Category Archives: Brown University

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 $z$.

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

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 $latex {0}$. 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 $latex {1}$. After all, adding $latex {1}$ 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 $latex {n}$, the $latex {n}$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 $latex {10\cdot10\cdot10\cdot26\cdot 26 \cdot 10 = 6760000}$ 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 $latex {u}$-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 $latex {u}$-substitution and integration by parts, and we might even use them here from time to time. This, after the fold.

(more…)

Posted in Brown University, Expository, Math 170, Mathematics | Tagged , , , , , , | Leave a comment

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

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

Then the big result concerning partial fractions is the following:

If $latex {R(x) = \dfrac{p(x)}{q(x)}}$ is a rational function and the degree of $latex {p(x)}$ is less than the degree of $latex {q(x)}$, and if $latex {q(x)}$ 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 $latex {R(x)}$ can be written as a sum of fractions of the form $latex {\dfrac{A}{(x-r)^k}}$ or $latex {\dfrac{Ax + B}{(x^2 + a_1x + a_2)^v}}$, where in particular

  • If $latex {(x-r)}$ appears in the denominator of $latex {R(x)}$, then there is a term $latex {\dfrac{A}{x – r}}$
  • If $latex {(x-r)^k}$ appears in the denominator of $latex {R(x)}$, 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 $latex {x^2 + ax + b}$ appears in the denominator of $latex {R(x)}$, then there is a term $latex {\dfrac{Ax + B}{x^2 + ax + b}}$
  • If $latex {(x^2 + ax + b)^v}$ appears in the denominator of $latex {R(x)}$, 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 $latex {A}$ and $latex {B}$ 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.

(more…)

Posted in Brown University, Expository, Math 170, Mathematics, Teaching | Tagged , , , , | Leave a comment

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.

(more…)

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

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.

Without further ado –

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!

Posted in Brown University, Math 100, Mathematics | Tagged , , , , , , , , , , | Leave a comment

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

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

And the other was

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

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

Some people used the wrong trig substitution, meaning they used $latex {x = \tan \theta}$ or $latex {x = \sec \theta}$, 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 $latex {x = 6 \cos \theta}$, which is perfectly doable and parallel to what I write below.

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

Substituting these into our integral gives

$latex \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 $latex {x}$ 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 $latex {\theta}$ to go from $latex {3}$ to $latex {6}$.

If you were to find the new limits, then you need to consider: if $latex {x=3}$ and $latex {\frac{x}{6} = \sin \theta}$, then we want a $latex {\theta}$ such that $latex {\sin \theta = \frac{3}{6}= \frac{1}{2}}$, so we might use $latex {\theta = \pi/6}$. Similarly, when $latex {x = 6}$, we want $latex {\theta}$ such that $latex {\sin \theta = 1}$, like $latex {\theta = \pi/2}$. 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 $latex {x}$ 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

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

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

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

And so the final answer is $latex {6 \pi – \dfrac{9 \sqrt 2}{2} \approx 11.0553}$. (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 $latex {\theta}$. Then, following the same steps as above, you’d be left with

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

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

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

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

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

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

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

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.

Posted in Brown University, Math 100, Mathematics | Tagged , , , , , , , , , , , | Leave a comment

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.

Posted in Brown University, Math 100, Mathematics | Tagged , , , , , , , , , | Leave a comment

An intuitive introduction to calculus

This is a post written for my fall 2013 Math 100 class but largely intended for anyone with knowledge of what a function is and a desire to know what calculus is all about. Calculus is made out to be the pinnacle of the high school math curriculum, and correspondingly is thought to be very hard. But the difficulty is bloated, blown out of proportion. In fact, the ideas behind calculus are approachable and even intuitive if thought about in the right way.

Many people managed to stumble across the page before I’d finished all the graphics. I’m sorry, but they’re all done now! I was having trouble interpreting how WordPress was going to handle my gif files – it turns out that they automagically resize them if you don’t make them of the correct size, which makes them not display. It took me a bit to realize this. I’d like to mention that this actually started as a 90 minute talk I had with my wife over coffee, so perhaps an alternate title would be “Learning calculus in 2 hours over a cup of coffee.”

So read on if you would like to understand what calculus is, or if you’re looking for a refresher of the concepts from a first semester in calculus (like for Math 100 students at Brown), or if you’re looking for a bird’s eye view of AP Calc AB subject material.

1. An intuitive and semicomplete introduction to calculus

We will think of a function $ {f(\cdot)}$ as something that takes an input $ {x}$ and gives out another number, which we’ll denote by $ {f(x)}$. We know functions like $ {f(x) = x^2 + 1}$, which means that if I give in a number $ {x}$ then the function returns the number $ {f(x) = x^2 + 1}$. So I put in $ {1}$, I get $ {1^2 + 1 = 2}$, i.e. $ {f(1) = 2}$. Primary and secondary school overly conditions students to think of functions in terms of a formula or equation. The important thing to remember is that a function is really just something that gives an output when given an input, and if the same input is given later then the function spits the same output out. As an aside, I should mention that the most common problem I’ve seen in my teaching and tutoring is a fundamental misunderstanding of functions and their graphs

For a function that takes in and spits out numbers, we can associate a graph. A graph is a two-dimensional representation of our function, where by convention the input is put on the horizontal axis and the output is put on the vertical axis. Each axis is numbered, and in this way we can identify any point in the graph by its coordinates, i.e. its horizontal and vertical position. A graph of a function $ {f(x)}$ includes a point $ {(x,y)}$ if $ {y = f(x)}$.

The graph of the function $ x^2 + 1$ is in blue. The emphasized point appears on the graph because it is of the form $ (x, f(x))$. In particular, this point is $ (1, 2)$.

Thus each point on the graph is really of the form $ {(x, f(x))}$. A large portion of algebra I and II is devoted to being able to draw graphs for a variety of functions. And if you think about it, graphs contain a huge amount of information. Graphing $ {f(x)= x^2 + 1}$ involves drawing an upwards-facing parabola, which really represents an infinite number of points. That’s pretty intense, but it’s not what I want to focus on here.

1.1. Generalizing slope – introducing the derivative

You might recall the idea of the ‘slope’ of a line. A line has a constant ratio of how much the $ {y}$ value changes for a specific change in $ {x}$, which we call the slope (people always seem to remember rise over run). In particular, if a line passes through the points $ {(x_1, y_1)}$ and $ {(x_2, y_2)}$, then its slope will be the vertical change $ {y_2 – y_1}$ divided by the horizontal change $ {x_2 – x_1}$, or $ {\dfrac{y_2 – y_1}{x_2 – x_1}}$.

The graph of a line appears in blue. The two points $ (0,1)$ and $ (1,3)$ are shown on the line. The horizontal red line shows the horizontal change. The vertical red line shows the vertical change. The ‘slope’ of the blue line is the length of the vertical red line divided by the length of the horizontal red line.

So if the line is given by an equation $ {f(x) = \text{something}}$, then the slope from two inputs $ {x_1}$ and $ {x_2}$ is $ {\dfrac{f(x_2) – f(x_1)}{x_2 – x_1}}$. As an aside, for those that remember things like the ‘standard equation’ $ {y = mx + b}$ or ‘point-slope’ $ {(y – y_0) = m(x – x_0)}$ but who have never thought or been taught where these come from: the claim that lines are the curves of constant slope is saying that for any choice of $ {(x_1, y_1)}$ on the line, we expect $ {\dfrac{y_2 – y_1}{x_2 – x_1} = m}$ a constant, which I denote by $ {m}$ for no particularly good reason other than the fact that some textbook author long ago did such a thing. Since we’re allowing ourselves to choose any $ {(x_1, y_1)}$, we might drop the subscripts – since they usually mean a constant – and rearrange our equation to give $ {y_2 – y = m(x_2 – x)}$, which is what has been so unkindly drilled into students’ heads as the ‘point-slope form.’ This is why lines have a point-slope form, and a reason that it comes up so much is that it comes so naturally from the defining characteristic of a line, i.e. constant slope.

But one cannot speak of the ‘slope’ of a parabola.

The parabola $ f(x) = x^2 + 1$ is shows in blue. Slope is a measure of how much the function $ f(x)$ changes when $ x$ is changed. Some tangent lines to the parabola are shown in red. The slope of each line seems like it should be the ‘slope’ of the parabola when the line touches the parabola, but these slopes are different.

Intuitively, we look at our parabola $ {x^2 + 1}$ and see that the ‘slope,’ or an estimate of how much the function $ {f(x)}$ changes with a change in $ {x}$, seems to be changing depending on what $ {x}$ values we choose. (This should make sense – if it didn’t change, and had constant slope, then it would be a line). The first major goal of calculus is to come up with an idea of a ‘slope’ for non-linear functions. I should add that we already know a sort of ‘instantaneous rate of change’ of a nonlinear function. When we’re in a car and we’re driving somewhere, we’re usually speeding up or slowing down, and our pace isn’t usually linear. Yet our speedometer still manages to say how fast we’re going, which is an immediate rate of change. So if we had a function $ {p(t)}$ that gave us our position at a time $ {t}$, then the slope would give us our velocity (change in position per change in time) at a moment. So without knowing it, we’re familiar with a generalized slope already. Now in our parabola, we don’t expect a constant slope, so we want to associate a ‘slope’ to each input $ {x}$. In other words, we want to be able to understand how rapidly the function $ {f(x)}$ is changing at each $ {x}$, analogous to how the slope $ {m}$ of a line $ {g(x) = mx + b}$ tells us that if we change our input by an amount $ {h}$ then our output value will change by $ {mh}$.

How does calculus do that? The idea is to get closer and closer approximations. Suppose we want to find the ‘slope’ of our parabola at the point $ {x = 1}$. Let’s get an approximate answer. The slope of the line coming from inputs $ {x = 1}$ and $ {x = 2}$ is a (poor) approximation. In particular, since we’re working with $ {f(x) = x^2 + 1}$, we have that $ {f(2) = 5}$ and $ {f(1) = 2}$, so that the ‘approximate slope’ from $ {x = 1}$ and $ {x = 2}$ is $ {\frac{5 – 2}{2 – 1} = 3}$. But looking at the graph,

The parabola $ x^2 + 1$ is shown in blue, and the line going through the points $ (1,2)$ and $ (2,5)$ is shown. The line immediately goes above and crosses the parabola, so it seems like this line is rising faster (changing faster) than the parabola. It’s too steep, and the slope is too high to reflect the ‘slope’ of the parabola at the indicated point.

we see that it feels like this slope is too large. So let’s get closer. Suppose we use inputs $ {x = 1}$ and $ {x = 1.5}$. We get that the approximate slope is $ {\frac{3.25 – 2}{1.5 – 1} = 2.5}$. If we were to graph it, this would also feel too large. So we can keep choosing smaller and smaller changes, like using $ {x = 1}$ and $ {x = 1.1}$, or $ {x = 1}$ and $ {x = 1.01}$, and so on. This next graphic contains these approximations, with chosen points getting closer and closer to $ {1}$.

The parabola $ x^2 + 1$ is shown in blue. Two points are chosen on the parabola and the line between them is drawn in red. As the points get closer to each other, the red line indicates the rate of growth of the parabola at the point $ (1,2)$ better and better. So the slope of the red lines seems to be getting closer to the ‘slope’ of the parabola at $ (1,2)$.

Let’s look a little closer at the values we’re getting for our slopes when we use $ {1}$ and $ {2, 1.5, 1.1, 1.01, 1.001}$ as our inputs. We get

$ \displaystyle \begin{array}{c|c} \text{second input} & \text{approx. slope} \\ \hline 2 & 3 \\ 1.5 & 2.5 \\ 1.1 & 2.1 \\ 1.01 & 2.01 \\ 1.001 & 2.001 \end{array} $

It looks like the approximate slopes are approaching $ {2}$. What if we plot the graph with a line of slope $ {2}$ going through the point $ {(1,2)}$?

The parabola $ x^2 + 1$ is shown in blue. The line in red has slope $ 2$ and goes through the point $ (1,2)$. We got this line by continuing the successive approximations done above. It looks like it accurately indicates the ‘slope’ of the parabola at $ (1,2)$.

It looks great! Let’s zoom in a whole lot.

When we zoom in, the blue parabola looks almost like a line, and the red line looks almost like the parabola! This is why we are measuring the ‘slope’ of the parabola in this fashion – when we zoom in, it looks more and more like a line, and we are getting the slope of that line.

That looks really close! In fact, what I’ve been allowing as the natural feeling slope, or local rate of change, is really the line tangent to the graph of our function at the point $ {(1, f(1))}$. In a calculus class, you’ll spend a bit of time making sense of what it means for the approximate slopes to ‘approach’ $ {2}$. This is called a ‘limit,’ and the details are not important to us right now. The important thing is that this let us get an idea of a ‘slope’ at a point on a parabola. It’s not really a slope, because a parabola isn’t a line. So we’ve given it a different name – we call this ‘the derivative.’ So the derivative of $ {f(x) = x^2 + 1}$ at $ {x = 1}$ is $ {2}$, i.e. right around $ {x = 1}$ we expect a rate of change of $ {2}$, so that we expect $ {f(1 + h) – f(1) \approx 2h}$. If you think about it, we’re saying that we can approximate $ {f(x) = x^2 + 1}$ near the point $ {(1, 2)}$ by the line shown in the graph above: this line passes through $ {(1,2)}$ and it’s slope is $ {2}$, what we’re calling the slope of $ {f(x) = x^2 + 1}$ at $ {x = 1}$.

Let’s generalize. We were able to speak of the derivative at one point, but how about other points? The rest of this post is below the ‘more’ tag below.

(more…)

Posted in Brown University, Expository, Math 100, Mathematics | Tagged , , , , , , , , , , , , , | 7 Comments

Twenty Mathematicians, Two Hard Problems, One Week, IdeaLab2013

July has been an exciting and busy month for me. I taught number theory 3 hours a day, 5 days a week, for 3 weeks to (mostly) devoted and motivated high school students in the Summer@Brown program. In the middle, I moved to Massachusetts. Immediately after the Summer@Brown program ended, I was given the opportunity to return to ICERM to participate in an experimental program called an IdeaLab.

IdeaLab invited 20 early career mathematicians to come together for a week and to generate ideas on two very different problems: Tipping Points in Climate Systems and Efficient Fully Homomorphic Encryption. Although I plan on writing a bit more about each of these problems and the IdeaLab process in action (at least from my point of view), I should say something about what these are.

Models of Earth’s climate are used all the time, to give daily weather reports, to predict and warn about hurricanes, to attempt to understand the effects of anthropogenic sources of carbon on long-term climate. As we know from uncertainty about weather reports, these models aren’t perfect. In particular, they don’t currently predict sudden, abrupt changes called ‘Tippling points.’ But are tipping points possible? There have been warm periods following ice-ages in the past, so it seems that there might be tipping points that aren’t modelled in the system. Understanding these form the basis for the idea behind the Tipping Points in Climate Systems project. This project also forms another link in Mathematics of Planet Earth.

On the other hand, homomorphic encryption is a topic in modern cryptography. To encrypt a message is to make it hard or impossible for others to read it unless they have a ‘key.’ You might think that you wouldn’t want someone holding onto an encrypted data to be able to do anything with the data, and in most modern encryption algorithms this is the case. But what if we were able to give Google an encrypted dataset and ask them to perform a search on it? Is it possible to have a secure encryption that would allow Google to do some sort of search algorithm and give us the results, but without Google ever understanding the data itself? It may seem far-fetched, but this is exactly the idea behind the Efficient Fully Homomorphic Encryption group. Surprisingly enough, it is possible. But known methods are obnoxiously slow and infeasible. This is why the group was after ‘efficient’ encryption.

So 20 early career mathematicians from all sorts of areas of mathematics gathered to think about these two questions. For the rest of this post, I’d like to talk about the structure and my thoughts on the IdeaLab process. In later posts, I’ll talk about each of the two major topics and what sorts of ideas came out of the process.

(more…)

Posted in Brown University, Expository, Mathematics, Story | Tagged , , , , , , , , , , , , , , | Leave a comment