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.

My Favorite One-Liners: Part 87

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 teaching Calculus I, I use the following mantra throughout the semester. I heard this from my calculus instructor back in 1984, and I repeat it for my own students:

There are two themes of calculus: approximating curved things by straight things, and passing to limits.

For example, to find a derivative, we approximate a curved function by a straight tangent line and then pass to a limit. Later in the semester, to find a definite integral, we approximate the area under a curve by the sum of a bunch of straight rectangles and then pass to a limit.

For further reading, I’ll refer to this series of posts on what I typically do on the first day of my calculus class.

My Favorite One-Liners: Part 86

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.

To get students comfortable with $i = \sqrt{-1}$ , I’ll often work through a quick exercise on the powers of $i$ :

$i^1 = i$

$i^2 = -1$

$i^3 = -i$

$i^4 = 1$

$i^5 = i$

Students quickly see that the powers of $i$ are a cycle of length 4, so that $i^5 = i \cdot i \cdot i \cdot i \cdot i$ is the same thing as just $i$ . So I tell my students:

There’s a technical term for this phenomenon: aye-yai-yai-yai-yai.

See also http://mentalfloss.com/article/52790/where-did-phrase-aye-yai-yai-come for more on the etymology of this phrase.

My Favorite One-Liners: Part 85

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 one-liner is one that I’ll use when I want to discourage students from using a logically correct and laboriously cumbersome method. For example:

Find a polynomial $q(x)$ and a constant $r$ so that $x^3 - 6x^2 + 11x + 6 = (x-1)q(x) + r$ .

Hypothetically, this can be done by long division:

However, this takes a lot of time and space, and there are ample opportunities to make a careless mistake along the way (particularly when subtracting negative numbers). Since there’s an alternative method that could be used (we’re dividing by something of the form $x-c$ or $x+c$ , I’ll tell my students:

Yes, you could use long division. You could also stick thumbtacks in your eyes; I don’t recommend it.

Instead, when possible, I guide students toward the quicker method of synthetic division:

My Favorite One-Liners: Part 84

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, I’ll show my students that there’s a difficult way to do a problem that I don’t want them to do for homework. For example, here’s the direct derivation of the mean of the binomial distribution using only Precalculus; this would make an excellent homework problem for the Precalculus teacher who wants to torture his/her students:

$E(X) = \displaystyle \sum_{k=0}^n k {n \choose k} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n k {n \choose k} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n k \frac{n!}{k!(n-k)!} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n \frac{n!}{(k-1)!(n-k)!} p^k q^{n-k}$

$= \displaystyle \sum_{k=1}^n \frac{n (n-1)!}{(k-1)!(n-k)!} p^k q^{n-k}$

$= \displaystyle \sum_{i=0}^{n-1} \frac{n (n-1)!}{i!(n-1-i)!} p^{i+1} q^{n-1-i}$

$= \displaystyle np \sum_{i=0}^{n-1} \frac{(n-1)!}{i!(n-1-i)!} p^i q^{n-1-i}$

$= \displaystyle np(p+q)^{n-1}$

$= np \cdot 1^{n-1}$

$=np.$

However, that’s a lot of work, and the way that I really want my students to do this, which is a lot easier (and which will be used throughout the semester), is by writing the binomial random variable as the sum of indicator random variables:

$E(X) = E(I_1 + \dots + I_n) = E(I_1) + \dots + E(I_n) = p + \dots + p = np$ .

So, to reassure my students that they’re going to be asked to reproduce the above lengthy calculation, I’ll tell them that I wrote all that down for my own machismo, just to prove to them that I really could do it.

Since my physical presence exudes next to no machismo, this almost always gets a laugh.

My Favorite One-Liners: Part 83

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 from calculus:

Let $f(x) = x^2 e^{3x}$ . Find $f''(x)$ .

We begin by finding the first derivative using the Product Rule:

$f'(x) = 2x e^{3x} + 3x^2 e^{3x}$ .

Next, we apply the Product Rule again to find the second derivative:

$f''(x) = (2 e^{3x} + 6x e^{3x}) + (6x e^{3x} + 9x^2 e^{3x})$ .

At this point, before simplifying to get the final answer, I’ll ask my students why the $6x e^{3x}$ term appears twice. After a moment, somebody will usually volunteer the answer: the first term came from differentiating $x^2$ first and then $e^{3x}$ second, while the other term came from differentiating $e^{3x}$ first and then $x^2$ second. Either way, we end up with the same term.