# My Favorite One-Liners: Part 28

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 when simple techniques get used in a complicated way.

Consider the solution of the linear recurrence relation

$Q_n = Q_{n-1} + 2 Q_{n-2}$,

where $F_0 = 1$ and $F_1 = 1$. With no modesty, I call this one the Quintanilla sequence when I teach my students — the forgotten little brother of the Fibonacci sequence.

To find the solution of this linear recurrence relation, the standard technique — which is a pretty long procedure — is to first solve the characteristic equation, from $Q_n - Q_{n-1} - 2 Q_{n-2} = 0$, we obtain the characteristic equation

$r^2 - r - 2 = 0$

This can be solved by any standard technique at a student’s disposal. If necessary, the quadratic equation can be used. However, for this one, the left-hand side simply factors:

$(r-2)(r+1) = 0$

$r=2 \qquad \hbox{or} \qquad r = -1$

(Indeed, I “developed” the Quintanilla equation on purpose, for pedagogical reasons, because its characteristic equation has two fairly simple roots — unlike the characteristic equation for the Fibonacci sequence.)

From these two roots, we can write down the general solution for the linear recurrence relation:

$Q_n = \alpha_1 \times 2^n + \alpha_2 \times (-1)^n$,

where $\alpha_1$ and $\alpha_2$ are constants to be determined. To find these constants, we plug in $n =0$:

$Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.

To find these constants, we plug in $n =0$:

$Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.

We then plug in $n =1$:

$Q_1 = \alpha_1 \times 2^1 + \alpha_2 \times (-1)^1$.

Using the initial conditions gives

$1 = \alpha_1 + \alpha_2$

$1 = 2 \alpha_1 - \alpha_2$

This is a system of two equations in two unknowns, which can then be solved using any standard technique at the student’s disposal. Students should quickly find that $\alpha_1 = 2/3$ and $\alpha_2 = 1/3$, so that

$Q_n = \displaystyle \frac{2}{3} \times 2^n + \frac{1}{3} \times (-1)^n = \frac{2^{n+1} + (-1)^n}{3}$,

Although this is a long procedure, the key steps are actually first taught in Algebra I: solving a quadratic equation and solving a system of two linear equations in two unknowns. So here’s my one-liner to describe this procedure:

This is just an algebra problem on steroids.

Yes, it’s only high school algebra, but used in a creative way that isn’t ordinarily taught when students first learn algebra.

I’ll use this “on steroids” line in any class when a simple technique is used in an unusual — and usually laborious — way to solve a new problem at the post-secondary level.

# My Favorite One-Liners: Part 27

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 an anecdote that I’ll share when teaching students about factorials:

$1! = 1$

$2! = 1 \times 2 = 2$

$3! = 1 \times 2 \times 3 = 6$

$4! = 1 \times 2 \times 3 \times 4 = 24$

$5! = 1 \times 2 \times 3 \times 4 \times 5 = 120$

The obvious observation is that the factorials get big very, very quickly.

Here’s my anecdote:

Many years ago, I was writing lesson plans while the TV show “Wheel of Fortune” was on in the background. And the contestant solved the puzzle at the end, and Pat Sajak declared, “You have just won $40,320 in cash in prizes. So I immediately thought to myself, “Ah, 8 factorial.” Then I thought, ugh [while slapping myself in the forehead, grimacing, and shaking my head, pretending that I can’t believe that that was the first thought that immediately came to mind]. [Finishing the story:] Not surprisingly, I was still single when this happened. # My Favorite One-Liners: Part 26 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 could appear early in a probability class: Let $P(A) = 0.2$, $P(B) = 0.4$, and $P(A \cup B) = 0.5$. Find $P(A \mid B)$. The standard technique for solving this problem involves first finding $P(A \cap B)$ using the Addition Rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.5 = 0.2 + 0.4 - P(A \cap B)$ $P(A \cap B) = 0.1$ From here, the Multiplication Rule can be used (or, equivalently, the definition of a conditional probability): $P(B \cap A) = P(B) \cdot P(A \mid B)$ $0.1 = 0.4 P(A \mid B)$ $0.25 = P(A \mid B)$ So far, so good. Now let me add a small twist to the original problem that creates a small difficulty when solving: Let $P(A) = 0.2$, $P(B) = 0.4$, and $P(A \cup B) = 0.5$. Find $P(A \cap B \mid A \cup B)$. Proceeding as before, we obtain $P( [A \cap B] \cup [A \cup B] ) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)$ The value of$P(A \cup B)\$ is obvious. But how do we evaluate the left side?

If I’m teaching an advanced probability class, I might expect them to use DeMorgan’s Laws. However, it’s a whole lot easier to reason out the left hand side: I’m looking for the probability that both $A$ and $B$ happen or else at least one of $A$ and $B$ happen. Well, that’s clearly redundant: if both $A$ and $B$ happen, then certainly at least one of $A$ and $B$ happen.

Here’s my one-liner, which I say, if possible, using only one breath of air:

Clearly, this is redundant. It’s like saying Dr. Q is my professor and he’s a total stud. It’s redundant. It’s obvious. There’s no need to actually say it.

After the laughter settles from this bit of braggadocio, the $A \cup B$ can be safely dropped from the left side:

$P( A \cap B) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)$

$0.1 = 0.5 \cdot P(A \cap B \mid A \cup B)$

$0.2 = P(A \cap B \mid A \cup B)$

However, I need to emphasize that dropping the term on the left side is a special feature of this particular problem since one set was a subset of the other, and that students shouldn’t expect to always be able to do this when computing conditional probabilities.

# My Favorite One-Liners: Part 25

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.

Consider the integral

$\displaystyle \int_0^2 2x(1-x^2)^3 \, dx$

The standard technique — other than multiplying it out — is using the substitution $u = 1-x^2$. With this substitution $du = -2x \, dx$. Also, $x = 0$ corresponds to $u = 1$, while $x = 2$ corresponds to $u = -3$. Therefore,

$\displaystyle\int_0^2 2x(1-x^2)^3 \, dx = - \displaystyle\int_0^2 (-2x)(1-x^2)^3 \, dx = -\displaystyle\int_1^{-3} u^3 \, du$.

My one-liner at this point is telling my students, “At this point, about 10,000 volts of electricity should be going down your spine.” I’ll use this line when a very unexpected result happens — like a “left” endpoint that’s greater than the “right” endpoint. Naturally, for this problem, the next step — though not logically necessary, it’s psychologically reassuring — is to absorb the negative sign by flipping the endpoints:

$\displaystyle\int_0^2 2x(1-x^2)^3 \, dx = -\displaystyle\int_1^{-3} u^3 \, du = \displaystyle\int_{-3}^1 u^3 \, du$,

and then the calculation can continue.

# My Favorite One-Liners: Part 24

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 could appear in my class in probability or statistics:

Let $f(x) = 3x^2$ be a probability density function for $0 \le x \le 1$. Find $F(x) = P(X \le x)$, the cumulative distribution function of $X$.

A student’s first reaction might be to set up the integral as

$\displaystyle \int_0^x 3x^2 \, dx$

The problem with this set-up, of course, is that the letter $x$ has already been reserved as the right endpoint for this definite integral. Therefore, inside the integral, we should choose any other letter — just not $x$ — as the dummy variable.

Which sets up my one-liner: “In the words of the great philosopher Jean-Luc Picard: Plenty of letters left in the alphabet.”

We then write the integral as something like

$\displaystyle \int_0^x 3t^2 \, dt$

and then get on with the business of finding $F(x)$.

# My Favorite One-Liners: Part 23

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 some sage words of wisdom that I give in my statistics class:

If the alternative hypothesis has the form $p > p_0$, then the rejection region lies to the right of $p_0$. On the other hand, if the alternative hypothesis has the form $p < p_0$, then the rejection region lies to the left of $p_0$.

On the other hand, if the alternative hypothesis has the form $p \ne p_0$, then the rejection region has two parts: one part to the left of $p_0$, and another part to the right. So it’s kind of like my single days. Back then, my rejection region had two parts: Friday night and Saturday night.

# My Favorite One-Liners: Part 22

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 example might be the most cringe-worthy pun that I use in any class that I teach.

In my statistics classes, I try to emphasize to student that a high value of the correlation coefficient $r$ is not the same thing as causation. To hopefully drive home this point, I’ll use the following picture.

Conclusion: If we want to stop global warming, we should all become pirates.

Obviously, I tell my class, there isn’t a cause-and-effect relationship here, even though there is a strong positive correlation. So, I tell my class, in my best pirate voice, “Correlation is not the same thing as a causation, even if you get a large value of ARRRRRRR.”

Without fail, my students love this awful wisecrack.

While I’m on the topic, this is too good not to share:

For further reading, see my series on correlation and causation.

# My Favorite One-Liners: Part 21

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.

Sometimes, just every once in a blue moon, something in mathematics doesn’t appear right to students at first glance. For example, take the common notation

$(a,b)$

What does this symbol mean? Sadly, it depends on the context.

Sometimes, it means a point in the Cartesian plane whose first coordinate is $a$ and whose second coordinate is $b$.

Other times, it could mean the set $\{x : a < x < b\}$, or the interval between $a$ and $b$ that does not contain the endpoints.

You’d think that, by now, mathematicians would’ve figure out a way to not denote these two completely different things with the same symbol. Indeed, I’ve seen textbooks that use $]a,b[$ to denote the open interval between $a$ and $b$ to avoid this duplication; however, this notation hasn’t been widely adopted by the mathematical community.

So here’s my quip when something like this comes up. Sometimes, a young child will come crying to her parents to complain about the injustices in the world, and the child may be right. But all the parent can say is, “Sorry, sweetheart, but sometimes life isn’t fair.” And I’ll act this out, talking to an imaginary child as I look down to the floor.

To complete the quip, I’ll then turn to my class and conclude, “Sorry, sometimes life isn’t fair.” It doesn’t make much sense, but we’re stuck with it.

# My Favorite One-Liners: Part 20

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.

Perhaps the world’s most famous infinite series is

$S = \displaystyle \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$

as this is the subject of Zeno’s paradox. When I teach infinite series in class, I often engage the students by reminding students about Zeno’s paradox and then show them this clip from the 1994 movie I.Q.

This clip is almost always a big hit with my students. After showing this clip, I’ll conclude, “When I was single, this was part of my repertoire of pick-up lines… but it never worked.”

Even after showing this clip, some students resist the idea that an infinite series can have a finite answer. For such students, I use a physical demonstration: I walk half-way across the classroom, then a quarter, and so on… until I walk head-first into a walk at full walking speed. The resulting loud thud usually confirms for students that an infinite sum can indeed have a finite answer.

For further reading, see my series on arithmetic and geometric series.

# My Favorite One-Liners: Part 19

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 quip that I’ll use when a theoretical calculation can be easily confirmed with a calculator. Today’s post is less of a one-liner than a story.

When I teach Algebra II or Precalculus (or train my  future high school teachers to teach these subjects), we eventually land on the Rational Root Test and Descartes’ Rule of Signs as an aid for finding the roots of cubic equations or higher. Before I get too deep into this subject, however, I like to give a 10-15 minute pseudohistory about the discovery of how polynomial equations can be solved. Historians of mathematics will certain take issue with some of this “history.” However, the main purpose of the story is not complete accuracy but engaging students with the history of mathematics. I think the story I tell engages students while remaining reasonably accurate… and I always refer students to various resources if they want to get the real history.

To begin, I write down the easiest two equations to solve (in all cases, $a \ne 0$:

$ax + b = 0 \qquad$ and $\qquad ax^2 + bx + c = 0$

These are pretty easy to solve, with solutions well known to students:

$x = -\displaystyle \frac{b}{a} \qquad$ and $\qquad x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

In other words, there are formulas that you can just stick in the coefficients and get the answer out without thinking too hard. Sure, there are alternate ways of solving for $x$ that could be easier, like factoring, but the worst-case scenario is just plugging into the formula.

These formulas were known to Babylonian mathematicians around 2000 B.C. (When I teach this in class, I write the date, and all other dates and discoverers, next to the equations for dramatic pedagogical effect.) Though not written in these modern terms, basically every ancient culture on the globe that did mathematics had some version of these formulas: for example, the ancient Egyptians, Greeks, Chinese, and Mayans.

Naturally, this leads to a simple question: is there a formula for the cubic:

$ax^3 + bx^2 + cx + d = 0$

Is there some formula that we can just plug $a$, $b$, $c$, and $d$ to just get the answer?  The answer is, Yes, there is a formula. But it’s nasty. The formula was not discovered until 1535 A.D., and it was discovered by a man named Tartaglia. During the 1500s, the study of mathematics was less about the dispassionate pursuit of truth and more about exercising machismo. One mathematician would challenge another: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” Then the second mathematician would solve it and challenge the first: “Here’s my cubic equation; I bet you can’t solve it. Nyah-nyah-nyah-nyah-nyah.” And so on. Well, Tartaglia came up with a formula that would solve every cubic equation. By plugging in $a$, $b$, $c$, and $d$, you get the answer out.

Tartaglia’s discovery was arguably the first triumph of the European Renaissance. The solution of the cubic was perhaps the first thing known to European mathematicians in the Middle Ages that was unknown to the ancient Greeks.

In 1535, Tartaglia was a relatively unknown mathematician, and so he told a more famous mathematician, Cardano, about his formula. Cardano told Tartaglia, why yes, that is very interesting, and then published the formula under his own name, taking credit without mention of Tartaglia. To this day, the formula is called Cardano’s formula.

So there is a formula. But it would take an entire chalkboard to write down the formula. That’s why we typically don’t make students learn this formula in high school; it’s out there, but it’s simply too complicated to expect students to memorize and use.

$ax^4 + bx^3 + cx^2 + dx + e = 0$

The solution of the quartic was discovered less than five years later by an Italian mathematician named Ferrari. Ferrari found out that there is a formula that you can just plug in $a$, $b$, $c$, $d$, and $e$, turn the crank, and get the answers out. Writing out this formula would take two chalkboards. So there is a formula, but it’s also very, very complicated.

Of course, Ferrari had some famous descendants in the automotive industry.

So now we move onto my favorite equation, the quintic. (If you don’t understand why it’s my favorite, think about my last name.)

$ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0$

After solving the cubic and quartic in rapid succession, surely there should also be a formula for the quintic. So they tried, and they tried, and they tried, and they got nowhere fast. Finally, the problem was solved nearly 300 years later, in 1832 (for the sake telling a good story, I don’t mention Abel) by a French kid named Evariste Galois. Galois showed that there is no formula. That takes some real moxie. There is no formula. No matter how hard you try, you will not find a formula that can work for every quintic. Sure, there are some quintics that can be solved, like $x^5 = 0$. But there is no formula that will work for every single quintic.

Galois made this discovery when he was 19 years old… in other words, approximately the same age as my students. In fact, we know when wrote down his discovery, because it happened the night before he died. You see, he was living in France in 1832. What was going on in France in 1832? I ask my class, have they seen Les Miserables?

France was torn upside-down in 1832 in the aftermath of the French Revolution, and young Galois got into a heated argument with someone over politics; Galois was a republican, while the other guy was a royalist. More importantly, both men were competing for the hand of the same young woman. So they decided to settle their differences like honorable Frenchmen, with a duel. So Galois wrote up his mathematical notes one night, and the next day, he fought the duel, he lost the duel, and he died.

Which bring me to the conclusion of this story: we have complete and total proof that tremendous mathematical genius does not prevent somebody from being a complete idiot.

For the present, there are formulas for cubic and quartic equations, but they’re long and impractical. And for quintic equations and higher, there is no formula. So that’s why we teach these indirect methods like the Rational Root Test and Descartes’ Rule of Signs, as they give tools to use to guess at the roots of higher-order polynomials without using something like the quadratic formula.

Real references: