CTK Insights

In the spirit of times the online store of the Mathematical Association of America moved into its new facilities hosted by amazon.com. Check it at

http://maa-store.hostedbywebstore.com/

This did not happen yesterday, and the original 10% discount is no longer available; neither is the member discount of long standing, but there is a long list of books on sale at a throw-away prices of $5-10. Some, like Mathematical Adventures for Students and Amateurs (edited by D. F. Hayes and T. Shubin) and Musings of the Masters (edited by R. G. Ayoub) make an enchanting reading. A Friendly Mathematics Competition (edited by R. Gillman) and Hungarian Problem Book I offer wonderful problem selections to ponder in spare time or for working out with kids and students. Hungarian Problem Book I is especially interesting as it covers years 1894 (when the Eötvös competition was conceived) through 1905. (There are additional volumes not currently on sale.) Hungary has long been a pioneer in ways of nurturing young mathematicians, of which problem solving is a necessary step stone. The book has an unusual structure. There are problems and solutions, of course, but there is also supplemental material that follows specific solutions and highlights the background necessary to understand the concepts involved and place them in a general setting. Such an organization makes the Solutions section a practically obligatory read.

Problems are interesting, but not evenly difficult. Here's a couple of similar problems:

1894/1. Prove that the expressions $2x+3y$ and $9x+5y$ are divisible by $17$ for the same set of integral values of $x$ and $y$.

1896/2. Prove that the equations

$$x^2 - 3xy + 2y^2 + x - y = 0$$

$$x^2 - 2xy +y^2 -5x + 7y = 0$$

$$xy-12x+15y=0.$$

Both come with 2 solutions. For the first problem I came up with a third and I believe the simplest of the three. It is apparent from the identity

$$4(9x+5y)-(2x+3y)=34x + 17y=17(2x+y).$$

I'd say it is natural to look for a combination of the two given expression that is divisible by 17.

Here's one of the published solutions to the second problem. The idea is nearly the same but the implementation requires by far greater effort and ingenuity:

$$(x^2 - 3xy + 2y^2 + x - y)(x-y-9)+(x^2 - 2xy +y^2 -5x + 7y)(-x+2y+3)=2(xy-12x+15y).$$

Well, there should not be any complaints. This is how you do that: progressing from easy to hard. (Descending from hard to easy happens much later in one's life.)

After a hiatus of several month, Carnival of Mathematics is back online. Do check the revived carnival at The Aperiodical page by Peter Rowlett, Katie Steckles, and Christian Perfect. The new edition is both edifying and entertaining.

I am reading an unusual book on an extraordinary weird subject. A Canadian mathematician Florin Diacu's The Lost Millennium collides two points of view on the existing chronology: one would shorten it by about a thousand years. Along the way, Diacu meticulously pursues the origin and evolution of chronology as a science. This is a fascinating book that deserves a careful reading and a detailed review. I hope to give it its due shortly. Meanwhile, here's a little piece of mathematics that was used by one of the fathers of the modern chronology Denis Pétau, a Jesuit theologian and philologist born in Orléans, France, in 1583. In 1627 Pétau - better known by his Latinized name of Dionysius Petavius - published a fundamental work on chronology, Doctrina Temporum (On the Doctrine of Chronology). Petavius

... laid more weight on astronomical phenomena than his predecessor (Joseph Justus Scaliger) had, coming up with ingenious ways of relating dates to the motions of celestial bodies. Petavius used the combined-cycles method extensively - lunar (19), solar (28), and indiction (15) a system Scaliger had developed from the twelfth-century work of Roger of Hereford.The rationale for those cycles can be discussed succinctly.

19 is the smallest number of full years the Moon takes to complete a full number of orbits (namely, 235) around Earth.

28 is the smallest number of years after which the calendar repeats itself, with the dates matching the days of the week.

15 represents the Roman indiction, a taxation cycle introduced by the Emperor Constantine, starting with AD January 1, 313. This period became a standard throughout the Eastern Roman Empire.

Every year, starting with 4713 BC which counts as 1, is assigned its Julian count. For example, the year 753 BC has the Julian count 3961 (= 4713 - 753 + 1). Every Julian count has a unique triplet of numbers which are remainders of the division of the count by 19, 28, and 15. Any such triplet uniquely determines a count up to 7980. (Do you see why?)

This is an instance of what is nowadays known as the Chinese Remainder Theorem. The applicability of the method depends of course on the presence of documents that relate a historical event with a year in a lunar cycle, another in the solar cycle, and yet another in the indiction cycle. I understand that there is a sufficent number of events with a complete set of triplet associations to make the combined-cycles method useful. Obviously, for the sake of an exercise, one may not need to refer to a well documented historical event.

Origami is an ancient Japanese art of paper folding, with adherents all over the world. Origami has a mathematical side to it and, as a tool of geometric construction, is more powerful than the Euclidean straightedge and compass. (For example, angle trisection is possible by paper folding.)

Is it silly to ask, How much folding is needed? Well, it may be a silly question, but only partially. I am working through Robert Lang's Origami Design Secrets - compendium of origami art. A tremendous book, it is. And one can't just read it; more than any math book it requires doing the folding with one's own hands. And there is much folding to accomplish. So my progress is very slow. But I was delighted at a few descriptive sentences already at the very beginning.

To return to the title question, What Is Origami? Lang quotes an architect and origami artists Peter Engel - origami is an art of suggestion.

Or put another way, it is an art of abstraction. The challenge to the origami designer is to select an abstraction that can be realized in folded paper.

The idea is of course that the abstraction needs to convey the image of the object being abstracted. Here's an elephant model by Dave Mitchell, with an instruction(!) for folding:

The instruction reads:

Begin with a sheet of writing paper. Fold the upper right corner along the edge.

Wonderful, is it not? Now try it. Do not be afraid! You can do that!

I often wondered why in the last decades educational reformers put so much emphasis on studying algebra. Until about 200 years ago studying geometry was the surest way on the road to mastering logical thinking. Personally, I do not believe the latter and doubt that emphasis on algebra will do any good to the current high school populations. But my views aside, I have recently come across a quote by Rafael Bombelli who (some time before the emergence of analytic geometry) wrote (Algebra, 1572):

I will end all my work here even though at first I wanted to prove all these arithmetic problems using geometric demonstrations. This because I know that these two sciences (that is, arithmetic and geometry) have such a strict relationship that the first proves the second and the second demonstrates the first. The mathematician who wants to be perfect cannot be such if he is not versed in both, although in our times many are those who believe differently.

Reference

1. F. La Nave, Deductive Narrative and Epistemological Function of Belief in Mathematics: On Bombelli and Imaginary Numbers, in Circles Disturbed, A. Doxiadis, B. Mazur (eds), Princeton University Press, 2012, p. 102

This is a short note for the record.

I've been preparing a page on the indivisibles (not ready yet), when my eighth grader boy came up to share his solutions to a couple of olympiad problems. We talked a few minutes about the olympiad and then, sensing his mood, I decided to catch the moment and expand a little of my own. So I told of Torricelli's paradox:

What do you think was the boy's reaction? He smiled and said:

In the previous post, The Joy of Homogeneity, I followed Gary Davis in establishing a statement observe by Ben Vitale. Ben's observation had to do with fractions in which both the numerator and denominator were sums of consecutive odd numbers. So that, for example,

$$\frac{1+3}{5+7}=\frac{1+3+5}{7+9+11}=\frac{1+3+5+7}{9+11+13+15}= \ldots = \frac{1}{3},$$

and, more generally,

$$\frac{1+2+\ldots +(2n-1)}{(2n+1)+(2n+3)+\ldots +(4n-1)}=\frac{n^2}{3n^2}=\frac{1}{3}.$$

Allen Pinkall left a commenet on the original page with an observation on a regularity in a somewhat modified fraction:

$$\frac{1+2+\ldots +(2n-1)}{(2n-1)+(2n+3)+\ldots +(4n-3)},$$

where the denominator starts with the last term of the numerator and not with the next one. The new observation concerns the sum of the numerator and the denominator of the reduced fraction. Let

$$A = 1+2+\ldots +(2n-1)=n^2$$

and

$$B = (2n-1)+\ldots +(4n-3)=(2n-1)^{2}-(n-1)^{2}=3n^{2}-2n.$$

Now then $\frac{A}{B}=\frac{n^{2}}{n(3n-2)}=\frac{n}{3n-2}=\frac{a}{b}$, where $a = n$ and $b=3n-2$. What we note is that the sum $a+b$ is twice the last term in the numerator (as well as the first term in the numerator.) This is always true. Furthermore, the fraction $\frac{a}{b}$ is irreduceable unless $n$ is even. For $n$ even, the fraction $\frac{a}{b}$ can be further reduced by just a factor of $2$:

$$\frac{n}{3n-2}=\frac{n/2}{3(n/2)-1}$$

such that the sum of the numerator and the denominator of the latter is $(n/2)+3(n/2)-1=2n-1$, exactly the last term of the sum in the numerator (as well as the first term in the numerator.)

In a recent blog A Lovely Observation Gary Davis (@RepublicOfMath) elaborated on an observation of Ben Vitale (@BenVitale) to the effect that

$$\frac{1+3}{5+7}=\frac{1+3+5}{7+9+11}=\frac{1+3+5+7}{9+11+13+15}= \ldots = \frac{1}{3}.$$

In the fractions both numerators and denominators are sums of successive odd numbers: the numerators start with 1, the denominators where the numerators leave off. Thus naturally derivation of the formula for the sum of successive odd numbers is the key to the explanation of Ben Vitale's observation. There are ways and ways to obtain the required expression. The most elementary one is to observe that

$$S(n) = 1 + 3 + \ldots + (2n - 1)$$

is the sum of an arithmetic sequence $a, a + d, a + 2d, \ldots, a + (n-1)d$,
where $a = 1$ and $d = 2$. With a nod to the young F. Gauss,

$$a+ (a + d)+(a + 2d)+ \ldots + [a + (n-1)d] = \frac{n(2a + (n-1)d)}{2},$$

which in the case of odd numbers gives

$$1+3+ \ldots + (2n-1) = \frac{n(2 + (n-1)2)n}{2}=n^2.$$

Another way to derive that formula is to noticee that the sum of odd numbers is the sum of all numbers less the sum of the even ones. It is convenient at this point to start using the symbol of $\sum$ to make the formulas shorter and more manageable:

$$S(n)=\sum_{k=1}^{n}(2k-1)=n^2$$

and, for the sum of all integers from $1$ to $n$,

$$N(n)=\sum_{k=1}^{n}k=\frac{n(n+1)}{2}.$$

Now a little of summation magic:

$$\begin{align}S(n) &= N(2n) - 2N(n) = \sum_{k=1}^{2n}k - 2\sum_{k=1}^{n}k \\ &= \sum_{k=1}^{2n}k - \sum_{k=1}^{n}(2k)\\& = \frac{2n(2n+1)}{2}-2\frac{n(n+1)}{2}=(2n^{2}+n)-(n^{2}+n)=n^2,\end{align}$$

with the same result. Let's note that here we made a subtle use of the homogeneity property of function $f(x) = x$, namely $f(2x)=2f(x)$. Simple as it appears, it was central to the derivation. For Ben Vitale's observation we'll have to do something very similar:

$$\begin{align}(2n+1)+(2n+3)+\ldots +(2n + (2n-1)) &= \sum_{k=1}^{2n}(2k-1)-\sum_{k=1}^{n}(2k-1) \\&= (2n)^2-n^2=3n^2.\end{align}$$

Ben Vitale's observation comes to

$$\frac{1+2+\ldots +(2n-1)}{(2n+1)+(2n+3)+\ldots +(4n-1)}=\frac{n^2}{3n^2}=\frac{1}{3}.$$

A generalization of this would be to take, say, twice as many terms in the denominator as in the numerator. This can be done in more than one way:

$$\frac{1+3+\ldots +(2n-1)}{1+3+\ldots +(4n-1)}=\frac{1}{4}$$

or

$$\frac{1+3+\ldots +(2n-1)}{(2n+1)+(2n+3)+\ldots +(6n-1)}=\frac{1}{8},$$

because $\sum_{k=1}^{3n}(2k-1)=(3n)^2-n^2=8n^2$, as before. Going further, there are at least three ways to have three times as many terms in the denominator as in the numerator:

$$\frac{\sum_{k=1}^{n}(2k-1)}{\sum_{k=1}^{3n}(2k-1)},\space \frac{\sum_{k=n+1}^{2n}(2k-1)}{\sum_{k=1}^{3n}(2k-1)},\space \frac{\sum_{k=2n+1}^{3n}(2k-1)}{\sum_{k=1}^{3n}(2k-1)}.$$

One can easily check that the three fractions are equal to $\frac{1}{9}$, $\frac{1}{3}$, and $\frac{5}{9}$, respectively.

Obviously, our success in getting a simple expression is due to the homogeneity of function $g(x)=x^2$: $g(ax)=a^{2}g(x)$. (There is a difference between the two functions $f$ and $g$. For the obvious reasons, the former is said to be homogeneous of order (degree) 1, the later of order 2.)

As Gary Davis notes at the end of his post, this kind of algebraic manipulations is accessible to middle and high school students. An additional example of the infinite series of the reciprocals of squares requires no more effort but a little hand-waving if not presented at the beginning Calculus class.

Assume we know that

$$\sum_{k=1}^{\infty}k^{-2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots =\frac{\pi^2}{6}.$$

Then of course, for the series that involves only even numbers,

$$\sum_{k=1}^{\infty}(2k)^{-2}=\frac{1}{2^2}+\frac{1}{(2^{2})(2^{2})}+\frac{1}{(2^{2})(3^{2})}+\ldots =\frac{\pi^2}{4\cdot 6}=\frac{\pi^2}{24}.$$

As a reward for the effort, we get an expression for the series of the reciprocals of the squares of odd numbers:

$$\sum_{k=1}^{\infty}(2k-1)^{-2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}.$$

Who would have thought that the series of the reciprocals of odd squares sums up to three times a similar series for even squares!

The bible takes six words to describe the evolution of the Hebrew tribe in Egypt from 70 souls to a people (Exodus 1.7). The closest translation I found among several is this: "... and the sons of Israel have been fruitful, and they teem, and multiply, and are very very mighty ..." In Hebrew, the four verbs are practically synonymous and the conveyed meaning is strengthened by the double "very very". What is the reason the bible found it necessary to use six words? Would not two or three suffice. Since everything in the bible has a meaning and a purpose, Rabbi Shlomo Yitzhaki (1040 - 1105) whose commentary became a standard fixture of the Hebrew Bible interprets the sentence to say that Hebrew women were producing six babies on every delivery. Bible commentaries and the rest of the vast Hebrew literature contain (if not overflow with) views and stories like that, legends that became an article of faith.

Rabbi Berel Wein quotes [Patterns in Jewish History, p. 167] the famous Rabbi Menahem Mendel Morgenstern of Kotzk who said

He who believes all of these tales is a fool and yet he who states that they could not have occurred is a non-believer.

and then proceeds to ponder that indeed the very existence of such legends reflects on historical events or persons. He corroborates that point of view with an anecdote:

Two men once exchanged stories about a great sage for whom fantastic claims of spirituality and piety were being made. One asked, "Do you really believe that story?" The other replied, "No, I do not. But no one tells such stories about the two of us.

This set me to thinking about the many legends - unconfirmed but widely told stories - which are abundant in math folklore. The story of Archimedes' death by a sword of a Roman soldier, irate for being ignored by Archimedes who was drawing math figures in the sand, is most likely untrue but sheds light on Archimedes' degree of concentration. Newton's conceiving of gravity due to a fall of an apple is a likely metaphor for Newton having insights that to other people might have appeared "out of the blue".

A quick search on the web brought to my attention an entertaining paper Life on the Mathematical Frontier: Legendary Figures and Their Adventures by Roger Cooke - a math professor at the University of Vermont - (Notices of the AMS, Volume 57, Number 4 (April 2010), 464-475). The paper was rather congenial to the idea I had in mind.

Cooke observes that legends arise also in modern times:

Who has not heard the "explanation" of the absence of a Nobel Prize in mathematics?

One of the modern legends is the story of George Bernard Dantzig (1914–2005) I have mentioned in a previous post. Dantzig confirms the story in an interview [More Mathematical People, p. 67], so what makes it a legend? In time, the story acquired a life of its own. Cooke recollects that the very same story was told to him by his roommate, but the central figure of it was John Milnor.

Some legends, Cooke says, have undoubtedly been lost. For example, L. A. Lyusternik reminisces that in Luzin’s Moscow school it was customary to invent exotic proofs of the infinitude of primes. Only a few have been documented. Among the surviving proofs is one ascribed to Khinchin. The proof is based on Euler’s formula

$$\prod_{p}(1-\frac{1}{p^{2}})^{-1}=\sum_{n=1}\frac{1}{n^{2}}=\frac{\pi^{2}}{6},$$

where the product on the left is over all the primes. If the set of primes were finite, the left-hand side of this formula would be a rational number, and hence $\pi^{2}$ would be rational. (The reciprocal of this expression gives the probability that two random integers are mutually prime.)

Actually, there are quite a few collections of mathematical anecdotes bordering on being legends. As H. Eves wrote of N. Wiener in his [Mathematical Circles Revisited, pp 174-175]

In time his brilliance and his eccentricity became woven into M.I.T. campus mythology, and a host of stories and legends sprang up about him.

This is the one I like best (355°):

It happened that the Wieners moved to a new house, in the same neighborhood as the former residence. Knowing her husband absentmindedness, Mrs. Wiener gave him careful and written instructions for reaching the new house. However at the close of the day, Professor Wiener could not find the written instructions and of course did not remember them. Hence, seeking something familiar, he set off in the direction of his former residence. Presently he spied a young child and asked her" "Little girl, can you tell me where the Wieners have moved to?" "Yes, Daddy," came the reply, "Mommy said you'd probably be here so she sent me over to show you the way home."

Rabbi Wein writes (p. 168)

The Jewish people as a whole possess a strong collective memory. ... This memory bank has been fed by stories about great people, significant events and terrible tragedies that have occurred over the millennia of Jewish life. These legends, whether completely accurate or not, help us recall the core event and/or person about which they revolve and, in so doing, keep our memory of the past alive enabling us to deal so much better with our present situations and challenges.

According to Philip Davis (the first chapter of the collection Essays in Humanistic Mathematics),

Mathematics, like literature, has metaphor. Mathematics, like poetry, has ambiguity. Mathematics possess aesthetic component ... Mathematics has paradox. Mathematics has mystery and can convey awe. Mathematics has: a sense of outcome, a feeling of rightness, a sense of catharsis. ... Mathematics has history. Like anthropology and literature, mathematics embodies mythologies.

Mathematics has collective memory that enlivens its subject and makes it a human endeavor. Sometimes I think that mathematics education, though, has no memory, see, for example, a blog by Julie Mack (with thanks to Gary Davis.) But this is a separate story.

For a prime $p$, two integers are both divisible by $p$ with the probability $p^{-2}$, because this only happens when the two integers have the residue 0 (one out of $p$ available residues) modulo $p$.

Two integers are mutually prime if they have no common nontrivial factors, prime facors in particular. Assuming divisibility by one prime is independent of divisibility by another, two integers are mutually prime with the probability

$\prod_{p}(1-p^{-2})=\frac{6}{\pi^{2}},$

where the product is over all prime $p$.

(There is an extended version of this argument - in Spanish.)

References

1. TOM M. APOSTOL, What Is the Most Surprising Result in Mathematics? Part II, Math Horizons, Vol. 4, No. 3 (February 1997), pp. 26-31