we can now. First, note that since $c$ is odd, we can write $c$ as $2z + 1$ for some $z$. But we can do more. We can actually write $c$ as either $4z + 1$ or $4z + 3$. (Can you prove this?)

Max conjectured that it is always the case that $c = 4z + 1$. So we might ask, “What if $c = 4z + 3$?”

Writing $a = 2x + 1$ and $b = 2y$, we get the equation

\begin{align} a^2 + b^2 &= c^2 \\ 4x^2 + 4x + 1 + 4y^2 &= 16z^2 + 24z + 3, \end{align}

which can be rewritten as
$$4x^2 + 4x + 4y^2 = 16z^2 + 24z + 2.$$
You can divide by $2$. Then we ask: what’s the problem? Why is this bad? (It is, and it’s very similar to some questions we asked in class.)

So Max’s Conjecture is true, and every number appearing as $c$ in a primitive Pythagorean triple is of the form $c = 4z + 1$ for some integer $z$.

## Three Conundrums on Infinity

In this short post, we introduce three conundrums dealing with infinity. This is inspired by my calculus class, as we explore various confusing and confounding aspects of infinity and find that it’s very confusing, sometimes mindbending.

Order Matters

Consider the alternating unit series $$\sum_{n \geq 0} (-1)^n.$$
We want to try to understand its convergence. If we write out the first several terms, it looks like $$1 – 1 + 1 – 1 + 1 – 1 + \cdots$$
What if we grouped the terms while we were summing them? Perhaps we should group them like so, $$(1 – 1) + (1 – 1) + (1 – 1) + \cdots = 0 + 0 + 0 + \cdots$$
so that the sum is very clearly ${0}$. Adding infinitely many zeroes certainly gives zero, right?

On the other hand, what if we group the terms like so, $$1 + (-1 + 1) + (-1 + 1) + \cdots = 1 + 0 + 0 + \cdots$$
which is very clearly ${1}$. After all, adding ${1}$ to infinitely many zeroes certainly gives one, right?

A related, perhaps deeper paradox is one we mentioned in class. For conditionally convergent series like the alternating harmonic series $$\sum_{n = 1}^\infty \frac{(-1)^n}{n},$$
if we are allowed to rearrange the terms then we can have the series sum to any number that we want. This is called the Riemann Series Theorem.

The Thief and the King

A very wealthy king keeps gold coins in his vault, but a sneaky thief knows how to get in. Suppose that each day, the king puts two more gold coins into the vault. And each day, the thief takes one gold coin out (so that the king won’t notice that the vault is empty). After infinitely many days, how much gold is left in the vault?

Suppose that the king numbers each coin. So on day 1, the king puts in coins labelled 1 and 2, and on day 2 he puts in coins labelled 3 and 4, and so on. What if the thief steals the odd numbered coin each day? Then at the end of time, the king has all the even coins.

But what if instead, the thief steals from the bottom. So he first steals coin number 1, then number 2, and so on. At the end of time, no coin is left in the vault, since for any number ${n}$, the ${n}$th coin has been taken by the king.

Prevalence of Rarity

When I drove to Providence this morning, the car in front of me had the license place 637RB2. Think about it – out of the approximately ${10\cdot10\cdot10\cdot26\cdot 26 \cdot 10 = 6760000}$ possibilities, I happened across this one. Isn’t that amazing! How could something so rare happen to me?

Amazingly, something just as rare happened last time I drove to Providence too!

Posted in Brown University, Mathematics, Teaching | Tagged , , | 1 Comment

## Trigonometric and related substitutions in integrals

$\DeclareMathOperator{\csch}{csch}$
$\DeclareMathOperator{\sech}{sech}$
$\DeclareMathOperator{\arsinh}{arsinh}$

1. Introduction

In many ways, a first semester of calculus is a big ideas course. Students learn the basics of differentiation and integration, and some of the big-hitting theorems like the Fundamental Theorems of Calculus. Even in a big ideas course, students learn how to differentiate any reasonable combination of polynomials, trig, exponentials, and logarithms (elementary functions).

But integration skills are not pushed nearly as far. Do you ever wonder why? Even at the end of the first semester of calculus, there are many elementary functions that students cannot integrate. But the reason isn’t that there wasn’t enough time, but instead that integration is hard. And when I say hard, I mean often impossible. And when I say impossible, I don’t mean unsolved, but instead provably impossible (and when I say impossible, I mean that we can’t always integrate and get a nice function out, unlike our ability to differentiate any nice function and get a nice function back). An easy example is the sine integral $$\int \frac{\sin x}{x} \mathrm d x,$$
which cannot be expressed in terms of elementary functions. In short, even though the derivative of an elementary function is always an elementary function, the antiderivative of elementary functions don’t need to be elementary.

Worse, even when antidifferentiation is possible, it might still be really hard. This is the first problem that a second semester in calculus might try to address, meaning that students learn a veritable bag of tricks of integration techniques. These might include ${u}$-substitution and integration by parts (which are like inverses of the chain rule and product rule, respectively), and then the relatively more complicated techniques like partial fraction decomposition and trig substitution.

In this note, we are going to take a closer look at problems related to trig substitution, and some related ideas. We will assume familiarity with ${u}$-substitution and integration by parts, and we might even use them here from time to time. This, after the fold.

## A bit more about partial fraction decomposition

This is a short note written for my students in Math 170, talking about partial fraction decomposition and some potentially confusing topics that have come up. We’ll remind ourselves what partial fraction decomposition is, and unlike the text, we’ll prove it. Finally, we’ll look at some pitfalls in particular. All this after the fold.

1. The Result Itself

We are interested in rational functions and their integrals. Recall that a polynomial ${f(x)}$ is a function of the form ${f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0}$, where the ${a_i}$ are constants and ${x}$ is our “intederminate” — and which we commonly imagine standing for a number (but this is not necessary).

Then a rational function ${R(x)}$ is a ratio of two polynomials ${p(x)}$ and ${q(x)}$, $$R(x) = \frac{p(x)}{q(x)}.$$

Then the big result concerning partial fractions is the following:

If ${R(x) = \dfrac{p(x)}{q(x)}}$ is a rational function and the degree of ${p(x)}$ is less than the degree of ${q(x)}$, and if ${q(x)}$ factors into $$q(x) = (x-r_1)^{k_1}(x-r_2)^{k_2} \dots (x-r_l)^{k_l} (x^2 + a_{1,1}x + a_{1,2})^{v_1} \ldots (x^2 + a_{m,1}x + a_{m,2})^{v_m},$$
then ${R(x)}$ can be written as a sum of fractions of the form ${\dfrac{A}{(x-r)^k}}$ or ${\dfrac{Ax + B}{(x^2 + a_1x + a_2)^v}}$, where in particular

• If ${(x-r)}$ appears in the denominator of ${R(x)}$, then there is a term ${\dfrac{A}{x – r}}$
• If ${(x-r)^k}$ appears in the denominator of ${R(x)}$, then there is a collection of terms $$\frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \dots + \frac{A_k}{(x-r)^k}$$
• If ${x^2 + ax + b}$ appears in the denominator of ${R(x)}$, then there is a term ${\dfrac{Ax + B}{x^2 + ax + b}}$
• If ${(x^2 + ax + b)^v}$ appears in the denominator of ${R(x)}$, then there is a collection of terms $$\frac{A_1x + B_1}{x^2 + ax + b} + \frac{A_2 x + B_2}{(x^2 + ax + b)^2} + \dots \frac{A_v x + B_v}{(x^2 + ax + b)^v}$$

where in each of these, the capital ${A}$ and ${B}$ represent some constants that can be solved for through basic algebra.

I state this result this way because it is the one that leads to integrals that we can evaluate. But in principle, this theorem can be restated in a couple different ways.

Let’s parse this theorem through an example – the classic example, after the fold.

## Math 100 Fall 2013: Concluding Remarks

This is a post written towards my students in Calc II, Math 100 at Brown University, fall 2013. There will be many asides, written in italics. They are to serve as clarifications of method or true asides, to be digested or passed over.

The semester is over. All the grades are in and known, fall 2013 draws to a close. As you know, I’m interested in the statistics behind the course. I’d mentioned my previous analysis about the extremely high correlation between first midterm and final grade (much higher than I would have thought!). Let’s reveal the statistics and distribution of this course, below the fold.

Posted in Brown University, Math 100, Mathematics, Teaching | | 2 Comments

## An Intuitive Overview of Taylor Series

This is a note written for my fall 2013 Math 100 class, but it was not written “for the exam,” nor does anything on here subtly hint at anything on any exam. But I hope that this will be helpful for anyone who wants to get a basic understanding of Taylor series. What I want to do is try to get some sort of intuitive grasp on Taylor series as approximations of functions. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) – so it’s a mathematical intuition. In this way, this post is a sort of follow-up of my earlier note, An Intuitive Introduction to Calculus.

PLEASE NOTE that my math compiler and my markdown compiler sometimes compete, and sometimes repeated derivatives are too high or too low by one pixel.

We care about Taylor series because they allow us to approximate other functions in predictable ways. Sometimes, these approximations can be made to be very, very, very accurate without requiring too much computing power. You might have heard that computers/calculators routinely use Taylor series to calculate things like ${e^x}$ (which is more or less often true). But up to this point in most students’ mathematical development, most mathematics has been clean and perfect; everything has been exact algorithms yielding exact answers for years and years. This is simply not the way of the world.

Here’s a fundamental fact to both mathematics and life: almost anything worth doing is probably pretty hard and pretty messy.

For a very recognizable example, let’s think about finding zeroes of polynomials. Finding roots of linear polynomials is very easy. If we see ${5 + x = 0}$, we see that ${-5}$ is the zero. Similarly, finding roots of quadratic polynomials is very easy, and many of us have memorized the quadratic formula to this end. Thus ${ax^2 + bx + c = 0}$ has solutions ${x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}}$. These are both nice, algorithmic, and exact. But I will guess that the vast majority of those who read this have never seen a “cubic polynomial formula” for finding roots of cubic polynomials (although it does exist, it is horrendously messy – look up Cardano’s formula). There is even an algorithmic way of finding the roots of quartic polynomials. But here’s something amazing: there is no general method for finding the exact roots of 5th degree polynomials (or higher degree).

I don’t mean We haven’t found it yet, but there may be one, or even You’ll have to use one of these myriad ways – I mean it has been shown that there is no general method of finding exact roots of degree 5 or higher polynomials. But we certainly can approximate them arbitrarily well. So even something as simple as finding roots of polynomials, which we’ve been doing since we were in middle school, gets incredibly and unbelievably complicated.

So before we hop into Taylor series directly, I want to get into the mindset of approximating functions with other functions.

1. Approximating functions with other functions

We like working with polynomials because they’re so easy to calculate and manipulate. So sometimes we try to approximate complicated functions with polynomials, a problem sometimes called
polynomial interpolation”.

Suppose we wanted to approximate ${\sin(x)}$. The most naive approximation that we might do is see that ${\sin(0) = 0}$, so we might approximate ${\sin(x)}$ by ${p_0(x) = 0}$. We know that it’s right at least once, and since ${\sin(x)}$ is periodic, it’s going to be right many times. I write ${p_0}$ to indicate that this is a degree ${0}$ polynomial, that is, a constant polynomial. Clearly though, this is a terrible approximation, and we can do better.

## Math 100: Before second midterm

You have a midterm next week, and it’s not going to be a cakewalk.

As requested, I’m uploading the last five weeks’ worth of worksheets, with (my) solutions. A comment on the solutions: not everything is presented in full detail, but most things are presented with most detail (except for the occasional one that is far far beyond what we actually expect you to be able to do). If you have any questions about anything, let me know. Even better, ask it here – maybe others have the same questions too.

And since we were unable to go over the quiz in my afternoon recitation today, I’m attaching a worked solution to the quiz as well.

Again, let me know if you have any questions. I will still have my office hours on Tuesday from 2:30-4:30pm in my office (I’m aware that this happens to be immediately before the exam – status not by design). And I’ll be more or less responsive by email.

Study study study!

## Math 100: Week 4

This is a post for my math 100 calculus class of fall 2013. In this post, I give the 4th week’s recitation worksheet (no solutions yet – I’m still writing them up). More pertinently, we will also go over the most recent quiz and common mistakes. Trig substitution, it turns out, is not so easy.

Before we hop into the details, I’d like to encourage you all to avail of each other, your professor, your ta, and the MRC in preparation for the first midterm (next week!).

1. The quiz

There were two versions of the quiz this week, but they were very similar. Both asked about a particular trig substitution

$\displaystyle \int_3^6 \sqrt{36 – x^2} \mathrm{d} x$

And the other was

$\displaystyle \int_{2\sqrt 2}^4 \sqrt{16 – x^2} \mathrm{d}x.$

They are very similar, so I’m only going to go over one of them. I’ll go over the first one. We know we are to use trig substitution. I see two ways to proceed: either draw a reference triangle (which I recommend), or think through the Pythagorean trig identities until you find the one that works here (which I don’t recommend).

We see a ${\sqrt{36 – x^2}}$, and this is hard to deal with. Let’s draw a right triangle that has ${\sqrt{36 – x^2}}$ as a side. I’ve drawn one below. (Not fancy, but I need a better light).

In this picture, note that ${\sin \theta = \frac{x}{6}}$, or that ${x = 6 \sin \theta}$, and that ${\sqrt{36 – x^2} = 6 \cos \theta}$. If we substitute ${x = 6 \sin \theta}$ in our integral, this means that we can replace our ${\sqrt{36 – x^2}}$ with ${6 \cos \theta}$. But this is a substitution, so we need to think about ${\mathrm{d} x}$ too. Here, ${x = 6 \sin \theta}$ means that ${\mathrm{d}x = 6 \cos \theta}$.

Some people used the wrong trig substitution, meaning they used ${x = \tan \theta}$ or ${x = \sec \theta}$, and got stuck. It’s okay to get stuck, but if you notice that something isn’t working, it’s better to try something else than to stare at the paper for 10 minutes. Other people use ${x = 6 \cos \theta}$, which is perfectly doable and parallel to what I write below.

Another common error was people forgetting about the ${\mathrm{d}x}$ term entirely. But it’s important!.

Substituting these into our integral gives

$\displaystyle \int_{?}^{??} 36 \cos^2 (\theta) \mathrm{d}\theta,$

where I have included question marks for the limits because, as after most substitutions, they are different. You have a choice: you might go on and put everything back in terms of ${x}$ before you give your numerical answer; or you might find the new limits now.

It’s not correct to continue writing down the old limits. The variable has changed, and we really don’t want ${\theta}$ to go from ${3}$ to ${6}$.

If you were to find the new limits, then you need to consider: if ${x=3}$ and ${\frac{x}{6} = \sin \theta}$, then we want a ${\theta}$ such that ${\sin \theta = \frac{3}{6}= \frac{1}{2}}$, so we might use ${\theta = \pi/6}$. Similarly, when ${x = 6}$, we want ${\theta}$ such that ${\sin \theta = 1}$, like ${\theta = \pi/2}$. Note that these were two arcsine calculations, which we would have to do even if we waited until after we put everything back in terms of ${x}$ to evaluate.

Some people left their answers in terms of these arcsines. As far as mistakes go, this isn’t a very serious one. But this is the sort of simplification that is expected of you on exams, quizzes, and homeworks. In particular, if something can be written in a much simpler way through the unit circle, then you should do it if you have the time.

So we could rewrite our integral as

$\displaystyle \int_{\pi/6}^{\pi/2} 36 \cos^2 (\theta) \mathrm{d}\theta.$

How do we integrate ${\cos^2 \theta}$? We need to make use of the identity ${\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}}$. You should know this identity for this midterm. Now we have

$\displaystyle 36 \int_{\pi/6}^{\pi/2}\left(\frac{1}{2} + \frac{\cos 2 \theta}{2}\right) \mathrm{d}\theta = 18 \int_{\pi/6}^{\pi/2}\mathrm{d}\theta + 18 \int_{\pi/6}^{\pi/2}\cos 2\theta \mathrm{d}\theta.$

The first integral is extremely simple and yields ${6\pi}$ The second integral has antiderivative ${\dfrac{\sin 2 \theta}{2}}$ (Don’t forget the ${2}$ on bottom!), and we have to evaluate ${\big[9 \sin 2 \theta \big]_{\pi/6}^{\pi/2}}$, which gives ${-\dfrac{9 \sqrt 3}{2}}$. You should know the unit circle sufficiently well to evaluate this for your midterm.

And so the final answer is ${6 \pi – \dfrac{9 \sqrt 2}{2} \approx 11.0553}$. (You don’t need to be able to do that approximation).

Let’s go back a moment and suppose you didn’t re”{e}valuate the limits once you substituted in ${\theta}$. Then, following the same steps as above, you’d be left with

$\displaystyle 18 \int_{?}^{??}\mathrm{d}\theta + 18 \int_{?}^{??}\cos 2\theta \mathrm{d}\theta = \left[ 18 \theta \right]_?^{??} + \left[ 9 \sin 2 \theta \right]_?^{??}.$

Since ${\frac{x}{6} = \sin \theta}$, we know that ${\theta = \arcsin (x/6)}$. This is how we evaluate the left integral, and we are left with ${[18 \arcsin(x/6)]_3^6}$. This means we need to know the arcsine of ${1}$ and ${\frac 12}$. These are exactly the same two arcsine computations that I referenced above! Following them again, we get ${6\pi}$ as the answer.

We could do the same for the second part, since ${\sin ( 2 \arcsin (x/6))}$ when ${x = 3}$ is ${\sin (2 \arcsin \frac{1}{2} ) = \sin (2 \cdot \frac{\pi}{6} ) = \frac{\sqrt 3}{2}}$; and when ${x = 6}$ we get ${\sin (2 \arcsin 1) = \sin (2 \cdot \frac{\pi}{2}) = \sin (\pi) = 0}$.

Putting these together, we see that the answer is again ${6\pi – \frac{9\sqrt 3}{2}}$.

Or, throwing yet another option out there, we could do something else (a little bit wittier, maybe?). We have this ${\sin 2\theta}$ term to deal with. You might recall that ${\sin 2 \theta = 2 \sin \theta \cos \theta}$, the so-called double-angle identity.

Then ${9 \sin 2\theta = 18 \sin \theta \cos \theta}$. Going back to our reference triangle, we know that ${\cos \theta = \dfrac{\sqrt{36 – x^2}}{6}}$ and that ${\sin \theta = \dfrac{x}{6}}$. Putting these together,

$\displaystyle 9 \sin 2 \theta = \dfrac{ x\sqrt{36 – x^2} }{2}.$

When ${x=6}$, this is ${0}$. When ${x = 3}$, we have ${\dfrac{ 3\sqrt {27}}{2} = \dfrac{9\sqrt 3}{2}}$.

And fortunately, we get the same answer again at the end of the day. (phew).

2. The worksheet

Finally, here is the worksheet for the day. I’m working on their solutions, and I’ll have that up by late this evening (sorry for the delay).

Ending tidbits – when I was last a TA, I tried to see what were the good predictors of final grade. Some things weren’t very surprising – there is a large correlation between exam scores and final grade. Some things were a bit surprising – low homework scores correlated well with low final grade, but high homework scores didn’t really have a strong correlation with final grade at all; attendance also correlated weakly. But one thing that really stuck with me was the first midterm grade vs final grade in class: it was really strong. For a bit more on that, I refer you to my final post from my Math 90 posts.

## Math 100: Week 3 and pre-midterm

This is a post for my Math 100 class of fall 2013. In this post, I give the first three weeks’ worksheets from recitation and the set of solutions to week three’s worksheet, as well as a few administrative details.

Firstly, here is the recitation work from the first three weeks:

1. (there was no recitation the first week)
2. A worksheet focusing on review.
3. A worksheet focusing on integration by parts and u-substitution, with solutions.

In addition, I’d like to remind you that I have office hours from 2-4pm (right now) in Kassar 018. I’ve had multiple people set up appointments with me outside of these hours, which I’m tempted to interpret as suggesting that I change when my office hours are. If you have a preference, let me know, and I’ll try to incorporate it.

Finally, there will be an exam next Tuesday. I’ve been getting a lot of emails about what material will be on the exam. The answer is that everything you have learned up to now and by the end of this week is fair game for exam material. This also means there could be exam questions on material that we have not discussed in recitation. So be prepared. However, I will be setting aside a much larger portion of recitation this Thursday for questions than normal. So come prepared with your questions.

Best of luck, and I’ll see you in class on Thursday.