Advertising for slide rules, from 1940

I’m about to begin a series of posts concerning how previous generations did complex mathematical calculations without the aid of scientific calculators.

Courtesy of Slide Rule Universe, here’s an advertisement for slide rules from 1940. This is a favorite engagement activity of mine when teaching precalculus (as an application of logarithms) as well as my capstone class for future high school math teachers. I have shown this to hundreds of college students over the years (usually reading out loud the advertising through page 5 and then skimming through the remaining pictures), and this always gets a great laugh. Enjoy.

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I’m about to do a series of posts concerning square roots and logarithms. So I thought that this picture would be an appropriate introduction to the topic. (One of these days, I’ll spring for the wall poster version of this picture to hang in my office.)

Source: http://www.xkcd.com/482/

Engaging students: Solving logarithmic equations

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Caitlin Kirk. Her topic: how to engage Algebra II or Precalculus students when solving logarithmic equations.

B. Curriculum: How does this topic extend what your students should have learned in previous courses?

Logarithms are a topic that appears at multiple levels of high school math. In Algebra II, students are first introduced to logarithms when they are asked to identify graphs of parent functions including f (x) = log_ax. Later in the same class, they learn to formulate equations and inequalities based on logarithmic functions by exploring the relationship between logarithms and their inverses. From there, they can develop a definition of a logarithm.

Solving logarithmic equations extends what students learned about logarithms in Algebra II. Once a proper definition of logarithms has been established, along with a graphical foundation of logs, students learn to solve logarithmic equations. Properties of logarithms are used to expand, condense, and solve logarithms without a calculator in Pre Calculus. Practical applications of the logarithmic equation also follow from previous skills. Students learn to calculate the pH of a solution, decibel voltage gain, intensity of earthquakes measure on the Richter scale, depreciation, and the apparent loudness of sound using logarithms.

C. Culture: How has this topic appeared in the news?

One application of logarithmic equations is calculating the intensity of earthquakes measured on the Richter scale using the following equation:

$R = \log(A/P)$

where $A$ is the amplitude of the tremor measured in micrometers and $P$ is the period of the tremor (time of one oscillation of the earth’s surface) measured in seconds.

Reports of earthquake activity appear in the news often and are always accompanied by a measurement from the Richter scale. One such report can be found here: http://www.bbc.co.uk/news/world-asia-20638696. As the story says, a 7.3 magnitude earthquake struck off the coast of Japan in December of 2012, and created a small tsunami. There were six aftershocks of this quake whose Richter scale measurements are also given. The article also explains how Japan has been able to enact an early warning system that predicts the intensity of an earthquake before it causes damage. All of the calculations given in this story, and almost all others involving earthquakes, involves the use of the Richter scale logarithmic equation.

D. History: What are the contributions of various cultures to this topic?

The development of logarithms saw contributions from several different countries beginning with the Babylonians (2000-1600 BC) who developed the first known mathematical tables. They also introduced square multiplication in which they simply but accurately multiplied two numbers using only addition and subtraction. Michael Stifel, of Germany, was the first mathematician to use an exponent in 1544. He developed an early version of the logarithmic table containing integers and powers of 2. Perhaps the most important contribution to logarithms came from John Napier in Scotland in 1619. He, like the Babylonians, was working with on breaking multiplication, division, and root extraction down to only addition and subtraction. Therefore, he created the “logarithm” $L$ of a number $N$ defined as follows:

$N = 10^7 (1-10^{-7})^L$

for which he wrote $\hbox{NapLog}(N) = L$ .

Napier’s definition of the logarithm led to the following logarithmic identities that are still taught today:

$\hbox{NapLog}(\sqrt{N_1N_2}) = \frac{1}{2} (\hbox{NapLog} N_1 + \hbox{NapLog} N_2)$

$\hbox{NapLog}(10^{-7} N_1 N_2) = \hbox{NapLog} N_1 + \hbox{NapLog} N_2$

$\hbox{NapLog} \left( 10^{-7} \displaystyle \frac{N_1}{N_2} \right)= \hbox{NapLog} N_1 - \hbox{NapLog} N_2$

Henry Briggs, in England, published his work on logarithms in 1624, which included logarithms of 30,000 natural numbers to the 14th decimal place worked by hand! Shortly after, back in Germany, Johannes Kepler used a logarithmic scale on a Cartesian plane to create a linear graph the elliptical shape of the cosmos. In 1632, in Italy, Bonaventura Cavalieri published extensive tables of logarithms including the logs of trig functions (excluding cosine). Finally, Leonhard Euler made one of the most commonly known contributions to logarithms by making the number $e = 2.71828\dots$ the base of the natural logarithm (which was also developed by Napier). While it is untrue, as is commonly believed, that Euler invented the number $2.71828\dots$ , he did give it the name $e$ . He was interested in the number because he wanted to calculate the amount that would result from continually compounded interested on a sum of money and the number $2.71828\dots$ kept appearing as a constant in his equation. Therefore he tied $e$ to the natural logarithm that was not as widely used because it did not have a base.

Logarithms were developed as a result of the contributions of many cultures spanning Europe and beyond, dating back over 4000 years.

Why does x^0 = 1 and x^(-n) = 1/x^n? (Part 2)

I distinctly remember when, in my second year as a college professor, a really good college student — with an SAT Math score over 650 — asked me why $x^0 = 1$ and $x^{-n} = \displaystyle \frac{1}{x^n}$ . Of course, he knew that these rules were true and he could apply them in complex problems, but he didn’t know why they were true. And he wanted to have this deeper knowledge of mathematics beyond the ability to solve routine algebra problems.

He also related that he had asked his math teachers in high school why these rules worked, but he never got a satisfactory response. So he asked his college professor.

Looking back on it, I see that this was one of the incidents that sparked my interest in teacher education. As always, I never hold a grudge against a student for asking a question. Indeed, I respected my student for posing a really good question, and I was upset for him that he had not received a satisfactory answer to his question.

This is the second of two posts where I give two answers to this question from two different points of view.

Answer #2. This explanation relies on one of the laws of exponents:

$x^n \cdot x^m = x^{n+m}$

For positive integers $n$ and $m$ , this can be proven by repeated multiplication:

$x^n x^m = (x \cdot x \dots \cdot x) \cdot (x \cdot x \dots \cdot x)$ repeated $n$ times and $m$ times

$x^n x^m = x \cdot x \cdot \dots \cdot x \cdot x \cdot \dots \cdot x$ repeated $n+m$ times

$x^n \cdot x^m = x^{n+m}$

Ideally, $x^0$ and $x^{-n}$ should be defined so that this rule still holds even if one (or both) of $n$ and $m$ is either zero or a negative integer. In particular, we should define $x^0$ so that the following rule holds:

$x^n \cdot x^0 = x^{n+0}$

$x^n \cdot x^0 = x^n$

In other words, the product of something with $x^0$ should be the original something. Clearly, the only way to make this work is if we define $x^0 = 1$ .

In the same way, we should define $x^{-n}$ so that the following rule holds:

$x^n \cdot x^{-n} = x^{n + (-n)}$

$x^n \cdot x^{-n} = x^0$

Being a good MIT freshman and using previous work, we see that

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

Dividing, we see that

$x^{-n} = \displaystyle \frac{1}{x^n}$

Why does x^0 = 1 and x^(-n) = 1/x^n? (Part 1)

He also related that he had asked his math teachers in high school why these rules worked, but he never got a satisfactory response. So he asked his college professor.

This is the first of two posts where I give two answers to this question from two different points of view.

Answer #1. Let’s recall the definition of $x^n$ for positive integers $x$ :

$x^4 = x \cdot x \cdot x \cdot x \cdot x$

$x^3 = x \cdot x \cdot x$

$x^2 = x \cdot x$

$x^1 = x$

Starting from the bottom, the exponents increase by $1$ with each step up, while an extra $x$ is multiplied with each step up.

Of course, there’s no reason why we can’t proceed downward instead of upward. With each step down, the exponents decrease by $1$ , while an extra $x$ is divided from the right-hand side. So it makes sense to define

$x^0 = \displaystyle \frac{x}{x} = 1$ .

We can continue decreasing the exponent — into the negative numbers — by continuing to divide by $x$ :

$x^{-1} = \displaystyle \frac{1}{x}$

$x^{-2} = \displaystyle \frac{1/x}{x} = \displaystyle \frac{1}{x^2}$

$x^{-3} = \displaystyle \frac{1/x^2}{x} = \displaystyle \frac{1}{x^3}$

$x^{-4} = \displaystyle \frac{1/x^3}{x} = \displaystyle \frac{1}{x^4}$

We see that, if $n$ is positive and $-n$ is negative, that $x^{-n} = \displaystyle \frac{1}{x^n}$ .

Technically, this only provides an explanation for this rule for negative integers. However, I haven’t met a student that didn’t believe this rule held for negative rational exponents (or negative irrational exponents) after seeing the above explanation for negative integers.

Unsolved problems: the Collatz conjecture

Students at all levels — elementary, middle, secondary, and college — tend to think that either (1) all the problems in mathematics have already been solved, or else (2) some unsolved problems remain but only an expert can understand even the statement of the problem.

There are plenty of famous unsolved problems in mathematics. And the Collatz conjecture is an easily stated unsolved problem that can be understood by most fourth and fifth graders.

Here’s the statement of the problem.

Start with any positive integer.
If the integer is even, divide it by $2$ . If it’s odd, multiply it by $3$ and then add $1$ .
Repeat until (and if) you reach $1$ .

That’s it. From Wikipedia:

For instance, starting with 6, one gets the sequence 6, 3, 10, 5, 16, 8, 4, 2, 1.

Starting with 11, for example, takes longer to reach 1: 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

The sequence for 27 takes 111 steps, climbing to 9232 before descending to 1: 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1

The longest progression for any initial starting number less than 100 million is 63,728,127, which has 949 steps. For starting numbers less than 1 billion it is 670,617,279, with 986 steps, and for numbers less than 10 billion it is 9,780,657,630, with 1132 steps.

Here’s the question: Does this sequence eventually reach $1$ no matter the starting value? Or is there a number out there that you could use as a starting value that has a sequence that never reaches $1$ ?

Like I said, this is an easily stated problem that most fourth graders could understand. And no one knows the answer. Every number that’s been tried by computer has produced a sequence that eventually reaches $1$ . But that doesn’t mean that there isn’t a bigger number out there that doesn’t reach $1$ .

I’ll refer to the above Wikipedia page (and references therein) for further reading about the Collatz conjecture. Pedagogically, I suggest that casually mentioning this unsolved problem in class might inspire students to play with mathematics on their own, rather than think that all of mathematics has already been solved by somebody.

Source: http://www.xkcd.com/710/

Math class needs a makeover

This is a thought-provoking TEDx talk by Dan Meyer (who also produced the Hollywood Hates Math video).

Taylor series without calculus

Is calculus really necessary for obtaining a Taylor series? Years ago, while perusing an old Schaum’s outline, I found a very curious formula for the area of a circular segment:

$A = \displaystyle \frac{R^2}{2} (\theta - \sin \theta)$

The thought occurred to me that $\theta$ was the first term in the Taylor series expansion of $\sin \theta$ about $\theta = 0$ , and perhaps there was a way to use this picture to generate the remaining terms of the Taylor series.

This insight led to a paper which was published in College Mathematics Journal: cmj38-1-058-059. To my surprise and delight, this paper was later selected for inclusion in The Calculus Collection: A Resource for AP and Beyond, which is a collection of articles from the publications of Mathematical Association of America specifically targeted toward teachers of AP Calculus.

Although not included in the article, it can be proven that this iterative method does indeed yield the successive Taylor polynomials of $\sin \theta$ , adding one extra term with each successive step.

I carefully scaffolded these steps into a project that I twice assigned to my TAMS precalculus students. Both semesters, my students got it… and they were impressed to know the formula that their calculators use to compute $\sin \theta$ . So I think this project is entirely within the grasp of precocious precalculus students.

I personally don’t know of a straightforward way of obtaining the expansion of $\cos \theta$ without calculus. However, once the expansion of $\sin \theta$ is known, the expansion of $\cos \theta$ can be surmised without calculus. To do this, we note that

$\cos \theta = 1 - 2 \sin^2 \left( \displaystyle \frac{\theta}{2} \right) = 1 - 2 \left( \displaystyle \frac{\theta}{2} - \frac{(\theta/2)^3}{3!} + \frac{(\theta/2)^5}{5!} \dots \right)^2$

Truncating the series after $n$ terms and squaring — and being very careful with the necessary simplifications — yield the first $n$ terms in the Taylor series of $\cos \theta$ .

Engaging students: Defining the words acute, right, and obtuse

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission again comes from my former student Jesse Faltys. Her topic: how to engage geometry students when defining the words acute, right, and obtuse.

E. TECHNOLOGY: How can technology be used to effectively engage students with this topic?

ACUTE, OBTUSE, and RIGHT Angles Song

This is a great video for the end of the lesson when first introducing acute, right, and obtuse angles. A little corny but it’s always helpful to link new knowledge to a song. Music brings back memories or in this situation recognition. By using creative things, you are helping the students reinforce new ideas. Just hearing words will not help us retain the information, but adding the words to a song help reinforce the reminder for the information. We can remember anything if we just put our minds to it. The kids in the video are singing lyrics about right, obtuse and acute angles to the song Old McDonald Had a Farm. The video helps the students to summarize their understanding of the three new terms and a way to retain it for future use.

http://www.watchknowlearn.org/Video.aspx?VideoID=2446

D. HISTORY: How have different cultures throughout time used this topic in their society?

In Egypt as far back as 1500BC, measurements were taken of the Sun’s shadow against graduations marked on stone tables. These measurements are just different angles used to show time with some degree of accuracy. Gromas were used for the purpose of construction in ancient Egypt. Gromas were right-angle devices that the ancient Egyptians used when they began construction project by surveying an area. They could sketch out long lines at right angles. The Romans will actually use the same tool to sketch out their roads. 1,713 years ago they were using right angles. This might be important.

http://www.fig.net/pub/cairo/papers/wshs_01/wshs01_02_wallis.pdf

C. Culture: How has this topic appeared in pop culture (movies, TV, current music, video games, etc.)?

Angry-Birds: “Use the unique powers of the Angry Birds to destroy the greedy pigs’ fortresses!“ Angry-Birds is an app that is played by a large percentage of children on a daily basis. Birds are positioned on a slingshot and launched at pigs that are resting on different structures. We create a zero plane from the bird sitting in the slingshot, releasing the bird, and mark the maximum height reached. We now have an angle. The bird has created an angle with its path. Can we classify the majority of these angles as acute, right or obtuse?

Bubble Shooter: A Puzzle game that will help you stay busy for a while!

The point of the game is to remove all the spheres by matching like colors. The “cannon” at the bottom of the page is your tool to directing the sphere were you want it to go. You can directly shot the sphere or you can bounce off the edge of the wall. Here is the trick, what kind of angle do you need to deliver your sphere. One of the helpful hints from the website, “you can use the left and right border to bounce new balls in more advanced angles.” These advanced angles can be denoted as acute, right or obtuse.

http://www.shooter-bubble.com/

Arctangents and showmanship

This story comes from Fall 1996, my first semester as a college professor. I was teaching a Precalculus class, and the topic was vectors. I forget the exact problem (believe me, I wish I could remember it), but I was going over the solution of a problem that required finding $\tan^{-1}(7)$ . I told the class that I had worked this out ahead of time, and that the approximate answer was $82^o$ . Then I used that angle for whatever I needed it for and continued until obtaining the eventual solution.

(By the way, I now realize that I was hardly following best practices by computing that angle ahead of time. Knowing what I know now, I should have brought a calculator to class and computed it on the spot. But, as a young professor, I was primarily concerned with getting the answer right, and I was petrified of making a mistake that my students could repeat.)

After solving the problem, I paused to ask for questions. One student asked a good question, and then another.

Then a third student asked, “How did you know that $\tan^{-1}(7)$ was $82^o$ ?

Suppressing a smile, I answered, “Easy; I had that one memorized.”

The class immediately erupted… some with laughter, some with disbelief. (I had a terrific rapport with those students that semester; part of the daily atmosphere was the give-and-take with any number of exuberant students.) One guy in the front row immediately challenged me: “Oh yeah? Then what’s $\tan^{-1}(9)$ ?

I started to stammer, “Uh, um…”

“Aha!” they said. “He’s faking it.” They start pulling out their calculators.

Then I thought as fast as I could. Then I realized that I knew that $\tan 82^o \approx 7$ , thanks to my calculation prior to class. I also knew that $\displaystyle \lim_{x \to 90^-} \tan x = \infty$ since the graph of $y = \tan x$ has a vertical asymptote at $x = \pi/2 = 90^o$ . So the solution to $\tan x = 9$ had to be somewhere between $82^o$ and $90^o$ .

So I took a total guess. “ $84^o$ ,” I said, faking complete and utter confidence.

Wouldn’t you know it, I was right. (The answer is about $83.66^o$ .)

In stunned disbelief, the guy who asked the question asked, “How did you do that?”

I was reeling in shock that I guessed correctly. But I put on my best poker face and answered, “I told you, I had it memorized.” And then I continued with the next example. For the rest of the semester, my students really thought I had it memorized.

To this day, this is my favorite stunt that I ever pulled off in front of my students.