My Favorite One-Liners: Part 97

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’ll save this tongue-in-cheek one-liner for the wonderful occasions when my students collectively ace an exam and they’re extremely happy after I’ve returned the tests:

I learned my lesson; clearly, I need to make the next test harder.

My Favorite One-Liners: Part 96

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

When assigning homework or a take-home project, my students may ask what the rules are for collaborating with their peers. As a general rule, I want my students to talk to each other and to collaborate on homework, even if that opens the possibility that some student may directly copy their answers from somebody else. (I figure that if any student abuses collaboration, they will get appropriately punished when they take in-class exams.) So, when students ask about rules for collaborating, I tell them:

To quote the great philosopher, “You go talk to your friends, talk to my friends, talk to me.”

My Favorite One-Liners: Part 95

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today’s quip is one that I’ll use in a statistics class when we find an extraordinarily small $P$ -value. For example:

There is a social theory that states that people tend to postpone their deaths until after some meaningful event… birthdays, anniversaries, the World Series.

In 1978, social scientists investigated obituaries that appeared in a Salt Lake City newspaper. Among the 747 obituaries examined, 60 of the deaths occurred in the three-month period preceding their birth month. However, if the day of death is independent of birthday, we would expect that 25% of these deaths would occur in this three-month period.

Does this study provide statistically significant evidence to support this theory? Use $\alpha=0.01$ .

It turns out, using a one-tailed hypothesis test for proportions, that the test statistics is $z = -10.71$ and the $P-$ value is about $4.5 \times 10^{-27}$ . After the computations, I’ll then discuss what the numbers mean.

I’ll begin by asking, “Is the null hypothesis [that the proportion of deaths really is 25%] possible?” The correct answer is, “Yes, it’s possible.” Even extraordinarily small $P$ -values do not prove that the null hypothesis is impossible. To emphasize the point, I’ll say:

After all, I found a woman who agreed to marry me. So extremely unlikely events are still possible.

Once the laughter dies down, I’ll ask the second question, “Is the null hypothesis plausible?” Of course, the answer is no, and so we reject the null hypothesis in favor of the alternative.

My Favorite One-Liners: Part 94

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s edition isn’t a one-liner, but it’s still one of my favorites.

When constructing a mathematical model, sometimes certain simplifying assumptions have to be made… and sometimes these simplifications can be less than realistic. If a student complains about the unreasonableness of the simplifications, I’ll share the following story (taken from the book Absolute Zero Gravity).

Once upon a time, a group of investors decided that horse-racing could be made to pay on a scientific basis. So, they hired a team of biologists, a team of physicists, and a team of mathematicians to spend a year studying the question. At the end of the year, all three teams announced complete solutions. The investors decided to celebrate with a gala dinner where all three plans could be unveiled.

The mathematicians had the thickest report, so the chief mathematician was asked to give the first talk: “Ladies and gentlemen, you have nothing to worry about. Without describing the many details of
our proof, we can guarantee a solution to the problem you gave us — it turns out that every race is won by a least one horse. But we have been able to go beyond even this, and can show that the solution is unique: every race is won by no more than one horse!”

The biologists, who had spent the most money, went next. They were also able to show that the investors had nothing to worry about. By using the latest technology of genetic engineering, the biologists could easily set up a breeding program to produce an unbeatable racehorse, at a cost well below a million a year, in about two hundred years.

Now the investors’ hopes were riding on the physicists. The chief physicist also began by assuring them that their troubles were over. “We have perfected a method for predicting with 96 percent certainty the winner of any given race. The method is based on a very few simplifying assumptions. First, let each horse be a perfect rolling sphere… “

My Favorite One-Liners: Part 93

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

This is a wisecrack that I’ll use in my probability/statistics classes to clarify the difference between $P(A \cap B)$ and $P(A \mid B)$ :

Even though the odds of me being shot by some idiot wielding a gun while I teach my class are probably a million to one, I’ve decided, in light of Texas’ campus-carry law, to get my concealed handgun license and carry my own gun to class. This is for my own safety and protection; after all, the odds of *two* idiots carrying a gun to my class must be absolutely microscopic.

My Favorite One-Liners: Part 92

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

This is one of my favorite quote from Alice in Wonderland that I’ll use whenever discussing the difference between the ring axioms (integers are closed under addition, subtraction, and multiplication, but not division) and the field axioms (closed under division except for division by zero):

‘I only took the regular course [in school,’ said the Mock Turtle.]

‘What was that?’ inquired Alice.

‘Reeling and Writhing, of course, to begin with,’ the Mock Turtle replied; ‘and then the different branches of Arithmetic — Ambition, Distraction, Uglification, and Derision.’

My Favorite One-Liners: Part 91

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Everyone once in a while, a student might make a careless mistake — or just choose an incorrect course of action — that changes what was supposed to be a simple problem into an incredibly difficult problem. For example, here’s a problem that might arise in Calculus I:

Find $f'(x)$ if $f(x) = \displaystyle \int_0^x (1+t^2)^{10} \, dt$

The easy way to do this problem, requiring about 15 seconds to complete, is to use the Fundamental Theorem of Calculus. The hard way is by multiplying out $(1+t^2)^{10}$ — preferably using Pascal’s triangle — taking the integral term-by-term, and then taking the derivative of the result. Naturally, a student who doesn’t see the easy way of doing the problem might get incredibly frustrated by the laborious calculations.

So here’s the advice that I give my students to trying to discourage them from following such rabbit trails:

If you find yourself stuck on what seems to be an incredibly difficult problem, you should ask yourself, “Just how evil do I think my professor is?”

My Favorite One-Liners: Part 90

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a typical problem that arises in Algebra II or Precalculus:

Find all solutions of $2 x^4 + 3 x^3 - 7 x^2 - 35 x -75 =0$ .

There is a formula for solving such quartic equations, but it’s very long and nasty and hence is not typically taught in high school. Instead, the one trick that’s typically taught is the Rational Root Test: if there’s a rational root of the above equation, then (when written in lowest terms) the numerator must be a factor of $-10$ (the constant term), while the denominator must be a factor of $2$ (the leading coefficient). So, using the rational root test, we conclude

Possible rational roots = $\displaystyle \frac{\pm 1, \pm 3, \pm 5, \pm 15, \pm 25, \pm 75}{\pm 1, \pm 2}$

$= \pm 1, \pm 3, \pm 5, \pm 15, \pm 25, \pm 75 \displaystyle \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{5}{2}, \pm \frac{15}{2}, \pm \frac{25}{2}, \pm \frac{75}{2}$ .

Before blindly using synthetic division to see if any of these actually work, I’ll try to address a few possible misconceptions that students might have. One misconception is that there’s some kind of guarantee that one of these possible rational roots will actually work. Here’s another: students might think that we haven’t made much progress toward finding the solutions… after all, we might have to try synthetic division 24 times before finding a rational root. So, to convince my students that we actually have made real progress toward finding the answer, I’ll tell them:

Yes, 24 is a lot\dots but it’s better than infinity.

My Favorite One-Liners: Part 89

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that might arise in my discrete mathematics class:

Find the negation of $p \Rightarrow q$ .

This requires a couple of reasonably complex steps. First, we use the fact that $p \Rightarrow q$ is logically equivalent to $\lnot p \lor q$:

$\lnot(p \Rightarrow q) \equiv \lnot (\lnot p \lor q)$ .

Next, we have to apply DeMorgan’s Law to find the negation:

$\lnot (p \Rightarrow q) \equiv \lnot(\lnot p \lor q) \equiv \lnot(\lnot p) \land \lnot q$

Finally, we arrive at the final step: simplifying $\lnot(\lnot p)$ . At this point, I tell my class, it’s a bit of joke, especially after the previous, more complicated steps. “Not not $p$ ,” of course, is the same as $p$ . So this step is a bit of a joke. Which steps up the following cringe-worthy pun:

In fact, you might even call this a not-not joke.

After the groans settle down, we finish the derivation:

$\lnot(p \Rightarrow q) \equiv \lnot(\lnot p \lor q) \equiv \lnot(\lnot p) \land \lnot q \equiv p \land \lnot q$ .

My Favorite One-Liners: Part 88

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In the first few weeks of my calculus class, after introducing the definition of a derivative,

$\displaystyle \frac{dy}{dx} = y' = f'(x) = \lim_{h \to 0} \displaystyle \frac{f(x+h) - f(x)}{h}$ ,

I’ll use the following steps to guide my students to find the derivatives of polynomials.

If $f(x) = c$ , a constant, then $\displaystyle \frac{d}{dx} (c) = 0$ .
If $f(x)$ and $g(x)$ are both differentiable, then $(f+g)'(x) = f'(x) + g'(x)$ .
If $f(x)$ is differentiable and $c$ is a constant, then $(cf)'(x) = c f'(x)$ .
If $f(x) = x^n$ , where $n$ is a nonnegative integer, then $f'(x) = n x^{n-1}$ .
If $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ is a polynomial, then $f'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + a_1$ .

After doing a few examples to help these concepts sink in, I’ll show the following two examples with about 3-4 minutes left in class.

Example 1. Let $A(r) = \pi r^2$ . Notice I’ve changed the variable from $x$ to $r$ , but that’s OK. Does this remind you of anything? (Students answer: the area of a circle.)

What’s the derivative? Remember, $\pi$ is just a constant. So $A'(r) = \pi \cdot 2r = 2\pi r$ .

Does this remind you of anything? (Students answer: Whoa… the circumference of a circle.)

Generally, students start waking up even though it’s near the end of class. I continue:

Example 2. Now let’s try $V(r) = \displaystyle \frac{4}{3} \pi r^3$ . Does this remind you of anything? (Students answer: the volume of a sphere.)

What’s the derivative? Again, $\displaystyle \frac{4}{3} \pi$ is just a constant. So $V'(r) = \displaystyle \frac{4}{3} \pi \cdot 3r^2 = 4\pi r^2$ .

Does this remind you of anything? (Students answer: Whoa… the surface area of a sphere.)

By now, I’ve really got my students’ attention with this unexpected connection between these formulas from high school geometry. If I’ve timed things right, I’ll say the following with about 30-60 seconds left in class:

Hmmm. That’s interesting. The derivative of the area of a circle is the circumference of the circle, and the derivative of the area of a sphere is the surface area of the sphere. I wonder why this works. Any ideas? (Students: stunned silence.)

This is what’s known as a cliff-hanger, and I’ll give you the answer at the start of class tomorrow. (Students groan, as they really want to know the answer immediately.) Class is dismissed.

If you’d like to see the answer, see my previous post on this topic.