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Math 100 Fall 2016: Concluding Remarks

It is that time of year. Classes are over. Campus is emptying. Soon it will be mostly emptiness, snow, and grad students (who of course never leave).

I like to take some time to reflect on the course. How did it go? What went well and what didn’t work out? And now that all the numbers are in, we can examine course trends and data.

Since numbers are direct and graphs are pretty, let’s look at the numbers first.

Math 100 grades at a glance

Let’s get an understanding of the distribution of grades in the course, all at once.

box_plots

These are classic box plots. The center line of each box denotes the median. The left and right ends of the box indicate the 1st and 3rd quartiles. As a quick reminder, the 1st quartile is the point where 25% of students received that grade or lower. The 3rd quartile is the point where 75% of students received that grade or lower. So within each box lies 50% of the course.

Each box has two arms (or “whiskers”) extending out, indicating the other grades of students. Points that are plotted separately are statistical outliers, which means that they are $1.5 \cdot (Q_3 – Q_1)$ higher than $Q_3$ or lower than $Q_1$ (where $Q_1$ denotes the first quartile and $Q_3$ indicates the third quartile).

A bit more information about the distribution itself can be seen in the following graph.

violin_plot

Within each blob, you’ll notice an embedded box-and-whisker graph. The white dots indicate the medians, and the thicker black parts indicate the central 50% of the grade. The width of the colored blobs roughly indicate how many students scored within that region. [As an aside, each blob actually has the same area, so the area is a meaningful data point].

So what can we determine from these graphs? Firstly, students did extremely well in their recitation sections and on the homework. I am perhaps most stunned by the tightness of the homework distribution. Remarkably, 75% of students had at least a 93 homework average. Recitation scores were very similar.

I also notice some patterns between the two midterms and final. The median on the first midterm was very high and about 50% of students earned a score within about 12 points of the median. The median on the second midterm was a bit lower, but the spread of the middle 50% of students was about the same. However the lower end was significantly lower on the second midterm in comparison to the first midterm. The median on the final was significantly lower, and the 50% spread was much, much larger.

Looking at the Overall grade, it looks very similar to the distribution of the first midterm, except shifted a bit.

It is interesting to note that that Recitation (10%), Homework (20%), and the First Midterm (20%) accounted for 50% of the course grade; the Second Midterm (20%) and the Final (30%) accounted for the other 50% of the course grade. The Recitation, Homework, and First Midterm grades pulled the Overall grade distribution up, while the Second Midterm and Final pulled the Overall grade distribution down.

Correlation between assignments and Overall Grade

I post the question: was any individual assignment type a good predictor of the final grade? For example, to what extent can we predict your final grade based on your First Midterm grade?

hw_vs_overall

No, doing well on homework is a terrible predictor of final grade. The huge vertical cluster of dots indicates that the overall grades vary significantly over a very small amount of homework. However, I note that doing poorly on homework is a great predictor of doing poorly overall. No one whose homework average was below an 80 got an A in the course. Having a homework grade below a 70 is a very strong indicator of failing the course. In terms of Pearson’s R correlation, one might say that about 40% of overall performance is predicted from performance on homework (which is very little).

Although drastic, this is in line with my expectations for calculus courses. This is perhaps a bit more extreme than normal — the level of clustering in the homework averages is truly stunning. Explaining this is a bit hard. It is possible to get homework help from the instructor or TA, or to work with other students, or to get help from the Math Resource Center or other tutoring. It is also possible to cheat, either with a solutions manual (which I know some students have), or a paid answer service (which I also witnessed), or to check answers on a computer algebra system like WolframAlpha. Each of these weakens the relationship between homework as an indicator of mastery.

In the calculus curriculum at Brown, I think it’s safe to say that homework plays a formative role instead of a normative role. It serves to provide opportunities for students to work through and learn material, and we don’t expect the grades to correspond strongly to understanding. To that end, half of the homework isn’t even collected.

m1_vs_overall

m2_vs_overall

The two midterms each correlate pretty strongly with Overall grade. In particular, the second midterm visually indicates really strong correlation. Statistically speaking (i.e. from Pearson’s R), it turns out that 67% of the Overall grade can be predict from the First Midterm (higher than might be expected) and 80% can be predicted from the Second Midterm (which is really, really high).

If we are willing to combine some pieces of information, the Homework and First Midterm (together) predict 77% of the Overall grade. As each student’s initial homework effort is very indicative of later homework, this means that we can often predict a student’s final grade (to a pretty good accuracy) as early as the first midterm.

(The Homework and the Second Midterm together predict 85% of the Overall grade. The two midterms together predict 88% of the Overall grade.)

This has always surprised me a bit, since for many students the first midterm is at least partially a review of material taught before. However, this course is very cumulative, so it does make sense that doing poorly on earlier tests indicates a hurdle that must be overcome in order to succeed on later tests. This is one of the unforgiving aspects of math, the sciences, and programming — early disadvantages compound. I’ve noted roughly this pattern in the past as well.

final_vs_overall

However the correlation between the Final and the Overall grade is astounding. I mean, look at how much the relationship looks like a line. Even the distributions (shown around the edges) look similar. Approximately 90% of the Overall grade is predicted by the grade on the Final Exam.

This is a bit atypical. One should expect a somewhat high correlation, as the final exam is cumulative and covers everything from the course (or at least tries to). But 90% is extremely high.

I think one reason why this occurred this semester is that the final exam was quite hard. It was distinctly harder than the midterms (though still easier than many of the homework problems). A hard final gives more opportunities for students who really understand the material to demonstrate their mastery. Conversely, a hard final punishes students with only a cursory understanding. Although stressful, I’ve always been a fan of exams that are difficult enough to distinguish between students, and to provide a chance for students to catch up. See Chances for a Comeback below for more on this.

Related statistics of interest might concern to what extent performance on the First Midterm predicts performance on the Second Midterm (44%) or the Final Exam (48%), or to what extent the Second Midterm predicts performance on the Final Exam (63%).

Homework and Recitations

hw_vs_m1As mentioned above, homework performance is a terrible predictor of course grade. I thought it was worth diving into a bit more. Does homework performance predict anything well? The short answer is not really.

Plotting Homework grade vs the First Midterm shows such a lack of meaning that it doesn’t even make sense to try to draw a line of best fit.

To be fair, homework is a better predictor of performance on the Second Midterm and Final Exam, but it’s still very bad.

rec_vs_hwHere’s a related question: what about Recitation sections? Are these good predictors of any other aspect of the course?

Plotting Recitation vs Homework is sort of interesting. Evidently, most people did very well on both homework and recitation. It is perhaps no surprise that most students who did very well in Recitation also did very well on their Homework, and vice versa. However it turns out that there are more people with high recitation grades and low homework grades than the other way around. But thinking about it, this makes sense.

These distributions are so tight that it still doesn’t make sense to try to draw a line of best fit or to talk about Pearson coefficients – most variation is simply too small to be meaningful.

Together, Homework and Recitation predict a measly 50% of the Overall grade of the course (in the Pearson’s R sense). One would expect more, as Homework and Recitation are directly responsible for 30% of the Overall grade, and one would expect homework and recitation to correlate at least somewhat meaningfully with the rest of graded content of the course, right?

I guess not.

So what does this mean about recitation and homework? Should we toss them aside? Does something need to be changed?

I would say “Not necessarily,” as it is important to recognize that not all grades are equal. Both homework and recitation are the places for students to experiment and learn. Recitations are supposed to be times where students are still learning material. They are to be inoffensive and safe, where students can mess up, fall over, and get back up again with the help of their peers and TA. I defend the lack of stress on grade or challenging and rigorous examination during recitation.

Homework is sort of the same, and sort of completely different. What gives me pause concerning homework is that homework is supposed to be the barometer by which students can measure their own understanding. When students ask us about how they should prepare for exams, our usual response is “If you can do all the homework (including self-check) without referencing the book, then you will be well-prepared for the exam.” If homework grade is such a poor predictor of exam grades, then is it possible that homework gives a poor ruler for students to measure themselves by?

I’m not sure. Perhaps it would be a good idea to indicate all the relevant questions in the textbook so that students have more problems to work on. In theory, students could do this themselves (and for that matter, I’m confident that such students would do very well in the course). But the problem is that we only cover a subset of the material in most sections of the textbook, and many questions (even those right next to ones we assign) require ideas or concepts that we don’t teach.

On the other hand, learning how to actually learn is a necessary skill, and probably one that most people struggle with when they first actually have to learn it. It’s necessary to learn it sooner or later.

Chances for a Comeback

The last numerical aspect I’ll consider is about whether or not it is possible to come back after doing badly on an earlier assessment. There are two things to consider: whether it is actually feasible or not, and whether any students did make it after a poor initial/early performance.

As to whether it is possible, the answer is yes (but it may be hard). And the reason why is that the Second Midterm and Final grades were each relatively low. It may be counterintuitive, but in order to return from a failing grade, it is necessary that there be enough room to actually come back.

Suppose Aiko is a pretty good student, but it just so happens that she makes a 49 on the first midterm due to some particular misunderstanding. If the class average on every assessment is a 90, then Aiko cannot claw her way back. That is, even if Aiko makes a 100 on everything else, Aiko’s final grade would be below a 90, and thus below average. Aiko would probably make a B.

In this situation, the class is too easy, and thus there are no chances for students to overcome a setback on any single exam.

On the other hand, suppose that Bilal makes a 49 on the first midterm, but that the class average is a 75 overall. If Bilal makes a 100 on everything else, Bilal will  end with just below a 90, significantly above the class average. Bilal would probably make an A.

In this course, the mean overall was a 78, and the standard deviation was about 15. In this case, an 89 would be an A. So there was enough space and distance to overcome even a disastrous exam.

But, did anyone actually do this? The way I like to look at this is to look at changes in relative performance in terms of standard deviations away from the mean. Performing at one standard deviation below the mean on Midterm 1 and one standard deviation above the mean on Midterm 2 indicates a more meaningful amount of grade fluidity than merely looking at points above or below the mean

m1_vs_m2_stddevs

Looking at the First Midterm vs the Second Midterm, we see that there is a rough linear relationship (Pearson R suggests 44% predictive value). That’s to be expected. What’s relevant now are the points above or below the line $y = x$. To be above the line $y = x$ means that you did better on the Second Midterm than you did on the First Midterm, all in comparison to the rest of the class. To be below means the opposite.

Even more relevant is to be in the Fourth Quadrant, which indicates that you did worse than average on the first midterm and better than average on the second. Looking here, there is a very healthy amount of people who are in the Fourth Quadrant. There are many people who changed by 2 standard deviations in comparison to the mean — a very meaningful change. [Many people lost a few standard deviations too – grade mobility is a two way street].

m1_vs_overall_stddevsm2_vs_overall_stddevs

The First Midterm to the Overall grade shows healthy mobility as well.

The Second Midterm to Overall shows some mobility, but it is interesting that more people lost ground (by performing well on the Second Midterm, and then performing badly Overall) than gained ground (by performing badly on the Second Midterm, but performing well Overall).

Although I don’t show the plots, this trend carries through pretty well. Many people were able to salvage or boost a letter grade based solely on the final (and correspondingly many people managed to lose just enough on the final to drop a letter grade). Interestingly, very many people were able to turn a likely B into an A through the final.

So overall, I would say that it was definitely possible to salvage grades this semester.

If you’ve never thought about this before, then keep this in mind the next time you hear complaints about a course with challenging exams — it gives enough space for students to demonstrate sufficient understanding to make up for a bad past assessment.

Non-Numerical Reflection

The numbers tell some characteristics of the class, but not the whole story.

Standard Class Materials

We used Thomas’ Calculus. I think this is an easy book to teach from, and relatively easy to read. It feels like many other cookie-cutter calculus books (such as Larson and Edwards or Stewart). But it’s quite expensive for students. However, as we do not use an electronic homework component (which seems to be becoming more popular elsewhere), at least students can buy used copies (or use other methods of procural).

However, solutions manuals are available online (I noticed some students had copies). Some of the pay-for sites have complete (and mostly but not entirely correct) provided solutions manuals as well. This makes some parts of Thomas challenging to use, especially as we do not write our own homework to give. I suppose that this is a big reason why one might want to use an electronic system.

The book has much more material in it than we teach. For instance, the book includes all of a first semester of calculus, and also more details in many sections. We avoid numerical integration, Fourier series, some applications, some details concerning polar and parametric plots, etc. Ideally, there would exist a book catering to exactly our needs. But there isn’t, so I suppose Thomas is about as good as any.

Additional Course Materials

I’ve now taught elementary calculus for a few years, and I’m surprised at how often I am able to reuse two notes I wrote years ago, namely the refresher on first semester calculus (An Intuitive Introduction to Calculus) and my additional note on Taylor series (An Intuitive Overview of Taylor Series). Perhaps more surprisingly, I’m astounded at how often people from other places link to and visit these two notes (and in particular, the Taylor Series note).

These were each written for a Math 100 course in 2013. So my note to myself is that there is good value in writing something well enough that I can reuse it, and others might even find it valuable.

Unfortunately, while I wrote a few notes this semester, I don’t think that they will have the same lasting appeal. The one I wrote on the series convergence tests is something that (perhaps after one more round of editing) I will use each time I teach this subject in the future. I’m tremendously happy with my note on computing $\pi$ with Math 100 tools, but as it sits outside the curriculum, many students won’t actually read it. [However many did read it, and it generated many interesting conversations about actual mathematics]. Perhaps sometime I will teach a calculus class ending with some sort of project, as computing $\pi$ leads to very many interested and interrelated project thoughts.

Course Content

I must admit that I do not know why this course is the way it is, and this bothers me a bit. In many ways, this course is a grab bag of calculus nuggets. Presumably each piece was added in because it is necessary in sufficiently many other places, or is so directly related to the “core material” of this course, that it makes sense to include it. But from what I can tell, these reasons have been lost to the sands of time.

The core material in this course are: Integration by Parts, Taylor’s Theorem, Parametric and Polar coordinates, and First Order Linear Differential Equations. We also spend a large amount of time towards other techniques of integration (partial fraction decomposition, trig substitution) and understanding generic series (including the various series convergence/divergence tests). Along the way, there are some seemingly arbitrary decisions on what to include or exclude. For instance, we learn how to integrate

$$ \int \sin^n x \cos^m x \; dx$$

because we have decided that being able to perform trigonometric substitution in integrals is a good idea. But we omit integrals like

$$ \int \sin(nx) \sin(mx) \; dx$$

which would come up naturally in talking about Fourier series. Fourier series fit naturally into this class, and in some variants of this class they are taught. But so does trigonometric substitution! So what is the rationale here? If the answer is to become better at problem solving or to develop mathematical maturity, then I think it would be good to recognize that so that we know what we should feel comfortable wiggling to build and develop the curriculum in the future. [Also, students should know that calculus is not a pinnacle. See for instance this podcast with Steven Strogatz on Innovation Hub.]

This is not restricted to Brown. I’m familiar with the equivalent of this course at other institutions, and there are similar seemingly arbitrary differences in what to include or exclude. For years at Georgia Tech, they tossed in a several week unit on linear algebra into this course [although I’ve learned that they stopped that in the past two years]. The AP Calc BC curriculum includes trig substitution but not Fourier series. Perhaps they had a reason?

What this means to me is that the intent of this course has become muddled, and separated from the content of the course. This is an overwhelmingly hard task to try to fix, as a second semester of calculus fits right in the middle of so many other pieces. Yet I would be very grateful to the instructor who sits down and identifies reasons for or against inclusion of the various topics in this course, or perhaps cuts the calculus curriculum into pieces and rearranges them to fit modern necessities.

A Parachute is only necessary to go skydiving twice

This is the last class I teach at Brown as a graduate student (and most likely, ever). Amusingly, I taught it in the same room as the first course I taught as a graduate student. I’ve learned quite a bit about teaching inbetween, but in many ways it feels the same. Just like for students, the only scary class is the first one, although exams can be a real pain (to take, or to grade).

It’s been a pleasure. As usual, if you have any questions, please let me know.

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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.

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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.

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Notes on the first week (SummerNT)

We’ve covered a lot of ground this first week! I wanted to provide a written summary, with partial proof, of what we have done so far.

We began by learning about proofs. We talked about direct proofs, inductive proofs, proofs by contradiction, and proofs by using the contrapositive of the statement we want to prove. A proof is a justification and argument based upon certain logical premises (which we call axioms); in contrast to other disciplines, a mathematical proof is completely logical and can be correct or incorrect.

We then established a set of axioms for the integers that would serve as the foundation of our exploration into the (often fantastic yet sometimes frustrating) realm of number theory. In short, the integers are a non-empty set with addition and multiplication [which are both associative, commutative, and have an identity, and which behave as we think they should behave; further, there are additive inverses], a total order [an integer is either bigger than, less than, or equal to any other integer, and it behaves like we think it should under addition and multiplication], and satisfying the deceptively important well ordering principle [every nonempty set of positive integers has a least element].

With this logical framework in place, we really began number theory in earnest. We talked about divisibility [we say that $latex a$ divides $latex b$, written $latex a mid b$, if $latex b = ak$ for some integer $latex k$]. We showed that every number has a prime factorization. To do this, we used the well-ordering principle.

Suppose that not all integers have a prime factorization. Then there must be a smallest integer that does not have a prime factorization: call it $latex n$. Then we know that $latex n$ is either a prime or a composite. If it’s prime, then it has a prime factorization. If it’s composite, then it factors as $latex n = ab$ with $latex a,b < n$. But then we know that each of $latex a, b$ have prime factorizations since they are less than $latex n$. Multiplying them together, we see that $latex n$ also has a prime factorization after all. $latex diamondsuit$

Our first major result is the following:

There are infinitely many primes

There are many proofs, and we saw 2 of them in class. For posterity, I’ll present three here.

First proof that there are infinitely many primes

Take a finite collection of primes, say $latex p_1, p_2, ldots, p_k$. We will show that there is at least one more prime not mentioned in the collection. To see this, consider the number $latex p_1 p_2 ldots p_k + 1$. We know that this number will factor into primes, but upon division by every prime in our collection, it leaves a remainder of $latex 1$. Thus it has at least one prime factor different than every factor in our collection. $latex diamondsuit$

This was a common proof used in class. A pattern also quickly emerges: $latex 2 + 1 = 3$, a prime. $latex 2cdot3 + 1 = 7$, a prime. $latex 2 cdot 3 cdot 5 + 1 = 31$, also a prime. It is always the case that a product of primes plus one is another prime? No, in fact. If you look at $latex 2 cdot 3 cdot 5 cdot 7 cdot 11 cdot 13 + 1=30031 = 59cdot 509$, you get a nonprime.

Second proof that there are infinitely many primes

In a similar vein to the first proof, we will show that there is always a prime larger than $latex n$ for any positive integer $latex n$. To see this, consider $latex n! + 1$. Upon dividing by any prime less than $latex n$, we get a remainder of $latex 1$. So all of its prime factors are larger than $latex n$, and so there are infinitely many primes. $latex diamondsuit$

I would also like to present one more, which I’ve always liked.

Third proof that there are infinitely many primes

Suppose there are only finitely many primes $latex p_1, ldots, p_k$. Then consider the two numbers $latex n = p_1 cdot dots cdot p_k$ and $latex n -1$. We know that $latex n – 1$ has a prime factor, so that it must share a factor $latex P$ with $latex n$ since $latex n$ is the product of all the primes. But then $latex P$ divides $latex n – (n – 1) = 1$, which is nonsense; no prime divides $latex 1$. Thus there are infinitely many primes. $latex diamondsuit$

We also looked at modular arithmetic, often called the arithmetic of a clock. When we say that $latex a equiv b mod m$, we mean to say that $latex m | (b – a)$, or equivalently that $latex a = b + km$ for some integer $latex m$ (can you show these are equivalent?). And we pronounce that statement as ” $latex a$ is congruent to $latex b$ mod $latex m$.” We played a lot with modular arithmetic: we added, subtracted, and multiplied many times, hopefully enough to build a bit of familiarity with the feel. In most ways, it feels like regular arithmetic. But in some ways, it’s different. Looking at the integers $latex mod m$ partitions the integers into a set of equivalence classes, i.e. into sets of integers that are congruent to $latex 0 mod m, 1 mod m, ldots$. When we talk about adding or multiplying numbers mod $latex mod m$, we’re really talking about manipulating these equivalence classes. (This isn’t super important to us – just a hint at what’s going on beneath the surface).

We expect that if $latex a equiv b mod m$, then we would also have $latex ac equiv bc mod m$ for any integer $latex c$, and this is true (can you prove this?). But we would also expect that if we had $latex ac equiv bc mod m$, then we would necessarily have $latex a equiv b mod m$, i.e. that we can cancel out the same number on each side. And it turns out that’s not the case. For example, $latex 4 cdot 2 equiv 4 cdot 5 mod 6$ (both are $latex 2 mod 6$), but ‘cancelling the fours’ says that $latex 2 equiv 5 mod 6$ – that’s simply not true. With this example in mind, we went about proving things about modular arithmetic. It’s important to know what one can and can’t do.

One very big and important observation that we noted is that it doesn’t matter what order we operate, as in it doesn’t matter if we multiply an expression out and then ‘mod it’ down, or ‘mod it down’ and then multiply, or if we intermix these operations. Knowing this allows us to simplify expressions like $latex 11^4 mod 12$, since we know $latex 11 equiv -1 mod 12$, and we know that $latex (-1)^2 equiv 1 mod 12$, and so $latex 11^4 equiv (-1)^{2cdot 2} equiv 1 mod 12$. If we’d wanted to, we could have multiplied it out and then reduced – the choice is ours!

Amidst our exploration of modular arithmetic, we noticed some patterns. Some numbers  are invertible in the modular sense, while others are not. For example, $latex 5 cdot 5 equiv 1 mod 6$, so in that sense, we might think of $latex frac{1}{5} equiv 5 mod 6$. More interestingly but in the same vein, $latex frac{1}{2} equiv 6 mod 11$ since $latex 2 cdot 6 equiv 1 mod 11$. Stated more formally, a number $latex a$ has a modular inverse $latex a^{-1} mod m$ if there is a solution to the modular equation $latex ax equiv 1 mod m$, in which case that solution is the modular inverse. When does this happen? Are these units special?

Returning to division, we think of the greatest common divisor. I showed you the Euclidean algorithm, and you managed to prove it in class. The Euclidean algorithm produces the greatest common divisor of $latex a$ and $latex b$, and it looks like this (where I assume that $latex b > 1$:

$latex b = q_1 a + r_1$

$latex a = q_2 r_1 + r_2$

$latex r_1 = q_3 r_2 + r_3$

$latex cdots$

$latex r_k = q_{k+2}r_{k+1} + r_{k+2}$

$latex r_{k+1} = q_{k+3}r_{k+2} + 0$

where in each step, we just did regular old division to guarantee a remainder $latex r_i$ that was less than the divisor. As the divisors become the remainders, this yields a strictly decreasing remainder at each iteration, so it will terminate (in fact, it’s very fast). Further, using the notation from above, I claimed that the gcd of $latex a$ and $latex b$ was the last nonzero remainder, in this case $latex r_{k+2}$. How did we prove it?

Proof of Euclidean Algorithm

Suppose that $latex d$ is a common divisor (such as the greatest common divisor) of $latex a$ and $latex b$. Then $latex d$ divides the left hand side of $latex b – q_1 a = r_1$, and thus must also divide the right hand side. So any divisor of $latex a$ and $latex b$ is also a divisor of $latex r_1$. This carries down the list, so that the gcd of $latex a$ and $latex b$ will divide each remainder term. How do we know that the last remainder is exactly the gcd, and no more? The way we proved it in class relied on the observation that $latex r_{k+2} mid r_{k+1}$. But then $latex r_{k+2}$ divides the right hand side of $latex r_k = q_{k+2} r_{k+1} + r_{k+2}$, and so it also divides the left. This also carries up the chain, so that $latex r_{k+2}$ divides both $latex a$ and $latex b$. So it is itself a divisor, and thus cannot be larger than the greatest common divisor. $latex diamondsuit$

As an aside, I really liked the way it was proved in class. Great job!

The Euclidean algorithm can be turned backwards with back-substitution (some call this the extended Euclidean algorithm,) to give a solution in $latex x,y$ to the equation $latex ax + by = gcd(a,b)$. This has played a super important role in our class ever since. By the way, though I never said it in class, we proved Bezout’s Identity along the way (which we just called part of the Extended Euclidean Algorithm). This essentially says that the gcd of $latex a$ and $latex b$ is the smallest number expressible in the form $latex ax + by$. The Euclidean algorithm has shown us that the gcd is expressible in this form. How do we know it’s the smallest? Observe again that if $latex c$ is a common divisor of $latex a$ and $latex b$, then $latex c$ divides the left hand side of $latex ax + by = d$, and so $latex c mid d$. So $latex d$ cannot be smaller than the gcd.

This led us to explore and solve linear Diophantine equations of the form $latex ax + by = c$ for general $latex a,b,c$. There will be solutions whenever the $latex gcd(a,b) mid c$, and in such cases there are infinitely many solutions (Do you remember how to see infinitely many other solutions?).

Linear Diophantine equations are very closely related a linear problems in modular arithmetic of the form $latex ax equiv c mod m$. In particular, this last modular equation is equivalent to $latex ax + my = c$ for some $latex y$.(Can you show that these are the same?). Using what we’ve learned about linear Diophantine equations, we know that $latex ax equiv c mod m$ has a solution iff $latex gcd(a,m) mid c$. But now, there are not infinitely many incongruent (i.e. not the same $latex mod m$) solutions. This is called the ‘Linear Congruence Theorem,’ and is interestingly the first major result we’ve learned with no proof on wikipedia.

Theorem: the modular equation $latex ax equiv b mod m$ has a solution iff $latex gcd(a,m) mid b$, in which case there are exactly $latex gcd(a,m)$ incongruent solutions.

Proof

We can translate a solution of $latex ax equiv b mod m$ into a solution of $latex ax + my = b$, and vice-versa. So we know from the Extended Euclidean algorithm that there are only solutions if $latex gcd(a,m) mid b$. Now, let’s show that there are $latex gcd(a,m)$ solutions. I will do this a bit differently than how we did it in class.

First, let’s do the case when $latex gcd(a,m)=1$, and suppose we have a solution $latex (x,y)$ so that $latex ax + my = b$. If there is another solution, then there is some perturbation we can do by shifting $latex x$ by a number $latex x’$ and $latex y$ by a number $latex y’$ that yields another solution looking like $latex a(x + x’) + m(y + y’) = b$. As we already know that $latex ax + my = b$, we can remove that from the equation. Then we get simply $latex ax’ = -my’$. Since $latex gcd(m,a) = 1$, we know (see below the proof) that $latex m$ divides $latex x’$. But then the new solution $latex x + x’ equiv x mod m$, so all solutions fall in the same congruence class – the same as $latex x$.

Now suppose that $latex gcd(a,m) = d$ and that there is a solution. Since there is a solution, each of $latex a,m,$ and $latex b$ are divisible by $latex d$, and we can write them as $latex a = da’, b = db’, m = dm’$. Then the modular equation $latex ax equiv b mod m$ is the same as $latex da’ x equiv d b’ mod d m’$, which is the same as $latex d m’ mid (d b’ – d a’x)$. Note that in this last case, we can remove the $latex d$ from both sides, so that $latex m’ mid b’ – a’x$, or that $latex a’x equiv b mod m’$. From the first case, we know this has exactly one solution mod $latex m’$, but we are interested in solutions mod $latex m$. Just as knowing that $latex x equiv 2 mod 4$ means that $latex x$ might be $latex 2, 6, 10 mod 12$ since $latex 4$ goes into $latex 12$ three times, $latex m’$ goes into $latex m$ $latex d$ times, and this gives us our $latex d$ incongruent solutions. $latex diamondsuit.$

I mentioned that we used the fact that we’ve proven 3 times in class now in different forms: if $latex gcd(a,b) = 1$ and $latex a mid bc$, then we can conclude that $latex a mid c$. Can you prove this? Can you prove this without using unique factorization? We actually used this fact to prove unique factorization (really we use the statement about primes: if $latex p$ is a prime and $latex p mid ab$, then we must have that $latex p mid a$ or $latex p mid b$, or perhaps both). Do you remember how we proved that? We used the well-ordered principle to say that if there were a positive integer that couldn’t be uniquely factored, then there is a smaller one. But choosing two of its factorizations, and finding a prime on one side – we concluded that this prime divided the other side. Dividing both sides by this prime yielded a smaller (and therefore unique by assumption) factorization. This was the gist of the argument.

The last major bit of the week was the Chinese Remainder Theorem, which is awesome enough (and which I have enough to say about) that it will get its own post – which I’m working on now.

I’ll see you all in class tomorrow.

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A proof from the first sheet (SummerNT)

In class today, we were asked to explain what was wrong with the following proof:

Claim: As $latex x$ increases, the function

$latex displaystyle f(x)=frac{100x^2+x^2sin(1/x)+50000}{100x^2}$

approaches (gets arbitrarily close to) 1.

Proof: Look at values of $latex f(x)$ as $latex x$ gets larger and larger.

$latex f(5) approx 21.002$
$latex f(10)approx 6.0010$
$latex f(25)approx 1.8004$
$latex f(50)approx 1.2002$
$latex f(100) approx 1.0501$
$latex f(500) approx 1.0020$

These values are clearly getting closer to 1. QED

Of course, this is incorrect. Choosing a couple of numbers and thinking there might be a pattern does not constitute a proof.

But on a related note, these sorts of questions (where you observe a pattern and seek to prove it) can sometimes lead to strongly suspected conjectures, which may or may not be true. Here’s an interesting one (with a good picture over at SpikedMath):

Draw $latex 2$ points on the circumference of a circle, and connect them with a line. How many regions is the circle divided into? (two). Draw another point, and connect it to the previous points with a line. How many regions are there now? Draw another point, connecting to the previous points with lines. How many regions now? Do this once more. Do you see the pattern? You might even begin to formulate a belief as to why it’s true.

But then draw one more point and its lines, and carefully count the number of regions formed in the circle. How many circles now? (It doesn’t fit the obvious pattern).

So we know that the presented proof is incorrect. But lets say we want to know if the statement is true. How can we prove it? Further, we want to prove it without calculus – we are interested in an elementary proof. How should we proceed?

Firstly, we should say something about radians. Recall that at an angle $latex theta$ (in radians) on the unit circle, the arc-length subtended by the angle $latex theta$ is exactly $latex theta$ (in fact, this is the defining attribute of radians). And the value $latex sin theta$ is exactly the height, or rather the $latex y$ value, of the part of the unit circle at angle $latex theta$. It’s annoying to phrase, so we look for clarification at the hastily drawn math below:

Screenshot from 2013-06-24 12:30:53
The arc length subtended by theta has length theta. The value of sin theta is the length of the vertical line in black.

Note in particular that the arc length is longer than the value of $latex sin theta$, so that $latex sin theta < theta$. (This relies critically on the fact that the angle is positive). Further, we see that this is always true for small, positive $latex theta$. So it will be true that for large, positive $latex x$, we’ll have $latex sin frac{1}{x} < frac{1}{x}$. For those of you who know a bit more calculus, you might know that in fact, $latex sin(frac{1}{x}) = frac{1}{x} – frac{1}{x^33!} + O(frac{1}{t^5})$, which is a more precise statement.

What do we do with this? Well, I say that this allow us to finish the proof.

$latex dfrac{100x^2 + x^2 sin(1/x) + 50000}{100x^2} leq dfrac{100x^2 + x + 50000}{100x^2} = 1 + dfrac{1}{100x} + dfrac{50000}{100x^2}$

, and it is clear that the last two terms go to zero as $latex x$ increases. $latex spadesuit$

Finally, I’d like to remind you about the class webpage at the left – I’ll see you tomorrow in class.

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Math 90: Concluding Remarks

All is said and done with Math 90 for 2012, and the year is coming to a close. I wanted to take this moment to write a few things about the course, what seemed to go well and what didn’t, and certain trends in the course. that I think are interesting and illustrative.

First, we might just say some of the numbers. Math 90 is offered only as pass/fail, with the possibility of ‘passing with distinction’ if you did exceptionally well (I’ll say what that meant here, though who knows what it means in general). We had four people fail, three people ‘pass with distinction,’ and everyone else got a passing mark. Everything else will be after the fold.

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Math 90: Week 8

Today, we had a set of problems as usual, and a quiz! (And I didn’t tell you about the quiz, even though others did, so I’m going to pretend that it was a pop quiz)!. Below, you’ll find the three problems, their solutions, and a worked-out quiz.

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