Posted by: Dave Richeson | August 30, 2012

## Ancient number systems in XeTeX

I am teaching a history of mathematics class this semester. We are beginning with a brief discussion of ancient number systems: Egyptian, Babylonian, Mayan, Chinese, Incan, GreekRoman, and Hindu-Arabic. As I was writing up the first homework assignment it occured to me that I should investigate whether these numbers could be typeset using LaTeX.

It quickly became apparent that, because fonts are involved, I would have to use XeTeX rather than LaTeX. It was a fun (although time consuming) exercise. In the end I was able to typeset Egyptian hieroglyphics, Babylonian cuneiform, and Chinese rod numerals. Because the syntax was often messy, I spent a while burying the complicated TeX in the headers so that the numbers would be easy to work with in the document.

For example, to generate the Egyptian hieroglyphics for 123 I write

\Ehun\Eten\Eten\Eone\Eone\Eone.

The fraction 1/123 is

\Efrac{\Ehun\Eten\Eten\Eone\Eone\Eone}.

To express 123 in cuneiform all I have to write is

\Bnum{123}.

To create a number board with the Chinese counting rods representing 123 I type

\Cnum{|x|x|x|}{\Cvone & \Chtwo & \Cvthree}.

If you would like to give this a try, download my .tex files:

I’d love to be able to do something similar with the Mayan numbers. I tried for a while, but couldn’t get them to work.

Disclaimer: I know my way around TeX pretty well, but I’m not a power user. It took me quite a while to get all this to work. I’m not sure I can offer much trouble-shooting advice if you can’t get this to work on your computer.

Posted by: Dave Richeson | August 16, 2012

## Mathematics departments at liberal arts colleges

I’m often curious about how other mathematics departments do things—how they structure their curriculum, run the Putnam Exam, handle research projects, etc. This invariably leads to a lot of web searching. So I decided to put together a collection of links to mathematics departments at schools like mine (a small liberal arts college). Because I thought others might like this resource, I decided to share the list here.

Here are links to the mathematics departments of top liberal arts colleges in the US (I used the top 60 schools as reported by US News in 2011).

Posted by: Dave Richeson | June 20, 2012

## Plato’s approximation of pi?

Today I came across an assertion that Plato used ${\sqrt{2}+\sqrt{3}}$ as an approximation of ${\pi}$. Indeed, it is not a bad approximation: ${3.14626\ldots}$ (although it is not within Archimedes’s bounds: ${223/71<\pi<22/7}$).

Not only had I not seen this approximation before, I had not heard of any value of ${\pi}$ attributed to Plato.

I investigated a little further and discovered that there is no direct evidence that Plato knew of this approximation. It was pure speculation by the famous philosopher of science Karl Popper! Here’s what Popper has to say (this is in his notes to Chapter 6 of The Open Society and its Enemies, Vol. 1, pp. 252–253).

It is a curious fact that ${\sqrt{2}+\sqrt{3}}$ very nearly approximates ${\pi}$… The excess is less than ${0.0047}$, i.e. less than ${1 \frac{1}{2}}$ pro mille of ${\pi}$, and we have reason to believe that no better upper boundary for ${\pi}$ had been proved to exist. A kind of explanation of this curious fact is that it follows from the fact that the arithmetical mean of the areas of the circumscribed hexagon and the inscribed octagon is a good approximation of the area of the circle. Now it appears, on the one hand, that Bryson operated with the means of circumscribed and inscribed polygons,… and we know, on the other hand (from the Greater Hippias), that Plato was interested in the adding of irrationals, so that he must have added ${\sqrt{2}+\sqrt{3}}$. There are thus two ways by which Plato may have found out the approximate equation ${\sqrt{2}+\sqrt{3}\approx \pi}$, and the second of these ways seems almost inescapable. It seems a plausible hypothesis that Plato knew of this equation, but was unable to prove whether or not it was a strict equality or only an approximation.

Popper then spends a couple of paragraphs tying this into an earlier discussion of Plato’s Timaeus. This is the work in which Plato discusses the four elements (earth, air, fire, and water) and associates them with four of the regular polyhedra (cube, octahedron, tetrahedron, and icosahedron, respectively). The connection between Timaeus and ${\pi}$ is the relation between the values ${\sqrt{2}}$ and ${\sqrt{3}}$, the 45-45-90 and 30-60-90 triangles which can be used to make the faces of the polyhedra, and the approximations to the area of a circle using these triangles.

Popper ends with the reminder/disclaimer:

I must again emphasize that no direct evidence is known to me to show that this was in Plato’s mind; but if we consider the indirect evidence here marshalled, then the hypothesis does perhaps not seem too far-fetched.

Note: if we take the unit circle and construct a circumscribed hexagon and an inscribed octagon, then the area of the hexagon is ${2\sqrt{3}}$ and the area of the octagon is ${2\sqrt{2}}$. So, it is true that the average of these areas is ${\sqrt{2}+\sqrt{3}}$.

Posted by: Dave Richeson | June 18, 2012

## Puzzler: a squarable region from Leonardo da Vinci

It is famously impossible to square the circle. That is, given a circle, it is impossible, using only a compass and straightedge, to construct a square having the same area as the circle.

I will let you read elsewhere about the exact rules behind compass and straightedge constructions. The punchline is that if you begin with two points 1 unit apart, then you can construct a line segment segment of length ${a}$ if and only if ${a>0}$ can be created from the integers using the operations ${+}$, ${-}$, ${\times}$, and ${\div}$, and by taking square roots.

Thus it is possible to construct a line segment of length ${(1+\sqrt{5})/2}$ (the golden ratio), but it is impossible to construct a segment of length ${\sqrt[3]{2}}$ (hence it is impossible to double the cube).

A circle with radius 1 has area ${\pi}$. A square having this same area would have side-length ${\sqrt{\pi}}$. Because ${\pi}$ is transcendental, this is not a constructible length. Thus it is impossible to square the circle.

Even though the circle is not squarable, some regions with curved boundaries are. For example, in the fifth century BCE Hippocrates of Chios (no, not that Hippocrates) showed that several lunes are squarable.

Let’s give a brief proof that the shaded lune shown below is squarable. Suppose the large circle has radius 1. Then triangle ${ACD}$ has area ${1/2}$ and sector ${ADCE}$ has area ${\pi/4}$. Thus segment ${ACE}$ has area ${\pi/4-1/2}$. The smaller circle has radius ${\sqrt{2}/2}$. So the semicircle ${ACB}$ has area ${\pi(\sqrt{2}/2)^{2}/2=\pi/4}$. It follows that the lune has area ${\pi/4-(\pi/4-1/2)=1/2}$. A square with the same area as the lune would have side-length ${\sqrt{2}/2}$, which is constructible. Thus the lune is squarable.

It turns out Leonardo da Vinci played around with squarable figures, and he discovered many beautiful examples. Below I’ve included one of Leonardo’s figures (the on the left). I’ve included the center and right-hand figures to give more information on how Leonardo’s design is created.

So here’s the puzzle: show that Leonardo’s figure is squarable.

(Hint: assume that the radius of the large circle is 1. Then find the total area of the shaded regions. Show that the area, and hence the square root of the area, is a constructible number.)

Have fun!

Posted by: Dave Richeson | June 1, 2012

## Angle trisection using origami

It is well known that it is impossible to trisect an arbitrary angle using only a compass and straightedge. However, as we will see in this post, it is possible to trisect an angle using origami. The technique shown here dates back to the 1970s and is due to Hisashi Abe.

Assume, as in the figure below, that we begin with an acute angle ${\theta}$ formed by the bottom edge of the square of origami paper and a line (a fold, presumably), ${l_{1}}$, meeting at the lower left corner of the square. Create an arbitrary horizontal fold to form the line ${l_{2}}$, then fold the bottom edge up to ${l_{2}}$ to form the line ${l_{3}}$. Let ${B}$ be the lower left corner of the square and ${A}$ be the left endpoint of ${l_{2}}$. Fold the square so that ${A}$ and ${B}$ meet the lines ${l_{1}}$ and ${l_{3}}$, respectively. (Note: this is the non-Euclidean move—this fold line cannot, in general, be drawn using compass and straightedge.) With the paper still folded, refold along ${l_{3}}$ to create a new fold ${l_{4}}$. Open the paper and fold it to extend ${l_{4}}$ to a full fold (this fold will extend to the corner of the square, ${B}$). Finally, fold the lower edge of the square up to ${l_{4}}$ to create the line ${l_{5}}$. Having accomplished this, the lines ${l_{4}}$ and ${l_{5}}$ trisect the angle ${\theta}$.

Let us see why this is true. Consider the diagram below. We have drawn in ${CD}$, which is the location of the segment ${AB}$ after it is folded, ${AC}$, the fourth side of the isosceles trapezoid ${ABDC}$, and ${AD}$, the second diagonal of ${ABDC}$. We must show that ${\theta=3\alpha}$, where ${\alpha=\angle DBE}$ and ${\theta=\angle CBE}$.

Because ${BE}$ and ${DF}$ are parallel, ${\angle DBE=\angle BDF}$, and because ${DF}$ is the altitude of the isosceles triangle ${ABD}$, ${\angle BDF=\angle ADF}$. Thus ${\alpha=\angle DBE=\angle BDF=\angle ADF}$. Now, ${ABDC}$ is an isosceles trapezoid and ${ABD}$ is an isosceles triangle, so ${ABD}$ and ${BCD}$ are congruent isosceles triangles. Thus ${\angle CBD=\angle ADB= \angle BDF+\angle ADF}$. It follows that ${\theta=\angle CBE=\angle DBE+\angle CBD=\angle DBE+\angle BDF+\angle ADF=3\alpha}$.

The geometric properties of origami constructions are quite interesting. Every point that is constructible using a compass and straightedge is constructible using origami. But more is constructible. As we’ve seen, it is possible to trisect any angle using origami (I’ll leave the obtuse angles as an exercise). It is possible to double a cube. It is possible to construct regular heptagons and nonagons. In fact, where the constructability of ${n}$-gons is related to Fermat primes, the origami-constructibility of ${n}$-gons is related to Pierpont primes. While the field of constructible numbers is the smallest subfield of ${\mathbb{R}}$ that is closed under square roots, the field of origami-constructible numbers is the smallest subfield that is closed under square roots and cube roots. In fact, it is possible to solve any linear, quadratic, cubic, or quartic equation using origami!

There are quite a few places to read about geometric constructions using origami, but a good starting point is this online article (pdf) by Robert Lang.

Posted by: Dave Richeson | April 18, 2012

## An interesting multivariable calculus example

Earlier this semester in my Multivariable Calculus course we were discussing the second derivative test. Recall the pesky condition that if ${(a,b) }$ is a critical point and ${D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-(f_{xy}(a,b))^{2}=0}$, then the test fails.

A student emailed me after class and asked the following question. Suppose a function ${f}$ has a critical point at ${(0,0)}$ and ${D(0,0)=0}$. Moreover, suppose that as we approach ${(0,0)}$ along ${x=0}$ we have ${f_{xx}(0,y)>0}$ when ${y<0}$ and ${f_{xx}(0,y)<0}$ when ${y>0}$. Is that enough to say that the critical point is a not a maximum or a minimum? His thought process was that if we look at slices ${y=k}$ we get curves that are concave up when ${k<0}$ and curves that are concave down when ${k>0}$—surely that could not happen at a maximum or minimum.

I understood his intuition, but I was skeptical. Indeed, after a little playing around I came up with the following counterexample. The function is

${\displaystyle f(x,y)=\begin{cases} x^{4}+y^{2}e^{-x^{2}} & y\ge 0\\ x^{4}+x^{2}y^{2}+y^{2} & y<0. \end{cases}}$

The first partial derivatives are

${\displaystyle f_{x}(x,y)=\begin{cases} 4x^{3}-2xy^{2}e^{-x^{2}} & y\ge 0\\ 4x^{3}+2xy^{2} & y<0, \end{cases}}$

${\displaystyle f_{y}(x,y)=\begin{cases} 2ye^{-x^{2}} & y\ge 0\\ 2x^{2}y+2y & y<0. \end{cases}}$

Clearly ${(0,0)}$ is a critical point. The second partial derivatives are

${\displaystyle f_{xx}(x,y)=\begin{cases} 12x^{2}-2y^{2}e^{-x^{2}}+4x^{2}y^{2}e^{-x^{2}} & y\ge 0\\ 12x^{2}+2y^{2} & y<0, \end{cases}}$

${\displaystyle f_{yy}(x,y)=\begin{cases} 2e^{-x^{2}} & y\ge 0\\ 2x^{2}+2 & y<0, \end{cases}}$

${\displaystyle f_{xy}(x,y)=f_{yx}(x,y)=\begin{cases} -4xye^{-x^{2}} & y\ge 0\\ 4xy & y<0. \end{cases}}$

Thus ${D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-(f_{xy}(0,0))^{2}=0\cdot 2-0^{2}=0}$. So the second derivative test fails. But observe that when ${x=0}$ we have

${\displaystyle f_{xx}(0,y)=\begin{cases} -2y^{2} & y\ge 0\\ 2y^{2} & y<0. \end{cases}}$

So ${f_{xx}(0,y)>0}$ when ${y<0}$ and ${f_{xx}(0,y)<0}$ when ${y>0}$. Yet it is easy to see that ${(0,0)}$ is a minimum: ${f(0,0)=0}$ and ${f(x,y)>0}$ for all ${(x,y)\ne (0,0)}$. A graph of the function is shown below. You can see the concave down cross sections for $x=0$ and $y>0$.

Posted by: Dave Richeson | February 21, 2012

## Parametric curve project for multivariable calculus

I’m teaching two sections of Multivariable Calculus this semester. Each class has 3 hours of lecture and a 1 hour 20 minute lab each week. Last week the students were learning about parametric equations. So in lab I wanted to give them some hands-on experience with 2-dimensional parametric curves. Their assignment was to create a work of art using Grapher (Apple’s free graphing program) and parametric curves. They worked on the projects in lab for about an hour (in pairs, mostly). Some finished in that time, but others finished outside of class. The results were fantastic, so I thought I’d share them here.

Here is a pdf of the lab assignment.

This lab was not my idea. It was written by my colleague Lorelei Koss. It was also classroom-tested by my colleague Jen Schaefer. This semester I took their feedback, changed the lab a little, and used it in my class. Lorelei said that she got the idea from Judy Holdener and Keith Howard at Kenyon College and Tommy Ratliff at Wheaton College. I’ve also found some similar ideas in the mathematical literature: there’s an article by Barry Tesman (also a colleague of mine) and Marc Sanders, “MATH and other four-letter words,” College Math Journal, Nov. 1998, and “Painting by Parametric Curves and Van Gogh’s Starry Night,” by Stephen Lovett, Matthew Arildsen, Jon Jones, Anna Larson and Rebecca Russ, Math Horizons, Nov. 2010.

Here are their amazing works of art (click to see a slide show):

Posted by: Dave Richeson | October 28, 2011

## Cantor set applet

I made this Cantor set applet for my Real Analysis class. It is nothing fancy, but it saves me from drawing it on the board.

Posted by: Dave Richeson | October 28, 2011

## Applet to illustrate the epsilon-delta definition of limit

Here’s a GeoGebra applet that I made for my Real Analysis class. It can be used to explore the definition of limit:

Definition. The limit of $f(x)$ as $x$ approaches $c$ is $L$, or equivalently $\displaystyle \lim_{x\to c}f(x)=L,$ if for any $\varepsilon>0$ there exists $\delta>0$ such that whenever $0<|x-c|<\delta$, it follows that $|f(x)-L|<\varepsilon$.

Posted by: Dave Richeson | October 7, 2011

## The danger of false positives

As I mentioned earlier, I’m teaching a first-year seminar this semester called “Science or Nonsense?” On Monday and Wednesday this week we discussed some math/stats/numeracy topics. We talked about the Sally Clark murder trial, the prosecutor’s fallacy, the use of DNA testing in law enforcement, Simpson’s paradox, the danger of false positives, and the 2009 mammogram screening recommendations.

I made a GeoGebra applet to illustrate the dangers of false positives. So I thought I’d share that here. Here’s the statement of the problem.

Suppose Lenny Oiler visits his doctor for a routine checkup. The doctor says that he must test all patients (regardless of whether they have symptoms) for rare disease called analysisitis. (This horrible illness can lead to severe pain in a patient’s epsylawns and del-tahs. It should not to be confused with analysis situs.) The doctor says that the test is 99% effective when given to people who are ill (the probability the test will come back positive) and it is 95% effective when given to people who are healthy (it will come back negative).

Two days later the doctor informs Lenny that the test came back positive.

Should Lenny be worried?

Surprisingly, we do not have enough information to answer the question, and Lenny (being pretty good at math) realizes this. After a little investigating he finds out that approximately 1 in every 2000 people have analysisitis (about 0.05% of the population).

Now should Lenny be worried?

Obviously he should take notice because he tested positive. But he should not be too worried. It turns out that there is less than a 1% chance that he has analysisitis.

Notice that there are four possible outcomes for a person in Lenny’s position. A person is either ill or healthy and the test may come back positive or negative. The four outcomes are shown in the chart below.

 Test result Ill Healthy Positive true positive false positive Negative false negative true negative

Obviously, the two red boxes are the ones to worry about because the test is giving the incorrect result. But in this case, because the test came back positive, we’re interested in the top row.

For simplicity, suppose the city that is being screened has a population of 1 million. Then approximately (1000000)(0.0005)=500 people have the illness. Of these (500)(.99)=495 will test positive and (500)(0.01)=5 will test negative. Of the 999,500 healthy people (999500)(.05)=49975 will test positive and (999500)(.95)=949525 will test negative. This is summarized in the following chart.

 Test result Ill Healthy Positive 495 49975 Negative 5 949525

Thus, 495+49975=50470 people test positive, and of these only 495 are ill. So the chance that a recipient of positive test result is sick is 495/50470=0.0098=0.98%. That should seem shockingly low! I wonder how many physicians are aware of this phenomenon.

You can try out this or other examples using this GeoGebra applet that I made.