For a San Francisco 49ers fan like myself, most of the last decade has been spent in a fairly constant state of disappointment.  But after ten years without a playoff appearance, the team blossomed this season under the influence of new head coach Jim Harbaugh, and came within one game of their first Super Bowl appearance since 1995.

This poster hangs proudly in our apartment.

Despite a great season, in which the team won 13 of their 16 games for the first time since 1997, many people have still voiced doubts about whether or not this team is for real.  What if this season was just a fluke?  After all, quarterback Alex Smith, long jeered by 49ers fans since being drafted in 2005, was the starting quarterback for every game this season.  And while his stats this year were his best ever, they don’t compare favorably to quarterbacks like Tom Brady, Aaron Rodgers, or Drew Brees.  While the offense might not be the flashiest, this year it was relatively error-free; combined with stellar defense and special teams units, the 49ers were able to grind out wins fairly consistently from week to week.

The question remains, though, no matter how much the fans may try to ignore it: how much of a role did simple luck play in the 49ers season?  To answer this question, of course, one needs a way to measure the team’s luck, and compare it to other teams.  Alternatively, one needs to look at some measures of skill, and show that in fact, the 49ers simply were a more skilled team this year, both compared to other teams in the league and compared to the 49ers of previous seasons.  Here I’ll talk a little bit about luck, and in a follow-up post I’ll talk more about skill.

In what sense could the 49ers be considered a lucky team during the 2011 season?  Well, one way to look for a team’s luck is to measure the strength of the opposing teams during the course of the season.  If your team plays all of its games against terrible opposing teams, then it’s more likely your team will have a favorable record, even if it’s only mediocre, simply by virtue of the fact that the teams it plays against aren’t very good.  Conversely, if your team plays all of its games against very strong teams, it will be much harder to get a winning record.  Bearing this in mind, we ask: how lucky were the 49ers in terms of the strength of the opponents they faced?

The answer, as many football fans could probably guess, is fairly lucky.  Part of this stems from the fact that the 49ers played six of their sixteen games against opponents from their own division, which is not known for its strength.  Here’s a list of all teams in the 2011 season, ranked by the strength of their schedules.

Table 1: 2011 strength of schedule. Green indicates teams that made it to the playoffs.

As you can see, the 49ers schedule is ranked near the very bottom, tied with New England and just above New Orleans.  So the schedule was relatively easy.  Having said that, I’ve heard many people attribute the success of the 49ers solely to the ease of their schedule, but have not heard the same criticism made for teams like New Orleans, New England, and Green Bay.  To remark on the ease of the 49ers schedule without pointing out the collective ease of all of these teams’ schedules seems a bit unfair.

Let’s ask another question: how does the strength of the 2011 schedule compare to the strength of the 2010 schedule?  Here’s a new table, just like the one above, but with data from the 2010 season:

Table 2: 2010 strength of schedule. Green indicates teams that made it to the playoffs.

In the 2011 season, the 49ers were much closer to the middle of the pack as far as strength of schedule is concerned.  Comparing year over year, based on this data, it would not seem unreasonable to attribute some of the 49ers’ newfound success to their relatively easy schedule.

But wait! the careful reader may exclaim.  There’s a chicken-and-egg argument to be made here: did the 49ers improve in 2011 because they had an easier schedule, or did they have an easier schedule because they had improved?  Indeed, San Francisco’s 13 wins in 2011 correspond to an extra 13 losses for their opponents, as compared to only 6 wins for the team in 2010.  Any conclusions drawn from the above data must keep this in mind, since the opponents wins for any team are not independent of the team’s performance.

One way to try and correct this is to remove each team’s performance when considering the strength of the opposing teams.  So, for example, in 2011 the teams the 49ers faced won 115 games and lost 141, for a win percentage of just 0.449.  However, if we remove games played against the 49ers, who had a record of 13 wins and 3 losses, then among all remaining games, the teams the 49ers faced won 112 games, and lost 129, for a slightly higher win percentage of .467.

If we make this adjustment across all teams for the 2011 season, the following picture emerges.

Table 3: 2011 Adjusted 2011 strength of schedule. Green indicates teams that made it to the playoffs.

While San Francisco’s schedule still ranks near the bottom after this adjustment, a few other teams have slipped just below it. How does this compare to the previous season?  To wrap things up, let’s look at the adjusted strength of schedule for 2010.

Table 2: Adjusted 2010 strength of schedule. Green indicates teams that made it to the playoffs.

Notice that in absolute terms, the difference in the 49ers’ schedule is quite small comparing the adjusted 2010 season to the adjusted 2011 season.  In 2011, the number of adjusted games opponents won was 112, versus just 115 in 2010.  So from this standpoint, much of the 49ers’ progress in 2011 appears to be due less to having a lucky schedule, and more do to with actual improvements of the team.  As promised above, next time I will dig a little deeper in an attempt to quantify the 49ers’ improvement during this season.  For now though, I must pick out an appropriate outfit to wear this Sunday, in mourning of the absence of this team from the Super Bowl.

(A final note for the more statistically minded: In 2011, the correlation between a team’s winning percentage and the opponents’ combined losing percentage is quite high – the correlation coefficient is roughly .766.  With the adjustment mentioned above, this drops to .529, which agrees with the notion that there should be a weaker correlation between one team’s record and the combined record of its opponents when the team’s record is removed from the combined record.  For 2010, the correlations are a bit weaker: .574 and .333, respectively).

A recent article on Wired’s website discusses the mathematics of Lego – more specifically, it highlights an article on the complexity of Lego systems.  As any child will tell you, Lego sets can vary from very simple, small sets, to much larger and more complicated ones.  As a simple corollary, smaller sets will have fewer pieces, and larger sets will have more pieces.  But how does the number of types of pieces grow as the size of the set grows?  For example, if a 100 piece set consists of 10 different types of pieces, is it reasonable to guess that a 1000 piece set will consist of 100 different types of pieces?

In a word, no.  Though the number of different types of pieces will grow as the size of a set grows, it will grow slower than the size of the set (in other words, it will grow sub-linearly).  To put it another way, as the size of the Lego set grows, rather than building more and more new types of pieces, the same types of pieces that are present in smaller sets tend to be used in new ways.  The effect is that the proportion of distinct piece types decreases as the size of the set grows.  From a mathematical standpoint, if we let y denote the number of different types of pieces, and x be the number of pieces, then this power law is giving us the following equation:

for some constants A and b, with b between 0 and 1.  While y grows as x grows, it does not grow as quickly as x itself.

Taken in a broader context, though, this should not be surprising.  Examples of similar phenomena are prevalent throughout nature, as well as in made-made phenomena such as urban planning.  One example cited in the Wired article is Kleiber’s Law, which states that the ratio of an animal’s metabolic rate to its mass tends to decrease as mass increases (in other words, larger animals are capable of metabolizing more efficiently).  Here’s an article that discusses an analogue of this power law in the context of brain development, and relates this to the development of cities.

So the next time you give a Lego set to a child, feel free to explain this connection – I’m sure any child will welcome the math lesson (at least, any child worth giving a Lego set to in the first place).  It’s also worth noting that this phenomenon is most likely not unique to Lego sets – I am eagerly awaiting a similar report on the mathematics of Tinkertoys – though unfortunately, in this case the number of piece types seems not to have increased in nearly a century.

Even XKCD is on the Car Talk bandwagon! (Click the image to go to the source)

Most significant to our present discussion, however, is Car Talk’s weekly diversion known as the Puzzler.  Each week, the brothers read a Puzzler (i.e. a brain teaser) to their listeners and request solutions.  The Puzzler’s solution is revealed the following week, and a new Puzzler is then presented; moreover, one of the correct listener submissions is chosen at random, and the winner receives a gift certificate for some Car Talk schwag.  While these puzzles are sometimes car-related, this is not a prerequisite, and indeed the puzzler I would now like to discuss makes no mention of cars.

Here’s the puzzle, with wording modified slightly from Car Talk’s website: Suppose there are 20,000 lights in a row, all turned off.  One person comes through and pulls the cord on every light, turning each of them on.  A second person then comes through and pulls the cord on every second light. A third person then pulls the cord on every third light, and so on. After the 20,000th person has gone through the room, which lights are turned on?

This problem also goes by the name of the Locker Problem, with open and shut lockers substituting for on and off lights.  But no matter how you contextualize the problem, the solution is the same.  To get some idea of what the answer should be, let’s consider what happens just for the first few (say, 10) light bulbs.  Feel free to think about this problem on your own before continuing on!

After the first person walks through the room, all the light bulbs are on.  So, the first 10 light bulbs will look like this:

After every second cord is pulled, half the lights will be on, and half will be off:

Now here’s where the fun begins.  When every third cord is pulled, off lights will turn on, and on lights will turn off.  This means that the arrangement of the first ten lights will look like this:

And, when every fourth cord is pulled, we get the following picture:

You can probably fill in the rest.  Note that when every 6th, 7th, 8th, 9th, or 10th cord is pulled, only one light in first 10 switches, so most of the work is already done.  In the end, the string of the first 10 lights will look like this:

Here we see that three of the first ten lights remain lit: the 1st, 4th, and 9th.  The astute reader will note that 1, 4, and 9 all share a common property; they are all perfect squares (1 = 1², 4 = 2², 9 = 3²).  The question to be asked, of course, is whether this pattern continues, and if it does, why?

To answer these questions we’ll need to think a bit more deeply about what’s going on.  First, when does a given light bulb in our string of 20,000 get its cord pulled?  Extrapolating from the first few cases, we see that if a person goes through the chain and pulls every dth cord, then the light bulbs with their cords pulled are precisely the ones whose numbers are multiples of d.  Therefore, the nth bulb’s cord is pulled once for every divisor of n.  More concretely, we see for example that the 12th light bulb will have its cord pulled 6 times: by the 1st, 2nd, 3rd, 4th, 6th, and 12th person, since 1, 2, 3, 4, 6, and 12 are precisely the divisors of 12.

What does this divisibility information tell us about the status of a given light bulb?  Since all the lights start in the off position, we see that a light will be off at the end if its cord is pulled an even number of times, and it will be on at the end if its cord is pulled an odd number of times.  In other words, the nth light will be on at the end of this process precisely when n has an odd number of divisors.

All we need to do now is convince ourselves that the numbers with an odd number of divisors are precisely the squares.  There are a couple of ways to do this.

Way 1: Pick your favorite whole number n.  For any divisor d of n, n/d is also a divisor (for example, if n = 30, the fact that 5 is a divisor immediately tells us that 30/5 = 6 is a divisor too).  In this way, we can naturally count up all the divisors of n in pairs.  Because of this, the only way we can have an odd number of divisors is if one of the pairs has the same number repeated twice, i.e. if for some divisor d, d and n/d are equal.  But if they are equal, this means that n = d², in other words, n is a square.

Way 2: By the fundamental theorem of arithmetic, Any positive whole number n > 1 can be written in an essentially unique way as a product of prime numbers, i.e. for any n > 1 we can write

,

where the numbers are primes, and the exponents are greater than zero (for example, ).  In particular, any divisor of n must be composed of the same primes as n itself, so that if d is a divisor, we can write

,

where each exponent is no larger than (for example, 12 is a divisor of 180 but 24 isn’t, since 2 goes into 24 three times but into 180 only twice).

How many divisors are there? Well, could divide d as few as 0 times, or as many as times – we have different ways to choose the exponent on , corresponding to the numbers 0, 1, 2, …, up to .  Similarly, we have different ways to choose the exponent on , and so on for each prime, so that there are different ways to choose the exponent on .  By the fundamental counting principle, this means that the number of divisors of n is equal to

Remember that our light will be on only if the number of divisors is odd, in other words, only if the above product is odd.  Notice that if any term in the above product is even, the product itself will be even – so in order for the product to be odd, each term in the product must be odd.  Since each must be odd, this means that each exponent must be even.  But if each is even, then is always a whole number, so

is a whole number whose square is equal to n.  Hence, once again we conclude n is a square.

There are plenty of variants to this problem worth pondering – for example, what if people only come in and pull every dth cord only for d prime?  Or, what if instead of the 2nd person pulling every 2nd cord, and the third person pulling every 3rd cord, the second person pulls every 2nd cord twice, the third person pulls every 3rd cord three times, and so on?  No doubt you can come up with some variants on your own as well.

If you prefer the locker version, here’s an interactive site where you can watch with satisfaction as lockers open and close.  No matter what model you use, though, this is a cute little problem on integers and their divisibility, and the result can be surprising for first time viewers.  So kudos to Car Talk for discussing this problem on a national stage! (Kudos also to Tom for drawing my attention to this particular episode of the program.)

This dude knows a thing or two about baked goods.

For my money, the best part of the baking process (aside from the delicious final act) is the careful and precise initial measurement of the ingredients.  Keeping an accurate account of the relative proportions of each piece of the recipe is a hallmark of baking, and reflects the nature of baking itself: one part art, one part science.  Unlike some other culinary arts, the measurements really do matter.  Screw up these proportions and those fudgy brownies you want to make will be too cakey (or vice versa).

But what in the recipe accounts for the qualitative differences we see and taste in the wide assortment of baked goods available at this time of year?  This question has been mulled over by electrical engineer and baking aficionado Michael Ohene, who in this article essentially considers all baked goods as a function of three parameters: Moistness Value, Butter Content, and Egg Content.

How are these values computed?  First, each wet and dry ingredient is assigned a value per cup – these values are used in the calculations to follow.  For examples of wet ingredients, buttermilk has a value of 1 per cup, applesauce has a value of .6 per cup, and so on.  With dry ingredients, things like flour and almond paste have a value of 1 per cup, while peanut butter has a value lightly less (2/3 per cup).  More detailed tables of values can be found at the link given above.  Note that flavorings, leavenings, seasonings, and food pieces (such as whole walnuts) are not included in the calculations.  You can think of these values as a weighing the ingredients according to their relative wetness or dryness.

Once the value of each ingredient is known, that value is multiplied by the volume of the ingredient appearing in the recipe, and these products are totaled up for both wet and dry ingredients – let me call these final values the wetness value and the dryness value, respectively.  Let’s see how this works in practice by analyzing my wife’s spectacular recipe for Pfeffernusse cookies (they are delicious bites of gingerbready goodness).  The ingredients are as follows:

3/4 cup molasses
1/2 cup butter
2 large eggs
4 1/4 cup all purpose flour
1 1/4 teaspoon baking soda
1/2 cup white sugar
1 1/2 teaspoon cinnamon
1/2 teaspoon clove
1/2 teaspoon nutmeg
dash of pepper
confectioners sugar

(In case you’re curious, here’s the recipe, cribbed from my wife’s 8th grade German class:

In a saucepan, combine molasses and butter. Stir until butter melts. Cool to room temperature. Mix in the eggs. In another bowl combine flour, baking soda, sugar, cinnamon, cloves, nutmeg, and a big dash of pepper. Add the flour mixture to the molasses mixture and mix well. Chill for 1-2 hours or overnight. Shape into 1”inch balls. (These can be frozen after this point and baked later if desired) Bake on greased cookie sheet at 350 degrees for 8-9 minutes, they should still be somewhat soft as they will continue to cook after they come out (underdone is better than overdone with these). Cool a bit and role in confectioners sugar. Makes 4 1/2 dozen cookies.)

Like a child who has eaten too much candy, these pfeffernusse cookies are the illest.

According to Mr. Ohene, there are three wet ingredients (butter, large eggs, and molasses) and one dry ingredient (flour).  The rest of the ingredients are not considered in the computations.  Butter has a corresponding value of .5, large eggs have a value of 1/6 per egg, and molasses has a value of 1.  Therefore, the wetness value is .5 times the volume of butter, plus 1/6 times the number of eggs, plus 1 times the volume of molasses, or

.5 x .5 + 1/6 x 2 + 1 x .75 = 1.33.

Similarly, flour has a corresponding value of 1, so the dryness value is simply 4.25, as the recipe calls for 4 1/4 cups of flour.  So the wetness value is 1.33, and the dryness value is 4.25.

The Moistness Value is the ratio of these two numbers, i.e. the ratio of wet to dry.  Meanwhile, the Butter Content is the ratio of the value coming from the butter to the dryness value, and the Egg Content is the ratio of the number of eggs to the dryness value.  In the example of Pfeffernusse, we see that the Moistness Value is 1.33/4.25, or about .3129.  Similarly, the Butter Content is .25/4.25, and the Egg Content is 2/4.25, or .0588 and .4706, respectively.

Varying these three values distinguishes different baked goods from one another.  Want a baked good with relatively high Moistness Value and Butter Content, but relatively low Egg Content?  Perhaps a coffee cake is what you’re after.  How about a moderately wet dough with high Egg Content and low Butter Content?  In this case, maybe a brioche is what you’re after.  A periodic table of sorts that plots baked goods according to these three values can be found here; note that the Pfeffernusse discussed above fall into a square marked “biscotti.”  It’s quite possible that the Pfeffernusse dough would make for good biscotti, though the process of baking is quite different for these two treats, since biscotti are typically twice baked.

You can do other cool things with this analysis.  For example, you can compute these values for a list of ingredients and make an educated guess about what type of baked good the ingredients will produce.  Also, one can automatically create recipes for baked goods based solely on the qualities one would like that baked good to have.  This service is provided here – you entire the baked good you want, the desired level of richness, and the desired level of sweetness, and it creates a recipe for you on the spot!  So if you can’t find just the right recipe for what you want, I’d encourage you to try this recipe creator on for size and see what comes out.

Whatever you or your family decide to bake this holiday, know that adding a little mathematics to the mix never hurt.  And, as evidenced by the work of Mr. Ohene, added analysis can provide some interesting and unexpected insights. Thanks go to him for sending me the links that I’ve shared with you here!

Needless to say, Cameron’s version of the story is much more embellished. In his rendition, their experiment was a success; as he puts it, the pumpkin flew across the field, “goal post to goal post.”

When I first heard him say this, my initial thought was “Is Cameron telling the truth?”  How likely is it that a pumpkin, launched from a slingshot at one end of a football field, could sail through the air to land on the other side?  Note that his story actually ends with the pumpkin falling through the sun roof of someone’s car – this outcome is, of course, highly implausible, and therefore casts a shadow of doubt upon the entire story.  While the sunroof claim would be difficult to verify, one can at least use some basic math and physics to test the plausibility of the first portion of the story.

We would like to be able to find a formula for the distance Cameron’s pumpkin should travel.  Key to our analysis is the conservation of energy principle, and Hooke’s Law.  In particular, we will assume that the slingshot behaves like a spring, i.e. the potential energy it stores is proportional to the square of the displacement (for example, stretching the slingshot two meters gives you four times the potential energy as stretching it one meter).  So, if Cameron and his friends stretched the slingshot a distance of x meters, the slingshot would then store a potential energy of , where k is the spring constant and depends on the physical properties of the slingshot.

Now let us invoke the conservation of energy principle: when the slingshot is released, suppose all that stored potential energy will be converted into the kinetic energy of the pumpkin.  The formula for the kinetic energy of an object is , where m is the object’s mass, and v is the object’s velocity.  So, when the pumpkin is released, by conservation of energy, this kinetic energy should be the same as the potential energy the system had before we chunked the pumpkin.  In other words,

,

which, with a bit of algebra, tells us that the initial velocity of the pumpkin when it comes out of the slingshot must be .

These poor little guys have no idea what's in store for them.

So we have an equation for the velocity of the pumpkin in terms of the spring constant k, the distance x we pull the slingshot, and the mass m of the pumpkin.  Let’s not forget our main goal, though – what we’re ultimately interested in isn’t a formula for the initial velocity of the pumpkin, but rather the distance the pumpkin travels.  With a little more physics, it’s not hard to get from one of these pieces of information to the other.

More specifically, we have a very good understanding of projectile motion.  As I’ve discussed before, if one ignores air resistance (as we will do for now), all the equations of projectile motion can be derived from the fact that the acceleration due to gravity is a constant, meters per second per second.

Using this fact, suppose you throw an object with an initial velocity at an angle to the horizontal.  Then if one sets the initial position of the object to have coordinates (0,0), the x-coordinate of the object at any time t will be given by , and the y-coordinate of the object at any time t will be given by .  Using these formulas and a bit more algebra, one can determine that the distance d an object will travel in the x direction can be written in terms of the initial velocity, the angle , and g, via the formula

Example trajectory of an object shot with an initial velocity v at an angle A. The distance the object travels in the x direction is given by the above formula.

To recap: we have the distance as a function of the initial velocity, the angle the pumpkin is shot from, and the acceleration due to gravity.  We also have the initial velocity in terms of the distance the slingshot is stretched, the mass of the pumpkin, and the spring constant.  Combining these two formulas, we find that

and we now have the desired formula for the distance the pumpkin travels.

Let’s check Cameron’s story against this formula.  If the pumpkin really went from goal post to goal post, the distance it traveled in the x direction must have been at least 120 yards (100 yards for the field of play, plus 10 yards for each end zone).  This is roughly 109.7 meters.  Therefore, we must have .

On the right side of the equation, the term is maximized when equals 45°, and in this case .  So after making this simplification, we see that in order for Cameron’s story to be true, it must be the case that

The acceleration due to gravity is known, but we need to provide estimates for m, k, and x.  The mass m is the easiest to estimate; let’s say the pumpkin is around 10 pounds (roughly 4.5 kilograms).  To be generous, we’ll even round down to 4 kilograms.  What about the distance the slingshot is stretched?  Based on the slingshot used in the episode, it’s unlikely the slingshot in Cameron’s story would have been stretched more than 2 meters or so, but let’s again be generous and say it’s stretched 3 meters.  Plugging in these values, we would have:

where the units on k are Newtons per meter.  Since one pound is approximately 4.45 newtons, this is saying that the spring constant is about 107 pounds per meter – in other words, for each meter you stretch the slingshot, you need to exert 107 pounds of force.  To put it another way, to stretch the slingshot 3 meters or more, you’d need to exert 321 pounds of force.

This seems like quite a lot, though Cameron is a large fellow.  But recall we were being generous, both in our computation of the spring constant and in our estimate of the pumpkin size.  We also neglected air resistance in this model, but air resistance probably has a non-negligible impact here – not only will it slow down the pumpkin’s forward motion, but it also decreases the optimal angle from 45°.  So in a real world situation, I’d have to remain skeptical of Cameron’s story.  On the other hand, these calculations don’t necessarily rule it out entirely (though a more sophisticated analysis might).

For more on the physics of pumpkin chunkin’, here‘s an article from last year courtesy of Wired.  For related trajectory issues, Angry Birds also provides plenty of fodder.

Unlike most Calculus textbooks, however, Oullette’s book has an extra helping of sympathy for its audience.  Oullette’s goal is not necessarily to make her readers expert mathematics students; instead, she focuses on unifying seemingly disparate types of problems under the umbrella of Calculus.  Included amongst these examples are applications of Calculus to the equations of motion, thermodynamics, surfing, and the spread of disease.  The wheel is not being reinvented here – most of these examples (with the possible exception of what Calculus tells us about the optimal strategy in the event of a zombie outbreak) should be covered in a Calculus course.  But since Oullette’s goal is to appeal to the people who never took Calculus, this overlap is likely intentional.

Writing about mathematics for a general audience can be a difficult balancing act.  If the language is too technical, the average reader may become confused and frustrated; on the other hand, if the language isn’t technical enough, the reader may not learn much of anything, so what’s the point of writing in the first place?  Unfortunately, I found Oullette’s book too frequently to be weighted towards the latter of these two extremes.  Though the book has the word “Calculus” in the title, there’s nary an equation to be found until the appendix, where all the scary mathematics has been relegated.  While the main body of the text offers well-written explanations for the main ideas, ultimately the writing feels too broad because Oullette refuses to get into the nuts and bolts of Calculus even a little bit.  I’m not saying we need another Calculus book in the market, but ultimately this book is extremely limited in its ability to describe what Calculus is actually like, because the very language of Calculus is barely used.

I realize, of course, that I am probably not the target audience for a book like this, so I asked my wife to read through the prologue and give me her thoughts.  While she said she found the beginning of the book interesting, she didn’t really seem compelled to continue reading.  There isn’t much I can infer from a sample size of one, but it seems to me like reading this book to try and overcome a fear of Calculus is a bit like reading books about French cooking and Louis XIV to try and overcome a fear of learning a foreign language.  Useful applications are presented, but the nature of the beast itself is never really tackled.

Having said that, if you’ve never taken a Calculus course, or have no idea what the word “Calculus” really involves, this book may be a good starting point for you.  It certainly won’t make you an expert, or give you many problems to solve, but if you have a fear of mathematics, you can think of this as dipping your toe in the water.

If you are a masochist like me, though, there are plenty of ridiculous articles floating around today to help you get your blood boiling.  One of my favorites comes from today‘s Philadelphia Inquirer.  It’s full of gems like:

• One may be the loneliest number, [La Salle University math teacher Stephen] Andrilli said, but 11 ranks among the most odd – and not just because it isn’t even. He sees 11 as sort of a netherworld number – one more than the familiar 10, one less than an even dozen. Uhh, what?  What does any of this even mean?
• In geometry, Andrilli said, an 11-sided polygon is called an “undecagon” – the shape of the one-dollar coin in Canada, whose people have a particular affinity for 11. Little known-fact: the Canadian people’s love for the number eleven has been the subject of intense research among anthropologists for centuries.  Oh wait – no, of course it hasn’t.  Because that would be stupid.
• Weirdly, the sum of 1,111 multiplied by 1,111 is 1,234,321, another numerical palindrome. Do editors no longer know the difference between the words “sum” and “product”?  As written, this sentence also makes no sense.

To its credit, the article does state that all of the listed coincidences involving 11 only show “the human propensity for seeing patterns where none exist.”
Not all articles may take the day with a healthy grain of salt, though, so make sure not to get caught up in all the 11/11/11 hubbub.  It will be difficult, I know.

Though smaller than the average cupcake, the macaron is also more labor-intensive, and is therefore frequently on the more expensive end of the confectionery spectrum.  The macarons at Bottega Louie, for example, will run you $1.75 each.

One of many delightful flavors

If you need a sweet fix, though, a single macaron may not be enough.  Anticipating such a first-world problem, Bottega Louie also offers boxes of macarons for purchase.  The boxes come in three sizes: the small holds five macarons, the medium holds thirteen, and for the true Francophiles, the largest box holds forty five.  If you buy a box, you can fill it with whatever flavors you like, and can then eat to your heart stomach’s content.

What does any of this have to do with mathematics?  As with so many things, a quantitative eye is useful when it comes time to look at the bottom line.  While the prices of most things decrease with scale – each donut in the purchase of a dozen is cheaper than the purchase of an individual donut, for example – in the case of these macarons, such scaling does not occur.

Let’s dig into some numbers.  The small box of macarons costs $10, or $2 per macaron.  This is more than a 10% increase in the price per macaroon; essentially, one is paying an extra $1.25 for a fancy little box.

What a tiny, fancy box.

Things are slightly better if you go bigger.  The medium box is $25, as compared to $22.75 for thirteen individual macarons.  That comes out to roughly $1.93 per macaroon, or a cost of $2.25 for the large box.  The largest box will run you $80, as compared to $78.75 for forty five individual macarons.  Equivalently, the cost is around $1.78 per macaron, and the additional cost of the box once again comes to $1.25.

In particular, it seems a little strange that the smallest and largest boxes both incur an additional $1.25 charge, while the one in the middle is an extra $2.25.  The mathematician in me would much rather see the box in the middle priced at $24, and the consumer in me would rather see all the boxes be cheaper per macaron than buying them individually.

Unfortunately, in the macaron game, this type of pricing is apparently not unheard of.  There is another small chain of macaron shops known as ‘lette, with several locations throughout the Los Angeles area.  I stopped in one this weekend and found the following prices displayed on the wall:

It’s not all bad news this time around.  While the smaller boxes cost more per macaron than buying individually ($2 each for the mini box, $1.75 for the box of six), the larger boxes are thankfully cheaper (around $1.63 each for the box of 12, around $1.58 for the box of 24).

In either case, though, the moral is the same: when it comes to macarons, make sure you do the math.  While we are all accustomed to the idea that larger purchases correspond to lower costs per unit, these examples show this is not necessarily the case.  If you’re only interested in stuffing your face, make sure to take a moment to crunch the numbers, as the best deal may not immediately present itself to you.

Without further ado, let’s begin!

1. Tony Stark

Yes, yes, I know – since Iron Man hit the screens in the summer of 2008, the titular character has become a popular costume idea, joining the ranks of comic book icons like Superman and Spiderman.  I’m not talking about dressing up as Iron Man, though.  Instead, I am recommending a costume based on the man inside the suit – Tony Stark, playboy billionaire and (more importantly) mathematical wünderkind.  All you really need is some delicately coiffed facial hair and a glowing circle on your chest.  Aside from that, the world is really your oyster.  You could go as classy Tony stark, for example:Or, if you’re looking for a more rugged look, you could try prisoner of war Tony Stark:

As with many things in life, the only limit is really your imagination.

2. Rubik’s Cube

I know, this bears a striking similarity to Rubik’s cube head from one of my earlier posts – but this one is even better, because not only does it drape over your body, it also explicitly states you are a Rubik’s cube, for those guests at the event you attend who have been living in a cave for the past thirty years.

RetroCRUSH has the costume on a list of the worst Halloween costumes of all time, an honor with which I must respectfully disagree.  I will concede, though, that if you are looking to score some ladies (or fellas), this may not be your best option.

While we’re on the subject, though, I’d like to send a quick shout out to Parks and Recreation for featuring Rubik’s Cube Head as a costume in their recent Halloween episode.  Here’s a picture of that fantastic party attendee standing next to Rob Lowe, who is sporting a less exciting Sherlock Holmes costume.

Rob Lowe next to Rubik's Cube Head

3. Human Calculator

This one requires some work, but the payoff may be worth it.  First, one must decide what type of calculator to become.  Then one must decide on the size – should it be a full body costume, or centered only on the torso, for example?  No matter what path you choose, however, the most important thing is getting the details right.  Nobody likes a costume made by an inferior craftsman (or craftswoman, for that matter).

Here is an example of a calculator costume gone right.  Note the pride this individual takes in his work.  No doubt he secured many digits on Halloween.

Click to go to the source link.

No matter what you ultimately decide, I hope your Halloween is a good one.  See you next year!