My Favorite One-Liners: Part 70

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 one really happened in one of my classes back in 1990 or 1991. My differential equations professor had written a very complicated theorem on the board that was true for all functions $g(x)$ ; unfortunately, thanks to the passage of time, I forget the exact statement of the theorem. I do remember that the theorem was quite complicated, and one of my classmates initially had a hard time understanding it. My instructor tried explaining for a couple of minutes, to no avail.

Finally, with some exasperation, my professor tried a different approach. The theorem was supposed to be true for all functions $g(x)$ , so he asked for a specific example to try. So he asked, “Give me a $g$ .”

One of my classmates immediate answered, as if a cheerleader at a football game had made this request instead of a college professor, and yelled out: “GEEEEEEEEEE!”, with his arms raised to the sky.

That definitely broke the tension in the room.

My Favorite One-Liners: Part 69

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 story, that I’ll share with my Precalculus students, comes from Fall 1996, my first semester as a college professor. I was teaching a Precalculus class, and the topic was vectors. I forget the exact problem (believe me, I wish I could remember it), but I was going over the solution of a problem that required finding $\tan^{-1}(7)$ . I told the class that I had worked this out ahead of time, and that the approximate answer was $82^o$ . Then I used that angle for whatever I needed it for and continued until obtaining the eventual solution.

(By the way, I now realize that I was hardly following best practices by computing that angle ahead of time. Knowing what I know now, I should have brought a calculator to class and computed it on the spot. But, as a young professor, I was primarily concerned with getting the answer right, and I was petrified of making a mistake that my students could repeat.)

After solving the problem, I paused to ask for questions. One student asked a good question, and then another.

Then a third student asked, “How did you know that $\tan^{-1}(7)$ was $82^o$ ?

Suppressing a smile, I answered, “Easy; I had that one memorized.”

The class immediately erupted… some with laughter, some with disbelief. (I had a terrific rapport with those students that semester; part of the daily atmosphere was the give-and-take with any number of exuberant students.) One guy in the front row immediately challenged me: “Oh yeah? Then what’s $\tan^{-1}(9)$ ?

I started to stammer, “Uh, um…”

“Aha!” they said. “He’s faking it.” They start pulling out their calculators.

Then I thought as fast as I could. Then I realized that I knew that $\tan 82^o \approx 7$ , thanks to my calculation prior to class. I also knew that $\displaystyle \lim_{x \to 90^-} \tan x = \infty$ since the graph of $y = \tan x$ has a vertical asymptote at $x = \pi/2 = 90^o$ . So the solution to $\tan x = 9$ had to be somewhere between $82^o$ and $90^o$ .

So I took a total guess. “ $84^o$ ,” I said, faking complete and utter confidence.

Wouldn’t you know it, I was right. (The answer is about $83.66^o$ .)

In stunned disbelief, the guy who asked the question asked, “How did you do that?”

I was reeling in shock that I guessed correctly. But I put on my best poker face and answered, “I told you, I had it memorized.” And then I continued with the next example. For the rest of the semester, my students really thought I had it memorized.

To this day, this is my favorite stunt that I ever pulled off in front of my students.

My Favorite One-Liners: Part 68

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 discussing the Laws of Logarithms, I’ll make a big deal of the fact that one law converts a multiplication problem into a simpler addition problem, while another law converts exponentiation into a simpler multiplication problem.

After a few practice problems — and about 3 minutes before the end of class — I’ll inform my class that I’m about to tell the world’s worst math joke. Here it is:

After the flood, the ark landed, and Noah and the animals got out. And God said to Noah, “Go forth, be fruitful, and multiply.” So they disembarked.

Some time later, Noah went walking around and saw the two dogs with their baby puppies and the two cats with their baby kittens. However, he also came across two unhappy, frustrated, and disgruntled snakes. The snakes said to Noah, “We’re having some problems here; would you mind knocking down a tree for us?”

Noah says, “OK,” knocks down a tree, and goes off to continue his inspections.

Some time later, Noah returns, and sure enough, the two snakes are surrounding by baby snakes. Noah asked, “What happened?”

The snakes replied, “Well, you see, we’re adders. We need logs to multiply.”

After the laughter and groans subside, I then dismiss my class for the day:

Go forth, and multiply (pointing to the door of the classroom). For most of you, don’t be fruitful yet, but multiply. You’re dismissed.

My Favorite One-Liners: Part 67

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 are a couple of similar problems that arise in Precalculus:

Convert the point $(5,-5)$ from Cartesian coordinates into polar coordinates.
Convert the complex number $5 - 5i$ into trigonometric form.

For both problems, a point is identified that is 5 steps to the right of the origin and then 5 steps below the $x-$ axis (or real axis). To make this more kinesthetic, I’ll actually walk 5 paces in front of the classroom, turn right face, and then walk 5 more paces to end up at the point.

I then ask my class, “Is there a faster way to get to this point?” Naturally, they answer: Just walk straight to the point. After some work with the trigonometry, we’ll establish that

$(5,-5)$ in Cartesian coordinates is equivalent to $(5\sqrt{2}, -\pi/4)$ in polar coordinates, or
$5-5i$ can be rewritten as $5\sqrt{2} [ \cos(-\pi/4) + i \sin (-\pi/4)]$ in trigonometric form.

Once this is obtained, I’ll walk it out: I’ll start at the origin, turn clockwise by 45 degrees, and then take $5\sqrt{2} \approx 7$ steps to end up at the same point as before.

Continuing the lesson, I’ll ask if the numbers $5\sqrt{2}$ and $-\pi/4$ , or if some other angle and/or distance could have been chosen. Someone will usually suggest a different angle, like $7\pi/4$ or $15\pi/4$ . I’ll demonstrate these by turning 315 degrees counterclockwise and walking 7 steps and then turning 675 degrees and walking 7 steps (getting myself somewhat dizzy in the process).

Finally, I’ll suggest turning only 135 degrees clockwise and then taking 7 steps backwards. Naturally, when I do this, I’ll do a poor man’s version of the moonwalk:

For more information, please see my series on complex numbers.

My Favorite One-Liners: Part 66

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.

Every once in a while, when I’m expecting my students to use a formula which is so long that I can’t reasonably expect my students to memorize it, I’ll let my students bring a 3-by-5 card to class. When I tell them this, I tell them the ground rules:

The units of the card should be in inches (as opposed to 3 meters-by-5 meters)
The card can have handwritten notes.
No magnifying glasses.

I then related an amazing anecdote that I heard when I was a teenager: for his German final, one of my acquaintances somehow used his Apple IIe computer and his Epson printer to somehow print in microfiche font. He got all of his notes from his thick German notebook onto one side of a 3-by-5 card and half of the other side, leaving half of the back side blank.

Amazing.

Presumably he brought a scanning electron microscope to class so that he could read his notecard.

Anyway, I tell my students, no magnifying glasses are permitted on exams.

My Favorite One-Liners: Part 65

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 use today’s one-liner just before I begin some enormous, complicated, and tedious calculation that’s going to take more than a few minutes to complete. To give a specific example of such a calculation: consider the derivation of the Agresti confidence interval for proportions. According to the central limit theorem, if $n$ is large enough, then

$Z = \displaystyle \frac{ \hat{p} - p}{ \displaystyle \sqrt{ \frac{p(1-p) }{n} } }$

is approximately normally distributed, where $p$ is the true population proportion and $\hat{p}$ is the sample proportion from a sample of size $n$ . By unwrapping this equation and solving for $p$ , we obtain the formula for the confidence interval for a proportion:

$z \displaystyle \sqrt{\frac{p(1-p)}{n} } = \hat{p} - p$

$\displaystyle \frac{z^2 p(1-p)}{n} = \left( \hat{p} - p \right)^2$

$z^2p - z^2 p^2 = n \hat{p}^2 - 2 n \hat{p} p + n p^2$

$0 = p^2 (z^2 + n) - p (2n \hat{p} + z^2) + n \hat{p}^2$

We now use the quadratic formula to solve for $p$ :

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{ \left(2n\hat{p} + z^2 \right)^2 - 4n\hat{p}^2 (z^2+n)}}{2(z^2+n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n^2 \hat{p}^2 + 4n \hat{p} z^2 + z^4 - 4n\hat{p}^2 z^2 - 4n^2 \hat{p}^2}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n (\hat{p}-\hat{p}^2) z^2 + z^4}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n \hat{p}(1-\hat{p}) z^2 + z^4}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm \sqrt{4n \hat{p} \hat{q} z^2 + z^4}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm z \sqrt{4n \hat{p} \hat{q} + z^2}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm z \sqrt{4n^2 \displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle 4n^2 \frac{z^2}{4n^2}}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + z^2 \pm 2nz \sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z^2}{4n^2}}}{2(z^2 + n)}$

$p = \displaystyle \frac{2n \hat{p} + 2n \displaystyle \frac{z^2}{2n} \pm 2nz \sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} +\displaystyle \frac{z^2}{4n^2}}}{2n \displaystyle \left(1 + \frac{z^2}{n} \right)}$

$p = \displaystyle \frac{\hat{p} + \displaystyle \frac{z^2}{2n} \pm z \sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z^2}{4n^2}}}{\displaystyle 1 + \frac{z^2}{n} }$

From this we finally obtain the $100(1-\alpha)\%$ confidence interval

$\displaystyle \frac{\hat{p} + \displaystyle \frac{z_{\alpha/2}^2}{2n}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } - z_{\alpha/2} \frac{\sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z_{\alpha/2}^2}{4n^2}}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } < p < \displaystyle \frac{\hat{p} + \displaystyle \frac{z_{\alpha/2}^2}{2n}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} } + z_{\alpha/2} \frac{\sqrt{\displaystyle \frac{ \hat{p} \hat{q}}{n} + \displaystyle \frac{z_{\alpha/2}^2}{4n^2}}}{\displaystyle 1 + \frac{z_{\alpha/2}^2}{n} }$ .

Whew.

So, before I start such an incredibly long calculation, I’ll warn my students that this is going to take some time and we need to prepare… and I’ll start doing jumping jacks, shadow boxing, and other “exercise” in preparation for doing all of this writing.

My Favorite One-Liners: Part 64

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.

From the category “I Really Don’t Recommend That Anyone Does This But It Sure Makes a Great Story Now”: I recently told my students about the time in Spring 2000 that, in the middle of class, I playfully threw an eraser at a wise-cracking student sitting in the back row…

…and I aimed about three feet above his head so that the eraser would richochet off the back wall…

…but the eraser kind of knuckleballed and inadvertently sailed barely over the head of the student sitting in front of him and then nailed him square in the chest…

…and I somehow kept a straight face as if I really had intended to peg him with a cloud of chalkdust…

…and, the next day, my students gave me a half-dozen new erasers for fresh ammuntion.

Ah, memories.

My Favorite One-Liners: Part 63

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 use today’s one-liner to explain why mathematicians settled on a particular convention that could have been chosen differently. For example, let’s consider the definition of $y = \sin^{-1} x$ by first looking at the graph of $f(x) = \sin x$ .

Of course, we can’t find an inverse for this function; colloquially, the graph of $f$ fails the horizontal line test. More precisely, there exist two numbers $x_1$ and $x_2$ so that $x_1 \ne x_2$ but $f(x_1) = f(x_2)$ . Indeed, there are infinitely many such pairs.

So how will we find the inverse of $f$ ? Well, we can’t. But we can do something almost as good: we can define a new function $g$ that’s going look an awful lot like $f$ . We will restrict the domain of this new function $g$ so that $g$ satisfies the horizontal line test.

For the sine function, there are plenty of good options from which to choose. Indeed, here are four legitimate options just using the two periods of the sine function shown above. The fourth option is unorthodox, but it nevertheless satisfies the horizontal line test (as long as we’re careful with $\pm 2\pi$ .

So which of these options should we choose? Historically, mathematicians have settled for the interval $[-\pi/2, \pi/2]$ .

So, I’ll ask my students, why have mathematicians chosen this interval? That I can answer with one word: tradition.

For further reading, see my series on inverse functions.

My Favorite One-Liners: Part 62

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 story that I’ll tell after doing a couple of back-to-back central limit theorem problems. Here’s the first:

The chances of winning a column bet in roulette is 12/38. The bet pays 2 to 1, meaning that if you lose, you lose $1. However, if you win, you get your $1 back and $2 more. If this bet is made 1000 times, what is the probability of winning at least $0?

With my class, we solve this problem using standard techniques with the normal approximation:

$\mu = E(X) = 2 \times \displaystyle \frac{12}{38} + (-1) \frac{26}{38} = - \displaystyle \frac{1}{19}$

$E(X^2) = 2^2 \times \displaystyle \frac{12}{38} + (-1)^2 \frac{26}{38} = \displaystyle \frac{37}{19}$

$\sigma = SD(X) = \sqrt{ \displaystyle \frac{37}{19} - \left( - \displaystyle \frac{1}{19} \right)^2} = \displaystyle \frac{\sqrt{702}}{19}$

$E(T_0) = n\mu = 1000 \left( -\displaystyle \frac{1}{19} \right) \approx -52.63$

$\hbox{SD}(T_0) = \sigma \sqrt{n} = \displaystyle \frac{\sqrt{702}}{19} \sqrt{1000} \approx 44.10$

$P(T_0 > 0) \approx P\left(Z > \displaystyle \frac{0-(-52.63)}{44.10} \right) \approx P(Z > 1.193) \approx 0.1163$ .

Next, I’ll repeat the problem, except playing the game 10,000 times.

The chances of winning a column bet in roulette is 12/38. The bet pays 2 to 1, meaning that if you lose, you lose $1. However, if you win, you get your $1 back and $2 more. If this bet is made 10,000 times, what is the probability of winning at least $0?

The last three lines of the above calculation have to be changed:

$E(T_0) = n\mu = 10,000 \left( -\displaystyle \frac{1}{19} \right) \approx -526.32$

$\hbox{SD}(T_0) = \sigma \sqrt{n} = \displaystyle \frac{\sqrt{702}}{19} \sqrt{10,000} \approx 139.45$

$P(T_0 > 0) \approx P\left(Z > \displaystyle \frac{0-(-526.32)}{139.45} \right) \approx P(Z > 3.774) \approx 0.00008$ .

In other words, the chance of winning drops dramatically. This is an example of the Law of Large Numbers: if you do something often enough, then what ought to happen eventually does happen.

As a corollary, if you’re going to bet at roulette, you should only bet a few times. And, I’ll tell my students, one Englishman took this to the (somewhat) logical extreme by going to Las Vegas and making the ultimate double-or-nothing bet, betting his entire life savings on one bet. After all, his odds of coming out ahead by making one bet were a whole lot higher than by making a sequence of bets.

Naturally, my students ask, “Did he win?” Here’s the video and the Wikipedia page:

My Favorite One-Liners: Part 61

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 story that I like to tell my probability and statistics students when we cover the law of averages.

One of my favorite sports is golf, and one spring afternoon in my senior year I went out to play a round. I was assigned a tee time with two other students (that I didn’t know), and off we went.

Unfortunately, the group in front of us were, as I like to say, getting their money’s worth out of the round. Somebody would be stuck in a sand trap and then blast the ball into the sand trap on the other side of the green. Then he’d go to blast the ball out of that sand trap, and the ball would go back to the original one.

Golf etiquette dictates that slow-playing groups should let faster groups play through. However, this group never offered to let us pass them. And so, hole after hole, we would wait and wait and wait.

On hole #9, a player walking by himself came up from behind us. I’m not sure how that happened — perhaps the foursome that had been immediately behind us was even slower than the foursome in front of us — and he courteously asked if he could play through. I told him that we’d be happy to let him play through, but that the group in front of us hadn’t let us through, and so we were all stuck.

As a compromise, he asked if he could join our group. Naturally, we agreed.

This solo golfer did not introduce himself, but I recognized him because his picture had been in the student newspaper a few weeks earlier. He was Notah Begay III, then a hot-shot freshman on the Stanford men’s golf team. Though I didn’t know it then, he would later become a three-time All-American and, with Tiger Woods as a teammate, would win the NCAA championship. As a professional, he would win on the PGA Tour four times and was a member of the 2000 President’s Cup team.

Of course, all that lay in the future. At the time, all I knew was that I was about to play with someone who was really, really good.

We ended up playing five holes together… numbers 10 through 14. After playing 14, it started to get dark and I decided to call it quits (as the 14th green was fairly close to the course’s entrance).

So Notah tees off on #10. BOOM! I had never been so close to anyone who hit a golf ball so far. The guys I was paired with started talking about which body parts they would willingly sever if only they could hit a tee shot like that.

And I thought to myself, Game on.

I quietly kept score of how I did versus how Notah did. And for five holes, I shot 1-over par, while he shot 2-over par. And for five holes, I beat a guy who would eventually earn over $5 million on the PGA Tour.

How did the 9-handicap amateur beat the future professional? Simple: we only played five holes.

Back then, if I shot 1-over par over a stretch of five holes, I would be pretty pleased with my play, but it wouldn’t be as if I had never done it before. And I’m sure Notah was annoyed that he was 2-over par for those five holes (he chili-dipped a couple of chip shots; I imagine that he was experimenting with a new chipping technique), but even the best golfers in the world will occasionally have a five-hole stretch where they go 2-over par or more.

Of course, a golf course doesn’t have just five holes; it has 18.

My all-time best score for a round of golf was a four-over par 76.; I can count on one hand the number of times that I’ve broken 80. That would be a lousy score for a Division I golfer. So, to beat Notah for a complete round of golf, it would take one of my absolute best days happening simultaneously with one of his worst.

Furthermore, a stroke-play golf tournament is not typically decided in only one round of golf. A typical professional golf tournament, for those who make the cut, lasts four rounds. So, to beat Notah at a real golf tournament, I would have to have my absolute best day four days in a row at the same time that Notah had four of worst days.

That’s simply not going to happen.

So I share this anecdote with my students to illustrate the law of averages. (I also use a spreadsheet simulating flipping a coin thousands of times to make the same point.) If you do something enough times, what ought to happen does happen. However, if instead you do something only a few times, then unexpected results can happen. A 9-handicap golfer can beat a much better player if they only play 5 holes.

To give a more conventional illustration, a gambler can make a few dozen bets at a casino and still come out ahead. However, if the gambler stays at the casino long enough, he is guaranteed to lose money.