Accommodating Blind Students Taking Mathematics

This was another interesting pedagogical article from the February/March 2014 issue of FOCUS, published by the Mathematical Association of America. I had never heard of the Nemeth code, the braille for mathematical notation.

http://digitaleditions.walsworthprintgroup.com/display_article.php?id=1639571

Flipping to Offer Low-Enrollment Courses

I just read a very interesting article about how an instructor at Ohio Dominican University is simultaneously teaching several upper-level mathematics courses with low enrollment by flipping the classroom (using the catchy title One-Room Schoolhouse). Here’s the article: http://digitaleditions.walsworthprintgroup.com/display_article.php?id=1639570

Approximating pi

I was recently interviewed by my city’s local newspaper about $\pi$ Day and the general fascination with memorizing the digits of $\pi$ . I was asked by the reporter if the only constraint in our knowledge of the digits of $\pi$ was the ability of computers to calculate the digits, and I answered in the affirmative.

Here’s the current state-of-the-art for calculating the digits of $\pi$ . Amazingly, this expression was discovered 1995… in other words, very recently.

$\pi = \displaystyle \sum_{n=0}^\infty \frac{1}{16^n} \left( \frac{4}{8n+1} - \frac{2}{8n+4} - \frac{1}{8n+5} - \frac{1}{8n+6} \right)$

Because of the term $16^n$ in the denominator, this infinite series converges very quickly.

Proof: If $k < 8$ , then we calculate the integral $I_k$ , defined below:

$I_k = \displaystyle \int_0^{1/\sqrt{2}} \frac{x^{k-1}}{1-x^8} dx$

$= \displaystyle \int_0^{1/\sqrt{2}} x^{k-1} \sum_{n=0}^\infty x^{8n} dx$

$= \displaystyle \int_0^{1/\sqrt{2}} \sum_{n=0}^\infty x^{8n+k-1} dx$

$= \displaystyle \sum_{n=0}^\infty \int_0^{1/\sqrt{2}} x^{8n+k-1} dx$

$= \displaystyle \sum_{n=0}^\infty \left[ \frac{x^{8n+k}}{8n+k} \right]^{1/\sqrt{2}}_0$

$= \displaystyle \sum_{n=0}^\infty \frac{1}{8n+k} \left[ \left( \frac{1}{\sqrt{2}} \right)^{8n+k} - 0 \right]$

$= \displaystyle \sum_{n=0}^\infty \frac{1}{2^{k/2}} \frac{1}{16^n (8n+k)}$

We now form the linear combination $P = 4\sqrt{2} I_1 - 8 I_4 - 4\sqrt{2} I_5 - 8 I_6$ :

$P = \displaystyle \sum_{n=0}^\infty \left( \frac{4\sqrt{2}}{2^{1/2}} \frac{1}{16^n (8n+1)} - \frac{8}{2^{4/2}} \frac{1}{16^n (8n+4)} - \frac{4\sqrt{2}}{2^{5/2}} \frac{1}{16^n (8n+5)} - \frac{8}{2^{6/2}} \frac{1}{16^n (8n+6)} \right)$

$P = \displaystyle \sum_{n=0}^\infty \frac{1}{16^n} \left( \frac{4}{8n+1} - \frac{2}{8n+4} - \frac{1}{8n+5} - \frac{1}{8n+6} \right)$

Also, from the original definition of the $I_k$ ,

$P = \displaystyle \int_0^{1/\sqrt{2}} \frac{4\sqrt{2} - 8x^3 -4\sqrt{2} x^4 - 8x^5}{1-x^8} dx$ .

Employ the substitution $x = y/\sqrt{2}$ :

$P = \displaystyle \int_ 0^1 \frac{4\sqrt {2} - 2\sqrt {2} y^3 - \sqrt {2} y^4 - \sqrt {2} y^5}{1 - y^8/16}\frac {dy} {\sqrt {2}}$

$P = \displaystyle \int_ 0^1 \frac{16 (4 - 2 y^3 - y^4 - y^5)}{16 - y^8} dy$

$P = \displaystyle \int_0^1 \frac{16(y-1)(y^2+2)(y^2+2y+2)}{(y^2-2)(y^2+2)(y^2+2y+2)(y^2-2y+2)} dy$

$P = \displaystyle \int_0^1 \frac{16y-16}{(y^2-2)(y^2-2y+2)} dy$

Using partial fractions, we find

$P = \displaystyle \int_ 0^1\frac{4 y}{y^2 - 2} dy - \int_ 0^1 \frac{4 y - 8}{y^2 - 2 y + 2} dy$

The expression on the right-hand side can be simplified using standard techniques from Calculus II and is equal to $\pi$ .

So that’s the proof… totally accessible to a student who has mastered concepts in Calculus II. But this begs the question: how in the world did anyone come up with the idea of starting with the integrals $I_k$ to develop an infinite series that leads to $\pi$ ? Let me quote from page 118 of J. Arndt and C. Haenel, $\pi -$ Unleashed (Springer, New York, 2000):

Certainly not by chance, even if luck played some part in the discovery. All three parties [David Bailey, Peter Borwein and Simon Plouffe] are established mathematicians who have been working with the number $\pi$ for a considerable time… Yet the series was not discovered through mathematical deduction or inference. Instead, the researchers used a tool called Computer Algebra System and a particular procedure called the “PSQL algorithm” to generate their series. They themselves write that they found their formula “through a combination of inspired testing and extensive searching.”

The original paper that announced the discovery of this series can be found at http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P123.pdf.

MyScript MathPad: Handwriting LaTeX generator

LaTeX offers a free way for typing mathematical equations (for example, all of the mathematical equations on this website are written in LaTeX). Here’s an app for iPhones and iPads that I just found that says it can convert hand-written equations into LaTeX: https://itunes.apple.com/us/app/myscript-mathpad-handwriting/id674996719?mt=8

The Law of Averages

Colloquially, the Law of Averages dictates that what ought to happen does happen if it happens long enough. If a gambler plays a casino for a very long time, he is almost certainly guaranteed to lose. If a weak sports team plays a stronger team in a multiple-game series, then it is almost certain to lose the series.

However, if the gambler plays in the casino for only a little while, then there is a realistic (though less than 50%) chance of coming out ahead. And a weak sports team may defeat a stronger one if only one game is played… hence the appeal of the NCAA basketball tournament and, on a larger scale, the knockout stages of the World Cup.

In my statistics class, I use a simple simple spreadsheet to illustrate that $\hbox{SD}(K) = \sqrt{n p(1-p})$ for a sample count, but $\hbox{SD}(\hat{p}) = \displaystyle \sqrt{ \frac{p(1-p)}{n} }$ for a sample proportion. Here is one image from the spreadsheet:

The user can change the bright green cell to be any positive integer up to 5000. This number represents the number of simulated coins that are flipped. In the above example, ten coins are flipped. Column B shows the results of the simulated coins, while column C shows a running count of the number of heads that have appeared. In the above example, 7 of the 10 flips are heads, for an observed error of +2 (two more heads than the expected number of 5) and a percentage error of 20%.

In class, I would run the spreadsheet several times, and students will see that the observed error usually is in the range of -2 to +2, and the percentage error is usually in the range of -20% to +20%.

By contrast, look what happens when the number of flips increases to a large number, like 5000.

There is now a larger absolute error — in this case, -28. Of course, an absolute error of that size is impossible with 10 coin flips or even 50 coin flips. However, at the same time, the percentage error is now significantly smaller (only -0.56%).

This example gives evidence for the counter-intuitive result that the absolute error grows like $\sqrt{n}$ while the relative error decreases like $\sqrt{n}$ .

Why do we still require students to rationalize denominators?

Which answer is simplified: $\displaystyle \frac{1}{2 \sqrt{2}}$ or $\displaystyle \frac{ \sqrt{2} }{4}$ ? From example, here’s a simple problem from trigonometry:

Suppose $\theta$ is an acute angle so that $\sin \theta = \displaystyle \frac{1}{3}$ . Find $\tan \theta$ .

To solve, we make a right triangle whose side opposite of $\theta$ has length $1$ and hypotenuse with length $3$ . The adjacent side has length $\sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$ . Therefore,

$\tan \theta = \displaystyle \frac{ \hbox{Opposite} }{ \hbox{Adjacent} } = \displaystyle \frac{1}{2 \sqrt{2}}$

This is the correct answer, and it could be plugged into a calculator to obtain a decimal approximation. However, in my experience, it seems that most students are taught that this answer is not yet simplified, and that they must rationalize the denominator to get the “correct” answer:

$\tan \theta = \displaystyle \frac{1}{2 \sqrt{2}} \cdot \frac{ \sqrt{2} }{ \sqrt{2} } = \displaystyle \frac{ \sqrt{2} }{4}$

Of course, this is equivalent to the first answer. So my question is philosophical: why are students taught that the first answer isn’t simplified but the second is? Stated another way, why is a square root in the numerator so much more preferable than a square root in the denominator?

Feel free to correct me if I’m wrong, but it seems to me that rationalizing denominators is a vestige of an era before cheap pocket calculators. Let’s go back in time to an era before pocket calculators… say, 1927, when The Jazz Singer was just released and stars of silent films, like Don Lockwood, were trying to figure out how to act in a talking movie.

Before cheap pocket calculators, how would someone find $\displaystyle \frac{1}{2 \sqrt{2}} ~~$ or $~~ \displaystyle \frac{ \sqrt{2} }{4}$ to nine decimal places? Clearly, the first step is finding $\sqrt{2}$ by hand, which I discussed in a previous post. So these expressions reduce to

$\displaystyle \frac{1}{2 (1.41421356\dots)}$ or $\displaystyle \frac{1.41421356\dots}{4}$

Next comes the step of dividing. If you don’t have a calculator and had to use long division, which would rather do: divide by $4$ or divide by $2.82842712\dots$ ?

Clearly, long division with $4$ is easier.

It seems to me that ease of computation was the reason that rationalizing denominators was required of students in previous generations. So I’m a little bemused why rationalizing denominators is still required of students now that cheap calculators are so prevalent.

Lest I be misunderstood, I absolutely believe that all students should be able to convert $\displaystyle \frac{1}{2 \sqrt{2}}$ into $\displaystyle \frac{ \sqrt{2} }{4}$ . But I see no compelling reason why the “simplified” answer to the above trigonometry problem should be the second answer and not the first.

Collaborative Mathematics: Challenge 07

My colleague Jason Ermer has posted his 7th challenge video, shown below. It’s both an experiment and an exercise in probability.

Video responses can be posted to his website, http://www.collaborativemathematics.org. In the words of his website, this is a unique forum for connecting a worldwide community of mathematical problem-solvers, and I think these unorthodox but simply stated problems are a fun way for engaging students with the mathematical curriculum.

Engaging students: Vectors in two dimensions

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 comes from my former student Derek Skipworth. His topic, from Precalculus: vectors in two dimensions.

A. How could you as a teacher create an activity or project that involves your topic?

While it may be a cop-out to use this example since I am developing it for an actual lesson plan, I will go ahead and use it because I feel it is a strong activity. I am developing a series of 21 problems that will be the base for forming the students’ treasure maps. There will be three jobs: Cartographer, the map maker; Lie Detector, who checks for orthogonality; and Calculator, who will solve the vector problems. The 21 problems will be broken down into 7 per page, and the students will switch jobs after each page. The rule is that any vectors that are orthogonal with each other cannot be included in your map. There are three of these on each page, so each group should end up with a total of 12 vectors on their map. Once orthogonality is checked by the Lie Detector, the Calculator will do the expressed operations on the vector pairs to come up with the vector to be drawn. The map maker will then draw the vector, as well as the object the vector leads to. Each group will have their directions in different orders so that every group has their own unique map. The idea is for the students to realize (if they checked orthogonality correctly) that, even though every map is different, the sum of all vectors still leads you to the same place, regardless of order.

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

Vectors build upon many topics from previous courses. For one, it teaches the student to use the Cartesian plane in a new way than they have done previously. Vectors can be expressed in terms of force in the $x$ and $y$ directions, which result in a representation very similar to an ordered pair. It gets expanded to teach the students that unlike an ordered pair, which represents a distinct point in space, a vector pair represents a specific force that can originate from any point on the Cartesian Plane.

Vectors also build on previous knowledge of triangles. When written as $\langle x,y \rangle$ , we can find the magnitude of the vector by using the Pythagorean Theorem. It gives them a working example of when this theorem can be applied on objects other than triangles. It also reinforces the students trigonometry skills since the direction of a vector can also be expressed using magnitude and angles.

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

The PhET website has one of the best tools I’ve seen for basic knowledge of two dimensional vector addition, located at http://phet.colorado.edu/en/simulation/vector-addition. This is a java-based program that lets you add multiple vectors (shown in red) in any direction or magnitude you want to get the sum of the vectors (shown in green). Also shown at the top of the program is the magnitude and angle of the vector, as well as its corresponding $x$ and $y$ values.

What’s great about this program is it puts the power in the student’s hands. They are not forced to draw multiple sets of vectors themselves. Instead, they can quickly throw them in the program and manipulate them without any hassle. This effectively allows the teacher to cover the topic quicker and more effectively due to the decreased amount of time needed to combine all vectors on a graph.

Collaborative Mathematics: Challenge 05

My colleague Jason Ermer is back from summer hiatus and has posted his fifth challenge video, shown below.

Thoughts on 1/7 and other rational numbers (Part 10)

In the previous post, I showed a quick way of obtaining a full decimal representation using a calculator that only displays ten digits at a time. To review: here’s what a TI-83 Plus returns as the (approximate) value of $8/17$ :

Using this result and the Euler totient function, we concluded that the repeating block had length $16$ . So we multiply twice by $10^8$ (since $10^8 \times 10^8 = 10^{16}$ ) to deduce the decimal representation, concluding that

$\displaystyle \frac{8}{17} = 0.\overline{4705882352941176}$

Though this is essentially multi-digit long division, most students are still a little suspicious of this result on first exposure. So here’s a second way of confirming that we did indeed get the right answer. The calculators show that

$8 \times 10^8 = 17 \times 47058823 + 9$ and $9 \times 10^8 = 17 \times 52941176 + 8$