Difference of Two Powers (Index)

I’m doing something that I should have done a long time ago: collecting a series of posts into one single post. The following links comprised my series on getting students to discover the formula for factoring $x^n - y^n$ .

Part 1: A numerical way of discovering the formula for $x^2 - y^2$ .

Part 2: A geometric way of discovering the formula for $x^2 - y^2$ .

Part 3: Pedagogical thoughts on the importance of students discovering the formula for $x^3 - y^3$ .

Part 4: A geometric way of discovering the formula for $x^3 - y^3$ .

Part 5: Guiding students to the formula for $x^n - y^n$ .

Lessons from teaching gifted elementary students (Part 7b)

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?

The guesses from the students ranged from the size of a small car to the size of a large pick-up truck. Here’s how I gave a reasonably accurate answer using only mental arithmetic. This kind of describes the way that I try to size up things when only an approximation (and not an exact answer) is necessary.

First, I had some real-world experience that quickly told me the answer was going to be deceptively small. When I was young — maybe 10 or 12 years old — I was getting ready for a picnic, and I was assigned cut a block of ice — maybe a cubic foot of ice, if memory serves — into smaller chunks. (The party organizer bought a block of ice instead of a bag of ice to economize.) I remembered how incredibly heavy that block of ice was even though it wasn’t much larger than a basketball… several of us kids had a lot of trouble lifting the block of ice as we prepared to chop it into pieces. So, for the sake of argument, if that cubic foot of ice weighed about 100 pounds, then 8 cubic feet would weigh 800 pounds. So, based on that chance encounter with a block of ice when I was a kid, my guess would have been that the hailstone would measure 2 feet across.

Back to the problem at hand.

First, I converted to metric. I knew that there are about 2.2 pounds in a kilogram, and so I knew that the block would weigh something like 400 or 450 kilograms. I knew that I would be making plenty of crude approximations, so I just went with 400 kilograms and didn’t worry too much about immediately calculating $1000/2.2$ .

Next, I knew that metric units were originally defined so that a cubic centimeter of water weighs a gram, so that a 10 cm-by-10 cm-by 10 cm cube of water weighs one kilogram. Ice (hail) is slightly less dense than water (after all, ice floats in water), but for crude approximation purposes, I ignored this.

So, if a cube of ice with a side length of 1 decimeter (10 cm) weighs 1 kilogram, then a cube of ice with a side length of $\sqrt[3]{400}$ decimeters would weigh about 400 kilograms.

How big is $\sqrt[3]{400}$ ? Well, I have memorized that $7^3 = 343$ and $8^3 = 512$ , so it’s between 7 and 8 someplace… say 7.5. So the answer would be a cube of side length 7.5 decimeters. Also, I have memorized that 1 decimeter (10 cm) is approximately 4 inches, so the cube would have side length $7.5 \times 4 = 30$ inches.

Finally, hailstones are more spherical in shape than cubic, and a sphere of diameter $d$ has less volume than a cube of side length $d$ . So the answer should be a bit larger than 30 inches, so I just rounded up to a nice even number: 36 inches (one yard).

This calculation took me about a minute to do in my head and another half-minute to re-do to make sure I didn’t botch the arithmetic. So I held my hands about a yard apart (perhaps the crudest part of this calculation), pretending to hold a ball of diameter a yard across, and announced, “The hailstone would be about this big.”

Of course, a more thoughtful analysis produces the actual answer. The density of ice at the freezing point is 0.9167 grams per cubic centimeter, and 1 pound converts to 0.453592 kilograms. So:

$\displaystyle \frac{4\pi}{3} r^3 \times 0.9167 \frac{\hbox{grams}}{\hbox{cm}^3} = 1000 \hbox{pounds} \times \displaystyle 0.453592 \frac{\hbox{kilograms}}{\hbox{pounds}} \times 1000 \frac{\hbox{grams}}{\hbox{kilograms}}$

$r^3 \approx 118127 \hbox{cm}^3$

$r \approx 49.1 \hbox{cm}$

$r \approx 49.1 \hbox{cm} \times \frac{1 \hbox{inch}}{2.54 \hbox{cm}}$

$r \approx 19.3 \hbox{inches}$

Therefore, the sphere would have a diameter of twice that, or 38.6 inches.

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

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)

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

$-x \epsilon \approx -0.693$ ,

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

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)

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,

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

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

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:

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

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

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:

Source: https://www.facebook.com/AmerMathMonthly/photos/a.250425975006394.53155.241224542593204/1021059947942989/?type=3&theater

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):

Engaging students: Classifying polygons
Engaging students: Congruence
Engaging students: Deriving the distance formula
Engaging students: Distinguishing between axioms, postulates, theorems, and corollaries
Engaging students: Distinguishing between inductive and deductive reasoning
Engaging students: Factoring quadratic polynomials
Engaging students: Finding x- and y-intercepts
Engaging students: Laws of Exponents
Engaging students: Multiplying binomials
Engaging students: Order of operations
Engaging students: Pascal’s triangle
Engaging students: Right-triangle trigonometry
Engaging students: Solving linear systems of equations by either substitution or graphing
Engaging students: Solving linear systems of equations with matrices
Engaging students: Solving one-step and two-step inequalities
Engaging students: Solving quadratic equations
Engaging students: Square roots
Engaging students: Translation, rotation, and reflection of figures
Engaging students: Using right-triangle trigonometry
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.