# Lessons from teaching gifted elementary school students (Part 7a)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprising depth of mathematical knowledge.

Here’s a question I once received after a really big hailstorm:

How big would a 1000-pound hailstone be?

My head hurts thinking about hail that large. After about a minute of thinking, without using a calculator or even a pencil, I gave my answer: about a yard across.

I’ll reveal how I got this answer — which turns out to be a lot close than I had any right to expect — in tomorrow’s post. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own without using a calculator.

# Lessons from teaching gifted elementary students (Part 6b)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprising depth of mathematical knowledge.

Here’s a question I once received:

255/256 to what power is equal to 1/2? And please don’t use a calculator.

Here’s how I answered this question without using a calculator… in fact, I answered it without writing anything down at all. I thought of the question as

$\displaystyle \left( 1 - \epsilon \right)^x = \displaystyle \frac{1}{2}$.

$\displaystyle x \ln (1 - \epsilon) = \ln \displaystyle \frac{1}{2}$

$\displaystyle x \ln (1 - \epsilon) = -\ln 2$

I was fortunate that my class chose 1/2, as I had memorized (from reading and re-reading Surely You’re Joking, Mr. Feynman! when I was young) that $\ln 2 \approx 0.693$. Therefore, we have

$x \ln (1 - \epsilon) \approx -0.693$.

Next, I used the Taylor series expansion

$\ln(1+t) = t - \displaystyle \frac{t^2}{2} + \frac{t^3}{3} \dots$

to reduce this to

$-x \epsilon \approx -0.693$,

or

$x \approx \displaystyle \frac{0.693}{\epsilon}$.

For my students’ problem, I had $\epsilon = \frac{1}{256}$, and so

$x \approx 256(0.693)$.

So all I had left was the small matter of multiplying these two numbers. I thought of this as

$x \approx 256(0.7 - 0.007)$.

Multiplying $256$ and $7$ in my head took a minute or two:

$256 \times 7 = 250 \times 7 + 6 \times 7$

$= 250 \times (8-1) + 42$

$= 250 \times 8 - 250 + 42$

$= 2000 - 250 + 42$

$= 1750 + 42$

$= 1792$.

Therefore, $256 \times 0.7 = 179.2$ and $256 \times 0.007 = 1.792 \approx 1.8$. Therefore, I had the answer of

$x \approx 179.2 - 1.8 = 177.4 \approx 177$.

So, after a couple minutes’ thought, I gave the answer of 177. I knew this would be close, but I had no idea it would be so close to the right answer, as

$x = \displaystyle \frac{\displaystyle \ln \frac{1}{2} }{\displaystyle \ln \frac{255}{256}} \approx 177.0988786\dots$

# Lessons from teaching gifted elementary school students (Part 6a)

Every so often, I’ll informally teach a class of gifted elementary-school students. I greatly enjoy interacting with them, and I especially enjoy the questions they pose. Often these children pose questions that no one else will think about, and answering these questions requires a surprising depth of mathematical knowledge.

Here’s a question I once received:

255/256 to what power is equal to 1/2? And please don’t use a calculator.

Answering this question is pretty straightforward using algebra:

$\displaystyle \left( \frac{255}{256} \right)^x = \displaystyle \frac{1}{2}$.

$\displaystyle x \ln \frac{255}{256} = \ln \displaystyle \frac{1}{2}$

$x \displaystyle \frac{ \displaystyle \ln \frac{1}{2} }{\ln \displaystyle \frac{255}{256}}$

However, doing this without a calculator — and thus maintaining my image in front of these elementary school students — is a little formidable.

I’ll reveal how I did this — getting the answer correct to the nearest integer — in tomorrow’s post. In the meantime, I’ll leave a thought bubble if you’d like to think about it on your own.

# A natural function with discontinuities (Part 3)

This post concludes this series about a curious function:

In the previous post, I derived three of the four parts of this function. Today, I’ll consider the last part ($90^\circ \le \theta \le 180^\circ$).

The circle that encloses the grey region must have the points $(R,0)$ and $(R\cos \theta, R \sin \theta)$ on its circumference; the distance between these points will be $2r$, where $r$ is the radius of the enclosing circle. Unlike the case of $\theta < 90^\circ$, we no longer have to worry about the origin, which will be safely inside the enclosing circle.

Furthermore, this line segment will be perpendicular to the angle bisector (the dashed line above), and the center of the enclosing circle must be on the angle bisector. Using trigonometry,

$\sin \displaystyle \frac{\theta}{2} = \frac{r}{R}$,

or

$r = R \sin \displaystyle \frac{\theta}{2}$.

We see from this derivation the unfortunate typo in the above Monthly article.

# A natural function with discontinuities (Part 2)

Yesterday, I began a short series motivated by the following article from the American Mathematical Monthly.

Today, I’d like to talk about the how this function was obtained.

If $180^\circ \le latex \theta \le 360^\circ$, then clearly $r = R$. The original circle of radius $R$ clearly works. Furthermore, any circle that inscribes the grey circular region (centered at the origin) must include the points $(-R,0)$ and $(R,0)$, and the distance between these two points is $2R$. Therefore, the diameter of any circle that works must be at least $2R$, so a smaller circle can’t work.

The other extreme is also easy: if $\theta =0^\circ$, then the “circular region” is really just a single point.

Let’s now take a look at the case $0 < \theta \le 90^\circ$. The smallest circle that encloses the grey region must have the points $(0,0)$, $(R,0)$, and $(R \cos \theta, R \sin \theta)$ on its circumference, and so the center of the circle will be equidistant from these three points.

The center must be on the angle bisector (the dashed line depicted in the figure) since the bisector is the locus of points equidistant from $(R,0)$ and $(R \cos \theta, R \sin \theta)$. Therefore, we must find the point on the bisector that is equidistant from $(0,0)$ and $(R,0)$. This point forms an isosceles triangle, and so the distance $r$ can be found using trigonometry:

$\cos \displaystyle \frac{\theta}{2} = \displaystyle \frac{R/2}{r}$,

or

$r = \displaystyle \frac{R}{2} \sec \frac{\theta}{2}$.

This logic works up until $\theta = 90^\circ$, when the isosceles triangle will be a 45-45-90 triangle. However, when $\theta > 90^\circ$, a different picture will be needed. I’ll consider this in tomorrow’s post.

# A natural function with discontinuities (Part 1)

The following tidbit that was published on the American Mathematical Monthly’s Facebook page caught my attention:

Here’s the relationship between $r$, $R$, and $\theta$ in case it isn’t clear from the description. The gray sector is determined by $r$ and $\theta$, and then the blue circle with radius $r$ is chosen to enclose the sector.

Unfortunately, there was typo for the third case; it should have been $r = R \sin \frac{1}{2} \theta$ if $90^\circ \le \theta \le 180^\circ$. Here’s the graph if $R = 1$, using radians instead of degrees:

As indicated in the article, there’s a discontinuity at $t=0$. However, the rest of the graph looks nice and smooth.

Here’s the graph of the first derivative:

The first derivative is continuous (and so the original graph is smooth). However, there are obvious corners in the graph of the first derivative, which betray discontinuities in the graph of the second derivative:

# One Day More: Back to School version

Where I live, public schools reopen to students next week… which means innumerable back-to-school organizational meetings for our teachers. Last year, the good folks of West Des Moines put together this flash mob at their district-wide back-to-school meeting. After the superintendent’s deliberately dry introductory remarks are cut off, the fun begins at the 1:20 mark. (Be sure to turn on closed captioning to see the lyrics.)

# 100,000 page views

I’m taking a one-day break from my usual posts on mathematics and mathematics education to note a symbolic milestone: meangreenmath.com has had more than 100,000 total page views since its inception in June 2013. Many thanks to the followers of this blog, and I hope that you’ll continue to find this blog to be a useful resource to you.

Twenty most viewed posts or series (written by me):

Twenty most viewed posts (guest presenters):

1. Engaging students: Classifying polygons
2. Engaging students: Congruence
3. Engaging students: Deriving the distance formula
4. Engaging students: Distinguishing between axioms, postulates, theorems, and corollaries
5. Engaging students: Distinguishing between inductive and deductive reasoning
6. Engaging students: Factoring quadratic polynomials
7. Engaging students: Finding x- and y-intercepts
8. Engaging students: Laws of Exponents
9. Engaging students: Multiplying binomials
10. Engaging students: Order of operations
11. Engaging students: Pascal’s triangle
12. Engaging students: Right-triangle trigonometry
13. Engaging students: Solving linear systems of equations by either substitution or graphing
14. Engaging students: Solving linear systems of equations with matrices
15. Engaging students: Solving one-step and two-step inequalities
16. Engaging students: Solving quadratic equations
17. Engaging students: Square roots
18. Engaging students: Translation, rotation, and reflection of figures
19. Engaging students: Using right-triangle trigonometry
20. Engaging students: Volume and surface area of pyramids and cones

If I’m still here at that time, I’ll make a summary post like this again when this blog has over 200,000 page views.

# Why the Math Curriculum Makes No Sense

As a follow-up to yesterday’s post, there’s a lot of wisdom in the Math With Bad Drawings post “Why the Math Curriculum Makes No Sense.” I recommend it highly.

# Quantitative Literacy

March 1 saw the publication of the book The Math Myth: And Other STEM Delusions, by Andrew Hacker. MAA members are likely to recognize the author’s name from an opinion piece he published in the New York Times in 2012, with the arresting headline “Is Algebra Necessary?

On page 48, Hacker presents a question he took from an MCAT paper. It provides some technical data and asks what happens to the ratio of two inverse-square law forces between charges of given masses when the distance between them is halved. The context Hacker provides for this question is that medical professionals needs to be able to read and understand the mathematics used in technical papers. His claim is that this requirement does not extend to the physics of electrical and gravitational forces. In that, he is surely correct… What this question is asking for is, Do you understand what a ratio is? Surely that is something that any medical professional who will have to read and understand journal articles would need to know. Hacker completely misses this simple observation, and presents the question as an example of baroque mathematical testing run amok.

On page 70, he presents a question from an admissions test for selective high schools. A player throws two dice and the same number comes up on both. The question asks the student to choose the probability that the two dice sum to 9 from the list 0, 1/6, 2/9, 1/2, 1/3. Hacker’s problem is that the student is supposed to answer this in 90 seconds. Now, I share Hacker’s disdain for time-limited questions, but in this case the answer can only be 0. It’s not a probability question at all, and no computation is required. It just requires you to recognize that you can never get a sum of 9 when two dice show the same number. As with the MCAT question, the question is simply asking, Do you understand numbers? In this case, do you recognize that the sum of two equal numbers can never be odd…

You get the pattern surely? Hacker’s problem is he is unable to see through the surface gloss of a problem and recognize that in many cases it is just asking the student if she or he has a very basic grasp of number, quantity, and relationships. Yet these are precisely the kinds of abilities he argues elsewhere in the book are crucial in today’s world. He is, I suspect, a victim of the very kind of math teaching he rightly decries—one that concentrates on learning rules and mastering formal manipulations, with little attention to understanding.

My favorite response came from a very perceptive high school students in the New York Times’ letters to the editor (http://www.nytimes.com/2016/02/19/opinion/maths-place-in-the-classroom.html?ref=topics&_r=1):

In “Who Needs Math? Not Everybody” (Education Life, Feb. 7), Andrew Hacker, who teaches quantitative reasoning at Queens College, says that since only 5 percent of people use algebra and/or geometry in their jobs, students don’t need to learn these subjects.

As a high school student, I strongly disagree.

The point of learning is to understand the world. If the only point of learning is job preparation, why should students learn history, or read Shakespeare?

And while your job may never require you to know the difference between a postulate and a theorem, it will almost certainly require other math-based skills, like how to prove something or how to understand a graph. If nothing else, people need math to understand finance, which is a part of everyone’s life.

I also disagree with the logic that if people are failing algebra, then they shouldn’t take algebra. If people approach life that way, they will get nowhere.

Algebra and geometry have a place in the classroom. If students are failing, then the way math is taught may need to change. But what is taught needs no alteration.

Which is crying shame, because Hacker does have good ideas about what a quantitative literacy course should look like (again, from Devlin):

The tragedy of The Math Myth is that Hacker is actually arguing for exactly the kind of life-relevant mathematics education that I and many of my colleagues have been arguing for all our careers. (Our late colleague Lynn Steen comes to mind.) Unfortunately, and I suspect because Hacker himself did not have the benefit of a good math education, his understanding of mathematics is so far off base, he does not recognize that the examples he holds up as illustrations of bad education only seem so to him, because he misunderstands them.

The real myth in The Math Myth is the portrayal of mathematics that forms the basis of his analysis. It’s the same myth you see propagated in Facebook posts from frustrated parents about Common Core math homework their children bring home from school.