Abstract. Continued fractions : introduction and applications. What is the connection between the following questions?

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1 Mandalay University November 6, 206 The Ninth International Conference on Science and Mathematics Education in Develoing Countries Continued fractions : introduction and alications Michel Waldschmidt Université Pierre et Marie Curie (Paris 6) France htt:// Abstract The Euclidean algorithm for comuting the Greatest Common Divisor (gcd) of two ositive integers is arguably the oldest mathematical algorithms : it goes back to antiquity and was known to Euclid. A closely related algorithm yields the continued fraction exansion of a real number, which is a very e cient rocess for finding the best rational aroximations of arealnumber.continuedfractionsisaversatiletoolfor solving roblems related with movements involving two di erent eriods. This is how it occurs in number theory, in comlex analysis, in dynamical systems, as well as questions related with music, calendars, gears... We will quote some of these alications. /80 2/80 What is the connection between the following questions? How to construct a calendar? How to design a lanetarium? How to build a musical scale? How to find two ositive integers x, y such that x 2 6y 2 =? How to rove the irrationality of constants from analysis? Answer : continued fractions. Number of days in a year What is a year? Astronomical year (Sidereal, troical, anomalistic...). A year is the orbital eriod of the Earth moving in its orbit around the Sun. For an observer on the Earth, this corresonds to the eriod it takes to the Sun to comlete one course throughout the zodiac along the eclitic. A year is aroximately days. A good aroximation is = with a lea year every 4 years. This is a little bit too much. A better aroximation is = /80 3/80

2 The Gregorian calendar The Gregorian calendar The Gregorian calendar is based on a cycle of 400 years : there is one lea year every year which is a multile of 4 but not of 00 unless it is a multile of 400. It is named after Poe Gregory XIII, who introduced it in 582. In 400 years, in the Gregorian calendar, one omits 3 lea years, hence there are =46097days. Since 400 = 4(33 3+), in 400 years, the number of days counted with a year of days is = = /80 6/80 Further correction needed Aroximating In years, the number of days in reality is Write = = while the number of days in the Gregorian calendar is = Hence one should omit three more lea years every years. The first aroximation is Next write =4+ The second aroximation is = /80 8/80

3 Relacing with Next write = The third aroximation is = Using.309 =+ writing = , one could continue by = /80 Continued fraction notation Write = Third aroximation : [365, 4, 7, ] = [365, 4, 8] = =[365, 4, 7,, 3,...] = / 80 References on calendars Descrition and design of a lanetarium Exercise ; During 4000 years, the number of Fridays 3 is 6880, the number of Thursdays 3 is 6840 (and there are 6850 Mondays or Tuesdays 3). V. Frederick Rickey, Mathematics of the Gregorian Calendar, The Mathematical Intelligencer 7 n (985) Automati lanetarii of Christiaan Huygens ( ) astronomer, hysicist, robabilist and horologist. Huygens designed more accurate clocks than the ones available at the time. His invention of the endulum clock was abreakthroughintimekeeing,andhemadearototyeby the end of 656. Jacques Dutka, On the Gregorian revision of the Julian Calendar, The Mathematical Intelligencer 0 n (988) / 80 2 / 80

4 Earth and Saturn The ratio between the angle covered by the Earth and the angle covered by Saturn is = Not to scale! Continued fraction of / The ratio between the angle covered by the Earth and the angle covered by Saturn is = = The continued fraction of this ratio is and [29, 2, 2,, 5,, 4,...] [29, 2, 2, ] = = = / 80 htt://lus.maths.org/content/chaos-numberland-secret-life-continued-fractions 4 / 80 The algorithm of continued fractions Let x 2 R. Euclidean division of x by : x = bxc + {x} with bxc 2Z and 0 ale {x} <. If x is not an integer, then {x} 6= 0. Set x =, so that {x} x = bxc + x with bxc 2Z and x >. If x is not an integer, set x 2 = {x } : x = bxc + bx c + with x 2 >. x 2 Continued fraction exansion Set a 0 = bxc and a i = bx i c for i. Then : x = bxc + [x ]+ bx 2 c +... = a 0 + a + a the algorithm stos after finitely many stes if and only if x is rational. We use the notation x =[a 0,a,a 2,a 3,...] Remark : if a k 2, then [a 0,a,a 2,a 3,...,a k ]=[a 0,a,a 2,a 3,...,a k, ]. 5 / 80 6 / 80

5 Continued fractions and rational aroximation Connection with the Euclidean algorithm For x =[a 0,a,a 2,...,a k,...] the rational numbers in the sequence k q k =[a 0,a,a 2,...,a k ] (k =, 2,...) give rational aroximations for x which are the best ones when comaring the quality of the aroximation and the size of the denominator. a 0,a,a 2,... are the artial quotients, If x is rational, x = q, this rocess is nothing else than Euclidean algorithm of dividing by q : = a 0 q + r 0, If r 0 6=0, x = q r 0 >. 0 ale r 0 <q. Euclide : ( -306, -283) 0 q 0, q, 2 q 2... are the convergents. q = a r 0 + r, x 2 = r 0 r 7 / 80 8 / 80 Harmonics The successive harmonics of a note of frequency n are the vibrations with frequencies 2n, 3n, 4n, 5n,... Octaves The successive octaves of a note of frequency n are vibrations with frequencies 2n, 4n, 8n, 6n,...The ear recognizes notes at the octave one from the other. Using octaves, one relaces each note by a note with frequency in a given interval, say [n, 2n). The classical choice in Hertz is [264, 528). For simlicity we take rather [, 2). Hence a note with frequency f is relaced by a note with frequency r with ale r<2, where f =2 a r, a =[log 2 f] 2 Z, r =2 {log 2 f} 2 [, 2). This is a multilicative version of the Euclidean division. 9 / / 80

6 The fourth and the fifth The successive fifths A note with frequency 3 (which is a harmonic of ) isatthe octave of a note with frequency 3 2 The musical interval ale, 3 2 endoints of the interval is 3 2 The musical interval is called fifth, the ratio of the ale 3 2, 2 is the fourth, with ratio 4 3. The successive fifths are the notes in the interval [, 2], which are at the octave of notes with frequency namely :,, 3, 9, 27, 8, , 9 8, 27 6, 8 64, We shall never come back to the initial value, since the Diohantine equation 3 a =2 b has no solution in ositive integers a, b. 2 / / 80 Exonential Diohantine equations We cannot solve exactly the equation 2 a =3 b in ositive rational integers a and b, but we can look for owers of 2 which are close to owers of 3. Exonential Diohantine equations Levi ben Gershon ( ), a medieval Jewish hilosoher and astronomer, answering a question of the French comoser Philie de Vitry, roved that 3 m 2 n 6=when m 3. There are just three solutions (m, n) to the equation 3 m 2 n = ± in ositive integers m and n, namely 3 2=, 4 3=, 9 8=. Levi ben Gershon Philie de Vitry 23 / / 80

7 3 m 2 n 6=when m 3. Assume 3 m 2 n =with m 3. Then n 2, which imlies that 3 m mod4, whence m is even. Writing m =2k, we obtain (3 k )(3 k +)=2 n, which imlies that both 3 k and 3 k +are owers of 2. Diohantine equations This question leads to the study of so-called exonential Diohantine equations, which include the Catalan s equation x y q =where x,y, and q are unknowns in Z all 2 (this was solved recently ; the only solution is =,as suggested in 844 by E. Catalan, the same year when Liouville roduced the first examles of transcendental numbers). Eugène Charles Catalan (84 894) But the only owers of 3 which di er by 2 are and 3. Hence k =, contradicting the assumtion m / / 80 Aroximating 3 a by 2 b Instead of looking at Diohantine equations, one can consider rather the question of aroximating 3 a by 2 b from another oint of view. The fact that the equation 3 a =2 b has no solution in ositive integers a, b means that the logarithm in basis 2 of 3 : log 2 3= log 3 = , log 2 which is the solution x of the equation 2 x =3, is irrational. Powers of 2 which are close to owers of 3 corresond to rational aroximations a b to log 2 3 : log 2 3 ' a b, 2a ' 3 b. Aroximating log 2 3 by rational numbers Hence it is natural to consider the continued fraction exansion log 2 3= =[,,, 2, 2, 3,, 5,...] and to truncate this exansion : [] =, [, ] = 2, [,, ] = 3 2, [,,, 2] = 8 5 =.6. The aroximation of log 2 3= by 8 =.6 means that =256 is not too far from 3 5 = The number = is close to 2 3 ; this means that 2 5 fifths roduce almost to 3 octaves. 27 / / 80

8 Aroximating log 2 3 by rational numbers The next aroximation is [,,, 2, 2] = It is related to the fact that 2 9 is close to 3 2 : 2 9 =524288' 3 2 =5344, = 9 = = is close to 2 7 =28. 2 In music it means that twelve fifths is a bit more than seven octaves. Pythagoras Pythagoras of Samos (about 569 BC - about 475 BC) The comma of Pythagoras is 3 2 = It roduces an error of about.36%, which most eole cannot ear. 29 / / 80 Further remarkable aroximations Leonardo Pisano (Fibonacci) =25' 2 7 =28 = ' 2 4 Three thirds (ratio 5/4) roducealmostoneoctave. 2 0 =024' 0 3 Comuters (kilo octets) Acoustic : multilying the intensity of a sound by 0 means adding 0 decibels (logarithmic scale). Multilying the intensity by k, amounts to add d decibels with 0 d = k 0. Since 2 0 ' 0 3, doubling the intensity, is close to adding 3 decibels. Fibonacci sequence (F n ) n 0 0,,, 2, 3, 5, 8, 3, 2, 34, 55, 89, 44, 233,... is defined by F 0 =0,F =, F n = F n + F n 2 (n 2). htt://oeis.org/a Leonardo Pisano (Fibonacci) (70 250) 3 / / 80

9 Fibonacci sequence and Golden Ratio The develoments [], [, ], [,, ], [,,, ], [,,,, ], [,,,,, ],... are the quotients F 2 F F 3 F 2 F 4 F 3 F 5 F 4 F 6 F 5 F 7 F 6 k k k k k k of consecutive Fibonacci numbers. The develoment [,,,,,...] is the continued fraction exansion of the Golden Ratio = which satisfies F n+ = lim = n! F n =+ Continued fraction of 2 The number 2= satisfies 2=+ 2+ For instance 2=+ = We write the continued fraction exansion of 2 using the shorter notation 2=[, 2, 2, 2, 2, 2,...]=[, 2]. 33 / / 80 A4 format The number 2 is twice its inverse : 2=2/ 2. Folding a rectangular iece of aer with sides in roortion 2 yields a new rectangular iece of aer with sides in roortion 2 again. Irrationality of 2 :geometricroof Start with a rectangle have side length and + 2. Decomose it into two squares with sides and a smaller rectangle of sides + 2 2= 2 and. This second small rectangle has side lenghts in the roortion 2 =+ 2, which is the same as for the large one. Hence the second small rectangle can be slit into two squares and a third smaller rectangle, the sides of which are again in the same roortion. This rocess does not end. 35 / / 80

10 Rectangles with roortion + 2 Irrationality of 2 :geometricroof If we start with a rectangle having integer side lengths, then this rocess stos after finitely may stes (the side lengths of the successive rectangles roduces a decreasing sequence of ositive integers). Also, for a rectangle with side lengths in a rational roortion, this rocess stos after finitely may stes (reduce to a common denominator and scale). Hence + 2 is an irrational number, and 2 also. 37 / / 80 Geometric roof of irrationality t = 2+= =2+/t Set t = 2+= The continued fraction exansion of t is [2, 2,...]=[2]. Indeed, from 2 = 2+, we deduce t =2+ t Decomose an interval of length t into three intervals, two of length and one of length /t. 39 / / 80

11 Decomose the interval of length New homothety /t 2+ t = t, 2 t + t 2 =. Intervalle (0, /t) enlarged : 2 t 2 + t 3 = t : We go from the first icture to the second by a homothety /t, because t =2+/t. 4 / / 80 Geometric roof of irrationality Decomosition of a rational interval An interval of length t = 2+is decomosed into two intervals of length and one of length /t. After a homothety /t, the interval of length is decomosed into two intervals of length /t and one of length /t 2. The next ste is to decomose an interval of length /t into two intervals of length /t 2 and one of length /t 3. Next, an interval of length /t 2 roduces two intervals of length /t 3 and one of length /t 4. At each ste we get two large intervals and a small one. The rocess does not sto. Start with a rational number u = a/b, say a>b>0 with a and b integers. Decomose an interval of length u into a whole number of intervals of length lus an interval of length less than. It is convenient to scale : it amounts to the same to decomose an interval of length a into a whole number of intervals of length b, lus an interval of length less than b, say c, which is an integer 0. This is the Euclidean division, again. The rocess stos after finitely many stes. 43 / / 80

12 The Golden Ratio The Golden Ratio satisfies = = =+ The Golden Ratio ( + 5)/2 = Golden Rectangle If we start from a rectangle with the Golden ratio as roortion of length sides, at each ste we get a square and a smaller rectangle with the same roortion for the length sides. 45 / / 80 Leonard Euler ( ) Continued fraction exansion for e Leonhard Euler De fractionibus continuis dissertatio, Commentarii Acad. Sci. Petroolitanae, 9 (737), 744, ; Oera Omnia Ser. I vol. 4, Commentationes Analyticae, e= lim n! ( + /n) n = e = =[2,, 2,,, 4,,, 6,...] =[2,, 2m, ] m. e is neither rational (J-H. Lambert, 766) nor quadratic irrational (J-L. Lagrange, 770). 47 / / 80

13 Lagrange (736 83) Irrationality of Joseh-Louis Lagrange was an Italian-born French mathematician who excelled in all fields of analysis and number theory and analytic and celestial mechanics. Johann Heinrich Lambert ( ) Mémoire sur quelques roriétés remarquables des quantités transcendantes circulaires et logarithmiques, Mémoires de l Académie des Sciences de Berlin, 7 (76), ; lu en 767 ; Math. Werke, t. II. tan(v) is irrational when v 6= 0is rational. Hence is irrational, since tan( /4) =. 49 / / 80 Lambert and King Frédérick II Continued fraction of Quesavezvous, Lambert? Tout,Sire. Etdequile tenez vous? Demoi-même! The develoment of = starts with =[3, 7, 5,, 292,,,, 2,, 3,, 4, 2,,,...] Oen roblem : is the sequence of artial quotients bounded? 5 / / 80

14 Continued fraction exansion for e /a Continued fraction exansion of tan(x) Starting oint : y =tanh(x/a) satisfies the di erential equation ay 0 + y 2 =. This leads Euler to e /a =[,a,,, 3a,,, 5a,...] =[, (2m +)a, ] m 0. tan(x) = i tanh(ix), tanh(x) =ex e x e x +e x tan(x) = 3 5 x 7 x 2 x 2 x 2 9 x 2 x / / 80 Padé aroximants Henri Eugène Padé, Claude Brezinski History of Continued Fractions and Padé Aroximants. Sringer-Verlag, Berlin, 99, 55 ages. htt://math.univ-lille.fr/~brezinsk/ 55 / 80 Diohantine aroximation in the real life Calendars : bissextile years Sokes Small divisors and dynamical systems (H. Poincaré) Periods of Saturn orbits (Cassini divisions) Chaotic systems. Stability of the solar system. Exansion of the universe. General theory of relativity. Cosmology. Black holes. Resonance in astronomy Quasi-cristals Acoustic of concert halls 56 / 80

15 Number Theory in Science and communication M.R. Schroeder. Number theory in science and communication : with alications in crytograhy, hysics, digital information, comuting and self similarity Sringer series in information sciences th ed. (2006) 367. Electric networks The resistance of a network in series is the sum R + R 2. r AAA R AAA R 2 The resistance R of a network in arallel satisfies r AAA R AAA R 2 R = R + R 2 r r 57 / / 80 Electric networks and continued fractions Decomosition of a square in squares The resistance U of the circuit r AAA R H H /S H /T H H H is given by U = S + R + T r 59 / 80 For the network r AAA R 0 AAA R AAA R 2 H H H /S H H /S 2 H H H /S 3 H the resistance is given by a continued fraction exansion [R 0,S,R,S 2,R 2,...] Electric networks and continued fractions have been used to find the first solution to the roblem of decomosing an integer square into a disjoint union of integer squares, all of which are distinct. r 60 / 80

16 Squaring the square Quadratic numbers The continued fraction exansion of a real number is ultimately eriodic if and only if the number is a quadratic number, that means root of a degree 2 olynomial with rational coe cients. For a ositive integer d which is not a square, the continued fraction exansion of the number d is d =[a0,a,a 2,...,a k,a,a 2,...,a k,a,a 2,...], which we write for simlicity d =[a0, a,a 2,...,a k ]. Hence 2=[, 2] and 3=[,, 2]. 6 / / 80 Connexion with the equation x 2 dy 2 = ± Problem of Brahmaguta (628) Let d be a ositive integer which is not a square. Consider the Diohantine equation () x 2 dy 2 = ± where the unknowns x, y take their values in Z. If (x, y) is a solution with y, then x (x dy)(x + dy) =, hence is a rational aroximation y of d and this aroximation is sharer when x is larger. This is why a strategy for solving Pell s equation () is based on the continued fraction exansion of d. Brahmashutasiddhanta : Solve in integers the equation x 2 92y 2 = If (x, y) is a solution, then (x 92y)(x + 92y) =, hence x y is a good aroximation of 92 = / / 80

17 Problem of Brahmaguta (628) Bhaskara II (2th Century) The continued fraction exansion of 92 is 92 = [9,,, 2, 4, 2,,, 8]. A solution of is obtained from x 2 92y 2 = [9,,, 2, 4, 2,, ] = 5 20 Indeed = =. Lilavati (Bijaganita, 50) x 2 6y 2 =Solution : x = , y = Cyclic method (Chakravala) of Brahmaguta. 6 = [7,, 4, 3,, 2, 2,, 3, 4,, 4] [7,, 4, 3,, 2, 2,, 3, 4,, 4,, 4, 3,, 2, 2,, 3, 5] = / / 80 Narayana (4th Century) Narayana cows (Tom Johnson) x 2 03y 2 =. Solution : x =227528, y = = =. 03 = [0, 6,, 2,,, 9,,, 2,, 6, 20] [0, 6,, 2,,, 9,,, 2,, 6] = Corresondence from Fermat to Brouncker our ne vous donner as tro de eine (Fermat) to make it not too di cult X 2 DY 2 =, with D =6and D =09. Solutions resectively : ( , ) ( , ) = ! / / 80

18 AreferenceontheHistoryofNumbers Farey dissection John Farey ( ) geologist André Weil Number theory. : An aroach through history. From Hammurai to Legendre. Birkhäuser Boston, Inc., Boston, Mass., (984) 375. MR 85c:0004 htt://www-history.mcs.st-andrews.ac.uk/pictdislay/farey.html Patrice Philion. A Farey Tail. Notices of the AMS Volume 59, (6),202, / / 80 Riemannian varieties with negative curvature Ergodic theory The study of the so-called Pell-Fermat Diohantine equation yield the construction of Riemannian varieties with negative curvature : arithmetic varieties. Nicolas Bergeron (Paris VI) : Sur la toologie de certains esaces rovenant de constructions arithmétiques Torus of dimension : R/Z ' S Gauss ma T : x 7! x x Deterministic chaotic dynamical system. This transformation is ergodic : any subset E of [0, ) such that T (E) E has measure 0 or. 7 / / 80

19 Birkho ergodic Theorem Let T be an ergodic endomorhism of the robability sace X and let f : X! R be a real-valued measurable function. Then for almost every x in X, we have as n!. n nx Z f T j (x)! j= George David Birkho ( ) fdm Connection with the Riemann zeta function For s of real art >, We have also with (s) = X n = Y s n (s) = s s T (x) = x Z 0 s T (x)x s x dx 73 / / 80 Generalization of the continued fraction exansion in higher dimension Jacobi Perron Partial answers are known, like the Jacobi Perron algorithm Simultaneous rational aroximation to real numbers is much more di cult than the rational aroximation theory for a single number. The continued fraction exansion algorithms has many secific features, so far there is no extension of this algorithm in higher dimension with all such roerties. Carl Gustav Jacob Jacobi (804 85) Oskar Perron ( ) Die Lehre von den Kettenbrüchen, / / 80

20 Geometry of numbers The LLL algorithm The geometry of numbers studies convex bodies and integer vectors in n-dimensional sace. The geometry of numbers was initiated by Hermann Minkowski ( ). Given a basis of R n, the LLL algorithm roduces a basis of the lattice they generate, most often with smaller norm than the initial one. Arjen Lenstra Hendrik Lenstra Laszlo Lovasz 77 / / 80 Parametric geometry of numbers Mandalay University November 6, 206 The Ninth International Conference on Science and Mathematics Education in Develoing Countries Recent work by Wolfgang M. Schmidt and Leo Summerer, Damien Roy Continued fractions : introduction and alications Michel Waldschmidt Université Pierre et Marie Curie (Paris 6) France htt:// 79 / / 80