In an earlier post, I closed by hinting at the mathematics of ranking. In this modern era, the topic is particularly relevant: the ranking algorithms are hard at work whenever you type something into a search engine, rate a movie on Netflix, or look at a product on Amazon. It’s also a popular area of study among sports enthusiasts, for whom accurate rankings of the relative strengths of teams can make all the difference in a fantasy league or a betting pool.
Because of all of these accessible applications, it should come as no surprise that the mathematics of ranking is the subject of a new book, titled Who’s #1? The Science of Rating and Ranking. Written by applied mathematicians Amy N. Langville and Carl D. Meyer, the book tackles a variety of methods used to extract ratings or rankings given some collection of input data.
Front Cover!
This book is chock full of information. The first part of the book discusses several different algorithms used to construct rankings. There are plenty of applications of the type I mentioned above, though the authors seem to be particularly fond of sports. They keep one running example from the 2005 NCAA football season running throughout; viewed through the lens of this common example, it’s easier to compare and contrast the ranking and rating methods they discuss.
They also devote a good amount of space to tweaking general ranking methods to suit your particular needs. Want a football ranking system that weighs late season games more heavily than early season games? Want to understand what you should do to break ties in your data (if anything)? The authors illustrate a variety of alterations that can be made to the basic framework they provide. These changes allow for a countless supply of ways to rank a given collection of data, each with its own strengths and weaknesses.
In the last part of the book, the authors focus on the ranked lists obtained from the methods they’ve previously discussed. How can you best aggregate information from a number of ranked lists to create one super ranked list? This is a question faced by metasearch engines, for example. Also, how can you measure the difference between two ranked lists, and is it possible to objectively declare one ranked list “better” than another? These are the types of questions addressed in the latter part of the book.
Examples are sprinkled liberally throughout, and the authors provide plenty of references for data junkies looking to apply the methods discussed to data that interests them. I will give one such example in a follow-up post, focused on my favorite baseball team, to give you a better sense of what this book is all about. Here’s Google’s explanation of how internet search works, if that’s more up your alley.
Some caveats: a bit of mathematical sophistication will help if you really want to appreciate this book. If the only matrix you’re familiar with is the one with Keanu Reeves in it, for example, you may be in trouble. But if you’ve taken a Linear Algebra course or two, you should have no trouble following the methods described in the text. And if you’re a teacher looking for a way to make an undergraduate course more relevant, this book may be able to help. Do you have students who are in to sports, for example? It’s no coincidence the book was released in time for March Madness – since ESPN offers $10,000 to the most accurate bracket each year, the right application of the methods in this book could bring in a tidy profit.
My one complaint is the book’s discussion of Arrow’s Impossibility Theorem, a result from social choice theory most commonly affiliated with voting systems. Since a voting rule is really just a rule to aggregate a large collection of rankings (i.e. ballots), it arises fairly naturally in their discussions a couple of times. I’ve discussed this theorem before, so won’t spend much time on it here, but the theorem essentially states that any reasonably constructed voting system can’t satisfy a set of common-sense criteria that one would like any voting system to satisfy. The theorem, however, only applies to ranked voting systems, i.e. systems in which the voters provide a ranking of the candidates (e.g. Candidate A in 1st place, Candidate B in 2nd place, Candidate C in 3rd place). What the authors fail to mention, however, is that Arrow’s theorem need not apply to rated voting systems, i.e. systems in which the votes provide a rating of the candidates (e.g. Candidate A 100 points, Candidate B 20 points, Candidate C 0 points). Since the authors discuss the importance of the distinction between ratings and rankings, it’s unfortunate that they overlook this point when discussing Arrow’s theorem.
Experts in the field may also be upset if their favorite method for ranking is not included, but the authors do a good job of providing honorable mentions in their epilogue. There’s plenty of material that they don’t discuss, but references are given throughout, so the interested reader can quite easily further pursue his or her own interests.
All in all, it’s quite an interesting read, so if you’re curious, I’d encourage you to check it out!
Some final remarks:
1. In the interest of full disclosure, this post is based on a review copy of the book I received from Princeton University Press.
2. If you’re interested in purchasing a copy of this book (or other books I’ve discussed on the site), please consider purchasing through the Amazon widget now displayed on the right hand side. This isn’t a big money making operation here, but if I could cover the annual site maintenance costs on the back of you building your library, it seems like a win-win!
If you follow me on Twitter, you may have noticed some activity over the past week in regards to a new Kickstarter project, Math52. From the creators of Mathalicious, this campaign has set the ambitious goal of raising $164,000 to help transform the way mathematics is taught in our schools. Every week for a year, they will release a video aimed at exploring mathematics through everyday questions – the types of questions that will immediately connect with students, and help motivate them to understand the math required to provide a reasonable answer. But don’t believe me, check out this video!
The team has raised over $12,000 in a week. This is amazing, but it’s not quite on track for them to reach their goal. So if you give a hoot, please consider donating to this most worthy of causes. Even better, if you have some wealthy friends, get them to donate too (friends of Bill Gates, I am speaking to you!).
If you can’t spare the change, then please consider spreading the word. If you’re on Twitter, tweet this project to your followers. If you’re on Facebook, share it with your friends. If you fear technology, deliver messages to your associates via carrier pigeon. (Though if you fear technology, what are you doing here? Leave now while you still can!)
I apologize for the solicitation, but I would not do it if I did not truly believe in this project. This approach to teaching and discussing mathematics dovetails quite nicely with what I have been doing on this blog for the past four years, so hopefully you will not be offended by this brief post.
If you’d like to donate, you can do it directly from the link below. Either way, thanks for reading.
This Saturday, folks from all over the country will be tuning in to the 138th Kentucky Derby. In fact, this year the Kentucky Derby falls on the same day as Cinco de Mayo; undoubtedly the result of this intersection will be a plethora of parties celebrating the melting pot that is America (tacos and mint juleps make for a wonderful combination, I’m sure).
Whenever racing comes up, mathematics can’t be far behind. Gambling is always a popular topic: how are the odds for the different racers determined, for example? But this is a question I will save for another time. Today, inspired by horse racing in particular, I’d like to discuss the following classic logic puzzle.
Suppose you have 25 horses and a 5 lane race track. You have no way to record the finishing times of the horses, but you can race up to 5 horses on the track at once and see how quickly the horses finish relative to one another. What’s the minimum number of races needed to find the three fastest horses?
If you had a stopwatch, you could find the three fastest horses by splitting the 25 horses into groups of 5, racing each group one at a time, and recording their times. In fact, this would provide you with a complete ranking of all the horses from the fastest to the slowest, rather than only an identification of the fastest three. Sadly, without any instruments to measure how quickly a horse completes a race around the track, all we can do is look and see which horses come in first when we race them 5 at a time.
If you haven’t seen this puzzle before, it’s fun to try and think through yourself. Beware, spoilers abound below!
Certainly, the number of races required is more than 5. The best thing to do, it seems, is still to split the horses up into five groups of five and race each group. This is the only way we can get each horse to race once during the first five races; otherwise, after five races there will still be some horses whose ability we know nothing about. So, let’s split the horses into groups of five and race each group. Suppose we then organize the results of the races in the following table:
25 horses, 5 horse races.
Here I’ve let each row represent a single race, and the horses are ranked by how they placed. For example, the first race consisted of horses in the top row. Within that race, the horse with the blue saddle came in first, followed by the horse with the yellow saddle, then the purple saddle, then the black saddle, with the horse in the dark blue saddle coming in last. We don’t know any of their individual racing times, and so we only know how a horse compares relative to the other horses in the same row.
After five races, what can we conclude? Well, since we want to identify the fastest three horses, we can easily eliminate the fourth and fifth place finishers in each race; each one of these horses is slower than at least three others. This allows us to narrow the field of horses from 25 down to 15.
Too slow!
Now within each row, we know faster horses are on the left and slower horses are on the right. But we have no way to compare horses across the columns. It may be that the first row consists of the 1st, 2nd, and 3rd fastest horses, but it’s just as likely that the first row consists of the 13th, 14th, and 15th fastest horses. Without being able to compare the columns, we can’t draw any further conclusions.
So, it makes sense to take a horse from each column and create a sixth race. It also makes sense to race the winners from all the previous races, because the winner among all the winners will be the fastest horse, and we will then only have to find the 2nd and 3rd fastest horses. After racing the first column of horses, suppose the ranking is as in the picture (i.e. blue horse in first, teal horse in second, orange in third, etc). If this is not the case, we can always interchange rows so that the first column is ranked according to the outcome of the sixth race. Our picture now looks like this:
The fastest horse is now in green.
We don’t need to race the horse in the upper left corner anymore, since we know it’s the fastest. In fact, we know even more: just as in the previous case, we can eliminate the fourth and fifth place finishers from the sixth race. But if we cross out the first horse in the fourth row, we can cross out all the horses to the right, since they are all slower! The same is true for the fifth row. This reduces the pool of horses from 15 to 9 (or rather, from 14 to 8, since we no longer need to race the fastest horse):
Only a few remain.
Are there any other horses we can eliminate? Thankfully, the answer is yes! We know that the horse on the left in the second row can, at best, be the second fastest horse. But this means the horses to the right of it can, at best, be the third and fourth fastest, respectively. Since we’re not interested in the fourth fastest horse, we can safely eliminate the third horse in the second row.
Similarly, the horse on the left in the third row can, at best, be the third fastest horse (since it was slower than the horses above it). This means the other horses in the third row can be no faster than the fourth and fifth fastest horses. Therefore, both horses can safely be eliminated. Removing these horses from contention leaves us with the following picture:
Almost there!
In other words, we have identified the fastest horse, and eliminated all but five remaining horses. There’s no way to determine the second and third fastest horses based on the information we have, but since our track has exactly five lanes, we can hold a seventh race among the five remaining horses. The first and second place finishers of this final race will be the second and third fastest horses overall.
So, the answer to this puzzle appears to be 7. Technically, I suppose I haven’t shown that it’s impossible to do in six races, but if you play around with these ideas you’ll quickly convince yourself that six races is insufficient. The nice thing about this problem is that it is easy to ask about generalizations: what happens if you change the number of horses, number of lanes, or number of rankings you wish to determine? Here‘s a discussion of a generalization to finding the best k horses out of horses total, if you have a race track with lanes. Though it doesn’t look like it, this is a fairly natural generalization of the case discussed here (in the above case, k = 3).
Of course, this is a fairly idealized situation, since it assumes the horses will finish the race with the same time whenever they run. In real life, there will be some variance in their finishing times, and in turn this will make the determination of the three “best” horses a bit more nuanced. This leads naturally into the mathematics of ranking, a topic I will turn my attention to in my next post!
In honor of this year’s Mathematics Awareness Month, titled “Mathematics, Statistics, and the Data Deluge,” I’ve contributed an article to CNN’s Light Years blog on how corporations might use big data to infer personal details about its customers. Mostly this was inspired by the recent New York Times investigation on how Target collects and uses customer data. Here’s an excerpt:
Whether you are trying to make the best decisions for your fantasy baseball league, looking to capitalize on an opportunity in a fluctuating stock market or simply filtering through the results of a Google search, it is hard to deny that we are surrounded by more data now than ever before. As such, the task of organizing and drawing conclusions from data can be a challenge, but thankfully mathematics can, in many cases, rise to the occasion.
My love of NBC comedies has, by now, beenwellestablished. Today I’d like to return to The Office, for although Steve Carrell’s absence may have hurt the ratings, it certainly has not diminished the potential for the show to inspire some mathematical thinking.
If you have not been watching recently, this season marked the debut of the company’s first ever tablet computer, dubbed the Pyramid. The Pyramid made its first appearance early in the season (and was also featured in on Wired), and has since been joined by a smartphone counterpart known as the Arrowhead. Here’s an image of Dwight touting the new tablet.
If you live in the States, you can also view the clip from which this image was taken:
On the face of it, the tablet is ridiculous (this fact is eventually sort of addressed later on in the season). Who wants to use a triangular shaped tablet to look at pictures or watch movies? The reason I bring this up is that from a strictly mathematical standpoint, there are reasons to view the design with quite a bit of skepticism.
Why is this the case? Well, let’s think about the practical matter of building a tablet like the one seen above. There is a certain cost involved in producing such a charming device; let us focus on the cost associated with the metal frame. Suppose the cost of the frame is in direct proportion to the amount of material used. If we want a large screen (and let’s face it, who doesn’t?), then we have a bit of a balancing act to play: we’d like to have a large screen, but we also want a small border so that we don’t have to spend a lot on material for the frame. In other words, if we have a certain amount of material for the frame, we would like to find the largest area possible that we could enclose with that particular length of material.
For ease of computation, let’s suppose the border of the pyramid is composed of a strip of metal one meter across, i.e. the perimeter of the pyramid is 1. If we are beholden to the “pyramid” gimmick, let’s ask: how can we make an isosceles triangle with the largest area if its perimeter must be 1 meter? I’m sticking to isosceles triangles only since we don’t usually think of pyramids as being lopsided.
Different isosceles triangles with perimeter equal to 1. The top angles are 30°, 60°, 90°, and 120°, respectively. Which do you think has the largest area?
If we really want to see which isosceles triangle with perimeter equal to one has the largest area, we need to find a formula for the area. One convenient way to organize such triangles is by the angle formed at the top of the triangle. For any angle between 0° and 180°, we can construct an isosceles triangle of perimeter 1 – what is the area as a function of the angle?
To answer this question, consider the following diagram:
A general isosceles triangle.
A general isosceles triangle has two sides of the same length, which we will denote x. Let’s cut the triangle in half, so that on the right hand side we have a right triangle with an angle . By the definition of sine and cosine, it follows that the line splitting the isosceles triangle in two has length , and half of the base of the isosceles triangle has length , so that the entire base of the isosceles triangle is twice as large, or .
We can use this to find the area of the triangle as a function of the angle . We know the perimeter of the triangle should equal 1, so from the above diagram this means
Solving for x, we then obtain
On the other hand, as any good geometry student should recall, the area of a triangle is equal to half the base times the height. Since the base and height can be read from the above figure, we see the area must therefore equal
So to find the largest area, we can plot the graph of this function and see where it attains a maximum (or, if you want to be fancy about it, you can use some calculus).
The graph of the above function looks like this (thanks, WolframAlpha):
The maximum occurs when equals 30° – in other words, the top angle of the isosceles triangle is 60°, which means the triangle itself is equilateral! So the largest area occurs for an equilateral triangle – in this case, the area equals , or roughly 0.0481.
But the Pyramid in The Office doesn’t look like an equilateral triangle. In fact, the top angle looks to be closer to a 90° angle. If this is the case, then would be 45°, and the area is but a meager , or around 0.0429. In other words, the designers of the Pyramid could have increased the area by roughly 12% with a better triangle design.
But why restrict to triangles in the first place? Let’s suppose the designers had instead decided to stick with convention and go with a tablet whose screen ratio was 4:3 (like the iPad, for example). Since the perimeter is 1, this would mean the length of the tablet would have to equal 4/14, and the height would have to equal 3/14. Since the area is the length times the width, this would give a total area of 12/196, or 3/49; this is around 0.612. In other words, this change would bring about a 27% increase in area compared to the optimal pyramid design, and a nearly 43% increase in area compared to the design featured on the show.
Of course, there’s nothing sacred about the 4:3 ratio – by considering an arbitrary rectangle, can you find the one that maximizes the area if the perimeter equals 1?
We can continue exploring this problem with a variety of geometric shapes. At some point, it is natural to ask: given a fixed perimeter of 1, what closed plane curve maximizes the area contained by the perimeter? This is the famous isoperimetric inequality, and it has a solution. Namely, if you have a closed figure of perimeter 1, the largest area you can enclose is , or approximately 0.0796. Moreover, you get this largest area precisely when the closed figure is a circle. Said another way, for a given perimeter, the circle encloses the largest possible area.
Had the designers of the Pyramid opted for a circular theme instead (e.g. the Orb), they could have increased the area of the screen by over 65% compared to using an equilateral triangle, and by over 85% compared to the designed used in the show! No wonder the folks on The Office are always worried about losing their jobs. Then again, it’s an open debate as to whether a triangular or circular tablet design is more useless.
I readily admit that this analysis is somewhat oversimplified. We’ve considered the cost of the tablet’s frame, for example, but completely ignored the cost of producing the screen itself. But sometimes in the interest of finding some nice mathematics, sacrifices must be made. I trust you will forgive me for this small indiscretion.
(Thanks to Karim from Mathalicious for the conversation which inspired this post!)
As we head into the final days of March, I’d like to share with you the following infographic sent to me by a reader. It collects some interesting (and depressing) data on women in STEM (Science, Technology, Engineering, and Mathematics) careers.
While I don’t necessarily put a lot of stock into the opening IQ numbers (see here and here for examples of problems with IQ testing), and I’m not sure if the data on course load in the first part is statistically significant, the data in the latter parts is quite compelling. I’ve discussed the psychological component of the gender gap before, but the data in the second section of this infographic provides more evidence to the claim that psychology and cultural influences, rather than biology, is behind the gender gap we see in the sciences.
Of course, anytime someone with the aptitude is dissuaded from pursuing a technical career, it is unfortunate. When a large number of people with the aptitude don’t pursue technical careers, we lose out on opportunities to innovate and discover. I’m no expert on what steps can be taken to try and close the gap, but if readers have any thoughts on the data or what should be done in response to it, feel free to sound off below.
Though we are still a few months away from the start of the summer blockbuster season, the scuttlebutt is that The Hunger Games, opening this weekend, is expected to do huge business (and by huge, I mean upwards of $100 million). Based on the 2008 Suzanne Collins book of the same name, this property is the hottest new thing in the realm of young adult fiction, and in this post-Harry Potter, nearly-post-Twilight era of cinema history, the timing could not be better for movie executives. The book is the first of a trilogy, so whether you like it or not, these films will be with us for the next few years.
Cover image for the book.
If you have not read the book, or have no idea what I’m talking about in general, a trailer for the film can be found here (sorry, embedding has been disabled for the video). The story takes place sometime in the future, many years after a war that has seemingly decimated the population. The United States is now broken into twelve districts, all under control of an totalitarian regime headquartered in the aptly-named Capitol. Every year the Capitol organizes the Hunger Games, in which one young man and one young woman between the ages of 12 and 18 is selected from each of the twelve districts to compete in a fight to the death (if this sounds a lot like a certain Japanese book/film combo to you, you’re not the only one). The story begins when 16 year old Katniss Evergreen volunteers to fight in the Hunger Games in place of her younger sister Primrose, who is selected at the age of 12.
This selection process is where the story intersects with mathematics. Take a look at this passage from the first chapter of the book.
You become eligible for the reaping the day you turn twelve. That year, your name is entered once. At thirteen, twice. And so on and so on until you reach the age of eighteen, the final year of eligibility, when your name goes into the pool seven times. That’s true for every citizen in all twelve districts in the entire country of Panem.
But here’s the catch. Say you are poor and starving as we were. You can opt to add your name more times in exchange for tesserae. Each tessera is worth a meager year’s supply of grain and oil for one person. You may do this for each of your family members as well. So, at the age of twelve, I had my name entered four times. Once, because I had to, and three times for tesserae for grain and oil for myself, Prim, and my mother. In fact, every year I have needed to do this. And the entries are cumulative. So now, at the age of sixteen, my name will be in the reaping 20 times.
The rules here practically beg for some mathematical analysis. Here are a few questions. What are the odds of being selected? How does the probability of a young adult being chosen over their lifetime increase with the number of tesserae she elects to receive?
To estimate values here, we need to know how many names are in the reaping. This is not such a trivial thing to estimate, since each child may have his or her name entered several times. To simplify our analysis, we will assume that (a) the number of names entered is the same for boys as it is for girls, and (b) this number is essentially the same from year to year. Our second assumption may not be terribly realistic, but since we’re analyzing a fictional game in a fictional dystopia, I hope you will suspend your disbelief with me for the moment.
We can estimate the number of female names entered by estimating the number of girls in District 12 between the ages of 12 and 18, and the average number of tessarae each girl requests. First let’s focus on the number of girls between 12 and 18. To start, it’s mentioned several times in the book that the population of District 12 is around 8,000 (see here for a fun estimate of the population of the entire country). According to this census data, a bit less than 20% of the US population was between the ages of 5 and 17 in both 2000 and 2010. But it’s a hard knock life in District 12, and the life expectancy is undoubtedly much lower than it is in present-day America. So let’s assume that the percentage of 5 to 18 year olds is closer to 35% (this article suggests such a figure is not unreasonable). That puts the number of 5 to 18 year olds in District 12 at 2,800. Cut that number in half to focus on 12 to 18 year olds, and cut it in half again to get rid of the boys, and we have approximately 700 girls in District 12 who are eligible to participate in the Hunger Games.
Assuming a roughly equal number of girls of each age (i.e. 100 12 year olds, 100 13 year olds, and so on), this would mean that each girl has her name entered an average of 4 times, ignoring tesserae (since (1 + 2 + 3 + 4 + 5 + 6 + 7)/7 = 4). But how many tesserae will each girl request on average? Though the number of upper-class citizens whose children don’t need to request tesserae is relatively small, it’s not the case that every poor child will request tesserae, since it makes the most sense for the oldest eligible child to shoulder the entire tesserae burden. Therefore, even in poor households, we should expect no more than one eligible child to request tesserae.
Throughout, we will consider three cases: where the average number of tesserae requested is 1, 1.5, and 2. With our assumptions and a bit of work, one can show that if the average number of requested tesserae is t, the average number of times each girl’s name is entered is 4(1+t). Therefore, the average number of times each girl’s name is entered in each case is 8, 10, and 12. Since the number of girls is 700, this means that the estimated number of names in each case is 5,600, 7,000, and 8,400. Call the number n, whatever it may be.
Now let’s suppose you are a girl and you plan on requesting t tesserae for your family every year you are eligible for the games. How does the probability of you getting picked increase with t? Well, when you are 12, the probability of being selected is , since there are n names entered, and t + 1 of them are yours. The probability of being in the games when you are 13 is equal to the probability of being selected when you are thirteen times the probability you were NOT selected when you were twelve (since, if you were selected when you were twelve, either you are dead or you were the winner of the previous Hunger Games – in either case, you won’t be playing again). Therefore, the probability is
Similarly, you can only be selected when you are 14 if you WEREN’T selected when you were 12 or 13, so the probability of being selected at 14 is
By now you should see a pattern emerging. Perform this calculation for every age from 12 to 18, add them all up, and in fancy math notation we find that the probability of being chosen is equal to
We can now use our above estimates for n to see how this number varies with t. For example, with n = 5600 and t = 0 (i.e. if we assume you never request tesserae), the probability you will be selected is 1/5600 + (5599/5600)*(2/5600) + (5599/5600)*(5598/5600)*(3/5600) + … . Adding up all these numbers gives a probability of roughly .499%.
Other combinations can be found in the following chart:
The x-axis represents number of tesserae requested. Red corresponds to n = 5600, green to n = 7000, blue to n = 8400.
The chart above tells us, for example, that if we use the estimate of n = 5600, then a girl requesting 3 tesserae every year has a roughly 2% probability of being selected for the Hunger Games over the 7 year span when she’s eligible. If n = 8400, that probability drops to around 1.3%.
In each case, we see the number of tesserae acts as a sort of multiplying factor. If you request one tesserae, you roughly double your chances of being selected. If you request two, you roughly triple your chances, and so on. It’s not hard to see why this should be the case, from what we’ve already done. For example, if n = 5600 and t = 0, the probability of being selected when you’re 13 is equal to (5599/5600)*(2/5600), but this is nearly the same as just 2/5600, since 5599/5600 is nearly 1. Similarly, the probability of being selected when you’re 13 is (5599/5600)*(5598/5600)*(3/5600), but this is again roughly equal to just 3/5600. By replacing all the probabilities of not being selected by 1, we see that the probability of being selected is approximately equal to just
or simply
This formulation is much simpler, and shows clearly how the probability increases with t. Note that this estimation only works when n is much larger than t; but since that always seems to be the case when it comes to the Hunger Games, we should be ok.
So, while Katniss increased the probability that she would be selected by a factor of 4 by planning to request 3 tesserae every year, in absolute terms, the probability that she would be selected is still relatively small. Even so, the probability that her younger sister would be selected in her first year is even smaller, and yet the whole series begins with the occurrence of this unlikely event. The moral here is clear: while you can try to hedge your bets, there is no surefire way to prevent a loved one from playing the Hunger Games. Unless, of course, you are willing to trade your fate for the one you love.
(If you’re interested in the raw data from the chart, you can find it below.)
In an attempt to spread the joy and cheer of mathematics to a broader audience, starting this month, I will occasionally be writing articles for CNN’s new science and technology blog, Light Years. Fear not, most of my content will still be appearing at Math Goes Pop, and every time one of my guest posts goes live, I will let you know about it here as well. Today the topic is voting systems, something I have discussed on this blog before. Here’s a piece of the intro to pique your interest:
When the results of an election (primary or otherwise) run counter to our desires, it is easy to scapegoat the political process. The right person didn’t win, we may argue, because the system itself is broken. The two-party system, for example, is sometimes cited as a leading cause of the current dysfunction in Washington. But perhaps much of what ails the political climate comes from an underlying mathematical dilemma in the way we determine the winners of our elections. The mathematics of voting highlights many problems with current systems, and also proposes some interesting solutions.
By now my views on Pi Day are well documented (see earlier posts from 2011 and 2009 if you’re curious). Recently, though, I’ve decided to try to be a little less curmudgeonly when it comes to math holidays. Consequently, while it would be easy to provide snarky commentary on articles with particularlyegregious mathematical errors, this year I will try to restrain myself.
As I’ve said before, one of my biggest problems with Pi day is that the activities are, for the most part, a little ridiculous, and don’t actually do anything to better the general populace’s understanding of mathematics. Last year, I explained why contests involving the recitation of digits of are silly, so this year I’d like to offer an alternative. Why not use the day as an opportunity to debate with students the relative merits of and ?
Of course, I’m talking about more than Greek letters here; I’m talking about what these letters represent, at least in certain circles (no pun intended). , the golden child of mathematical constants, known by all (though adored by few), represents the ratio of a circle’s circumference to its diameter. , on the other hand, has recently taken on a new identity as the ratio of a circle’s circumference to its radius - in other words, .
In the last few years, a debate has emerged over which constant is more fundamental. Some argue that should be viewed as the true circle constant, rather than . I’ve written about this subject here and here, so you can view the links if you want more details, but good primary sources for the argument in favor of include the Dr. Michael Hartl’s Tau Manifesto and Pi is Wrong! (the latter, in fact, helped to inspire the former).
On the other hand, last year a rebuttal by the name of the Pi Manifesto emerged, written by an author who identifies him or herself only as MSC. This article attempts to reclaim ‘s status as top dog by arguing, essentially, that the Tau Manifesto cherry picks formulas in which naturally appears, but ignores other situations in which itself seems to be the star. Both manifestos are worth reading, and for students in Geometry or Trigonometry, a debate about the relative merits of these constants could prove quite instructive.
I’d like to point out, though, that there are essentially two questions are being asked, and only one of them is really interesting. In my mind, the two questions are:
(1) Which constant is more canonical?
(2) Which constant is more pedagogically valuable?
The first question is typically the one mathematicians answer, but the question of vs. isn’t much of an interesting question, mathematically speaking. After all, any argument involving can be made into an argument involving by a simple substitution . Similarly, any argument involving can be made into an argument involving . So, from a strictly mathematical standpoint, the question of which notation to use doesn’t really matter.
This point is frequently overlooked when this debate is mentioned in the media. Gimmicky titles like “Pi is Wrong!” don’t really help. After all, there’s nothing wrong with the mathematical definition of – the issue is whether the ratio of a circle’s circumference to its diameter is a more or less natural thing to consider compared to the ratio of a circle’s circumference to its radius, but either ratio gives rise to a perfectly well-defined mathematical constant.
Regardless of your response the first question, I think the second question is the more interesting one, and based on my own teaching experience, this is where I’d give a slight advantage. The most compelling reason for using as a teaching tool comes from trigonometry – specifically, determining the sines and cosines of angles is much easier to understand for first timers if things are written in terms of rather than . These trig values are notoriously difficult for students to remember, but thinking in terms of makes things more intuitive. These pictures, taken from the Tau Manifesto, provide a clearer picture:
Angles as fractions of a circle.
Angles as fractions of tau.
Many students have trouble with trigonometry; if using instead of makes things easier to understand, then it’s certainly worth discussing.
So this day, let’s set aside the silly contests and ridiculous news articles, and instead think about some actual mathematics. Please note, though, that I still consider it acceptable to commemorate the holiday with a slice of pie (or two!).
A while back I was asked to contribute an essay to a book on mathematics and popular culture. I’m pleased to announce that this book is now available for purchase! There are some great essays in this book – I’ll let you decide how mine stacks up with the rest – and it also features a foreword by Keith Devlin, a Stanford University mathematician who you may know as NPR’s Math Guy.
I suggested they use my face instead, but they respectfully declined.
The price of entry is a little steep ($45), but if you’re someone interested in buying many copies (maybe you are a teacher, or maybe you just have a huge crush on David Krumholtz), I can get you a discount on bulk orders.
To whet your appetite, the title of my essay is Counting with the Sharks: Math-Savvy Gamblers in Popular Culture. Here’s the abstract:
While mathematicians in pop culture are often portrayed as misanthropic savants who may or may not be insane, films like 21 have helped usher in a new stereotype: the math-whiz card shark. We explore the origin and development of this stereotype through films such as Rain Man and 21, as well as its appearance in TV shows such as Numb3rs and 2 Months, $2 Million. We also discuss the pros and cons of this stereotype relative to the more traditional one.
In the coming months, I may use some of the other essays in this collection as a springboard for a couple of posts here. So stay tuned!
horses total, if you have a race track with 
lanes. Though it doesn’t look like it, this is a fairly natural generalization of the case discussed here (in the above case, k = 3).
. By the definition of 
, and half of the base of the isosceles triangle has length 
, so that the entire base of the isosceles triangle is twice as large, or 
.
, or roughly 0.0481.
, or around 0.0429. In other words, the designers of the Pyramid could have increased the area by roughly 12% with a better triangle design.
, or approximately 0.0796. Moreover, you get this largest area precisely when the closed figure is a circle. Said another way, for a given perimeter, the circle encloses the largest possible area.
, since there are n names entered, and t + 1 of them are yours. The probability of being in the games when you are 13 is equal to the probability of being selected when you are thirteen times the probability you were NOT selected when you were twelve (since, if you were selected when you were twelve, either you are dead or you were the winner of the previous Hunger Games – in either case, you won’t be playing again). Therefore, the probability is
are silly, so this year I’d like to offer an alternative. Why not use the day as an opportunity to debate with students the relative merits of 
?
.
naturally appears, but ignores other situations in which 
. Similarly, any argument involving 
