Division by Zero

Posted by: Dave Richeson | August 24, 2010

A game for budding knot theorists

Thanks to Sam Shah for introducing me to this fascinating online game: Entanglement.

The rules are simple. You are given hexagonal tiles, one at a time, each adorned with six short segments of rope. Use them to construct the longest possible knot (measured in segments) before running into a wall. Entanglement is fun and addicting!

How high can you go? A quick analysis shows that the highest possible score is 169. How do we come up with that value?

Each tile has six segments on it—two ends on each side of the hex. There are spaces for 36 hexagons. Thus a full board will have 6*36=216 strands. However, some of these strands will end at a wall. To be precise, there are 48 boundary sides. One of those boundaries is the starting wall (and the ending wall of a “perfect game”). So a perfect game must contain at least 47 unused strands (such as the strand shown above that starts and ends at the central hex). Thus it is impossible to get a score higher than 216-47=169.

Sure, that is a theoretical upper bound. Is it attainable? It turns out that it is! A player named “atomic” got a perfect 169 and there is a screenshot to prove it.

Now the only question is, what knot is that?

Update: Thank you to commenter “Evan” for pointing out a very similar board game, Tantrix, which came out in the late ’80′s. Also, I want to mention KnoTiles which was given to me by my friend Gene Chase—also very fun.

7 Comments

Posted in Math, Puzzle | Tags: Puzzle, knot theory, game

Posted by: Dave Richeson | August 18, 2010

Mathematical surprises

I’m interested in compiling a list of “mathematical surprises.” The best possible example would be a mathematical discovery that no mathematician saw coming, but after it was discovered it changed mathematics in some fundamental way—Cantor’s discovery of the nondenumerability of the continuum is such an example. But I’ll settle for any surprise—Andrew Wiles surprised everyone with his proof of Fermat’s Last Theorem, the solution of the Monty Hall problem surprised many capable mathematicians, etc.

I’ve spent a couple days brainstorming and I’ve come up with the following list. Some are better than others, and they’re listed in no particular order. Please add your surprises in the comments below!

Gödel’s incompleteness theorems
The discovery of irrational numbers by the Pythagoreans
Cantor’s theorems—nondenumerability of the continuum and the cardinality of the power set of A is greater than the cardinality of A
The rational numbers are countable
The continuum hypothesis can neither be proved nor disproved in ZFC
The existence of a continuous nowhere differentiable function
Euler’s solution of the Basel problem
The existence of non-Euclidean geometries
The insolvability of quintic equations
The Monty Hall problem
Fermat’s non-prime (Euler proved that $2^{2^5}+1$ is composite)
The shape of a hanging chain is a catenary
The existence of space filling curves
The Banach-Tarski theorem
The relationship between the complex numbers and the primes (E.g., Riemann zeta function)
The prime number theorem
Aperiodic tilings
Arrow’s impossibility theorem
Ulam’s spiral of primes
Andrew Wiles’ proof of Fermat’s Last Theorem
The use of a computer to prove the four color theorem
Russell’s paradox
The Cantor set
Euler’s polyhedron formula
The five Platonic solids
The Brachistochrone problem
Noncircular figures of constant width
0.999…=1
Lorenz’s “butterfly effect”
Period 3 implies chaos (and Sharkovsky’s theorem)
The fundamental theorem of calculus
Descartes’ discovery of analytic geometry
Discovery of complex numbers (and their real-world applications)
Hamilton’s discovery of the quaternions
There exists a flow in 3-space with closed orbits of every knot and link type
19-year-old Gauss’ ruler-and-compass construction of a 17-g0n (and its relation to Fermat primes#mce_temp_url#)
Proving the impossibility of squaring a circle, trisecting an angle, and doubling a cube
The Euler line
A complex function that is once differentiable on a disk is infinitely differentiable
Liouville’s theorem—a function that is bounded and differentiable at every point in the complex plane is constant
Thomae’s function—a function that is continuous at every irrational number, discontinuous at every rational number
The elementary linear algebra behind Google’s pagerank
Kuratowskis closure-complement theorem

23 Comments

Posted in Math

Posted by: Dave Richeson | August 16, 2010

Goodstein’s unprovable theorem

Recently I learned about a family of sequences of nonnegative integers (called Goodstein sequences) and two remarkable theorems about these sequences.

Begin with any positive integer ${m_{1}}$ . This is the first term in the sequence. For example, suppose we begin with ${m_{1}=18}$ .

The first step in computing the second term of the sequence, ${m_{2}}$ , is to write ${m_{1}}$ in hereditary base 2 notation. That is, write ${m_{1}}$ using only powers of 2 (the base 2 expansion of ${m_{1}}$ ): ${m_{1}=\sum a_{i}2^{i}}$ . Then write all of the exponents in their base 2 expansions. Then write all of the exponents of the exponents in that way, etc. Our number written in hereditary base 2 notation is ${m_{1}=18=2^{4}+2^{1}=2^{2^{2}}+2^{2^{0}}}$ .

To obtain ${m_{2}}$ we change all of the 2′s to 3′s, then subtract 1. For our example, ${m_{2}=3^{3^{3}}+3^{3^{0}}-1}$ . Notice that ${m_{2}}$ is very large number; it is approximately ${7.63\times 10^{12}}$

We continue in the same way. We obtain ${m_{n}}$ from ${m_{n-1}}$ by writing ${m_{n-1}}$ in hereditary base ${n}$ notation, changing all of the ${n}$ ‘s to ${(n+1)}$ ‘s, and subtracting 1.

${m_{1}=2^{2^{2}}+2^{2^{0}}=18}$

${m_{2}=3^{3^{3}}+3^{3^{0}}-1=3^{3^{3}}+2\cdot 3^{0}\approx 7.63\times 10^{12}}$

${m_{3}=4^{4^{4}}+2\cdot 4^{0}-1=4^{4^{4}}+4^{0}\approx 1.34\times 10^{154}}$

${m_{4}=5^{5^{5}}+5^{0}-1=5^{5^{5}}\approx 1.91\times 10^{2184}}$

${m_{5}=6^{6^{6}}-1\approx 2.66\times 10^{36305}}$

To obtain ${m_{6}}$ we must express ${m_{5}}$ in hereditary base 6 notation. First notice that

${m_{5}=6^{6^{6}}-1=5\cdot 6^{6^{6}-1}+5\cdot 6^{6^{6}-2}+\cdots+5\cdot 6^{1}+5\cdot 6^{0}.}$

(An easy way to see this is that written in base 6, ${6^{6^{6}}}$ is ${100\cdots 00_{6}}$ , 1 followed by ${6^{6}}$ zeros. So ${6^{6^{6}}-1}$ is ${55\cdots 55_{6}}$ , ${6^{6}}$ fives.) But this expansion of ${m_{5}}$ is not in hereditary base 6 notation yet. We must express the towers of exponents in base 6, etc. When that is done, we replace the 6′s with 7′s and subtract 1. In the end, ${m_{6}}$ is a gigantic number.

Now here are two amazing theorems about this sequence of integers. In 1944 Reuben Goodstein proved:

Theorem. There is a ${k}$ such that ${m_{k}=0}$ .

That is, this sequence, which looks like it is rocketing to infinity, will eventually become zero and terminate. Wow! The proof of this theorem is very sophisticated and uses the theory of ordinal numbers.

I’ll have to file this sequence away as an example that shows why we can’t use the behavior of the first few (or first few million) terms of a sequence to determine the limiting behavior of a sequence.

Then, in 1982 Laurie Kirby and Jeff Paris proved the following theorem.

Theorem. Goodstein’s Theorem is not provable using the Peano axioms of arithmetic.

In other words, this is exactly the type of theorem described in 1931 by Gödel’s first incompleteness theorem!

Recall what Gödel’s theorem says. If there is an axiomatic that is rich enough to express all elementary arithmetic (such as that formed from the Peano axioms), then it must be incomplete. In other words, there must be a true statement about arithmetic that cannot be proven from the axioms. In his proof Gödel produces an explicit example of a true, but unprovable statement. But it is complicated to grasp and more reminiscent of a logical paradox than a mathematical statement.

The first nice mathematical example of such a statement was presented in 1977 by Paris and Harrington (in a field called Ramsey theory). Then in 1982 Kirby and Paris proved that Goodstein’s theorem was unprovable and they gave another elementary example, called “Hercules versus the hydra,” which relates to the growth of the hydra (a tree) and its destruction by Hercules.

11 Comments

Posted in Math | Tags: axiomatic systems, Gödel's incompleteness theorem, Goodstein's theorem, Peano axioms, sequences

Posted by: Dave Richeson | August 6, 2010

An island on an island on an island

A few weeks ago I wrote about an island on an island on a landmass. Today I found this website which shows an island on an island on an island. Pretty cool! Zoom out on the map below to see the nested islands.

View Larger Map

There is a puddle…

…on an island…

…in a lake (Crater Lake)…

…on an island (Volcano Island)…

…in a lake (Taal Lake)…

…on an island (Luzon, Philipines)…

…in the ocean.

2 Comments

Posted in Humor

Posted by: Dave Richeson | August 5, 2010

The left-handed boy problem

A few months ago Gary Foshee was scheduled to speak at the Gathering for Gardner. He got up and gave a presentation that was all of three sentences. He said:

I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

This deceptively simple problem quickly made the rounds. The knee-jerk answer is ${1/2}$ , of course—the gender of one child doesn’t change the probabilities for the second child. People with a little training in probability know that that reasoning isn’t a valid—it is an exercise in conditional probability. The probability of having two boys given that one child is a boy is ${1/3}$ . Surely, that’s the correct answer, right? How could the day of the week matter?

It turns out, however, that the probability is an unexpected ${13/27}$ .

[Note: The real correct answer is 0 or 1 since Gary Foshee either has two boys or doesn't have two boys. The actual question should be "If a randomly-selected family has two children and at least one of them is a boy born on Tuesday, what is the probability that they have two boys?" But that doesn't sound nearly so slick. I'll be consistent and stick with his less-than-perfect wording.]

I’d like to pose a question similar to the Tuesday boy problem, then describe how to compute the probabilities for a whole class of problems like these.

I have two children. One is a left-handed boy. What is the probability I have two boys?

You may assume that the probability of being left-handed is 1/10 and that left-handedness is not a genetic trait.

Stop reading here if you want to think about the problem on your own.

Before we answer the left-handed boy problem, lets look at an easier question.

I have two children. One is a boy. What is the probability I have two boys?

We must think of this in terms of conditional probability. We can compute the probability using this formula:

${q=P(\text{A family has two boys, given that they have at least one boy})}$

$=\frac{P(\text{A family has two boys})}{P(\text{A family at least one boy})}$ .

There are four equally likely options for any two-child family: the first child is a boy and the second child is a boy (written ${BB}$ ), the first child is a boy and the second child is a girl (written ${BG}$ ), ${GB}$ , and ${GG}$ . These are shown in the chart below with their probabilities.

${\begin{array}{|c|c|}\hline BB \,\, (1/4) & BG \,\, (1/4) \\\hline GB \,\, (1/4) & GG \,\, (1/4) \\\hline \end{array}}$

However, in our case we can ignore the ${GG}$ case because we know that at least one child is a boy.

${\begin{array}{|c|c|}\hline BB \,\, (1/4) & BG \,\, (1/4) \\\hline GB \,\, (1/4) & \\\hline \end{array}}$

Observe that:

${P(\text{A family at least one boy})=1/4+1/4+1/4=3/4}$ , and

${P(\text{A family has two boys})=1/4}$ .

So, $q=\frac{P(\text{A family has two boys})}{P(\text{A family at least one boy})}=(1/4)/(3/4)=1/3$ .

This problem is particularly easy because all four outcomes are equally-likely. The Tuesday boy problem is trickier. The probability of being born on a Tuesday is ${1/7}$ and the probability of a non-Tuesday birthday is ${6/7}$ .

Now there are 16 possibilities. The notation ${G_{N}B_{T}}$ means that the first child is a girl born on a non-Tuesday and the second child is a boy born on Tuesday. The probability of this particular occurance is ${(1/2)(6/7)(1/2)(1/7)=6/196}$ . The probabilities of the others are calculated similarly:

${\left.\begin{array}{|c|c|c|c|}\hline B_NB_N \,\, (36/196)& B_NG_N \,\, (36/196)& G_NB_N \,\, (36/196)& G_NG_N \,\, (36/196)\\\hline B_NB_T \,\, (6/196)& B_NG_T\,\, (6/196) & G_NB_T \,\, (6/196)& G_NG_T \,\, (6/196)\\\hline B_TB_N \,\, (6/196)& B_TG_N \,\, (6/196)& G_TB_N \,\, (6/196)& G_TG_N \,\, (6/196)\\\hline B_TB_T \,\, (1/196)& B_TG_T \,\, (1/196)& G_TB_T \,\, (1/196)& G_TG_T \,\, (1/196)\\\hline \end{array}\right.}$

In the chart below we remove the cases in which there is no Tuesday-born boy. ${\left.\begin{array}{|c|c|c|c|}\hline & & & \\\hline B_NB_T \,\, (6/196)& & G_NB_T \,\, (6/196)& \,\,\,\,\,\,\, \\\hline B_TB_N\,\, (6/196)& B_TG_N \,\, (6/196)& & \\\hline B_TB_T \,\, (1/196)& B_TG_T \,\, (1/196)& G_TB_T \,\, (1/196)& \\\hline \end{array}\right.}$

We see that:

$P(\text{A family has at least one boy born on a Tuesday})$
$=4(6/196)+3(1/196)=27/196$ .

$P(\text{A family has two boys with at least one born on a Tuesday})$
$=2(6/196)+1/196=13/196$ .

So the answer to the question is:

$\displaystyle q=\frac{P(\text{A family has two boys with at least one born on a Tuesday})}{P(\text{A family has at least one boy born on a Tuesday})}$
$=(13/196)/(27/196)=13/27$ .

Let us investigate the general case.

I have two children. One is a boy with a trait which occurs with probability ${p}$ . What is the probability that I have two boys?

We build the chart just as we did in the Tuesday boy problem. Here ${G_{N}B_{Y}}$ means that the first child is a girl who does not have the trait and the second child is a boy who has the trait.

${\left.\begin{array}{|c|c|c|c|}\hline B_NB_N \,\, ((1-p)^2/4)& B_NG_N \,\, ((1-p)^2/4)& G_NB_N \,\, ((1-p)^2/4)& G_NG_N \,\, ((1-p)^2/4)\\\hline B_NB_Y \,\, (p(1-p)/4)& B_NG_Y\,\, (p(1-p)/4) & G_NB_Y \,\, (p(1-p)/4)& G_NG_Y \,\, (p(1-p)/4)\\\hline B_YB_N \,\, (p(1-p)/4)& B_YG_N \,\, (p(1-p)/4)& G_YB_N \,\, (p(1-p)/4)& G_YG_N \,\, (p(1-p)/4)\\\hline B_YB_Y \,\, (p^2/4)& B_YG_Y \,\, (p^2/4)& G_YB_Y \,\, (p^2/4)& G_YG_Y \,\, (p^2/4)\\\hline \end{array}\right.}$

Again, we focus on those that have a boy with the trait.

${\left.\begin{array}{|c|c|c|c|}\hline & & & \\\hline B_NB_Y \,\, (p(1-p)/4)& & G_NB_Y \,\, (p(1-p)/4)& \,\,\,\,\,\,\, \\\hline B_YB_N\,\, (p(1-p)/4)& B_YG_N \,\, (p(1-p)/4)& & \\\hline B_YB_Y \,\, (p^2/4)& B_YG_Y \,\, (p^2/4)& G_YB_Y \,\, (p^2/4)& \\\hline \end{array}\right.}$

Calculating as before we obtain the probability:

$q=\frac{2p(1-p)/4+p^{2}/4}{4p(1-p)/4+3p^{2}/4}=\frac{2-p}{4-p}$

Now we can solve a whole class of these problem. To solve the left-handed boy problem we simply plug in ${p=1/10}$ to obtain an answer of ${q=19/39}$ . That is, if a family has two children and one of the children is a left-handed boy, then the probability that they have two boys is ${19/39\approx 48.7\%}$ .

If we had asked the “right-handed boy” problem, then we’d plug in ${p=9/10}$ to obtain ${q=99/279}$ . That is, if a family has two children and one of the children is a right-handed boy, then the probability that they have two boys is ${99/279\approx 35.5\%}$ .

Notice what happens in the limiting values of ${p}$ .

If the trait is very common, like “has two eyes,” then ${p\approx 1}$ and we’re essentially in the “I have two children and one of them is a boy” case. Accordingly we have ${q\approx 1/3}$ .

On the other hand, if we have an extremely rare trait, like “has climbed Mt. Everest” ( ${p\approx 0}$ ), then it is very unlikely that both children have this trait. We’ve essentially uniquely identified one of the children. If we looked at all the two-child families in the entire world that have a son who climbed Mt. Everest, very few of them will have another child who also climbed Mt. Everest. Most of them are a boy who climbed Mt. Everest, and one other child. The chance of the other child being a boy is 1/2. (It is like asking “My first born child is a boy. What is the probability that I have two boys?”) Accordingly, ${q\approx 1/2}$ .

I’d like to thank my colleagues Jeff Forrester and Barry Tesman for their helpful comments.

28 Comments

Posted in Math, Puzzle | Tags: conditional probability, probability, Tuesday boy problem

Posted by: Dave Richeson | August 2, 2010

Mathematical magic tricks for kids

My six-year-old son loves the website ActivityTV.com, especially their science, origami, cooking, and magic videos.

I watched a few of the magic how-to videos with him and was pleasantly surprised to see that some of them had a distinctly mathematical feel to them. For example:

Jumping rubber bands: topological properties of circles and linked circles
Magic knots: knot theory
Ring escape: topology/knot theory
Magic compass: reflections of a square
Mathemagic: properties of integers

When I stared to watch the “mathemagic” video I was expecting one of those formulaic tricks involving the integers which are simply basic algebra in disguise. While not earth shattering, the trick turned out to be slightly more subtle than that. In case you don’t want to watch the video, here’s the trick:

Pick a 3-digit number with all three digits different. Reverse the digits and subtract to get another 3-digit number (if the difference is negative, take its absolute value, if it is less than 100, add leading zeros (e.g., 71 becomes 071). Reverse the digits of the difference and add it to the difference. Your sum will be 1089.

For example, start with 845. Then 845-548=297 and 297+792=1089.

Usually “Ryan” explains his tricks afterward. For this one he doesn’t. So why does it work?

SPOILER: Suppose you pick the number $abc$ (that is, $100a+10b+c$ ). Furthermore, suppose $abc>cba$ (the other case is handled similarly). This implies that $a>c$ . If we reverse the digits and subtract we obtain

$(100a+10b+c)-(100c+10b+a)=$

$=100(a-1-c)+10\cdot 9+(10+c-a)$ .

This is perhaps easier to see if we line the digits up in columns (note that $a>c$ so we must “carry” twice):

$\left.\begin{array}{cccc}&a & b & c \\-&c & b & a \\\hline &(a-1-c) & (9) & (10+c-a)\end{array}\right.$

Finally, if we reverse the digits and add we get

$100(a-1-c)+10*9+(10+c-a)+100(10+c-a)+10*9+(a-1-c)=$

$=1089$ ,

as promised.

(Actually, if you examine the trick more closely you see that the three digits need not be distinct. But the number cannot be a palindrome; that is, $a\ne c$ .)

[Aside: it is too bad that "Ryan" starts this video with his comments about math not being fun.]

2 Comments

Posted in Links, Math, Puzzle | Tags: topology, knot theory, Mathemagic, magic, integers

Posted by: Dave Richeson | July 30, 2010

Thirteen mathematically-inspired products

Just for fun, here is a list of 13 mathematically-inspired products that I’ve stumble upon recently.

I’m sure there are many more. Add your own discoveries in the comments…

5 Comments

Posted in Humor, Links, Math

Posted by: Dave Richeson | July 27, 2010

Tricks for easily creating BibTeX files

I wrote my last book (my only book, that is) using LaTeX. I had a large bibliography with close to 400 entries. I stored all of the bibliographic items in a BibTeX file (a text file ending in .bib). Each item looks something like this:

@book {Richeson:2008, AUTHOR = {Richeson, David S.}, TITLE = {Euler's gem: The polyhedron formula and the birth of topology}, PUBLISHER = {Princeton University Press}, ADDRESS = {Princeton, NJ}, YEAR = {2008}, PAGES = {xii+317}, ISBN = {978-0-691-12677-7; 0-691-12677-1} }

The beautiful thing about BibTeX is that after you cite works in your LaTeX document (for example \cite{Richeson:2008}), the references are automatically generated in whichever bibliographic style you specify (APA, MLA, Chicago, etc.).

However, as you can see above, it is a pain to enter the bibliographic information by hand—the syntax is rather cumbersome. Fortunately there are a number of places on the web to get pre-formated BibTeX entries. You can pull them up in your browser, then copy-and-paste them in to your .bib file. Here are a few useful sites for mathematicians (I learned about a few of these from this MathOverflow conversation).

MathSciNet (subscription required)—this is the web front-end for the extensive Mathematical Reviews database. It currently has over 2 million items. It gives bibliographic information, reviews, and more.
MR Lookup—this is the free version of MathSciNet. It gives bibliographic information only.
Zentralblatt Math database (subscription required)—similar to MathSciNet, it has over 3 million items.
OttoBib—search for books by ISBN
Lead2Amazon—searches six different Amazon sites
Google Scholar—this is a well-known scholarly search site run by Google. If you click the “Scholar preferences” link you’ll see that under Bibliography Manager, one of the options is to “Show links to import citations into: BibTeX.” Now each Google Scholar search result will have a link to the BibTeX entry (see below).

Mac-only goodness: BibDesk + TexShop

I could create the .bib files using a text editor, but instead I use the great, free, Mac-only program BibDesk (see screenshot below).

I was a huge fan of BibDesk while writing my book, but it turns out I was using only a small fraction of its capabilities. I used it to create, edit, and organize bibliographic entries. It has a simple interface that allows you to enter bibliographic information into the fields provided. The file it creates is a .bib file, not a BibDesk file, so there is no need to export the information for use in LaTeX.

Even though I spent many, many hours using BibDesk, I never explored its other features (and now I’m kicking my self). I want to share a few of them with you now.

Importing BibTeX entries. Above I gave several locations on the web to go to find BibTeX entries for mathematical works. I’ve known for a while if I highlighted a BibTeX entry, then dragged-and-dropped the text onto the BibDesk program that it would create a new entry with the fields filled in. I thought that was cool. That was nothing. Recently I discovered that you can use BibDesk like a mini web browser and import bibliographic entries with one click.

On the side-bar click External>Web. The screen splits vertically into thirds. You get a little browser in the top third with links to MathSciNet, Zentralblatt Math, the arXiv, etc. You can also type in addresses (such as the ones I gave above) or create bookmarks. When you perform a search using one of these websites BibDesk uses the middle screen to list all of the bibliographic items it finds. (The bottom third is used to preview the contents.) Each entry has an “import” button next to it. When you find the item you want, click the button and you’re done (see below).

I also like that when you import an entry from MathSciNet it saves a direct link back to the page that contains the review.

(I see that you can also search databases, like the Library of Congress—these databases are found in the Search menu.)

Paper archive. If you have an electronic copy of an article (a pdf for instance), you can drag-and-drop it onto the BibTeX entry for the article. Then BibDesk will place the article into a dedicated folder of your choosing (and a subfolder named for the first author). Then you can access your articles straight from BibDesk. Not only that, you can use BibDesk’s search feature to search inside the articles!

I’ve written before about my love for DropBox. I have it set up so that my bibliographic files are in sync and so are my pdfs. Since the folder structures are the same, I can access all my articles through BibDesk on any of my computers.

TexShop integration. I’m also a die-hard TexShop user. This front-end for LaTeX is another free and elegant Mac-only program (which I may write about in another blog post). It integrates with BibDesk beautifully. If you type the beginning of a citation in TexShop, like \cite{Ri, and then press F5 it will pull up a contextual menu listing all of the entries from your bibliography that could match that fragment. With a couple keystrokes you’re done.

Enjoy!

6 Comments

Posted in Academic Technology, Math | Tags: latex, bibtex, TexShop

Posted by: Dave Richeson | July 26, 2010

Algebra: the Faustian bargain

I came across this great quote from Sir Micheal Atiyah. It is in Barry Mazur’s foreword to Tobias Dantzig’s book Number: the language of science (the 2005 reprinting of the 1930 classic).

Algebra is the offer made by the devil to the mathematician. The devil says: I will give you this powerful machine, it will answer any question you like. All you need to do is give me your soul: give up geometry and you will have this marvelous machine. —Sir Michael Atiyah, 2002

6 Comments

Posted in Math | Tags: quotes, Sir Michael Atiyah

Posted by: Dave Richeson | July 20, 2010

Neat facts from Euler 2010

I had the wonderful honor of being the keynote speaker at the 9th annual meeting of the Euler Society. I spoke today about my book. It is now the end of the second day of this 2.5 day conference. I thought I’d post a few of the many interesting things that I learned.

1. Larry D’Antonio shared this quote from Kant (Physical Monadology, 1756):

But how in this business can metaphysics be reconciled with geometry [mathematics], when it appears easier to mate griffins with horses than to unite transcendental philosophy with geometry [mathematics]?

(Apparently the expression “mating griffins and horses” goes back to Virgil’s 8th Ecologue and is supposed to signify the impossible, since griffins view horses as prey. When the two do mate their offspring is a hippogriff—a creature that was recently reintroduced in the Harry Potter series.)

2. Let $S=\{m^n:m,n\in\mathbb{Z},m,n>1\}=\{4,8,9,16,25,27,32,36,\ldots\}$ be the set of all nontrivial powers (listed without repeats). Christian Goldbach discovered the following summation:

$\displaystyle \sum_{j\in S}\frac{1}{j-1}=\frac{1}{3}+\frac{1}{7}+\frac{1}{8}+\frac{1}{15}+\frac{1}{24}+\frac{1}{31}+\frac{1}{35}+\cdots=1$ .

Euler gave a “proof” of this in Variae observationes circa series infinitas and extended it in a number of interesting directions. Bruce Burdick gave a talk in which he showed Euler’s slick proof—it begins with Euler taking $x$ to be the sum of the harmonic series (which we all know is infinite). You can read Euler’s short proof on the first page of this English translation or on Ed Sandifer’s How Euler Did It site. Then Bruce showed how to make Euler’s proof rigorous.

3. The zeta function has the following property:

$\displaystyle\sum_{k=2}^\infty(\zeta(k)-1)=1$ .

4. As tweeted by Erik Tou:

Quote of the Day from Bruce Petrie: “In the 18th century, existence proofs didn’t exist.”

5. Tom Osler spoke about oblique angle diameters for curves. The definition is a little wordy, so I’ll describe it for a parabola. Take any line parallel to the axis of symmetry of a parabola (this line is an oblique angle diameter). Draw the tangent line to the parabola where it meets the line. Then draw any line parallel to this tangent line that meets the parabola twice. This line segment is always bisected by the oblique angle diameter (in the diagram below FD is the same length as DC).

It turns out that every conic section has an infinite family of oblique angle diameters. For the parabola it is any line parallel to the axis of symmetry. For an ellipse and a hyperbola it is any line through the origin.

Euler had a lot to say about this, but he had a very slick proof that if a curve has two oblique angle diameters $a$ units apart, then it is possible to find infinitely many simply by repeatedly translating one of the given ones by $a$ units.

6. In this same paper Euler discusses a neat fact about triangles with vertices on a conic section. I’ll describe it for the ellipse. If you pick a point on an ellipse ( $A$ on the ellipse below) and draw the tangent line to the ellipse through $A$ , then draw a line from the center ( $B$ below) to the ellipse parallel to the tangent line and it meets at the point $C$ , then the area of triangle $ABC$ does not depend on the point $A$ . In particular, if the ellipse has the form $x^2/a^2+y^2/b^2=1$ , then the area is $ab/2$ .