Bring an end to the rationalization madness

In less than a month, we’ll be hosting the one and only James Tanton at our school. We’re so excited! I’m especially excited because he’s totally going to help me rally the troops in this fight:

He posted this a few years ago, but I only stumbled on it recently. I’ve been looking for Tanton videos to use in our classes so we can get all psyched up about his visit! Needless to say, I was loving this video :-).

For more on why I’m not such a big fan of ‘rationalizing the denominator’ see this post.

Composing power functions

I presented the following example in my Precalculus classes this past week and it bothered students:

Let f(x)=4x^2 and g(x)=x^{3/2}. Compute f(g(x)) and g(f(x)) and state the domain of each.

 

As usual, I’ll give you a second to think about it yourself.

..

 

 

..

 

Done yet?

.

 

Here are the answers:

f(g(x))=4(x^{3/2})^2=4x^3, x\geq 0

g(f(x))=(4x^2)^{3/2}=8|x|^3, x\in\mathbf{R}

The reason that the first one was unsettling, I think, is because of the restricted domain (despite the fact that the simplified form of the answer seems not to imply any restrictions).

The reason the second one was unsettling is because they had forgotten that \sqrt{x^2}=|x|. It seems to be a point lost on many Algebra 2 students.

 

Mindset List for incoming High School class of 2017

Flickr, Creative Commons License

Happy first day of school! For us, today marks the first day of the school year and we’re welcoming students into our midst. What are the new kids (the freshmen) going to be like?

Each year Beloit College describes the incoming college freshmen class with its now famous “Mindset List.” I looked around and couldn’t find a high-school equivalent. So here’s one I came up with. This is a description of the incoming high school freshmen class (class of 2017). Note that all of descriptions on Beloit’s college freshmen mindset list apply also to high school freshmen. So here’s my own “high school class of 2017 mindset list.” Enjoy!

  1. The Euro has always existed. So has Sponge Bob Square Pants. And Google, Inc. And the iMac. And Viagra.
  2. Bill Gates has always been worth over $100 billion.
  3. You can talk to them about the Sandy Hook shooting or the Virginia Tech massacre, but they won’t remember anything about Columbine, which happened the year they were born.
  4. Star Wars Episode 1: The Phantom Menace is as much of an ‘old-school’ classic as any of the original Star Wars movies. The movies Fight ClubThe MatrixAmerican PieSaving Private RyanArmageddon, and The Sixth Sense also came out in the years they were born.
  5. East Timor has always been a sovereign nation.
  6. George W. Bush and Barack Obama are the only presidents that they really know. Clinton left office when they were just 2.
  7. Exxon and Mobil have always been the same company.
  8. Movies have always been reviewed by Ebert & Roeper .(Gene Siskel died the year they were born; Roger Ebert just died this past April.)
  9. They won’t have any memories of John F. Kennedy Jr, Dr. Spock, Frank Sinatra, Roy Rogers, or Alan Shepherd, all of whom died just as they were being born.
  10. They have always had their music in mp3 format and used mp3 players (invented in 1998). CD players? SO passé.
  11. Seinfeld closed up shop before they were born.

——–

For more events that happened in 1998 and 1999, visit the wikipedia articles. Please feel free to correct any of my above information or suggest additions!

Why I hate the definition of trapezoids (part 3)

Yes it’s true. I’m writing about trapezoids again (having written passionately about them here and here previously). I’ve been taking a break from blogging, as I usually do in the summer. For us, school starts in just two weeks. So I thought I’d come out of my shell and post something…and of course I always have something to say about trapezoids :-).

Let’s start with the following easy test question. Don’t peek. See if you can answer the question without any help.

Which of the following quadrilaterals are trapezoids?

which of these is a trapezoid

Before giving the answer, let me first just remind you about my very strongly held position. I believe that instead of this typical textbook definition (the “exclusive definition” we’ll call it) that reads:

“A quadrilateral with one and only one pair of parallel sides.”

the definition should be made inclusive, and read:

“A quadrilateral with at least one pair of parallel sides.”

So the test question above was easy, right? Quadrilaterals (A) and (C) are trapezoids, I hear you say.

Not so fast!! If you’re using the inclusive definition, then the correct answers are actually (A), (B), (C), (D), and (E). But it gets better: If you were using the the exclusive definition, then NONE of these are trapezoids. In order for (A) and (C) to be trapezoids, under the exclusive definition, you must prove that two sides are parallel AND the two remaining sides are not parallel (and you can’t assume that from the picture…especially for (C)!).

Can you see the absurdity of the exclusive definition now?

I finish by offering the following list of reasons why the inclusive definition is better (can you suggest more reasons?):

  1. All other quadrilaterals are defined in the inclusive way, so that quadrilaterals “beneath” them inherit all the properties of their “parents.” A square is a rectangle because a square meets the definition of a rectangle. Likewise, parallelograms, rectangles, rhombuses, and squares should all be special cases of a trapezoid.
  2. The area formula for a trapezoid still works, even if the legs are parallel. It’s true! The area formula A=\frac{1}{2}h(b_1+b_2) works fine for a parallelogram, rectangle, rhombus, or square.
  3. No other definitions break when you use the inclusive definition. With the exception of the definition that some texts use for an isosceles trapezoid. Those texts define an isosceles trapezoid has having both legs congruent, which would make a parallelogram an isosceles trapezoid. Instead, define an isosceles trapezoid as having base angles congruent, or equivalently, having a line of symmetry.
  4. The trapezoidal approximation method in Calculus doesn’t fail when one of the trapezoids is actually a rectangle. But under the exclusive definition, you would have to change its name to the “trapezoidal and/or rectangular approximation method,” or perhaps ban people from doing the trapezoidal method on problems like this one: Approximate \int_0^4(4x-x^2)dx using the trapezoidal method with 5 equal intervals. (Note here that the center trapezoid is actually a rectangle…God forbid!!)
  5. When proving that a quadrilateral is a trapezoid, one can stop after proving just two sides are parallel. But with the exclusive definition, in order to prove that a quadrilateral is a trapezoid, you would have to prove two sides are parallel AND the other two sides are not parallel (see the beginning of this post!).

Arithmetic/Geometric Hybrid Sequences

Here’s a question that the folks who run the NCTM facebook page posed this week:

Find the next three terms of the sequence 2, 8, 4, 10, 5, 11, 5.5, …

Feel free to work it out. I’ll give you a minute.

Done?

still need more time?

..

give up?

Okay. The answer is 11.5, 5.75, 11.75.

The pattern is interesting. Informally, we might say “add 6, divide by 2.” This is an atypical kind of sequence, in which it seems as though we have two different rules at work in the same sequence. Let’s call this an Arithmetic/Geometric Hybrid Sequence. (Does anyone have a better name for these kinds of sequences?)

But a deeper question came out in the comments: Someone asked for the explicit rule. After a little work, I came up with one. I’ll give you my explicit rule, but you’ll have to figure out where it came from yourself:

a_n=\begin{cases}6-4\left(\frac{1}{2}\right)^{\frac{n-1}{2}}, & n \text{ odd} \\ 12-4\left(\frac{1}{2}\right)^{\frac{n-2}{2}}, & n \text{ even}\end{cases}

More generally, if we have a sequence in which we add d, then multiply by r repeatedly, beginning with a_1, the explicit rule is

a_n=\begin{cases}\frac{rd}{1-r}+\left(a_1-\frac{rd}{1-r}\right)r^{\frac{n-1}{2}}, & n \text{ odd} \\ \frac{d}{1-r}+\left(a_1-\frac{rd}{1-r}\right)r^{\frac{n-2}{2}}, & n \text{ even}\end{cases}.

And if instead we multiply first and then add, we have the following similar rule.

a_n=\begin{cases}\frac{d}{1-r}+\left(a_1-d-\frac{rd}{1-r}\right)r^{\frac{n-1}{2}}, & n \text{ odd} \\ \frac{rd}{1-r}+\left(a_1-d-\frac{rd}{1-r}\right)r^{\frac{n}{2}}, & n \text{ even}\end{cases}.

And there you have it! The explicit formulas for an Arithmetic/Geometric Hybrid Sequence :-).

(Perhaps another day I’ll show my work. For now, I leave it the reader to verify these formulas.)

Women & Math

Here’s a thoughtful TED talk from Laura Overdeck of bedtime math. I’ve highlighted this website before and I think it’s such a brilliant idea for kids! Laura was the lone girl in her astrophysics undergraduate studies, so she has a great vantage point from which to give this stellar talk (pun intended! :-) ).

She has lots of great points. In particular, I liked these three simple recommendations for women–and for everyone:

  1. Don’t play the lottery.
  2. Don’t say you hate math.
  3. Don’t ask others to calculate the tip.

 

Improper integrals debate

Here’s a simple Calc 1 problem:

Evaluate  \int_{-1}^1 \frac{1}{x}dx

Before you read any of my own commentary, what do you think? Does this integral converge or diverge?

image from illuminations.nctm.org

Many textbooks would say that it diverges, and I claim this is true as well. But where’s the error in this work?

\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_a^{1}\frac{1}{x}dx\right]

= \lim_{a\to 0^+}\left[\ln(a)-\ln(a)\right]=\boxed{0}

Did you catch any shady math? Here’s another equally wrong way of doing it:

\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^+}\left[\int_{-1}^{-a}\frac{1}{x}dx+\int_{2a}^{1}\frac{1}{x}dx\right]

= \lim_{a\to 0^+}\left[\ln(a)-\ln(2a)\right]=\boxed{\ln{\frac{1}{2}}}

This isn’t any more shady than the last example. The change in the bottom limit of integration in the second piece of the integral from a to 2a is not a problem, since 2a approaches zero if does. So why do we get two values that disagree? (In fact, we could concoct an example that evaluates to ANY number you like.)

Okay, finally, here’s the “correct” work:

\int_{-1}^1 \frac{1}{x}dx = \lim_{a\to 0^-}\left[\int_{-1}^{a}\frac{1}{x}dx\right]+\lim_{b\to 0^+}\left[\int_b^{1}\frac{1}{x}dx\right]

= \lim_{a\to 0^-}\left[\ln|a|\right]+\lim_{b\to 0^+}\left[-\ln|b|\right]

But notice that we can’t actually resolve this last expression, since the first limit is \infty and the second is -\infty and the overall expression has the indeterminate form \infty - \infty. In our very first approach, we assumed the limit variables a and b were the same. In the second approach, we let b=2a. But one assumption isn’t necessarily better than another. So we claim the integral diverges.

All that being said, we still intuitively feel like this integral should have the value 0 rather than something else like \ln\frac{1}{2}. For goodness sake, it’s symmetric about the origin!

In fact, that intuition is formalized by Cauchy in what is called the “Cauchy Principal Value,” which for this integral, is 0. [my above example is stolen from this wikipedia article as well]

I’ve been debating about this with my math teacher colleague, Matt Davis, and I’m not sure we’ve come to a satisfying conclusion. Here’s an example we were considering:

If you were to color in under the infinite graph of y=\frac{1}{x} between -1 and 1, and then throw darts at  the graph uniformly, wouldn’t you bet on there being an equal number of darts to the left and right of the y-axis?

Don’t you feel that way too?

(Now there might be another post entirely about measure-theoretic probability!)

What do you think? Anyone want to weigh in? And what should we tell high school students?

.

**For a more in depth treatment of the problem, including a discussion of the construction of Reimann sums, visit this nice thread on physicsforums.com.

Half-your-age-plus-seven rule

Looking for a great application of systems of linear inequalities for your Algebra 1 or 2 class? Look no further than today’s GraphJam contribution:

You might just give this picture to students and ask THEM to come up with the equations of the three lines.

There’s also a nice discussion to be had here about inverse functions, or about intersecting lines. And there might also be a good discussion about the domain of reasonableness.

Here are the three functions:

f_{\text{blue}}(x)=x

f_{\text{red}}(x)=\frac{1}{2}x+7

f_{\text{black}}(x)=2x-14

This is especially interesting because I never think of the rule as putting boundaries on a person’s dating age range. Usually people talk about it in the context of “how old of a person can I date?” not “how young of a person can I date?” Or rather, if you’re asking the second question, it’s usually phrased “how young of a person can date me?” (All of these questions relate to functions and their inverses!) But in fact, the half-your-age-plus-seven rule puts a lower and and upper bound on the ages of those you can date.

As far as reasonableness, is it fair to say that my daughter who is 1 can date someone who is between the age of -12 and 7.5? I don’t think so! I’m definitely going to be chasing off those -12 year-olds, I can already tell :-).

For my daughter, the domain of reasonableness might be x\geq 18!