Infraction

While I can’t take credit for this one-liner, I’m more than happy to share it.

A colleague was explaining his expectations for simplifying expressions such as

\displaystyle \frac{\displaystyle ~~~\frac{2x}{x^2+1}~~~}{\displaystyle ~~~\frac{x}{x^2-1}~~~}

Of course, this isn’t yet simplified, but his students were balking about doing the required work. So, on the spur of the moment, he laid down a simple rule:

Not simplifying a fraction in a fraction is an infraction.

Utterly brilliant.

Math education: an international comparison

A quote from Prof Brian Butterworth, an Emeritus professor from the Centre of Educational Neuroscience at the University College London:

The UK is not very good at maths. We are about average looking at all [Organization for Economic Cooperation] countries. So, we are significantly worse than Canada and Australia and much worse than China and Japan although we are a bit better than Germany and significantly better than the United States.

Credit: Learn maths to boost the economy, scientist advises, The Guardian, November 18, 2010.

Greek letters

One evening, I was watching the Philadelphia Eagles play the Chicago Bears on NBC’s Sunday Night Football telecast. The bottom of the screen showed the score: “PHI 14 CHI 7.” As my wife walked by, she innocently asked, “Why are there Greek letters on the screen?”

We’ve been fans of \phi and \chi ever since.

Student misconceptions about PEMDAS

Simplify 6/2*(1+2).

A Common Incorrect Answer. According to PEMDAS, we should handle the parentheses first. So 6/2*(1+2) = 6/2*3. Next, there are no exponents, so we should proceed to multiplication. So 6/2*3 = 6/(2*3) = 6/6. Finally, we move to division, and we obtain the answer 6/6 =1.

The above answer is incorrect and (even worse) arises from a natural but unfortunate misconception of the way that children are commonly taught order of operations. If you don’t see the misconception, please give it some thought before continuing.

green line

The mnemonic PEMDAS, commonly taught in the United States, stands for

Parentheses

Exponents

Multiplication

Division

Addition

Subtraction

I personally never learned this memorization trick when I was in school. What I do remember, from learning BASIC computer programming around 1980, was the mnemonic My Dear Aunt Sally. I’m told that in the United Kingdom (and perhaps elsewhere in the English-speaking world) schoolchildren are taught BIMDAS, where B stands for Brackets and I stands for Indices.

Unfortunately, all of these memorization devices suffer from a common flaw: they do not indicate that multiplication and divison have equal precedence, and that addition and subtraction have equal precedence. In other words, the order of operations really are

Parentheses

Exponents

Multiplication and Divison (left to right)

Addition and Subtraction (left to right)

Therefore, the correct answer to the above problem is

6/2*(1+2) = 6/2*3 = (6/2)*3 = 3*3 = 9.

In brief, though not intended by teachers, PEMDAS and BIMDAS perhaps promote the misconception that multiplication takes precedence over division and addition takes precedence over subtraction. To avoid this misconception, one of my colleagues suggests that PEMDAS be taught more visually as

P
E
MD
AS

so that students will have a better chance of remembering that MD and AS should have equal precedence.

Fun with Dimensional Analysis

The principle of diminishing return states that as you continue to increase the amount of stress in your training, you get less benefit from the increase. This is why beginning runners make vast improvements in their fitness and elite runners don’t.

J. Daniels, Daniels’ Running Formula (second edition), p. 13

In February 2013, I began a serious (for me) exercise program so that I could start running 5K races. On March 19, I was able to cover 5K for the first time by alternating a minute of jogging with a minute of walking. My time was 36 minutes flat. Three days later, on March 22, my time was 34:38 by jogging a little more and walking a little less. During that March 22 run, I started thinking about how I could quantify this improvement.

On March 19, my rate of speed was

\displaystyle \frac{5000 \hbox{~m}}{36 \hbox{~min}} = \frac{5000 \hbox{~m}}{36 \hbox{~min}} \times \frac{1 \hbox{~min}}{60 \hbox{~sec}} \approx 2.3148 \hbox{~m/s}.

On March 22, my rate of speed was

\displaystyle \frac{5000 \hbox{~m}}{34 \times 60 + 38 \hbox{~sec}} \approx 2.4062\hbox{~m/s}.

That’s a change of 2.4062 - 2.3148 \approx 0.0913 \hbox{~m/s} over 3 days (accounting for roundoff error in the last decimal place), and so the average rate of change is

\displaystyle \frac{0.0913 \hbox{~m/s}}{3 \hbox{~d}} = \frac{0.0913 \hbox{~m/s}}{3 \hbox{~d}} \times \frac{1 \hbox{~d}}{24 \times 60 \times 60 \hbox{~sec}} \approx 3.524 \times 10^{-7} \hbox{~m/s}^2.

By way of comparison, imagine a keg of beer floating in space. The specifications of beer kegs vary from country to country, but I’ll use the U.S. convention that the mass is 72.8 kg and its height is 23.3 inches = 59.182 cm. Also, for ease of calculation, let’s assume that the keg of beer is a uniformly dense sphere with radius 59.182/2 = 29.591 cm. Under this assumption, the acceleration due to gravity near the surface of the sphere is the same as the acceleration 29.591 cm away from a point-mass of 72.8 kg. Using Newton’s Second Law and the Law of Universal Gravitation, we can solve for the acceleration:

ma = \displaystyle \frac{GMm}{r^2},

where G \approx 6.67384 \times 10^{-11} \hbox{~m}^3/\hbox{kg} \cdot \hbox{s}^2 is the gravitational constant, M = 72.8 \hbox{~kg} is the mass of the beer keg, r = 0.29591 \hbox{~m} is the distance of a particle from the center of the beer keg, m is the mass of the particle, and a is the acceleration of the particle. Solving for a, we find

a = \displaystyle \frac{6.67384 \times 10^{-11} \times 72.8}{0.29591^2} \approx 5.5487 \times 10^{-8} \hbox{~m/s}^2.

Since this is only an approximation based on a hypothetical spherical keg of beer, let’s round off and define 1 beerkeg of acceleration to be equal to 5.5 \times 10^{-8} \hbox{~m/s}^2.

With this new unit, my improvement in speed from March 19 to March 22 can be quantified as

\displaystyle 3.524 \times 10^{-7} \hbox{~m/s}^2 \times \frac{1 \hbox{~beerkeg}}{5.5 \times 10^{-8} \hbox{~m/s}^2} \approx 6.35 \hbox{~beerkegs}.

I love physics: improvements in physical fitness can be measured in kegs of beer.

I chose the beerkeg as the unit of measurement mostly for comedic effect (I’m personally a teetotaler). If the reader desires to present a non-alcoholic version of this calculation to students, I’m sure that coolers of Gatorade would fit the bill quite nicely.

For what it’s worth, at the time of this writing (June 7), my personal record for a 5K is 26:58, and I’m trying hard to get down to 25 minutes. Alas, my current improvements in fitness have definitely witnessed the law of diminishing return and is probably best measured in millibeerkegs.

Entrance exam at MIT

Here’s a story that I’ll tell my students when, for the first time in a semester, I’m about to use a lemma to make a major step in proving a theorem. (I think I was 13 when I first heard this one, and obviously it’s stuck with me over the years.)

At MIT, there’s a two-part entrance exam to determine who will be the engineers and who will be the mathematicians. For the first part of the exam, students are led one at a time into a kitchen. There’s an empty pot on the floor, a sink, and a stove. The assignment is to boil water. Everyone does exactly the same thing: they fill the pot with water, place it on the stove, and then turn the stove on. Everyone passes.

For the second part of the exam, students are led one at a time again into the kitchen. This time, there’s a pot full of water sitting on the stove. The assignment, once again, is to boil water. Nearly everyone simply turns on the stove. These students are led off to become engineers. The mathematicians are ones who take the pot off the stove, dump the water into the sink, and place the empty pot on the floor… thereby reducing to the original problem, which had already been solved.

Laws of Logarithms

One of the most common student mistakes with logarithms is thinking that

\log_b(x+y) = \log_b x + \log_b y.

When I first started my career, I referred to this as the Third Classic Blunder. The first classic blunder, of course, is getting into a major land war in Asia. The second classic blunder is getting into a battle of wits with a Sicilian when death is on the line. And the third classic blunder is thinking that \log_b(x+y) somehow simplfies as \log_b x + \log_b y.

Sadly, as the years pass, fewer and fewer students immediately get the cultural reference. On the bright side, it’s also an opportunity to introduce a new generation to one of the great cinematic masterpieces of all time.

One of my colleagues calls this mistake the Universal Distributive Law, where the \log_b distributes just as if x+y was being multiplied by a constant. Other mistakes in this vein include  \sqrt{x+y} = \sqrt{x} + \sqrt{y}  and  (x+y)^2 = x^2 + y^2.

Along the same lines, other classic blunders are thinking that

\left(\log_b x\right)^n  simplifies as  \log_b \left(x^n \right)

and that

\displaystyle \frac{\log_b x}{\log_b y}  simplifies as  \log_b \left( \frac{x}{y} \right).

I’m continually amazed at the number of good students who intellectually know that the above equations are false but panic and use them when solving a problem.

Interdisciplinary Studies (Part 1)

An eccentric investor hired a biologist, a mathematician, and a physicist to design and train the perfect racehorse. After studying the problem for a couple of weeks, they returned to the investor to present their results.

The biologist said, “I’ve come up with a plan to breed the perfect racehorse. It’ll just take a thousand or two generations of breeding.”

The mathematician said, “I haven’t been able to solve this problem yet, but I’ve made some preliminary findings. So far, I’ve been able to show that for each horse race there will exist a winner, and furthermore that winner will be unique.”

So the investor’s hopes were pinned on the physicist, who began, “I think I’ve solved this race horse problem, but I had to make a few simplifying assumptions. First, let’s assume that each horse is a perfect frictionless rolling sphere…”