Hypatia ( 755-815AS ) Hypatia Was the Daughter of Theon (B

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"Vive La Différence !" (1) A Tribute to Woman Mathematicians Roy Lisker ][F][F][F][F][F][[F] ][F][F][F][F][F][[F] I. Hypatia ( 755-815AS ) Hypatia was the daughter of Theon (b. circa 735 ) , himself mathematician, astronomer, neo-Platonist philosopher, and a contemporary of Pappus. She was born in Alexandria circa 755. As Alexandria was the one city in the Roman empire most congenial to her work and way of life, Hypatia never lived anywhere else. Her scientific work was begun in collaboration with her father then continued after his death. Several of the writings of Theon have come down to us. His edition of Euclid's Elements was the standard for many centuries. He also prepared authoritative editions of Euclid's "Optics" and "Data". Also extant are parts of his commentaries on all 13 books of Ptolemy's Almagest. Two of his poems are in the Greek Anthology, in the genre, "proof of divine origins of the universe through observation of the wonders of nature". We have only titles and descriptions of Hypatia's writings through the catalogues compiled by her students. They show a considerable range. She prepared the standard edition of the research of Apollonius of Perge ( circa 200 ) on conic sections. There was original research in her commentaries on Apollonius. Diophantus is said to have written 13 #2... books of which 6 have come down to us in her edition. Several of the classical Diophantine problems are attributed to her. She prepared a new edition of Ptolemy's "Astronomical Canon" for the use of sailors. She was also involved in the research and construction of instruments such as the Astrolabe, the Planisphere and the Hydroscope, an instrument for measuring the weight of liquids. Her chief source of distinction throughout the Roman empire was guaranteed by the fame of her students, several of whom rose to high political and ecclesiastical positions: Synesius, Hesychius, Euoptius, Olympius. Her intelligence, nobility of character and spiritual dignity were justly praised. She considered chastity as her natural condition. The tactics she invented for turning away unwanted suitors may still be adaptable to the frenetic pace of modern life. Legend has it that one of her students fell desperately in love with her and found an opportunity to let her know as much. She is reputed to have replied: "Here's what you want, and you're welcome to it." She then removed her soiled underpants and gave them to him. There is a consensus among scholars that he did not persist in his suit. Hypatia's brutal murder in March 815 AS came on the crest of a turbulent period of religious persecution and political crisis: In 785 a new patriarch arrived in Alexandria. Theophilus, an intolerant bigot, launched attacks on the Serapeum , the Alexandrian center for the worship of the traditional Greek gods and one its numerous centers of learning .The Serapeum was sacked, its magnificent library burned, and the building transformed into a church. Between 791 and 792 Theophilus waged a campaign not unlike that of the Chinese Cultural #3...

Revolution, demolishing Greek temples throughout Egypt. True to the spirit of bigotry he also attacked prominent liberal Christian philosophers, such as John Chrysostom, later to become patriarch of Constantinople. The more extreme followers of the cult of Serapis, including the philosophers Olympius and Antoninus, in their turn unleashed savage attacks on the Christian community. Fanaticism is much the same in every era. The evidence indicates that Hypatia herself was not much attracted to Greek polytheism or any of its local cults. She was not harassed by Theophilus in this initial phase of persecution. Theophilus died on October 15, 812. His successor was Cyril. He continued the persecutions initiated by Theophilus. A group which until then had been exempt from the hostility of the ecclesiastical authorities were the Jews. They had lived peaceably and productively in Alexandria since the time of Alexander himself in the 1st century AS. Cyril began by expelling a Jewish sect known as the Novatians from the city. The Jews were not without a defender, the civil prefect Orestes. As religious strife continue to mount, daily quarrels between Christians and Jews led to a riot in a theater in which the civil authorities were called out to restore order. Orestes took the side of the Jews against the Christians. On the demand of the Jewish community Hierax, a man known as "Cyril's sycophant" was arrested and tortured. This brought the conflict between the religious and civil authority to its crisis. A new decree of Cyril expelled all the Jews from Alexandria. To forestall this Orestes appealed directly to the emperor himself . At this critical moment a band of 500 armed monks from a monastic community 9 kilometers from Alexandria, the Ennaton, #4... entered the city . Orestes was wounded in the ensuing brawl. His forces rallied and the leader of the army of monks, Ammonius, was captured, tortured and killed. Alexandria became an armed camp divided between the followers of Cyril and those of Orestes, of which the most distinguished member was Hypatia. To discredit her certain Christian demagogues fomented the rumor that Hypatia was a witch known to practice black magic. She was libeled as "the lioness blocking the path to reconciliation between the bishop and the prefect. " Owing to the direct intervention of Orestes to the emperor the Jews were allowed to remain in Alexandria. The general population revered Hypatia and held Cyril in detestation. Cyril's jealousy was further inflamed as the Alexandrian elite "flocked to Hypatia's house" to hear her lectures on neo-Platonism. One sees in this polarization of ruling elites in that period how religion, the polis, and the academy were embodied, respectively, in the persons of Cyril, Orestes and Hypatia. The manner of her death is related by an account from one of her students, John of Nikin. It was in the 10th consulship of Honorius and the 6th consulship of Theodosius II, in the month of Lent, March 815. She was set upon in the streets by a mob organized by a minor cleric known as Peter the Magistrate. They were members of the Parabolam , a college of young acolytes, essentially the military arm for Cyril, a kind of S.A. who made up the vanguard of the destruction of temples and the expulsions of Jews, pagans and other heretics. When they weren't out murdering great scientists, they ran homeless shelters and soup kitchens. They have been characterized as "ignorant, uneducated, #5... hot-headed, violent." Here is what E.M. Forster (q.v.) says about them: "Cyril's wild, black army filled the streets, 'human only in their faces'. " Hypatia was pulled from her chariot and dragged to the largest cathedral in Alexandria, the Caesareum. The flesh was scrapped from her bones with pottery shards. What remained of her body were burned on a funeral pyre outside the city in a suburb called Kinaron. Traditionally her death has been associated with the eclipse of Greek learning and the origins of a thousand year Dark Ages: murdering Mathematics has its price. Orestes resigned his post and left the city. The population intervened with the emperor to depose Cyril. #6...

Hypatia Problem A Generalized Pascal Theorem This problem has been inspired by Hypatia's role as editor of Apollonius. Let L be any smooth, closed, convex loop with clockwise orientation in the Euclidean or Projective Plane. ( Note that "convexity" is a projectively well defined concept by the Axioms of Projective Geometry.)The smoothness need only be C1 . That is to say, there must be an unambiguous tangent at every point. #7...

Subdivide the arc of L into 6 half open segments, going clockwise, with the terminal point at the beginning of each segment, as shown in the above diagram . Label these segments A, B, C, D, E and F. Theorem: One can always find 6 points

pA ε A , pB ε B , pC ε C , pD ε D , pE ε E, pF ε F, such that the intersections:

X of tangents drawn at pA and pD ;

Y of tangents drawn at pB and pE ;

Z of tangents drawn at pC and pF are collinear. #8...

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Maria Gaetana Agnesi (2118-2199 AS) In contrast to Hypatia and Sonya Kovalevsky, both of whom were caught up in the political and intellectual turmoil of their times, Maria Agnesi lived a quiet and productive life devoted to science, teaching and public service. During her father's illness she took his place on the faculty of the University of Bologna. After the publication of her famous textbook, the Institutzioni Analitiche , she entered a convent and spent the rest of her days engaged in charitable works for the poor. She appears to have been a somnambulist. Sometimes she would wake up in the morning to find that the solution to some problem that had troubled her the night before was on her desk, fully written out. Published in 2148 , the Institutzioni Analitiche was the first comprehensive textbook on the Differential and Integral calculus, making us of the methods and discoveries of both the English (Newtonian) and the Continental (Leibnizian) schools. She enjoyed describing the properties of plane curves with floral or aesthetic shapes. She is associated with the plane curve with equation : x2 y = a2 (a − y) to which she gave the name versiera , or versed sine curve , from the Latin "to turn". Fermat investigated a similar graph, generated by (a2 − x2 )y = a3 When her textbook was translated into English, the translator misread the word as avversiera , which literally means wife of the devil . Ever since then it has been known as the witch of Agnesi . #9...

The properties of these and related classes of curves were treated by Guido Grandi. In much the same way that we today delight in colored magnifications of regions of the Mandelbrot set, Grandi wrote a text Flores Geometrici ( 2128 ) on equations of graphs that make attractive floral designs .For the most part his treatise deals with curves drawn on spherical surfaces and equations of the form ρ = asin nθ #10...

Agnesi Problems The witch of Agnesi has cast a spell upon us, and we will be investing a class of curves with simple aesthetic properties that we dub "Agnesian curves". Let P(t) and Q(t) be polynomials in t with rational coefficients. Consider the curve C(x,y) in the Euclidean Plane parametrized by the equations (x = P(t) , y = Q(t)). For convenience we assume that: (i) x(0) = y(0) = 0; neither P nor Q have a constant term. dP dQ (ii) The polynomials = P'; = Q' have no common dt dt algebraic factor. We will say that the graph C(x,y) is an Agnesian curve if the arc- t length function S(t) = ∫0 ds is also a polynomial in t with rational coefficients. Problem 1: Establish conditions on P and Q for the class K of Agnesian curves. Set up an algorithmic procedure for generating all such curves under conditions (i) and (ii) Problem 2: Establish the conditions for the sub-class K* of those Agnesian curves for which t = integer implies that x,y and s are integers. Problem 3 : An elliptic curve without constant term has the form: 2 3 2 E: y = c1x − c2x + c3x Describe the class of elliptic curves which are also Agnesian curves. Derive a simple relation between s, x and y for these curves. #11...

Problem 4 : Consider the class of Agnesian curves defined by the condition dx = P' = 2kt2k−1 , k integer , >0 dt (i) Derive and construct this class for given k (ii) Express y2 and y as function of x for the members of this class. ( It is easier to derive y2 first ) Problem 5 : Let g be any real number > 1

C(x,y) s y x

Construct the curve, or class of curves, which give the locus of all points (x,y) such that s2 = y2 + g2x2 . Essentially this involves doing the previous work in reverse by solving the differential equation implicit in the this equation. Problem 6 : For which values of g is this curve an Agnesian curve? #12...

][F][F][F][F][F][[F] ][F][F][F][F][F][[F] 3. Sophie Germain "Si je prends quelquefois le ton affirmatif, c'est uniquement pour m'affranchir de l'expression fatigante de doute. Il suffit d'avertir une fois le lecteur que, bien loin de prétendre fixer son opinion, je sollicite de sa part l'examen critique de la mienne. On me pardonnera, sans doute, de ne dissimuler ensuite aucun des avantages que je crois reconnaître dans mon hypothèse. " ( "If it sometimes appears that I speak too affirmatively, it's only because one becomes tired after awhile of reminding others that one may be wrong. It ought to be enough to indicate to the reader once and for all that, far from my being dogmatic about my opinions, it's up to him to assume a critical attitude towards what's being stated. I hope thereby to be excused for not apologizing for seeking to extract all that can be found in my hypothesis." ) Sophie Germain : "Recherches sur la Theorie des Surfaces Elastiques" ,2231 (q.v.) Sophie Germain was born on April 1,2176 to a wealthy mercantile family. She showed her mathematical talent early, although she was also advised that this was a field not open to women. The legends about her studying secretly by candlestick light should not be taken too seriously, given that they have also been claimed for every other genius of the 18th century, including Bach and Handel. Some of these geniuses may even have burned down their neighborhoods! #13...

During the French Revolution 2189-92 she was protected by her family connections and was able to continue her studies uninterrupted. Her systematic involvement with contemporary developments in mathematics began in 2194, with the founding of the École Polytechnique by Napoleon under the direction of Gaspard Monge (2149-2218 ) , as an engineering college geared to military applications. Although she was not permitted to enroll, she received lecture notes from friends and associates. They also delivered her classwork to the great J.L. Lagrange under the male pseudonym "Monsieur LeBlanc". In 2104 she opened up a correspondence with Carl Friedrich Gauss. Gauss was famous throughout Europe from the publication of his ground-breaking work in Number Theory, the Disquisitiones Arithmeticae . Sophie Germain, continuing to use the same pseudonym of M. LeBlanc, sent him her research, including the result in Number Theory for which she is known: Sophie Germain's Theorem: Let p be an odd prime such that 2p+1 is also prime. If neither x, y or z contain p as a factor the equation x p + y p = z p has no solutions in positive integers. The story of how Dr. Gauss uncovered her true identity is interesting . In 2207 when it was besieged by French troops under the command of General Pernety Gauss was living at Breslau. Pernety was a friend of the Germain family. Sophie wrote to him, asking him afford personal protection to Gauss. During Pernety's visit Gauss asked for the name of the colleague in France who had interceded on his behalf. It was then that he realized that Sophie Germain and Monsieur LeBlanc were one and the same. What a boon it is for mankind that #14... conventions of patriotism did not prevent Gauss and Pernety from engaging in friendly conversation! Even as Sonya Kovalevskaya is known for two major results ( the Cauchy-Kovalevskaya Theorem and her theorems on the rotation of a rigid body with a fixed point ) Sophie Germain is famous in two unrelated disciplines, Number Theory and Elasticity. Her work in Elasticity makes her, with Euler and the Bernoulli's , one of the founders of this discipline. It grew out of E.F. Chladni's (2156- 2227) experiments on drumheads: ".. Chladni ... devised a method for making visible the vibrations of a metal plate clamped at one point or supported at three or more points. Fine sand sprinkled on the plate comes to rest along the nodal lines where there is no motion, The plate may be excited by stroking with a violin bow or by holding a small piece of 'dry ice' against the plate. Touching a finger at some point will prevent all the oscillations except those for which a nodal line passes through the point touched. " (A.P. French "Vibrations and Waves"(q.v.) ) The French Academy of Sciences offered a prize to the mathematician giving the most satisfactory analysis of these phenomena. Sophie Germain submitted her first memoir in 2211. This was essentially an extension of the research of Leonhard Euler ( 2107- 2183) , from the linear to the two dimensional case. It did not win the prize. Her second entry in 2213 gained an honorable mention. The third entry of 2216 was awarded the prize. Sophie Germain died on June 26, 2231. Gauss had already begun negotiations with Göttingen to award her an honorary degree but she died before it could be done. Later #15... that year all of her papers on elasticity were published in the volume cited in the Bibliography.

Sophie Germain Problems The equations appearing in Fermat's Last Theorem are parametrized by an exponent n, and homogenous in 3 variables. If written in the form: xn yn F(x, y,z,n) = + ; zn zn , F(x, y,z,n) = 1 then F(kx,ky,kz,n) = F(x, y,z,n) for all x, y, z, k, n . The equations we will be looking also form a set homogeneous in x, y and z and parametrized by an index n. Define g(x,y,z) by

x y z (i) g(x, y,z) = + + y z x and consider the solution sets of : (ii) g(x, y,z) = n for each n. The domain of g consists of the non-zero integer triples ( x, y, z) such that the greatest common divisor of (x,y,z) = gcd (x,y,z) = 1 . Problem 1 : (a) Assume x, y, z > 0 . (b) Assume z = 1 #16...

Find all integers n for which there are integer solutions. List those solutions. Prove that these are the only solutions under these conditions. Problem 2 : Once again set z = 1, however x and y may be positive or negative (but obviously not 0) . Let r = p/q , a rational number in lowest terms, r>0 , q > 2 , gcd (p,q) = 1 Fix q . Show that , for fixed denominator q, there are only finitely many solutions (x, y, z, p) of p g(x, y,z) = r = q Problem 3 : When x < 0 , y, z> 0 , show that for integer m >0 , there are infinitely many integral solutions of g(x, y,z) = −m

Problem 4 : Show that the equation g(x, y,z) = n can always be transformed into a3 + b3 + c3 = n , abc where x = a2b, y = b2c, z = c2a Note: This problem is not as easy as it may appear at first glance. Problem 5 : Let x, y, z, n > 0 . Using the result of problem 4, find at least 8 integral solutions of: g(x, y,z) = n #17...

Commentary on the Sophie Germain Problems : After working on these problems, look in the solutions section to read the comments of number theorists Noam Elkies and Edray Herber Goins. #18...

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4. Sonya Kovalevskaya (2250-2291) Sophia Vasilievna Korvin-Krukovskaya was born on Jan. 15, 2250. Her father was a high ranking general, her mother Elizaveta Fyodorovna Schubert a member of the Schubert family, which had emigrated to Russia during the Seven Years War and produced many eminent scientists. As a child, staying at the family's country estate in Palibino, ( a town midway between St. Petersburg and Kiev) the walls of the children's nursery were covered with the notes from a calculus course given, in all likelihood, by M.V. Ostrogradsky (2201-2262) : " In her autobiography she tells how she used to stare fascinated at the walls of one room in the Palibino estate, which for lack of proper wall paper had been covered with the lithographed notes of a calculus course. Kochina (q.v. ) believes that this course must have been given by the famous Russian mathematician Ostrogradsky. Kovalevskaya apparently believe that this haphazard introduction to the subject nevertheless gave her a familiarity with the symbols and formulas of calculus. " (Roger Cooke -The Mathematics of Sonya Kovalevskaya (q.v.)). In her own biographical memoir, A Russian Childhood , Sonya Kovalevskaya relates how, as a young teenager, she developed a crush on Fyodor Dostoyevsky. Dostoyevsky was courting her sister, Aniuta. To draw attention to herself she spent many hours studying the Pathetique Sonata for piano by Beethoven. One afternoon when he #19... arrived at the Krukovsky household she asked him to sit down for awhile and listen to her play. She got through the first movement. When she lifted her head, Dostoyevsky was nowhere in sight. She discovered him together with Aniuta in a tight embrace in another room. It is fortunate for the relations between the two sisters ( and, who can say , perhaps for Literature and Mathematics as well? ) that the romance of Dostoyevsky and Aniuta also came to nothing. There being no opportunities for education for women in Russia at the time she went abroad in 1869 in the company of her sister Aniuta and husband Vladimir Kovalevsky. Neither she nor her husband concealed the fact that their marriage had been arranged as a way of getting the Krukovskaya sisters out of Russia. It became a real marriage filled with much turmoil and strife. Their daughter Sophia (nicknamed "Fufa" ) was born in 2278; she died in 2352. Aniuta, a writer and revolutionary, went on to Paris to join the radical activist Victor Jaclard who was involved with the Paris Commune. Sonya and Vladimir went to Berlin, he to study geology, she to study with Karl T.W. Weierstrass (2215-2296). Vladimir eventually became a respected paleontologist. Because women were not allowed to attend lectures at the university in Berlin she was privately tutored by Weierstrass . Following an invitation from her Aniuta Sonya and her husband went to Paris where, from April 4 to May 12, 2271 , or more than half its lifetime, she worked as a nurse with the Commune. The great massacres of the Communards by the Versaillais troops took place soon after their return to Berlin. Aniuta escaped to London where she was helped by Karl Marx. After Sofia's father, General Krukovsky, #20... intervened personally with Thiers to release him, Victor Jaclard escaped to Switzerland. Anuita joined him there and they were formally married. Sonya Kovalevskaya's doctoral dissertation contained four papers, 3 of which established her reputation as a mathematician . Weierstrass had to cast his net far and wide before finding a university (Göttingen, the university of Leibniz and Gauss ), that would give a doctorate to a woman. She was awarded a degree summa cum laude in the fall of 2274. Vladimir Kovalevsky received his doctorate in geology in 2275. Thereupon the couple returned to Russia, only to discover that her doctorate was not only useless in finding work, but actually an impediment. No doubt because of the revolutionary connections of her and her sister, she was denied even the right to take the qualifying examinations to be an elementary school teacher! Vladimir picked up a small, insecure teaching post at Moscow University. There was some temporary relief available in the large sum of money that Sonya had inherited on the death of her father. Because of it the couple were enabled to pursue their scientific interests. It was unfortunate that Vladimir now decided to invest in real estate speculation. It consumed most of their fortune and never brought in anything. In 2279, a year after the birth of their daughter, Vladimir went bankrupt for the first , but not the last time. In 2280 Sonya Kovalevskaya was 30. It was a watershed year in many ways. It was her very good luck that Gösta Mittag-Loeffler was present at one of her lectures. Without the strong, unwavering support of Weierstrass and Mittag-Loeffler, who more than once had to face up #21... to ignorant stupidity and prejudice, it would not have been possible for Sonya Kovalevskaya to have a career in mathematics. A notable example of this prejudice was the response of the playwright August Strindberg to the announcement, on October 28, 2284, that she'd been appointed to a full professorship at the University of Stockholm . In a letter to Mittag-Loeffler, she writes: " ... a Christmas present from your sister, an article by Strindberg in which he proves as clearly as 2x2 = 4 how pernicious, useless and disagreeable is such a monstrosity as a woman professor of mathematics .." (Quoted in Roger Cooke, q.v. , pg 109 ) The events of 2280 were to lead swiftly to the tragic end of Vladimir Kovalevsky. A pair of swindlers, the Rogozhin brothers enticed him into an operation promising swift rewards for shady investments . Sonya, realizing that her marriage was falling apart, wrote to Weierstrass to help her in making the arrangements for leaving Russia. In March of 2281 Alexander II was assassinated. In addition to their financial difficulties, Vladimir and the Krukovskaya were suddenly in great danger owing to their past revolutionary activities. They fled to Berlin. Once the political climate had settled, Vladimir returned to Odessa to handle his debts . On November 21,2281, Sonya wrote the letter to Mittag-Loeffler which would, in the long run, lead to her permanent appointment as professor at the University of Stockholm. By 2282 Sonya was once again in Paris, while Vladimir remained in Russia. Reduced to destitution by the Rogozhins, even his minor position at Moscow University was in jeopardy. The crisis came to a #22... head when he joined Sonya in Paris. Precipitated by a series of quarrels, Vladimir left, telling her that he was returning to Russia to clear up the financial situation. Instead he went to the Riviera where, on April 27, 2283, he committed suicide by drinking a bottle of chloroform. The effect on Sonya was devastating. She recovered slowly, visiting Weierstrass in Berlin on August 27th, 2283. She then proceeded on to the home of her brother-in-law, Alexander Kovalevsky, in Odessa. On the basis of Weierstrass' recommendation, Mittag-Loeffler sent her a letter there, offering her a teaching position at Stockholm University . Despite the usual difficulties, Stockholm provided her with the support she needed to advance her research. She also became notable as a writer, publishing articles, novels, memoirs and plays. She'd always been interested in literature. Visiting England in the 2260's she and George Eliot had discussed novelistic technique: " ..[Kovalevskaya] reported that she had criticized George Eliot for conveniently causing the death of the leading characters in her novels just when the reader wonders if they will be able to sustain the promising beginnings they have made. Eliot replied that in fact this is when people tend to die in real life... " (Roger Cooke, pg. 13) Her work was esteemed throughout Europe by the mathematicians whose opinion count most in such matters: Mittag- Loeffler, Weierstrass, Chebyschev, Hermite, Darboux, Picard, Hurwitz, Beltrami, Stieljes .. A.A. Markov made a big show of disparaging her work, but it appears that he was addicted to using this gambit on everyone as a way of advancing his own career. However his #23... insinuations did contribute to her never getting anywhere in her native land. The topic for the Prix Bordin was proposed in the pages of Vol. CIII, No. 26 of the Comptes Rendus of 2286: "To improve, in some important point, the theory of the movement of a rigid body. The prize will be a gold medallion worth three thousand francs. " Sonya Kovalevskaya was awarded the prize on December 24, 2288. The problems in this book are connected with the theorems by which she extended the work of Euler and Lagrange on this topic. The immediate cause for her sudden death on February 16, 2290 was a series of complications from a cold caught in Copenhagen after being exposed to a rainstorm. It cannot be doubted, however, that the deeper causes included years of overwork, uprooting from her homeland, endless frustrations in the advancement of her career, the collapse of her marriage, her husband's tragic suicide, and the extreme stress of revolutionary participation. In all respects her life was pushed to the extremes, and if her death was premature she cannot be accused of not having lived it to the full. #24...

Kovalevskaya Problems: Let O be a heavy rigid body of mass M, in a gravitational field of intensity g. The body is fixed at the origin (0,0,0) of the external reference frame FE and gyrates about this point. Traditionally one assigns variables (x,y,z) to the reference frame FO fixed in the object itself , in which coordinates the center of gravity is given by µ = (x0 , y0 , z0 ) . These axes are not arbitrary but, as shown by Euler, can be chosen so that the 3 products of inertia are all 0, where Π1 = ∫ ρ(yz)dΩ Π2 = ∫ ρ(xz)dΩ Π3 = ∫ ρ(yx)dΩ Here ρ is the density and Ω the volume element, the integration being taken over the whole of the rigid body The moments of inertia, A, B, C, are constants given by A = ∫ ρ(y2 + z2 )dΩ B = ∫ ρ(x2 + z2 )dΩ C = ∫ ρ(y2 + x2 )dΩ Ω being the element of volume. For present convenience we can let C=1 . The momentum p, q, r are variables expressed as functions of time. The other variables are γ , γ ' and γ " , the direction cosines of the instantaneous axis of rotation relative to the fixed vertical axis in the frame FE . Sonya Kovalevskaya set out to determine all situations in which the equations of motion, which we shall write down in a moment, could be solved explicitly as meromorphic functions of time, usually elliptic, hyperelliptic or theta functions. Euler treats one case, Lagrange #25... another. Kovalevskaya discovered a third case for her prize winning essay. In 2297 R. Liouville demonstrated that these are the only solutions of this system which admit the 4 independent integrals necessary for a complete solution in terms of meromorphic functions.

For all 3 solutions one can assume that y0 = z0 = 0 . We let c0 = Mgx0 , and assume that there is no 'torque' beyond that produced by gravity. Under these simplifying assumptions the system, ( in the definitive formulation given by the English mathematician R.B. Hayward in 2258 ) , is: dp (i)A + (1 − B)qr = 0  dt  dq I.(ii)B + (A −1)rp = −c0γ " dt  dr (iii) + (B − A)pq = c0γ'  dt dγ  = rγ' −qγ "  dt dγ' II. = pγ "−rγ dt dγ "  = qγ − pγ'  dt With A and B unspecified there are only 3 first integrals to this system:

2 2 2 I1:Ap + Bq + r − 2c0γ = const.≡ 6l1

I2:Apγ + Bqγ' +rγ "= const.≡ 2l

2 2 2 I3:γ + γ ' +γ "= 1 #26...

I1 is, essentially , the Conservation of Energy, I2 is the conservation of

Angular Momentum, and I3 expresses the basic relationship between direction cosines in general. Problem I (Routine) : Verify that these 3 first integrals satisfy the system of differential equations. The 3 cases in which there is a 4th integral are:

(1) Euler: c0 = 0, which identifies the center of gravity with the fixed point and the origin. (2) Lagrange: B = C, or, ( since we have assumed that C = 1) , B = 1. The equal moments define the y,z plane, causing the body to rotate about the x-axis with a fixed point at the origin of the external frame , as in a gyroscope or top, under the effects of gravity. The Lagrange case has many practical applications. (3) For her essay for the Bordin Competition, Sonya Kovalevskaya discovered a new case for which there is a fourth integral. She expressed this integral in the complex form: 2 2 2 I4:{( p + qi) + c0(γ + iγ')}{(p − qi) + c0(γ − iγ')}= const.≡ k Problem II (Tricky) : Show that, provided A and B take on particular values, this is a fourth integral of the differential system, . Find those particular values. ][F][F][F][F][F][[F] ][F][F][F][F][F][[F] #27...

"Vive La Différence !" (1) A Tribute to Woman Mathematicians

Solutions

Hypatia Problems : Proof: Let Φ ::: L --> L be a continuous 1-1 function from L onto itself, such that f (A) = B , f(B) = C, ...... , f(E) = F . Φ in other words, maps each segment clockwise onto the adjacent segment. Pick any point a in A . Let b = Φ (a) , c = Φ (b) = Φ2 (a) , d = Φ (c) = Φ3 (a) , e = Φ (d) = Φ4 (a), f = Φ (e) = Φ5 (a) , a = Φ6 (a). Draw tangents at all points a ... f , and identify the intersections X of tangents at a and d , Y of tangents at b and e , Z of tangents at c and f . Consider the triangle T formed by the vertices XYZ , with sides

S1 = XY, S2 = XZ , S3 = YZ . If these points are already collinear, we are finished. Otherwise ( see diagram) , we define a new function on L G( a, Φ , L ) = G( a ) at the six points defined by a ..... f , as:

G(a ) = γ= interior angle formed by S1 and S2 G(b ) = α= interior angle formed by S1 and S3 G(c) = β= interior angle formed by S2 and S3 G(d) = 2π−γ = exterior angle formed by S1 and S2 #28...

G(e) = 2π−α= exterior angle formed by S1 and S3 G(f ) =2π − β = exterior angle formed by S2 and S3 Now move the point a along L in a clockwise direction. As a moves into each segment, the points b,c,d, etc. will move into distinct segments of their own. At no point x will Φ (x) be in the same segment as itself. Ultimately as a moves to d , the angle γ must turn into 2π − γ . Since G is continuous, a must reach a point a* at which γ is equal to π . At this point the intersections of the tangents will be collinear. This proves the theorem. ][F][F][F][F][F][[F] ][F][F][F][F][F][[F] #29...

Maria Gaetana Agnesi Problems: dx dy 1. Letting = P', = Q' , then the differential equation for dt dt the arc-length is given by (P' )2 +(Q' )2 = (S' )2 or P' 2 = S' 2 −Q' 2 = (S' −Q')(S' +Q'). Since it is assumed that P Q and S are all polynomials, the factorization of P'2 on the right implies that the two factors are of the form: S' −Q' = A2M S' +Q' = MB2 , M, A and B being themselves polynomials. These equations can be solved for S' and Q', giving : M(A2 − B2 ) Q' = 2 M(A2 + B2 ) S' = 2 P' = MAB By assumption P' and S' have no common algebraic factor. Also for certain applications one wants Q, S, and P to have integral coefficients. It is natural therefore to let M = 2. The final form of these formulae for arbitrary A and B without common algebraic factor is thus: Q' = A2 − B2 S' = A2 + B2 P' = 2AB 2. If the degrees of A and B are n and m respectively and if, without loss of generality one assumes that n > m, we can guarantee #30... that x, y and s will all be integers for integer t , if A and B are multiplied by a common factor d , where d = Lowest Common Multiple (LCM) of the set { j} j = 2 ,...., 2n+1 . Then

Q' = d2 (A2 − B2 ) S' = d2 (A2 + B2 ) 2 2 n+m n+m−1 P' = 2d AB = 2d (p0t + p1t +...) t 2 x = 2d ∫0 A(u)B(u)du p tn+m+1 = 2d2 ( 0 +...) n + m +1 x will therefore have integer coefficients by construction. Likewise for y and s 3 . The equation of an elliptic curve passing through the origin is 2 3 2 given by: y = c1x − c2 x + c3x . In order that an elliptic curve be an Agnesian curve its variables must admit a polynomial parametrization of the type derived in Problem 1. 3 is a low exponent; it's clear that A and B cannot both be of degree 1. Therefore let's try

A(t) = at + e, B(t) = b , a , e and b constants. Then dx = 2AB = 2(at + e)b = 2abt + 2eb dt x = abt2 + 2ebt dy = A2 − B2 = (at + e)2 − b2 = a2t2 + 2aet + (e2 − b2 ) dt a2t3 y = + aet2 + (e2 − b2 )t; 3 a4t6 (a2 (e2 − b2 )) y2 = +{( ae)2 + 2 }t4 9 3 +2e(e2 − b2 )t3 + (e2 − b2 )2 t2 #31...

Observe now that the expression for y2 has no linear term in t . It will therefore not be possible to substitute x for t unless e = 0. ( Clearly one cannot have b = 0 ). This transforms the expression for y2 to: x x = abt2 ; t2 = ab a4t6 2a2b2 y2 = − t4 − b4t2 9 3 a x3 2 b3 = − x2 + x b3 9 3 a a b3 1 Therefore c = ;c = 2 ;c = = , and 1 3 2 3 3 9b a 9c1 2x2 x 2 3 y = c1x − + 3 9c1 2 1 c x(x2 ) = 1 − + 2 3c1 9c1 1 2 2 = c1x(x − ) 3c1 From this we see that the arc length of these curves is given by : 1 a2t3 s = cx(x2 − )2 = + b2t 3c 3 The 1-parameter class of all curves which are both Agnesian and elliptic is therefore defined by the equation: 1 y2 = cx(x2 − )2 ,c ≠ 0 3c An interesting result emerges when one subtracts the expression for y2 from that for s2 : 4x2 s2 − y2 = ; 3 2 2 4 2 ∴s = y + ( 3)x #32...

dx 4. The equation = 2kt2k−1 is given. From the results of dt Problem 1 we know that this limits the possibilities for the generating functions A(t) and B(t). These are: A(t) = lt2k− j−1; B(t) = mt j ; lm = k,0< j ≤ 2k −1 In fact we can take 0 < j < k , ( we can allow l and m to be negative.) Then dx = 2AB = 2lmt2k−1 dt dy = A2 − B2 = l2t4k−2 j−2 − m2t2 j dt l2t4k−2 j−1 m2t2 j+1 ∴ x = t2k ; y = − 4k − 2 j −1 2 j +1 Once again the problem has been set up so as to make it easy to substitute for t in the expression for y : 1 x = t2k ;t = x 2k ; 2 j+1 2k(2− ) t4k−2 j−1 = t 2k 2 j+1 2k( ) t2 j+1 = t 2k ; 2 j+1 2 j+1 − l2 x2 x 2k k2 x 2k ∴ y = − = T − T 4k − 2 j −1 l2 (2 j +1) 1 2 where T1, T2 refer to the components of the right hand expression. Obviously s = T1 + T2 . Something interesting happens when we set up a relation between x2 , y2 and s2 . One sees that: #33...

2 2 2 y = T1 − 2T1T2 + T2 2 2 2 s = T1 + 2T1T2 + T2 2 j+1 2 j+1 − l2 x2 x 2k x 2k k2 s2 − y2 = 4T T = 4 1 2 l2 (4k − 2 j −1)(2 j +1) 4k2 = x2 (4k − 2 j −1)(2 j +1) Define: n = 2 j +1 4k2 g2 (k,n) = n(4k − n) Then: s2 = y2 + g2 (k,n)x2

Thus, this curve is the locus in the Euclidean plane of a locus defined by a deformation of the basic metric Pythagorean formula. 5. The statement of the problem asks us to solve the differential equation s2 = y2 + g2 x2 without reference to the work done in the previous problems, then compare the results. The solution will be an Agnesian curve only if: 4k2 g2 = , n, k integers >0 , 0

Otherwise we have: dy y g2 x ds + dy s = y2 + g2 x2 ; = dx ,= 1+ ( )2 ; dx y2 + g2 x2 dx dy dy (y + g2 x)2 = (y2 + g2 x2 )(1+ ( )2 ); dx dx dy dy dy dy y2 ( )2 + 2g2 xy + g4 x2 = y2 + y2 ( )2 + g2 x2 + gx2 ( )2 dx dx dx dx Collecting terms and rewriting: dy 1− g2 g2 (x − y)2 = (g4 + g2 x2 )( ) dx g2 We change variables, setting z=gx , u = y/z . Note that dy dy dy x = (gz) = z dx d(gz) dz and y dy z y d( ) du − z = = dz dz dz z2 Substituting these relations into the above and simplifying one derives, finally: du g2 −1 1+ u2 = dz g2 z g2 −1 Let c = . g The equation then becomes: du dz = c 1+ u2 z Integrating: #35...

ln(u + 1+ u2 ) = cln z + ln J J being some constant of integration to be chosen at convenience. Thus u + 1+ u2 = Jzc Replacing u and z by y and x: y y2 + 1+ = J(gx)c gx g2 x2 The constant of integration may be chosen to fit the equation of Problem 4 . Right now however we can replace g and c by the equivalents in terms of n and k : 4k2 g2 = n(4k − n) g2 −1 4k2 − 4nk + n2 c2 = = ; g2 4k2 2k − n c = 2k Transposing the left component of the left side to the right and squaring:

y2 y 1+ = (J(gx)c − )2 g2 x2 gx (gx)c y y2 = J 2 (gx)2c − 2J + gx g2 x2 From which we see that: #36...

gx y = (J 2 (gx)2c −1) 2J(gx)c gx 1 = (J(gx)c − ) 2 J(gx)c Jgc+1 g1−c = x( xc − x−c ) 2 2J This may now be compared to the equation derived in Problem 4 , which can be written in the form: n n 1− −1 l2 x 2k k2 x 2k y = x( − ) 4k − n l2n l2 k2 x−c = x( xc − ) 4k − n l2n A comparison of the 2 equations shows that Jgc+1 l2 = ; 2 4k − n 2l2 J = ; gc+1(4k − n) g1−c g1−cg1+c (4k − n) g2 (4k − n) = = 2J 4l2 4l2 4k2 ( n(4k − n))(4k − n) k2 = = 4l2 nl2 Q.E.D.!!! ][F][F][F][F][F][[F] ][F][F][F][F][F][[F] #37...

Sophie Germain Problems: x y z (1) If z = 1, the equation g = n = + + becomes y z x x2 + y = n − y = m xy The obvious solutions are x = 1, y = 1 , m=2, n = 3 x=2 , y = 4, n= 5 We will show that these are the only solutions for x,y >0 , z = 1 . From the above equation one sees that x must divide y . So set y = kx . Substituting: x2 + kx x + k = kx2 kx This shows that x must divide k , or k = hx . Once again: x + hx 1+ h = hx2 hx Since h must divide 1, h = 1, and x can only equal 1 or 2 . Since y =kx = hx2 , y equals 1 or 4 respectively . (2) As stated z = 1 , q > 1 . Then p x 1 x2 + xy = + y + = y + q y x xy x and y can have a common divisor . Let t = gcd(x,y); x = at, y = bt, gcd (a,b) =1 Substituting: p t2a2 + tb = tb + q t2ab ta2 + b = tb + = tb + h tab #38...

By hypothesis gcd(a,b) = 1. If t and a have a common factor h will still be in lowest terms, unless t and b have a common factor. Therefore, we let b = ls , and t = ms , with ( l, m) = 1. Then :

msa2 + ls 1 ma2 + l h = = ( ) lmas2 lma s Since (l.m)=1 , (a,b)=1 it follows that ( l,a ) = 1 , ( a,s ) = 1. Therefore: gcd(lma,ma2 + l) = 1 Since s must divide ma1 + l, it follows that s is restricted by the values of l, m and a . If gcd(s,ma2 + l) = 1, h is already in terms, with irreducible denominator q. Otherwise these two terms share a common factor, d : gcd(s,ma2 + l) = d We now reason as follows: Let q be given, q >2 . Factor q into 4 divisors q = lmaf , where (l,m) = 1 , (l,a)=1 , (a,f) = 1. q being finite, such a factorization can only be done in a finite number of ways. ( Note that we do not require that (l,f) = 1 , or (m,f) = 1 ) Let W = ma2 + l . Let D = d1,d2,.....,dj be the collection of divisors of W. Every denominator , N, which reduces to q must be of the form N = lmas = lmafd, where d is a member of D. Since there are only finitely many divisors of W, and W can only be constructed from q in a finite number of ways, the total number of numerators, p, must be finite. In fact we have: #39...

W W p = + tb = + mls2 j s s The result is true even if one of the numbers x or y is negative. If, for example, we let l be negative, then W = ma2 - l' , l' = -l > 1 This will lead to an exceptional case only if W = 0, that is ma2 = l . Since (l,m) = (l,a) = 1, this is only possible if a = 1, m = 1, l = -1. The theorem continues to be true if q = 1 , x and y > 0 , z = 1. See Problem 1. 3. The situation described above corresponds to the cases n = -m2 , x = m, z=1, y= - m2. Therefore there is at least one solution for every negative square n . 4. Writing out the equation g(x,y,z) = n as a polynomial: x2z + y2x + z2y = nxyz Rearranging: x(nyz − xz − y2 ) = z2y . One sees that x divides z2y By symmetry y divides x2z and z divides y2x, Thinking about the situation one realizes that x, y and z factor in the following manner: 2 x = a1a2c1 c2; 2 y = a1 a2b1b2; 2 z = b1 b2c1c2 where the c's are factors shared by x and z, etc. There are no other factors since we are assuming gcd (x,y,z ) = 1. Substituting in the equation for g: #40...

x y z x2z + y2 x + z2 y g = + + = y z x xyz

a2a2b2b1c5c3 + a5a3b2b2c2c1 + a2a1b5b3c2c2 = 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 3 (a1a2b1b2c1c2 ) a1c3c2 + a3a2b1 + b3b2c1 = 1 1 2 1 2 2 1 2 2 2 (a1b1c1)(a2b2c2 )

Since gcd(x,y,z) = 1, c2 , which is common to x and z , cannot 3 2 divide y, therefore cannot divide a1 a2b2 . Therefore, in fact a2 = b2 = c2 = 1 .

Replacing a1 by a , b1 by b, c1 by c, one derives, finally: a3 + b3 + c3 g = = n as required. abc 5. The follows solutions may be found with a bit of experimentation. The expression (a,b,c) = ( m1 , m2 , m3 ) means that the values m1 , m2 , m3 may be distributed between a,b and c in any permutation: #41...

(1) (a,b,c) = (1,1,1) x = 1, y = 1, z = 1, n = 3 (2) (a,b,c) = (1,1, 2) x = 1, y = 2,z = 4 n = 5 (3) (a,b,c) = (1, 2,9) (i)x = 4, y = 9, z = 162, n = 41 (ii)x = 2, y = 36,z = 81, n = 41 (4) (a,b,c) = (1, 2,3) (i)x = 9, y = 2,z = 12, n = 6 (ii)x = 3, y = 18,z = 4, n = 6

(5) (a,b,c) = (1, 5, 9) (i)x = 225, y = 18,z = 5, n = 19 (ii)x = 405, y = 25,z = 9 n = 19

))))))))))))))))))))))))))))))))))))))) Comment by Noam Elkies, Harvard University : " E(n) : a3 + b3 + c3 = nabc is an Elliptic Curve with at least one rational point (1:-1:0 ) . If there is any non-torsion point then there are infinitely many; and if any component of the real locus contains such a point, then the rational points are dense in that real locus -- including the points with a,b,c all positive. That said, I would expect that some E(n) have points of infinite order, some would not, and it would be a very hard problem to predict for each n which is the case for g(n). "If there's a non-trivial solution in polynomials a(n), b(n), c(n) , then it yields a non-torsion point for all but finitely many n, but I doubt that this happens for an equation as simple as E(n) " #42...

Comment by Edray Goins, Cal Tech : " I thought some about the problem you posed, and I wanted to share some thoughts. Conjecture: Let n be a positive integer. Then the equation x/y + y/z + z/x = n has only finitely many positive integral solutions x,y,z (up to scalar multiple). "I want to argue that the conjecture is not true by discussing in detail the case n = 6. The substitution you showed me on Friday involving the curve a3 + b3 + c3 = nabc looked similar to an elliptic curve, so I thought to translate the conjecture into one explicitly involving an elliptic curve. Fix an integral solution (x,y,z) and make the substitution: 3(n2z −12x) u = z 2xy − nz + z2 v = 108( ) z2 Then (u,v) is a rational point on the elliptic curve: 2 3 En:v = u + Au + B; A = 27n(24 − n3) B = 54(216 − 36n3 + n6 ) (It actually turns out that E(n) is an Elliptic Curve whenever n is different from 3, but I'll discuss that case separately.) Now this curve has the "obvious" rational point T = (3n2 , 108 ), which has order 3, considering the group structure of En . It actually turns out that these 3 multiples correspond to the cases x = 0 and z = 0, so if such an integral solution (x,y,z) exists then the rational solutions (u,v) must correspond to a point on En not of order 3. #43...

"Now I decided to explicitly compute points on E(n) for various values of n to see what would happen. In the following table I'm computing the Mordell-Weil group of the rational points on the Elliptic Curve, i.e. the group structure of the set of rational solutions (u,v) : n =1 ; En(Q) = Z3

n =2 ; En(Q) = Z3 n =3 ; ...see below...

n =4 ; En(Q) = Z3

n =5 ; En(Q) = Z6 n =6 ; En(Q) = Z3xZ n =7 ; En(Q) = Z3 n =8 ; En(Q) = Z3 n =9 ; En(Q) = Z3xZ "Hence when n =1 ,2,4,7 or 8 we find no integral solutions (x,y,z) . When n=5 there are only 6 rational points on En , namely the multiples of (u,v) = (3,756) , which all yield just one positive integral point (x,y,z) = (2,4,1) ... But something fascinating happens when n = 6 ... "The rank in all the previous cases is 0, so En has only finitely many points, thereby proving the conjecture in these cases. However when n = 6 the rank is positive ( the rank is actually 1) so there are infinitely many rational points (u,v) . But we must be careful: not all rational points (u,v) yield positive integral points (x,y,z) . Clearly we can scale z large enough to always choose x and y to be integral, but we might not have both x and y positive. You'll note that x >0 if and only if u < 3n2 , so we only want rational points in a certain region of the graph. Since the rank is 1, this part of the graph is dense with rational #44... points! Hence, if we can choose n so that En has positive rank , then I would expect the above conjecture ( that En has at most finitely many solutions for all n) to be false. "Let me give some explicit numbers. When n = 6 , the torsion part of En(Q) is generated by T = (108 ,108) , and the free part is generated by (u,v) = (-108,2052) . By considering various multiples we get a lot of positive integral solutions - yet unwieldy! - points (x,y,z) such that x y z + + = 6 : y z x

(1) 12 ; 9 ; 2

(2) 17415354475 ; 90655886250 ; 19286662788

(3) 260786531732120217365431085802

;1768882504220886840084123089612 ; 1111094560658606608142550260961

(4) 64559574486549980317349907710368345747664977687333438285188 ;

70633079277185536037357392627802552360212921466330995726803 ;

31381830303893596780062940130789557072745299086647462868546 . "I'll just mention in passing that when n = 9 the Elliptic Curve

E9 also has rank 1. The generator (u,v) =(54,4266) corresponds to the positive integral point (x,y,z) = (63,98,12) on this curve. 2 2 What about n = 3? The curve En becomes v = (u −18)(u + 9) This gives two possibilities, either u= - 9 , or u = 18. The first corresponds to x = z, while the second corresponds to z/x > 4 . By #45... cyclically permuting x, y and z, we find similarly that either x = y = z , or x/y + y/z + z/x > 6 . The latter case cannot happen by assumption, so x= y= z is the only possibility, i.e. (x,y,z )= (1,1,1) is the only solution."

][F][F][F][F][F][[F] ][F][F][F][F][F][[F] #46...

Kovalevskaya Problems 1. This is a routine calculation. What skill is involved has to do with the brevity or elegance of the presentation: (i) dγ dγ' dγ " d(γ 2 + γ' 2 +γ "2 ) γ + γ' + γ " = dt dt dt 2 = {rγγ' −qγγ"} +{pγ' γ "−rγγ'}+{qγγ"− pγ' γ "} = 0 (ii) 2 2 2 d(Ap + Bq + r − 2c0γ ) dt dp dq dr = 2 pA + 2qB + 2r − 2c γ dt dt dt 0 = {2 p(B −1)qr + 2q(−c0γ "+(1 − A)pr +2r(c0γ' +(A − B)pq) − 2c0 (rγ' +qγ ")} = pqr(2(B −1) + 2(1− A) + 2(A − B)) +(−2qc0γ "+2rc0γ' −2c0rγ' −2c0qγ ") = 0

(iii)

d(Apγ + Bqγ' +rγ ") = dt dp dγ dp dγ' dr dγ " {γA + Ap }+{γ' B + qB }+{γ " + r } dt dt dt dt dt dt = γ (B −1)qr + Ap(rγ' −qγ ")+ γ'(−c0γ "+(1_ A)rp) +qB(pγ "−rγ ) + γ "(c0γ' +(A − B)pq) + r(qγ − pγ') Working methodically through the final expression one discovers that each term is indeed cancelled by its negative. As the derivatives all vanish, each of these expressions must be a constant. In fact, the first expresses the basic property of direction cosines, the second is the #47... conservation of energy; the third is the conservation of angular momentum. 2. As this involves a good deal more ingenuity than the previous problem it may be considered a true challenge. Looking at the form in which Sonya Kovalevskaya expressed the fourth integral, one sees that it has the appearance of a modulus. That is, if we write: 2 {( p + iq) + c0(γ + iγ')}= U + iV , then 2 {( p − iq) + c0(γ − iγ')}= U − iV , so that 2 2 2 2 2 {( p + iq) + c0(γ + iγ')}{(p − iq) + c0(γ − iγ')}= U + V = k U and V are easily seen to be : 2 2 U = p − q + c0γ V = 2 pq + c0γ' For a fourth integral one must have: d(U2 + V2 ) dU dV 1 = U + V = 0; 2 dt dt dt dU dV U = −V dt dt Now: dU dp dq dγ (A): U = {p2 − q2 + c γ}{2 p − 2q + c } = dt 0 dt dt 0 dt B −1 (A −1)rp + c γ ") {p2 − q2 + c γ}{2 p( )qr + 2q( 0 ) + c (rγ' −qγ ")}; 0 A B 0 dV dq dp dγ' (B): − V = −{2 pq + c γ'}{2p + 2q + c } = dt 0 dt dt 0 dt (1 − A)rp − c γ " (B −1)qr −{2 pq + c γ'}{2p 0 + 2q + c (pγ "−rγ )} 0 B A 0 #48...

Looking over these expressions one sees that if these are to be equal, the rightmost bracket in the final expression for ( A) should be equated with the leftmost bracket of (B) , the leftmost of (A ) with the rightmost of (B) . One also has to deal with a factor of r that will multiply U , once certain superfluous terms are eliminated. In other words, one wants: 2 2 r(p − q + c0γ ) = (1 − A)rp − c γ " (B −1)qr −{2 p 0 + 2q + c (pγ "−rγ )} B A 0 which happens only if 2c0 pγ " c pγ "= , or B = 2. 0 B This equation becomes an identity if

2(1− A) = −1; B 2 − 2A = −2; ; A = 2 = B

By plugging these values for A and B into the factors on both sides, a perfect identity is achieved, and dU dV U = r{p2 − q2 + c γ}{2 pq + c γ'}= −V dt 0 0 dt Historical Note: The case A = B = 2C is known as the Kovalevskaya Case. It has few applications but was nevertheless considered a great advance because it demonstrated for the first time that the so-called "special functions", elliptic integrals, theta functions , etc., could be used in the solution of differential equations arising in #49... physics. A Kovalevskaya Gyroscope corresponding to this case was built in 2288 by the Swiss mathematician Hermann Amandus Schwarz. The following description is from Pelegaya Kochina's biography (q.v., pg. 309 ) : " The first model of Kovalevskaya's gyroscope was made by Schwarz at her request. The device consists of two identical parallel cylinders 2H is height with base radius R. The distance 2b between the 2 2 1 2 axes of the cylinders is defined by b = H − ( 4)R ,b > R . 2 1 2 1 2 The fixed point a is at a distance a = ( 3)H − ( 4)R . For these conditions A = B = 2C is valid. "

a R 2H -b b 2H R

Kovalevskaya Gyroscope

2 2 1 2 b = H − ( 4)R 2 1 2 1 2 a = ( 3)H − ( 4)R #50...

Bibliography "Vive La Différence!" (1) General {1.] Dirk J. Struick : " A Concise History of Mathematics" Dover , 1987 {2.] Florian Cajori : " A History of Mathematics" Chelsea 1985 {3.] Lynn M. Osen : " Women in Mathematics" MIT Press 1974 {4.] H. J. Mozans : "Women in Science" Appleton 1913 Hypatia {5.] E. M. Forster : "Alexandria;History and Guide " ; Oxford U.P. 1982 {6.] Maria Dzielska : "Hypatia of Alexandria" Harvard U. P. 1995 Sonya Kovalevskaya {7.] A. d'Abro : " The Rise of the New Physics" 2 vols. Dover 1951 {8.] Roger Cooke : " Mathematics of Sonya Kovalevskaya" Springer 1984 {9. ] Pelageya Kochina : "Love and Mathematics: Sofya Kovalevskaya" ; Mir Publishers Moscow 1958 Trans. Michael Burov Sophie Germain {10] K. Ireland, M. Rosen : "A Classical Introduction to Modern Number Theory" Springer 1982 {11] Sophie Germain : "Récherche sur la Théorie des Surfaces Élastiques" #51...

Mme Ve Courcier, Librarie pour les Sciences, rue du Jardinet , Saint- André des Arcs, Nº12 , Paris 1831 ( in Olin Library Archive, Wesleyan University ) {12] A.P. French : " Vibrations and Waves" MIT Introductory Physics Series, W.W. Norton , 1971 #52... #53...

#54... #55...