Uniformly at Random

Concerning the machines of Archimedes

leave a comment »

Below is an excerpt from Plutarch’s Life of Marcellus (the famous Dryden translation) concerning the mechanical work of Archimedes in devising war machines for the defense of Syracuse.  What is notable to me is the mention of Plato’s indignation at the application of mathematics to practical ends.

These machines he had designed and contrived, not as matters of any importance, but as mere amusements in geometry; in compliance with King Hiero’s desire and request, some little time before, that he should reduce to practice some part of his admirable speculation in science, and by accommodating the theoretic truth to sensation and ordinary use, bring it more within the appreciation of the people in general. Eudoxus and Archytas had been the first originators of this far-famed and highly-prized art of mechanics, which they employed as an elegant illustration of geometrical truths, and as means of sustaining experimentally, to the satisfaction of the senses, conclusions too intricate for proof by words and diagrams. As, for example, to solve the problem, so often required in constructing geometrical figures, given the two extremes, to find the two mean lines of a proportion, both these mathematicians had recourse to the aid of instruments, adapting to their purpose certain curves and sections of lines. But what with Plato’s indignation at it, and his invectives against it as the mere corruption and annihilation of the one good of geometry, which was thus shamefully turning its back upon the unembodied objects of pure intelligence to recur to sensation, and to ask help (not to be obtained without base supervisions and depravation) from matter; so it was that mechanics came to be separated from geometry, and, repudiated and neglected by philosophers, took its place as a military art. Archimedes, however, in writing to King Hiero, whose friend and near relation he was, had stated that given the force, any given weight might be moved, and even boasted, we are told, relying on the strength of demonstration, that if there were another earth, by going into it he could remove this. Hiero being struck with amazement at this, and entreating him to make good this problem by actual experiment, and show some great weight moved by a small engine, he fixed accordingly upon a ship of burden out of the king’s arsenal, which could not be drawn out of the dock without great labour and many men; and, loading her with many passengers and a full freight, sitting himself the while far off, with no great endeavour, but only holding the head of the pulley in his hand and drawing the cords by degrees, he drew the ship in a straight line, as smoothly and evenly as if she had been in the sea. The king, astonished at this, and convinced of the power of the art, prevailed upon Archimedes to make him engines accommodated to all the purposes, offensive and defensive, of a siege. These the king himself never made use of, because he spent almost all his life in a profound quiet and the highest affluence. But the apparatus was, in most opportune time, ready at hand for the Syracusans, and with it also the engineer himself.

The manner of the death of Archimedes is quite famous, and is described by Plutarch (Marcellus):

But nothing afflicted Marcellus so much as the death of Archimedes, who was then, as fate would have it, intent upon working out some problem by a diagram, and having fixed his mind alike and his eyes upon the subject of his speculation, he never noticed the incursion of the Romans, nor that the city was taken. In this transport of study and contemplation, a soldier, unexpectedly coming up to him, commanded him to follow to Marcellus; which he declining to do before he had worked out his problem to a demonstration, the soldier, enraged, drew his sword and ran him through. Others write that a Roman soldier, running upon him with a drawn sword, offered to kill him; and that Archimedes, looking back, earnestly besought him to hold his hand a little while, that he might not leave what he was then at work upon inconclusive and imperfect; but the soldier, nothing moved by his entreaty, instantly killed him. Others again relate that, as Archimedes was carrying to Marcellus mathematical instruments, dials, spheres, and angles, by which the magnitude of the sun might be measured to the sight, some soldiers seeing him, and thinking that he carried gold in a vessel, slew him. Certain it is that his death was very afflicting to Marcellus; and that Marcellus ever after regarded him that killed him as a murderer; and that he sought for his kindred and honoured them with signal favours.

Written by uncudh

November 8, 2009 at 3:00 pm

Posted in history, math

Tagged with , , , ,

Napkin thieves in ancient Rome

leave a comment »

Apparently Romans got pretty upset when people would steal napkins from their dinner tables.  Catullus, at any rate, didn’t care for the practice.  He wrote, not one, but two poems berating someone for swiping napkins at dinner.  From Catullus 12 (Green’s translation):

Your left hand, friend Asinius, you provincial,
works its michief while we drink and gossip,
snitching napkins from distracted guests.  You
think this trick is smart?  So dumb, you can’t see
just how dirty your game is, how unlovely?
[...]
Either, then, you give me back my napkin,
or else you’ll get a scad of scathing verses.
It’s not so much the price that’s made me angry:
this was a gift, a memento from my comrade,
top-line real native hand-towels, that Fabullus—
and Veranius—sent me all the way from
Spain: so I must love them just as much as
sweet Veranius and my dear Fabullus.

Written by uncudh

October 25, 2009 at 8:40 pm

Posted in literature

Tagged with , ,

Sigurd as the “chosen one”

leave a comment »

We have previously made some remarks concerning Tolkien’s recently published Lays of Sigurd and Gudrun.  One thing I neglected to mention was Tolkien’s one major departure from his sources: namely, his portrayal of Sigurd as the “chosen one” of the gods, whose presence at the Last Battle will determine its outcome and the future of the world to follow.  The Norse Poetic Edda begins with the famous poem Voluspa, which describes the prophecy of the Seeress concerning the Ragnarok, or the Doom of the gods.  Tolkien begins his Lay of Sigurd with his own version of the Voluspa; however, he inserts several stanzas describing the role of Sigurd in the Last Battle:

If in day of Doom
one deathless stands,
who death hath tasted
and dies no more,
the serpent-slayer,
seed of Odin,
then all shall not end,
nor Earth perish.

This conception of Sigurd is not at all present in any of the Norse texts.  One wonders why Tolkien felt it necessary to introduce such an element into the mythology.  Christopher Tolkien points out in his commentary to the Lay the connections to his father’s own imagined mythology, in particular to the tale of Turin Turambar:

This mysterious conception [...] reappeared as a prophecy in the Silmarillion texts of the 1930s: so in the Quenta Noldorinwa, ‘it shall be the black sword of Turin that deals unto Melko [Morgoth] his death and final end; and so shall the children of Hurin and all Men be avenged.’

In general, the parallels between Sigurd and Turin are of course quite clear:  Turin as the Dragon-slayer, the wearer of the Dragon-helm, etc.

Written by uncudh

October 24, 2009 at 2:38 pm

Posted in literature

Tagged with , , , , ,

The death of love

with one comment

Arguably the greatest long poem (mahakavya) in classical Sanskrit is Kalidasa’s Kumarasambhava (or “The Birth of Kumara”), which tells the tale of how the warrior god Skanda (the “Kumara” of the title) came to be born.  Once, the gods were suffering greatly from the attacks of the demon Taraka.  Unable to defeat Taraka, the gods approached the creator-god Brahma to ask for his help.  Specifically, the gods asked for a general who could lead them to victory against Taraka.  Brahma told the gods that they must find a way to convince Shiva the Destroyer to marry Parvati, the daughter of the mountain god.  The child of Shiva and Parvati would be the general that they were looking for.  However, Shiva was deeply absorbed in meditation and would not easily be tempted into marriage.  The gods approached Kamadeva, god of love (armed, like his Greco-Roman counterpart, with bow and arrow), and requested that he use his unique abilities to make Shiva fall in love with Parvati.  Kamadeva agreed, and went to the mountaintop where the god Shiva was engaged in meditation.  Shiva, however, sensed the intrusion of the love god.  Kalidasa describes what happened next (the Clay Sanskrit Library edition 3.69-72):

Then Three-eyed Shiva,
through his self-control
powerfully suppressing
the disturbance of his senses,
wished to see the cause
of his mind’s disturbance
and sent his gaze in all directions.

He saw Self-born Love ready to attack,
his lovely bow drawn right back
to form a circle,
his fist resting
at the corner of his right eye,
shoulder hunched,
left foot arched.

Enraged by the violation of his penance,
his frown made his face
dreadful to behold,
and from his third eye
a sparkling, blazing fire
suddenly flew forth.

“Lord, hold back your anger,
hold back!”—
even as the cries of the wind-gods
crossed the sky,
that fire born from the eye of Shiva who is Being,
reduced to ashes Intoxicating Love.

His corporeal form having been disintegrated by the fire emanating from the mystical third eye of Shiva the Destroyer, Kamadeva, the god of love, is henceforth known as Ananga, the Bodiless God.

Written by uncudh

September 7, 2009 at 7:39 pm

The General Burnside Problem

leave a comment »

In 1902 William Burnside posed the following question concerning finitely generated groups:

(Bounded Burnside Problem) If G is a finitely generated group and there is an integer n such that g^n = 1 for every g\in G, then must G be finite?

The problem also has the following variant:

(General Burnside Problem) If G is a finitely generated group and every element of G has finite order, then must G be finite?

The answer to both questions was expected to be “yes”’; the solution to the General Burnside Problem was therefore anticipated to be somewhat harder than that of the Bounded Burnside Problem. However, the answer to both problems turned out to be “no”. A counterexample to the General Burnside problem was given by a beautiful and elegant construction of Golod and Shafarevich in 1964; and a counterexample to the Bounded Burnside Problem was given by Novikov and Adjan in 1968.

The Burnside Problem has the following ring-theoretic analogue:

(Kurosh’s Problem) If A is a finitely generated algebra over a field F and every element of A is nilpotent, then must A be nilpotent?

(A element a \in A is nilpotent if a^n=0 for some n; and A is itself nilpotent if there is an m such that a_1a_2\cdots a_m = 0 for all a_1,a_2,\ldots,a_m \in A.) The Golod-Shafarevich construction also provides a counter-example to Kurosh’s Problem. We sketch the construction below.

The Golod-Shafarevich Theorem

Let F be a field and let T = F\langle x_1,x_2,\ldots x_d \rangle be the free non-commutative algebra over F generated by the variables x_1,x_2,\ldots,x_d. Let T_n denote the subspace of T consisting of all linear combinations of monomials of degree n. The elements of T_n are the homogeneous elements of degree n. Let I be a two-sided ideal of T generated by a set A of homogeneous elements, each of degree at least 2. Suppose that A has at most r_i elements of degree i for i \geq 2. The following (which we do not prove here) is the main computational result of the Golod-Shafarevich construction:

Theorem. The quotient algebra T/I is infinite dimensional over F if the coefficients in the power series expansion of (1 - dz + \sum_{i=2}^\infty r_i z^i)^{-1} are nonnegative.

Using this theorem, one constructs a counterexample to Kurosh’s Problem as follows.

Counterexample to Kurosh’s Problem

Let T = F\langle x_1,x_2,x_3 \rangle be the free algebra over a countable field F. Let T' be the ideal of T consisting of all elements of T without constant term. Enumerate the elements of T' in increasing order by degree as t_1, t_2, \ldots. Choose an integer m_1 \geq 2 and write t_1^{m_1} = t_{1,2} + t_{1,3} + \cdots + t_{1,k_1}, where each t_{1,j} \in T_j. Choose another positive integer m_2 sufficiently large so that t_2^{m_2} = t_{2,k_1+1} + t_{2,k_1+2} + \cdots + t_{2,k_2} for some k_2 > k_1. Continue in this way for sufficiently large powers of t_3, t_4, \cdots. Now let I be the ideal generated by the t_{i,j} defined in the process above. Consider the quotient T'/I. The construction of I guarantees that each element of T'/I is nilpotent; but the theorem above ensures that T'/I is infinite dimensional over F (and hence not nilpotent). Thus T'/I is a counterexample to Kurosh’s Problem. From this construction, we can in turn build a counterexample to the General Burnside Problem.

Counterexample to the General Burnside Problem

Let us suppose now that p is a prime number and F is the field with p elements. Let T and I be as defined above. Let a_1, a_2, a_3 be the elements x_1 + I, x_2 + I, x_3 + I of the quotient T/I respectively. Let G be the multiplicative semigroup in T/I generated by the elements 1+a_1, 1+a_2, and 1+ a_3. An element of G has the form 1+a for some a \in T'/I. The element a is nilpotent by the construction of T'/I, and so for sufficiently large n, we have a^{p^n} = 0. Since we are in characteristic p we have (1+a)^{p^n} = 1 + a^{p^n} = 1. It follows that 1+a has an inverse, whence G is a group. Moreover every element 1+a of G has finite order (indeed order a power of p). Thus G satisfies the conditions of the General Burnside Problem. It remains to show that G is infinite. If G were finite, then the linear combinations of its elements would form a finite dimensional algebra B over F. Moreover, since both 1 and 1+a_i are in G, the linear combination (1+a_i) - 1 = a_i is in B. Thus, 1, a_1, a_2, a_3 are all in B. But 1, a_1, a_2, a_3 suffice to generate T/I, which was previously shown to be infinite dimensional. The algebra B is therefore also infinite dimensional, a contradiction. Thus G is infinite, as required.

Written by uncudh

September 4, 2009 at 12:20 am

Roger Loomis on source study

leave a comment »

Roger Sherman Loomis, in the preface and first chapter of his The Development of Arthurian Romance, makes some interesting remarks in defence of so-called “source study” as an approach to literary criticism:

If I have stressed origins and sources, it is because I believe with Vergil: ‘Felix qui potuit rerum cognoscere causas.’  For, without understanding the forces which went to the making of a work of art, one cannot understand the work of art, and right understanding is the basis of all true appreciation.

[...]

The reaction against the study of sources has led to the discovery in medieval literature of meanings and intentions which we may feel sure were never realized by the author.  Professor C. S. Lewis has been impressed by this phenomenon in the case of his own fictions:

Some published fantasies of my own have had foisted on them (often by the kindliest critics) so many admirable allegorical meanings that I never dreamed of as to throw me into doubt whether it is possible for the wit of man to devise anything in which the wit of some other man cannot find, and plausibly find, an allegory.

[...]

Practical-minded persons and aesthetically sensitive critics may well demand whether it is worth while to follow up studies so exotic and intricate in order to attain results so uncertain.  Is it not enough to read the Mabinogion or Malory’s prose epic or Gottfried von Strassburg’s Tristan or Tennyson’s Idylls or T. H. White’s The Once and Future King for pure enjoyment?  Is not the search for origins and influences really a ‘flight from the masterpiece’?

Now the raison d’etre of any work of art lies in the responses of those who see, hear, or read it, but its value lies only in the responses of those who are attuned to it; and in the realm of literature that means those who understand it.  In its fullest sense, understanding means knowing its creator, the materials he has worked with, what he did with them, and why.  When one cannot know the author, one must be content with some knowledge of his milieu and his age.  The reader who is content with his subjective reactions alone runs the risk of interpreting the work of art in a fashion which would evoke peals of laughter from the author.  He also shows his lack of interest in the creative process which produced the work of art.

Written by uncudh

August 30, 2009 at 1:48 am

Borges’ Theorem

leave a comment »

No doubt one of the most delightful parts of writing a book of mathematics is the search for clever literary quotations to place at the beginning of chapters or to scatter throughout the book as appropriate (perhaps also, in the eyes of some, one of the greatest wastes of time).  Flajolet and Sedgewick, in Analytic Combinatorics, make reference to the following result, which they call “Borges’ Theorem”, namely that given any finite set P of words, a random text of length n will contain every word of P with probability tending to 1 exponentially fast as n tends to infinity.  The reason for the appellation “Borges’ Theorem” is due to Borges’ story The Library of Babel, which describes a library that contains

Everything: the minutely detailed history of the future, the archangels’ autobiographies, the faithful catalogues of the Library, thousands and thousands of false catalogues, the demonstration of the fallacy of those catalogues, the demonstration of the fallacy of the true catalogue, the Gnostic gospel of Basilides, the commentary on that gospel, the commentary on the commentary on that gospel, the true story of your death, the translation of every book in all languages, the interpolations of every book in all books.

Written by uncudh

August 15, 2009 at 1:57 pm

A quotation from Weierstrass

leave a comment »

I recently came across the following quotation from Weierstrass, which I found to be interesting.

A mathematician who is not also something of a poet will never be a complete mathematician.

It reminded me a little bit of the following quotations from Hardy.

A mathematician, like a painter or a poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas.

Beauty is the first test: there is no permanent place in the world for ugly mathematics.

Written by uncudh

August 8, 2009 at 3:31 pm

Posted in math

Tagged with ,

Matrix mortality

leave a comment »

The matrix mortality problem is the following:

Given n \times n matrices M_1, M_2, \ldots, M_k, is there some product M_{i_1}M_{i_1}\cdots M_{i_\ell} = 0?

Remarkably, this problem is undecidable for n=3.  Paterson first proved this in 1970; we follow the treatment of Halava and Harju in the Amer. Math. Monthly, Aug.-Sep. 2001.  The proof is by a reduction from the Post Correspondence Problem, which is well-known to be undecidable.  Let A and B be finite alphabets and let A^* and B^* denote the sets of words over the alphabets A and B respectively.  The Post Correspondence Problem is the following:

Let g,h be maps from A \to B^*, which extend by concatenation to maps from A^* \to B^*.  Does there exist a word w such that g(w) = h(w)?

This problem is undecidable, even when the alphabet B consists of only two symbols.  Before proving the undecidability of the mortality problem, we first sketch a proof that the following problem is undecidable for n=3:

Given n \times n matrices M_1, M_2, \ldots, M_k, is there some product M_{i_1}M_{i_1}\cdots M_{i_\ell} that has its upper left entry equal to 0?

Let A = \{1,2,3\} and let B = \{2,3\}.  Define a map \sigma: A^* \to Z^{3 \times 3} by \sigma(a_1a_2\cdots a_\ell) = \sum_{i=1}^\ell a_i 3^{\ell-i}, for all words w = a_1a_2\cdots a_\ell, where a_i \in A.  The map \sigma is injective and satisfies \sigma(uv) = 3^{|v|}\sigma(u) + \sigma(v) for all words u,v \in A^*, where |v| denotes the length of the word v.

Define the matrix T = \left( \begin{array}{ccc} 1&0&1\\ 1&1&0\\ 0&0&1 \end{array} \right).

Define the map \gamma: A^* \times A^* \to Z^{3\times 3} by \gamma(u,v) = T \left( \begin{array}{ccc} 3^{|u|}&0&0\\ 0&3^{|v|}&0\\ \sigma(u)&\sigma(v)&1 \end{array} \right) T^{-1}.

It is easy to show that \gamma is injective and is a homomorphism (i.e., \gamma(u'u'', v'v'') = \gamma(u',v')\gamma(u'',v'')).  Furthermore, the upper left entry of \gamma(u,v) is equal to 3^{|u|} + \sigma(u) - \sigma(v).

Consider now an instance g,h of the Post Correspondence Problem: i.e., g,h:A^* \to B^*.  Define matrices N_a = \gamma(h(a),g(a)) and N_a' = \gamma(h(a),1g(a)) for all a \in A.  Consider a product M = M_{a_1} M_{a_2} \cdots M_{a_\ell}, where w = a_1a_2\cdots a_\ell is a word with a_i \in A, and M_{a_i} is either N_{a_i} or N_{a_i}'.  The upper left corner of M equals 3^{|u|} + \sigma(u) - \sigma(v), where u = h(w) is a word over the alphabet B and v is some word over the alphabet A.  One then shows that this upper left corner is 0 if and only if 1u = v if and only if v = 1g(w) if and only if g(w) = h(w).

To show that the mortality problem is undecidable for n=3, let S = \left( \begin{array}{ccc} 1&0&0\\ 0&0&0\\ 0&0&0 \end{array} \right).  For any matrix M, the matrix SMS consists of all 0’s except for its upper left corner, which is equal to the upper left corner of M.  Thus if given matrices M_1, M_2, \ldots, M_k, some product M has upper left corner 0, then SMS = 0.  On the other hand, one can show that if some product is 0, say wlog of the form, S A_1 S A_2 \cdots S A_\ell S = 0, where A_i is a product of the M_j’s, then some A_i has its upper left corner equal to 0.

The decidability of the mortality problem for 2 \times 2 matrices remains open.

Written by uncudh

July 9, 2009 at 3:41 pm

The necklace splitting problem

leave a comment »

The necklace splitting problem may be formulated (somewhat fancifully) as follows.

Suppose that two thieves have stolen a necklace.  The necklace is open at the clasp and consists of some number of jewels of k different types.  There are an even number of jewels of each type.  The thieves wish to cut the necklace into as few pieces as possible and share the pieces between the two of them so that each thief has a fair share of each type of jewel.  What is the minimum number of cuts that suffices?

Alon and West (Proc. Amer. Math. Soc. 98 (1986) 623-628) showed that k cuts always suffices for a fair division of the necklace, regardless of the total number of jewels.  Remarkably, the (very short!) proof of this combinatorial result uses a result from topology, namely the Borsuk-Ulam theorem.

Written by uncudh

June 9, 2009 at 3:04 pm

Posted in math

Tagged with