FUNCTIONS OF A COMPLEX VARIABLE BY JAMES PIERPONT, LL.D. PROFESSOR OF MATHEMATICS IN YALE UNIVERSITY GINN AND COMPANY BOSTON ~ NEW YORK * CHICAGO * LONDON ATLANTA * DALLAS * COLUMBUS ~ SAN FRANCISCO COPYRIGHT, 1914, BY JAMES PIERPONT ALL RIGHTS RESERVED PRINTED IN THE UNITED STATES OF AMERICA 933.1 The Itbennunm Aredm GINN AND COMPANY. PROPRIETORS BOSTON U.S.A. SEINEM IIOCIIVER-,,EHITEN FREUNDE HERRN PROFESSOR I)R. G. VON ESCHERICH IN HERZLICHER DANKBARKEIT ZUGEEIGNET VOM VERFASSER PREFACE The present volume has arisen from lectures on the Theory of Functions of a Complex Variable which the author has been accustomed to give to juniors, seniors, and graduate students of Yale University during the last twenty years. As these students often do not intend to specialize in mathematics, many topics which might properly find a place in a first course in the function theory have not been treated; for example, Riemann's surfaces. On the other hand the author, having in mind the needs of students of applied mathematics, has dwelt at some length on the theory of linear differential equations, especially as regards the functions of Legendre, Laplace, Bessel, and Lame. As a splendid application of the principles of the function theory and also on account of their intrinsic value, three chapters have been devoted to the elliptic functions. The author wishes to acknowledge in a general way his indebtedness to the works of C. Jordan, H. Weber, C. de la ValleePoussin, W. Jacobsthal, and others, but to L. Schlesinger his debt is especially great for the treatment of linear differential equations here given. The author wishes also to express his deep appreciation of the assistance so cheerfully given by his colleagues, Professor W. A. Wilson of Yale University and Professor E. L. Dodd of the University of Texas, and by his pupils P. R. Rider and G. H. Light. Last but not least the author fulfills a very pleasant duty in tendering his thanks to the house of Ginn and Company for the great care they have given to the make-up of the book and for the generous manner in which they have met his every wish. JAMES PIERPONT NEW HAVEN v CONTENTS CHAPTER I ARITHMETICAL OPERATIONS ARTICLES PAGE 1. Historical Sketch............... 1 2. Arithmetical Operations............. 3 3. Critical Remarks.................. 8 4. The Polar Form................ 11 5. Inequalities. Geometrical Correspondence......... 13 6. Moivre's Formula............... 17 7. Extraction of Roots................ 18 8. Casus Irreducibilis.......... 22 CHAPTER II REAL TERM SERIES 9. Introduction............ 25 10. Definitions.................. 25 11. Geometric Series................. 27 12. Harmonic Series.......... 28 13. Fundamental Postulate................ 29 14. Hyperharmonic Series................ 29 15. Alternating Series................ 30 16. The E Notation.................. 32 17. Necessary Conditions for Convergence......... 34 18. Adjoint Series................... 35 19. Remainder Series.............. 37 POSITIVi, TERilMi SI:IlES 20. Theorems of Comparison............. 39 21. D'Alembert's Test............... 42 22. Cauchy's Integral Test......... 44 23. Logarithmic Scale............. 46 24. Kummer's Test............. 48 25. Raabe's and Cahen's Tests............ 49 26. Gauss' Test................. 51 28. Binomial Series................ 52 29. Hypergeometric Series.............. 54 vii ,o, V111 CONTENTS CHAPTER III SERIES WITH COMPLEX TERMS ARTICLES 30. Introduction and Definitions.. 31. Absolute Convergence... 32. Addition and Subtraction... 33. Multiplication.......... 34. Cauchy's Paradox..... 35-36. Associative and Commutative Properties. 37. Riemann on Simply Convergent Series PAGE........ 56.............58....... 59....... 60.............62....... 64....... 69 POWER SERIES 38. Introduction.......... 39. Circle of Convergence.... 40. Two-Way Series.... 41. Double Series.... 42. Row and Column Series.... 43. Application to Power Series..... *....*..................* *...... *...... ~ ~ ~ ~ 0 6 ~ ~ ~ ~ ~ ~ 71 72 75 77 80 84 CHAPTER IV THE ELEMENTARY FUNCTIONS 44. Introduction.............. 45. Integral Rational Functions... 46. Rational Functions..... 47. Algebraic Functions..... 48. Explicit Algebraic Functions.... 49. Study of z......... 50. Study of /(z - a) (z - b).... 51. The Elliptic Radicals.... 52. Study of --.. 53. One- and Many-Valued Functions...... 86.... 87.... 88.... 89.... 90.... 91... 94.... 97.... 98.... 100 EXPONENTIAL FUNCTION 54. Addition Theorem.............. 102 55. Euler's Formula................ 106 56. Period of e.................. 108 57. Graphical Study of ez.......... 109 CONTENTS ix CIRCULAR FUNCTIONS ARTICLES 58. Addition Theorem..... 59. Zeros and Periodicity..... 60. Graphical Study of sin z... PAGE...... 111...... 116...... 117 OTHER FUNCTIONS 61. Hyperbolic Functions..... 62. Logarithmic Function... 63. The Function z...... 64. Inverse Circular Functions........ 118...... 120...... 123..... 125 CHAPTER V REAL VARIABLES 65. Introduction............ 66. Notion of a Function... 67-68. Limits............ 69. Continuity............ 70. Geometric Terms..... 71. Uniform Continuity.... 72. Differentiation........... 73. Law of Mean........ 129... 129... 131 ~... 134... 138 ~... 139... 140... 143 74. 75. 76. 77-79. 80. INTEGRATION Elementary Properties... Surface Integrals..... Curvilinear Integrals..... Illustrations from Physics... Stokes' Theorem....... 146... 149... 150... 153... 158 CHAPTER VI DIFFERENTIA TION AND INTEGRA TION 81. Resume................ 82. Definition of a Function of z.... 83. Limits, Continuity....... 84. Elementary Theorems in Differentiation.. 85. Differentiation of Power Series... 86. Cauchy-Riemann Equations....... 163... 164 *.. 167... 170... 172... 175 x CONTENTS ARTICLES PAGE 87. Derivatives of Elementary Functions..... 176 88. Inverse Functions...............178 89. Function of a Function..........181 90. Functions having a Derivative........... 183 INTEGRATION 91. Definition.................186 92. Properties of Integrals.............187 93. Fundamental Theorem............. 189 94. Examples.................. 190 95. The Indefinite Integral.............192 96. Change of Variable..............193 97. Integration by Parts...........195 98. Differentiation with Respect to a Parameter...... 196 FUNCTIONS DEFINED BY SERIES 99. Steady Convergence..............197 100. Continuity of Series............. 200 101. Termwise Integration............. 202 102. Calculation of r................ 205 103. Termwise Differentiation.......... 206 CHAPTER VII ANALYTIC FUNCTIONS 104. Definitions..................210 105. Cauchy's First Integral Theorem.......... 211 106. Cauchy's Second Integral Theorem.......214 107. Derivatives................. 215 108. Termwise Differentiation........... 216 109-112. Taylor's Development.............. 218 113-114. Analytic Continuation............. 224 115-116. Undetermined Coefficients........... 229 117. Laurent's Development............ 234 118. Zeros and Poles............... 237 119. Essentially Singular Points............ 244 120. Point at Infinity................ 247 121. Integral Rational Functions...........249 122. Rational Functions..............251 123. Transcendental Functions......... 253 124. Residues................ 256 125. Inversion of a Power Series.............. 259 126. Fourier's Development........... 262 CONTENTS xi CHAPTER VIII INFINITE PRODUCTS ARTICLES PAGE 127-128. Introduction, Definitions........... 266 129. Fundamental Theorem............. 269 130. Associate Logarithmic Series....... 271 131. Absolute and Steady Convergence........ 271 132-133. Examples............... 273 134. Normal Form............. 276 135. Arithmetical Operations............ 279 136-139. Circular Functions................ 281 140. Weierstrass' Factor Theorem........... 289 CHAPTER IX B AND r FUNCTIONS, ASYMPTOTIC EXPANSIONS 141. Introduction................. 295 142. B (x, y) expressed by T Functions......... 297 143-148. Properties of r(z)............... 299 149. I' expressed as a Loop Integral........... 305 150. B expressed as a Double Loop Integral....... 309 ASYMPTOTIC EXPANSIONS 151. Introduction...................... 309 152-154. Bernoullian Numbers and Polynomials........ 310 155. Euler's Summation Formula........... 318 1 1 1 156. Asymptotic Expansion of 1 + + + +......322 1 2 h, 157. Stirling Formula for n!........... 324 158. Asymptotic Expansion of P(x)........... 325 159. Asymptotic Series.............. 328 CHAPTER X FUNCTIONS OF VTEIERSTRASS 160. Limiting Points................. 333 161-162. Periodicity................. 334 163. Various Periodic Functions.......... 339 164. The Elliptic Functions........... 342 165. General Properties of Elliptic Functions....... 344 xii CONTENTS ARTICLES PAGE 166-168. Elliptic Functions expressed by a, p, and...... 354 169. Development of ao, 5, p in Power Series........ 363 170. Addition Formulke........ 366 171. The o,, g2 g................ 369 172. The Co-sigmas................ 370 173-174. The Inverse p Function............ 373 175. The p Function defined by g2, g.........379 176. The Radicals p(-l.............. 381 CHAPTER XI FUNCTIONS OF LEGENDRE AND JACOBI 177-178. The Elliptic Integrals............. 383 179. Linear Transformations.......... 388 180. Legendre's Normal Forms.......... 393 181. Real Linear Transformations........ 396 182. Quadratic Transformations......... 398 183-184. Rectification of the Hyperbola and Lemniscate.... 399 185. Integrals of the First Species......... 401 186. Inversion.................. 409 187. The sn, cn, dn Functions.......... 413 188. Addition Formlulae.............. 417 189. Differential Equation for K and K'........ 421 CHAPTER XII THE TA FUNCTIONS 190. Historical................. 423 191. The i's as Infinite Products......... 426 192. The,'s as Infinite Series............ 428 193. Zeros of the 6's............... 429 194. The i's with Zero Argument.......... 431 195. Definition of sn, en, dn as a Quotients...... 433 196. Numerical Calculation.......... 437 197-199. The ~ and Z Functions............ 439 200-201. Integrals of Second and Third Species....... 444 202. Relation between sn and p.......... 447 203. The el, e2, e3................. 449 204. Relation between ao's and a's........... 450 CONTENTS CHAPTER XIII xiii ARTICLES 205. 206. 207-208. 209. 210. 211. 212. 213. 214. 215. 216. 217. 218. 219. 220. LINEAR DIFFERENTIAL EQUATIONS Introduction........ Existence Theorem..... Fundamental Systems..... Simple Singular Points.... Hypergeometric Equation... Bessel's Equation..... Logarithmic Case..... Method of Frobenius........ Logarithmic Case of Hypergeometric Equation. Logarithmic Case of Bessel's Equation.. Differential Equation for K, K'... Regular Points..... Equations of the Fuchsian Class... F(a, /, y, x) as an Integral.... Loop Integrals...... PAGE.....453.....455.....459.....462.....466.... 469.....470.... 474.... 476.....480.... 483.....484.... 486.... 488.....489 CHAPTER XIV FUNCTIONS OF LEGENDRE AND LAPLACE 221. The Potential.... 222. Legendre's Coefficients... 223. Development of P,. in Multiple Angles. 224. Differential Equation for P..... 225. Integral Properties of P..... 226. Rodrigue's Formula.... 227. Development of f(x) in Terms of the P, 228. Recurrent Relations.... 229-231. Legendre's Functions of the Second Kind, Q,...... 493..... 495.....497..... 498..... 499..... 502...... 505........0505..... 506 LAPLACE'S F UNCTIONS 232. Laplace's Equation u =0............508 233. Theorems of Gauss and Green........... 511 234. Potential expressed in Terms of Boundary Values.... 513 235. Outline of Solution of Au = 0.......... 516 236. Case of Axial Symmetry............ 518 237. Spherical Harmonics............. 521 238.- Relations between Ym and the P...........524 xiv CONTENTS ARTICLES PAGE 239. Development off(O, 4) in Terms of the Ym. 525 240. Fundamental System of Harmonics of Order n. 525 241. Integral Relations between Pn, k and Pn.... 528 242. Development of f(0, 4) in Terms of the Pn,....... 530 243. Expression of Pn (cos w) in Terms of Pn, P,,..... 532 244. Solution of Dirichlet's Problem for the Sphere....... 532 CHAPTER XV BESSEL AND LAMEl FUNCTIONS 245. Integrals of Bessel's Equation...... 5.33 246. Relations between the J,, and J,.......... 534 247. Integral Relations............... 536 248. The Roots of J,,(x).............. 538 249. Bessel's Functions as Loop Integrals......... 541 250. Other Loop Integrals for x>0........... 545 251. Relations between u. and us, up.......... 547 252. Asymptotic Solutions............. 548 253. Asymptotic Development of J,............ 552 254. An Expansion in Series of Bessel Functions.. 553 255. An Integral Expression of Jn............. 554 256. Solution of Kepler's Equation............. 556 257. Development of f(x) in Terms of the J,,......... 556 258. Development of f(?, 4<) in Terns of the n.... 557 259. Solution of A () = 0 for the Cylinder......... 558 LAME FUNCTIONS 260. Confocal Quadrics.............. 561 261. Elliptic Co6rdinates................ 564 262. Transformation of A(u) = 0......567 263. Lam6 Equation............... 568 264. Lame Functions............... 570 265. Integral Properties of Lame Functions......... 575 266. Solution of AV= 0 for an Ellipsoid.......... 579 Index............ 581 List of Symbols.................... 583 NOTE As a book of this kind is meant to appeal to a variety of readers, any one who wishes to study only the more elementary parts of the Theory of Functions may omit with advantage the following sections: 8, 23, 34, 37, 42, 43, 71, 77, 78, 79, 108, 125, 151-159, 189, 190; all of Chapters XIII, XIV, XV, if not interested in Differential Equations, otherwise sections 212-220, 249-253, 260-265 inclusive. FUNCTIONS OF A COMPLEX VARIABLE CHAPTER I ARITHMETICAL OPERATIONS 1. Historical Sketch. In elementary mathematics we use for the most part only real numbers. There is however a branch of elementary mathematics, viz. algebra, where a wider class of numbers,. the complex numbers, are employed almost from the start. The quadratic equation 22 ax + 6 0 (1 has the two roots a 2 a + Va2 - b, (2 provided a2 > b. If a2 < 6, there is no real number which satisfies 1). As long as we restrict ourselves to the system of real numbers, the expression 2) is devoid of meaning. In fact the square root of a number c, in symbols V/e, is a number d such that d2 = c. But there is no real number d whose square is negative. Thus Va2 — b does not exist in the real number system when a2 < 6. The older algebraists found it extremely convenient to enlarge their number system, in order that the equation 1) should have two roots even when a2 < b. The new numbers are denoted by a+ bV-1, or setting i=V/-1, by a+ bi. When = 0, they reduce to the real numbers a. Thus the new class contains the class of real numbers as a subclass. With these new numbers it was found that not only the roots of the quadratic, but also of the cubic, the biquadratic, in short the roots of all algebraic equations could be expressed. By their introduction, the theory of algebraic equations attained a simplicity and comprehensiveness quite impossible without them. Complex numbers are to-day indispensable in algebra. 1 2 FUNCTIONS OF A COMPLEX VARIABLE In other branches of mathematics, the importance of complex numbers was perceived much later. By means of a formula discovered by Euler ei = cos ~ + i in, ( an intimate relation was established between exponentials and analytic trigonometry. Indeed, a good part of this subject may be developed from this formula. In the system of real numbers, the logarithm of a negative number -a does not exist. In the system of complex numbers it does. We have, in fact, log (- a) =log a + (2 n + 1)Tri, (4 where n is any integer, including zero. It is thus infinite valued like the inverse circular functions. A great discovery made nearly a century ago by Abel rendered the complex numbers as necessary to analysis as they long had been in algebra. He found that the elliptic functions, whose properties had been carefully studied by Legendre, admit a second period, when one passes from the real to the complex number system. Possessed of this fact, Abel and his contemporary Jacobi were able to develop the theory of elliptic functions in a manner undreamed of before. About the same time the illustrious French mathematician Cauchy began to show what great advances could be made in the theory of differential equations when the variables are allowed to take on complex values instead of being restricted to real values alone. By the year 1850 complex numbers had proved to be of incalculable value in many and widely separated branches of mathematics, and before long the theory of functions of a complex variable sprang into existence. To-day this theory has grown to gigantic size. It forms the foundation on which much of modern mathematics is built. Without a knowledge of its elements, a student of mathematics finds himself somewhat in the position of a traveler in a strange land; every one is using a language which he does not comprehend. Even the physicist and astronomer find that the masters in these subjects are using freely the function theory. It is thus becoming daily more important for them to gain some familiarity with this theory. ARITHMETICAL OPERATIONS 3 The present work is intended to give the general reader an account of some of the elementary parts of the theory of functions of a complex variable. For further study we add the following list. The easier books are placed first. The last two are intended for the specialist. E. B. Wilson, Advanced Calculus. Ginn and Company, Boston, 1912. G. Humbert, Cours d'Analyse. 2 vols., Paris, 1904. E. Goursat, Cours d'Analyse Mathematique. 3 vols., 2d edition, Paris, 1910: H. Burkhardt, Einfuhrung in die Theorie der Arialytischen Funktionen. Leipzig, 1903. R. Fricke, Analytisch-Funktionen-theoretische Vorlesungen. Leipzig, 1900. Durege-Maurer, Theorie der Funktionen einer komplexen veranderlichen Gr6sse. Leipzig, 1906. E. Picard, Traite d'Analyse. 3 vols., 2d edition, Paris, 1905. E. Whittaker, A Course of Modern Analysis. Cambridge, 1902. A. Forsyth, Theory of Functions of a Complex Variable. 2d edition, Cambridge, 1900. W. Osgood. Lehrbuch der Funktionentheorie. 2d edition, Leipzig, 1912. 2. Arithmetical Operations. 1. As the reader has already studied the arithmetical operations on complex numbers, we treat this topic but briefly. The complex numbers are represented by the symbol a + a'i, where a, a' are real numbers and i is a symbol to be defined later. The plus sign between its two parts has, of 4 FUNCTIONS OF A COMPLEX VARIABLE course, no meaning as yet. It is convenient to denote a+ ati by a symbol as a. We have then a = a + a. (1 With 1) we will associate a point A whose abscissa and ordinate are respectively a and a' as in Fig. 1. Conversely to a point whose abscissa is x and whose ordinate is y we will associate a complex number x+yi. From this correspondence we are led to.. a A call a the abscissa of the complex a, and a' 'a' its ordinate. We write I x a= Aba, a' = Ord a. When a' = 0 in 1) we assign to a the value FIG. 1. a, and denote a+ 0 i more shortly by a. The associated points lie on the x-axis which we call the real axis. When a= 0, we say a is purely imaginary; we denote 0 +ai more shortly by a'i. The associated points lie on the y-axis, which we call the imaginary axis. 2. Two complex numbers a=a+a'i, =b + bti are equal, when a =b at= b (2 or in symbols Abs a= Abs, Ord a = Ord/3; (3 that is, when the associated points are coincident. In particular a = a + ai = O only when a= a'_ 0. The sum of a and / is a +/3= (a + b) + (a'+ b') i; (4 their difference is a-/3 = (a- b) +(a'-b)i. (5 Let us show how the points corresponding to 4) and 5) may be found. ARITHMETICAL OPERATIONS 5 Let A, B in Fig. 2 be the points associated with a, 8. Let us construct the parallelogram whose two sides are OA, OB. Then OA' =a AA = a', c OB' = b BBt = b'. Obviously OC = OA' + A' C' = a + b, COC = C' C" + C" C= a' + b'. Thus C has the coordinates a + b, A' a' + b and hence is the point assoFiG. 2. ciated with a + 3 in 4). To find the point associated with a - 3 in 5), let A, B in Fig. 3 be the points associated with a, /. We produce OB backward so that OC= OB. Obviously C has the coordinates - b, - '. Thus if we set B = - b - bi, \ A we see that a - 3= a + y, 0 since both are (a -)(a' - ) b')i. \ But we have seen how to plot the c point corresponding to a+ y. It is FIG. 3. in fact the vertex D in the parallelogram two of whose sides are OA, OC. 3. We note that in adding and subtracting a, 3 ill 4), 5) we have treated i as if it were a real number. We do the same in defining multiplication, except that we agree that i2=-1 (6 Thus ~Thus o~a/, = ab + a6bi + a' 6i ~- a' bi2, or using 6) a/3 = (ab - aib') + (ab' + a'b)i. (7 We take 7) as the definition of multiplication. From 6), 7) wevehave i i =4 1 (8 and hence i4n+m= im, (9 where m, n are positive integers or zero. 6 FUNCTIONS OF A COMPLEX VARIABLE 4. To define the quotient of two complex numbers we recall that in real numbers bx-= a has one and only one solution when b 0, this we call the quotient of a by b and denote by -. Let us consider the analogous equation 3 t= oa (10 where a, 3 are complex numbers a = a + ai, 3=b i. We shall define the quotient of a by 38 to be the number or numbers f which satisfy this relation. Let us first suppose I3 = O. Then b, b' are not both = 0. Let X =+- x'i be a solution of 10). Putting this in 10) gives (bx - lbx') + i(bx' + bx) = a + a'i (11 This relation yields by virtue of 3) two equations to determine x, l, viz.': bx-bx' = a, bx' + b= a'. (12 From these we get ab + a'b a -ab b2 + b 2 ' b2 + b2 Thus the solution of 10) is a ab + a ab - ah'- (13 3 bz b+ b'2 2 2~ b'12 whenS m 0. Suppose 3 = 0. Then b=b'=0, and 12) requires that a=d =0, or a = 0. Putting these values in 12) we see that the equations are satisfied however x, x' are chosen, that is, for every value of I. We have thus the following result: When 1 4: O, the equation 10) admits one and only one solution 4 which is given by 13). When 3 = 0, the equation admits no solution unless a = 0. In this case it is satisfiedfor every value of ~. For this reason division by 0 is excluded in modern mathematics. As some students have not been trained in accordance with this law, we wish to emphasize its inviolate character. ARITHMETICAL OPERATIONS 7 5. We prove now the important theorem:,If a product a/3 = 0, then either a or / must = 0. This theorem we know is true for real numbers. On effecting the multiplication of a by 3 we get the relation (ab - ab') + (ab' + a'b)i = 0. Hence by 3), a0 + =. (14 ab-a ba=- O, abU + alb = O. (14 We show that these equations cannot hold if both a and 3 = 0. For since a 0 either a or at 0 0; suppose a 0. Also since,3 0, either b or b't 0; suppose b * 0. From the first relation of 14) we get ab This put in the second equation of 14) gives ab'2 + ab2 = 0, or a(b2 + b'2) = 0 As a = 0, we must have b2+ b12 = 0. This requires that both b and b = 0, and this is contrary to hypothesis. In a precisely similar manner we may treat the other cases. In each case we are led to a contradiction. Thus the assumption that a. / may = 0 without either a or 3 being = 0 is untenable. In an elementary work like the present, it would be out of place to demonstrate that the formal laws governing the arithmetical operations on real numbers go over without change to complex numbers. Thus the reader knows that a3 = 3a, that 3+ ) + that ~cc~a(,8 + 7) = a,8- +cy, etc., just as if a, 1, 7 were real numbers. 6. Two complex numbers a=a+ ib, 3= a- ib as in the figure are called conjugate numbers. We note that a+/,=2a 8 FUNCTIONS OF A COMPLEX VARIABLE is real, and that a- = 2 ib a - / = 2 ib is purely imaginary. ib Also the product 0 (,3 = a2 + b2 is real and positive. ia-ib 3. Critical Remarks. 1. We have now defined the four rational operations on complex numbers, viz. the operations of addition, subtraction, multiplication, and division. At this point we may return to the symbol (1 a +i a which we have employed to represent a complex number. We can show what its component parts mean. We begin by considering the two numbers a, y The rule for multiplication, 2, 7) gives /3. y = a'i. Thus the term a'i in 1) may be regarded as a' times the number i. The number i may be called the imaginary unit in contradistinction to the real unit 1. Again let /3=a, y=a'i. The rule for addition, 2, 4) gives / + ry = a + a'i. Thus the complex number 1) may be regarded as the sum of a real units and a' imaginary units. We may call a the real part and a'i the imaginary part of 1). When we introduced the symbol 1) in 2, we did not say that it was the sum of these two parts for the reason that we had not defined addition and multiplication of these new numbers. Indeed we expressly stated that the symbols + and i had at that stage no meaning. They acquired meaning only after the arithmetical operations on the symbols 1) had been defined. 2. Instead of denoting the new numbers by the symbol 1), let us denote them by some other symbol, say by (a, a'). (2 ARITHMETICAL OPERATIONS 9 We may proceed then as follows: Two numbers a=(a, a'), 3=(b b') are equal when and only when a= b, a'= b. The sum of a and / is defined to be the number (a +, a' + '); (3 their difference is defined as the number (a - b, a' -'). Their product shall be (ab - a'b', ab' + atb). (4 When 13 = 0 the quotient of a by / shall be the number fab + a'b a'b - ab b2 +b2 ' b2 b12 The number (a, O) shall be the real number a, and for brevity we will write ( ) (a, 0) = a. The number (0, 1) we will denote more briefly by i. Then 4) gives (a', 0). (0, 1) = (0 a')= a'i. (5 Also 1) and 5) give (a, a')=(a, 0)+(0, a') = a + a'i. Thus another representation of the complex number (a, a') is a + a'i, and we have reached the standpoint taken in 2. In both cases we start with a symbol; in one case with a + a'i, in the other with (a, a'). These at the start are mere marks to indicate that the new numbers are a complex of two real numbers a, a'. These marks take on a meaning when we give them the above arithmetical properties. The complex then becomes a definite concept which we call a number. 10 FUNCTIONS OF A COMPLEX VARIABLE 8. We wish now to make another remark of a critical nature. The complex numbers are often called imaginary numbers, and we have in the present work followed usage as far as to call the numbers a'i purely imaginary, the number i the imaginary unit, and the axis of ordinates the imaginary axis. For the beginner the term imaginary is most unfortunate; and if it had not become so ingrained in elementary algebra, much would be gained if it could be dropped and forgotten. The use of the term imaginary in connection with the number concept is very old. At first only positive integers were regarded as true numbers. To the early Greek mathematician the ratio of two integers as -- was not a number. After rational numbers had been accepted, what are now called negative numbers forced themselves on the attention of mathematicians. As their usefulness grew apparent they were called fictitious or imaginary numbers. To many an algebraist of the early Renaissance it was a great mystery how the product of two such numbers as - a, - b could be the real number ab. Hardly had the negative numbers become a necessary element to the analyst when the complex numbers pressed for admittance into the number concept. These in turn were called imaginary, and history repeated itself. How many a boy to-day has been bothered to understand how the product of two imaginary numbers ai and bi can be the real number - ab. As well ask why in chess the knight can spring over a piece and why the queen cannot. The pawns, the knights, the bishops, etc. are mere pieces of wood till the laws governing their moves are laid down. They then become chessmen. The symbols (a, at) or a + a'i are mere marks until their laws of combination are defined, they then become as much a number as 2 or -5. The student must realize that all integers, fractions, 3 and negative numbers are imaginary. They exist only in our imagination. Five horses, three quarters of a dollar, may have an objective existence, but the numbers 5 and 3 are imaginary. Thus all numbers are equally real and equally imaginary. Historically we can see how the term imaginary still clings to the complex numbers; pedagogically we must deplore using a term which can only create confusion in the mind of the beginner. ARITHMETICAL OPERATIONS 11 4, The Polar Form. 1. Let A be the point associated with a = a + ai (1 Let OA = p and angle A OB =. We call p the modulus or absolute value of a. We have p = /a2 + a12 the radical having the positive sign. The modulus of a is also denoted by It is the distance of A from the origin. The angle 0 is called the argument of a; we have at tan6= -, a =# 0. a We often write 0 = Arg a. Making use of p and 0 we have a=pcos0 a = p sin and a = p(cos 0 + i sin 0). (2 This is called the polar form of a. The form 1) may be called in contradistinction the rectangularform. 2. The rule for multiplication and division of complex numbers is particularly elegant when the polar form is used. In fact let 3 = b + b'i = o(cos f + i sin qb). (3 Then a/3 = pa I cos 0 cos q - sin 0 sin q + i(cos 0 sin q + sin 0 cos q) )1, or ac = par{cos(0 + b)+ i sin (0 + -) I. (4 From this we have (5 I - = P1 = I I. i l. (5 Arg (a/3)= 0 + b = Arg a + Arg/3. (6 This may be expressed as follows:The modulus of the product is the product of the moduli, and the argument of the product is the sum qf the arguments. 12 FUNCTIONS OF A COMPLEX VARIABLE The above enables us to plot the product ap\ a/3 very easily. Having plotted a and /, we compute the product pa and describe a \ I circle about the origin 0 with the radius. We then lay off on this circle the angle t 0+ 8+"4. The resulting point is that asso- \ ~ ciated with the number a/9, as in the figure.. 3. Let us now turn to division. The quotient 8=, /,Ao is defined by 2, 13). This expression is complicated and not easy to remember. If, however, we use the polar forms 2), 3) of a, /, we can readily prove that =P Icos (0 - )+ i sin (0-; (7 P aor that a= = (8 13 a /( and Arg = Arg a- Arg/3. (9 This may be expressed as follows: The modulus of the quotient is the quotient of the moduli, and the argument of the quotient is the difference of the arguments. The proof of 7) may be effected by replacing a, a', b, b' in 2, 13) by their values a = p cos 0, a = p sin, b = a cos q, b' = asin, and performing the necessary reduction. A more instructive way is the following: Since /3 0 by hypothesis, the equation /3=cc (10 admits one and only one solution. The relation 7 states that this solution is a-=Pscos ( - 0)+i sin ( 0)) (11 ARITHMETICAL OPERATIONS 13 This is indeed so, for by the foregoing rule of multiplication Arg (/3) = k (0 - 0)= 0Arg a. Thus is having the same modulus and argument as a is in fact a. Thus 11) satisfies 10). The above enables us to plot the quotient a//l very easily. Having plotted a and i, we compute the quotient p/a and describe a circle about the origin with this radius. We then lay off on this circle the angle 0 -. The resulting point is that associated with a/l8. 4. We have seen that two complex numbers a, f3 are equal when and only when their associated points coincide. Let us suppose a, 83 are expressed in their polar forms 2), 3). Then from the relationa a=/ we may conclude at once that p=a1. (12 We cannot conclude, however, that 0= b. We can only conclude that 0 can differ from q by a multiple of r, or that O= +2n7 (13 0 = qb + 2 nmr (13 where n is an integer or 0. In particular we have: For a to = O, it is necessary and sufficient that its modulus = 0. 5. Some Inequalities. Geometrical Correspondence. 1. Let us plot the points A, B, C corresponding to a, fi, a+/ as in Fig. 1. Then a+ l=-OC, a very useful relation. Since from geometry we have OC < OA+AC 14 FUNCTIONS OF A COMPLEX VARIABLE we conclude a relation of utmost imu- portance B la~131~ a~~I3L-PI (1 The < sign holds unless 0, A, B are collinear, and A, B are on the A same side of 0. From 1) we have 0 + )3 + ry + 1 8 + I ry (2 FIG. 1. and so on for any finite number of terms. To prove 2) we note that a+ + y r=a+(y 3 y). Hence 8 + y + ()+ + ~y)1< ~ C<~ ~ +yI, by 1) <Ja +K 3 + Yj, also by 1). 2. Let us now consider a - ) i. Let A, B in Fig. 2 be the points associated with a, 3. Then, following the construction given in 2, the point D is associated with a - p3 Thus 0-D -/3 'A oo = I a:- P i1t~I But obviously AB = OD, hence D cea - p8 AB. FIG. 2. This gives us a result we shall often use: If A, B are the points associated with ae, 3, then the length of AB is I a- /3i. 3. Let us also note the relation la-L ~ al+ 1 11. (3 For a -13=+(-/3). But Thus Ia- I=iaI+(- )l< a I+ -/1= I a I+ I 1. ARITHMETICAL OPERATIONS 15 4. Let a be a real positive number and / any complex number. We wish to find the points Z associated with the complex numbers ' which are subjected to the relation 1-31 <a.. (4 If B in Fig. 3 is associated with /3, the condition 4) says that ZB must be < a. Thus Z is restricted to the interior of a circle whose center is B and whose radius is a. The condition / I -/3|=a ( I J states that Z must lie on the circumference of this circle. The condition -~ I -C 1 <a FIG. 3. requires that Z lie within or on the circumference of the circle. In the following pages we shall make great use of the geometrical interpretation of complex numbers by points in a plane. The point associated with a complex number a may be called its image. Moreover, instead of using another letter as A to denote this point, we shall usually denote it by the same letter a. This will not produce any ambiguity and is shorter. We shall introduce another change. Up to the present we have usually denoted real numbers, angles excepted, by Roman letters as a, b, c o.. and complex numbers by Greek letters a, /3, r *.. in contradistinction. There is no further need of this; any letter may denote a complex number. It may be well to recall that real numbers are merely a special case of complex numbers, just as integers are a special case of rational numbers. Thus when we say let a be a complex number we do not at all mean that it may not be real. As in algebra and the calculus, so in the theory of functions we deal with constants and variables. The former are usually denoted by the first letters of the alphabet, the latter by the last. Let us consider a few examples of a variable: Example 1. Let z = a + yi, where a is a real constant and y is real and ranges from - o to + O. Then the point associated with 16 FUNCTIONS OF A COMPLEX VARIABLE z in Fig. 4 ranges over the right line AA'; or more shortly we say z ranges over AA'. A Example 2. Let z = x + yi, where x, y are real numbers satisfying the inequality.z= a+iy x2 y2 a" + __a< 1 a2 b2- a a, b real. Then z ranges over the interior and the edge of the ellipse x y2 1 A' a2 62 FIG. 4. Example 3. Let z - a = r(cos 0 + i sin 0) where a is any constant, r > 0, and 0 < 0 < 2 vr. Then z ranges over a circle whose center is a and whose radius is r. Example 4. Let z = x + yi, where x, y are real and satisfy the relation y=f(x). (5 Then z ranges over the curve whose equation is 5). 5. Let us note the following relations: a -bh ]= ib-al (6 Also if Alsoif la-bl<e and \lb-c <7 then i + (7 [a- c\<~+Vr (7 To prove 7) we observe that a - c = (a -b) + ( -c) Hence by 3) ac-ej< a-b +b-cl < [e + D 6. Let us recall from algebra the notation n am = al + a2 +... +- a, (8 m=l an, = a a2. a.a3... a. (9 m=l= ARITHMETICAL OPERATIONS 17 Thus the left side of 8) is read:the sum from m 1 to n of a,, is, etc. The left side of 9) is read the product f rom m = 1 to n of a.. is, etc. 6. Moivre's Formula. In 4 we sawv that any complex number a can be written a=pcs0+?sn0.( Then by 4, 4) a2 = p2(cos 2 0 ~ i sin 2 0), and in general n~ = pn(cos n0~+isin nO). (2 Let us takeI a= p = 1. Then 1), 2) give (cos 0~+isin0O)n= cos n0+i sin nO (3 which is ]Joivre's Formula. Now, the Binonmial Ttieorem for a positive integral exponent n is (a + 6)n an + ()an-lb ~ (n)an-262 +... + (jba-i + bn~. (4 Here are the binomial coefficients. Let tis make use of the relation 4) to develop the left side of 3). It becomes cosn 0 + i nCosn — 0 Sinl 0 - C()oSn2 0 sin-2 0 -incoSna3 0 sin3 0 +. (5 Thus 3) may be written Cos nO + i sin nO = co-sn 09- (~coSn2 0 sin.2 0 + Equating the real and imagin ary parts of this relation by 2, 2) we get Cos nO = cos', 0 -- nCOS n2 0 sin2 0 + G)coSn4 0 sin14 0 -..( sin nO (= n C5~n- 0 Sinl 0 - nCOSn-3 0 sin13 0 + n csn-50 sin.5 0 -.. (7 18 FUNCTIONS OF A COMPLEX VARIABLE Giving n the values 2, 3, 4... in these equations, we find cos 2 0 = cos2 0 - sin2 0 cos 3 9 = cos3 - 3 cos 0 sin2 O (8 cos 4 0 = cos4 0 -6 cos2 0 sin2 0 + sin4 0 sin 2 0 = 2 cos 0 sin 0 sin 3 0 = 3 cos2 0 sin - sin3 0 (9 sin 4 0 = 4 cos3 0 sin - 4 cos 0 sin3 0 In the relation 8) we notice that sin 0 occurs only in even powers. Since sin2 0 = 1 - os2 0, we see that: cos nO can be expressed as a rational integral function of cos 0 of degree n. Making this substitution, we get cos 2 0= 2 cos2 0-1 cos 3 0 = 4 cos3 0 - 3 cos (10 cos 4 0= 8 cos4 0-8 cos2 + 1 In equations 9) we notice that cos 0 enters in even powers when n is odd, and in odd powers when n is even. Thus we see that: sin n is an integral rational function of sin 0 of degree n when n is odd. When n is even, it is the product of cos 0 and a rational integral function of sin 0 of degree n - 1. Making the substitution cos2 = 1- sin2 0 in 9) we get sin 2 0= 2 cos 0 sin 0 sin 3 0 = 3 sin 0 - 4 sin3 0 (11 sin 4 0 = cos 0(4 sin 0 - 8 sin3 0) 7. Extraction of Roots. 1. In the domain of real numbers xn = a a > 0 has one root if n is an odd positive integer, and two roots if n is even. The equation x-n=-a, a>0 ARITHMETICAL OPERATIONS 19 has one root if n is odd, and no root if n is even. Let us now pass to the domain of complex numbers. We ask how many roots has n = a (1 where a is any complex number, and n is a positive integer. Let us write a and z in polar form a = o(cos 0 + i sin 0) (2 z = ((cos + + i sin f). (3 Then if 3) satisfies 1) we must have C'(cos no + i sin n)) = c(cos 0 + i sin ). (4 Then from 4), 12), 13) we have n = a (5 and ~and no = + 2 k7r, k an integer or 0. (6 From 5) we have. n- (7 and from 6) we have 0 2o =- + IC - (8 n n Thus the modulus ~ and the argument q of any number z which satisfies 1) must have the form 7), 8). On the other hand by actual multiplication we see at once that Zk {cos(+k )+ isin +k ) (9 n n n XJ is a solution of 1). In fact ZkA is a number whose modulus is the nth power of {/a and whose argument is n times - + k2. But n n this number has therefore the modulus a and the argument 0+2k7r, or neglecting 1 multiples of 2 7r, the argument 0. It is 2 thus a. Hence the nth power of 9) is a. \ To plot the numbers 9) we describe about o the origin as in the figure a circle whose 3 \ radius is Voa. On this circle we lay off the angle - Let us call this point z0. Start- 4 20 FUNCTIONS OF A COMPLEX VARIABLE ing from zo we now divide the circle into n equal parts which give the points z1, Z2, *.. z,1 in the figure, corresponding to k = 1, 2, n n-t in 9). If now we give to i any other integral value, we will get one of the values z0, z1,... z,-1 already obtained. Thus 1) has just n roots whose values are obtained by giving k in 9) the values 0,1, 2, *..,n - i. Example 1. = 1. 2 7r1 )~ Here n = 3, a = 1, a =1, =0,- = =1200. n Thus 9) gives Xg = 1' z1= cos 120' + i sin 120- - 1~iV+ 2 z = cos 240' + i sin 240= -IV3 2 Example 2. Z = - 8. Here n=3, a=-8, a=8, 0=7r, 600. n Thus 9) gives zo =2 ~cos 60~ + i sin 600 1 + i V3, z 2 eos (600 + 1200) + i sil (600 + 1200) = - 2, Z2 2 cos (60O + 240')~ i'sin (600 + 2400) 1 - iV3. Example 3. Z4 = 1. Here n=4, a=l, 0-0, 2 =90 o n mvl- ONh - - xJIuo U.) glvuo 0o = 1' zi = cos 900 + i sin 900 = i, Z2 = Cos 1800 + i sin 1800 = - 1, Z3 = cos 2700 + i sin 2700 = - i. 2. The n roots of (10 ARITHMETICAL OPERATIONS 21 are called unit roots. They are of great importance in algebra, and occur in other branches of mathematics. Their values are given by 9) on settinga = 1, = 0. We get o00 = 1, 27r.. 27r 1 = COS + sin-, n n 02 = cos 2 27+ i sin2 2, (11 n n *... * *..., 7, Wl-1 = cos (n - 1) + i sin(n- 1)27r. n n We notice that Wo has the property that 02 — ' - n —1 602 = Ct12 - 603 = ()13 1... Wn- = l-1 X co -= 601n that is all the roots of 10) are merely powers of wl. Such a root is called a primitive unit root. It is easy to show that: If m is relatively prime to n, then Co = -m is a primitive root; that is, that4 n (0, 02, to03, (04... 0n (12 are all roots of 10) and are all different. For- 27r 27r F~or (08 = e8= (cos m - + i sin m = co s ~ *- + i sin ms. 2. n n Let now ms = ln + p. Then 27.. 2 r C0' = cos p * -- + i sinp - = n n Thus d5 is a root of 10). To show that the roots 12) are all different let us suppose that r = s 2 =r 27r) Then their arguments rm- 27, sm2 —7 can differ only by a multiple n n of 2 r. Hence, e denoting an integer, 27r 27r rm - - sm — =e 2 7r n n or m(r - s) = en. (13 22 FUNCTIONS OF A COMPLEX VARIABLE As m and n are relatively prime, they have no factor in common; e must be divisible by m as 13) shows. Thus if we set e = gm 13) gives r - s- = gn or r-s is a multiple of n. This is impossible, as r, s are both <n. Hence no two of the roots 12) are equal. 3. Let us now return to 9). We set 2r.. 27r o= cos-+ i sin, n n and notice that Z1 =o z, 2= %0,... n- = o lzo (14 wlhere 0 0 zo= q/a cos - + i sin- ). (15 9n n This may be easily generalized as follows: All the roots of 1) may be obtained from any root by multiplying this root by the n roots of unity. Hence in particular the two roots of z2 = a = a (cos 0 + i sin 0) are are = OV (cos_+ isin ), (16 Z1=-Z-. (17 The three roots of z3 = a are 3/- 0..0\ zo = V/y cos -+ isin, (18 Zl = WZO, Z2 = -o 2Z (19 where o is the first imaginary cube root of unity, viz.: -l + i~/2 2-~V (20 2 8. The Casus Irreducibilis. As an application of the foregoing let us consider the irreducible case of Cardan's solution of the cubic ^ x~ - px + q = u, p. q real. (1 ARITHMETICAL OPERATIONS 23 The roots of 1) have the well-known form X= - -2 q + VR/+ - - q-V where R = - q2- 123 =- A. (3 Now when the roots of 1) are all real, it is shown in algebra that A is positive, hence R is negative, and VR is purely imaginary. Thus 2) expresses the real roots x as the sum of imaginaries. To Cardan and his contemporaries, who had no idea how such cube roots could be found, this case was highly paradoxical. Since that time mathematicians have attempted to present these real roots as sums of real radicals. As their efforts were unsuccessful, this case, that is, the case when A > 0, was called the casus irred'ucibilis. It is only recently that a proof has been given that this case is indeed irreducible.* Let us see how the roots 2) may be computed, using our new complex numbers. We set -2 +i\/ A=r(cos + sin l ). Then < —q = Os + k 7r +isin O+k2 3 1 -------- - 2 L 3 27 X- q -/i +VA=/r { cos (+ - 2k- i sin( + k -) }, L \J o \o o Thus the three roots of 1) are Xk =2 / r cos ( + k 120~), k = 0, 1, 2. (4 Example. Let us take the equation X3- 2 2 - + 2 = (- l)(x+ 1)( - 2)= 0, (5 whose roots are, - x = 1, 2, -- 1. To reduce this to the form 1) we set = I-+. (6 Then 5) goes over into - Y + 2- =0, (7 whose roots by 6) are 1 4 5 Y 3' a 3 3 * Hilder, Math. Annalen, vol. 38 (1891), p. 307. 24 FUNCTIONS OF A COMPLEX VARIABLE tLU. 2 0 A 243 3i, q=~, A 272, 1 A= 10. V-243 2 q + = - + i 27 1 3= 343 Hence 3 \343 27 Also 243 tan 0 =.. tanf - gq 10 log 243 = 2.3856, log V243 = 1.1928, log(-tan 4)=0.1928, = 12' as 4 lies in the second quadrant by 8). Hence 1. A,0, A, (8 i~ 41', 3 p = 4 01 4', log 343 = 2.5353, log /343 = 1.2676, log 27 = 1.4314, log r= 9.8362, log /r = 9.9454, log 2 = 0.3010, log cos 1 = 9.8784, log yo = 0.1248 o = 1.333 =. + 120~ = 160~ 54', cos 160~ 54' -cos 19 6', log cos 19~ 6' = 9.9754, log 2 4/r = 0.2464, log(-yl)=0.2218, y =-1.666=- a Q + 240~ = 280~ 54', cos 280~ 54' = sin 10~ 54', log sin 10~ 54'= 9.2767, log 2 /r = 0.2464, logY2 =9.5231, Y2= 333=. CHAPTER II REAL TERM SERIES 9. The reader is already familiar with infinite series. An important chapter in the calculus treats of Taylor's development f(a + x)= f(a) + f (a)+ ft'(a)+... By its means we find for example that x2 x4 x6 cos x= -! +... (1 X X3 X5 x7 sin x= x + - x - (2! 3! 5! 7! X Xe=l+ 2 X3 ex= + x + +.. (3 1! 2! 3! Infinite series were first used to compute the values of a function. Later it was found that they could be used to great advantage to study the analytical nature of a function in the vicinity of a given point. They are still used for the purpose of computation especially in constructing tables; but their chief value to-day in the theory of functions is the aid they afford us in establishing existence theorems, and in studying the properties of functions. We propose in this chapter to develop only as much of the theory of infinite series as is necessary for our immediate purpose. Later we will give further details. 10. Definitions. 1. Let al, a2, a3... be an infinite sequence of real numbers. The symbol a + a2 + a3 + (1 25 26 FUNCTIONS OF A COMPLEX VARIABLE is called an infinite series. We may also denote it by 00 ~am or by sam, = 1,2, 3.. We call An a, a 2+.an -= am,, 177=1 the sum of the first n terms of 1). Suppose that as n increases indefinitely, An converges to a definite value. Then we say 1) is convergent and assign this value to the series; we call it the sum of 1). If An does not converge to some definite value as n increases indefinitely, we say the series 1) is divergent. Whether 1) converges or diverges, it is often convenient to denote it by a single letter, as A; we may write A = a + a2+ a3 + * (2 When this series converges, it is customary to denote its sum by the same letter A. This notation may be slightly confusing at first, but the reader will soon recognize in which sense A is used in any given case. Associated with the series 2) is the series An = an+l + a+2 + *. (3 It is called the deleted series, or the remainder after n terms. It will be convenient to denote the sum of the first s terms of the series 3) by An,; thus An, s= a,1+l + a7,2 + + aA.- 4) Let us now recall a notation with which the reader is already familiar. When An converges to A as n increases indefinitely, we write lim A = A n==, and read it: "the limit of An for n = oo is A." The same fact may be expressed by the notation A, A as n oo. The symbol ' is read "converges to." REAL TERM SERIES 27 2. Let us establish here the obvious theorem: LetA=a1~a9i- A=. a aThe series B = ka1 + ka2+... where k c z 0, converges or diverges simultaneously with A. When convergent, B =,A. For B= If 11ow A or B is convergent, we have lim Bn- k~ lim AnI or B=kA. 11. The Geometric Series. Th is is G = I ggy + g3~2 (1 Let g * — 1. Then by elementary algebra 1 1~qg+ 2 ~... +q1~ gn (2 1 —g 1. -g This identity is often useful and the reader should memorize it. Then using the notation of 10 Ca n 1 + q + q2 + _.. + gn-1 11g i-g ( by 2). Now when g <.1 lim fn 0. n=00, Hence in this ease I r1n glq =nC 0 Thus 3) gives n l~-g The series 1) is therefore convergent when. lq <1 and in this case 1-g If g=1, a=1+1+i~.. Hence an= n and linhum, = + ~0. n= t g Thus G is divergent when g = 1. 28 FUNCTIONS OF A COMPLEX VARIABLE When ~When G_-1, G= -1+-1+.- +.. Hence Gn, = 0 when n is even, = 1 when n is odd. Thus Gn does not converge at all as n = oo. Hence G is divergent in this case. When the series A = a1 + a2+ as+ * is such that lim An = +X0 n=oo it is sometimes convenient to indicate this fact by the notation A = +oo. Similarly if lim A,= -O, n=0o we may write A Returning to the geometric series, we see at once that when g > 1, Ga + o; while when g <- 1, Gn oscillates between ever larger limits. We have thus the theorem:The geometric series 1) is convergent when lg < 1. It diverges when g I > 1. When convergent, its sum is 1-g 12. The Harmonic Series. This is H=+1+1+ i+.. (1 We show that H= H+ x (2 and is therefore divergent. In fact 1 + > 9 1Thus > + > 16 etc. Thus >, 2 > m. +, 23 > + + and in general H2m > m * I REAL TERM SERIES 29 Thus, however large the positive number G is taken Hn >, if n is greater than some integer v. Thus lim H= -+ 00o, which establishes 2). 13. Fundamental Postulate. In order to go on with our work we need to use a fact which the reader will admit as soon as understood. Suppose a variable v steadily increases G as in Fig. 1, and yet always remains less v than a fixed number G. Then obviously v must tend to a limit V. This limit may o be less than G but it certainly cannot be greater than G. We have then FIG. 1. V= limv< a. (1 Similarly suppose a variable w steadily decreases as in Fig. 2, and yet always remains greater than a fixed number Gr. Then manifestly w must tend to a limit W and and W= limw > G. (2 o We take it that these two facts are selfevident and require no proof. 2. By means of this postulate we can w establish a theorem of great importance FIG. 2. in the theory of series: Let A = al + a2 +... be a positive term series. If An < some fixed number G, however large n is taken, then A is convergent andA < G. For as A, < G, lim An exists by the above postulate and this limit is < G. But then A is convergent and A=lim A, < G. 14. The Hyperharmonic Series. This is X I 1' S= 1 + + + +... ( 2s 313 43 30 FUNCTIONS OF A COMPLEX VARIABLE We establish now the following theorem: The series S is convergent if s > 1, and divergent if s < 1. For when s = 1, S becomes the harmonic series H, which is divergent as we saw in 12. When s < 1, each term - is greater than 1 ~~~~~~ns the corresponding term in -ff. n Thus H As 1-, + x, so does S, + 0, and S is divergent in this case. Let now s > 1. Then 1 1 2 1 +_< _= =.= say. 28 3s 28 2. 28 2 g-1 As s > 1, g is < 1. Similarly 1 1 1 1 114 1 2 4s 5s 6s 7s 4s 4s 4 4S 4S 4S-1 11 1 8 1 + I +... + 1 > = I =g3,etc. 8s 9s 15s 88; 8-1 Thus however large m is taken there exists an integer n such that Sm < 1 + g + + qn-1 = Gn- (2 As here g < 1, the geometric series G is convergent and G,< G. Thus 2) gives m < G for any m. Thus by the theorem in 13, S is convergent, and moreover S <. (3 15. Alternating Series. 1. Let a, > a2 > a03> >. 0. Then the series A=a -a 2~a3- a4+ is an alternating series. Examples. 2 3 4 X2 X4 O<X<~ 2! 4! X X3 X5 1 3! 5! (1 REAL TERM SERIES 31 The last two series are the developments of cos x, sin x as we observed in 9. We prove now the theorem: The alternating series 1) is convergent. Its sum A is > 0, and the remainder after n terms An is numerically < al. For A2n+l = (al - a) + (a3 - a4) + *.. + (a2n-1 - a2n) + a2n+l1 Thus A9n+1 > al - a > 0. We also have A2n+l = a, - (a2 - 3) - (a4 - a5) - Thus A2n+1 is steadily decreasing and < al-(a -a3)< a1. Hence by the fundamental postulate 13, lim A2n+1 n=ao exists and is < a1 - a2 and < a1 - (a - a3). Next we note that+ a2 A2n+l -= A2n.+ a2ntl. As lim a2n+l = 0, by hypothesis we have lim A2n = lim A2n+l. Hence ~Hence ~ lim Am exists. Thus A is convergent and 0<A<a1. (2 Finally we note that the series P = an+l -an+2 + an+ - is an alternating series; it is therefore convergent and therefore analogous to 2), 0 <P < a (3 U < P < a,,,3n+1 But obviously the series P and the residual series An differ at most by their sign; hence Ai AThus= using 3) < P. Thus using 3) I AnI< a,+,. 32 FUNCTIONS OF A COMPLEX VARIABLE 2. The fact that the remainder after n terms in an alternating series is numerically less than the next term enables us to estimate the error in calculating such a series and stopping the summation at the nth term. Example. Let us compute sin 10~, using the development X X3 X5 sin — x X3 X5 (4 We first convert 10~ into circular measure and find x=.1745329. log x = 9.2418774. log x = 7.7256322, x3 =.0053165. log5= 6.2093870, x5=.0001619. X3 3=. 0008861. 3! x=.00000135. 5! X7<.00000001. 7! Thus the first two terms in 4) give sin 10~ correct to 5 decimals, and the first three terms to 7 decimals. We have in fact x X3 =.1736468. 1 3! X X3 X5 -- -+ = -. 1736482. 1 3! 5! From the tables we find sin 10~ =.1736482. 16. The E Notation. 1. Sooner or later the student must learn to use the e notation. We propose to introduce it gradually, so that it will not seem difficult to him. The object of the notation is to enable one to think more easily and accurately when dealing with limits. Suppose we have a sequence of real numbers e1 C2,t C3 '" (1 REAL TERM SERIES 33 What do we mean when we say c, = or lim c, = c. (2 n=zoo Let us plot the numbers 1) and the limit c on an axis. Let e be a small positive number. The points c-e, e+e determine an interval E of,E length 2e as in the Figure. Then the - c c+e limit 2) simply means that the e, eventually lie within E no matter how small e is taken. Put in more precise language, the limit 2) means that taking e > 0 small at pleasure and then fixing it, there exists an index m such that em+l, Cm+2, Cm+3 ". all lie within E. The fact that these lie in E is expressed by the inequalities. I c - C, < e n > m. (3 For the relation 3) merely states that whenever the index n is > m, the distance of en from c is < e. It will be convenient to adopt a standard notation. To express that c is the limit of the ce we shall write > O, m, Ic -c <E, n > m. (4 This we will read as follows: For each positive e there exists an index m, such that l c - c < E for all n > m. Conversely if 4) holds, we know that 2) does. This may sound elaborate and formidable to the beginner and quite unnecessary to express a very simple fact. This is indeed so if we never deal with but very simple limits; or never employ but very simple reasoning on limits. Now the fact is that the function theory is founded on the notion of limits. We are constantly reasoning on limits. The same is true in the calculus. But in a first course in the calculus the student is too immature to pay much attention to a rigorous treatment of limits. His main object should be to seize the spirit of the methods of the calculus and to learn how to use them easily. Then as he becomes more mature he can pay more attention to the demonstrations on 34 FUNCTIONS OF A COMPLEX VARIABLE which these methods are founded. In the present work we have no intention of insisting on rigor. Being a first course in the function theory, we shall endeavor to avoid all topics which require delicate handling. A demonstration of such matters is quite out of place in a first course. On the other hand, the student has advanced in maturity since his calculus days, and has reached the point when the subject of limits may be treated appropriately with more care. 2. Let us note that if e- 0, then 4) becomes >0, 0, c|,e <, n>m. (5 Conversely, if 5) holds, lim c =- 0. 3. A simple reflection will show that if lirm n = c then not only does 4) hold, but we also may write > 0, m, c e-e I <, n>m, (6 where 8 may be any fixed positive number < e. For the relation 6) merely says that we have replaced the interval E above, of length 2 e by another smaller interval of length 2 8. We frequently have to deal with several inequalities of the type 6). In such cases we shall see that it is convenient to take 6 _or -3, etc. 2 3 17. Necessary Conditions for Convergence. 1. When dealing with infinite series our first care is to see if the series in hand is convergent. As we never deal with divergent series in the elements of the function theory, if a series is found to be divergent it must be discarded. The following theorem is often useful: For the series A = a, + a2 + -. to converge it is necessary that a-, =0. For suppose A is convergent. Then by 16, 3 we have E > 0O m, IA- Anl < <, n>m. Also < A - An, +l < REAL TERM SERIES 35 Hence by 5, 7) A-,+l - A, I < e. But An+l - An = an+}l Thus I,n+l I < C. Hence by 16, 5) lim an+1 = 0, or what is the same, lim an = 0. 2. Although it is necessary for a ' 0 when A = aI + a2 +... is convergent, this condition is not sufficient as the following example shows. The harmonic series H= 1+ + + is divergent as we saw in 12. Yet here a, =1 - 0. n 3. Let m be an arbitrary but fixed index. The two series A, A, converge or diverge simultaneously. When convergent A = Am + Am. For when A is convergent A = lim A,. Let n =m + s. Then An = Am + Am,. When n ' oo, so does s. As Am is a constant, we see that when lim An exists, so does limn An,s, and conversely. n=o3 s=r 4. If A is convergent, lim An = 0. n=x3 For any n we have A = An + An. As A is convergent, lim A, = A. Thus An=A-A,0. 18. Adjoint Series. 1. In studying the convergence of a series A= al+a2+a3 + *-. (1 it is convenient to consider the series obtained by replacing each term an by its numerical value an = an. The resulting series 2= Cl + a2 + a3 +.. (2 36 FUNCTIONS OF A COMPLEX VARIABLE is called the adjoint of A. We write 3= Adj A. In the function theory we often have to deal with the numerical or absolute values of numbers as a, b, c.... It will often be convenient to denote them by the corresponding Greek letters a, y,.... Sometimes the Greek letter is so much like the Roman letter that the reader is apt to mistake it. We will replace it by the corresponding German letter. Thus Greek A, M look like Roman A, M; we therefore replace them by 1, E9t. The following examples will illustrate the notion of a series and its adjoint. Example 1. A= 1- + 3 - +... Its adjoint is =l+ +I + *+ Example 2. A _ x2 X4 X6 2! 4! 6! Its adjoint is 1 = 1 + 4 + 6. + + 2! 4! 6! where according to our notation = xl. Should the terms of a series A be all positive, then A and 1 are identical. 2. We prove now the fundamental theorem: If 21 converges, so does A. For let For let B=bl+b2+ b + ** be the series formed of the positive terms of 1) taken in order, and C = c1 + 2 + C3 + *.. be the series formed of the negative terms of 1) taken, however, with positive signs. Then Bn < ca + +a3 2 + *a = since B, contains only a part of the terms of 21. Hence B is convergent by 13, 2. Similarly C. < 21 and hence C is convergent. REAL TERM SERIES 37 Suppose An contains r positive terms and s negative terms. Then An= Br- C,, r+ s =n. Let n oo, then B - B, C, C, and hence lim A = lim Br- lim C0, ~or ~A=B- C. Hence A is convergent. 3. A series may converge, although its adjoint does not. Example. A=-1 +-... is convergent because it is an alternating series, by 15. Its adjoint =1+-+ + 1+ +.. is divergent since it is the harmonic series, by 12. A series whose adjoint is convergent is called absolutely convergent. If A converges while 2 does not, we say A is simply convergent when we wish to indicate that A does not converge absolutely. The greater part of the series employed in the elements of the function theory are absolutely convergent. We shall therefore have little to do with simply convergent series. 4. The following theorem is very useful in ascertaining if a given series is absolutely convergent: Let B = bl + b2 + 43 + **. converge and have all its terms > 0. Then the series A = al + a2 + a3 + *. is absolutely convergent if an < bn. Moreover I A I < B. For passing to the adjoint of A, we have n= 0a + +. + n < bl + b2 + '...+- X < B. Thus 21 is convergent by 13, 2. As As JIA, < A n, B we have A < B. 19. The Remainder Series. 1. Suppose we wish to compute the value of the convergent series A = a1 + a2 + a3 + - (1 (1 38 FUNCTIONS OF A COMPLEX VARIABLE correct to a certain number of decimals, say to p decimals. We compute successively Aj A1 - al A2 = al + a2 A3 = a1 a2 + as etc. In order to know if we may stop at An we must know if the remainder An affects the pth decimal in An. We must know, therefore, if 1G-P In case that A is an alternate series the theorem of 15 shows that we may take n so that 1 < - l | i < 10-p. For we showed that I A, I < a,+,. When the series 1) is not alternate, it is not so easy to estimate the magnitude of the remainder. The theorem of 18, 4 may sometimes be applied with advantage to An. In fact if an<bn we have IA < B < B<B. 2. Example. Let us use the theorem of 18, 4 to show that exponential series 3 x x2 X23 1! 2 E=1 +.,+2~ +:, +.. (2 is convergent for x > 0, and to estimate the magnitude of the remainder E,. Let us take x large at pleasure and then fix it. We next take m so large that m + 1 > x. Then < 1. (3 m+ 1 Let us set Xm M mI As x and m are fixed, Mis a constant. Then Xm+s Xm Xs = ---. < Mg. (m + s)! m! (m + 1) (m + s) Thus each term of - xm Xm+1 m —2 - (m + - + +.. POSITIVE TERM SERIES 39 after the first is less than the corresponding term of the convergent geometric series MA - M g + M g2 +... = I( +, + 2 +.) i= 1 —g Hence the remainder series E is convergent. Thus E is convergent and M Em< - < (4 where g is given by 3). Positive Ternm Series 20. Theorems of Comparison. Series whose terms are all positive are of especial importance for deducing tests of convergence. To ascertain if a given positive term series A is convergent it is generally advantageous to compare it with some other positive term series B whose convergence or divergence is known. We begin therefore by establishing two theorems of comparison. 2. Let A = a a, +a2 *.., B = b61 + +. be positive term series. Let r, s be positive constants. If 1~ a<s, n=l, 2, 3... bn or 2~ lim an exists and is = 0, bn then A and B converge or diverge simultaneously. For on the 1~ hypothesis a < sb,; hence if B converges, A < sBn <sB. Thus A converges by 13, 2. Also a > rbh; hence A,> rB,. Thus if B is divergent, so is A. On the 2~ hypothesis, let an _ and I > 0. bn Then as n increases, an gets nearer and nearer 1. Hence for a n sufficiently large m, there exist two positive numbers r, s such that a( r < I < a, n > em. 40 FUNCTIONS OF A COMPLEX VARIABLE Thus the terms of the series Am= am+l + a.+2 + **satisfy condition 1~ above. Thus this 2~ case is reduced to the preceding. Example 1. A 1 +. + A= +- + +^ 1.2 2.3 3.4 _ 1 1 Here an= n(n+l) 2 n(n +1) nP Thus each term of A is less than the corresponding term of the convergent series 1 1 12 22 32 Hence A is convergent. Example 2. A os x cos 2 x A= - F... X >0. ex e2X The adjoint series is = Icosx I Lcos xI +... ex e2x As I cos u I< 1, each term of this series is < the corresponding term of the convergent geometric series 11 1 l+ I +... ex e2x e3x Hence l1 is convergent, and thus A is absolutely convergent. Example 3 A= anE log(l+ -+n), r>l n nr where, is a constant and O I < some G. By the calculus we have, setting r = 1 + s, Ift = 0, we have a ( O\ On nO) POSITIVE TERM SERIES 41 which is comparable with the convergent series X-, r >1. Thus A is convergent in this case. If 4 # 0, we see that nal ' ~. Thus A is comparable with the divergent series + + + ET = ~ + _L2 $- 4.... When g > 0, we see the terms of A finally become positive 'and A = + o. When au < 0, the terms finally become negative and A =- co. Example 4. 0 0 11 1 E 4 l = nX en This series is convergent. For if n > 1, by the law of the mean, logf +)= + —+ 2, IXnI< some M. nT n n Thus M cl <- 2 n The adjoint of C is thus comparable with the convergent series M 12 n2 The series C is therefore absolutely convergent. Its sum called the Eulerian constant. By calculation we find C =.57721566... 3. The second theorem of comparison is: Let A = a + a2 —3 +.., B = 1b2 + b3 + *. be positive term series. If B is convergent and an+l< n n=1, 2,. an - n A is convergent. If B is divergent and an > - A is divergen n n A is divergent. is 42 FUNCTIONS OF A COMPLEX VARIABLE For on the 1~ hypothesis, anl < an. = q, say. bn+l bn - b Thus an bnand we may apply 2. On the 2~ hypothesis we have an ->q and may again apply 2. 21. D'Alembert's Test 1. As an application of the second theorem of comparison 20, 3, we will establish a test for convergence or divergence of a positive term series which is perhaps more often used than any other. It is called D'Alembert's test. The positive term series A = al + a2 + *.. converges if there exists a constant r < 1 for which anl< r, or liman-= r. an an The series A diverges if an+ > 1, or if lim a-l > 1. an an Let us suppose that n+l <r. an We compare A with the convergent geometric series R = 1 + ++ r2 +.. and apply 20, 3. Let us next suppose that liman+l = r an Then we may choose e>0 so small that s=r+e is also <1. Then 1) states that there exists an m such that ls, n>m. an Thus we are led back to the former case. In a similar manner we may treat the divergence part of the theorem. POSITIVE TERM SERIES 43 Example. A a+2 a2+3 a3+... is convergent if 0< a< 1. For the ratio of two terms a,,l +. 1 an n Example 2. Let us show that the exponential series E=1+ x! 2+. 1 + 2+... converges absolutely for any x. To this adjoint + 2 1! 2! end we consider its The ratio of the n + 1st term to the nth is 5. 5 - O! (n - 1)! n for any given:. Thus in this case r= 0 in D'Alembert's test. Hence D converges, and thus E converges absolutely for any x. Example 3. Let us consider the convergence of the series which are the developments of the cosine and sine, viz.: X2 X4 C= 1 +... 2! 4! ry X Xs,3 X5 1! 3! 5! The adjoints of these are -=+2+!+ +4! =: +3 + +. 1! 3! 5! The terms of these series form a part of the in Example 2. Thus < i and hence A, ~ converge for any ~ since e S converge absolutely for any x. series ( considered does. Thus C and 44 FUNCTIONS OF A COMPLEX VARIABLE This result may also be obtained directly from D'Alembert's test. For the ratio of two successive terms of D is _ 2n _2n-2 ~2 -2,-2 - 2 0. (2n)! (2 n- 2)! 2n(2 n - 1) Hence E converges for any ~, and a similar result holds for S. Example 4. Let us show that the logarithmic series x x2 x3 L=~2+~... converges absolutely for any I x I < 1 and diverges for I x I > 1. In fact the adjoint series is t +2 + + V= 4 +!+ +... The ratio of two successive terms is fn+l kn n n + 1 n n + 1 Here the limit r in D'Alembert's test is 4. 2. We must note that when in D'Alembert's test the limit lim an+ = 1, an we can neither conclude that A converges or that it diverges, as the following example shows. Example. 1 1 1 Let A = + ~ + s +. 18 28 3s Here an+1 ns 1 an (n T 8= (I)=1 n Now when s > 1, A is convergent, while when s < 1, A is divergent. 22. Cauchy's Integral Test. 1. This is a test of great power; it is expressed in the theorem: Let f(x) be a steadily decreasing positive function such that f(n) > an. POSITIVE TERM SERIES 45 Then the positive term series is convergent if J= = f(x)dx 0m is convergent. For on the ordinates x = n, in Fig. of an. Then An = area of the shaded But the curve belonging to y=f(x) lies above this shaded region. Thus Am,n = am+1 + am+2 -+. + am+n < J. Hence Am is convergent, and hence A is. 2. Similarly we have a divergent test: Let f(x) be a steadily decreasing positive function such that an >f(n). 1, let us lay off the values region from x = 0 to x = n. FIG. 1. Then the positive term series is divergent if A = a + a2 + +. K= f(xf)dx is divergent. For consulting Fig. 2 we see that - m+n+l Am, n = am+l + am+2 + * am+n > f (x) dx cm+l Let now n c'o. The integral on the right ' + oo by hypothesis; hence Am is divergent, hence A is divergent. 3. In the last section the student might be tempted to reason as follows. Am is the area of the rectangles from x = m to oo. This is greater than the area of the curve from x = m to oo. Thus one would have at once A, > K. a, a2 a3 FIG. 2. (2 As the integral K= oo, so is Am, hence A = oo. 46 FUN(TIONS OF A COMPLEX VARIABLE Against this form of reasoning one can urge the objection that one is dealing with oo as if it were an ordinary number. It is true that in a first course in the calculus the student often falls into this habit. At times this is quite convenient, at other times it can create great confusion. To avoid such loose reasoning mathematicians to-day do not operate on infinite quantities as if they were finite. For example in the present case we wish to show that Am is divergent. To this end we have compared two infinite areas in 2) and asserted that one is larger than the other. The modern mathematician avoids this; instead he would reason as in the foregoing section 2. The relation 1) compares finite areas. In this relation the variable n is allowed to increase indefinitely. Since Am,n increases indefinitely, the series Am is divergent by definition. Hence also A is divergent. The reader will perhaps think this a very small point. In the present case it is indeed trivial. We have chosen it however to illustrate a great principle: The student must avoid operating on infinite quantities as if they were finite. All operations must be performed on finite quantities, except in the single operation of passing to the limit. 23. The Logarithmic Scale. 1. As we have already remarked, the convergence or divergence of a positive term series A = a1 + a + a + a ** may often be determined readily by comparing A with some series whose convergence or divergence is known. Two such series we have already found. The geometric series a= 1 +gg+ +... (1 and the hyperharmonic series s= 1+ +S1+ 4;+... (2 28 38 4s We propose now to use Cauchy's integral test to show that the series POSITIVTE TERMT SERIES 47 1 ( ~nlsn ( l1nl2 sn (4 Y'vlnlnlnlsn (5 all converge when s > 1 and divej~qe whten s < 1. For brevity we have set l~n=logn,12n=log(logn) We must note that in the domain of real numbers, log x does not exist for x < 0. Thus the summation in the series 3), 4).. must begin with a value of n for which lmn exists. Let us consider the series 3), or 1 1 1 L = - ~ — ~ (6' 2 logs 2 3 log"34lg4 when s> 1. 3 4o4 From the calculus we have d log'-sx 1 -Thus ~~dx x logs X f dx _ 1f 1 J~~~~x~~~~o0 -~0 <8 X 0r8 X s- I logs-'a logs-'3J Hence dx 1d1 J X xlog8x 8 1 log-S-a is convergent. Hence by Cauchy's test 6) is convergent when S > 1. Let us now take s = 1.' From the calculus we have dl2X = log(logx) = dx dx x log x Hence dx -log(log,8) - log(log a),0 < a <f. JX log x 48 FUNCTIONS OF A COMPLEX VARIABLE Thus Ifo dx I =+oo. J xlog x Hence by Cauchy's test 6) is divergent for s= 1. Hence d fortiori it diverges for s < 1. To treat the general case we would employ the function f() = 1 xllxl2x... 1I,,-lXl x 2. The series 3), 4), 5), *.. form a scale. That is when s > 1 each converges more slowly than the foregoing. When s= 1 each diverges more slowly than the foregoing. To apply this scale to test the convergence or divergence of a given positive term series A we begin by comparing the terms of A with those of 3). If no test results, we next employ the series 4), and so on. 24. Kummer's Test. 1. This is embodied in the following theorem: Let A = al + a2 + **. be a positive term series. Let k1, k2... be a set of positive numbers chosen at pleasure. A is convergent if for some constant k > 0. K,=k* an ka+l>k, n=l 1,2,... (1 an+l A is divergent if 1 ( D = 2 "(2 is divergent and K, < 0, n = 1, 2, *.. For on the first hypothesis a2 (klal - k2a2). a3 I (k2a2 -k a3) an < (kn_1a- - knan). a k<_~ ~_ POSITIVE TERM SERIES 49 Hence adding these 0 < An<~ al+ 1 kl,-k,,a,, Thus A is convergent by 13, 2. On the Second Hypothesis -an-<k1 an+1 klCn or ani > n+1 Thus A is divergent by 20, 3. 2. We shall call the divergent series 2) Kummer's series. 25. Raabe's and Cahen's Tests. 1. From Kummer's test we may deduce a set of tests of great usefulness. Thus if we take ki, = k2==~~ we get D'Alembert's test 21. If we take k k, =291k =3... we get: ]iaabe's Test. The positive term series A = al + a2 +... is convergent if A is divergent For here if 1) holds. X, ( =n 1 ~1>1. (1 an~~~~~~~~~~(,K. =nX-(n)<1. (2> an+ ( On the other band K,<0 if 2) holds. 2. In tbe foregoing we have used the divergent series L,=i + +-I+... and 1 2 35~~ 50 FUNCTIONS OF A COMPLEX VARIABLE to get D'Alembert's and Raabe's tests. If we use the scale of divergent logarithmic series considered in 1 ~nlix 1 we get a set of tests which may be stated as follows: Let A = a, + a2 +... be a positive term series. Let X (n)= I nln [= a.-11 \n+j ( X (n)=ln in f a I 1 2n JX,(n) - 1 1. Then A converges if there exists an s such that X,,(n) > > 1 for some n > m; A diverges if X0(n)~l for n>m. Let us prove the first test 3) in this set. The others are proved similarly. We take here ikn= n log n.. Then A converges if En = n log nn~-~ - (n + 1)log (n + 1)> k > 0. an+ As n+1=n(1 +i) K Xj=n). -log log - 1 + X,( -log I / \ n+1 5 Xj(1Zn (1 a) a>O. POSITIVE TERM SERIES 51 Thus A converges if Xl(n)> 8 > 1 for n > some m. In this way we see also that A diverges if Xi(n) < 1 for n > some m. 3. From 3) we deduce Cahen's Test. If the positive term series A = a, + a2 + ** is such thatfor every n Cn n l n(:-1 -1 l} < some G, then A is divergent. For l(n) 0n G. n n Here the right side 0. Hence Xl(n)< 1 for n > some m, and A is divergent by 2. 26. Gauss' Test. Let A = a1 + a2 + 4.. be a positive term series such that an _ nS + xln8s1 + ** + /c, (1 an+l ns + aln —l +.** + s, where s, acc 2 *..,.. *. do not depend on n. Then A is convergent if a1 - /3i > 1, and divergent if ac - /1 < 1. This may be deduced from 25 as follows. Here / - 1/31 + 2 - /32+ - 0(n) = na. -1 (2 an+il l 1 +1+- '" 1 Thus lim Xo(n) = a - /31. Hence if a, -l / > 1, certainly there exists some I > 1 such that nan- 1> 1 for all n > some m. van+i 52 FUNCTIONS OF A COMPLEX VARIABLE Thus Raabe's test shows that A is convergent. If l - 1-= 1, Raabe's test does not always apply. To dispose of this case we may apply the Xl(n) test of 25, 2. Or, more simply, we may apply Cahen's test. We find at once lim C = a2 - a 2 - /31 Thus C, < some G and A is divergent. 27. A test similar to Gauss' test in 26 is the following Let A = al + a2 +... be a positive term series such that a, a 3n = 1++ + an+l n WI where ps > 1, and /n, < some G. Then. A is convergent if a > 1, and divergent if a < 1. For here X(n)=( an1 = + a an+l n/ -t Thus A is convergent if a >1, and divergent if a < 1. If a= 1, we have 1 Xl(n) = 11nX0o(n) - 1} = -1 *n- and A is divergent. 28. Binomial Series. This is B = 1 +L fx+ ' L- - 12 - 1 --- 2- 3 +... 1.2 1.2.3 I + (/)x + ( 2 + P X3+* This series arises when we develop (1+x)g by Taylor's theorem; here we wish merely to consider the convergence of the series as an application of the foregoing tests. If /. is a positive integer, B is a polynomial of degree A. If / = 0, B = 1. We now exclude these exceptional values of P. Applying D'Alembert's test to the adjoint of 1), we find an+l_ n + | _ |. | an n Thus B converges absolutely for Ix I < 1, and diverges if | x > 1. POSITIVE TERM SERIES 53 Let x = 1. Then B= +1 +~ + t l+... 1.2 Then O(n= /LH-n +l -. an n As D'Alembert's test gives us no information in this case, we apply Raabe's test. Here \enl - 1 - /+ n for n sufficiently large. Thus 0(n) ' 1 + 1. Hence B converges absolutely if L > 0, and its adjoint diverges if /u< O. But in this case we note that the terms of B are alternately positive and negative. Also ann+1 I1 1+/X aXn n so that an form a decreasing sequence from a certain term, provided Iu>-1, when an 0. Thus B converges when > >- 1 and diverges when A - 1. Let x=-1. Then B= 1- +' 1... 1.2 If u > 0, the terms of B finally have one sign and Xo(n) 1 + w. Hence B converges absolutely. If / < 0, let / = - X. Then B becomes 1.+ X+ +1 +X.X+ 1.X+2+. 1+ 1.2 1. 2.3 Here XX0 (n) = 1 - X, 1+ Xn and B therefore diverges in this case. To sum up, we have the theorem: 54 FUNCTIONS OF A COMPLEX VARIABLE The binomial series 1) converges absolutely for x I < 1, and diverges for IxI >1. When x= 1, it converges for u > - 1 and diverges for s < - 1; it converges absolutely only for a > O. When x =- 1, it converges absolutely for p > 0 and diverges for p < 0. 29. The Hypergeometric Series. This is _(a, a Y, ) =1 + + '-+ a + X2 )1+ 4/31-~ +l./3./3+l2 I~ I 'Y 1.2.y7y l 7 I(1 +a a + 1 ~ a+ 2. /3 /3 + 1 3 + 2x +.. Let us find when this very important series converges. Passing to the adjoint series, we find an+a (a+n)(/3fn) =n+2 = (it + ) () +. ) 1i I i. (2 an+l (n + )(7+) i Thus F converges absolutely for | x I< 1 and diverges for x > 1. Let x = 1. The terms of F finally have one sign and a,+l _ n2+ n( + ly) + y a,+2 n2 + n(a + 3) + oa, Applying Gauss' test, 26, we find F converges when and only when a+13-7<0. Let x =-1. The terms finally alternate in sign. We may write F= al - a2 + a3- **.. Let us find when an = 0. We have a -, (, + 1)... (a + n)(/s + 1)... (/3 + n) 2- * ( + 1)... (1 n)(7 + 1)... (7 +n) Now a+m=n = m l a), I +m=m (1+, 1+m=m(l1+-), 7+m=m(l+ -). Thus I+ + n qt (1+ (1+ n+21m/\ m/ POSITIVE TERM SERIES.55 But =1 1 m rn2 m n M2 where m 1,T 2 8S 7O Hence n (a~/3-y-l an+2 = slog+ 2+ - ) 1+ a+ Y- n~~~MI Hence log a,, = ilo g 1 ( a cr-P-~y + 13m Thus L= lim log a,+2 = Ylm. 1 Now for a. to 0 it is necessary that Ln -. In 20, Ex. 3, we saw that this takes place only when a ~ / - ey - 1 < 0. Let us now see if L is an alternating series. If so, we must also have an> an~1 <.... From 2) we have a n +n2 ~1 n+1~~~~~~~/ Thus when a + / - ry - 1 < 0 the L series is alternating. Summing up, we have the following theorem The hypergeometric series F converges ahsolutely when x < 1, and diverges when x > 1. When x = 1, P converges only when a + / - ry< 0, and then absolutely. When x = - 1, P converges only when a + 3 -7fy-1<<0, and absolutely if a + - 7y < 0. CHAPTER III SERIES WITH COMPLEX TERMS 30. 1. Having discussed series whose terms are real, we now consider those whose terms are complex numbers. As heretofore such series will be represented by A = al + a2 + a3 + (1 the sum of the first n terms by An, and the residual series by An. If we replace each term of A by its numerical value C, = a an, the resulting series. = al + a2 + a3 +... will be the adjoint series. Before defining the sum of 1) we must define what we mean by the phrase "An converges to a number L as n increases indefinitely," or in symbols An L as n oo, or lim An= L. (2 n=-oo Suppose we plot the points associated with the complex numbers, Al, A2, A... and L. Then when we say An L, we mean that these points get nearer and nearer L. More precisely this idea may be expressed as follows: About L describe a circle of radius e as small as we choose. Then all the points A+l A A +ml1, Am+2 1 m+31 fall within this circle for some m, as in the figure. In other words, there exists an index m such that IL-An <e for alln>m. (3 / If the reader will turn to 16, he will see that A^ this is a natural extension of the term limit when Am the numbers considered were real. 56 SERIES WITH COMPLEX TERMS 57 We are now ready to give a final definition. We say 2) holds when for each positive e there exists an m such that 3) holds. This definition applies to the limit of any sequence of complex numbers as C1, C2, C3 To express that c is the limit of cn we shall write e>0, m, IC —Cn<e, n>m. (4 This we read as in 16, viz.: For each positive e there exists an index m, such that c - c, < e for all n > m. Having now defined the term limit we may extend the terms convergent, divergent, sum, defined in 10, without further comment to the series 1) whose terms are complex. Thus when lim A, exists, we say A is convergent. The limit of An is the sum of 1). If lim An does not exist, A is divergent. A number of results established in the last chapter hold for series whose terms are complex. In fact the reader will see that the demonstration applies equally well to complex terms. For the convenience of the reader we state some of them here. 2. Let A = al + a2 +... be a series with complex terms. Then A and the residual series Am both converge or both diverge. If A is convergent, A - 0, also a = 0 as n oo. If A converges, B = kaI + ka2 +... converges and B = kA, k = 0. 3. We have just noted that when A is convergent, an must - 0. From this we draw the obvious yet important conclusion: If A = a1 + a2 +... is convergent, then Ian, <some G, n =1, 2, 3, (5 For, describe a circle C about the origin. Then since an 0, all the terms am+l, am+2 *" lie within C for some definite m. Let us now describe another circle D about the origin so large that it contains the m points a1, a2... am and also C. If G is the radius of D, the relation 5) holds obviously. 4. The reader should note that although the terms al, a2... of the series 1) are complex, it does not follow that they may not be real. The class of complex numbers contains the class of real 58 FUNCTIONS OF A COMPLEX VARIABLE numbers as a subclass. It follows therefore that any theorem established for series with complex terms must necessarily hold when the terms of the series are all real. 31. Absolute Convergence. 1. The terms of the series A = a, + a2 q as +... (1 being complex, let us set an=bn+icn, n=1,2,-.. (2 Let LetB = bI + b2+ b3+ (3 = c + C2 + C +. (4 Then An=Bn + iC. (5 We show now that: If B, C converge, so does A, and A = B + i. Conversely, if A converges, both B and C converge. For if B, C are convergent, we have from 5) ca An lim An = lim B, + i lim C, A or A=B+iC. 0 Conversely, let 1) be convergent. Let its Bn.... I-. A t r_. T v' I - I -..l uuiLL e Ut = p -- z7y. I iel rig. - snows tiat as A,-=A, then B, -=f and C = y. 2. As already remarked the adjoint of 1) is = 41 + C 2 + a3 + *t where I a, | = an. From Fig. 2 we see that n = |bnl < an a 7 = |C n | < CnSimilarly the adjoints of B and C are 8 = 8 +/3+... = /1+ 72+ 2~ We now prove the important theorem: If the adjoint of A converges, A is convergent. FIG. 1. 0 dn/a bn FIG. 2. SERIES WITH COMPLEX TERMS 59 For obviously 3, < ~1 < <I, hence i3 is convergent, and therefore B converges absolutely. Similarly (S< 3n < 32 and hence C converges absolutely. The theorem now follows from 1. 3. When the adjoint of A converges, we say A converges absolutely. The great importance of the last theorem is obvious. It enables us in nearly every case in practice to reduce the problem of determining whether the series A is convergent or not to the same problem relative to th, adjoint series W. But the terms of 1 are real positive numbers, and the convergence, of such series was treated in the last chapter. 4. Having established the last theorem, the reader will note that the reasoning of 18, 4 holds for complex terms. Hence the theorem: If each term of A = al + a2 +. is numerically < the corresponding term of the converz/ent positive term series B= b- + b2 +- * then A is absolutely convergent and IAI_< B. 5. Returning to 2, let us note that the reasoning there shows that: For 1) to converge absolutely, it is necessary and sufficient that the two real series 3), 4) converge absolutely. 32. Addition and Subtraction. From the two series A = a, + a2 + as- + B=b1 +b2+b3+ + let us form the series C= ( a, + b) + (2 +b2) + (a3 + 3) + *" We now show that: If A, B are convergent, C is convergent and its sum is A + B. For Cn = (a + bl) + * + (an + bn) = An + Bn, Now lim A =A, lim Bn= B. Hence lim =A C= lira Cm = A + B. 60 FUNCTIONS OF A COMPLEX VARIABLE Similarly we prove: The series D = (a, - bl) + (a - b) + -.. converges if A, B converge and D = A- B. 33. Multiplication. 1. Suppose we have two polynomials P =Pi + P2 + * p = Pi, Q = q1+ q2 + + q = qj. Then from algebra we know that PQ = q1P1 + q1P2 + ** + qlpm + q2P1 + q2P2 + + q21fPm + qnPl + qnP2 + ** + 4qPm' The general term of the product is piqj. We may thus write == 1, 2,...m i j=, =12,. n. (1 Instead of two polynomials P, Q let us take two infinite series A = al + +., B = b + b2+... (2 and from them form the series C = Eabi (3 i,j which contains all possible terms ajbi without repetition. We prove now the theorem: If the series A, B are absolutely convergent, so is C and C=A * B. We begin by considering the adjoint series a2 = Eai 3 = 2Si @ = Eai/3A. Let us look at the product fm93,; it contains all terms ai3i whose indices i, j are both < m. Let us now take n so large that the sum of the first n terms of 7, that is (, contains all the terms of 9SI,3*m. In general (~ contains other terms of the type ar3, SERIES WITH COMPLEX TERMS 61 where r, s are not both < m. On the other hand let no term ar/, have an index > v. Then n - 2mJ, < alPm+1 + al}/m+2 + +... al/ + 0A2/m+1 + X2/3m+2 +... + a21/v +..... + av/38n+l + av/m+2 + '* + a3 + /lam+l + 13arn+2 + * + 31a + ** + /am+1 + /3vra,+2 + '*" + I3Vy For every possible term ca,/S which (E - W1nm3 can contain is to be found among the terms on the right. Moreover all the terms involved are positive numbers. Now the first row on the right gives a1(mn+1 + - m+2 + "' + /v) < alm, and a similar relation holds for the other rows. Thus Cgn -mom < -ren + + UAto + l3m- + - + 3lm < (Qk1 + ' m + v)-+ (3i +,8v)Tm < tam + ~ M. Let now m oo. Then [m ' 0, m =' 0 by 17, 4. Thus the left side - 0. But limr 2fmm = lim WM lim m -= a * t' Hence ( is convergent and S = *.S3. This shows that the C series is absolutely convergent. To show that C = A. B, let m, n have the same meaning as before only now referred to the A, B, C series. Then Cn- AmBm is numerically < the sum of the corresponding terms in, - m3m. Hence I C- AmBma <(n —mA3m 62 FUNCTIONS OF A COMPLEX VARIABLE Now when m = co, the right side 0. Thus limn (C,- ABm) = 0. As lim Am.Bm = AB, this gives C = AB. 2. In forming the product series 3) it is well to have a definite law in order that no term aibi is omitted, and no term is repeated. Such a law is expressed as follows: C = alb6 + (alb2 + a2bl) + (alb3 + a2b2 + a3b1) + (alb4 a + a3 + b+ la4) + ** (4 We notice that the sum of the indices i + j is 2 in the first term, it is 3 in the second term, 4 in the third term, etc. Also in each term the index i increases while j decreases. In this way it is possible to form all the terms aibi in 3) without repetition or omission. Of course there are many other simple ways of doing this, but this is in general the most convenient. 34. Cauchy's Paradox. 1. At this point we are face to face with a paradox. One would expect that if the series A and B converge, the series C in 33, 4) would converge and have as sum A ~ B. In case that A, B converge absolutely, we have just seen that this is indeed true. We now exhibit an example due to Gauchy which shows that if A, B are convergent but not absolutely convergent, then the series C may not even converge. In fact, let 1 - 1 1 A= —+ +- +... 1 2 V 14 B= _ +._.. A. V1 V'2 V3 V/4 The series A being an alternating series, is convergent by 15,.s Its adjoint is divergent by 14 since here s= 2. Let us now form the series C in 33, 4). We have 1 1 1 11 A/I/ /2 I2 V/ VV = 2 + C3 + C4 + =^+^+?4-! ** SERIES WITH COMPLEX TERMS 63 Here 1 1 1 1 1 1 c_-_-+ + '"+ ' =/+r I V /V/n-1 2/ 2Vn-2 Vn-1 IV Now from algebra we have Vm(n- m) n, n >m > O. Hence 1 >2 2(n- 1) Vm(n - m) n n Thus C is divergent since c, does not 0, as it must if C were convergent, by 17, 1. 2. The foregoing paradox arises from the tacit assumption that the earlier mathematicians made and which students to-day are too prone to make; viz. that infinite series have the properties of finite sums. The sum of an infinite series is the limit of a sum of a finite number of terms, in symbols A = lim An. n=co Now it does not follow that the properties which each A, may possess also hold in the limit. In other words we must learn to discredit the dictum: what is true of the variable is true of the limit. In general this dictum is valid; there are, however, countless cases where it is not. In particular it is true that infinite series have many properties in common with finite sums, but they do not have all their properties, witness the foregoing paradox. It is helpful indeed in our reasoning to remember that in general infinite series do behave as finite sums. It is also extremely helpful to remember that very often what is true of the variable is true of the limit. Such partial truths are of great value in exploring the way and in seeking for proofs that are really rigorous. Their value is heuristic and every student should employ them freely. He must, however, learn to replace reasoning founded upon them by proofs of a more binding character. 3. Let us note a few cases where the student is apt to go astray unless warned. 64 FUNCTIONS OF A COMPLEX VARIABLE Example 1. Let us plot a finite number of real positive numbers, al, a2 *.. am, no two of which are equal. Then there is always one point which is nearest the origin. This is true for any m. Is it true for an infinite sequence of such numbers a1, a2, a3 Not always, as the sequence 1 1 1 1,, 3 -. shows. Obviously there is no a =- which is nearest the origin. 1 For a+1= - 1 is nearer 0 than as. s + 1 Example 2. Similarly in any finite set of different numbers there is always one which is greatest. In an infinite set this is not always true. Thus the set 3 7 15 31... has no greatest. Example 3. In the interval (0, 1) formed of the point x such that 0 < x < 1 there is a first point x = 0 and a last point x = 1. On the other hand, in the set of points x such that O<x<l there is no first point, and no last point. 35. Associative and Commutative Properties. In any sum of a finite number of terms as S= a+(b +c)+ d +e, (1 we may leave out parentheses or put them in wherever we choose. This is called the associative property of sums. Thus the sum 1) may be written S= (a + b) + (c + d)+ e = a + b + c 4- (d + e), etc. Also the value of 1) is not changed when its terms are rearranged in any way. Thus S=b + a + d + e = e +c d + a +b, SERIES WITH COMPLEX TERMS 65 etc. This is called the commutative property. The student is so used to making these transformations that he does it almost without thought. It is natural for him to extend these properties to infinite series. Yet simple examples will show him that this is not always permissible. Example 1. Let A= 1+(1- )+(1-)+(1 - 1)+... (2 = al + a2 + a3 + a4+ + Here, A,= a +.. + a =1+(1 -1)+... +(1 -1), nterms =1. Hence lim A, = 1. Thus 2) is convergent and its sum is 1. If we remove the parentheses from 2), we get the series B=1 1 + -+-1 + 1 -1 -+ =b + b2 + b3+ * Here, B =2, B2 = 1 B2n 2 B2n+l 1 -Hence lim Bm does not exist and B is not convergent. t==oo Example 2. (Dirichlet.) Let A= l-i -++ -4 + (3 This we saw is convergent. We shall show directly that we may group the terms of A by twos or by fours without changing its value. Let us admit this fact for a moment. Then we have A=(1 —+~-+-)+0-+-)+.- (4 = ( 1- +1 3 ) + — 1 + 1) + U*- (5 From 4) we have 2 A = ( - ) + (1-) + ( - ) +... Adding this to 5) gives 3A=( + 3-) + ( + - )+. (6 2 1 3 2 \+ T I ~~( 66 FUNCTIONS OF A COMPLEX VARIABLE We shall show directly that it is permissible to remove the parentheses in 6) without changing the value of the series. Thus Let now co th tw ri (7 Let us now compare the two series 3) and 7). We notice that 7) is obtained from 3) by taking two positive terms of 3) to one negative. Each term of 3) is to be found somewhere in 7) and no term is repeated. Thus the series 7) is merely a rearrangement of 3). If now all infinite series enjoyed the commutative property, the rearrangement of the terms of 3) would not affect its value. But the left side of 7) shows that the sum of 7) is - times greater than the sum of 3). Thus not all infinite series are commutative. 36. Since it is often convenient to put in or to leave out parentheses in a series and also to rearrange its terms, it becomes necessary to ascertain when this is permissible. To this end we establish the following theorems: 1. Absolutely convergent series are commutative. For let A = al+ a2+a2 + 3. be absolutely convergent. Let B = b1+ b2+4 +.. be a series obtained from A by rearranging its terms. We wish to show that B is convergent and that its sum is A. Since the adjoint series a/= a1 + a2 + a3 + *.. is convergent, we may take m so large that 2m <e. (1 We may then take n so large that B, contains all the terms of Am, and v so large that Av contains all the terms of Bn. Then.. A;, -n (2 SERIES WITH COMPLEX TERMS 67 contains no term of index < m and the terms of the sum 2), each taken in absolute value, lie among the terms of the residual series Im. Hence I n < m or, using 1), < e Thus B is convergent and lim B A. 2. Let us now turn to the associative property. We begin by showing that we may insert parentheses at pleasure in any convergent series, a fact we embody in the following theorem: Let A = a, + a2 + a3 +... be convergent. Let b1 = al + + a, b2 = am1+1 +l + am" 2, Then the series B = (al +... + aml) + (aml+ + + am2) +. = bl + b2 + is convergent and A = B. Moreover the number of terms which bn embraces may increase indefinitely with n. For B= A (3 Since A is convergent, li Am. lim Amn- =A. n=ox Thus passing to the limit n = o in 3) gives B=A. 3. The next theorem relates to removing parentheses from a series. Thus if we remove the parentheses from the series B =(a + a2 + *'" -+ am1) + (am1+l + * * + m2) + *~ = bl + b2 + "- we get the series A = a+ a + as + (5 We show now that in the following three cases the parentheses may be removed from the series 4). 1~ If A is convergent, B converges and A = B. 2~ If A is a positive term series and B converges, then A is convergent and A = B. 68 FUNCTIONS OF A COMPLEX VARIABLE 3~ If the number of terms in each parenthesis in 4) is < a fixed number p, and if a,, 0, then A converges if B does and A =B. For on the 1~ hypothesis, we have only to apply 2 to show that B converges and A = B. On the 2~ hypothesis, we have E < 0, m, B <, n > m. (6 If now we take s > mn, B - A will contain only terms in the residual series BP. As the terms an are positive, we have from 6) B -A,< e. Thus A8' B or A= B. On the 3~ hypothesis, we note that the terms of An will embrace a certain number of terms of B, say Bm, and in general a part of the next term of B. We may therefore write An = Bm + 1, (7 where bm,+ is a part of bm+l. Since bm+1 contains at most p terms an and as by hypothesis an 0, we see that b' 0. Passing to the limit in 7) we see that A, B. 4. Let us now return to. verify the statements made in 35, Ex. 2. Since the series 3) in that article is convergent, we may indeed group its terms by twos or by fours without changing its value. In the series 6) we see that p = 3 and that in 3) a =(- 1)n+l.1. n Hence this series falls under the 3~ case of the theorem 3 above. Hence if we remove the parentheses from 6), the resulting series 7) has the same value as 6). Thus the series 3) is not commutative. It is also not absolutely convergent and the theorem 1 does not apply. SERIES WITH COMPLEX TERMS 69 37. Riemann on Simply Convergent Series. 1. It will interest the reader to see that a simply convergent real term series may be rearranged so as to give a series whose sum is any desired real number. Let the given series be A= a + a2 +3+ (1 Let B= bl+ b2 + b3+ (2 be the series formed of the positive terms of 1), keeping their relative order in 1). Let C(= c1 + 2 + ~+ C (3 be the negative terms of 1) with their signs all changed. We begin by establishing the following theorem: If A is a real term simply convergent series, both B and C are divergent, i.e. B= +, C=+ 30. For in the first place B and C must both have an infinite number of terms. Otherwise some residual series Am would have terms with only one sign. As A is convergent, Am would converge absolutely. Hence A would be absolutely convergent, which is contrary to hypothesis. Let us thus suppose that An contains r terms of B and s terms of C. Then n = Br+ C, n=r+s. If now B and C converge, we see that 2 also converges and thus A is absolutely convergent. On the other hand An = Br,shows that if B or C were convergent, both would converge, since An ' A by hypothesis. 2. We can now establish Riemann's Theorem. If A is a simply convergent series with real terms, it is possible to rearrange the terms of A forming a series S for which lim S, is any prescribed number 1, or ~ o. 70 FUNCTIONS OF A COMPLEX VARIABLE To fix the ideas let 1 be a positive number; the demonstration of the other cases is similar. Since by 1, B, + +o, there exists an m such that b + b2 + ** + bl > 1. (4 Let ml be the least index for which 4) holds. Since also Cn + oo, there exists an index m such that (bl + b"' +b - n)-(1i+ +cm2)<. (5 Let m2 be the least index for which 5) holds. Continuing, we take just enough terms, say m3 terms, of B so that (b1 +... + b,,l) - (c1 *- + Cm2) + (bm+l + + bm,+?n) > 1. In this way we may form the series Tr= (b1 + * + bm)- (c1 +. + Cm~)+( )-( )+ *" It is easy now to show that lim Tl= i. For since A is convergent, a -0. Moreover we choose our terms in 1 so that TI differs fronm by an amount < some a, of A. Thus T7 -l-0. Let now S be the series T with the parentheses removed. Since the terms in the parenthesis are positive, the series S is convergent and has T as sum. 3. The foregoing theorem shows that Dirichlet's example considered in 35 does not illustrate an exceptional case, but the rule. This remarkable behavior of non-absolutely convergent series should make the reader more careful in dealing with infinite series. On the other hand, it would be a great misfortune if he became afraid of them. Let him consider infinite series just as if they were finite sums when striving to prove a theorem or solve a problem. Only he must not neglect at the end to go back over his steps and justify them carefully. POWER SERIES 71 4. Let us make an obvious extension of the foregoing result to series whose terms are complex. If A = 1 +- a2+ a +.. where a = bn + ic,, we saw in 31, 1 that A= B+ i C when A is convergent; while we saw in 31, 5 that if A does not converge absolutely, at least one of the series B, C is not absolutely convergent. Suppose the series B is simply convergent, then B is not commutative. Hence A cannot be commutative. Thus we have the theorem: No simply convergent series of complex terms can enjoy the commutative property. Power Series 38. The series A = a + a2 33 + A = ao +ai2z 4 a2 + a3.. (1 is a power series. Here the coefficients a0, al, a2... and z may be complex numbers. Such series are of utmost importance in the function theory. Indeed one is tempted to say they form the most important class of series. Special cases of such series are the series afforded by Taylor's development in the calculus. In fact Taylor's series f (O) + f + f O + x is only a power series as is seen by setting a0=f(0), a,=f'0) Thus the developments of sin x, cos x, ex, etc., given in 9 are power series. The variable x is there real, of course. A slightly more general form of 1) is ao + a(z - a) + a2(z- a)2 + a3(z- a)3 + *. (2 Since the series 2) goes over into 1) on replacing z - a by z we may reason on 1) without loss of generality. 72 FUNCTIONS OF A COMPLEX VARIABLE 39. Circle of Convergence. 1. A fundamental theorem in the theory of power series is the following: Let the series A = ao a^ z + a22 + (1 converge for z =b. Then it converges absolutely for any c within the circle Kthrough b with the origin as center. If A divergesfor z=b, it diverges for any point d without K. For the adjoint series corresponding to z = c is 81 = ao + axv + o2Vy2 + *.. (2 where an= an, i = i c To show that d this converges we observe that by hypothesis b a + alb + a62 +... (3 converges. Thus by 30, 3, = Ca0, ec, a2/2, \ are all < some g. We now write a thus: a = (o% + ) + a2/2 + a8() +... (4 Comparing this with the convergent geometric series aG.g (,^,(\.. J< (5 we see each term of 4) is < the corresponding term of 5). Thus 2) is convergent and A converges absolutely for z = c. Suppose now A diverges for z = 6. Then it diverges at any point d without K. For if it converges at d, it must converge, as we have just seen, at all points within a circle f passing through d and having 0 as center. Thus A would converge at z = b, which is contrary to hypothesis. 2. If the circle C whose center is z=O and whose radius is R is such that 1) converges for every point within C and diverges for every point without C, this circle is called the circle of convergence of the power series 1). Nothing is said about the convergence of 1) at points on C. It may or may not converge at a given point on C. POWER SERIES 73 3. Let us note bef ore passing on that the series 4), 5) enable us to give a rough estimate of the numerical value of the series 1) at a point z = c. For let A, denote the sum of 1) for the point Z =C. Then IC 2 by 31, 4. But we saw that 21, or what is the same the sum of 4), is less than the sum of 5). But the sum of this series is g Thus IA, < g( /3r ( which is the relation we had in view. 4. Let us find the circle of convergence of certain series which. we shall employ later. The value of the radius B is placed at the right. 1) Z Z~~~~ ~ ~~2 Z3 R 1! 2! 3! z2 z4 2) C~os z= 1-+ --- = 2) ~~~~2! 4! z z3 z5 3) sinz= ---+-~ — 1! 3! 5! 42 + R 5) ~~~~~~~Z2 Z3 Z4 z 3 4 6) sinh z= z+-+-~.. B-c 3! 5! 2! 4! 8) F~z~i3,ry~z)=1+ z~ l 2 1 24 7~ 2nn 2(2 n + 2 2. 4(2n-f-2)(2 n —4)~ 74 FUNCTIONS OF A COMPLEX VARIABLE For convenience of reference we have added on the left side their values as functions of the complex variable z. For the present the reader should consider the series on the right merely as series whose circles of convergence are to be found. Since these circles all have z = 0 as center, it is the radius R which we seek. Suppose now that the adjoint of one of these series, call it A, converges for ] z I = y. Then R is certainly as great as ry. If, on the other hand, A diverges for z \ =, R is certainly no greater than y. Finally, if A converges for I x <ry while it diverges for I x I > y, then the radius of convergence R is = y. Now the series 1), 2), 3), 4), 5), 8) we have already considered for real values of z. In 21, Ex. 2, we saw that 1) converges for any real x. Thus for this series R = c. Similarly 21, Ex. 3, shows tlat R-= o for the series 2), 3). In 28 we saw that 4) converges for real x such that I x < 1 and diverges when Ix I> 1. Thus R =1 for this series. Similarly 21, Ex. 4, shows that R = 1 for the series 5). Finally in 29 we saw that 8) converges for real x such that i x i < 1 and diverges for Ix I > 1. Thus R = 1 for this series. Thus there remain only the series 6), 7), 9). The first two are at once disposed of. For the terms of their adjoint series form a part of the adjoint series of 1). Thus R=x for both 6) and 7). As to 9), the ratio of two successive terms of its adjoint is a, 22(s+1)(n +s+ ) for any given. Thus R = cc for this series. 5. The following development we shall use later 1 = {l+:z-a~+(zi-a )+ } (10 - z u- a u- a \u- valid for Iz-a I< u- al. To prove it we note that u - z = (u - a) - (z - a) =(u-a)j 1 -- - a - )., =(u - a)(1 - v). POWER SERIES 75 Now 1 -l+v+v2+..., Ivl<1. (11 1 - v + 2 V < Thus 1 1 u-z u-a 1-v gives 10) on using 11). 40. Two-way Series. 1. In the series considered up to the present a al + a2 + a... =. an 1 the index takes on only positive values. It is sometimes convenient to consider series in which n takes on both positive and negative values. This leads to the symbol ** + a —3 + a-2 + al + a0 + a1 + a2 + as +. * (1 or an. We call 1) a two-way series. Example i. We shall see that in certain cases a function of z can be developed in the form ao + a1z + a2z2 + * +I I + b2 + z z2 z3 If we set b, = a_, this can be written -00 Example 2. In the elliptic functions we consider series of the type 1 + qe2riz q- q4e2.2iz + 9e3a'2fiz... + 2i+ 2'2niZ 32 e%. e22+ e3+. which may be represented by qn2e2rinz. (3 -2. 2. With the series 1) we associate the two series B=E C= a_.,, 0 I (4 76 FUNCTIONS OF A COMPLEX VARIABLE If these two series converge, we say A is convergent and its sum i~s A~ A=B+ C. If either or both the series 4) are divergent, we say A is divergent. Thus the theory of the two-way series is made to depend on tlhe two one-way series 4). Instead of the series 4) we could use any two other series obtained from 1) by breaking it at any other index m. Obviously the same results would be obtained with these as with the series 4). If the adjoint series -00 converges, we say that A converges absolutely. Thus if B and C converge absolutely, A is also absolutely convergent. 3. Another definition of convergence and sum of the series 1) is the following. Let Am,n a-m + a_+l + a* - +al+ + a al+ + a (5 Suppose that as m, n-oo independently of each other, Am,n converges to some fixed number which we denote by lim Am, n (6 m, n=oo or more briefly by I; that is, suppose that for each e > 0 there exists an r such that Am, differs from I by an amount < e for all m and n > r. In this case we say that A is convergent and its sum is the limit 6). This definition leads to that given in 2, but we do not wish to urge this point. 4. As an example let us show that 3) converges absolutely for any given z = x + iy when r = q < 1. For, assuming for the moment that e+iv = eve ^ we have e2rinz = e27rinx e-27rny Hence Hence I e27rinz = e-2rny The adjoints of the B and C series defined in 4) are, here 3 = e nr 2rrny re 0 I POWER SERIES 77 The ratio of two successive terms in l is?(n+1)2 e-2try = ry2n+le-27ry _ 0 n2 r as n-oo. Thus B converges absolutely; similarly C also. 5. Two-way Power Series. Let us consider the series a,+ a32 a+... (7 z z2 z3 If we set 1 U z =-. it becomes albu + a2u2 + au3 +... If this series converges for u = c, it converges absolutely for all | I < c I. Hence if 7) converges for z = b, it converges absolutely for all 1z I > b I. Let C be a circle about the origin such that 7) converges for every z without C and diverges for every z within C. Then Cis called the circle of convergence of 7). Let us now consider the two-way series a0 + aZ - a2+z2 + *. + C + + +...=P + Q(8 Z22 where P is the series in the first line. If C is the circle of convergence of P, and D that of Q, the ring R = - D lying between these two circles is called the ring of convergence of 8. The radius of U may be infinite. 41. Double Series. 1. A point whose coordinates x, y are integers or zero is called a lattice point. Any set of such points is a lattice set. Let am, be given numbers, the indices m, n corresponding to points of some lattice set. The symbol A= Eam, (1 is called a double series. With 1) we may associate a series B = b, (2 78 FUNCTIONS OF A COMPLEX VARIABLE where b, is some term am, of 1) and where each term ap, of 1) is some b. of 2). If 2) converges absolutely, all these B series, being merely rearrangements of one of them, have the same sum. In this case we say 1) is convergent and its sum is that of 2). As heretofore it is often convenient to denote the sum of the series A when convergent by the same letter. When the series 2) does not converge absolutely, we shall say 1) is divergent. The seriesn am, n = am, n ^ == 2^ ^,, n t | is called the adjoint of 1). From our definition of convergence it follows that 81 converges when A does, and conversely. 2. Let us note that the multiplication of two simple series 00 co A=lam, B = bn 1 1 leads to double series. In fact let us set Cm, n = ambnThen C = em,n (3 is a double series, and when A and B are absolutely convergent, we saw that Cis convergent and A ~ B = C. In the series 3) the indices m, n range over the lattice points in the first quadrant, excluding those on the x or y axes as for these m or n would have the value 0. 3. We have seen that 1= 1+a+a2+... a<l, a = 1 + a'+ a2 + I*, I a I < 1, 1-a 1 +b+b+..:, Ib<. 1-b Thus 1 = ambn = cm,, (4 (1l-a)(1 - ) m,n=O where m, n range over all lattice points in the first quadrant, including those which lie on the x and y axes. POWER SERIES 79 4. In studying the double series 1) it is often convenient to suppose the terms a,, placed at the lattice points m, n. From this point of view any simple series A= am may be converted into a double series as follows. Choose an infinite lattice set V at pleasure, e.g. the points in the first quadrant. Put each term am at some lattice point r, s, so that each point of 8 bears one term of A. If am lies at the point r, s, we may denote it by b,,, so that brs is only another symbol for am. In this way we get a double series B= br,,,. 5. Example. Let a, b be any two complex numbers, such that the three points 0, a, b are not collinear. If m, n range over all lattice points, the origin excluded, ma +nb will be the vertices of a set of congruent parallelograms, as in the figure, which completely cover the plane. The series A (ma + nb)P (5 is important in the elliptic functions and will be employed later. We now establish the theorem: The series 5) converges when p > 2 and diverges when p < 2. For brevity let us set. mn = ma + nb. The adjoint of 5) is thus / 3l=S 1-. (6 Then by definition 5) and 6) converge simultaneously. 80 FUNCTIONS OF A COMPLEX VARIABLE We replace 6) by the simple series where 8i denotes the sum of the terms of 6) whose indices m, n correspond to points on the first parallelogram P1 whose center is the origin O; 12 is the sum of the terms of 6) whose indices lie on the second parallelogram P2 about 0, etc. Let d and D be the least and greatest distances of the sides of P1 from 0. Then each of the 8 numbers om,n which lie on P1 satisfy the relation d < l amn I < D. Similarly each of the 2.8 numbers Womn which lie on P2 satisfy the relation 2 d < I CI <2 D, etc. Thus 8 </ < 8 DP-1 d' 2.8 2.8 2 <,etc. (2D)P- 2(2 d) Thus -< < - d l (sD)P~ - (sd) or 8 1 3 8 1 VDP SP — 'dp S-i As 1 Xp-1 converges when p > 2 and diverges when p < 2, the theorem is proved. 42. Row and Column Series. 1. Let us consider the double series A = all - a12 + a1... + a21 + a22 + a23 + +..... (1 = Tam,n m, n = 1, 2,... The mth row of A gives a series co rm = =amn = aml + aCn2 + '* n=1 (2 POWERE SERIES 81 ani from these we caii form a series Rz=ri + r2 +. =r (3 rn=] = 0 0! am, n, m=1 n=1 putting in the value of r,, as given by 2). We say the series 1 is obtained by summing 1) by rows. Similarly the nth Colunl, of A gives a series cn I amn= al. + a2n + (4 mn=1 annd f rom these we can form a series (= c+c2+..2 =+ c h nn=i =~ Ca~m (5 n2=1 rn==1 We say the series C is obtainet by summingq 1) by columns. 2. To sum a double series A which is known to be convergent, we may often use the following theorem with advantage If A is convergent, each series rm, c. is absolutely convergent. The series R and (I both converge absolutely and A= R=C. (6 For let -B = bl + bb2+.. = 2Yb8 (7 be one of the simple series associated wvith the double series A. Since A is convergent, -B converges absolutely by definition, and A = -B. Let a,,,,, = ar The adjoint of A is = CC,, Let us denote the series formed from W analogous to rM Cm,, I C! b, by To show that rm is absolutely convergent we observe that we can take s so large that each term of pm, p m Cem, 1 i+an, 21 + am,P 82 FUNCTIONS OF A COMPLEX VARIABLE lies in QO8. Hence PM~p < Q08 ~<~ Thus PM = urn PM, p < S8. Hence the series Pm is convergent and rm is therefore absolutely convergent. The same holds for c~. To show that 1? is absolutely convergent let us consider But = lim P', = liM (aml + am2 + + am.). If theref ore we set Wfm, n= ( 11+ C12 + + aln + (21~+ 22~ ~L+ 2n + (Xml + am +.. amn9 we get lirn W~mn =9. and hence 9?= Irn li11n 2fm, n. Now let us take s so large that each term of W., n lies in 23.Then 2fn~Q:~ 0s< Q0. Passing to the limit n = ocx gives Passing to the limit m = oo 'shows that 9Y? < Q3. Hence T is convergent. As each ~ M PM, we see that B is absolutely convergent. To show that BR = A we begin by taking s so large that POWER SERIES 83 Next we take p, q so large that every term of f not in 9p, a lies in ~. Then 3- Wmn contains only terms of 33 when m > p, n > q. Thus - 2'fmn < E. Now the numerical value of B - Amn is < the sum of the numerical values of the terms of this series. Thus B - Amn <i - mn < e. Letting now n o, we get IB - Rm <. Letting m co, we get B- B <c. As e > 0 is small at pleasure, this gives B = R... A = R. Q.E.D. Similarly we show that C is absolutely convergent and A = C. 3. Let us now show conversely: If the 9? or the ( series converges, A is convergent. Let us suppose that 9 is convergent. Taking s at pleasure, we may take m, n so large that the terms of 5, lie in Sn. Hence ~8 < 3Ln < 9m < 9. Thus 3 is convergent by 13, 2. Hence A is convergent by definition. 4. The following example will show that double series cannot be treated as if they were finite sums. They are the limits of such sums and often illustrate the fact that what is true of the variable is not necessarily true of the limit. Consider the series a2 a3 A=l-a+a a+ *+ 2!- 3! +1 12-2a+ (2 a)2 (2 a)+ 2! 3! + 1-3a+ ( a) (3 a)+... 2!where a > 03! + ~ where a > O. 84 FUNCTIONS OF A COMPLEX VARIABLE The mth row has here the sum r = e-ma. Thus summing A by rows we get R = rl + r2 + = e-a + e-2a + e-3a +.' This is a geometric series and converges absolutely since a > 0. We cannot infer, however, that A is convergent or that if it were its sum = R. In fact A is divergent. For if it were convergent each ec series must be convergent by 2. This is not so, for c= 1 + 1 +1 +... is divergent. 43. Application to Power Series. We wish to apply the foregoing theorem to obtain a result which we shall use later. Let the power series - P(z) = a0 + alZ + a2z2+.. (1 have ( as a circle of convergence. About any point z within 6 let us describe a circle c of radius p which also lies within G. The point z + h will lie in c if / I h < p. Hence the series 1) converges absolutely when we replace z by z + h; that is o P(z + h) = ao + a1(z + h) + a2(z + )2+... (2 is an absolutely convergent series. Let us expand the terms of 2) and write the result as a double series. We get A= ao+ 0 + +0 0+... + a1z + alh + 0 + 0 +... + a2z2 + 2 a2zh + a2h2 + 0 +.. + a3z3 + 3 a3z2h + 3 a3zh2 + h3 +.. If we sum 3) by rows, we get the absolutely convergent series 2). From this we cannot infer that 3) is convergent as we saw in 42, 4 POWER SERIES 85 The series A is, however, convergent, as we may easily see as follows. Let us set I z = r. Then 1) converges absolutely for z = r + p since this point lies within (. Thus % + Ya(r + p) + Y2(r + p)2 + * (4 (4) = ao + air + alP+ a2r2 + 2 a2rp + a2p2 + *.* is convergent. Thus the simple series B = ao + a1z + a1h + a1z2 + 2 a2zh + a2h2 + *.. is absolutely convergent since each of its terms is numerically > the corresponding term of 4). Thus A is convergent and we can sum it by rows or by columns. Summing by rows gives A= P(z+h). Summing by columns we get, since the result is the same as before, P(x + A) = P(z) + hPl(z) + 2 h2P2(z + 1h3Pz) +. (5 where P1 = a1 + 2 a2z + 3 a3z +... P2= 2a2 + 2. 3 a3 + 3 4 a4z2 + CHAPTER IV THE ELEMENTARY FUNCTIONS 44. 1. The functions employed in elementary mathematics are the following: Integral rational functions. Exponential functions. Rational functions. Hyperbolic functions. Algebraic functions. Inverse circular functions. Circular functions. Logarithmic functions. Except in case of the algebraic functions the independent variable x is usually real. We propose in this chapter to define these functions for complex values of the variable, and to study a few of their simplest and most useful properties. 2. The reader is perfectly familiar with all these functions, the variable being real, except possibly the hyperbolic functions. For such as have not used these functions in the calculus we add the following. They are defined by eX + e- eX - e-~X cosh x = ex, sinhx= ex- - * (1 2 2 The left sides are read " hyperbolic cosine of x " and " hyperbolic sine of x." We see that they are merely linear combinations of ex and e-x. These functions have been computed and tabulated, so that one is as free to use them as sin x, cos x. The reader is referred for example to B. O. Peirce, A Short Table of Integrals. Ginn and Company, Boston, 1899. Jahnke and Emde. Funktionentafeln. Teubner, Leipzig, 1909. 86 THE ELEMENTARY FUNCTIONS 87 By means of these tables we may draw the graph of these functions which we herewith give. The graph of cosh x is the familiar catenary, that is, the form of a chain supported at y y two points on the same level. It is important to note that: cosh x never vanishes, while sinh x vanishes just x~ x 0 ~ once, viz: for x = 0. y=cos hx Y /sin hx We note also that cosh x is symmetric with respect to the y-axis, and sinh x with respect to the origin. If we express ex, e-$ as series we find x2 x4 cosh x=l1+ 2 +4+** (2 2! 4! x X3 X5 sinh x= - + 3+!+ (3 From 1) we find at once that cosh2 x - sinh2 x = 1. (4 45. The Integral Rational Function. These functions have the form a0 + az + a22 ~... + amzm, (1 where the coefficients ao, a1, *.. am are any given complex numbers and the independent variable z is free to take on all complex values, or as we say is free to range over the whole z-plane. The exponent m is an integer > 0, and is called the degree of 1). Such functions are called polynomials in algebra. Since 1) involves only the operations of addition and multiplication of complex numbers, its value can be calculated for any given value of z. In algebra we learn that 1) vanishes for just m values of z, say Z1, Z2 "' * mZ (2 88 FUNCTIONS OF A COMPLEX VARIABLE some of which however may be equal. We call 2) the roots or zeros of 1). Knowing the roots 2) and denoting the expression 1) by w, it is shown in algebra that w = am(z - Z) (z - Z2) "' (Z - Z). (3 The theorem which states that: Every polynomial of degree m has m roots is called the fundamental theorem of algebra. As often given in algebras, its demonstration is long and difficult. Few students really comprehend it. It is a luxury which most students are willing to dispense with. And quite rightly, for it is far beyond their powers at that time. Later we shall give two proofs of this important theorem which the student will comprehend; one is so simple that he will not need to set pen to paper to follow it. 46. Rational Functions. The quotient of two integral rational functions of z is a rational function. Such functions have the form ao 4- + a1z.. - amz" (1 bo + blz +. * - b,zn' where the coefficients a0, al... bo, b1 *.. are constants and m, n are integers > 0. The expression 1) involves division by 0 for those values of z for which the denominator vanishes. Let these be these be z1, z2... z,. (2 For any value of the complex variable z not included in 2) the value of the expression 1) may be computed by rational operations. These values of z form the domain of definition of the expression 1). We may thus state: The domain of definition of a rational function of z is the whole z-plane excepting the zeros of the denominator. The degree of 1) is the greater of the two exponents m, n, supposing of course that a,, bn are = 0. When 1) is of the first degree, it is said to be linear. The type of a linear rational function of z is therefore a + bz c + dz THE ELEMENTARY FUNCTIONS 89 The rational functions include the integral rational functions as a subclass. In fact let the numerator of 1) be exactly divisible by the denominator; then 1) reduces to a polynomial. This takes place in particular when the denominator reduces to a constant, that is when n = 0. 47. Algebraic Functions. 1. We say w is an algebraic function of z when it satisfies an equation of the type wn + Rx(z)wun- + R2(z)w.n-2 + * + Rn(z) = 0, (1 where the coefficients Rr, R2... are given rational functions of z, and n is a positive integer. The degree of w is n. Let us give to z a definite value, real or complex, say z = a. If a is a zero for one of the denominators of the coefficients, we shall say that the point corresponding to this value of a is an exceptional point. Obviously 1) has no meaning at such a point. Suppose now that a is not such a point. Then all the coefficients in 1) can be calculated, and 1) reduces to an equation with constant coefficients. But such an equation admits n roots, which in general are unequal, w,.. w. (2 Thus the equation 1) defines an n-valued function w of z for all values of z not included among the exceptional points of the coefficients. These values of z, or, using our geometric language, the points in the z-plane corresponding to these non-exceptional points, constitute the domain of definition of the algebraic function w. The number of exceptional points is finite. For the highest degree of any coefficient R1, R2... being say h, no coefficient has more than h exceptional points. As there are only n coefficients, there are at most hn exceptional points. Hence: The domain of definition of an algebraic function of z embraces the whole z-plane, excepting possibly a finite number of points. 2. Let us note in passing that the class of algebraic functions embraces the rational functions as a subclass. For let n = 1 in 1); it reduces to w + Rl(z) = 0, or w = - 1(z), a rational function. 90 FUNCTIONS OF A COMPLEX VARIABLE 3. We have said that the roots 2), are in general unequal. Let us denote the equation 1) by F(w, z)= 0. If we eliminate w from the two relations = 0, =0 Ezc we will get an algebraic equation in z, say G(z) = 0, (3 of degree m, let us say. It is shown now in algebra that the roots 2) are all distinct at a point z = c when c is not a root of 3). We may call the roots of 3) critical points; they are finite in number. The exceptional and critical points together may be called singular points. All the other points may be called ordinary. 48. Explicit Algebraic Functions. 1. The two roots of the algebraic equation Wbraic eq n 2 + R1()W + R2(z) = 0, (1 where R1, R2 are rational functions of z, are w=- R1 ri ~ — V-R_ 4 R2. (2 Since 2) satisfies 1), it is an algebraic function of z. To calculate this function we have to perform, besides the rational operations, the operation of extracting a square root of a known quantity. 2. The three roots of the algebraic equation w3 + R1(z)w + R(z) = 0, (3 where R1, R2 are rational functions of z, are given by w -\- K2 + R2 -22 + - 13, + -- I~ R V R22 + 2-1 R,1 (4 Since 4) satisfies 3) it is an algebraic function of z. The right side of 4) exhibits this function by means of roots of quantities which can be successively calculated by rational operations. We say 2) and 4) are explicit algebraic functions of z. In general we say w is an explicit algebraic function of 2) when its expression involves the extraction of roots of rational functions of z, or the extraction of roots of such roots, or the THE ELEMENTARY FUNCTIONS 91 rational operations on such roots, each operation performed only a finite number of times. This definition is clumsy, but its idea is very simple. The expressions 2) and 4) will serve as illustrations. 3. It is shown in algebra that every explicit algebraic function of z is a root of an equation of the type 47, 1). The demonstration is simple but will not be given here. On the other hand it is not true that every algebraic function of z is an explicit algebraic function. The demonstration of this fact is anything but simple. The solution of the cubic and biquadratic equations was effected by the Italian algebraists in the first half of the sixteenth century. The algebraic solution of the quintic was then sought. This became one of the celebrated problems of the seventeenth and eighteenth centuries. The greatest mathematicians of their time sought its solution, but in vain. At last Abel in 1826 demonstrated that the roots of the general equation of the fifth and higher degrees cannot be expressed as explicit algebraic functions of their coefficients, in other words that these equations do not admit an algebraic solution. If then the roots of the general equation of fifth degree cannot be expressed in terms of radicals, in terms of what functions can they be expressed? In 1858 the illustrious French mathematician Hermite showed that these roots may be expressed in terms of the elliptic modular functions. This will be referred to again when we take up the study of elliptic functions. 49. Study of Vz. 1. Let z = r (cos + i sin 0), then as we saw in 7, 3 the two values of Vz are wl=V/r cos + i sin), W2= —w. (1 If we let z describe a curve in the z-plane, the two roots wr, %w2 will describe curves in the w-plane. For example, let z describe a circle @ of radius r as in Fig. 1. When z is at A, =0, hence wl = r, w = -Vr. Thus w is at a and w2 is at 8. Let z describe the quadrant AB. Then f increases from 0~ to 90~ while r remains constant. From 1) we see that the argument - p of w1 increases only half as fast while 92 FUNCTIONS OF A COMPLEX VARIABLE its modulus /r remains constant. Thus when z describes the quadrant AB, w1 describes the octant a/3. Since w2 =- w1, we see that at the same time w2 describes the octant 8e. Let now z describe the second quadrant BC. At C, 1= 180~, hence I q = 90. ^'B w plane Z plane aC r A ( o ) D FIG. 1. Thus when z has reached C, w1 has arrived at 7, while w2 has got to g. Continuing, we see that when z has gone all around the circle, = 360~, hence - - = 180~, and hence wl is at 8. Meantime w2 has moved from g to a. Now when we began, wl was at a, and w2 at 8. After the circuit, w1 is at 8 and zv2 at a. The two roots have been interchanged. 2. We will now let z describe any closed curve S about the origin as in Fig. 2. To any point P on t whose polar coordinates are r, b will correspond a value of w1 given by 1), and the point in the w-plane corresponding to this value w1 has the polar coordinates /r, 2- b. K Bi 0 Al FIG. 2. FIG. 3. As z describes + starting from A, 0 will increase steadily from = 0~ to ( = 360~ when z will have returned to A. The modulus r varies continuously from its original value, say r = a, sometimes increasing, sometimes decreasing, but finally returning to its original value a. From this we see that the argument I + of w1 will increase steadily from 0~ to 180~ while the modulus /Vr THE ELEMENTARY FUNCTIONS 93 will vary continuously. Thus w1 will describe a continuous curve as in Fig. 3, whose end points Al, B1 lie on the real axis at the distance /a from the origin in the w-plane. Moreover, as w2= - wl we see that w2 will describe a symmetric curve on the other side of the origin. Hence when z has completed its circuit about the origin of the z-plane, wl and w2 have interchanged positions in the w-plane. Thus we see that this more general case behaves in an entirely analogous manner to the simple case considered in 1. The main facts to remember are these: 1~ When z describes a circuit about the origin, w1 and w2 each describes a curve, but not a closed curve. 2~ At the end of the circuit, w1 and w2 have interchanged values. In other words, a circuit about the origin effects the substitution Wl W2. W2 w1 For this reason the point z = 0 is called a branch point. 3. Let now z describe a closed curve, EF as in Fig. 4, which does not enclose the origin 0. Let the polar coordinates of A B C D E c be a a; b, 3; c, 7; d, 8; e, e. The value of w1 at the point A is thus, by 1), FIG. 4. W1 = Va (cos. a + i sin 1 a). The point Al in Fig. 5, corresponding to this, has the polar coordinates Va, g a. Let now z move along the D1 arc ABC.... Its argument 4 steadily 0 1i increases till the radius vector becomes Al B, tangent to the curve at E; that is k FIG. 5. increases from = a to e = c. Thus the argument lq of w1 steadily increases from 2 a to 2 e, as w1 moves from Al to E1 in Fig. 5. The modulus r of z increases from r = a to r = d as it moves from A to a point D in Fig. 4, when it begins to decrease. 94 FUNCTIONS OF A COMPLEX VARIABLES Similarly, if to fix the ideas we suppose r > 1, the modulus Vr of w1 will increase from Va to V/d as w1 moves from Al to a point D1 in Fig. 5. As z moves from E back to A and so completes the circuit, f decreases from e back to its original value a. The modulus r also assumes its original value r = a, at the close of the circuit. Thus w1 also comes back to its point of departure A1. Hence: When z describes a closed curve not including the origin, w1 describes a closed curve also. Since w2 =- w1 we see that w2 will describe a symmetric closed curve on the other side of the origin. 4. When z describes a circuit about z = 0 we have seen that the two values of w = Vz are permuted; on the other hand, we have seen that if z describes a circuit about any other point, which does not include the origin, the two values of w are unaltered at the end of this circuit..We therefore say that z= 0 is the only branch point of w. 50. Study of w = (z - a)(z -b). 1. Let z - a = a (cos 0 + i sil 0), z - b =, (cos + i sinll ). Let uLet =/ (cos 0 + i sin - 0), u2= — u (1 1 = V\/(cos i+ sin ), =-v1. (2 Then the two values of w are W1 1 =l1 W2 = - ulV1. We note that the expressions 1), z 2) defining ul, v1 have precisely the same form as that defining w1 in 49. a From this it follows that when z describes a circuit 3I about a, u1 will go over into us = - u. On the o other hand the curve ~2 lies outside of b, and hence by 49, 3, when z describes Af, v1 returns to its original value at the close. Thus v1 is unchanged. Hence the effect of the circuit 2 on wl is to change THE ELEMENTARY FUNCTIONS 95 it into - ulv1 or w2. As w2 = - w, we see that the same circuit converts w2 into wl. Thus a circuit of z about a effects the substitution (w W2) W2 W^u Similarly if z describes a circuit B about b, but excluding a, the roots w, w2 are also permuted. For this reason the points z = a, z = b are called branch points. The same reasoning shows that if z describes a circuit about a or b, but in the opposite direction, the two roots w1 w2 are permuted also. Next let z describe a closed curve d which does not include either a or b. Then 49 shows that both u,, v1 return after the circuit to their original values and hence wl = ulv1 also returns to its original value. 2. Let z move from c to d over the path V. If it describes the same path in the opposite direction, i.e. from d to c, we may denote it by V-1. Let 9J denote the other path from c to d as in the figure. Then 9?-1 will denote the closed Yjd curve from c over V to d and back to c over the curve ). c At each point z, the algebraic function w = V(z - a)(z - b) has two values. The values of w for z = let us call y and - y. When z ranges over the curve V, the different values that w has group themselves into two curves or branches which we may call L1 and L2. An end point of one of these curves, say L1, is y. Let 8 be the other end point of L1. If V does not pass through one of the branch points z = a, z = b, the two values that w has for each value of z are distinct. In this case the two branches L1, L2 have no point in common. We may thus distinguish the two branches L1, L2 by giving one of their points. Thus the branch L1 is determined by the fact that it passes through y, or through 8. Suppose now we allow z to range from c to d over V. If we start with w = y, what value will we have when z reaches d if, as 96 FUNCTIONS OF A COMPLEX VARIABLE we pass over 8, we always choose that determination of w which will form a continuous set of values? Now at z = d, w has two values w = 8, and w = - 8. From the foregoing the value we must choose is obviously w= S. Let us indicate this by the notation 'y = $, whereby we mean that if we start from z = c with that value of w which is = y and allow z to range over the path 8, the value that w has at the other end of 8 is 8. Suppose that we next allow z to move from c to d over the path 1J. We prove now the important fact: If 8, 9N do not pass through or enclose a branch point, 72 = 7= a. In other words, if we start with the same determination of w at z = c we arrive at the same value of w at z = d, whatever path we choose, provided no two paths pass through or enclose a branch point. The proof is very simple. For by 1, 7W-1 = 7. Hence 'V-19w= y7. But m2-W = 72. Hence t72= 'Y 3. Suppose we start at z = c with the determination of w = y. We allow z to describe the circle ( as in the figure. We ask what is the value of w when z returns to c? As w has only two values at c we have 7,y,= y or Y,= —7y. To determine which, we introduce the paths de and ce. Then 'Y(S = cd * dc ec' As the two paths de and de ce do not include a branch point, 7Y( = 7cd d de ce * ec' THE ELEMENTARY FUNCTIONS 97 Now ed. dc is a circuit about the branchpoint z = a. Hence Y/cd -de = - YI a and thus = Also ce. ec is a circuit about the 0,\ branch point z = 6. Hence (- rY)ce-ec = 7. Thus finally 'Y_ - 51. The Elliptic Radicals. 1. When we take up the elliptic functions, we shall find that two radicals w =V4(z- e1)(Z - e)(Z- e3) (1 and w =-\(1 - Z2)(1k- k2Z2) (2 figure very prominently. Let us consider first 1). If we set Z - em = rm(cos On + i sin Om) and Am'b(Cos Om +iSil Om)2,3, IUM rM 2 2 "/ the two values of w in 1) are W1= U1UU3, W2=-W1. Now the radicals Urn =.V —e have the same form as those considered in 50 and we may therefore conclude at once The branch points of 1) are el, e2, e5. When z describes a circuit about one of them, w, and w2 are interchanged. A circuit which includes two of the branch points leaves w1, W2 unchanged; a circuit which includes all three branch points interchanges w,, wT. Pinally a circuit which includes none of the branch points leaves wl, w2 unaltered. 2. Let us now turn to the radical 2). Since (1- Z2)(1 - 1 k)= 1(- 1)(Z+1)(Z- + k we set z-1=rl(cos61~isinl) x z+1=r2(cosO02 + i sin.O2), 1 1 z- I=r(cos 3+ isinO3), +-= r4(cos 4+ i sin 04). k 3(I 3+c il0) Z CS0 98 FUNCTIONS OF A COMPLEX VARIABLE Finally we set Urn Vr uM r,,(COS -1 0,,$ i sinl 1 0,) Then the two values of w in 2) are w1= ku 1u2u3u4, W2=-W1. From this we conclude: M = 11,2, 3, 4. The branch points of the radical 2) are 1, - 1 -. When x describes a circuit which includes no branch point or an even number of them, the values of w1, w2 are each the same at the end of the circuit as at the start. If the circuit contains an odd number of the branch points, the values of wl, u'2 are interchanged after the circuit. 52. Study of w = -a* The method we adopt to study this radical is the same as that employed in 49, 50, and 51. We set z-a=a(cos6~isin6), z-b=fl(cos0~isin0) and introduce the cube root of unity ocos + i sin2. 3 3 We also set = 3 Cos~i sin, u (DoU, 2 v1= /'8(cos + isin) v2- =wv1 v =3=o2V Then 3t 2 1 3w Then the three values of wr are u w 1 1 1 (1 Let now z describe a circle C about Z = a, as in the figure. Then 9 increases from say 0 = t to 9 = t + 2 7r while c returns to its original value. At the beginning of the circuit u = -xvz(cos t1 + i sint) 1 t. (2 1 3 1V b6" 0 ' z 0 CL at 0(I THE ELEMENTARY FUNCTIONS 99 At the end of the circuit, u,1 has acquired the value = WUU1. Thus (u2) C = 0(u1) C = (A2 = U3, (u3)C = -w(2 U)c a 3u = UThe effect of C! is to convert U1, U2, U3 into U2,, It1. As vI remains unaltered by the circuit C, the relation 1) shows that after the circuit Wi, W2, 3 go over into 2, W3,, The circuit thus effects the substitution Ww " W A(1 w2 ) W32 W3 1 Let z now describe a circle D about z = b in the positive direction. The same considerations sbow that ~ increases from say p = p to 0 =p + 2 7. On the other hand, 0 returns to its original value. Thus at the beginning of the circuit D, =V (cos 1p ~ i sin 1p), At the end of D, v1 has acquired the value (vl)D = V/3(cos 1 (p + 277) + i sin 1-(p + 27 )) = (0V1. Similarly ( (V2)D = 6OV2 = V3;V3 = WCV3 = V1. As u1 remains unaltered by the circuit D we have () =()D == 2=2 - =60 Lil= 6 W = W3' D W1 V 100 FUNCTIONS OF A COMPLEX VARIABLE Similarly w2 goes over into w1 and w3 into w2 after D. Thus the circuit D effects the substitution B=l w2 W3) W3 W1 W2 We notice A2= B; that is, going around z = a twice in the positive direction produces the substitution B. If we go around a three times, w1 w2 w3 take on their original values, or A- (W1 W2 W3) wU1 2 7W3 A substitution which effects no change in the roots is called the identical substitution and is denoted by 1. Thus A3= 1. Let us now see what happens when z describes the circuit C in the negative direction. This path according to our agreement is represented by C-1. In this case 0 decreases from say 0 = t to 0 = t - 2. At the end of the circuit C-1, uz as given by 2) has acquired the value (u=) c-_ = -(cos l(t - 2 7r) + i sin -(t - 2 r)) = C-1U= o 02U = -U3. Similarly (u2)c_- = l and (u3) - = u2 As vi is unaltered by this circuit, we see that C-1 produces the substitution A-1 (wl w2 W3 B. w3 W1 w22 Since the circuit C-1 just undoes what C does, we should have AA-1 = 1 and this we see is indeed so. Similarly the circuit D-1 produces the substitution B-' = A. We notice thatA, B combine as products. 53. 1. One and Many Valued Functions. The integral rational function, w = aO + alZ +... + amZm, assigns for each value of z, a single value to w. It is a one-valued function of z. Let w= b for z = a. If we allow z to describe THE ELEMENTARY FUNCTIONS 101:>ome curve in the z-plane returning to its point of departure z = a, w will describe a curve in the w-plane which starts from w = b, and returns to this point. 2. On the other hand the function w = Vz - a assigns to w two values for each value of z except z = a where w has only one value w = 0. This function is a two-valued function. For a similar reason s a three-valed function. is a three-valued function. 3. Since the equation wn + Rlwn-1 +... + Rn = 0 (1 considered in 47 has in general n distinct roots, the function w of z defined by this relation is called an n-valued function. Let z= a be an ordinary point [47, 3]. If now z describes a curve C which does not pass through a singular point, the n values which w has at each point of C can be grouped together z=a z=b =a Z plane w plane so as to form n curves or branches. If w = a is one of the roots of 1) for z = a, one and only one of these branches will pass through the point w = c. It thus serves to characterize this branch. Now when dealing with many-valued functions we very often have to solve this problem: We take one of the values as w = a which w has at the ordinary point z= a and ask what value of w do we get when z describes some curve C leading to z = b and avoiding all singular 102 FUNCTIONS OF A COMPLEX VARIABLE points. The many values which w has for the points of C will be distributed over certain continuous curves or branches and the value of w we always want is that value of w for z = b which lies on the branch passing through the point w = a. This general problem we have studied in several simple cases in 49-52. If this value of w is w = /3, we say that branch of w which takes on the value w = a for z = a, has the value w = / at z = b, when z describes the curve C. We have used the notation to denote this fact. If z describes some other curve D not passing through a singular point and leading from z= a to z = b, the end value of this branch will not be in general /. In any case it must be one of the many values which w takes on for z = b. The Exponential Function 54. Addition Theorem. 1. In the foregoing articles we have considered the algebraic functions which include as special cases the rational and the integral rational functions. All functions which are not algebraic are called transcendental. The first such function we shall study is the exponential function. It is defined by the series Z z2 Z3 +j- a+ - + 2 + (1 1! 2! 3I In 39, 4 we saw that this converges absolutely for every z. Thus it defines a function of z which is denoted by the symbol ez or Exp z. The domain of definition of this function is the whole z-plane. When z has a real value x, 1) reduces to the well-known exponential function 2 3 x x2 x3 1! 2! 3! studied in algebra and the calculus. A most important property of ex is the addition theorem, as it is called, viz. ezey = e.+y THE EXPONENTIAL FUNCTION 103 Let us show that the relation holds for complex values. Let u, v be complex numbers. Then by definition U U2 3 eu _= - +. - + +... 1! 2! 3! V V2 V3 ev = 1 + -- + v, +-v +... If we multiply these absolutely convergent series, as shown in 33, 2, we get euev - 1 + u + + + ~ l^ ll! l! 1I 1 2!y /U3 LU2 V 1 V2 V3 3- 2 1721 32y1 + (+ ++1! (+ 3+...! 3! = e Thus euev = eu+v (2 holds for any complex numbers u v. 2. From the addition theorem we can show how to calculate ez for any z = x + iy, x, y real by using our ordinary logarithmic tables. We have in fact ez = e+iy - exeiy. ( Now i ++ 3!i + L+ -.... 1! 2! 3! 2! 4! Hence ei = cos y + i sin y. (4 Thus from 3) we have ez ex(co y + isin y (5 ezx e(cos y + i sin ) (5 104 FUNCTIONS OF A COMPLEX VARIABLE The relation 5) is an expression of the complex number ez, which is nothing but the sum of the series 1), in its polar form r(cos 0 + i sin 0). Thus the modulus of ez is ex, and its argument is y. In symbols i I ez = ex, Arg e= y. (6 Thus to plot the value of w = ez = exei we first describe a circle of unit radius about the origin z of the w-plane and then lay off on this an arc of length = y as in the figure. Y \ On a radius through the end of this arc we now lay - o X off a length = ex. The end point of this segment is w. z plane w plane To calculate ez by the tables we first compute r = ex. Then we convert the arc y into degrees, getting an angle 0. Then w = ez = r (cos 0 + i sin 0). (7 If we wish to reduce w to the rectangular form w = u + iv we have, comparing with 7), u = r cos, v=rsin. (8 Let us note in passing the important theorem: The exponentialfunction e vanishes for no value of z. For ez cannot = 0 unless its modulus ex = 0. But ex vanishes for no real x. 3. As an example let us compute w = ez for z =- 1.6- 2. 8 i. Here x =-1.6, y=- 2.8. Hence Hence r = ex= e-1-6 Let us set =. then, -. 0., log s = 1.0 log e = ', say. THE EXPONENTIAL FUNCTION 105 Here the symbol log stands for the logarithm whose base is 10. log e = 0.43429 log (log e) = 9.63778 log 1.6 = 0.20412 log t = 9.84190 log s = 0.69487 log r = 9.30513, r =.20190 We now convert y into degrees. From our tables we have 2r.a = 114~ 35' 30" 0.8rad = 45 50 12 Hence 0 = - 160~ 25' 42", or adding 3600 0 = 1990 34' 18"/ To simplify our calculation let us take 0 = 199~ 34' log (- cos ) = 9.97417 log (- sin 0)= 9.52492 log r = 9.30513 log (-u) = 9.27930 u = -.19024 log (- v) = 8.83005 v =.067616 As a check for our work we should have tan 0 = -. U But ~~But ~logv = 9.55075 U or 0 = 199~ 34' as before. As a final result we have therefore w =-.19024-.06762 i w =.20190, Argw = 199~ 34'. 4. Since the function ex for real x often occurs in calculations, tables of this function have been prepared. We mention those of B. O. Peirce and those of Jahnke and Emde already referred to in 44. 106 FUNCTIONS OF A COMPLEX VARIABLE From Peirce's Tables, p. 114, we have log e-1 = 9.56571 log e —6 = 9.73942. Hence log e-1-6 = 9.30513, e-6 =.20190, which agrees with its value obtained in 3. In the Tables of Jahnke and Emde we may take out the value of e-1'6 directly from the table on p. 6. 55. 1. Euler's Formula. Let us suppose the real angle q is expressed in circular measure. Then 54, 4), gives Euler's celebrated formula... i e~ = cos ~ + i sin 4, (1 which we referred to in 1, 3). The point in the z-plane associated with this number lies on the unit circle, i.e. a circle about the origin of unit radius. It lies on the radius making an angle q with the real axis. Since every complex number can be expressed in polar form z = r (cos ) + i sin q), (2 we have, using 1) =r (3 We call this the exponential form of z. Thus we have three ways of representing a complex number: the rectangular form x + iy, the polar form given by 2), and the exponential form given by 3). Each way of expressing z is useful at times. From 1) we have 7ri.7r e~=i, e = - 1 e2 =-i, e2i = 1. (4 The n roots of unity are represented by '2 ri '2 7ri 2 i 2 iri 2. *;-.2 (n-l)2 1,, e e, e, (5 Th-e n roots of a = r(cos 0 + i sin 0) <IrQ ^s /0_L 27\ e + 2(n I \ ~are ~ ~n - - (67ere n (6, ro = Vr' +(fl) ) (6 Or if ro is the first imaginary nth root in 5).0 =z0 e7 = /Z,... Z 1= n-lz. (7 THE EXPONENTIAL FUNCTION 107 2. In plane trigonometry the two formulae cos ( + )=cos 0cos - sin sin4 (8 sin(O + )= sin O cos + cos sin (9 are of fundamental importance. They express the addition theorem of the cosine and sine functions. Let us show how they may be derived from Euler's formula. From 1) we have e- c - i sin (10 From 1), 10), we have, adding and subtracting, cos = ~(e4 e-0) (11 and sin 1 = (ei~ - e (12 2 i These last two formulae expressing the cosine and sine as exponentials are often useful. Let us now multiply 1) by e = cos 6 + i sin 9. We get ei-ei = ei(0+0) - cos (0 + +)~ i sin (0 + ~) (13 = cos 0 cos 0 - sin 0 sin 0 + i(cos 0 sin ~ + sin 0 cos Equating the real and imaginary parts of tbis equation gives 8) and 9) at once. 3. Let us show how the powers of sin 0, cos 0 may be expressed in terms of the sine and cosine of multiple angles. To this end we set ue, Then 11) and 12) give (2 cos 0)nt = (u + v)lfl = + (n 2 ( + = (UM~ + v4) ~ (?)UM72 ~ VZn2 )UV ~(M (um-4 + vm-4 U2V2... (14 Now _ Tmy = etke- 1. Also us?, + vm = em4~ + e-4i~ = 2 eos mo. 108 FUNCTIONS OF A COMPLEX VARIABLE Thus 14) gives for m = 2, 3,... 2 cos2 4 = cos 2 + 1 4 cos3 = = cos 3 ) + 3 cos O (15 8 cos4 = cos 4 + 4 cos 2 + 3 Similarly (2 i sin O)m = (U - V)" gives -2 sin2 = cos 2 - 1 - 4 sin3 4 = sin 3 4 -3 sin 4 + 8 sin4 = cos 4 -4 cos 2 +3 (16 56. Period of ez. We are familiar with the fact that sin (x + 2 7r) = sin x for aly real x. We say sin x is periodic and has the period 2 tr. It is easy to show that ez admits the period 2 7ri, that is, ez+2n7i = ez for all values of z. For ez+2ri = ez e2ri by 54, 3) = ez. In the same way we see that ez admits 2 m7ri as period, where m is an integer. We say any complex number a is a period of ez when ez+a = e (1 holds for all values of z. Let us show that ez has only the periods 2 m7ri just given. For let = a +ib be a period. Then 1), holding for every value of z, will hold for z = 0, and we have ea e = 1, or putting in the value of a ea+ib = eaeib. Thus ea = 1, b = 2k7r, k an integer. Hence a= 0. Thus any period must have the form 2 kri. But these we have already seen are periods. We call 2 wri the primitive period of ez, since all its periods can be expressed as multiples of this period. THE EXPONENTIAL FUNCTION 109 57. 1. Graphical Study of ez. In the calculus the student has become thoroughly familiar with the notion of the graph of a function, and has seen on many occasions how useful it may be. In the function theory of a complex variable, the graph of a function is also most serviceable at times. Let us study the graph of w = ez = exeiy = ex(cos y + i sin y) (1 Z plane w plane We have seen that when the variable z describes some curve C in the z-plane w will describe a curve ( in the w-plane, and we call ( the image of C. Let z range over a line parallel to the x-axis. Then z = x + ib, where x ranges from - oo to + oo, and b is constant. Let us call this parallel I. The value of w corresponding to such a z is, by 1), W = exeib As ~~As lwl=ex, Arg. w= b, as moreover lim ex = 0, lim ex = + 0, a==-_0 3=+oO we see that w describes a straight line or ray r issuing from w= 0, and making the angle b with the real axis in the w-plane. Thus to each point on I corresponds some point on r. As b increases from 0 to 2 r, that is, as I moves parallel to itself through the distance 2 wr, the corresponding r rotates through an angle 2 7r. Let z now range over a parallel X to the y-axis. Then z = a + iy, where a is constant and y ranges from -ao to + o. From 1) 110 FUNCTIONS OF A COMPLEX VARIABLE the value of w corresponding to such a z is w = eaeiy As w= ea, Arg. w=y we see that the corresponding points in the w-plane lie on a circle c. When z moves over a segment of length 27r on the line X, y moves over an angle of 2 wr, and hence w goes once around the circle c. The parallels 1, X divide the z-plane into a set of rectangles R. To each such Rm corresponds in the w-plane a curvilinear rectangle 9m bounded by the rays r and the circles c. When z lies in Rm, the corresponding value of w lies in 9m. In this way the relation between z and the corresponding value of w is roughly given. The smaller we take the rectangles R the more accurately we will know the value of w corresponding to a given z. Suppose now that z describes a curve C. To find approximately the curve ( which w describes we have only to note the points al, a2... where z enters and leaves the rectangles R. Then ( will enter and leave the corresponding rectangles 9? at points which may be estimated roughly by proportion. The smaller we take the R, the more accurately we will be able to plot S. 2. Let us draw the lines y=2m7r, m= l, ~2,... 4 parallel to the x-axis. This divides the B z-plane into a system of bands B, B1, B_1, B B2I B ak... -or 1 If we take a point z = x + iy in B, the point = z + 2 mri will have the same position relative to Bm as z does to B. We say Zm is congruent to z. On account of the periodicity of ez erm = ez+2mlri = ez. Thus w has the same value at Zm as it has at z. For this reason we call these bands, bands of periodicity. All the values that w can take on at any point, it takes on in any one of these bands as B. Let us show that: The function w = ez does not take on the same value twice in B. THE CIRCULAR FUNCTIONS 111 For say z1=a+ib, Z a+i2 are two points in B for which e = ez2. Then ea+ib = ea4i3 or ieaeib - eaei This requires that e= e or a= also that b=/3+2m7r. As z1, z2 both lie in B we must take m = 0. Thus b =,. Hence Z1 = Z2. Thus to each point z in B corresponds one point in the w-plane, and conversely to each point w in the w-plane corresponds just one point in B, if we agree to reckon only one of its two edges as belonging to B. For this reason B is called a fundamental domain of ez, that is, a domain in which ez takes on every value it can take on, once and only once. The Circular Functions 58. 1. Addition Theorem. We wish now to extend the definition of the circular functions to complex values. In the calculus we learn that the developments X2 X4 cos x= 1 - -. X X3 X5 sin X + sn 1 - 3! 5! are valid for all real x. If we replace x by the complex variable z, the series on the right converge absolutely for every value of z, as we saw in 39, 4. This affords a natural extension of the circular functions when we wish to pass from the domain of real to complex numbers. We thus set z2 z4 cosz= 1- + —... (1 2! 4! Z Z3 Z5 sin z -= t — + -(2 1! 3! 5! 112 FUNCTIONS OF A COMPLEX VARIABLE The domain of definition of these functions is the whole z-plane. When z becomes real, cos z, sin z reduce to the familiar cos x, sin x. For real values of z therefore, cos z, sin z have the properties of cos x, sin x. We wish now to show that these properties still hold for complex z. The most important of all these is the addition theorem sin (u + v)= sin u cos v + cos u sin v, (3 cos (u + v) = cos u cos v - sin u sin v. (4 Here u, v are any complex numbers. Let us establish 3). The reader may prove 4) in the same way as an exercise. From 1), 2) we have on multiplying *U3,V2\ / U5 U3V2 2UV4 sin u cos v = u - + + +) + — I+2! )-05! +.3! 2: 1! 4! ( u7 u 52 u3V4 UV6 - + V + + V7! 5! 2 3! 4! 1! 6! * V3,2V\ / V5 v3f2 VU4 \ cosusinv=v —.+ u +!+3! 2!+1 (3 2!) 1.2! 1!4! f( 7 v 5U2 V3u4 vu6 7! 5!2! 3!4! 1!6!J Adding we get sin u cos v + cos u sin v = (u +) - v)- u + )3 + - ( + v)5-... = sin (u + v) which is 3). 2. Since we have now defined ez, cos z, sin z for all values of z, let us note that the relation 54, 4) holds whether y is real or not. We therefore have eiu = os u + i sin u (5 for any complex u. Hence also eu+iv = eu(cos v + i sin v) (6 holds for all complex u and v. fiHE CIRCULARI FUNCTIONS 11.3 We note that 5) is a generalization of Euler's Formula, 55, 1). Since 5) holds for complex u, we see that we may also generalize 55, 11), 12). We thus have eiu - eiu ( sin u= ei=,-u( 2 i and these relations hold for anly complex u.. Let us set u z+7riu 8); then 2 sin ~ i(z~!r) — ~~~ 1(2g 7ri =- e 2 e e z - e Yeiz 2i L - l-(eiz + eiz), by 55, 4) = COS z, by 7). 7Thus sin~~ + =cos z,sin(z + w)= -sinz. (9 etc., as for real values of z. 3. From the addition theorem we can show how to calculate cos z, sin z for any complex Z = X + ig by using ordinary logarithmic tables. We have in fact from 3). sin z =sin(x + iy)= sin x cos ig1 + cos xsin is, (10 Now coi+1(y2i) -+ V!+ +... = cosh y by 44, 2). Thus cos iy = cosh y (11 Similarly sin iy = i' (yi + (iy) 1! 3! 5! =i sinh y, by 44, 3) Thus sin iy = i sinh y. (12 114 1 FUNCTIONS OF A COMPLEX VARIABLE Putting 11), 12) in 10) we get sin z = sin (x + iy) = sin x cosh y + i cos x sinh y. (13 Similarly we get cos z = cos(x + iy) = cos x cosh y - i sin x sinh y. (14 The formulas 13) 14) express sin z, cos z in the rectangular form u + iv, where u, v are known. 4. As an example let us compute w = COS Z = U + iv for z = -1.6002-2.8 i. Here x=-1.6002, y=- 2.8. We first reduce x to degrees. From our tables we have prad = 570 17' 4511.6002 = 34 23 20 Thus - x = 91' 41' approximately. log I cos x I = 8.4680 log i sin xI= 9.9998 log cosh y = 0.9166 log I sinh y = 0.9134 log I u = 9.3846 u = -.2424 log v = 0,9l32 v = 8.188 Hence cos z = -.2424 + 8.188 i, approx. 5. As an exercise in multiplying series, let us show that sin2 z+ C+os2 z = 1. From 1) and 2) we have Sjn2 Z - z1 -+-J+ Z-+ - — + -!" \3! 3!/t \5! 3!3! 51/ f-! t-3! 51-5 3 7!T' CosZ Z2 + 11 1 + \2!2!/\4! 212! 4!,! (6! 412 24! 6-! ( 8! + 12 1 612 8 8 6! 2! 4! 4! 6 1.2! 8j" (15 THE CIRCULAR FUNCTIONS 115 Thu s 2l sin2z +cos2"z= 1 2 +4{l -(4)~(2)-(4)~ 1 (16 -6! | l-(1 +() 6)+ 6_()(16 Now from elementary algebra we have for a positive integer m (a + b)m=am (am-lb+(M am2+*+( )abm-l+bm. (17 \ j-/ \l~'t+(~)~,~ ~l~z+~~~+( tj-1 q- \m ' (17 If we set here a= 1, b =- 1, it gives 0 = 1 -() + ()(m)+... + (_ 1)-. (18 This relation shows that all the.." } in 16) vanish. Thus the right side of 16) reduces to its first term, and establishes 15). 6. When a function f(z) is such that f(-Z)= -f(z) (19 for all values of z for which f(z) is defined we say f is an odd function of z. Similarly if f(- z)=f(z) (20 we say f is an even function. Since 1) involves only even powers, and 2) involves only odd powers, we have: The function cos z is an even function, and sin z is an odd function of z. 7. Let us now define the other circular functions for complex z. This we do as in trigonometry. We set sinl z Cos z 1 1 tanz=-, cot z - sec z = --, cosec z = ---- cos z sin z cos z sin z These functions are defined for all values of z for which their respective denominators do not vanish. 116 FUNCTIONS OF A COMPLEX VARIABLE 59. Zeros and Periodicity. 1. Let us find for what values sin z=. (1 We already know that such values are z=m7r, m=, ~1, ~2, *.. (2 We show that there are no others. Suppose in fact that sin (a + ib) = 0. Then by the addition theorem 58, 13) sin a cosh b + i cos a sinh b = 0. Hence sin a cosh b = 0, (3 cos a sinh b = 0. (4 Now when a product a/3 = 0, either a or / must = 0. Thus in 3) since coshb does not vanish at all [44, 2], we must have sin a = 0. Thus a = m7r. Putting this value of a in 4) it gives sinh b = 0. But this requires [44, 2] that b = 0. We thus get the important result: The function sin z has the same zeros as the real function sin x; they are given by 2). From 58, 9) we see that the zeros of cos z are - +T-, m=0, ~1, ~2. 2 as for real cos x. 2. Let us now investigate the periodicity of w = sin z. From 58, 9) we have sin(z+2m7r)=sinz m=~1, 2, Thus 2 mrr are periods. Let us show that there are no others. For say a + ib were a period. Then for all values of z we would have. sin (z + a + ib) = sin z. THE CIRCULAR FUNCTIONS 117 In this relation set z = 0; it gives sin (a + ib) = 0. But this equation is satisfied, as we saw in 1, only when a = nr, b=0. Thus all the periods of sin z must be included among the numbers n7r. But among these we know only those are periods for which n is even, or n = 2 m. We have thus established: The periods of sin z are 2 mwr. m = ~1, ~ 2,... Since all of these are multiples of 2 vr, this is called the primitive period. 60. Graphical Study of w= sin z. Let z describe a parallel I to the x-axis. Then z = x + ib where b is constant. Then w = sin(x + ib) = sin x cosh b + i cos x sinh b = z + iv. Thus to the point z whose coordinates are x, b corresponds in the w-plane a point whose coordinates are u= sin x cosh b, v = cos x sinh b. (1 0 Thus the image e of the line z is the curve whose equations in parameter form are 1). As sin x, cos x have the period 2 7r, we see that when z describes a segment of 1 of length 2 rT, w comes back to its original position in the w-plane. The curve 1) is thus closed. It is in fact an ellipse. For 2 2 u. + -- = sin2 x + cos2 x Cos1h2b sinh2 b =1.I (2 118 FUNCTIONS OF A COMPLEX VARIABLE Next we let z range over a parallel X to the y.axis. Then z = a+ iy, where a is constant. As before we have w = sin(a + iy) = sin a cosh y + i cos a sinh y = u + iv so that u sin a cosh y, v = cos a sinh y. As z ranges over X, w will describe a curve t whose parametric equations are 3). As before we have U2 V2 2 - 2 = osl2 y - sinh2 y sin2 a cos2 a =1, by 44, 4). (4 Thus t) is a hyperbola. The ellipses 2) and the hyperbolas 4) are confocal and hence cut each other orthogonally. The parallels 1, X divide the z plane into a set of rectangles R, to which corresponds a set of curvilinear rectangles W9. When z lies in some R, the corresponding value of w will lie in the image 9,, of R,. Thus the smaller the rectangles R are taken the smaller the rectangles W become and therefore the more accurate is our knowledge of the true value of w. The Hyperbolic Functions 61. For any complex z we define cosh z = e - + e- + 4 + z4 (1 2 2! 4! ez - e- z Z3 Z5 sinh z = e- = z +. + (2 2 1! 3! 5! From these we further define sinh z. cosh z tanhz = sin coth z = (3 cosh x sinh z sech z= --, cosech z (4 cosh z sinh z THE HYPERBOLIC FUNCTIONS 119 We saw in 39, 4 that the series 1), 2) converge absolutely for every value of z. The domain of definition of sinh z, cosh z is therefore the whole z-plane. That of the four functions 3) 4) is the whole z-plane except those points for which their respective denominators vanish. Since we have now defined sin z, cosz, sinhz, cosh z for all values of z, the relations 58, 11), 12) are not restricted to real y as we easily see, but hold for any complex value. We have therefore the important relations connecting the circular and hyperbolic functions, viz.: cos iz = cosh z, (5 sin iz = i sinh z, (6 tan iz = i tcaih z, etc. (7 These relations show that all the analytical properties of the hyperbolic functions may be deduced from the corresponding relations between the circular functions. Thus the addition theorem sin (u + v)= sin u cos v + cos u sin v gives sin i (u + v)= i sinh (u + v) = sin iu cos iv + cos iu sin iv = i sinh u cosh v + i cosh u sinh v, or dividing by i, sinh (u + v)= sinh u cosh v + cosh u sinh v. (8 Similarly cos (u + v) c os u cos v - sin u sin v gives cosh (u + v)= cosh u cosh v + sinh u sinh v. (9 The formulae 8), 9) express the addition theorem of. the sinh and cosh functions. Again the relation sin2 u + cos2 = 1 becomes on replacing u by iu sin2 i + cos2 iu = 1, or ^oah2 /, _ i;nh2 / - 11 (1 VVOAL W __I111 A '.I - 1 k — and so on. 120 FUNCTIONS OF A COMPLEX VARIABLE In a similar manner, when we have considered differentiation and integration, we shall see that all formulae involving these operations on circular functions go over into corresponding formulae for the hyperbolic functions on using the relations 5), 6). A student of a new subject is naturally eager to see its usefulness made manifest as soon as possible. We submit the results just obtained as an example. As another example we recall the treatment of analytic trigonometry founded on Euler's formula which we indicated in 55. We adduce also the elegant manner of establishing the formulae of 6. Other examples will occur as we advance. Logarithms 62. 1. In the equation ew = (1 let us find the values of w corresponding to a given z. We set w=u+iv, z=rei. (2 Here r, 0 are known, and we are to find u, v. Putting 2) in 1) gives eueiv= rei0. (3 Equating moduli, we have eu = r, whence u= log r. (4 Equating arguments in 3) gives v= 0 +2 rm, an integer. (5 This shows that all values of w which satisfy 1) must have the form w = log r + i + 2 mnri. (6 If we set these values of w in 1) we see that they do in fact satisfy it. Thus 6) is the solution of 1). We call it the logarithm of z, and write w = log Z. (7 There is a slight ambiguity which custom and usefulness sanction. In 6) the symbol log r means the logarithm of algebra. For each LOGARITHMS 121 positive r, log r has one and only one value. On the other hand the symbol log z has an infinity of values, viz. those given in 6). As the reader is never seriously in doubt which of the two meanings the symbol log has, there is no need of denoting w by a new symbol, as Log z for example. 2. In algebra y = log x is defined by ey = x. (8 But here x is restricted to positive values. By enlarging our number system, the equation 8) admits a solution whatever value x has, the value x = 0 excepted. In doing this we find, however, not one but an infinity of solutions as given in 6). If we allow z to range over a curve, the corresponding values of w will trace out an infinity of congruent curves or branches. Each value of m = 0, ~, ~ 2,... in 6) will give one of these branches. The branch corresponding to m = 0 we will call the principal branch. The branch belonging to a special value of m as n, say, we may denote by log, z. Thus wn = logn z = log r + iO + 2 nri. (9 3. The relation log xy = log x + log y (10 established in algebra for positive x, y is called the addition theorem of logarithms. It is an immediate consequence of exey - ex+y A similar relation holds in the complex domain. For let u = rei, v =sei. Then i(0+6) Then~ uv = rse0+*. Hence by 2) and 6) log uv = log (rs) + i(O + )) + 2 nri (11 = (log r + iO + 2 n!'ri) + (log s + ikb + 2 n"7ri) when n', n" are any two integers such that n' + n"t = n. Thus 11) may be written, using 9) log1(uv) = logn,u + logn,,v. (12 122 FUNCTIONS OF A COMPLEX VARIABLE If we leave the symbol log undetermined, we may write 11) log uv = log u + log v. (13 The relations 12), 13) express the addition theorem of the logarithm in the complex domain. They are the generalization of 10). 4. Let us find the zeros of log z. If we set w = log z =- + iv, we have w I = /u2 + v2 = /log2r + ( +2 m7r)2. This being the sum of two squares cannot vanish unless log r = 0, 0+ 2 r = 0, or unless r1, -2m, r ' —1 = - 2 mnr, that is, z=1. Since log 1 = 0 this shows that: - The function log z has one and only one zero, viz.: z = 1. 5. Branch Points. Let z describe in the positive direction a circle of radius r about the origin. Then r in 2) remains constant while 0 increases from say 0 = 0Q to 0 = 0o + 2 r. If we start with the value of w corresponding to 0- = 0 in 9), w, will acquire at the end of the circuit the value Wn = log r + i(0o + 2 r)+ 2 niri = log r + i0o + 2(n + 1)7ri = Wn+1 Thus a circuit about the origin in the positive direction converts each branch Wn into w+1. Suppose now that z passes around C in the negative direction. Then 0 decreases from say 0 = 0 to = 0- 2 7r. Thus w, will have at the end of the circuit the value w = log r + i(80 - 2 7r) + 2 nri = log r + iO0 + 2(n- 1) 7ri = Wn-1. LOGARITHMS 123 Thus a circuit about the origin in the negative direction converts each branch w, into w,_1. Finally, let z describe a closed curve @ which does not include the origin as in the Figure. In this case both r and 0 vary as z moves over (. But if z starts from the point z z = z for which r= r0, = 00 as in the ~ Figure, obviously r and 0 acquire these/ same values when z has returned to z0. z rof Thus if we start out with the value of w corresponding to r = ro, 0 = 00 in 9), w" o will have this same value when z has returned to z0. Hence each w, remains unaltered when z describes a closed curve which does not enclose the origin z = 0. Since the branches w, permute when z describes a circuit about the origin z = 0, this point is called a branch point. There is no other point in the z-plane having this property; that is, z= 0 is the only branch point of the logarithmic function w = log z. Since logx in the real domain is not defined for x = 0, it follows that the formula 6) has no meaning for r = 0; that is, when z = 0. Thus the domain of definition of w is the whole z-plane excepting z = 0. 63. The Function zo. 1. Letting,a denote any complex number, we define the symbol by e z'L by zL_ eIlogz (1 Let z = rei, then log z = log r + ic + 2 srri, s an integer. Hence 1) gives z = eliogreje2sl, (2 Let us consider special cases of the exponent A. 1~ Jt a positive integer m. As e2msri = 1, 2) becomes zn_ emlogremi. But by algebra, emlogr - rm Hence z= rmemi (3 == z *... z, m times, which agrees with 4, 4). 124 FUNCTIONS OF A COMPLEX VARIABLE 2~ s a negative integer. Let = - n. Then we find — n e r-ne-ni = _zn as in algebra. 3~ M= m, afraction. Here n e2s e n =co an nth root of unity by 55, 5). Also geIlogr -_ elogr _= -/r Hence z m (4 z n=/rmV. 0 ~ e, as in algebra. Here the radical has the value heretofore assigned to it; that is, it is the positive real root of the positive real number rm. One often denotes the right side of 4) by 4~ ta a real number not a fraction, and z real and positive. Let us denote this value of z by a. Here = 0, r = a. Then eI logr - elloga = a'l, as is shown in algebra. Then 2) gives Z =_ ae2si O, s 1, ~ 2.. (5 This shows that in this case zI has an infinity of values, each differing from the others by a factor of the form e2S7ii. For s=0, z = a,. 2. From the foregoing it follows that the function 1) is in general infinite-valued, as the logarithm is. If we give to log z a definite one of its many values, the exponent on the right side of 1) takes on a definite value, as v, say. Then 1) gives z as the absolutely convergent series v v2 v3 Z1 = + +1~ + + + + 1! 2! 3!t Its value is thus completely determined. INVERSE CIRCULAR FUNCTIONS 125 3. Suppose /J is not a positive or negative integer or 0. Then z = 0 is a branch point of z". For if we start at z = a with one of the many values of w= zm, w will not have this value when z describes a circuit about the origin. Since z = 0 is the only branch point of log z, the only branch point of w is also z = 0. That is to say, if we start at z = a with one of the many values of w, as w,, and allow z to describe a closed curve, which does not enclose the origin, w will return to its original value w, when z returns to z = a. Inverse Circular Functions 64. 1. In trigonometry we learn that the equation sin y = x (1 has two sets of solutions y when 0< x < 1. If y = yo is one solution of 1), all solutions are given by y = yo + 2 m7r,, n = 0,1, 2,... (2 y = 7r - Yo + 2 nr, Thus 1) defines an infinite-valued function which is denoted by y = sin-ix or y = arc sin x. Of these two, we shall employ the latter only. Let us now pass to the domain of complex numbers. We seek the solution w of sin ut = z (3 where z is any given complex number. Now by 58, 8) eiweiw sin w = 2. 2i This in 3) gives e2iw _ 2 izeiw = 1 or setting t eiw, we get t2 - izt =. Solving this for t, we get ei = iz + /1 - z2 (4 126 FUNCTIONS OF A COMPLEX VARIABLE where the radical may have either of its two values. From 4) we have, taking the logarithm of both sides, w = log {iz + V1 - z2}. (5 z As the log is infinite-valued and as the radical may have either sign, we see that 3) admits a twofold infinity of solutions. If we denote these solutions by arc sin z, we have w = arc sin z = log 1iz + V1 - z2}. (6 2. Let us show that the relations 2) still hold for complex values. Let us set iZ + /1I- Z2 = Z1 iz - -2 = Z. Then if log L1 = log Z1 is one of the determinations of log Z1, all the other determinations of log Z1 are given by L1 + 2 mw7i, as we saw in 62. Thus if u =- log.{z + V - z2} = log Z1 is one of the values of 5) when the radical is taken with the positive sign, all the other values of 5) for this determination of the radical are given by u + 2 mr, m an integer. (7 Similarly if = 1l - 2 lo v= - log{izz -/1 }- z2f = log Z2 g Z is one of the values of 5) when the radical has the negative sign, all the other values of 5) for this determination of the radical are given by vv + 2 mr, m an integer. (8 Now 1 u + v =- log Z1Z2, (9 Z INVERSE CIRCULAR FUNCTIONS 127 the logarithm on the right being properly determined. But zz ==- 1 and log (- 1) = 2ri + 2 nri. Thus 9) gives u + v = + 2 nr, or v = =r - u + a multiple of 2 7r. Putting this value of v in 8) it becomes 7r -u + 2 sr, s an integer. (10 Thus all the values 6), that is, of arc sin z, are given by 7) and 10). We have therefore shown: If w = wo is a solution of 3), all solutions of 3) are given by w=wo + 2 mr, w = -w + 2 mr, (11 where m=O, ~ 1, 2. 3. Since z = 1, z =- 1 are branch points of 1 z2= _ —(- 1)(Z + 1), we see that when z describes a small circuit C about one of them, Z1 and Z2 permute. Thus if at z= zo we start out with one of the values of w at this point, say w = w0, and allow z to pass around C, w will nbt return to its original value w0 when z returns to z0. Thus z = 1, and z =- 1 are branch points of w. Let us see if there are any other branch points for w = arc sin z. Since z = 1, z =- 1 are the only branch points of Z= iz +/1 Z 2, the only other branch points of w must be branch points of log Z, that is, points z = a for which Z = 0. But the relation za + -/1 -a2 = 0 gives -a2= 1 - a2 or 1 = 0, which is absurd. We have thus shown that: The branch points of the function w = are sin z are the points z = = 1 = - 1, and only these. 128 FUNCTIONS OF A COMPLEX VARIABLE 4. In a precisely similar manner we may define and study the other inverse circular functions arc cos z, arc tanz, etc. We will not take space to do this, as it is all too obvious. We note, however, that the solution of the equation tan w = z (12 is 1 1 + iz iw = -log - = arc tan z. (13 2i 1- iz Its branch points are z = i and z = -i. CHAPTER V REAL VARIABLES 65. Many readers of this book have studied the calculus chiefly with the view of learning its technique and of applying it to geometry, mechanics, physics, etc. Such students have little time to spend on demonstrations, and it is natural that their ideas of the limiting processes which lie at the base of all the principles and method of the calculus are often vague. It seems best, therefore, to insert a chapter at this point whose object is to briefly treat such topics of the calculus as we shall need in the course of this work. It will be our purpose rather to refresh the reader's memory and to illuminate the subject than to repeat demonstrations or to discuss delicate points. One may therefore, turn over the following pages, reading such parts as are not quite familiar; or he may pass at once to the next chapter and return to this one when further explanations are necessary. 66. Notion of a Function. 1. The functions used in the calculus are usually made up of simple combinations of the elementary functions, as y - sin x + - (! log x eX+y2 sin xy _ = -sin- - Vxy sin y. (2 tan (x + y) Or they are defined implicitly by equations between such functions, as for example, X2 ]2 ~ 2+ _1= 0. (3 a2 62 The equation 1) defines y as a function of the independent variable x.:But when x < 0, log x has no sense, the variables being 129 130 FUNCTIONS OF A COMPLEX VARIABLE real. Also V1 - x is not real when x2 > 1. Finally, when x = 1, log x=0, and the denominator in 1) vanishes. Thus 1) defines y only for 0 < x < 1. These values of x form the domain of definition of y. 2. Let us turn now to 2). This relation defines u as a function of two variables x, y. We note that the denominator either = 0 or is not defined when the point x, y lies on one of the lines, '7r x +y=m., m=0, ~1, 2... (4 Also the radical V/sin y is not real when sin y is negative, that is, when en (2n-1)7r <y < 2 nr, n = 0, ~ 1, 2.. (5 The domain of definition of u embraces thus that part of x, y plane after removing the lines 4) and the bands 5) which are parallel to the x-axis and of width 7r. 3. It is convenient to generalize our definition of a function. To fix the ideas, let us take a single independent variable x. Let us mark a certain set of points on the x-axis and denote them by W. At each value of x in W, let us assign to y one or more values according to some law. Then we call y a function of x, and we call W its domain of definition. To illustrate this let us take the function 1). For the point set W1 we take the values of x such that 0 < x < 1, that is, the interval (0, 1) except the end points. The value we assign to y for a given x in W is the value that the right side of 1) has for this value of x. Another illustration would be the temperature y at a given place at a given time x. If we were concerned only with temperatures from the time x = a, to the time x = b, these would define y as a function of x in the interval 51 =(a, b), and this would be its domain of definition. This would be an example of a function which is not defined by an analytic expression as in 1). As an illustration of a function of two variables not defined by an analytic expression we may take the following. Let x denote the latitude and y the longitude expressed in circular measure of a point on the earth's surface. Let u denote the temperature REAL VARIABLES 131 at the earth's surface at one and the same instant. As the lati7rtude is restricted to lie between -7r and 7r, and the longitude 2 2 between - 7 and 7r, we see the domain of definition of u is the rectangle 7r. r < x <- - r < y < 7r. 2 2 67. Limits. 1. We have already discussed the notion of a limit of a sequence of numbers al, a2, a3,.. (1 when studying series. In the calculus we use the notion of limit also in another way. Thus in defining the derivative of a function y =f(x) we form the difference quotient ay f (x + Ah)-f(x) (2 Ax h and allow the increment h = Ax to converge to 0 by passing over all values near h = 0. The value h = 0 is excluded, since this would make the denominator in 2) vanish. What do we mean by the limit of 2) as h 0? The value of x being fixed, the quotient 2) is a function of h; let us denote this by q(h). If now q(h) converges to some fixed value I as h 0, we mean that q may be made to differ from I by an amount as small as we please, say by < e, provided h remains numerically < some positive number s. Graphically we may state this as fol- - I I_ lows. Let us plot the values of q for values of h near = 0, on an axis which I Ie we may call the q-axis. With I as a center q we may lay off an interval of length 2 e. Then if q I as h 0, there must exist an interval (-, 8) on the h-axis such that q falls within the e interval when h = 0 is restricted to lie within the 8 interval. This graphical formulation of the notion of a limit may be put in analytic form. By the phrase q(h) 1 as h 0 we mean that for each e > 0, there exists a 8 > 0 such that Iq(h)-1 < e provided O < Jh < a. 132 FUNCTIONS QF A COMPLEX VARIABLE The reader should think this over carefully and see that this analytic formulation exactly represents the above graphical formulation. 2. This notion of the limit is used in many other parts of the calculus. We will therefore state the definition quite generally. Let the function y =f(x) be one-valued about the point x = a. When we say f (x) I as x a or otherwise expressed that the limit of f(x) = I for x = a, in symbols limf(x) = 1, (3 x=a we mean this: For each e > 0 there exists a 8 > 0 such that If(x) - 1 < e provided 0 < x- a <8. The last sentence will be expressed more briefly by a line of symbols, e > > 0,, If)- 1 <, < Ix- al< 8, (4 and such a line of symbols is to be read as the sentence above in italics. 3. The reader will note the similarity of this definition and that employed in 16. Almost all students dislike this e form of the definition when first presented to them. It seems so much easier to think off (x) converging to 1 as x converges to a. Why bother about these e's and 8's? In reply we refer the student to the remarks made in 16. Fortunately the intuitive form of the definition of f converging to its limit is usually quite sufficient, and we shall avoid the e's as much as possible. When we do employ them, it will be to aid clear thinking. When the reader can think clearly without the e's, let him do without them. 4. The reader should note that when the limit 3) exists, f(x)l when x converges to a from the right side, or when it converges to a from the left. For by the definition given in 2, the only restriction on x is to remain in the 8 interval, excluding, of course, the value x=a. It can therefore approach a from either side, andf (x) must I in either case. REAL VARIABLES 133 5. From the definition of a limit we conclude that when 3) holds, we may write f () = l+ E (5 and know that I' I < e provided x lies in some 8 interval. We may also say in this ease that e' 0 as x - a. 68. Limits for Two Variables. Let us now consider limits of a function of two variables. Suppose the one-valued function is defined in a certain domain 21. Let x = a, y = b be a point a of P2 Then u(a, b) is the value of u at the point a. More briefly we may denote this value by u(a). Let us describe a circle of radius 8 about a. The points x, y within this circle may be called the domain of this point of norm 8, and denoted by any one of the symbols,(a,), I(a, b), D(a), I)(a). The simpler forms may be used when no ambiguity can arise. When the center a of this domain is excluded, we indicate this fact by a star, thus D*. What now do we mean when we say: u converges to I as the point i = x, y converges to a = a, b? In symbols u l, as -a. We mean just this: Let us plot the values ) of u on an axis, the u-axis. With I as a center we lay off the e interval as in the figure. - About the point a = a, b we describe the 8 circle in the x, y plane of radius 8 as in the figure. Then for each e interval there must u' axis i -e i e exist a 8 circle such that u remains within this interval when the point x, y remains within this circle, the center a, b excluded. Analytically we may formulate this as follows: The limit of u(x, y) for x = a, y = b is 1, or in symbols, lim u(x, y)= l x=a, y=b 134 FUNCTIONS OF A COMPLEX VARIABLE when for each e > 0, there exists a 8 > 0 such that I(x, y)-II <e when x, y lies in Ds*(a, b). The reader will note that the definition of a limit for a single variable is a special ease of the one just given, the domain of the point a= a, b reducing to an interval. He should also notice that in passing to the limit the point x, y is never allowed to become a = a, b; that is, x, y ranges in D*-(a) and not in l)(a). 69. Continuity. 1. Let y=f(x) be a one-valued function in the interval a = (p, q) whose graph is given shows that y is continuous in lS except at the point x = b. Let us formulate the notion of continuity analytically. To this end we note how f(x) behaves at a point of continuity as x = a, and about the point o of discontinuity x= b. Letf(x) have the value a at x = a. Then as x a, we see that y ' a; in syml in Fig. 1. The graph _,, A X 7- B p) a FIG. 1. )ols u (I lim f(x) =f(a) = a. x=a On the other hand, at the point x = b, the ordinate does not converge to a definite value. For when x= b on the left y - '; when x _ b on the right, y f R". But if the reader will turn back to 67, 4, he will see that when lirm f(x) x=b exists, f(x) must converge to the same value whether x' b on one side or on the other. Another case of discontinuity is illustrated by Fig. 2. Here y =f(x) is a onevalued function whose value at x = a is defined to be y =a. The figure shows that lim f(x) = a'. x=a a a' iJ (. FIG. 2. Y REAL VARIABLES 135 Here the limit exists, but its value is not the value that f(x) has assigned to it at this point which is a. These considerations lead us to say: The function f(x) is continuous at x = a when and only when f(x) converges to the value off at a, that is, when lim f(x) =f(a).:x=a Therefore if lim f(x) does not exist at x = a, or if it exists but has a value different from the value f(a), then we must say that f is discontinuous at x = a. If f(x) is continuous at each point of an interval 21 = (p, q) we say f is continuous in W. 2. When f(x) is continuous at x = a, we know that the value off(x) differs from its value at x=a by an amount as small as we please if x is only kept sufficiently near a. In symbols f(x) = f(a) +' (1 where et - 0 as x - a. This is the same as saying that I e I<e, providing x-a l< some 8. (2 3. Let f(x) be continuous in 2 = (p, q). Let us take e>0 at pleasure and fix it. At any point x = a we lay off the 8 interval about a as in Fig. 3 such that the c' in 1) satisfies the inequality 2). Let us pass to another point b in S1 and lay off the corresponding 8 interval about b, such that el again satisfies 2). At the point x = b the curve is steeper than at a, and therefore the 8 interval at b is shorter than at a, that is, the value of 8 at b is less _ than at a. The reader will note, how- - l a __ ever, that as a ranges from p to q the P at b q value of 8 corresponding to the dif- FIG. 3. ferent values of a will never sink below some positive number 7. In other words, for the value of e that we have been using, there exists an r > 0 such that f(x) differs from f(a) by an amount <e when Ix-a | < 7; and here a is any point in W. 136. FUNCTIONS OF A COMPLEX VARIABLE If we expressed this in symbols, we would say that for each e>0, and for any and all points a in A, there exists a fixed r >0 such that If(x)-f(a) <e, provided I x-a <7. (3 Or what is the same, (x) =f(a) + e', where ie <e provided x -al<?. (4 This important property of a continuous function in an interval f is expressed by saying that f(x) converges uniformly to f(a) in 2; or in symbols f(x) f(a) uniformly in?2. The reader should remember that here a is a variable point in the interval 2. 4. When f(x) has a point of dis- continuity in = (p, q) as at x= e in Fig. 4, the reader will see at once that taking E > 0 small enough and then fixing it, there is no corresponding p a e q 7 > 0 such that 3) or 4) holds wher- FIG. 4 ever a is taken in W. To make this perfectly clear we have taken e as in Fig. 4, and laid off the corresponding 8 interval at a point a. The reader will see at once that for a point b very near e, the length of the 8 interval will be determined by the fact that it cannot contain the point x = e. For in any interval containing this point f(x) could differ from f(b) by an amount - -L - far greater than the small quantity e, as FIG. 5. Fig. 5 shows. 5. Finally let us consider the function f(x) = tan x. This is defined for all x not included in the point set i7 3 5 ~ - ~ 7 - ~ 7t... ~ 2 2 REAL VARIABILES 137 Let x range over the point set 2 defined by 0>.x<;. This may be denoted by (0, -), the star * indicating that the end point 2 is omitted. We call such a set an incomplete interval. Obviously f(x) is continuous at each point x = a of 2, for lim tan x = tan a. x=-a It is not uniformly continuous in W. For in the relation 2) we see that 8 must 0 as the point a approaches.. There exists therefore no V >0 such that 3) or 4) holds. 6. The analytic definition of continuity can be extended at once to functions of any number of variables. For clearness let us take two. Let the one-valued function u(x, y) be defined in some domain D about the point x = a, y = b. Then u is continuous at this point when u(x, y) converges to u(a, b) as the point x, y converges to the point a, b; in symbols when limn u(x, y) = u(a, b). x=a, y=b If u(x, y) is continuous at each point belonging to some point set 2, we say u is continuous in 21. 7. Let u(x, y) be continuous at the point a, b. If c = u(a, b) 4 0, there exists a domain D,(a, b) about a, b such that in it u(x, y) has the same sign as c. For since u is continuous at a, b there exists a 8 >0 such that I (X, y) -l <. (5 when x, y is restricted to D,(a, b). But 5) is equivalent to - e < u(x, y) < c + e. (6 Obviously as c is = 0 we may make e > 0 so small that c- e and c + e have the same sign as c. 138 FUNCTIONS OF A COMPLEX VARIABLE 8. If u(x, y) is continuous at a, b ju(x, y) <some G (7 in D)(a, 6) for a sufficiently small 8 > 0. This follows at once from 6). 70. Geometric Terms. 1. At this point it will be convenient to introduce some geometric terms which we shall need constantly. We begin by considering the point set 21 formed of the points within a circle f, that is, in the interior of R. Any point a of 2[ is such that we may describe a circle c about it as center, and all the points within c form a part of [. In other words, 21 is such that any point a of it is surrounded by some domain D)(a) which also lies in W. As a approaches 9, -- 0. But for each given a, 8 is >0. We may now generalize. Let 21 denote any set of points in the plane. If 1 is such that each of its points a has a domain D(a) which also lies in 21, we call the point set 2S a region. For example, the two curves C1, C2 in Fig. 1 define a ring T whose boundary or edge e is formed by them. The set of points 21 in the ring t but not on its edge e form a region. \ Let us look at the set of points 3 formed of 1 and the curve Cl, in symbols the point set =Fi. 1. e = = + C]. The set 3 is not a region. For let e be a point of C0 as in Fig. 2. Then however c is taken it will contain points of e and points not in ~3. As another example of a region, let 1 be the point set formed of all the points of the x, y plane except the points lying on a finite num- c ber of ordinary curves, and also a finite number of isolated points. Obviously W2 is a region. We say a region 21 is connected when any two of its points can be joined by a curve lying in 1. If the boundary of a connected region is a closed curve without double point, we call it a simple region. REAL VARIABLES 139 2. Let 1 be a connected region whose boundary or edge ( is formed of a finite number of ordinary closed curves or points. The point set ( formed of 2 and (, that is, G = 2 + A, we call a connex. If 2 is a simple region, the corresponding connex Q is called simple also. 3. The reader will note that the definition of continuity given in 69, 6 requires that if u(x, y) is continuous in a connex S, it must be continuous at each point of its edge, and this requires that uis defined as a one-valued function for all points in some D(e) of e. 4. Let 2 be a connex or a connected region. Let P be any point; it may or may not lie in 2. Let r be the radius vector joining P to a point z of 1. If z can describe a C continuous curve lying in 21 such that r describes N a complete revolution about P, we say that 2 is cyclic relative to P, otherwise acyclic Thus in Fig. 3 let 21 be the ring-shaped figure bounded by C1, C2. Then 2 is cyclic relative to L and M, but acyclic relative to N. FIG. 3. 71. Uniform Continuity. Let us now show that if u(x, y) is continuous in a connex ( it is uniformly continuous. By this we mean the following. Since u is continuous at a point a, b of 21 we have we have (x, y) = u(a, b) + e', El < E, (1 if only x, y lies in some domain D8(a, b) of the point a, b. We say now that the point set (S being a connex, 8 cannot sink below some minimum value?} > 0, as the point a, b ranges over G. For say that as a, b converges to some point a, /3 of (, 8 0. Since u is continuous at ao, /, e'l < 2' if x, y lies in some D,(a, 3). But then 8 cannot -0, for as the point a, b converges to the point a, /3, the figure shows a \ s a, and not to 0. Since therefore 8 cannot a 0 O as a, 6 ranges over (, it follows that there \\ exists an q? > 0 such that U(x, y)= u(a, ) + el, ' I < e (2 140 FUNCTIONS OF A COMPLEX VARIABLE wherever the point a, b is taken in (, provided x, y lies in D,(a, b), the norm s/ being the same wherever a, b may be in S. We may thus state the following theorem, which for our further development is of utmost importance: If u(x, y) is continuous in a connex L, it is uniformly continuous in; or in other words the relation 2) holds in G. 72. Differentiation. 1. We touched on this subject by way of illustration in 67. We wish now to discuss it more fully. Let y =f(x) be a one-valued function in the interval I = (a, b) whose graph is given in Fig. 1. If we give to x an increment h = Ax, 9 the function receives an increment P _y Ay =f(x + h) -f(x). The quotient Ax a x x+h b Ay =f(x + ) -f() (1 FIG. 1. Ax h is called the difference quotient. From the figure we see Y= tan. (2 Ax As Ax 0, the secant PQ converges to a limiting position, viz. the tangent at P. We call the limit of 1) when h ' 0 the differential coefficient at the point x and write dy ='(x) = lin f(x + h) -f(x) (3 dx h=-o ht Iff has a differential coefficient at each point of some interval A, we sayf has a derivative in A; the value of this derivative at a-point x of A is given by 3). 2. Let us consider the function y=(x- 1). (4 The graph of this function is given in Fig. 2. It has a point of inflection with a vertical tangent at x= 1. We see here that REAL VARIABLES 141 the secant PQ converges to a vertical posi- tion. The difference quotient Ay_ Ay AX (Ax)3 0 is always > 0 and increases indefinitely as Ax ' 0. We say its limit is + o and write d- =f(z)=+oo for x=l. dx FIG. 2. To say that a variable q has the limit + oo as x -a is only a short way of saying q has no limit at x = a, but that it increases without limit as x' a. Strictly speaking the function 4) has no differential coefficient at x = 1. Usage, however, permits us to say its differential coefficient is + oo at this point. We also say the derivative f'(x) is definitely infinite at this point, meaning thereby that the difference quotient does not change its sign about x = 1. 3. Let us consider the function y = (x - 1)l, radical with positive sign. (5 The graph of this function, Fig. 3, has a ver- y tical cusp at x = 1. At this point y A) (6 (6 Ax Ax 0 1 x It is thus positive for positive Ax, and negative when Ax is negative. Thus FIG. 3. Ax _ + o as Ax 0 on the right, Ax _ — --. *... left. Here we cannot say that the difference quotient Ay/Ax converges to any value, not even to an improper limit as + o, or -oo, since it changes its sign in any interval about x = 1. 4. Let us now consider the differentiation of a function of several variables. For clearness let us take a function u(x, y) of two 'variables which we suppose is one-valued in some domain D 142 FUNCTIONS OF A COMPLEX VARIABLE of the point a, b. Then, as the reader knows, lim u(a + h, b)- u(a, b) = u (a 6)= (7 h=O h xa ' Oa is the partial differential coefficient of u with respect to x at the point a, b, and a similar definition holds for au uy (a, 6) = o The values of 7) as the point a, b ranges over some set of points.I define the first partial derivative of u with respect to x; this we denote by ao u (cx, y) or u I or - - (8 Ox A similar definition holds for u/(x, y) or u' or -. (9 Oy The derivatives 8), 9) are also functions of x, y and so in general possess partial derivatives if a2U aI d2U " - u -, etc. XX Ox2, txyc. 5. Suppose u possesses first partial derivatives for the points x, y of some point set 21. The expression au On du = h + k (10 ax ay is called the first differential of u. Similarly, if u has second partial derivatives, 2 U O2, O+2U ks2 d2u = - h2+ 2 On Iek + 22 (11 dx2 Odxy Oy2 is the second differential of u, etc. We note that the right side of 11) may be written symbolically (h -+k+)- u. (12 To deduce 11) from 12) we expand 12) as in algebra and replace )a by2 e02u x) u by - etc. Wx 'Ox2 REAL VARIABLES 143 In general, suppose u has all partial derivatives of order <n for all points x, y in some set W. Then dnu=(h -+ k qu, (13 when after expanding the right side we replace ( aAm(O8n-m by — U k\x, \Oyy) / y axm/yn-m 73. Law of the Mean. 1. Let the graph of the one-valued function y =f(x) be as in the figure. Let the secant PQ make the angle ( with the x-axis. At each point x Q let us draw the tangent to the curve. It makes an angle 0 with the x-axis. Let now x start at a and move toward b. The tan- p O gent changes its inclination from point to point. If the reader will reflect a few moments, he will see that it is altogether im- a c b possible to pass from a to b without somewhere the tangent being parallel to the secant PQ. Let this be the case for x = c. Now tan = b)-f(a) b-a tan. =dy= f' (x). dx Since c = 0 at the point e, we have, on equating these two expressions, f(b) - (a) = (b - a)f(c), a<c < b. This is the celebrated Law of the Mean. It is one of the most important theorems in the whole calculus. The foregoing considerations do not form an analytic proof of this law. They do, however, make the reader feel in the most convincing manner that this law is true, and this is all that he needs at this stage. On account of its importance let us formulate it as follows: If f(x) is one-valued and continuous in the interval a =(a, b) and if f (x) is finite or definitely infinite within Af, then for some point a<c<b6 aer_ ', /\-t_ -,f 1 J VtI) — / ) = ku- - v)./ 1 144 FUNCTIONS OF A COMPLEX VARIABLE 2. Let- a< x <.b, f(x + h) -f(x)= Af. Then the law of the mean 1) gives, setting h =Ax, -=f'(u), x<u<x+h. (2 Ax Suppose now that f'(x) is continuous in the interval 21. Then f'() = f'(lX)+ E' ' I < 6, (3 provided Ax I< some fixed 8, wherever the point x is chosen in W. This in 2) gives A = ff/x +e (4 Ax and e' - 0 uniformly in 1. 3. From the law of the mean it is easy to establish another very important theorem, viz.: Taylor's Development in Finite Form. In the interval 2 = (a, b) let f(x) and its first n- 1 derivatives be continuous. Let f()(x) be finite or definitely infinite within 21. Then for any x within 1 x - a, (x - a)2... f(x) = f(a) + -Tjfl (a) + (a)+ + ( - ) f(~(a) + ( - )f( )(v) a <v <. (5 (n - 1) n ' As this is not a work on the calculus we do not intend to prove this theorem; we have quoted it in order to state precisely conditions for its validity. 4. From the law of the mean we can draw an important conclusion which we shall need later. Suppose u(x, y) has continuous first partial derivatives about the point a, b. When we pass from this point to the neighboring point a + A, b + k, u receives the increment Au = u(a + h, b + k)- u(a, b). But we have Au = lu(a + h, b + k)-, b+)+(a, + k)+ a, k)- u(a, b) = 1 + A2. REAL VARIABLES 145 By the law of the mean A = hu (c, b + k), c between a and a + h A= ku(a, e) e e between b and 6 + k. But ux, u' being continuous, u(c, b +k)= ux(a, b) a (a, e) = u (a, b) + where a 0, 8O 0 as h and k 0. Hence we may state the theorem: Let u(x, y) have continuous first partial derivatives about the point a, 6. Then the increment Au differs from the differential du by a quantity of the form ah + /3k where a and /3 O as h and k 0. Thus Au = O h+ k + ah + /k. 5. Suppose now that ux and ur are continuous functions of x, y in a connex (. Then, as observed in 71, a and /3 will converge uniformly to 0 in C. Hence in particular I a < C, 3 1< G, provided Ih I and I k I are < some 8, and here 8 is independent of the position of the point a, b in C. 6. Taylor's development 5) may be extended to a function of any number of variables. For clearness let us take a function u(x, y) of two variables. Suppose u and all its partial derivates of orders < n are a+h, continuous in some domain D about the point a, b. Let a + h, b + k be any point D L/ y in D. Let L be the segment joining these \,b two points. Then any point x, y on L o has the coordinates x=a+sh, y=b+sk, 0<s<1. 146 FUNCTIONS OF A COMPLEX VARIABLE When s ranges over the interval 5 = (0, 1), the point x, y ranges over L. Then u(x, y)= u(a + sh, b + skc)= f(g) is a function of 8 defined for values of s lying in 25. But f/(s) = h + k = du(x, y), Ox ay a2U a2 a222 O8tt h2+ 2 Ot + h+ k2= du(x, y), 4Y()ax2 Oxay Oy2 Hence 4'(s), p" (s). 4(nl)(s) are continuous in 25 and we may apply 2 to the function f(s). Doing this and then setting S = 1, we get u(a + h, b + k)=u(a, b)+ 1du(a, b) + d2(a, 6)+... 1! 2!~~ 1 +- du(a + 0h, b + Oh) (6 where 0 < 9 < 1. For convenience -of reference we note that du(a, 6) auh+Ou k, aa Ob Oct 02 d'2u(a, b) h2 + 2 u h/ + Ouk2. Oc2 aaOb 062 Integration 74. 1. The integral Jlb f(x)dx may be defined in connection with the notion of area as follows. Let the graph of f(x) be as in Fig. 1. In the interval of A = (a, 6) we interpolate the points a1, a2... If no interval (a,,,-, a,m) has a length greater than 8, we say these points a a1 a, a3 b effect a division of I of norm. FIG.. We set RN A Vm z-_C- am "m-1 l 4A.&M INTEGRATION 147 and form the sum f(a1) AxI +f(a2) x2 +/-... = f (am)Axm. (2 The value of this sum is the area of the shaded rectangles in the figure. If now 8 0, this sum obviously converges to the area under the curve. Thus when f(x) is a one-valued continuous function of x in the interval X, the sum 2) has a limit as 8 0. This limit is the value of the symbol 1). This symbol we also denote sometimes by ff (x) dx. 2. These geometric considerations enable us to take a more general definition of 1). In the intervals 8, let us take at pleasure a point am, If f(am) = mm let us construct the rectangles of base 8m and height/3m, a tla2ao ct3 as in Fig. 2. We now form the sum FIG. 2. f(Ao) Ax1 +f (2) AX2 + *. = if (am) Ax,. (3 The value of this sum is the area of the shaded rectangles. If now 8 0, the sum obviously converges to the area under the curve and therefore has the same value as before. Let us state this in a theorem: If f(x) is continuous in the interval (a, 6), liam f(a(n)Axm (4 6=0 exists and this limit is the value of the integral 1). 3. With this definition we can establish the following fundamental theorem: In the interval (a, b) let F(x) be one-valued and have the continuous derivative f (x) Then fb f(x) dx = F(b) -(a). (5 148 FUNCTIONS OF A COMPLEX VARIABLE For, using the points a,, a.. an_1 introduced in 1, we have by the law of the mean E )(a) - 17(a) =f (aj)Ax 1(a2)- (a1) =f( c2)Ax2 where am is some point in the interval Axm. Adding these equations gives ](b) - F(a) = /f(am)Axm. Now by 2 the limit of the right side as 8 0 exists and equals the integral in 5). 4. From the definition of an integral given in 1) follows an important property which is useful in estimating the numerical value of an integral. Since I f (am)Axm < E I f(am) Ax,,, we have, on passing to the limit 8 = 0 f(x)dx < ~ f(x)ldx, a<b. (6 Also let If (x) < (x) in the interval (a, b). Then similarly we have f(x) dx < (x)dx, a<b. (7 5. Another property of importance is that: J(x) = f(x)dx, a < x < b considered as a function of its upper limit x is a continuous function of x such that dJ d = f (x), the integrand f( being continuous in (a, ). the integrand f(x) being continuous in (a, 6). INTEGRATION 149 75. Surface Integrals. 1. Let u =f(x, y) be a one-valued con-, tinuous function of x, y in a field f bounded by a finite set of ordinary curves. As x, y ranges over Af, the end points of the ordinate through x, y will describe a surface S, while the points of the ordinates m \ will constitute a solid of volume V lying between S and the x, y plane X. under S. \ Let us draw a set of parallels to the x, y axes in the x, y plane as in Fig. 1. This effects a division of the plane into rectangles R. If their F.. sides Ax, Ay are all < 8, we say the norm of this division is 8. Let us now form the sum analogous to 74, 2),;f(am,,b) AxmAyn, (1 extended over all rectangles containing a point of WI. The points am, b, are the vertices of these rectangles. The sum 1) is obviously the volume of the set of prisms whose bases are the rectangles AxmAyn and whose heights are the ordinates of the surface S at the points x = ax, y = b,. We take it as geometrically evident that the sum 1) converges to the volume V as 8 0. This limit we use to define the symbol f f(, y)dxdy. (2 2. To calculate this integral it is usual to express it as an iterated integral dx rf(x, y)dy. (3 c/H J( Here the symbol 3 denotes the projection of the field I1 on the x-axis. Let x be a point of 3. The line through x parallel to they-axis will partly lie in lW. This section we denote by,x or more shortly by (. Thus to calculate 3) we give x a fixed value in 3 and calculate f(x, y)dy, (4 150 FUNCTIONS OF A COMPLEX VARIABLE the field of integration being the section 6 of S9 corresponding to the value of x chosen. This integral itself is a function of x. This we now integrate relative to x over the field 3, getting in this way 3). To illustrate, suppose the field of integration y / 1 in 2) is bounded by the three outer curves \ and the two inner curves of Fig. 2. Then the projection ~3 on the x-axis consists of the segments l AB, CD. For a value A E B C F D of x corresponding to FIG. 2. E, the section ( of 2 is formed of the two segments marked ( in the figure. At F the section ( is made up of three segments, also marked (. It is shown in the calculus that: The two integrals 2) and 3) are equal. The following geometrical considerations will make this apparent: That slice of Vwhich lies between the two planes x=am, x - m+1 in Fig. 3 has approximately the volume Axm f(am, y)dy,. as is seen from Fig. 3. The _ am a1 +l sum of these slices is A1xm,f (am, y)dy. (5 Ax C/t^~~~~~~~~ zFIG. 3. Thus the volume V is the limit of 5) or the iterated integral 3). Thus 2) and 3) are equal as they both = V. 76. Curvilinear Integrals. 1. Let us suppose that f(x, y) is a one-valued continuous function of x, y at the points of a curve C INTEGRATION 151 whose end points are a, b, as in Fig. 1. If we interpolate a set of points al, a2, a3. such that the arcs aal, ala2, a2a3.. are of length <8, we say these points al, a2.. effect a division of norm 8. Let Xm, ym be the coordinates of am; let b AXm Xm+l - Xm. We now form the sum /f(Xm, Ym,)Xn (1 1. a3 If we let 8- 0, this sum converges to a FIG. 1. definite limit which we denote by /-6b f(x, y)dx or by f(x, y)dx (2 and call it the x-curvilinear integral of f(x, y). When the curve C reduces to a segment of the x-axis, the integral 1) reduces to the ordinary integral considered in 74, since y is now constant. Let us prove that the limit of 1) exists for the simple case that the equation of C is y= ~(x), (3 C being one-valued and continuous, the end points of C corresponding to x = a, x = /3. Then f(x, y) = f[,,(x)] = (x), and 1) becomes E #(x,.)Ax. (4 But g(x) is continuous, hence the limit of 4) exists and is g(x)dx. (5 Hence the limit of 1) exists and has the same value. We note that this form of proof not only establishes the existence of the limit of the sum 1) but determines its value. Because 2) and 5) are equal, we may extend the properties of ordinary integrals to curvilinear integrals. Thus if c is some point on C between the end points a, b, we have ff(x y)dx = fA(x y)dx + ~ f(x y)dx. (6 fca ^a (c 152 FUNCTIONS OF A COMPLEX VARIABLE Let us return to the sum 1). The factorf(x,,, ym) denotes the value of f at the end point of the arc amlam. The same geometric considerations used in 74, 2 would show that 1) has the same limit when xmym denote any point in the arc a,_iam. We shall make use of this fact in 3. 2. The foregoing proof applies to the Q case when C is the arc PQ of the circle of radius r in Fig. 2. For then R _ -s 0 P y = + Vr2-x2 = +(x), (7 the radical being taken with the pils sign. It does not apply immediately if C is the arc PQS. For if - s is the abscissa of S, FIG. 2. y is two-valued in the interval (-r, -s). In this case we have only to break C into two arcs CQ= PQR and C = RS. Then on C1 we have ) determined as in 7), while on C2 ( is determined by y = - Vr2 - _X2 = (x), the radical now having the negative sign. Corresponding to this we would break 2) into two integrals j and f. To each of these our proof applies. 3. In 2 we have taken the equation of the circle as x2 + y2 = r2, (8 which defines y as a two-valued function ~ V/r2- x2. Instead of the equation 8) we may use the equations of the circle in parameter form x=rcos, y=rsin u. (9 If we do this, we can avoid the radical and so deal from the start with one-valued functions. In general, let x. = (l r 1n - _- V ly, - irY \.^ INTEGRATION 153 be the parametric equations of a curve C; that is, when u ranges from u = a to u = /3, the point x, y defined by 10) describes the curve C. We suppose of course that d, r are one-valued continuous functions in U = (a, /3), with continuous first derivatives. Then f(x, y) = f f~ (u), ' (a) f = g (U) is a continuous function of u in U. Let us effect a division of U of norm 8 by interpolating the points a,, a2 **. To them will correspond certain points a, a2... on C. Then by the law of the mean AXm = 0(0om) - K(a-l) = 00(V.)Az,, where vm is some point in the interval (a,1_l, a,^). To this point Vm will correspond xm, ym in the arc (am-i, am) on the curve C. Thus f (X, m Ym)Axm = 1g (vm) ~f (vm)aum. If we let the norm - 0 in this relation, we get in the limit f (x, y) dx = g (u) ' (u) du. (11 4. In precisely the same manner the y-curvilinear integral gives f (xI y) dy = g(u)' (u)du. (12 77. Work. 1. Let us show how the notion of curvilinear integrals presents itself naturally in mathematical physics. Suppose a particle is acted on by a force { whose components are X and Y, as in Fig. 1. The work b done in passing from P to a point Q near by/ T on the curve C is dW= cos. ds, (1 a x where 0 is the angle between a and the tan- FIG. 1. gent T. If a makes the angles a, 3 with the x and y axes, and T the angles ac, /3 with these axes, we have, from analytic geometry, cos 0 = cos cos a' + cos 3 cos /'. (2 154 FUNCTIONS OF A COMPLEX VARIABLE Now acos =X, acos/ = Y, dx = cdos ds ds Thus 1) and 2) give dW= Xdx + Ydy. (3 Thus the work done by the particle in passing from a to b along the curve C is W = (Xdx + Ydy) (4 =J x +fY. = Xd.r +\ Ydy. a/~i a (5 It is therefore the sum of two curvilinear integrals. 2. The relation 3) may be obtained more yt simply by referring to Fig. 2. The work performed in passing from P to a point Q very near is in general the same as if the particle p took the route PR, RQ. The work done along PR is Xdx; the work done along RQ is Ydy. The total work dWis the sum of these or 3). dx FIG. 2. 78. Potential. 1. In physics we often deal with forces a whose components are the partial derivatives of some one-valued function V(x, y), that is, T T-7' - V Ox Y= - _O ay (1 In this case 77, 3) gives - dW= a dx+0 dy = d, ax ay and the element of work is the total differential of the function V, with sign reversed. Let us suppose that v, - are continuous. We can then Ox Oy show that the work done in passing from a to b is independent of the path. For let us effect a division of norm 8 of C by inter INTEGRATION 155 polating points a,, a2,... between a and 6. Then by 73, 4 v=OV Ax + y + aAX +, ay ax ay =- xax - YAy + aCx~ + Ay where a, 8 0 with Ax, Ay. Thus we have V(a) - VF(al) = Xjzxl + Ylay1 ~ alAx, + I3yl V(al) - V(a2) = X2AX2 + Y2zy2 + a-A2+ 1f842 V(an-1)- V(6)= XA n + Ylyn, ~ aCnxn + /,Ax,. Now by7.,5 7mi and 3lf are all < some e for all norms 8 < some 80. Thus, adding the foregoing equations, we get V(a) - V(b) = S (X,,x, + Ym,/m) + el where El 0 as 8 _ 0. Hence, letting 8 1 0, the last relation gives IY= V(a)- T(6)=f(Xdx~+ Ydy). (2 As V(x, y) is one-valued, the value of the work done by the particle in moving from a to 6 is independent of the path taken, and depends thus only on the end points a, b. 2. It is sometimes useful to know that the relation 2) holds, in a certain sense, when Vis not one-valued. In fact the foregoing reasoning is valid if V(x, y) is only one-valued about each point of the curve Cand possesses continuous first partial derivatives as before. Suppose then that V,, is the determination of V(x. y) with which we start at a. If Vb denotes the value that V,,, acquires on reaching 6, passinig over C, we see that we may write 2) W=f(Xdx i Ydy) = T7 - V. (3 3. The simplest case of a potential function is presented by several particles of masses mi, in,25 156 FUNCTIONS OF A COMPLEX VARIABLE A unit mass at P is attracted by rm by P(xy) a force whose components are, as seen by the figure, 01 X - "1cosO01 Yj = mini01. 712 r2 r2 1 2 1 1 2 1*o1t Similar forces are exerted by i2, i3, *... Thus the total force ~ exerted on a unit mass at P has the components X Mk cos k, r Mk sin 0k' (4 Let us set JT = (5 rk From = x2 ~ y2 we have rdr = xdx ~ ydoy, or dr = - dx +~ydy. r r Hence Or x 0=os; Or _ysin 0 Ox r ay r Thus a1 a OOh\O 1 cos 9, xrax/ Or\rJax r2 a 1I 1:G)= -2 sin 0. ay rr r2 Hence a F a ink COS 2k 2 1Mk, - Mk L1 Ox ax AK) We have therefore Ox ' ay and hence the function Vis a potential function for the force ~. 79. Electric Current. 1. Suppose a constant current of electricity is passing along the wire PQ. The lines of force generated by this current are circles C as in Fig. 1. The intensity of the force ~ is given by P= -, r INTEGRATION 157 where a is proportional to the strength of the current. We have here, as shown in Fig. 2, Q 1' X=- Fsin60= -c1 r2 Suppose a unit mass of electri the circle C. Then x=rcosO = x Y=- c r2 [city to describe r sin 0, 0 C dx=-rsin dd, dy = r cos MO. The work is W=f(Xdx + Ydy) = cf2 (sin2 0 + cos2 0)dO = f dO G1 0 xY 04 = 27 e. (1 2. Let us now suppose that the unit mass is restricted to move in a connex 21 acyclic with respect to 0, as in Fig. 3. Let us set V= - c. arc tan ff. x Then av aX2 Ox x2 + y2 x2 r2 aov= - C = Y. ay r2 (2 IC FIG. 2. xy 00 (3 (4 Thus V is one-valued in W and has continuous first partial derivatives. We can therefore apply 78, 2), which gives as the vork done moving from a to b, W=e {arctg 7 - arctg j FIG. 3. by the unit charge (5 where x'y', xfit" are the codrdinates of a, 6. The work WV in moving from a to 6 is independent of the path between these points, provided only it lies in W. 158 FUNCTIONS OF A COMPLEX VARIABLE 3. Let the convex a be cyclic with respect to 0 as in Fig. 4. The origin 0 is the point where the current pierces the x, y plane. It is excluded from 1 by a small circle. The partial derivatives of V are onevalued and continuous in f as 3), 4) show, but T is no longer one-valued in f. / For when the point x, y makes a circuit about 0, V has increased by -2 err. Thus, if Va is the determination of V chosen at the point a, after the circuit Va has acquired the value - 2 er. FIG. 4. Thus, if we apply 78, 3), the work done by a unit charge moving around the circuit C is W= 2c7r. This agrees with the result found directly in 1). 80. Stokes' Theorem. 1. The theorem we now wish to prove is a special case of a theorem due to Stokes and which is much used in mathematical physics. For our purposes it may be stated thus: Let F(x, y), G(x, y) be one-valued functions having continuous first partial derivatives in a connex 9E whose edge we denote by A. Then (Fdx + Gdy) = ffaa- 2)dxdy. (1 In calculating the curvilinear integral on the left we let the point x, y run over e in the positive sense, that is so that the region bounded by ( lies to the left of the direction of motion as in Fig. 1. Let S be the projection of a on the x-axis, and ( the section of a at a point x of 3. Then Jet By J Jady as we saw in 75, 2. At a point x = a in Fig. 1 f dy = (2 - F3) + - (4 ) = (2 - 1=) ( - F) (F4 - F3), Jsay INTEGRATION15 159 where F1 denotes the value of -F(x, y) at the point 1, etc. At a point as x =/3 in Fig. 1 the right side becomes F2 - F,. Thus in any case2 2 and hence 3 J F dxdy dxY-(-F,+-F.). (.2 Let us show that the right side4 is equal to11 -fdx. (3 o In fact to calculate 3) we FIG. 1. break (Y into a number of arcs such that for each arc y is a onevalued function of x. Along the lower arc AB of Fig. 2 let y = Yi' along the upper arc let Y = 1/2* Along the lower arcs C-D, BEF let y = Y4, and along the upper arcs let Y =y3. Thus frdx fF(xl y1) dx- 2 F,y2)dx d 3d y4)dx. FIG.d2 + fdxx(,~ -F) (4 160 FUNCTIONS OF A COMPLEX VARIABLE as in 2). Thus 2), 4) give / Lddy=- dx. (5 In the same way if C denotes the projection of 21 on the y-axis and B a section of 2 parallel to the x-axis, we have a -- dxdy =fdy j G dx Ji ay J JB Ox = fdyY;(Csl-(e). (6 On the other hand, taking account of the positive direction of e G dy = fdxwX(Gs+- - Gs) (7 Hence from 6), 7) we have f a dxdy = Gdy. (8 On subtracting 5) and 8) we get 1). 2. As a physical application of Stokes' theorem let us return to our line integral == (Xdx + dy), (9 ta which expresses the work done by a unit mass moving from a to b along some curve C. in a field of force ~ whose components are X, Yas explained in 77. We saw that when j has a one-valued potential V(x, y) whose first partial derivatives are continuous in some connex 2, the value of W is the same for all curves C in 1 leading from a to b. This condition is sufficient to make the value of W independent of the path C. In this case X V - V ( x=-a= - ' = - — y-. (10 Ox ay If X, Y have continuous first partial derivatives, Stokes' theorem shows that if C,, C2 are two paths leading from a to b anrd E the INTEGRATION 161 connex they bound, the work done in running over the boundary ( = C1. + -1 of is W=f(Xdx + fdy) f(aX - adY)dxdY (11 \y OxY Now from 10) OX aY 02 Oy ax axay Thus 11) gives W=f -_ = 0, or the work performed along C1 is the same as the work done along C2, as it should be. This gives us nothing new. But let us reverse our reasoning. Let us suppose that X, Y have continuous one-valued first partial derivatives in a certain connex I. We ask what condition must X, Y satisfy in order that Wis independent of the path C? The answer is that ax aY =X _dY (12 Oy Ox must hold at each point of 3W. For suppose it did not hold at a point c, in Fig. 3. Then within some domain D about c, oX OY y Ax ' being a continuous function of x, y, must have one sign, by 69, 7. Let y be a circle / b with center c and lying in D. Then by 2 Stokes' theorem / (Xd + Ydy) = ( - )dxdy, a Jay ar\^ ) where F is the region bounded by y. FIG. 3. Now the right side cannot = 0 since the integrand has one sign in r. Thus the work done in going around ry is not 0, or (13 162 FUNCTIONS OF A COMPLEX VARIABLE Let us now go from a to b along opposite sides 7'y 7 of y. Suppose the work We We2 for these two paths C1, 2 were the same. Then Wc C' = O. CBut1 = aay1. * /b b. * rlaa. Also Waa =- Waa, Ib - Wb Wy 7-1 =WY which contradicts 13). Thus WC - WC, Cf2= WI = O, which contradicts 13). Stated in mathematical language these considerations give: Let t, G be one-valued functions having continuous first partial derivatives in the connex W. In order that the value of f(Fdx + Gdy) shall be the same for all paths in a leading from a to 6, it is necessary and sufficient that in 1. (14 ay ax CHAPTER VI DIFFERENTIATION AND INTEGRATION 81. Resume. Before going further, let us take a look back and see what we have accomplished so far. In Chapter I we have established the arithmetic of complex numbers. It is thus possible at this point to define algebraic functions of a complex variable z, since the definition involves only rational operations. The reader will recall that a rational function of z is defined by an expression of the type R(z)= ao + alz + - + abz", m, n positive integers, which obviously involves only rational operations. An algebraic function w of z was defined by an equation of the type wn + Rlwn-l + *..1w + R_ + R -0, where the coefficients are rational functions of z. Thus the definition also only involves the rational operations of addition, subtraction, multiplication, and division on the variable z. The transcendental functions ez, sinz, logz, sinh z, arc sinz.* (1 cannot be defined in this simple manner. The definitions we have chosen as the most direct and simple employ infinite series. We have therefore developed the subject of series. Now the convergence of a given series A whose terms are any real or complex numbers is of prime importance because divergent series are not employed in elementary mathematics. To test the convergence of A we pass to the adjoint series aI when possible, because the terms of W are real and positive. Thus we are led to consider first the theory of series whose terms are real, and especially those which 163 164 FUNCTIONS OF A. COMPLEX VARIABLE are positive. This we did in Chapter II. In the next chapter we studied series whose terms are complex, and in particular the important subject of power series. Having developed the theory of infinite series as much as needful we were in position to study in Chapter IV the elementary transcendental functions 1). At the same time we took a brief survey of the algebraic functions. The next topic in order would be the calculus of these functions, that is, we should learn to differentiate and integrate these functions just as is done for a real variable x. In order to treat this subject clearly we have inserted a chapter, the foregoing one, whose object is to furbish up the reader's knowledge of the calculus and to emphasize certain points of theory which are usually passed over hurriedly in a first course. We also developed the notion of a curvilinear integral which is the foundation of the following chapters. These matters having been looked after, we are now in a position to take up the differentiation and integration of functions of a complex variable z. But first let us define more explicitly a function of z. 82. Definition of a Function of z. 1. We have already defined a number of functions of the complex variable z, viz.: the algebraic functions, ez, sin z, log z, etc. These we called the elementary functions. From these we can form more complicated functions of z as w = 2 + sin zVl + z8. 1 - Z All such expressions will be called functions of z just as they would be in the calculus if z were replaced by the real variable x. Any such relation establishes a relation between z and w as follows. For each value of z which belongs to a set of points 2f in the z-plane, one or more values are assigned to w. We now generalize as in 66 in this manner. Let I be a point set in the zplane. Let a law be given which assigns to the variable w one or more values for each value of z in W. Then we say w is a function of z in W2. If w has but one value for each z in 21, w is a one-valued function in 2l, otherwise many-valued. DIFFERENTIATION AND INTEGRATION 165 For example, the relation /z2_ 1 w== ---- =1 sin z assigns to w two values for each z not among 0, ~77r, ~ 27 r, * (2 except at z = ~ 1, where w has but a single value, w = 0. Thus w is a two-valued function in the point set 3 which embraces the whole z-plane excluding the points 2). The branch points of this function are z = 1, that is, when z describes a circuit about one of these two points, the two values of w permute. By means of this two-valued function we can define a one-valued function of z. In fact let 23 be an acyclic part of P1 relative to the points z = + 1, and z = - 1. For example, let ~3 denote the points of I\ which lie to the right of the parallel to the y-axis, x = 1. At the point z = 2, w has two values / and - sin z sin z Each of these may be used to define a branch of 1) and this branch is a one-valued function of z in S3. If instead of 3 we take a cyclic set ( relative to + 1 or - 1, the function of z just defined, which is one-valued in S3, is two-valued in E. Thus a function which is one-valued relative to one domain may be many-valued in some other. Conversely by taking on a part S of the domain of definition 3f of a many-valued function we may employ one of its branches to define a one-valued function of z relative to S. 2. It is important to remember that the functions we deal with in the following are one-valued in the domain 2f under consideration unless the contrary is stated, or unless it is obvious from the matter in hand. We make also another limitation. The domain for which a given function w is defined will always be a region [70, 1], unless the contrary is stated. For example, the domain of definition f of the function 1) is a region. For z = a being any point of 1 we may obviously describe a circle c about a such that all points within c belong to 7I. The 166 FUNCTIONS OF A COMPLEX VARIABLE reader will also note that the domain of definition of all the elementary functions defined in Chapter IV are regions. For example, the domain of w = log z is the point set 1 formed of the whole z-plane excluding the origin z = 0. This function is infinite-valued in 1; but any one of its branches is a one-valued function in a connex acyclic relative to z= 0. As another example, the domain of definition of sin z w = tan z = — cos z is a set 1 embracing the whole z-plane excluding the points r + mr. Obviously, 2 is a region. 3. Let w be a function of z defined over some point set 2. To each point z = x+iy in 32, w will have one or more values, w= u+ iv. (3 The values of u, v will depend on the position of z in 21, that is, on the values of x, y. Thus u, v are real functions of the two real independent variables x, U y. If w is one-valued, u so are u and v. z —=+iy, Conversely, let o x v (x, y), v(, y) be two real functions of the real variables x, y defined over some domain W. If we set z = x+ iy, (4 then to each point x, y of 21 will correspond a value of z. By means of 3) we can now define a function w of z by stating that at the point z, w shall have the value 3) when u and v are given the values that they have at the point x, y corresponding to this value of z as defined in 4). Example 1. Let u=ex cosy, v = e sin y. (5 DIFFERENTIATION AND INTEGRATION 167 To a given value of z, correspond a pair of values x, y determined by 4). For these values of x, y, the relations 5) define the values of u, v. These put in 3) determine the value w has for this value of z. Thus w = ex cos y + iex sin y is a function of z. It happens to be the exponential function ez [54, 5)1. Example 2. Let u-=2 +y2, V=-2xy. Then we must consider from the foregoing definition = (X2 + y2) - i. 2 xy as a function of z. 4. Images. Let w =f(z) (6 be a function of z defined over a point set Wf. When z ranges over X, w will range over some point set, call it A, in the w-plane. It is convenient to call 9 the image of.I afforded by 6). We write This we may read: 3 is the image of 31, or 3 corresponds to 1. The relation 6) establishes thus a relation between the points of 1 and 3. If f is a one-valued function in A1, to a point z = a in 21 will correspond but one point wz = b inl. If f is on the other hand a many-valued function in X, to z = a will correspond more than one point in 3, as b', bV, b"'...* If b""') is one of these points, we may write b(al)' which we read: b(m) corresponds to a. When to each point a in 1, there corresponds but one point b in 9, and to each b in 3 but one a in 21, we say the correspondence between 32 and 3 is one to one or unipunctual. This we may write 23 1 3, unipunctually. 83. Limits, Continuity. 1. Let w be a one-valued function of z defined about z = a. Suppose as z - a, that the values of w converge to some value 1. We say I is the limit of w for z = a and write lim w=l; or w I as z= a. (1 z=a 168 FUNCTIONS OF A COMPLEX VARIABLE Geometrically this means that having described an e circle about the point I in the w-plane there exists a 8 > 0 such that when z is restricted to lie in a 8 circle about z = a, the plane wplane center excluded, the corresponding values of w fall in i) the e circle. ~ o Expressed in e language the relation 1) means that for each e > 0, there exists some 8 > 0 such that Iw-11 < e for all 0 < z -a < $. (2 The reader will note the perfect analogy of this definition with the definition of a limit given in Chapter V where the variables are real. From this follows that the ordinary properties of limits employed in the calculus will also hold here. Thus if f(z) r, g(z) s as z a, then ~then ~ lim(f+ g) = r + s, z=a limf.g= r. s. z=a If Ig(z)l > some y > 0 near z = a, limf = r, z=a g s etc. 2. Suppose we write w=u+iv, = X + i, a = a + i, where u, v are one-valued functions of x, y about the point a, /. Obviously if uX, V (3 as x, y converges to the point a, i, then w =X+ i= I as z a. Conversely if necessarily 3) holds. necessarily 3) holds. DIFFERENTIATION AND INTEGRATION 169 3.. Continuity. Let z describe a continuous path P in the z-plane; if the corresponding values of w define a continuous curve in the w-plane, we say w is continuous. To obtain an analytic form-ulation of this we have only to repeat the considerations of 69, with slight modifications. This leads us to define as follows: Let w be a one-valued function of z defined about the point z= a. Let w have the value a at z =a. If lirn w = a, z=a we say w is continuous at a. If w is continuous at each point of some domain I, we say w is continuous in W. 4. As in 2, let w = u + iv. The same considerations show that for w to be contilnuous at z = a = a + i13 it is necessary and sufficient that u(x, y), v(x, y) be continuous at the point a, 3. 5. If iw is continuous and /= 0 at z = a, w is =7 0 in some circle c described about a. For let wv = a at z = a. Then for each e > 0 there exists a c such that Iw (z) - a < e for any z in c,, or what is the same, o a - e < () < a +. (1 If we take e such that e < I a\ = R, this relation shows that Iw(z) I> r where 7 = al - E >0 (2 for all z in c. 6. If w is continuous at z = a, I (z) < some G (3 for any z in some circle c about a. This follows at once from 1). 7. The inequalities 1), 2) may be extended to any connex G as follows: If w is continuous and -- 0 in the connex A, the numerical value of w never sinks below some positive constant 7 in S, or w(() I > n > 0 in. (4 I I - 170 FUNCTIONS OF A COMPLEX VARIABLE For suppose w 0 as z ranges over a set of points al, a2.. in ( which converge to a. In symbols suppose lim w(a,) = 0. (5 fs = 00 Now t being continuous, w (a) = lm w (a,). in = oo Thus w = 0 at z = a by 5), and this contradicts the hypothesis that w = 0 in G. 8. If w is continuous in the connex C, w(z) I < some G, in C. (6 For if not, suppose w (z) I + o as z ranges over some set of points a1, a2... in ( which - a. But w being continuous at z = a, the relation 3) holds in c. But then lim I w (a) \ cannot be + oo. n=oo Thus if 6) does not hold, we are led to a contradiction. Differentiation 84. 1. Let w be a one-valued function about the point z = a. When the independent variable z passes from z = a to z = a + h, that is, when z receives an increment h Az, the function w(z) receives an increment Aw = w(a + A) - wv(a). The quotient Aw w(a + h)- w(a) Az h is called the difference quotient as in the calculus. If lim w(a + h)-w(a) (2 h-=0 h exists, we say w has a differential coefficient at z = a, whose value is the limit 2). It is denoted by w'(a). If the limit 2) exists DIFFERENTIATION AND INTEGRATION 171 for each point z of a region X, it defines a function of z denoted by dw dz or by w'(z) and called the derivative of w(z). The value of w'(z) at z=a is of course w' (a). 2. The reader will note that this definition of the differential coefficient w'(a) is entirely analogous to the definition when the variable is real, given in 72. The only difference lies in the fact that in the calculus, h is restricted to move on an axis about the point h = 0, while in 2) h is any complex number = 0, in some circle about the point h = 0. 3. Let us note that if w has a differential coefficient at z = a, w must be continuous at a. For by hypothesis the limit 2) exists and is finite. As the denominator h ' 0, the numerator must also ' 0. But then w(a + h) ' w(a), which is the definition of continuity, 83, 3. 4. By reasoning exactly as in the calculus we can show that d(f + ) =f'(z) + '() ( (3 dz d (fq ) =r A' + Jf', (4 dz d(f 9=gf'-fg' (5 hold under the same conditions as when the variable is real. To illustrate this let us show that 5) holds in any region A1 in which g*0. For let us set h = Az, w=f, g=g(z+ h) Then Aw gAf- fAg lAf f. 1A ( Az g#Az jAz y Az 172 FUNCTIONS OF A COMPLEX VARIABLE If Az is taken so small at a given point z that z + h lies in f[, g is = 0. Next we note that lim g = lim g(z + h) = (z), az=o h=0 since g is continuous by 3. Passing now to the limit Az = 0 in 6), we get 5). 5. By the aid of the foregoing we can find the derivative of a rational integral function f= a0 + alz + a2z2 + * + am. For as in the calculus we show that dz n dz Thus by 3), df dz a + 2 az + *.. + mamz-l. Also the derivative of a rational function k=ao+alz+ ~.+a f ho + lz+ + 6 bnz g can be found by 5). 6. Let us prove here a theorem we shall need later. If w =f(z) has a differential coefficient f'(a) = 0 at x = a, there exists a 8 > 0 such that Aw does not vanish when z =a + Az is restricted to D^(a). For as Aw lim - =fl (a), at z = a, Az=O Az we have Aw = f' (a) + et Az (7 where | ' I < e if only 0 < |z I < some 8. If now we take 0 < e < f'(a) I we see that f'(a)+ e' cannot vanish when 0 <[ Az < 8. Thus Aw cannot vanish under this restriction, as 7) shows. 85. The Derivative of a Power Series. 1. Let the power series P(z) = ao + alz + 2 +... (1 DIFFERENTIATION AND INTEGRATION 173 have ( as a circle of convergence. We show that P has a derivative within (, viz.: P(z) = al + 2 a2z + 3 a32 + (2 For by 43, 5) P(Z + A) =P(z) + hPi(Z) + h2P,(z) + (3 where where P1 = a1 + 2 a2z + 3 a3z2 + (4 which is the series on the right side of 2). As z is an arbitrary but fixed point, let us write 3) P(z + h) = b0 + b1h + b2h2+. (5 This converges absolutely as long as the point z + h lies within L, that is as long as h I ome.=h1I< some a. The adjoint of 5) is S 3 0 + = 3, +213 2 + and as this converges for 7 = 8, /0 + /18 +4 22 -+..- /+ - 1 + 82-2 + 238 + 4 -82+.. is convergent. Hence - =,2- +38 + 482 +- (6 is convergent. From 3) and 5) we have Ap P(z h)-P(z) = pl(z)+h hb2 + b3h + Az h = P(z) + Q. (7 Now each term of Q = h + h2 +. is numerically < the corresponding term of the series 6) when \hl< 8. Thus | QI <, a constant. Hence hQ- 0 as h 0. Hence, passing to the limit h= 0 in 7), we get 2). We have thus this result: The function of z defined by a power series 1) has a derivative within its circle of convergence, which is obtained by differentiating 1) term by term. 174 FUNCTIONS OF A COMPLEX VARIABLE 2. Let us show that: The derivative series 2) has the same circle of convergence ( as the series 1). For let z be any point within 6; let b be any point within 6 such that H<1. (8 Since 1) is convergent at b, gan/3 < some M n =0, 1,2... by 30, 3. Let us now look at the adjoint of 2); it is +2 +3^ +... (9 0c, + 2 U2~ + 3 + 3P + * (9 Its mth term is mam -= f(-l: mamPm- m-L~ Cm-1 Thus each term of 9) is < the corresponding term of the series Dim (U~m-l (10 This last series is convergent by 21, Ex. 1 by virtue of 8). Hence 9) is convergent and hence 2) is absolutely convergent. The series 2) cannot converge for any z without (. For then 9) would converge for some 1 > the radius of (. Thus (a + lr + 2 a232 + 3 +... is convergent for this value of '. Hence afortiori ao + alg + a2 2 ++ 3+3 +.* "+ is convergent, and thus 1) converges at a point without (, which is impossible. 3. Since =1 +++.. 1! 2. z 3 Z5 sin z = - + - 1! 3! 5! Z2 Z4 cosz= 1 - -- + 2! 4! DIFFERENTIATION ANID INTEGRATION 175 etc., we have, differentiating these series terrwise, dez 2 dz 1! 2! d sinz z2 ~ * -= - X = COS ZI dZ 2! 4! deosz z z3 dz 1!! Similarly dl. sinh z h d. cosh s cosh z z sinh z. dz 86. The Cauchy-Riemann Equations. 1. In the foregoing article we have been able to find the derivatives of ez, sin z, sinh z... because these functions are defined by means of power series. In other eases the following theorem is of great service; it also has a deeper significance froom a theoretical point of view. Let wv -- u ~ iv be a one-valued function- of z = x + iy in the domain 2f. Let u, v considered as 'functions of the real variables x, y have continuous first partial derivatives which satisfy au= av auu -v Ox~~y ' Oy~~~x (1 ax ay ay ax in WI. Then w has a derivative in W and dw au.Ov Ov 1 au + - = -+ (2 dz ax Ox Oy i ay For aw au AV = + - -- Az Az Az But by 73, 4, A au au Au=-Ax + -Ay + anx +IAy, (3 Ox ay Ov Ov Av = ax +av Ay + yAx + 8Ay, (4 Ox ay where 4are all< 176 FUNCTIONS OF A COMPLEX VARIABLE if 0 < Az I < some 7. Thus, using 1), - (Ax + iAy) + i (Ax + iAy) Aw ax Ox Azs Ax + iAy = + ~ + E, (5 Ax Ax where, ax= c +,Ay + i(yAx + SAy). Az Now |A|x, lay[ are < azl; hence hence!E' II I+ + I I + + II<e. This says that E'=0 as Az-O Hence, passing to the limit Az = 0 in 5), we get 2). 2. The equations 1) play a very important part in the theory of functions. They are called the Cauchy-Riemann equations. From 5) we have, on using 2), Aw = w'(z) + I Az, (6 where e ' 0 with Az. For later use we note here an important property of C': Let w(z) be one-valued about each point of a connex (, and let wt (z) be continuous in C. Then e - 0 uniformly in Q; that is, for each e > 0, there exists a s > 0 such that I1 <e provided O < I Az I <; moreover the same 8 holds wherever z is taken in G. For by 73, 5, a, /, y, - 0 uniformly in G. 87. Derivatives of the Elementary Functions. 1. Let us apply the theorem of 86, 1 to find the derivatives of the elementary transcendental functions. We have w = ez = ex(cos y + i sin y) so that here u -= ex cos y v = ez siny. We have at once - = ex cos y = ax dy Ou. Ov -= - ex sin y= --- ay ax DIFFERENTIATION AND INTEGRATION 177 As these are continuous functions of x, y in the whole x, y plane, the Cauchy-Riemann conditions are satisfied. We have, therefore, by 86,2) dw du. dv d = a +f z- = ex cos y + iex sin y dz ax ax = ex(cos y + i sin y) = ez. (1 This result agrees with that already found in 85, 3 by another method. The method just employed may be used to prove the more general relation = aea (2 dz 2. Similarly, we can show that d. sin z -- =cosz. (3 dz For w = sin z = sin (x + iy) = sin x cosh y + i cos x sinh y, by 58, 13) = + iv. Here - =c cosh = -, dx Oy Ou.. dv = sin x sinh y = Oy ax These derivatives are continuous and satisfy the Cauchy-Riemanin equations 86, 1). Hence d ~ sin z --- = cos x cosh y- i sin x sinh y dz = cos z, by 58, 14), which is 3). Another way to establish 3) is to start from =eiz e-iz sin Z= --- 2i which we derived in 58, 8). Then by 2), d sin z ieiz + ie-iz eiz + e-iz ---- == ----- == ---- == — COS Z dz 2i 2 by y58, 7)). 178 FUNCTIONS OF A COMPLEX VARIABLE 3. Let us now show that d * log z 1 ( dz z taking one of the branches of log z, say log z = log r + i) + 2 mri. Hence Hence qu=logr, v= + 2m7r, where r = Vx2 + y2, -=arctg. Now at any point different from the origin Ou Ou Or 1 x x Ov Ox Or Ox r r r2 Oy Ou _ Ou Or 1 y _y _ Ov Oy Or Oy r r r2 Ox Thus the Cauchy-Riemann conditions are satisfied, and d * log z au v x dz Ox Ox r2 r2 = - (cos < - i sin qb) r 1 which is 4). 4. In a similar manner we find: d - arc sinz 1 darctgz 1 dz/- V 2 ' dz + 2' d * sinh cosh z d cosh z dz dz In the first equation of 5) we must choose the right branch of V/1- z2 for the particular branch chosen for arc sin z, just as in the calculus. 88. Inverse Functions. 1. Let w be a one-valued function of z in the domain WI. As z ranges over 2, let w range over a domain 3, in such a way that to each point w in 03 corresponds but a single DIFFERENTIATION AND INTEGRATION 179 point z in T1. Then the relation =f() (1 may be used to define a one-valued function of w, z = g(w) (2 in the domain 3. We call this the inverse function. If, on the other hand, to several values of z in 2 corresponds the same value of w in 3, the relation 1) may be used to define a many-valued inverse function. We have already had examples of inverse functions. Thus w = ez defines the logarithmic function. We note that S is the image of 1 afforded by 1). When the inverse function 2) is one-valued in A3, 23 1 I, unipunctually. 2. Let us now consider the derivative of the one-valued inverse function 2). We have the theorem: If is Z 0 in 21, the inverse function has a derivative dz dz 1 dw dw in (3 dz For Az Aw Aw' (4 Az provided Aw O. Now by 84, 6, Aw:= 0, if we take 0 < Az < some S. Thus, passing to the limit Az = 0 in 4), we get 3). 3. We have already found the derivative of log z directly from its analytic expression log r + i + 2 mari. It may, however, be found much more easily from the theorem2 above. We start from w = ez. We have seen in 62 that log w is one-valued in any connected region 3, acyclic relative to the branch point w = 0. While w 180 FUNCTIONS OF A COMPLEX VARIABLE ranges over 3, let z range over W. Since dw= cez dz never vanishes for any value of z by 54, 2, we have at once from 3), dz 1 1 dw ez w ~or ~ d. log w 1 dw w which is the result obtained by another method in 87, 3, the letters w, z being of course interchanged. 4. Let us find the derivative of the arc sin function considered as the inverse of w = sill Z. We saw, 64, 3, that the branch points of z = arc sin w are= i 1 w = ~ 1. Thus in any connected region 3 in the w-plane, which is acyclic relative to both of these points, any branch of the arc sin function, call it z, is a one-valued function of w. While w ranges over A, let z describe the set W. Then in ~2 d-= cos z dz does not vanish. For cos z vanishes only for 77 or ~, ~.... Z~i2 i 2 But for these points w = 1, and these points are by hypothesis excluded from the region A3. Thus all the conditions of the theorem in 2 are satisfied. We have therefore t darc sin w dz 1 1 dw dw dw cos z dz -=- l 2 (5 Vl - w2' DIFFERENTIATION AND INTEGRATION 181 where the radical must have the sign of cos z at B the point z which corresponds to the value of w in question. 5. In the calculus we have A d arc sin x 1 _ ----— 1 -/ l 1 dx -x2 As the radical is two-valued, the sign to be taken depends on the branch of the function we employ. Thus if we take the branch which passes through A in the figure, we must take the + sign. If we take the branch which passes through B, the figure shows that we must take the - sign. 89. Function of a Function. 1. Let us now extend the familiar relation dw dw dz dt dz dt for complex values, under certain restrictions. Let z be a function of t in some domain X. When t ranges over Z let z range over a domain 3 in the z-plane. Let w be a function of z in 3. Then w may be considered as a function of the variable t in T. Example 1. Let =si, w z = sin t, w = e2. While t ranges over the whole t-plane X, z ranges over the whole z-plane 3. Thus w int is a function of t in ~. Example 2. Let V Z = A/1 - t2, taking that branch which corresponds to z= 1 for t = O. Then z is a one-valued function in any connected region Z, which, as in the figure, is acyclic relative to the branch points t= ~ 1 of the radical. When t ranges over Z, let z range over 3. Let U)o =- lor i, 182 FUNCTIONS OF A COMPLEX VARIABLE taking that branch which = 0 for z = 1. Then w is a one-valued function of z in 3. Hence w log/ w = log V1 - t2 is a one-valued function of t ill. 2. We now prove the following theorem: Let z have the derivative dzin, and w the derivative in dzdt dz If d does not vanish in X, then dt dw dw dz in ~n -- m. (1 dt dz dt For A A A Aw Aw At 'A^~'A t ' ^' (2 provided Az z 0. But by 84, 6, this condition is satisfied if we take take ~ ~0 < I At I < some > 0 since by hypothesis d = 0 in T. Let now At _ 0, at the same dt time Aw - 0. Thus, passing to the limit At = 0 in 2), we get 1). 3. We may use the relation 1) to calculate the derivative of complicated expressions, just as we do in the calculus. Thus, let W = esiinz We set - w=eu, u=sin z. Then d =.. eui -- COSe. =eu= eSil_1z; d= cosz. du dz Hence d = esin Cos (3 dz for all z for which cos z = 0. For these exceptional values of z it is easy to show directly from w that dw= 0, so that the relation 3) holds even in this case. Az dz 4. Let us find the derivative of w = (1 + z), (4 where / is a constant. Then by 63, W = e0lg(l+z) DIFFERENTIATION AND INTEGRATION 183 Let us set uL = u log(1 + z), w=eu. The only branch point of u is z =- 1. Let then 3 be any connected acyclic region relative to this point. Let now u denote one of the branches of A log (1 + z); it is one-valued in 3, and du L dz +z As this does not = 0 in 3 we have, from 1), dw = (1 ( + )1 (5 d — ' l+z 5. We have proved the important relation 1) on the hypothesis dz that is = 0. This condition is imposed by the fact that our dt reasoning requires that Az cannot = 0 as At 0. In 118, 9 we shall see that the relation 1) holds even when dz= 0, provided dt dz. t - s a continuous function of t. dt 90. Functions having a Derivative. 1. Let us return to 86 and prove the important converse theorem: Let f (z) = u + iv be one-valued in the domain I and have a derivative f'(z). Then u, v satisfy the Cauchy-Riemann equations du dv Au v Ox ay ay Ox at each point of 21. For at any point z of 2 f Au_.Av (2 Az Az Az Since f'(z) exists at z the left side of 1) z+iAy - z+Az must converge to f (z) however z'= z + Az I Z+AX converges to z. Suppose we allow z' to z by making it approach z along a 0 parallel to the x-axis. As in general Az = Ax + iAy, 184 FUNCTIONS OF A COMPLEX VARIABLE we see here that Ay = 0. Then 2) becomes Af = A f Au + iV Az Ax Ax Ax Passing to the limit, we get f/ (Z) Au iv. (3 ax ax Let us now allow x' z z by making it approach z along a parallel to the y-axis. Then Ax = 0 and hence Az = iAy. Thus 2) becomes Au Af _ Af _ Au AV Az iAy i Ay Ay Passing to the limit, we get 1 au av f'(z) = - (4 i ay ay Comparing 3), 4) gives au.+ iv. + av ax ax ay ay Equating the real and imaginary parts gives 1). 2. Conformal Representation. Let w = f(z) be a one-valued function having a derivative in the connex W. Let eCl C2 be two curves within I which meet at z = a, making the angle 0 with each other. Iff' (a) = O, their images (S1-2 will cut at the same angle 8, at the point a o a. For let al, a2 be points on C1, C2 near a, as in the figure. Let a1 = f(a), a2 = f(a2). Then l - a = f'(a) + el}(al - a), a2 - a = f'(a)+ 2}(a2 - a). Since f (a) # 0, a2 - a is: 0 if a2 is sufficiently near a. Hence a_1-a _ a1_-a f' (a)+ ex (5 a2-a a2- a f'(a) + e2 Now the argument of the left side is the angle (> between the chords ala and a2a. The argument of the first factor on the right DIFFERENTIATION AND INTEGRATION 185 of 5) is the angle ( between the chords ala and a2a. Since E1, e2 are numerically small, the argument of the second factor on the right of 5) is a small number S. Thus, taking the arguments of both sides, we have = e + i on choosing e properly. Now as al and a2 a, O 0 and 8 - O. Hence 0 also. al z plane a2 a 7ia: W plane 3. This property of the representation of the z-plane on the w-plane afforded by a function w =f(z) having a derivative is of great importance in many applications of the function theory. We see that if f'(z) $ 0 in circle ( about z.= a, to any little triangle T in 6 will correspond a triangle X in the w-plane which is the more nearly similar to T, the smaller T is. This we may state briefly by saying: The image of an infinitesimal triangle T, in which f' (z) = 0, is a similar infinitesimal triangle X in the w-plane. For this reason the representation of the z-plane afforded by the function w =f(z), is said to be conformal, wheref'(z) 0. We have had examples of this conformality in studying the representations afforded by the exponential and the sine functions in 57 and 60. -Thus in the case of w = ez, we divided the z-plane into a set of rectangles and found that their images are a set of circles and their radii which, of course, cut each other at right angles. In the case of w = sin z, the rectangles had as images a set of confocal ellipses and hyperbolas which also cut orthogonally. 4. The reader should note that if f(z) is not one-valued or if f (z) is 0 or does not exist at z = a, the reasoning in 2 breaks down. We cannot say the representation is conformal at this point. 186 FUNCTIONS OF A COMPLEX VARIABLE For example, w= Vz which we studied in 49, is not one-valued at z = 0. Two radii in the z-plane passing through this point and making an angle 0 with each other have as images two radii going through the point w = 0 and making an angle 2 0 with each other. Thus the representation is certainly not conformal at this point. Integration 91. Definition. 1. Let f(z)= u + iv be one-valued and continuous on the curve C whose end points are a, b. We will suppose the equations of C are given by x = (t), y = (t) ( as t ranges over an interval: = (cc, /). We will suppose that 0f(t), #'(t) are continuous in; also that the correspondence between C and T is unipunctual. Let us effect a division of Z of norm }q by interpolating the points \ tP t2, t3 b 0 To these points will correspond the l points Z1, 2 Z... (D \ on C which effect a division D of norm 8, say, of C. Moreover,,., a tl t2 P 3-0, aas 7-0. Let us now calculate the sum Ef(Zm) AZm =f(z1)(z1 - a)+ f(z2) (2 - ) +... (2 Since Az = Ax + iAy, we have f (z)Az = (u + iv)(Ax + iAy) = uAx - vAy + i(uAy + vAx). Thus Ef (z)Az = 2(uAX - vAy) + i(UAy + vAx). The sum 2) has, therefore, the value; (UmAXa - VmA/ym) + i (UmAym + VmAXm). (3 INTEGRATION18 187 Suppose now we let 8 cSI 0, the sums in 3) converge to curvilinear integrals. Thus the limit of 2) exists; we denote it by ff(z)dz or by ff(z) dz. (4 We have, theref ore, ff(z) dz =lim If (zm,)Azm, f(udx - vdy) ~ f(udy + vdx). (5 2. Example. Let us evaluate,Z 2dz, where C is an arc of the circle x=rcost, y= rsin t. Here f (Z)= Z2=(X +iy)2 =X2 -y 2+2 xyi, hence -,v2y dx =- rsin tdt,dy= r costdt, udx =-r3 (cos2 t - sin 2 t) sin tdt =r3(sin3 t - cos2 t sin t)dt, vdy - 2 r~cos2t sin tdt, udy - r3(cos2 t - sin2 t) COS tdt, vdx =- -2 r3 cos t sin2tdt. Thus ffdz - r3~ (sin t 3COS2 t Sinl t)dt + ir j 3cost - 3 sin 2t COS t)dt. (6 In particular, we note that if C is the whole circle, call it (Y, fZ2dz = 0. (7 92. Properties of Integrals. 1. The definition ffz) dZ I I 1 If (Z.)AZm.( 188 FUNCTIONS OF A COMPLEX VARIABLE of an integral of a function of a complex variable is entirely analogous to the definition when the variable is real. The only difference is the path of integration; in one case it is a piece of the x-axis, in the other it is a curve C in the z-plane. From this we conclude that many of the properties of integral developed in the calculus can be extended to the integral 1). Thus we have f f()dz = - f(z)dz (2 Mb rc b f/ (z)dz =f fdz+ ~ fdzs (3 where c is a point on C. r*b ab rb (f + )dz =f fdz + gdz. (4 Ca a a/ 2. As an exercise let us prove the very important relation f (z)dz < GO, (5 where |If(z) I, on C, (6 and C on the right of 5) stands for the length of the path of integration C. We have at once I Ef(zm)Azm < f (zm) | A | <_ E I AZm | (7 on using 6). As AZm = Zm -Zmwe see that IAzml is the length of the chord joining the points m-1, Zm on C. Thus, referring to the figure in 91, I AzmI (8 in 7) is the length of all the chords corresponding to the division D of norm 8. Now by definition the length of the curve C is the limit of 8) as 8 = 0. Thus passing to the limit 8= 0 in 7), we have 5). 3. From 5) we have the useful relation f)d <2 G n an integer (9 (z- a)n — Rn-1 INTEGRATION 189 where C is a circle of radius R about z = a, and If(z) I <G on. For the integrand is here f(Z) G f/z) which is numerically < (z - a)n RA and the length of C is 2 wrR. 93. Fundamental Integral Theorem. 1. In 91, 5) we have seen how the calculation of an integral may be reduced to that of two line integrals. In a great many cases it may be effected by a far simpler formula, as the following theorem shows. Let F(z) be one-valued about each point of a connex 7f, and have a continuous derivative f (z). Then f (z)dz = - (b) - (a), (1 where if F7(z) is many valued in Xf, F(b) is the value which F(a) acquires as z rancges over the path of integration C(. For let us effect a division of C of norm 8 by interpolating the points z1, 2 *.. Zn-. Then by 86, 6), -F(Z) - I(a) =f(zl)Azl + E1iAZ _(z2) - F(z1)f (Z2)/AZ2 + -2AZ2 '(b) - E(zn-1) =f/(Z)AZn + enAn. Adding, we get E(b) - F(a) = f (zm)Azm + C6mAZm (2 Now by the theorem in 86, 2, the lEml are all < e for any 8 < some 80. Thus the last term in 2) is numerically - EyI Az ~ I < eC where C is the length of C. This shows that the last term in 2) has the limit 0 as 8 0. Thus passing to the limit 8 = 0 in 2), we get 1). 190 FUNCTIONS OF A COMPLEX VARIABLE 2. The relation 1) is merely an obvious extension of the similar relation in the integral calculus. It is just as useful in the function theory as it is in the calculus. 3. A particular case of the theorem 1 and one of especial value is: Let F(z) be one-valued in the connex I and have a continuous derivative f(z). Then fTf(z)dz O (3 for any closed curve (C in W. 94. Examples. To make the reader feel perfectly at home with integration in the complex domain, we give now a number of examples. -Example 1. Let us evaluate zmdz m a positive integer. Here 1 f(z)= Zm E(z) = m+l. m+l Thus F(z) is one-valued and f(z) continuous in any connex. Hence for all a, b we have f (z)dz= -- bm+ - am+l. (1 / 'a m+l I The reader will note that the integral considered in 91, 2 is a special case of 1). Example 2. [ dzt m a positive integer > 1. Here 1 f (z) = is continuous in any connex W which does not contain the origin, as for example the ring in Fig. 1. FIG. 1. INTEGRATION 191 Also I)= 1 1 A F(z) 1 - m zmis one-valued and hasf(z) as derivative in 2i. Hence fbd i{ 1}* 1(2 Z l-mbm-1 a Jm-1 Example 3. bd fbJa cz (3 Here 1 Here f(z)= 1 F,(z) = logz. Thus f(z) is continuous in any connex aI which does not contain the origin O. Unless S is acyclic relative to O, P(z) is manyvalued. In fact if we start from z = a with one of the determinations of loge at this point which we call b Fa and allow z to describe a circuit / about 0 in the positive sense, F will acquire the value Fb = Fa + 2 7ri at the end of R. /c Thus Thus dz 27ri. (4 - i. Let now J1 be the value of 3) for the path C1 in Fig. 2, and J2 for the path C2. Then J 2ri. ( FIG.2. For Cl 2 forms a circuit 2 about 0. Hence by 4) 2 7ri= c,- =J J = -+ = - wnich proves 5). Finally let e3 be any connex, acyclic relative to the origin. Then any one of the branches of the logarithmic function is one-valued in Q. Denoting this by log z, we have bdz log b- log a, (6 Za and this integral is independent of the path of integration, provided of course it remains in S. 192 FUNCTIONS OF A COMPLEX VARIABLE 95. The Indefinite Integral. 1. Returning to the relation 1) in 93, let us write it f(z)dz = F (z)-F(a) (1 Let G(z) be any other function of z which has the continuous derivative f(z). Then similarly f (z)dz = G(z) - G(a). (2 Comparing 1) and 2), we have G(z) = F () + where C is a constant. The functions E, ( are called primitive functions of f (z). They are denoted by f (z)dz, no limits of integration appearing in this symbol. Primitive functions are also called indefinite integrals. 2. Every formula of differentiation as dF(z) = (Z), dz where PF(z) is one-valued andf (z) is continuous in some connex [, gives rise to a formula of integration, f(z)dz = F (z). Thus any table of indefinite integrals given in the calculus may be extended to the complex variable z, provided z is restricted to a connex in which F (z) is one-valued and f (z) is continuous. 3. Let the one-valued continuous function f (z) be such that G(z) =jf(z)dz (3 is also one-valued in the connex 2W. Then dz dz=f (>~ ). INTEGRATION 193 For let z = u be some point of Wt. Then rtu+h -*u AG G(u + h)- G(u)= f fdz- fdz u a </a fu+h =J fdz. As f (z) is continuous, f (z) =f (u) + e', and e' <e for h < some 8. Thus A n- 1 /I U+h * nAG=J +f (u) + e'}dZ. As u is a fixed value of z, f (u) is constant. Hence =G f(u) Cf+hdz + 1 r+h'dz = J+ K. (5 Az JA hJh Obviously J=f (). Also +h, < Eh j d'az < e]ht. Hence Ii <E. Thus K- 0 as h = 0. Hence letting h = Az - 0 in 5), we get G'(u)= f (u), which is 4). 96. Change of Variable. 1. Every student of the calculus knows that a change of variable is often of great assistance in calculating an integral. It is equally useful in the function theory. To this end we establish the following theorem: Let f (z) be continuous on the curve C. When z ranges over C, let u = -(z) range over a curve D which corresponds to C unipunctually. Let the inverse function z = -f(u) have a continuous derivative on D. Then f (z)dz =. f f'(u) u)du. (1 For let us effect a division of norm 8 of D, by interpolating the points ul, u2.... To these points on D will correspond points 194 FUNCTIONS OF A COMPLEX VARIABLE zl, z2 *.. on C which effect a division of norm, say y, of C. Also Zm - Zm-i = AZm = '(Um)AUm + e/aUm, where Ieml< m= 1, 2,.. provided 8 < some 80. Thus /f (zm)AZrn = If2 (Um) (um)Aun + IfemAUm. (2 Now the last term on the right is numerically < eG I Au. < eGD, (3 where D is the length of the D curve, and If(z) I < G. But 3) states that the last term of 2) has the limit 0 as 8 -0. Thus passing to the limit in 2), we get 1). 2. Example. Let us calculate b dz J= 2 -,:e/O0 (4 Ja V/z2 -_ 2 along a curve C lying in a connex 2I which is acyclic relative to the branch points ~ c of the radical. We change the variable, setting su = (Z) = z + Vz2_ 2. (5 Then =c2 +) (62 2u if u * 0. But u cannot vanish, for if it did, 5) gives Z+-\/Z2=-2o0, or z2 =2-_2, which requires c= 0, and this is contrary to hypothesis. From 6) we see that z is a one-valued function of u. To the end points of C correspond = (a), =(b a= ~(a), 3 = ~(b) on the curve D C. From 6) we have U2 C2 dz = ' (u)du = - du. 2 ut2 From 5), 6) we have /z2 - C2 = U - z =.2u INTEGRATION `195 Thus 4) becomes, on using 1), b + -0b2 - C2 = log (7 a + 'a2/62 -97. Integration by Parts. 1. This is another important method for evaluating integrals., Analogous to the calculus we have the following theorem: In the connex 2f, let the one-valued functions f(z), g (z) have continuous derivatives. Then f~ldz [(z)g() fgfd (1 For let us set (z) f Then Z + Then ~~~~~h' (z) = fg' ~ gcf'. Hence by 93, 1), ffg' c+ gf')dz = h(b) - h(a)=q[fj orJ 'dz + qf = gf'dz = which gives 1). 2. The relation 1) still holds when f and g are many-valued in the connex W, provided we take the right determinations of f, g and their derivatives along the path of integration as 91, 1 shows. Example. Let us evaluate J=fzlog zdz. (2 We set f(z)=logz, g'(Z)=z. Then 1 fI(Z (Z = _.x Z 2 Thus J= 1 Z2 log Z- Adz 2 Z = 1 Z2 (log Z 1). (3 196 FUNCTIONS OF A COMPLEX VARIABLE 98. Differentiation with Respect to a Parameter. 1. Let g(z, u) be a one-valued continuous function of z on a curve C for each value of u in some connex U. Then the integral (u) =cg(z, u)dz (1 Jc defines a function of u over U. For example, let g(z, u) = (z, n an integer, (Z - U) n where f(z) is a continuous function of z alone. Also for purposes of illustration let C be a simple closed curve, as a circle or an ellipse, and let the connex U lie within C as in the figure. Then z- u does not = 0 as z ranges over C, for any point u in U. Then C f (z)dz (2 Jc (Z - ( defines a function of u over U. We shall see that the integrals of the type 2) are very important. Returning to the general integral 1), we prove the following theorem, which will be of great service later. If -_ is a continuous function of u and z for each u in U and z on au C, we have (3 ' (U) = q dz. (8 uc a For by definition (u) = lim AZu=O Au But AO = (u + h)- ((u), h= Au = g(z, u + h)dz- Jg(z, u)dz c (4 = (z, u + h)- g (z, u) dz. (4 Jc FUNCTIONS DEFINED BY SERIES 197 Now as in 86, 2 g(z, u + A) -g(z, u) = gf(z, u) + E' AU, where el < e provided 0 < Au I < some 8; moreover this holds for every z on C. Thus 4) gives - = g u (z, ) dz + e'dz. (5 But< -e, e'dzl< -c, as we have often seen. But this states that the last term of 5) has the limit 0 as 8 - O. Hence passing to the limit 8 = 0 in 5), we get 3). Functions Defined by Series 99. Steady Convergence. 1. Let us consider series of the type P=f=(Z) +f(z) +f3(z) +. = fm(z), (1 whose terms fm(z) are one-valued functions of z in some point set 1, which may be unlimited. The simplest case of such series is power series power sere a + a(z - a) + a2(z - a)2 +... (2 or changing the variable by replacing z - a by z, ao + a1z + a2z2 +.. (3 By means of such series we defined the functions e2, sin z, etc. If the series 1) converges in 92, it will define a one-valued function of z in 21, which we denote by FP(z). We wish to study such functions relative to continuity, differentiation, and integration. To this end we introduce the notion of steady convergence. Suppose for all z in 21 the mth term is such that I f(z) <gm m=l, 2,... (4 where g1, g2.** are positive constants. Let the series G = gl +g2 + + (5 converge. Obviously the series 1) converges absolutely for each z in 1J, by virtue of the relations 4). In this case we have com 198 FUNCTIONS OF A COMPLEX VARIABLE pared a series whose terms are functions of z with a convergent series whose terms are constant. This kind of convergence we call steady. We therefore define: The series 1) converges steadily in 3f when each term of 1) satisfies the relation 4) for all z in X1, and when the corresponding constant term series 5) is convergent. 2. An important property of steadily convergent series is the following: If the series 1) converges steadily in XI, the remainder after n terms F, is numerically <e for any n > some m, for every z in 21. For the a series being convergent, Gm < e, for some m. Hence as the gm are positive Gn < a fortiori for any n >m. But from 4), 1Fn < for any z in 1. 3. Power Series. Let the circle of convergence of 3) be C. Let D be a circle lying within C, and having the origin as center. Then 3) converges steadily in D. For, let z = / > 0 be a point lying between the two circles C, D on the real axis. Then as 3) / / \ converges absolutely at this point, the constant o term series a + al, + a22 + (6 ao+ a13+~a232 + - (6 is convergent. But for any z in D I amZ| I< =m' m= 0, 1, 2... (7 Thus the terms of 3) satisfy the relations 4) for every z in D, and 3) converges steadily in D. 4. The definition of steady convergence may be extended at once to two-way series oo E fm(z), (8 and to double series E z, s(Z). (9 FUNCTIONS DEFINED BY SERIES 199 Thus if for every z in some point set 2 Ifr. (Z) I ~gros, and if If, is convergent, then 9) converges steadily in W. 5. Example. Let us consider the series F = ~~1 (10 (Z - mn) (10 where p is an integer > 2 and mn = a +nb (11 as in 41, 5. These series, as we shall see, are very Important in the elliptic functions. Let us describe a circle S about the origin and consider the series H formed only of terms of 10) for which the points 11) lie without R. We show that the H series converges steadily in R. For _Z <k<1 (mn for any z in M and for any om in H. Now Now 1 1 -1 Z - mn o mn 1- (Omn Hence 1 C Z- mn I \mn where C is some constant > 0. Thus each term of the H series is < the corresponding term of the positive constant term series 1 n i, which we saw converges in 41, 5. Thus H converges steadily in the circle R. 6. Let us show that: A two-way power series ao + alz + 2z2 +.. (12 b l+b+ z Z2 200 FUNCTIONS OF A COMPLEX VARIABLE is steadily convergent in any ring T9 lying within its ring of convergence R. For let R = C- D, and = - -. Then the series in the first line of 12), call it P, converges steadily inl G. Let now z = 8 > 0 be a point on the x-axis between the two circles D and Z. Then the series Q formed of the second line in 12) converges absolutely for z =, that is, the positive term series L1 + 132 + 83 + as 1 2 b = 3n (13 8 82 83 is convergent. Let now z be any point on the circle ) or without it. Then I z = > 8; hence each term of Q bl+ 6b2,... z z2 z3 is numerically < the corresponding term in 13). Thus Q converges steadily for all points on and without 3. Hence 12) converges steadily in the ring 9t formed by ( and S. 100. Continuity. 1. It is quite important at times to know if the sum of a series of continuous functions is itself continuous. The following theorem is often useful. Let the terms of the series.= f(z) +f2(z)+f3(z)+.. (1 be continuous and one-valued about z = a. If 1) converges steadily in some circle G about a, F is continuous at a. To prove this we have only to show that lim F(a + h)= F(a). (2 h=O Let /AF= F(a + h) - (a) then 2) is equivalent to lin F =0. (3 h=O Let us write 1) =m+ m thlelnl ATJ = AFm A+ T thlen A T A TY 'i A ' V/1 A7 = Li- xm-t- Li 7a M. I ~ FUNCTIONS DEFINED BY SERIES 201 Since F converges steadily in C, i Fm(Z)1 < 2' for some m and for any z in ( by 99, 2. Thus, in particular, I| J(a)I|<6, |J-(a+h)l< m a+h in G. 2 2 Hence subtracting, Am | < e, or lim AFm = 0. (5 h=0 Since /m is the sum of m continuous functions, it is itself continuous. Hence lim AF = 0 lirm AFm = 0. (6 h=O Thus letting h = Az 0 in 4), we get 3) on using 5), 6). 2. The power series = a z + az a2z2 +...(7 is a continuous function of z at any point within its circle of convergence d. For let z = a be a point within (. Let E be a circle about z = 0 which contains a in its interior. Then we can describe about z= a a circle c which lies in R. As 7) converges steadily in R by 99, 3, it does in c also, since this is a part of M. Hence F is continuous at z = a by 1. 3. A property of power series often used is this: Let the series 2 p ao + axz + a2Z +... converge about the origin. If ao - 0, P does not vanish in some circle c about the origin. This is an immediate consequence of 83, 5, since P is continuous at z = 0. 4. Closely connected with the property of continuity is the following theorem; it embraces 1, in fact, as a special case. Let r^,., _ _ r\, rx \ I /o. —d, lW 1: ) -I tJ2 ~.~ ) **- r ko 202 FUNCTIONS OF A COMPLEX VARIABLE converge steadily in some circle f about z = a. Let each f,(z) C7L )asz= a. Irf _Tf C CI~C~ c + C2 + ~ is convergent, we have lim F(z) = E linf,(z) = C. (9 z=a z=a For we may take m so large that IFm(z)1<- C <- z ing. (10 3 3 Also we may take 8 > 0 so small that -Fm(z) - Cm I E 0 < Iz-al < 8 (11 since by hypothesis, P(Z) Ur asz a. Then the relation F- C=Prn Cm~ + Fn - Cm gives FC! <I.FmCmI~I mI+j mI < 6 E by 10), 11), 3 3 3 and this establishes 9). 101. Termwise Integration. 1. In order to integrate a series P(z)=fl(Z)~f2(Z) ~ (1 it is -usually most convenient to treat it as we would a finite sum and integrate it term by term, or as we say termwise. This method, which suggests itself at once to the reader, is permissible as follows: Let each term of 1) be one-valued and continuous in a connex 9f. If 1) converges steadily in f, we may integrate it termwise over any curve C in At; that is, fP(z)dz fdz +ff2dz+.. (2 FUNCTIONS DEFINED BY SERIES 203 For being steadily convergent, F(z) = +f, (3 and In|<E (4 for all n > some m, and for all z in W1. By 100, 1 F is continuous. As F, is the sum of n continuous functions, it is continuous. Hence Fn is continuous. Thus 3) gives Fdz =j dz + fEndz. (5 By 92, 2, < C, where C is the length of (i. Thus the last term in 5) has the limit 0 as n - o. Thus letting n c in 5) we have Fdz = lim JFndz c no=oo C = lim { rfdz+... + Jfdz} (6 n=0 t/ C Now for the series on the right of 2), that is, /fidz + f 2dz+ **z to converge it is necessary that the sum of its first n terms should converge to some limit. The relation 6) shows that this sum does converge and has as limit the member on the left. Thus 2) holds. 2, From the foregoing we can show that log(l- z)=- {t -2.3 + (7 is valid within the unit circle, that is, for z zl< 1. For =l+z+z2+... if Izl<1. 1-z 204 FUNCTIONS OF A COMPLEX VARIABLE Hence z d z z Hnz c f - =-lo-g(l z)= dz + zdz + Jo z Jo Jo = z + 1 z2 + z3 +... which is 7). 3. We can show similarly that Z3 Z5 arctgz= z- - + -. IZ[|< 1. (8 3 5 For 1 =1 2 + z4z6 +... I<1. Hence dz rz rz z- -+ JO 1+z2 arctgz o z + l3 =Z —+... which is 8). 4. The reasoning in 1 shows that 2) holds provided each term of 1) is continuous on the curve C and the series 1) converges steadily on C. 5. Since a two-way power series 00 F = an(Z - a)n (9 -0o converges steadily in any ring 9 lying within its ring of convergence, we have for any curve C in E fJdz = Jaodz + aI (z - a)dz +. + a- dz lz + " z -- a V z a-a)2 = a (z -a)ndz. (10 — 00 C( FUNCTIONS DEFINED BY SERIES 205 Let now C be a circle R. Since (z - a)ndz = n -1 we have /^ -rr-y o * ^ 1 we have f Jdz = 2 7ria_. (11 102. Calculation of ir. Let us use the relation 8) in 101 to calculate 7r; it will serve as an exercise in infinite series. Putting z= = in that relation we get ==arctg +.=. -..5+5.. (1 5 5 3 53 5 55 7 57 The error committed in breaking off the summation of any term is less than the next term, as we saw in 15, 1. From trigonometry we have tan 2 2 tan a tan 2 a = 1 - tan2 a' which gives here an 2 tan 2 ao= ~'2. Similarly tan4 a = 12. Let /3=4 7?x- (2 Then n tan 4 a- 1 1 1 + tall 4 a 239 Thus 101, 8) gives 1 1 1 1 239 3 2393 5 2395-* and the error committed in breaking off the summation at any term is less than the next term. Thus from 1), 2), 3) we get Zr T I 4|-1 - r 1+ 1..1....l 4 5 3 53 5 55 J 239 3 2393 We have now 12 1 1 - _='.2 I. ~ =.002666667 I.1 =.000064 3 5 5 55 1 1 1.1=.000001829 - - =.000000057 7 57.000001829 9 59 1 1 1 1 1. 53 =.0000000001. 000000002. 13 513 11 511 206 FUNCTIONS OF A COMPLEX VARIABLE Thus a=.200064057 -.002668497 =.197395560 is correct to 9 places, and 4 a =.78958224 is correct to 8 places. Also we have 1 =.004184100 1 1. -=.000000024. 3 2393 Thus /3=.004184076 is correct to 8 places. Thus 7r Thus?- -=.78539816 4 or 7r= 3.1415926... is correct in the last decimal. In fact a more elaborate calculation gives 7r = 3.14159265358.. 103. Termwise Differentiation. 1. From the theorem on termwise integration of a series given in 101, 1 we can deduce a useful theorem on termwise differentiation: In the connex Xf, let each term of the convergent series = f(Z) + f(Z) +.. (1 be one-valued and have a continuous derivative. If =fl(Z) +f2(Z)+ "- (2 converges steadily in X1, - = dE in 1. dz For by 101, 1 f Gdz = f(z)dz + 2(z)dz+... = Ifl(z)-fl(a) + If2(z)-f2(a) + * =F(z)- (a). (3 Thus by 95, 3 we get, on differentiating 3), d/a dz FUNCTIONS DEFINED BY SERIES 207 2. A power series may be differentiated termwise at any point within its circle of convergence. For let FP= a +a1z + a2z2 +.. (4 have the circle of convergence (. Then by 85, 2 the series G = a1 + 2 a2z + 3 a3z2 +.* obtained from F by differentiating it termwise has the same circle of convergence C. Thus if c is a little circle about a point z = a within C, the series P converges in c, and G converges steadily in c by 99, 3. Thus the condition of the theorem 1 holding, we may differentiate 4) termwise, or dz d? a= G=al +2 a2z+3 a3z2+ Remark. We note this theorem was proved in 85, 1, making use of double series. 3. A two-way power series PF = a,zn may be differentiated - 00oo termwise at any point within its ring of convergence; that is, dF _ nanz-1. dz -oo To prove this we need to consider only the special case F =- + b2+.. circle of convergence. z 2 If we set 1 U u we get =b1u + b2+... and d b1 +2 b2u + du If now we apply 89, 2, we get dF dF du du 1 dz - = - since dz - =- 0 without 9 dz du dz dz z2 1 bl + 2 b2u - * *" zwhout. without S. 208 FUNCTIONS OF A COMPLEX VARIABLE 4. Differential -Equation for F(a/3yz). As an exercise in differentiating power series let us find the derivatives of the hypergeometric series Q ) a ~a/3z+a. aa+t././+lz2+ (5 ((a3ryz)=I+ a z+ aa1.. 3+1 + (5 1. 2' 7 1 2y y+i introduced in 39, 4 and show that it satisfies the differential equation d2F dF Z )2 + + + 1)z - v +/~=o. (6 On differentiating 5) termwise we get (ay)= l a 1... c n -...y.y +...yn + n —1 a a+.+l. *a+n./3 /3 +... +n n o__ 1 ~ 2... n ~ - 1... ry +r- 1+ ~y 1.2 n... n+ y + 1 - y+ = ( + 1,/3 + 1, + 1, z). (7 Hence "l(a3ryz) = y 'P(a + 1, / + 1, v + 1, z) + F(a + 2, 3 + 2, ry + 2, z), (8 etc. To prove that 5) satisfies 6) let us set p _ a a+ 1... a -n-1. /3 31. 3+n-1 1 ~ 2... n. ry.7+1.. r +nThen the coefficient of zn in Z2F" is n(n -1)P FUNCTIONS DEFINED BY SERIES 209 in -yE'" it is -ya (c n) (i~+n)~ 7+n in cq3E' it is(a+n 8 n Adding all these gives the coefficient of n in the left side of 6>. We find it is 0. CHAPTER VII ANALYTIC FUNCTIONS 104. Definitions. 1. At this point we begin the study of the theory of functions of a complex variable. The functions considered in this theory are not the general functions considered in Chapter VI, but a subclass of these, viz. those functions which have a continuous derivative. To be more specific, let w have assigned to it a definite value for each point z of the connected region WI such that w has a continuous derivative in W. We call w a one-valued analytic function of z in W. Suppose, on the other hand, that w has in general more than one value assigned to it for the points of 2. We will call it a many-valued analytic function if its values can be grouped in branches, each of which is a one-valued analytic function about each point of I. 2. From this definition it follows that ez, cos, sin z, cosh z, sinh z are one-valued analytic functions in the entire plane. For they each have a continuous derivative for any z. Similarly 1 - 1' g2 1 -for example, is a one-valued analytic function for the region 21 formed of the whole z-plane after deleting z = i 1, the zeros of the denominator, while tan z is analytic for the region W formed of the whole plane after deleting the infinite point set 7r r 7r 7 7 5 -5-., 7 77 3, - 2' 2 2 2 2 ' 2 On the other hand, / + Z 210 + 210 ANALYTIC FUNCTIONS 211 is a two-valued function for the region 2f formed of the whole plane after deleting the branch points ~ i, while log z is an infinite-valued analytic function in the region f formed of the whole plane after deleting the origin z = 0. 3. A power series P(z) = ao + a(z- a) + a2(z -a)2+ is an analytic function within its circle of convergence G. For by 85, P has a derivative within ( which is a power series and therefore continuous within G. 4. The quotient of two power series Q= + aIl(z - a)+ ". bo + b(z -a)+ having a common circle of convergence C, is an analytic function of z within some circle c about the point z = a. For the denominator does not vanish at z = a, since by hypothesis bo / 0. Thus by 83, it does not vanish in some c lying in (. Hence by 84, Q has a continuous derivative within c and is therefore analytic within c. 5. A two-way power series X = a(z- a)n -00 is an analytic function of z within its ring of convergence. This follows at once from 103, 3. 6. We propose now to study the general properties of analytic functions and shall rest our treatment on twotheorems of a grand importance due to Cauchy, and called his first and second integral theorems. 105. Cauchy's First Integral Theorem. 1. Let f(z) be one-valued and analytic in the simple connex 9A. Then f(z)dz= 0 (1 for any simple closed curve C in W. 212 FUNCTIONS OF A COMPLEX VARIABLE For let F t f(z) =u+iv, = x + iy. Then since f has a continuous derivative in 2, the first partial derivatives of u, v are continuous functions of x, y which satisfy the Cauchy-Riemann equations au av O1u av O_ =. ~-= (2 Ax ay ' y Ox' by 90. On the other hand, let us express 1) as line integrals, using 91, 5). Then fdz = (udx - vdy) + if (udy + vdx). (3 We now apply Stokes' theorem 80, 1) and get f(udx - vdy) = -(J + au) dxdy = 0, by 2), Jc (EAJAx 9y/ J(udy + vdx) = a - a) dxdy= 0, by 2). These in 3) give 1). 2. From this we conclude that: f fdz = fdz, (4 where C1, C2 are two simple curves in 21 having the same end points but no other points in common. For C1C'1 is a closed curve. Hence by 1) fdz = 0 C2 =which gives 4). 3. The restriction that C1, C2 should have only their end points in common is obviously not necessary. For if C1, C2 have other points in common, we can break them up into arcs which have only their end points in common. Similarly C1, C2 may have multiple points. ANALYTIC FUNCTIONS 213 4. In the connex PW in which f(z) is one-valued and analytic, let C1, C2 be two simple closed curves forming the complete boundary of a ring-shaped connex, as in _Fig. 1. Then I f(z)dz= f(z)dz, (5 the curves C1, C2 being passed over in the same sense. For let us join C1, C2 by two adjacent curves 34, 16, as in Fig. 1. Then z C=123.34.456.61 2 is a closed curve forming the edge of a simple connex in 1. Thus FIG. 1. 0= fdz=J3 + f+ i + (6 Now the value of f(z) at a point z on 34 does not differ by an amount greater than e from a point near by on 16. Thus i>4 A6 4 fdz and f differ by an amount as small as we please as the curve 34 is made to approach 16. {'6 >1 As fdz=- fdz, we see that under these circumstances f fdz + ffdz _ 0. t/3 t 6 Similarly z fandz ffdz and f /dz - Z.o 1/456 J/ C2 Thus passing to the limit in 6) we get 5). 214 ~FUNCTIONS OF A COMPLEX VARIABLE 5. The little strip 1643 taken out of Fig. 1 and whose edges are then allowed to approach indefinitely near we call a cross cut. 6. Let C, C2, *.. Cm be simple closed curves as in Fig. 2, each exterior to the others and all interior to a simple closed curve C. Let these curves form the com- \ cl c0 2 plete edge of a connex 2 in which f(z) is one-valued and analytic. Then if all these FIG. 2. curves are described in the same sense, fdz =ff d +ffdz+... + ffdz. (7 CG JCl 2 By JCm To prove 7) we need only to put in the cross cuts 71, 72, * 7m, as in the figure. This produces a simple connex L with edge (. Asf is one-valued and analytic in (, we have fdz-= 0. We may now reason as we did in 4. 106. Cauchy's Second Integral Theorem. 1. Let f(z) be onevalued and analytic in a simple connex whose boundary is C. Then for any point z within C I (urfw^du( f (Z) = 2 i fJ u (1 For by 105, 4 we can replace C by a circle R of radius r and center z. Then for a point u on 9, f(u) =f(z) + 'E I eI < for all u on 9 if the radius r is sufficiently small. Thus asf(z) is constant du du - rf (u)u f( r( + e du= J+ K. (2 J~ ~u-z J -z. u-z But J= 2 7if(z) while I KI < 27re. Thus lim K= 0. r=O ANALYTIC FUNCTIONS 215 On the other hand, the left side of 2) does not depend on r. Hence letting r 0 in 2) we get 1) in the limit. 2. This theorem of Cauchy brings to light a tremendous difference between analytic functions of z and the general function of z. For suppose we know of a function f(z) that it is one-valued in a simple connex ( and has a continuous derivative in G. Suppose also that we do not know the values of f within L but only on the edge C. Then the relation 1) says that to learn the value of f at some interior point z we need only to calculate the integral in 1). In other words the values of an analytic functionf(z) are completely determined when its values on the boundary C are given. This is not at all the case for the non-analytic functions of z. 3. At first it seems strange to students that the values of an analytic function f(z) should be fixed for all points within C when its values are assigned on the curve C. Here the study of nature reveals many cases of just this phenomenon. For example, the stationary flow of water, heat, or electricity in a body are all determined by the flow at the surface. In case the body assumes the form of a thin plate, it may be treated as a plane figure E bounded by a curve C, provided the flow is parallel to the plane. 107. Derivatives. 1. The integrand in 106, 1) f(U) U- z is an analytic function of z for each value of u on the curve C, provided z is restricted to lie in a connex bounded by a curve C' which lies within C. We may, therefore, apply 98 and get 1 Of(u)du f' (z) = (. ( Z)21 f" (zz)- - i C (f>) (3 f (c(7iU - Z)n f(") (z)= _it2,rie (U- z)nl ' 216 FUNCTIONS OF A COMPLEX VARIABLE From this we conclude: An analytic function of z has derivatives of every order. We also have the result: Iff(z) is a one-valued analytic function, so is each of its derivatives. 2. From 3) we get at once an inequality called Cauchy's Inequality, which is of great service in theoretical work, viz.: If('n)(Z) Ina _Rn where If < G on a circle R of radius R one-valued and analytic in M. For we need only to replace the curve and apply 92, 9). 108. Termwise Differentiation of Series. integrals of Cauchy we can establish a than that given in 103. Let........ (4 with center z, and f is C in 3) by the circle A 1. By the aid of these more general theorem fm(Z). (1 We saw that if this series converges and the series G(Z) =/ Z) + - f = E Z) /2) (2 converges steadily in the connex 31, then G represents the derivative of F. Let us now prove the more general theorem: Let the terms of 1) be one-valued analytic functions in the connex S. If F converges steadily in t, F is an analytic function within S and d dF G. dz In other words under these conditions we may differentiate 1) termwise. For let c be a circle of radius r and center z, lying within R. From 1) we have U-_Z u (3 ZG-X f 1A — ANALYTIC FUNCTIONS 217 and since 1) convergo —s steadily on c, the series on. the right converges steadily on c sinceI u - z = r a constant on c. Thus by 101, 4 we may integrate 3) termwise fF(u)du =ff (U)duh( Now by 106, 1) ()d Hence 4), 5) give__ _____ u~ f)d( 27i u I.Z)6 But by 1) the series on the right is F(z); thus 6) gives F(z) = 2J1 ~u)du (7 Now by 98, 3xud F'z=27rij (U - Z)2 ( Reasoning as before, we have (T~u~du - rfm(u) du -Z)2 ~e!(u ) -2 ilfjIz). From this and 8) we have 2. From 8) we get, using 98, F z) 2! F(u)du FHz~27ri c(u -zy3 Proceeding as in 1, we get and so on for higher.derivatives. 218 FUNCTIONS OF A COMPLEX VARIABLE 109. Taylor's Development. 1. Let f(z) be a one-valued analytic function in a circle ( of radius R about the point z = a. Then at any point z within S ) - a (Z - a)2If f(Z) =f (a) + f (a) ++ 2! ( This theorem is a direct extension of the corresponding theorem in the calculus. It is of transcendental importance in the function theory, as we shall see at every turn. Its demonstration may be conducted very simply by resting it on the termwise integration of steadily convergent series. Let c be a circle about z lying within ( as in the figure. Then by 106, 1), ( f( ) 21 r f(u)du (2 a But f (u)du f (u)du J^u- 2 c U -Z - We now develop - in a power series about the z - a, getting as in 39, 10), u-z ~1 if ~z-a z - a2 1+ - + + ~. (3 u-z u-a u - a \u - a This series converges for any u on ( since z -al=r Iu-a =R and r<R. Hence f (u) f (u ) f ( + (z- a)2 f (u)... (4 u-z u-a (u- a)2 (u - a)3 This series converges steadily on (. For f being continuous, f(u) I < some M on S. Hence each term of 4) is numerically less than the corresponding term of the constant term convergent series Lfr\ M r \2 M r \3 r _R +r _R 9 ANALYTIC FUNCTIONS 219 We may thus integrate 4) termwise and get f (u)du _ rf(u)du f (u)du f -U I ------ _+- (z- _ - - + " /, U- - a. (u - a)2 -27 ri (a) + (z - a)f(a) + ( - a)2 f"(a) - *** 2! on using 106, 1) and 107, 1), 2),.. Replacing the first member of the last equation by 2) we get 1). 2. If we set z = a + h in 1), it takes the form f(~ + h)= f(a) + Uf (a) + h f(a)+... (5 This is the form of Taylor's development usually given in the calculus. If we set a= 0 in 1), we get f(Z)= f(0) + 2g, (0) +..f. I.(0) (6 which is often called Maclaurin's development. The coefficients in 1) and 6) are constants. These series are thus power series. We say that 1) is a development of f(z) about the point z = a. Thus Maclaurin's development 6) is merely the development of f(z) about the origin. The two series on the right of 1) and 6) are called Taylor's and Maclaurin's series respectively. 3. Let f(z) be one-valued and analytic in the region A. Let a be any point of AI. About a as a center describe a circle o which contains no point of the frontier of WA. Then by the theorem 1, f(z) can be developed in a power series valid in R, f(z) = a0 + al( - a) + a2( - a)2 + * (7 whose coefficients are those in 1). We may also proceed thus: With a as a center describe a circle ( which passes through a point of the frontier of 2f but contains no frontier point within (. Then the development 7) holds for all points within G. For let z be any given point within C. We can describe a circle 9 with a as center which contains z but no point of tlhe frontier of W. Thusf(z) is analytic in Q and we can apply 1. 220 FUNCTIONS OF A COMPLEX VARIABLE 110. Examples of Taylor's Development. 1. Example 1. Let f () = sin z. This function is analytic in any region. If we levelop about the origin z = 0, we have, exactly as in the calculus, f' () = os f' () = 1 /f(z) = - sin f"(0) = 0 f/"(z) -cos z /f" (0) =-1 fi"(z) = sin ==f(z) fiv(O) = 0, etc. Thus 109, 5) gives z = Z5 sn z =+ _1! 3! 5! which is exactly the series we used to define sin z. Similarly we may develop e, cos z, sinh z, etc. Example 2. Let f(z) = logz. This is an analytic function in the region ~ formed of the whole plane after deleting the origin. The frontier of 21 is thus a single point z = 0. It is one-valued in any connex which is acyclic relative to this point. Thus by 109, 3 we may develop log z about z = 1 and the development will be valid within the circle having z = 1 as center and passing through the frontier point z = 0. Proceeding as in the calculus, we have: f(1) o f' (Z) = f'(1)= 1 Z f"(Z) = - f"() =-1 2 f"'(z)= f'"'(1) -2, etc. Thus (Zogz=z - )2 (-1)23 I-lI log z == Z - 1 - + ( 8 ) - - |Z| <1. 2 3 If we set z = 1 + u, this gives i,\ u2 u this gives log (1 + u) = u — 2 + - <, (1 h is the development given in the alculus. w\hich is the development given in the calculus. ANALYTIC FUNCTIONS 221 If the reader asks why one develops about the point a = 1 instead of about a = 2, the answer is that the values of the coefficients ) ' (a) f"(a) J(a) '! ' 2! are simpler for a = 1 than for any other value of a. Example 3. The Binomial Formula. Let f(z) = (1 + Z)I. (2 From 89 we know that any branch of f is a one-valued analytic function in any connex acyclic relative to the point z = 1. It thus admits a development about z = 0 which is valid for all points within the unit circle. Proceeding as in the calculus, we have, choosing that branch of 2) which reduces to 1) for z = 0, f() =1 f',() = (1 + z)-l0 f'(0) = f,, () = Z (a - 1)(1 + z)~-2 / (0) = f (I - l), etc. Thus (1 + Z) = I + -z + ( -1),2+ /( )( j - 2) +... 1.2 I. z <1 2. Let us make use of 1) to develop a formula which we shall need later. If we set z = - rei, it gives for r < 1 ~,2 e2i q- 3 eiq -log ( 1 - u) = - lo(1 -ei 2 + r3 e + *2t 3 But eni = cos no + i sin no. ence _ log (1 - rei) = r cos i rn sin (4 n n Let us write the left side of 4) in rectangular form, log (1 - rei) = A + iB. 222 FUNCTIONS OF A COMPLEX VARIABLE To determine A and B we set 1 - rei = sei. Now 1 - rei =1 - r (cos + i sin ) Tus = (1 - r cos 4)- ir sin 4 Thus s2 (1 - r cos 40)2+ r2 sin2 4 = 1 - 2 r cos ) + r2, and ~~~and ~ — r sin b tan = -- 1 - r cos ' Hence A = - log s2 = 1 log (1 - 2 r cos + r2), rsin4) B = = - arctg r sin 1 - r cos ) Thus, equating the real and imaginary parts in 4), we get Xrn COs 4 = - log (1 -2rcosb + r2), (5 n 2 r-si nn 4= arctg rsin. (6 n 1 - r cos < The relations 5), 6) hold for 0 < r < 1 as we have just shown; a more delicate analysis shows that they hold for r= 1. Without establishing this important fact we shall set r= 1 in these formule, getting, replacing 0 by 2 7rx, s 2 n7x = - log (2 sin 7rx), (7 1 ^ n 00 sin 2 n rx Sin 2 nf-X =arctg (cot 7rx) = r (- - x). (8 111. Critical Remarks on Taylor's Development. We are now in a position to point out another great advantage which we reap from the theory of functions. Let us compare Taylor's development as here presented and as given in the calculus. To make use of the development f( + h)=f(x)+ f'(x)+ f2 f (x + h) =f (x) + hfy' (x) + -fi (x) +.. ANALYTIC FUNCTIONS 223 in the calculus we must first assure ourselves that the remainder fhn(n(x + h) 0 < 0 < 1, n! or one of its equivalent forms, converges to 0 as n oo. This is an easy matter for ex, sin, cos, but it is far from easy for most functions; for example, (l + x), tan x. How difficult it is to show that the remainder for (1 + x)' converges to 0, the reader may see by turning to a good work on the calculus. As to the remainder for tan x, no one, as far as we know, has ever shown that it - 0. These considerations show that the applicability of Taylor's development in the calculus is crippled by the fact that we cannot show that the series on the right of 1) really has as sum the function of the left. How differently we are situated in the function theory. Take, for example, tan z. We know without putting pen to paper that this can be developed about z = 0, and that the development is valid for all z I< 7, since tan z is one-valued and analytic within this circle. Thus if we wish to restrict ourselves to real values, the development holds for — < x < r 2 2 Let us look similarly at +. (1 + z). This we know is one-valued and analytic for all points within the circle of unit radius about z= - 1. Thus the validity of the binomial formula for real values of x for which -2 < x < 0 is again established without any calculation whatever. 112. Remainder in Taylor's Development. Let us write 109, 1) f(z)= f(a) + Z - a f' (+ (- f() +.. (~)+ 2! + (z a)I f(a) + R, (1 where (z- (2 s=n+ -l 224 FUNCTIONS OF A COMPLEX VARIABLE Let r be the radius of a circle c about z= a, lying within the circle (, for which 1) holds. Let If(z) <G on c. Let z be any point within c; we set z - a = p. Then by Cauchy's inequalities, 107, 2, s! y If(s)(a) < * rs Thus 2) becomes IR, ] < G ) { + +e+ (+... Pn+l 1 < &(P) 1 (3 r 113. Analytic Continuation. 1. We have seen in 106, 2 that a one-valued analytic function is completely determined in a simple connex ( when its value is known along its edge. We now wish to generalize this result. Suppose 1~ it is known that f(z) is one-valued and analytic in a connected region W. 2~ the values of f(z) are given along some curve C in A, as, for example, a small segment of the x-axis. WVe show that under these conditions the value of f may be found at any point of z / in W; that is, the value of f at this point is / determined by the above data. / Suppose in Fig. 1 that C is the arc a, 6. Join a and z by a curve D lying in W. Since t f(z) is analytic about a, it can be developed b by Taylor's series (z)=f(a)+ (z - a)f'(a) + f (a)- (1 2! The value off will be known for all values of z within a circle 2 whose center is a and which extends as near the frontier e of 21 as we choose. Let now u be an arbitrary but fixed point on C. Then f' (u) is the limit of f( f(z-f(u (2 X-/U~~~( ANALYTIC FUNCTIONS 225 as z u. Moreover f(z) being analytic, this limit is the same, however z approaches u. Let us suppose that z approaches u by running along the curve C as in Fig. 2. Then for a each such value of z the difference quotient 2) is c known by hypothesis. Thus its limit is known, that is, f' (z) is known for each z on C. As now)- f(u f"t(u) = lim f (Z) - (U, (3 z=-u Z- U we may reason onf'(z) as we did onf(z). Thus f"(z) is known for each z on C. In this way we see that the values of the derivatives of every orderf(^)(z) are known for all values of z on U. In particular they are known for z = a. Hence all the coefficients of 1) are known. Thus 1) gives us the values off(z) for all points in R. Let t cut D in a1. About this as a center we can describe a circle M1, which extends as near the frontier @ as we choose. Since f(z) is now known on the arc 1 = ala, we can reason on C1 as we did on C. If R1 cuts D in a2, these considerations show that f is now known for all z in 1l, and in particular on the arc U2 = ala2. Continuing in this way we may finally reach z, when the value off will be known. 2. This process of finding the value of an analytic function f(z) at a point z, when its value is known at the points of some curve C, is called analytic continuation. It has little or no practical value as a means of actually computing f at the various points of W; but it has an inestimable value in many theoretic investigations. 3. In the foregoing we have supposed f(z) to be one-valued in W. This is not necessary; we made this assumption merely for clearness. The same considerations apply if we suppose thatf (z) is many-valued in Wt, but such that each branch is analytic, and onevalued about each point of I. 4. The foregoing reasoning shows that: If the analytic function f(z) = a, a constant on the curve C, then f (z) = a everywhere in the region fI. 226 FUNCTIONS OF A COMPLEX VARIABLE For the difference quotient 2) has the value 0 and hence f'(z) = O on a Similarly f"(z) f"'(z),. 0 on C. Thus 1) shows that f (z = (4 in the circle Q, and the remainder of the reasoning in 1 shows that 4) holds for any z in W. 114. Application of Analytic Continuation. 1. For the reader to realize the immense power of this process let us show how most of the analytic relations of plane trigonometry and the calculus are valid when the variable is complex. For example, suppose we wish to show that si2 z + cos2 z 1 (1 holds for any complex z. This we have already proved in 58, 5 by the lengthy method of series. To this end we consider f(z) = sin2z + cos2z. As sin z, cos z are one-valued analytic functions in the whole plane, so are their squares and therefore f(z) is analytic. For the real axisf(z)= 1. Hencef(z)= 1 for all values of z by 113, 4. This reasoning is so simple that with a little experience the reader may do it in an instant. The same is true in the following examples. 2. Let us show by this method that d. tan z - = sec2 z (2 dz holds for all values of z for which tan z, see z are defined, that is, for the region W formed of the whole z-plane after deleting the points z = ~ (2 n+1)7 Since f(z) = tan z is analytic in X, its first derivative, call it g(z), is analytic by 107, l. Thus 107, 1. Thus h(z) = g(z) - sec2z is an analytic function in l. For real x in 72, h=0. Thus h(z) = 0 everywhere in W. Thus 2) holds in W. ANALYTIC FUNCTIONS 227 3. In the calculus it is shown that fsin3 xdx = - cos x (sin2 x - 2). (3 From this we can show at once that.fsin3 zdz - cos z (sin2 z + 2) (4 for every z. For let us set f (z) = sin3 z, F (z) =- cos Z (sin2 z + 2). Then the relation 4) means that ' () = f (). (5 Let us set G(z) = '(z)- f(z). As F(z) is analytic, its derivative F1 is also. Hence G is'ana. lytic. As G = 0 for real values of z by 3), it is 0 for all z. Thus 5) holds for all z, and hence 4). Of course the relation 5) is easy in this case to verify by direct differentiation. But for a more complicated formula this labor of differentiation might be considerable. The method of analytic continuation enables us to avoid this operation. 4. In 61 we saw how relations in circular trigonometry go over into relations in hyperbolic trigonometry by using sin iz =i sinhz, cosiz =coshz, etc. (6 Let us show that relations between circular functions in the calculus give us corresponding relations between hyperbolic functions. For example, from d sec x = tan x sec x (7 dx we infer by the method of analytic continuation that dz 228 FUNCTIONS OF A COMPLEX VARIABLE Setting now z = ix, this gives Id -i x sec ix = tan ix sec ix, or, using the relations 6), i dx or d ~~or d sech x = - tanh x sech x, (8 dx which is the formula in hyperbolic trigonometry corresponding to 7). 5. To illustrate integration let us start with 3). We have seen that the method of analytic continuation shows that we may replace x in 3) by ix. It becomes then i sin3 ixdx =- cos ix (sin2 ix + 2). Using the relations 6) this gives fsinh3 xdx = cosh x (sinh2 x- 2), (9 which is the formula corresponding to 3). 6. Let us show by the method of analytic continuation that the addition theorem sin (u + v) = sin u cos v + cos u sin v (10 holds for any complex u, v. This we established in 58, 1 by infinite series. We may now do it without putting pen to paper by the following simple reasoning. Let us give to v a real value as v = a; we consider f(u) = sil (u + a)- sin u cos a- cos u sin a. (11 This is an analytic function of u which = 0 for real u. Hence f= 0 for all u. Let us now give to u an arbitrary but fixed, real, or complex value, and consider g(v) = sin (u + v)- sin u cos v - cos u sin v. (12 As g = 0 for any real v, it = 0 for all v and hence 10) holds for any u and v. ANALYTIC FUNCTIONS 229 115. Undetermined Coefficients. 1. A very useful method to develop a function in a power series is that of undetermined coefficients. Before explaining it let us develop a theorem on which it rests. If P= a + alz + az2 +... (1 vanishes for a set of points bl, b, 3 (2 21 L ~ y~ ' b 3..(2 which are all different and 0 O and which 0, then all the coefficients in 1) are 0; that is, P = 0 for every z, or as we say, it vanishes identically. For P being a continuous function, lim P(bn) = P(0), since b, 0. But each P(bn) = 0, hence P(O) = 0. (3 Setting z = 0 in 1), we see that 3) requires a, = 0. Thus P = z(a1 + a2z + a3z2 + )= P1. As P1= 0 (4 for the same set of points 2) and as z - 0 for these points the relation 4) requires that P1 = 0 for the points 2). Thus we can reason on P1 just as we did on P. This shows that al = 0. Continuing in this manner we show that each a,n=, n = 0,1,2,... 2. A special case of 1 is this: If the series P = a + alz + a2Z2 + ' vanish for the points of any curve ending at the origin, it vanishes identically. 3. If P = ao + alz + a2z2 + *., and Q = bo + blz + b2z2 + * are equal for a set of different points e1, C2, c3 " which - 0, then the 230 FUNCTIONS OF A COMPLEX VARIABLE coefficients of like powers in P and Q are equal; that is, P and Q are the same series. For For R = P- = (a0- 60) + (al - b1)- + (a2 - 62)2 + vanishes at the points ce. Hence by 1 all the coefficients are 0. Thus an= bn == 0, 1, 2,... 4. From 3 we have the important theorem: If f(z) admits a development f(z)= a0 + al(z - a)+ a2(z - a)2 + ". the serieson the right must be Taylor's series, that is an =. f (n)(a). n = In other words, Taylor's development is unique. 5. The labor of calculating the coefficients of a development may be materially lessened when the following theorem applies: Let f (z) = ao + alz + a2 + *.. (5 be the development off about the origin. If f is an odd function, the coefficients of all the even powers are 0; if f is an even function, all the odd power coefficients are 0. For suppose that f(z) is odd. Then f(- z) = -f(z) by definition. ~~~But f(- z) = ao - a1z + a2z2 - az +. IHence 0 =f(z) +f(- z) = 2(ao + a2z2 + a4z4 + ). As this series = 0 for all values of z near the origin, all its co. efficients are 0, or = a O = a4 = a a= 6. The method of undetermined coefficients will be best understood if we illustrate it by two or three examples. This we now do. 116. Example 1. Let us develop tan z in a power series about the origin. Such a development we saw is possible and the de ANALYTIC FUNCTIONS 231 velopment is valid for I z I < '7r Moreover, tan z being ll anodd 2 function, its development will contain oniy odd powers. We set therefore tanl z = a z + az +5z+ (1 where the coefficients al, a3... are to be determined. To do this we use the fact that z3 z5 --- -- sinl 8!3 5! tan z = ~~ - - 2 4 (2 2! 4! Let us equate 1), 2) and clear of fractions. We get z2 z4 - + - (alz a~z+ a3Z3 + ( X If we multiply out the two series on the left by 33, 2 we get the series aiz+(a z2! S2(a5 -! + Comparing the coefficients of this series with the series on the right side of 3) gives a I a a1 1 1_ a3! 3! a3 3 ~I 33 3 __ a 2 a 15 -3 + a a =- 2! 4! 5! 15 a a a,-1 a 7 - -3 a7 2! 4! 6! 7! 315 Thus tan z=z~+ -3- z5 17 z7+ (4 3 15 315 -valid for z <7. 232 FUNCTIONS OF A COMPLEX VARIABLE Example 2. Let us develop 1 cosec z = s sin z about the origin. At first sight it would seem that our method would not apply. For the very first thing to do is to assure ourselves that the development is possible. The conditions of Taylor's theorem, 109, are not fulfilled here, since cosec z is not even defined at z = 0, and for the theorem to hold it should be analytic in some circle about this point. However, a slight consideration enables us to proceed. We have sin z z { 1 - - + - _ z Q Now since P converges for all values of z, so does Q= 1-. + - -...(5 3! 5! As P = sin z = 0, for z = ~ r, zQ=0, forz= ~iqr. Hence Q = 0 for z= ~ 7r. On the other hand, Q = 0, within the circle ( about the origin of radius 7r. For Q u= 0 for z = 0 as 5) shows. If now Q =0 for some z =- 0 within C, P = zQ would = 0 also. But P=sinz does not vanish at this point. Thus by 104, 4, ' is an analytic function within d. It may therefore be Q developed in a power series by Taylor's theorem, about z= 0. Moreover Q being an even function, this development will contain only even powers of z. We may therefore set - 2 4 — = aO + a 2z2 4 + *4 1 + --... 3! 5! or clearing of fractions, 1= -- + —... )(aO+ a2+ a44+ ) ANALYTIC FUNCTIONS 233 The multiplication of the two series on the right by 33, 2 gives + (a _a4+ a + _ (6 -3! 5! 7( Here the left side is to be regarded as a series co + c1z + C2Z2 +.. all of whose coefficients = 0 except the first. Thus equating coefficients of like powers on both sides of 6) gives a0 = 1. a2- -=.. a 1 % -=-0+A. "a4=60 a4 _a2- + a-O =.. a- = 31. 6 3! +5! 7-! 6 37 Thus 1 7 31 -- ++ -. -- 4- (7 sinz z 6 360 37.! valid for 0 < I z< 7r. Example 3. Division by Power Series. In the two foregoing examples we have divided by a power series. As this operation is not infrequent, let us state the following theorem: Let P = =a + alz +a2z2 +.., a0= 0 converge within a circle 9 about the origin and be 0 within R. Then the reciprocal of P can be developed in a power series -= C + C1z + C2Z2 + valid within R and the first coefficient 1 co = —. (8 ao 234 FUNCTIONS OF A COMPLEX VARIABLE For P being 0 within R is analytic within M and can be developed in a power series valid within f. The coefficients a0, al... are found by the method of undetermined coefficients, and this shows that c0 has the value given in 7). Suppose the first coefficient a0 = 0. In general let us suppose P = amzm + am+lzm+ + am: O0. (9 We write = zm(am + am+lz + ) = zQ. Suppose now that P does not = 0 within S except at the origin. Then Q + 0 within R. Therefore by what we have just seen 11 -= -1 + c1z + c22 + *. z within S. Q am Hence 1 e _= a z.+ 1 _ + + m - + m + m+lz +-** (10 P amzm zm-l Z for any z = 0 within R. This gives the theorem: Let the series P in 8) converge within the circle M about the origin. If P does not vanish within S except at z = 0, the reciprocal of P can be developed in a series of the form given in 10). 117. Laurent's Development. 1. When f(z) is one-valued and analytic within some circle c about z = a, Taylor's theorem asserts thatf can be developed in a power series about this point. f(z) = aO + al( - a) + a2(Z -a)2 +.. (1 We call the point a a regular or ordinary point and we say f(z) is regular at a. If f(z) cannot be developed in a series of the form 1) about the point z = a, we call it a singular point. Let us consider for example VZ ++ log. (2 z —1 Here z= 0, = 1, z =-1 are singular points. For suppose 2) could be developed in a power series P(z) about one of these points. Now P being a power series is defined at every point within its circle of convergence f, is one-valued, and has a contin ANALYTIC FUNCTIONS 235 uous derivative. But the function 2) is two-valued about the point z = - 1, and is not defined at the points z = 0 and z= 1. Thus 2) certainly cannot be developed in a power series about these points; they are therefore singular points. When Taylor's development is not applicable at a point z =a, we may often use another development due to Laurent, as we now show. 2. Letf(z) be one-valued and analytic in the ring R determined by the circles E, F, whose centers are z = a. Then f can be developed in an ascending and descending integral power series f(z) = ao + al(Z - a) + a2( - a)2+ *.. + - +_ b2 + (1 z - a (z - a) valid within R. For let z be any point within R as in the figure. Then by Cauchy's integral theorem, 106,1 f ( F) f(=) C a u-z But by 105 4, 9= +, R, Hence -\Hence =__1 7f(u)du 1 rf(u)du (2 f(z) = - = (2 2 riJ u - 2 *7W F - Z We now develop in a power series as in 109, 3). We have U —z for u on F -a <1, Z —a uonL7 z-a- < 1. u-a Thus by 39, 10) we have for any u on E 1 1 __ _a z - -a+ +- a, u- z u- a u - a ( )- a } while for any u on F =- l+U-7 —a + +... u-z z-a z- a \z - a/ 236 FUNCTIONS OF A COMPLEX VARIABLE If we multiply these relations by f(u), the series so obtained are steadily convergent with reference to their respective circles E, F. Thus we may integrate termwise and get j f(u)d =f- Jd- _ du + (z - ) f (u)du E U u- a E (u - a) Jf(u)d 1_ aff()du+ - a)f(u)du +. u u-z z- a (z - a) F Putting these values in 2), we get 1) where I1f f (u)du =n 2 riJ (Bu-a)n ( n =1 (u - a)n-f (u)du. (4 2 7rF 3. By 105, 4 we note that the circles E, F in 3), 4) may be replaced by any circle S in the ring R. For the integrand of 3) is analytic in the ring E- i, and that of 4) is analytic in i - F. 4. Let us now prove the important theorem: If f(z) can be developed in a two-way power series 00 f (z) = a (z - a)(, (5 -oo this series must be the series of Laurent. For the function defined by the series 5) satisfies the conditions of Laurent's theorem in 2. Thus f admits the development 00 f(z) = In (z- a)' (6 -00oo where the coefficients In are the coefficients of Laurent given in 3), 4). Subtracting 5) and 6) we get 0 = b.(z-a)n, bn=a - l. (7 -00O ANALYTIC FUNCTIONS 237 Let us multiply 7) by (z - a) —('"'+ and integrate around a circle C lying within the ring of convergence of 7). Then by 101, 11) 0= 2 ribm..'. b,, = O, m = 0, 1, ~ 2... Thus am =- m and the coefficients in 5) are the coefficients of Laurent. 118. Zeros and Poles. 1. Let f(z) be a one-valued and analytic function within a circle 9 about z = a. Then Taylor's development is valid within t and we have f(z)= a0 + al(z - a) + a2( - a)2 + (1 For z = a, this gives f(a) = a0. If ao = 0, f vanishes at z = a. We say z = a is a root or a zero of f(z). Suppose ao = aI = *. = am-1 = O am -- O. Then 1) becomes f(z) = (z - a)m(am + am+l(z - a) + a+2 (z - a)2 +...) = ( - a)mg (). Here g(z) is analytic within M and does not vanish at z = a. We say z = a is a root or zero of order m. Since g does not vanish at a it cannot = 0 in some circle about this point. We have thus the theorem: Let f(z) be one-valued and analytic about the point z = a, and vanish at this point, but not identically. Then there exists a positive integer m such that () = (z a)(z) (2 f(z)- (z - y)mg(z) (2 where g(z) is analytic about z = a and does not vanish in some domain about a. 2. Suppose now that f(z) is one-valued and analytic within a circle M about z = a except at the center itself. Such functions are 1 Z2 —1, tanz. In the first a = ~ 1, in the second a = (2n + 1)2. If we describe a little circle ~ of radius r about a, we get a ring - - ( and for 238 FUNCTIONS OF A COMPLEX VARIABLE all points within this ring Laurent's development will hold. Thus f(z)= ao + a( -a) + a2 (- a)2 + *. + b +- b2 + (3 z-a ( - a)2 =P+ Q. We call b ( Q = ~ - +.+... (4 z-a (z-a)2 the characteristic of f (z) at z = a and write Q = Charf(z). z=a The coefficient bI is of great importance in some investigations. It is called the residue of f(z) at z = a and we write b1 = Res f(z). z=a Since r may be taken just as small as we choose, the development 3) holds for all points within R except its center. Looking at the characteristic 4) all its coefficients may be zero after the mth. In this case, which is very important, we have Q= b- - +. + bm z- a ( - a)m bm + bm (z -a)+ + bl(z - a)-1 (z _ a)m -a(z) (z - a)m where p is a polynomial of degree < m - 1. Thus Q is a rational function of z. From 3), we have f (z) b + bm_l(z- a)+.. + bl(z- a)m-1 + a(- a)m _-&~~ ( - a)m(z ( (Z), g(a):, 0 (6 (z - a)ln ANALYTIC FUNCTIONS 239 where g(z) is an analytic function in which does not vanish at z = a. Since lim (z- a)m = 0, limg(z) = bm 0, z=a z=a the expression 6) shows that lim If (z) I = + oo, (7 Z=a that is, as z approaches a, f recedes indefinitely from the origin. This we will indicate by the symbolic equation lim f ()= co, (8 z=a so that 8) is only another way of writing 7). On the other hand, the reciprocal of f is 1 (z-a)m (9 f(z) g(z) As g does not vanish at a, its reciprocal is an analytic function, call it h(z), about this point, and h does not vanish at a, as shown in Ex. 3, 116. Thus, we may write 9) 1 (z- a)m(), h(a) 0. (10 This shows that tlle reciprocal of f has a zero of order m at z = a. The function f(z) and its reciprocal behave thus in opposite manners like the poles of a magnet. As z a, f _ o while its reciprocal ' 0. For this reason we say, that when the characteristic of a function has the form 5), that z = a is a pole of f(z), and in fact a pole of order m. We thus have this result: If f(z) has a pole of order m at z = a, it has the form f (z)=g (z) (Z) f(=(z- a)m where g is analytic about z = a and does not vanish at this point. The reciprocal of f(z) has a zero of order m. Conversely, if f has the form 11), z = a is a pole of order m. 3. Let f(z) be one-valued about z =a and analytic except at z = a. If the reciprocal of f has a zero of order m at a, this point is a pole of order mfor f(z). 240 FUNCTIONS OF A COMPLEX YAP,,IABLE For by hypothesis, 1 fZ)= (z ~a) mgz, and g(z) does not vanish about z= a. Hence, g b) 6+ 61(z - a)+ b2(Z- a)2+ and thus f 6)b + h(z - a~)+.* (Z - a)m - b0 ~ +... ++bb~m + m+K(z -a) +. (z -a)m z- a Thus the characteristic of f has the form 5). 4. If z =a is a pole of order m o f f(z), it is a pole qf order m+1 off,(z). For about z = a we have f (z) (Z )m z+- a here g is analytic about z =a, and hm#0. Hence mbm b__ ________ - - 1 + g'1(Z). (z - a)m+l (z - a)2 As bm =# 0, z = a is a pole of order m + 1 for f' (z). -Example 1. f(z) =211 About any point z = a for which the denominator does not =0, f(z) is analytic. For z = 1 we have z 1 g(z) Now g is analytic about z = 1. Hence z = 1 is a pole of. the first order. Similarly f = z I z- 1 z+1 and this shows that z =-1 is a pole of order 1 also. ANALYTIC FUNCTIONS 241 Example 2. f(z)=in z cos z This is analytic except at the points (2 n + 1) Let us call one of these a. We have for z = a + u cos z = cos a cos u - sin a silln u = (- 1)n+l sin u =(_ 1)n+l U- +... =(-l)^n+1 (Z) z {a-(5 ^)2+... } Thus 1 1 (_ 1)n+ cosz z-a _(z-a)2 - g(z) z-a where g is analytic about z = a and / 0. Thus tan _si z. g(z) h(z). - a - a But sin z does not vanish at a. Hence h(z) is analytic about z =a and does not vanish at this point. Hence z=(2n + 1) is a pole of order 1, for tan z. 5. Let us note that no point z = a which is a pole can belong to the domain of definition of an analytic function. For by definition f'(a) must exist, and this requires that f (z) is continuous at a, by 84, 3. Thus by 83, 6 If(z) < some G (12 in Ds(a), 8 sufficiently small. On the other hand, if a is a pole of z=a as we saw in 2. This contradicts 12). 242 FUNCTIONS OF A COMPLEX VARIABLE 6. If z = a is a zero or pole of order m of f(z), it is of order mn for the function g = \f (z),n, n an integer > 0. For about z = a we have f(z) = (z - a)mc(z), where m > 0 for1 a zero and < 0 for a pole, and where q(a) # 0. Thus, qn(z) = c(z) ins a analytic function which does not vanish at z= a. lencCe g = (z - a)mnr(Z), which proves the theorem. 7. Tf f(z) is a one-valued analytic function in the connected region 1, the poles of its derivative are also poles of f(z) and the residues of f(z) are all 0 in WI. For at a pole f (X)= (a + + la <13 m(Z) " - a. m +-1 + — g(Z), (13 (z - a)m z-a where g is regular at a. If now we integrate, we get f1 -a -l+ + + alog(z-a)+ h(z), (14 where h = f(z)d is regular at a. As f (z) is one-valued in 2 the logarithmic term cannot appear so that a_ = Res f' (z) = 0. z=a 8. As an example let us find the singular points and the residues of the function hWz), S^z^f'() (15 h (z) =g (z) f (z)' (5 which we shall employ later. Here g (z) is regular in the corinex s and has no zero in common with f (z), which latter has certain [)oles z = a, b,... in l but is otherwise regular, Let z = c be a regular point of f and not one of its zeros. Obviously c is a regular point of h. Let z = c be a zero or a pole of order m of f (z). Then f = (Z - m(), ANALYTIC FUNCTIONS 243 where m is a positive integer if c is a zero, and negative if c is a pole. O(z) is regular at c and: 0. Then = m(z - C)m-l0 + (z - C)mcp. Hence m f (Z) Z- C f x) -c = + +f(Z), where 4 is regular at c. On the other hand, by Taylor's theorem g(Z) = g(c)+ cjz - C) + C2(Z - )2 + Hence h(z) - rgtc) ~ k(z), (16 z-c where k is regular at Z = c. From 16) we see that z - c is a pole of order 1 and that Res h(z) = mg(c). (17 Z=C At a zero m is a positive integer, at a pole it is negative. 9. Before laving this topic let us show that the relation 89, 1 or dw dw dz dt dz dt dz holds even when d = 0, provided z is an analytic function of t. dt As we observed in 89, 6 we have only to show tbat Az # 0 as A = At -o0. But z = 0(t) being an analytic function of t, Az = 0(t + h) - 0(t) considered as a function of I is regular at the point h = 0. It therefore does not vanish for by 1. 0 < I h I < some 8 10. Let u = a0o ~a1z + a2+2 +..I, ao 0. Then w = log u considered as a function of z is regular at z = 0 and dw 1 du a, + 2 a2Z + (18 dz u dz ao + a.z+ + 2.AA W-M 24 FUNCTIONS OF A COMPLEX VARIABLE For we -may take, 8 so small that u remains in -D,(a0) if z remains in D1(0). If now r, < Iao 1, the origin u = 0 will not lie in DR. Thus w considered as a function of z is one-valued in -Da and thus by 9, dw dw du dz du dz' which gives 18). Since wv is reg-u Here Hence 11. Let Then Llar at z = 0, we have, by Taylor's theorem, W = w(0)~+zw(0) +... w(0) =log ao, ()=a.. a0 w =log a. + ~iz +... ao (19 u = ao ~ a1z + a2z2 +... w = V a0#:_-0. is regular at z = 0, and 1 1 n ao n (20 For 1 logu w =e As log u is regular at z = 0 by 10, so is w. Hence by Taylor's theorem wwO —'0z 1 1 ~1-du 0 ~~n.dz etc., which gives 20). 1 1-,w'(0)=-a& na... n 119. Essentially Singular Points. 1. Let z = a be a singular point of f (z). If fP is one-valued about this point which is not a-pole, we say z =_ a is an essentially singular point. It is easy to construct functions having such singular points. For example, let the infinite series, ao + a1z + a2z2 ~. (1 converge for all values of z. Such series we considered in 39, 4. From 1), we can form the series f~)a ~ai a2~ 0 z Z (2 ANALYTIC FUNCTIONS 245 which converges obviously for all z * 0. Thus, the function f(z) defined by 2) is an analytic function of z for all z 4= 0, by 104, 5. If we develop f about the point z = 0 in Laurent's series, we get the series 2) again by 117, 4. Thus, by 118, 2, Charf(z)= + a 2+~. (3 z=O z z2 As the an are not all 0 after some m, the point z = 0 is not a pole. It is, therefore, an essentially singular point. From this function f (z) we can form an infinity of other functions having z = 0 as an essentially singular point. For let g(z) be regular about z = 0. Then +(z) = f (z) + g(z) has z = 0 as an essentially singular point. 2. That z = a is an essentially singular point may often be seen by the aid of the following theorems. We exclude, of course, the trivial case that the functions considered are constants. If f(z) is regular at z = a, f(z) cannot have the same value c at a set of distinct points a, a2, a... which a. For then g(z)= f(z)- c is regular at z = a and vanishes at each an. As g is continuous at a, lim g(an) = g(a). n= 00 As each g(a,) = 0, we see g(z) = 0 at a. But then, by 118, 1, g(z) (z-a)mh(z) where h does not vanish in some domain about z= a. As g(a<) = 0, it follows that h(a) = 0. Thus, h(z) vanishes in any circle about z = a, however small. 3. If f(z) is regular at each point of a circle R about z = a, the center a excepted, and if f has the value c at a set of distinct points, a,, a2,... which ' a, then z = a is an essentially singular point. For we saw in 2 that a is not a regular point. If it is not an essentially singular point, it must be a pole. Then its reciprocal 246 FUNCTIONS OF A COMPLEX VARIABLE g(z) is regular at z = a, and takes on the value - at the points an. This contradicts 2. Thus, z = a is not a pole of f(z). Example.. 1 w = sin -* z This function is regular at each point except z = 0. Now w = 0 for z=-, n=1, 2, 3... nfr and these values- 0. Thus by 3, the origin is an essentially singular point. 4. Let f(z) be regular about z = a except at a and at a set oJ points al, a2.** which a. If each an is a pole, the point z = a is an essentially singular point. The point z = a cannot be regular, for f(z) is infinite in any domain D(a). It cannot be a pole, for the reciprocal of f (z) would be regular at z = a and vanish at the points a., which is impossible by 2. Example. 1 f (z)=.sin - z Let us set g(z) = sin z Then if we set 1 z =-, z COS - dq dg du z dz du dz z about the point a = * Thus g'(z) is continuous about z a mnr and hence g(z) is regular at a. Hence by Taylor's theorem g(z) = g(a) + (z - a)g(a) + (z - a)2g"(a) + 2. 2! ANALYTIC F'UNCTIONS 247 Here g(a)=O, gr(a)=(-l)m+lm27r2. Thus g(z) = (z- a) (- 1)+lm2r2 + *. } Hence (_ )ni+l 1 ) ( - = + o+ - l( - a) +- ~ m2q7r2 z-a where we do not care to know the values of the coefficients c. This shows that the points are poles of order 1. Hence m7r z = 0 is an essentially singular point. 120. Point at Infinity. 1. In seeking to characterize an analytic iction of z, it has been found extremely important to study its dhavior for large values of z. Let us change the variable by setting z=1. (1 U Then a function as 1) + z2 f (Z)= (2 goes over into a function of u,, g(u) = + u3 (3 To learn how f behaves for large values of z, we need only to see how g behaves about the point u = 0. We see it has a zero of order 1. Let us look at the geometrical side of the transformation 1). If we set z= rei, we have - -r- e i r This shows that as z describes a unit circle U in the positive sense, u describes the unit circle U in the u-plane in the negative sense. To a point a = peio within U corresponds the point a =- e-i withP out U. As z 0 along a radius as Oa, u - oo along the corresponding radius Oa. To each point in the z-plane except z =0 corresponds a single point in the u-plane, and conversely to each point except u = 0 in the u-plane corresponds one point in the z-plane. 248 FUNCTIONS OF A COMPLEX VARIABLE To complete the correspondence,, mathenaticians adjoin to the plane an ideal point called' the point at infinity and denoted by the symbol o. They ud plane say that to z = 0 shall z plane correspond u = oo, and to u = 0 shall corre- G spond z = o. Instead then of asking how a function f(z) tM behaves for large values of z, they ask how it behaves about the ideal point z = o. By such a question one means nothing more than this: Change the variable from z to u as in 1). Then if f considered as a function of u is regular at u = 0, we say f is regular at z = so. If f considered as a function of u has a pole of order m or an essentially singular point or a branch point at u = 0, we say f has this same property at z= 0o. 2. We must caution the reader to note that we do not introduce the symbol oo as a number; we do not define any arithmetical operations on this symbol. Also when we say f (z) has a certain property for every z we always mean for finite z unless the contrary is stated. 3. Let us note that the theorems in 119 may at once be extended to the point z = oo. For all we have to do is to replace z by, and reason on the behavior of f considered as a function of u, about u = 0. We may thus state: Ilff(z) is one-valued about z = oo and analytic for large values of z except at a set of points al, a2.. which o; or if f is analytic also at the points an and has the same value at these points, then z = XC is an essentially singular point off(z). 4. It is sometimes convenient to speak of the domain of the point z = oo. By this we mean all the points in the z-plane without some circle ( about the point z = 0. We may denote it by D(oo). ANALYTIC FUNCTIONS 249 If we apply the substitution 1), this domain goes over into the points within some circle about the point u = 0 in the u-plane. 121. Integral Rational Functions. 1. Let us show how the rational and integral rational functions can be characterized from the standpoint of the function theory. We begin by proving a theorem of great value. Let f(z) be regular for every finite z. If I f(z) < some G, (1 however large z is taken, f is a constant. For let us develop f (z) about the origin, we have f(z =f(0) + zf'(0)+ f(0) + - (2 f (Z) = f (~) + 0 ft(~) + 2Z! "(0) +... ( Now by Cauchy's inequalities, 107, 2, f(n)() a n! < Rn This relation holds however large R is taken. As the right side -0 as R oo we see that each coefficient in 2) is 0. Thus f (z) =f(O), a constant. 2. Iff (z) is regular for every z including z- =, it is a constant. For let us describe a circle ( about z= 0. Then, f being continuous in:, we have If (z) | < some Gl1 in (S. Let us now set u = -; this converts G into some circle S about z u = 0. But by hypothesis f considered as a function of u is regular at u = 0. Thus f is a continuous function of u in l, and hence I f < some G2 in R. Thus if G is > both G1, G2, if (z) < for every finite z. Hence f (z) is a constant by 1. 250 FUNCTIONS OF A COMPLEX VARIABLE 3. Let f(z) be regular for every z. If it has a pole of order m at z c, f(z) = a alz + a2 +... + a., a. 0, (3 and conversely. 1 To see how f behaves for z = oo we set z= - and get U a=m + aXm-lu- +. + ao0um f Urnm Thus f has a pole of order m. To prove the other part of the theorem: Since f has a pole of order m, f considered as a function of u has a pole of order m at u = 0. Hence by 118, f= el 2 +02... + + g(U), Cm 0o, (4 U U2 Urn where g is regular at u = 0. We show g is a constant. For f has no singular points except at u = 0. Hence g has no singular point u z 0. But by hypothesis u = 0 is not a singular point. Hence g(u), having no singular point, is a constant co by 2. Thus 4) becomes C eCm f= c-+... + +, or going back to z, f has the form 3). 4. We now establish the fundamental theorem of algebra: Every polynomial of degree m has m roots al, a2... am, some of which may be equal. In other words: If f= a+ az + a2+ *** + az", am O, (5 there exists m numbers al, *.. am such that f= am(z - a)... (z - an). (6 For since lim (z) = Z=0o there exists a circle C about z = 0 such that If (x) for every z outside C. ANALYTIC FUNCTIONS 251 Hence f= 0 outside (7. Thus if f has any roots at all, they lie in C, that is on or within it. Suppose f were - 0 in C. Then If(z) > some r > 0 in, since f is continuous in C, by 83, 7. Let 1. f Then |g < for z outside (, <- for z in C. Thus Igl<some f for every z. Hence g is a constant by 1, which is absurd. Thus f= 0 for some z in C, say for z = /i. Then by 118, 1, f () = (z - l)3,ifx(z). By 3, fi must be a polynomial. The method of undetermined coefficients, 115, shows that it is of degree m - mi. We may now reason on fA(z) as we did on f(z). We thus get f = am(z - l1)ml ( - 32)m2... (Z- _3), (7 where m = m1 + *2... ns. The factor am on the right of 7) is due to the fact that the coefficients of like powers on both sides of 7) must be equal. 122. Rational Functions. 1. These have the form f(z)= b~ + az + '"+ am, b = 0 (1 am(z- a)ml.. ( -- Oa)mr. (2 b(Z- 1)nl..* *(Z - /s)ns We will suppose that numerator and denominator do not have a zero in common. Each zero of the denominator is a pole of f. For example, 1 a,(Z-3l)ml -... f 1 (Z -l)nl b(3n(Z- 2)n2....2 9-:() (3 (Z- nl1)h 252 FUNCTIONS OF A COMPLEX VARIABLE where g is regular at 81 and does not vanish. Thus z = /3 is a pole of f(z) of order n1. Similarly at a zero of the numerator as a, we have f= (z - a)mh(z), where h = 0 at a1. Thus z = ac is a zero of order mi. At any point z = c not a zero of the denominator, f is regular. 1 2. Let us now see how f behaves at z = co. Setting z = -, we have u af= o + + am nm = Un-mh(U). (4 bonu + * + bn As bn= 0, h is regular at u = 0 by 1. From 4) we have: The rational function 1) is regular at z = oo if n > m. It has a zero of order n- m if n > m. It has a pole of order m -n if m> n. If we count the zeros or poles at z = oo with their proper order, we have: The rational function 1) has p zeros and p poles, where p is the degree of 1), that is, p is the greater of the two integers m, n. 3. Let us now establish the converse theorem: If a one-valued analytic function has only a finite number of poles, taking into account z = co, it is a rational function of z. For let z = a, b,.** be these poles. About z = a we have (z)- (Z am + - +" e -- + fl(Z) = kl(z) + fl(z), (z - a)mz — where fl is regular at z = a. As f has a pole at z = b and as k1(z) is regular at this point, fA must have a pole at b. Thus fi(z) ( - b)n +' + f+2() = k2(z) + f2(Z) Here f2 is regular at z = a, and at z = b. Thus we may continue for all the poles of f in the finite part of the plane, getting, say, f(z) = kl() )+ k2(z) +... + k8(z) +f,(z), (5 ANALYTIC FUNCTIONS 253 where the last term has no pole at z = a, b,... that is, has no pole in the finite part of the plane. We now consider the point z = oo. Let us first suppose this is not a pole of the original function f(z). Then 5) shows it is not a pole of f8(z). Thus f,, having no pole even at infinity, is a constant by 121, 2. Suppose now z = oo is a pole of f(z), then 5) shows it is a pole of f,. Thus fS is an analytic function whose only pole is z = oo; hence by 121, 3 it is a polynomial: fs = Po P1Z+ **~ + PiZ' (6 4. The foregoing section shows that we can write the fraction 1) or 2) as follows: 6f=0+p + 6* + +,, +. + b AZ -2 (Z (-)2n2 Z-, (Z- /s)ns where 1 = m- n. When f is written as in 8), we say it is decomposed into partial fractions. Knowing that f can be written as in 8), the coefficients which enter in this expression can be determined by the method of undetermined coefficients. 123. Transcendental Functions. 1. The foregoing articles show us how the rational and integral rational functions are completely characterized by the nature of their singular points. All one-valued analytic functions which are not rational functions are called transcendental. Every transcendental function must have one essentially singular point by definition. The simplest transcendental functions are those which have only one singular point, and that an essentially singular point at oo. Such onevalued functions are called integral transcendental functions. 2. It is easy to show that ez, sinz, cosz, sinhz, coshz 254 FUNCTIONS OF A COMPLEX VARIABLE are integral transcendental functions. For being defined by power series which converge for every z, they have no singular points in the finite part of the plane. On the other hand, if sin z = c for z = a, it. = c for a+27, a+46 a+6. But these values -- o. Thus z = co is an essentially singular point by 120, 3. 3. The same reasoning shows that any one-valued periodic analytic function which has no poles in the finite part of the plane must be an integral transcendental function. For if such a function has the period o, it takes on the same value at z z+, z+2),... which *oO. 4. A one-valued transcendental function which has only poles in the finite part of the plane is called a rational transcendental function. Such functions are tan z, sn sin z 5. As an example of rational transcendental functions let us consider the following, which occur in the elliptic functions. Let o, wo2 be any two numbers which are not collinear with the origin z = 0. With these we form the series 1 (1 (z- ml11 -m2c2) where mr m = 0, ~ 1, ~ 2... and p is a fixed integer > 2. We show now that the function defined by 1) is a one-valued analytic function for every z except at the points t) = COmim = 10.1+ m202 (2 which are poles of order p. To show that P is regular at a point z = a not included in 2) we describe a circle 9 about the origin exterior to a. We now break the series 1) into two parts F =- F, + Fe (3 ANALYTIC FUNCTIONS 255 where Fe contains all the terms of 1) corresponding to values of 0m.m2 which lie exterior to St, and Fj contains the other terms. In 99, 5 we saw that Fe converges steadily in A. Hence by 108, 1, Fe is an analytic function of z in R. On the other hand, E, consists of a finite number of terms of the type (4 (z - (0)p But each such term is regular except at z = c. Hence FJ is regular except at points included in 2). Thus 3) shows that F is regular at z = a. To show that F has a pole of order p at the point z = b = ro1 + SW2, we take S so large that the point b lies within it. Then as before Fe is regular at z = b, while FEt_ 1= +a(z). b (z- b)P + () Now a is the sum of a finite number of terms of the type 4), each of which is regular at b. Thus FS has a pole of order p at z = b, and hence F has also by 3). Thus the points 2) are poles; as these points oo the point z = o is an essentially singular point by 120, 3. 6. Let us show that coI is a period of the function defined by 1). The same reasoning will then show that Co2 is also a period and hence the numbers 2) are also periods. We have from 1) ( (z + w1 - mol^ - m2c2)P (5 (z - (m -1) o1 - m22)P As ml,m2 run over all integral values 0, ~ 1, ~ 2, *.. we see that mI - 1, m2 run over the same values. Thus the terms in 5) are identical with those in 1). As the series 1) is convergent, its sum is independent of the order of its terms and hence 5) has the same sum as 1). The points 2) are the vertices of a set of parallelograms as in the figure. Any one of them as P is called a 256 FUNCTIONS OF A COMPLEX VARIABLE parallelogram of periods. In P, the function F(z) takes on every value it can take on anywhere. For any point z lies in one of these parallelograms as Q. Let z1 be the point in P which is situated in P as z is in Q. Obviously,. = z + mll + m2w02 (6 But then F(Z1)= F(z). Two points z, z1 which are related as in 6) are said to be congruent; we write z z mod o), o2 (7 which we read z1 is congruent to z with respect to the periods w1 Wc2. When no ambiguity can result, we do not need to mention the periods and we write simply Z1-Z. 7. The series 1) define therefore an infinity of periodic functions corresponding to p = 3, 4,... The reader will note that they differ from the periodic functions heretofore considered as e2, sin z in this important particular. Their periods do not all lie on one line, but are spread out over the whole plane, as in the figure. 124. Residues. 1. We saw in 117 that if f is one-valued and regular about z = a, but not regular at a, it can be developed in Laurent's series: f(z) = ao + al(z - a) + a( - a)2 +... + +al 2 +... (1 z - a (z- a)2 The coefficient al we said, in 118, 2, is the residue of f at the point a. These residues are of great importance in certain investigations, for example in the elliptic functions. A fundamental theorem is the following: ANALYTIC FUNCTIONS 257 Let f(z) be regular in the simple connex G except at the points z = al, a2 ". am which we suppose do not lie on the edge e of C. Then E Resf(z) = 2i f(z)dz. (2 For simplicity suppose there are only two singular points a and b in S. Then by 105, 7) 7 fdz ==fdz + fdz. (3 Let 1) be the development of f about z = a. Then fAdz = 2 ria1 by 101, 11) = 2 wri Res f(z). z=a Similarly 1,, J fdz = lIes f(z). 2 rJB z=b These values in 3) give 2). 2. From 1 we may deduce the following general theorem from which we shall draw important conclusions, especially in the elliptic functions. Let f(z) be regular in the simple connex ( except for certain poles. On the edge d of L let f be regular and =: 0. Let g (z) be regular in ( and have no zero in common with f. Then 1-ig (z)d log f (z) = Emrg (ar) - Ensg (as) (4 where a, a, are the zeros and poles of f(z) of orders mr, ns respectively. The integrand in 4) is h(z) =g(z) f () f(z) Its singular points, as we saw in 118, 8, are the zeros and poles of f(z). The formula 118, 17) shows that at a zero z = aar Res h = mrg (ar), (5 258 FUNCTIONS OF A COMPLEX VARIABLE and at a pole z= as Res h = - ng (). (6 If we now put 5), 6) in 2), we get 4). 3. If we agree to count a zero or pole of order m, as m simple zeros or poles, we can write 4) thus: 2 ri gdlogf = g (an) - g (a). (7 4. From 4) or 7) we have as corollary: Letf(z) be regular in the simple connex ( exceptfor certain poles. On the edge @ of ( let f be regular and = 0. Then.d logf (z)= M- N, (8 where M; N are the number of zeros and poles off in g each counted as often as its order. This follows from 7) on setting g(z) = 1. 5. As a corollary of 8) we may prove the fundamental theorem of algebra, viz.: f(z) = azm + alzm-1+ * + a, a0 0 has just m roots, a multiple root of order s being counted as s simple roots. For as lim (z)= we can take a circle C about the the origin of radius R so large that no root of f lies on C or without it. As f has no poles in C, Vin 8) is 0. Thus f 2 dlog f = M. But d logf = ma-zm- + (m - l)alm-2 + - dz a0zm + alzm-1- +. 1 1 m z 2 z I I z 1+xl+ 2+... z z = _mI1 + (z)} Z^+^ ANALYTIC FUNCTIONS 259 where I < e on C if only R is taken > some p. Thus 1 ~mdzm dzJK (9 M= fi +2r -=J+ K. (9 2 wi~J Q2z 2riJ0 z Now by 94, 4) J= m, while [K < *- -.2 R, by 92, 2 < me. But this says that lim K= 0. R=~0 Hence passing to the limit R = oo in 9), we get M= m, that is, f vanishes m times. 125. Inversion of a Power Series. 1. If w = w(z) (1 is regular at z = a, it can be developed in a power series: w = ao + a1(z- a) + a2(z- a)2 +.* (2 It is sometimes convenient to develop the inverse function z in a series whose terms depend on w. This is called inversion of the series 2). If we replace z - a by z and w - a0 by w, the series 2) may be written w = alz + az2 + az3 + 3 *. (3 and without loss of generality we may suppose this is the developnent of 1) instead of 2). In inverting the series 3) there are two cases which must be distinguished. Case 1. a1 =- 0. This condition expresses the fact that w' (z) * 0 for z = 0. Let z range over some circle S about z= 0; then w as given by 3) ranges over some connex R whose edge does not pass through w = 0 if S is sufficiently small by 118, 1. Also, if ( is sufficiently small, w will not take on the same value twice in ( by 119, 2. 260 FUNCTIONS OF A COMPLEX VARIABLE Thus it follows by 88, 2 that the inverse function z is regular at w = 0 and hence can be developed in the power series z = b1w + b2w2 + b3W3 +. (4 valid in some Z about w = 0. As dz 1 dw dw' dz we have for = o h, =1~ (5 =a1 2. Case 2. a1 = 0. In the series 3) suppose that am is the first coefficient i 0. Then W = zm(am + am+lz + a,,) m =#(6O. Let us set = tM. (7 Then 1 Te = Z(a a,,, + am ~Z~.) = z(cO + C1Z + C2z2 + CO.) C0 = amm by 118, ii. We are thus led back to case 1. Inverting, we get 1 2 + arnm Putting in the value of u in 7), we get finally 1( z = wm { + d2wm +. j (8 3. Let us show how to get the coefficients of the inverse series using tie method of undetermined coefficients. Let us suppose that the original series is v = bo + bl(t - b) + b2(t - b)2~... bk rf#0. (9 Let us set W = V -- bo x = bi(t - b). Then 9) takes the form w = z - a2Z2 - a3Z3 - (10 ANALYTIC FUNCTIONS 261 where we have introduced the minus signs for convenience later. Then the inverse series has the form Z = W + C2W2 + 3W3+ (11 Raising 11) to successive powers gives z2=W2+ 2 C2W3 + (c~ 2C3) W4 + (2 c4~ 2 c2C3)W 5 + zw 3~3cw4 + (3C2 + 3C8)W ~ (12 z =w4~4 c2w5 + All these series may be denoted by Zm = Cm1W +~cW +. (13 Putting these series for z, z2 z3... in 10) gives rise to a double series D=~W+ ~W2 C3W + - a2c21w - a202w2 - a2c2w3 - *.. (14 - a,c31w - a3032w2 - a3c39w3 - If we sum this series by rows, we fall back on 10). The sum of D by rows is thus w. If the series 14) be summed by columns, we get a power series in w, and this is what we want. Now by 42, 2 if 14) is convergent, its sum is the same whether summed by rows or by columns. To show that 14) is convergent we shall show that its adjoint is convergent. Let us denote this adjoint series by W+ 72 W2 + '13 W3 +. 11%2 + y3IY+ + acvy2l W+ az227 2 + a2Y23 W3 +... (15 Now 11) is valid in some circle f, it thus converges absolutely for all W < some TV0. Then Z = W+ y W2 + y3 W3 + converges for W ~ TlVO. The series 10) converges absolutely for all Z < some Zo. Thus the adjoint of 10) Z + a 22 + X3Z3 +... (16 converges for all Z ~ Z0. 262 FUNCTIONS OF A COMPLEX VARIABLE Returning to the adjoint series 15), we see that if we sum this by rows we get 16). As this converges, D is convergent for all Iw I <- W by 42, 3. We may thus sum D by columns, getting a power series. As the sum of D by rows is w as we saw, the sum of this power series is also w by 42, 2. Thus we get, replacing the Cmn by their values in 12), the identity w = w + (c2 - 2)2 + (3 - 2 a22 - a3)w3 + ~* Hence all the coefficients on the right are 0, except the first. The resulting equations give C2 = a2, 3 = 2 a22 + a3, e4 = a2 (c + 2 C3) + 3.a3C2 + a4, (7 c =2a2(e4 - e2e3)+ 3 a3(c2 +e3) + 4 a4c2 + a5, etc. 4. Example. We saw that Z2 z3 Z4, w=log(l+z)=z- +=- + *-... <1 (18 Here3 4 1 a2 3 4 - 4 These values in 17) give -1 1i =A 2 2 3 6 4 2 Hence, inverting the series 18), we jet w2 W3 W4 z= w $+ 3 +..* (<19 2! 3! 4! Now from 18) we have W2 W3 1+z == ew += + + ++... 2! 3! which agrees with 19). 126. Fourier's Development. 1. When the real function f(x) has the period 2 7r, Fourier showed that in many cases it can be developed in a trigonometric series f(x)= a0 + aI cos x + a2 cos 2 x +* + bi sin x + ba sin 2 x + * (1 ANALYTIC FUNCTIONS 263 This development is of extraordinary importance in mathematical physics and in some branches of pure mathematics. Let us show how this development appears in the function theory. We begin by proving the theorem: Let f(z) be a one-valued function having the period w. If it is analytic in a band B whose sides are parallel to Oo, we have 2 lriz f () = ame (2 where 1 + 2 am - J f(v)e — dv (3 and b is any point in B. For let us set 2 iz U= e, (4 then u has co as period. To find the image of the band B in Fig. 1, let us begin by finding the image of a line la parallel to Ow and cutting the real axis at z = a. When z lies on such a line, we have z = a + rwo, where r ranges over all real values. Let us set o) Then 4) gives U = e +ro)= e2 r(a'+a"i)e2 rtr = e2e2 i(r+a") / /. plane z plane o / < \ XI /_._. O \ 1 b S- o / o / - V '-J-~ FIG. 1. FIG. 2. Thus when z ranges over l, u moves over the circle C, in Fig. 2 of radius e2a'. When r increases from r=0 to r = 1, u has 264 FUNCTIONS OF A COMPLEX VARIABLE moved once around this circle. Hence when z moves on la over a segment of length =I (o I, u has moved once around Ca. Similarly when z ranges over is in Fig. 1, u moves over the circle C0 in Fig. 2. Thus the image of a line L going from b to 6 + o is a circle lying between C0 and C~. The image of the parallelogram 3 = (ABA'B') is the ring R. To a point within $3 corresponds a single point within B, and conversely. Asf(z) has the period co, f takes on every value in q3 that it can take on anywhere in B. Since f(z) is a onevalued analytic function in 3, it is considered as a function of u, a one-valued analytic function of u in R. Hence, by Laurent's theorem, 117 ao f= 2am (5 where 1 fdu ^^ijc^u ~(6 2 77i ucm+l and C is any circle in R whose center is u = 0. Now from 4) 2 7ri du = - udz. o.) Hence u 2 z 2 2 riz du 2 iTz dz 2 7ri -m- 7 - -= -- = e to dz. Um+l O Um Oe) Thus 5) goes over into 2), and 6) into 3). To avoid confusion we have changed the variable of integration from z to v. 2. The development 2), which is known as Fourier's development, may also be written as follows: O + 2" 2w m f ()=1 J (v)dv + ()o (z- v)dv. (7 For we may write 2) 2 riz 2 riz 2 triz 2 7riz (z)=a+2.e -2 -f(z)= ao + (ale " + ale )+(a2e + a_2e " )+.. (8 ANALYTIC FUNCTIONS 265 Now from 3) 2 riz 2 tVri 2 7riz m. L a-me w A. w to = -f(v)ew dv. Similarly a em._2 1, f -m f. (z-v). a_mTe =- I f(v)e v. Their sum is m 2{i(z-v) -m27r2(z-) -~ If (V) e o +e A fLf (v) {lm 3(z-v) }dv or using Euler's formula, 55, 11) 2 0 27r =-2 f(v) cos m -(z v) dv. This in 8) gives 7). CHAPTER VIII INFINITE PRODUCTS 127. Introduction. In the theory of the gamma function and especially in the theory of the elliptic functions, both of which will be treated later, infinite products play an important part. They are also useful in other parts of analysis. We therefore propose to give a brief account of them here. It is easy to see how mathematicians were led to consider them. Every polynomial, aoaz + + a + + a,z" = am, (1 can be written in the form n an(Z - a)(Z - a2)... (Z - a t) = an, (z - m), (2 where a1, a2... a are the zeros of 1). Since a power series, 00 f(z) = at + a a1z a2z2 + * a = zm, (3 is the limit of a polynomial of the type 1), it is natural to expect that the function f(z) defined by the series 3) can be expressed as the limit of the product of type 2), that is, as an infinite product C(z - al)(z - "2) '.. = C(Z - m), where the al, a2... are the zeros of f(z). As an illustration let us take Z z3 z5 f(z) = sin z = - + -... 1! 3! 5! whose zeros are 0, ~ wr, ~ 2 7r,.... We shall show directly that sinz=z( 1-)( -222)... f 22) (4 266 INFINITE PRODUCTS 267 We notice that each factor 1 2 = (Z - n7r)(z + n7r), n1 n2w-2 vanishes at two of the zeros of sin, viz. at ~ nr. If we set z = 7 in 4) we get r 2 2 4 4 6 6 (5 2 1 3 3 5 5 7 ( one of the earliest infinite products considered, due to Wallis. As examples of other infinite products we notice Q=I(l+q2f), n==1,2, 3... (6 e-Cz Pr(Z) = (7 \ n zII (1 + -)e n 1 0(z) =2 q Q sin 7rzI1(1 - 2 q2 cos 2 7rz+ q4). (8 Here Q is an elliptic modular function, r is the celebrated gamma function, and 0 one of the theta functions which are so fundamental in the elliptic functions. All these we shall consider in the course of this book. 128. Definitions. 1. Let us now define infinite products more precisely. Let al, a2 as... be a sequence of complex numbers. The symbol A=al a2. a *... = nlla (1 is called an infinite product. As in infinite series we set An = al a2 an, A,-= an+1 *a an+2 (2 If lim A, (3 n=ao is finite or definitely infinite, we call it the value of the product 1). As no ambiguity need be feared, we denote an infinite product and its value, when it has one, by the same letter. When 3) is finite and =/ 0, or when one of its factors am = 0, we say A is convergent, otherwise divergent. 268 FUNCTIONS OF A COMPLEX VARIABLE Let us consider 1 1 2 3 4 A= -. -....(4 1 2 3 4 5 Here 1 n and hence lim A, = 0. n==o Thus according to our definition the value of 4) is 0, although no factor of this product is 0. For this reason we do not care in this book to consider infinite products which = 0 although no factor is 0. We have therefore put them in the class of divergent products. The infinite product A. in 2) is called the co-product. Obviously if An is convergent, so is A, and conversely, when zero factors are not present. In a similar manner we define infinite products whose factors are functions of z. Thus if fi(z), f2(z).. are functions of z defined over some point set 21, F=AfA(z)f2(). =-fn(Z) (5 is such a product. Giving z a value in 1 as z = a reduces 5) to a product of the type 1), the factors being now constants. If 5) converges for this value of z, we say it converges for z = a, etc. 2. Just as we have double series;am, (1 so we can have double products Hia,,. (2 With 2) we associate a simple product Ha,, (3 where each factor am of 2) is some factor a, of 3), and conversely. Analogous to double series we will say 2) is convergent when 3) converges absolutely, otherwise 2) is divergent. When 2) is convergent, its value shall be that of 3). From these definitions we may build up a theory of double products in much the same way as we have developed the theory of double series in Chapter III. INFINITE PRODUCTS 269 129. Fundamental Theorem. In the infinite product A= a,. a2. a3, a 0 (1 let us set (2 a = Pmei~~ (2 where we will agree to choose Om so that - 7T < Om < 7r. (3 We now introduce the real series = + 01 + + 02 - 03 + (4 and the real product R = P' P2 P3 (5 R=~PI 'P2 'P "...(5 and prove a theorem on which our treatment of infinite products will rest: For A to converge it is necessary and sufficient that e and R are convergent. When A is convergent, A = ReiO. (6 For A-a.a An al2 n '" a P=... ptnei(t1+ +On) or An = RBeiAn. (7 If now R and e are convergent, lim An = lim Rn. lim eiOn or A = Rei~. Hence A is convergent and its value is given by 6). Conversely, if A converges, Rn and eion must obviously converge to finite values = 0. Thus in the first place R is a convergent product. As ei9n converges to some number - 0, we can denote it by eiT; we have therefore lir ei~n = ei. (8 Now from this we cannot say at once that lim e( = T since ei(u+2r) = eiu 270 FUNCTIONS OF A COMPLEX VARIABLE The relation 8) however shows that | eiT eiOt I < e for all n > some m, or that eiTI 1 ei(n- T) <. This requires that aside from multiples of 2 7r, On shall T, ~that is lim (On - 2 kqr) = T k an integer. n=oo Thus Oh = T+ 2 kn7r + Wqn (9 and however small /7 > 0 is taken, |I In <q for all n > some m. (10 From 9) we have =( -= - = 2 7r(kn - kn_1) + (7n -- 7n-). (11 Now Now k = kn -kn-1 is some integer or 0, while?7 = - n - rn- is as small as we choose. Thus 11) shows that = k27r +. Hence the value of 1 n 1 is not far from I k I 2 a. But from 3) I I < 7r. To reconcile these two facts we must take k = 0, since i k is 0 or a positive integer. From this it follows that all the k, in 9) after some k, are equal. Hence denoting the constant k, by c we have n = T+ 2 Kc7r + n n > s. As i7,, 0 by 10) we have, passing to the limit n = oo in 9), lim O, = T+ 2 c7r. n/=oo This shows that 0 is convergent. We have thus shown that when A is convergent, so are R and 0. INFINITE PRODUCTS 271 130. The Associate Logarithmic Series. To study the convergence of the infinite product A=a1.a a3..., a,~0 (1 we introduce the series log a + log a +. ( L = log a, + log a2 + *~ (2 where using the notation of 129 we take log a = log pn + i,, (3 that is, the principal branch of log an. We call L the associate logarithmic series. Let us prove the theorem: For A to converge it is necessary and sufficient that L converges. When A is convergent, A eL (4 *7 A = e2'. (4 In fact Infact L = log Pi -.. + log pn + i (1 +...+ n) = log Rn + ion (5 = log An + 2sr (6 where s, is some undetermined integer. From 6) we have An= eL Thus, when L is convergent, A is convergent and its value is given by 4). Conversely, suppose that A converges. Then by 129 we know that R and () converge. Hence passing to the limit il 5) we have have L = log R + i~ (7 and L is convergent. 131. Absolute and Steady Convergence. l. In analogy to series one would be tempted to say that A is absolutely convergent, if the product = PP2 formed of the absolute values of the factors of A= ala2a.., an~, (1 is convergent. This is not admissible, as the following example shows. Let A= I(-l)-. (2 1 272 FUNCTIONS OF A COMPLEX VARIABLE The product formed of the absolute values of the factors is R= 1.1.1.... As RA = 1, we see that R converges and has the value 1. On the other hand, the product of the first n factors of 2) is An =(- 1)n which has no limit as n co. Thus 2) is divergent. We could thus have divergent products which converge absolutely. Such a definition is therefore useless. We shall therefore say: The product 1) converges absolutely when the associate logarithmic series log an (3 L = X log a, (3 is absolutely convergent. Hence if L converges absolutely, L is afortiori convergent and thus A converges by 130. From this it follows that when an infinite product converges absolutely, its convergence may be determined by considering the convergence of a positive term series, viz. the adjoint series of 3). For L to converge, it is necessary that log an 0. As ion an -= pne this requires that Pn (4 We have already seen in 129 that o,0. (5 2. If the factors of an infinite product F=J l(Z) ~ fA(z).. (6 are functions of z defined over a point set 2f, we shall say that F converges steadily in SI when the associate logarithmic series L = logf,(Z) (7 converges steadily in. L Thus when L converges steadily, the factors fn(z) all differ from 1 by an amount < e for n > some m. Thus if each fn(z) is onevalued and analytic in some circle S about the point z, Lm(z) will INFINITE PRODUCTS 273 be one-valued and analytic when L converges steadily in R. Hence: F(z) = eL(z)=fj z)f2jz)... (8 is a one-valued analytic function in 9, whose logarithmic derivative is F~z) L (z). (9 P(z) 132. 1. Example 1. Let us consider the analytic character of F~= zHKIx n2r2) (1 We shall prove in 136 that F= sin z. Let Q be a circle of radius B described about z = 0. We take the integer m so large that mw r > B. Then for any z in R, z2 n272 < 1 if n > m. (2 qn 7 r~r We consider now the co-product Fml= K1 I n~w2) (3 where m is now fixed. Obviously if F, converges absolutely in R, so does F. The associate logarithmic series of 3) is Lm =1log(1 _- n =r2l)n- (4 As 2 2 +111z2 Nt2 log 1l2_ ) n2n2 2 7ir2 2\nr22 we have q n 1-< n+ A 'it qn g 274 FUNCTIONS OF A COMPLEX VARIABLE Now qn < qm since n,> m. Hence |l|< qn or as m is fixed 1-m is a constant, and thus l l CR2 1 1 \ln I < Cqn< — - = C C 7W2 nL2 n2 Thus each term of the adjoint of 4) is < the corresponding term of the convergent series C Thus Lm is absolutely and steadily convergent in S. Hence by 131, 2 the product 1) defines a one-valued analytic function of z for any z. This function vanishes for z =, ~ 7r, ~27,... (5 and for no other z. For being convergent, the product 1) cannot vanish unless one of its factors vanishes. Each of the zeros 5) are simple. For we have F= (z -mr) G, where G = m-2(Z + m ) 2) where the dash indicates that the index n does not take on the value n = m. Now G, being a convergent product, does not vanish for z = m7r, since none of its factors vanishes at this point. 2. Let us note that the foregoing reasoning establishes the theorem: The series L=Zlog(- n2) n=] converges absolutely for all values of z = nwr. It converges steadily in any connex not containing any of the points nwr. INFINITE PRODUCTS 275 133. Example 2. The P function is defined, as we shall see, by F (X> = e n=1,2,.. (12 Where GO 1 wherec lolo g(1~I+}=.57 7215... (2 Lnn is the Eulerian constant considered in 20, Ex. 4. We show that the infinite product in the denominator r= j+ n)\ (3 is an analytic function of z which has z —15 -2, - 3,. (4 as zeros of order 1. To this end we describe a circle 9 of radius R about z = 0. We take the integer m so great that rn > B. Then n <,1 ~n>m n for any z in R. We now consider L, O fllog I+ Z Z/xi c7( which is the associated logarithmic series of the co-product Fm, of 3). Now 1 Z2 1 Z3 in -- - - - -. 2 n2 3 n3 Hence I i< q2 + +s ci < I n' 1+ fl Hence In n2,1<2 276 FUNCTIONS OF A COMPLEX VARIABLE This shows, as in Ex. 1, that the series 5) converges steadily in R. Hence, as before, F is an analytic function which has 4) as simple zeros. Thus the function r defined by 1) is a one-valued analytic function having z= 0, - 1, - 2, *.. as simple poles. By 120, 3 z = 0o is an essentially singular point. 134. Normal Form. 1. We have seen that if the infinite product A = al ' a2 ~ a3 *' is convergent, then an -. It is natural, therefore, that many infinite products present themselves in the form A= (1 + bi)(1 + b2). = I(1i + bn). (1 We call this the normal form of an infinite product. Since we can always set always set a= 1 + (an - 1) = 1 + b we can always reduce an infinite product to the normal form. We prove now: For the product 1) to converge absolutely, it is necessary and sufficient that the series B=b + b+ (2 B = - + -1 + '- ' (2 is absolutely convergent. Suppose that A is absolutely convergent. Then the associate logarithmic series of 1) is absolutely convergent, that is, = I log (1 + b-) = X, is convergent. Thus Xn 0 and hence bn - 0. Thus I| b, < 1 for n > some m. Thus log(1 + b,) = bn n + * n >m. 2 3 Hence log(1 + b,) n l bn + T h u s T ^ n T l o g~ l+nn) 3 Thus lim n =im l o (1 + b,)= 1 (3,= oo n n=oo bn Hence by 20, 2 11 + 32 + (4 is convergent, that is, 2) is absolutely convergent. INFINITE PRODUCTS 277 Conversely, suppose B converges absolutely; then 4) is convergent. Then 3) holds once more and hence by 20, 2, 2 is convergent. But then by definition the product 1) is absolutely convergent. Definition. The series 2) is called the normal series of the product 1). 2. From 1 we conclude that if 1) is absolutely convergent, the series 2 log (1 +/ 3), /= b (5 is also absolutely convergent, and conversely. For when the product A converges absolutely, the series 9,3, converges. But this series and 5) converge simultaneously as lim log(l + ) = 1 n=oo jn 3. In 131, 2 we have seen how the analytic nature of F(z) = nff(z) may be determined from that of the associate logarithmic series. Let us now show how it may be inferred from the analyticity of the normal series. We prove in fact: The product P= H(1 + f(z)) (6 is a one-valued analytic function of z within a circle S about z = a, if the corresponding normal series F= Ifn(z) (7 is steadily convergent in & and each f,(z) is one-valued and analytic in R. For 7) being steadily convergent in f, each term f, is numerically < some c, for any z in S, and the series Ec is convergent. Thus c, 0, or c,< e for n > some m. Hence o *y Ifn(z) <e n>m for any z in i. We show now that the logarithmic series L = log (1 + f n()) (8 m+1 278 FUNCTIONS OF A COMPLEX VARIABLE converges steadily in S. For fn fn fn 1 2 3 Hence lnll< If nl+If 2+ IfIn 13. <Ifn | l+ +e+... < n ryen. 1-e Thus each term of 8) is numerically < the corresponding term of the convergent series Example 1. The product A = 1. *. 3 * 6 **= ala2a** (9 is convergent. For consider the product ( 1 22 —1 42 —1 62 —1 P= II1 - ln2)221. 42. 62 \ 22 642 62 = (' 2)( 6 6)(. 6 The normal series belonging to P is -n2 As this converges, P is convergent. Now A=P P 2m = pm. p A -a.P -^ 2m+l.Ip p A2m+ = a2m+l 'Pm 2m + 2 2m+2 Thus hlim A = P, n=ao and A is convergent. Example 2. Q () = n (1 + Z2n) (10 converges steadily in any circle S about z = 0 of radius R < 1. For the normal series corresponding to 10) is z2. (11 INFINITE PRODUCTS 279 But each term of 11) is < the corresponding term of R2, which converges since R < 1. Thus Q converges steadily in R. The proof also shows that 10) converges absolutely for any zI|<1. The product 10) is the product 6) in 127, q being replaced by z. The function Q (z) is thus an elliptic modular function. It is a most extraordinary function, since every point on a unit circle ( about z = 0 is an essentially singular point. It admits therefore no analytic continuation outside G. Here then is an analytic function whose domain of definition, instead of being the whole z-plane, certain isolated points excepted, as is the case with all the elementary functions, is the interior of C. 135. Arithmetic Operations. 1. Let us now see whether the usual transformations of finite products hold for infinite products. We have in the first place: Let A = Ia,, B= Hbn be convergent. Then the products C7= Hann, D I= Han (no b, = 0 in D) bn are convergent and = A B D A B Moreover if A, B converge absolutely, so do C and D. For For - Cn = AB.nHence letting n oo we have lim Cn = lim An * lim Bn = AB, etc. To show that C is absolutely convergent when A and B are, we set a, =l+ a, b,=1+b l al=an, ib|=/,.3 Since A and B converge absolutely, 1: log ( + n), log (1 + 3,) 280 FUNCTIONS OF A COMPLEX VARIABLE converge absolutely by 134, 2. Hence log (1 + aj) + log (1 + ~s) = E log (1 + a) (1 + /,) is convergent. Hence C is absolutely convergent. In the same way we may reason on D. 2. An absolutely convergent product is commutative. For let A =la be absolutely convergent. Then L = - log a. is absolutely convergent. As by 130, A = eL and as we may permute the terms of L without changing its value, we may do the same with the factors of A. 3. A convergent infinite product is associative, that is, we may insert parentheses at pleasure. For let A = aa. be convergent. Let us consider B = (a1... aml)(aml+,l a) am = b, b2... Now Bn = *... bn = al. a2 a. amn = Am. As n = o, Am A, hence lim Bn = A. n=00 Example. The following infinite products occur in the elliptic functions ( + q2n) Qx= 11(1 + -29 Q2 -= I( + 2n-1) n = 1, 2,... Q3 = HI(1 - q2n-1). They are obviously absolutely convergent for [q < 1. As an exercise, let us prove an important relation which we shall need later, viz.: _ - 1 r( - - 6Y1T2Z3 - -' - - INFINITE PRODUCTS 281 For P = 11(1 ~ q2n)(l + y2n-1)(1 - q2n-1) = 71(1 + q2n)(l - q4nl2) by 3. Now all integers of the type 2 n are of the form 4 n - 2 or 4 n. Hence 11(1 - q2n) = 1(1 - q4n)(l - q4n2) = 11(1 - q4n)rH(l q4n-2), or H (1 - q4n-2) q2) H(1 - q4) Thus ( 11(1 + q1lq(1 fq2n) 4n 1 - q4n l-q4n Circular Functions 136. The Sine and Cosine Products. 1. Let us show how sin z may be developed in an infinite product. This product is useful in various transformations and gives rise to many useful relations. We wish to show that sin z = zTT T~2X n = 1 21 39 ~ — / z2 2N We begin by showing that 1) holds for real x lying in the interval (=(a, b) 0 < a< b <; 2 it will then be easy to show that it holds for complex z. In 6 we saw that sin nx is a polynomial of degree n in sin x when n is odd, or sin nx = ao sinnx ~ a, sin'ln x + + an,.1 sin x. If we set t = Sill x, (2 this gives sin nx = P(t) = a tn + a tn1+ ~.. + a' t. (3 rhere is no constant term here, since when t = 0, F = 0 also. Now F, considered as a function of t, has n roots. On the other hand, considered as a function of x, it vanishes when sin nx = 0, 282 FUNCTIONS OF A COMPLEX VARIABLE that is, when x=0, ~+O n 9 7r n- - 7r ~ -... ~ ---- n2 n Putting these values in 2), we see that F = 0 when t=0, ~-sin7r, ~sin2.7,.. ~sinnn n 2 n Thus F(t) = at(t - sin ) t + Si 7.. 2 rr )...n — 1 7 =aot(t2- sin2 ) (t2- sin2 1. (4 Dividing through by sin2. sin22.sin2 -1 7r n n n n and denoting the new constant by C, we get sin nx = Csin x 1 sin2x sin2 J... 1 - *4i- sin2x].2n-1 7r sin2 *2 ni (5 To find C we have from 5) sin nx i 1 sin2 x sin x [ s n sin" 72r n Let now x ' 0, the last relation gives n = C which in 5) gives, on replacing x by, n sin x = n sin x P(x, n) n where sin2 X n (6 (7, r =, 2,.. n-1 2 INFINITE PRODUCTS 283 As n oo we have sin-. x n. nsin-= -. n 1 n Also sin2X n x2 s 2 r7r r2rr2 n It seems likely, therefore, that on letting n oo in 6) and 7) we shall get 1) for values of x lying ill 9. To prove this let us set P(x) = i( - 2) (8 sin2 x r=1 sin2 r 2 n L(x) = log P(x) = log( 1 - 12) (10 We then have lim P(x, n) = lim eL(x,n) = eL()= P, 7n=00 n=oo provided lim L(x, n) = L(x). (11 n=oo Thus we need only to prove 11), which we easily do as follows. Let us denote the sum of the first m terms of 9) by Lm(x, n) and the rest of the sum by Lm(x, n). Then from L (x, n) = Lm(x, n) + Lm(x, n), L (x)= Lm(x) + Lm(X), we have IL(x, n)-LL(x)l I L(x, n)-Lm(x)| + I Lm(x, n)l + Lm(x). (12 Now for < r< 7 2 we have x < sinx < x. 2^ 284 24FUNCTIONS OF A. COMPLEX VARIABLE Thus for all r > some m, and for any x in 21, sin2 X n <4 x2<1(3 n Also for 0< <' we have, by the law of the mean, 0< -log(1- ca)< -log(1-/3)<fl+ GI32, a some constant. (14 Thus 13), 14) give for any value of n 0 < IL,(x, n)j _ rn2l m+1lr 3 if m is taken sufficiently large. Also by 132 on taking m still larger if necessary, I Lm(x)l Finally if n is > some Thus 12) gives which establishes 11). Thus 1) holds for z in 21.' To extend it to all values of z we need only to observe that the right side of 1) is an analytic f unction of z, as we saw in 132. Thus by the principle of analytic continuation 1) holds for any complex z. 2. Let us show that sin z has the period 27w by using the product 1). From 1) we have Sin z = urn Q"(z) (13 where ='(z z ~,mw M= 0 excluded. m =-n Mr Thus Q,(z + w) z~(n+l)w v s~cc Q'(z) z - nw7 INFINITE PRODUCTS 285 Hence Hence rn(lim Qn(z + r) = - lilm Q,(z). Thus 15) gives Thus 15) gives sin (z + r) = - sil z. Thus 2 7r is a period of sin z. 3. From the product expression 1) we may derive cos z = (-(2 -) (16 K (2 n - 1)2w21 For from sin 2 z = 2 sin z cos z we have z2 COSZ — 2 1 2z " nK2r2w2 2n f 4 ( 2 4 z2 (2 m)27r2)( (2m —1)27r2 - / s r 2 \ II( 2 ( -2 ) u(l 2) K2 (2m-1)2 ' which gives 16) at once. 137. Infinite Series for tan z, cosec z, etc. 1. From 136, 1), 16) we have log sin z = log z + 2 log (1 - -22) (1 1 n27r2 log cos z = log (1 4-2 ) (2 Differentiating these, we get cot z =- + 2 Z2 nr2 (3 tanz=2: 2n-1 3r2 (4 1 )r 2 - 286 FUNCTIONS OF A COMPLEX VARIABLE valid for all z for which tan z, cot z are defined, that is for all z which do not cause a denominator to vanish. The relations 3), 4) exhibit cot z, tan z as a series of rational functions whose poles are precisely the poles of the given functions. They are analogous to the representation of a rational function as the sum of partial fractions as shown in 122, 4. 2. As an exercise let us show the periodicity of cot z from 3). We have cot z = lim iF,(z)= lim, z m7r. n=oo n==-o -Z + n7r Now 1 1 Now ~Fn(z + r) = F(Z)+ - Z +( + 1)r Z- nr Letting n oo we get lim F,(z + 7r) = lilm F,(z), or cot (z + 7r)= cot z, and thus cot z admits the period 7r. 138. Infinite Series for sec z, cosec z. 1. From the relation cosec z = tan z + cot z we have, using 3), 4) of 137 cosec z2= 2 22 Z2 1 _+( — 44 2 1 l272 Z24 z (2 n - 1) 2r2 - z2 n27r2 - z2 _1 + ~ ( —1)n+12 z 2. To get sez we use te r 2. To get sec z we use the relation cosecr -z = sec z. From 1) we have cosec z = 1 + E ( _- 1) n +l 1 z i [ n7r-Z n7r + z INFINITE PRODUCTS 287 Hence cosec( -z - - --— 1) f 1 = S Z lfl1+Zn r- +z n7r+ —z 2 2 2 Let us regroup the terms of S, forming the series T=. + _+_ +.. 7r rr 3 7r 3 r T + Z + L 2 2: 2 2 As 1 I S. -TI|= -- 0 as n oo, we see that T is convergent and = S. Thus 100 ~ (2 n - l)7, sec z= n (-_ n 1)2 (2 2 n -1 2 valid for all z for which sec z is defined. 139. Development of log sin z, tan z, etc., in Power Series. 1. From 137, 1) we have log s-= log 1 - ) (1 zq;1 \ / If we agree to give sn its limiting value 1 as z 0, the relaz tion 1) holds for I zl < 7r. For such z 2 1 z4 log l- 1 -2 --- - -- +... 2)~ n2 r2 2 n4 7 Hence _SiHence sn z 1 z4 1 z6 - z 7w2 27 r4 3..r6 z2 z4 1 z6 22 72 2 24 7r4 3 26 7r6 (2 z2 1 Z4 1 z6 32r2 2 34 7r4 3 36 r6 + * **. 288 FUNCTIONS OF A COMPLEX VARIABLE provided we sum this double series by rows. As this series has all its terms positive for a real positive value of z < wr, say for z = r, and as this series summed by rows is convergent, since the relation 2) holds for this r, we may sum 2) by columns for all I z I < r, by 42, 2, 3 Doing this, we get =Sin z 2 1 z4 I1 z6 -log siz + / /H2++ 4 + Z + < 7r, (3 where as usual 1 +1 + + H- +-+-+ *** (4 in 2n 3n 2. In a similar manner we find 22 Z2 1 24 4 6 26 7 5 -log cos z = G2,,2 + 4- 4 + 6 60 + z. < (5 Wr2 2 7w6 3 7T 2 where 1 1 1 Gn = + - + '"+ in 3n 5n We observe that 2n - 1 (n = ~ H. - 2n This il 5) gives -log cosz = (22-1) + i(24-l) + (26-)6 +. (6 valid for I z < 7 If we differentiate 3) and 6), we get tan z = 2(22- 1)H + 2(24 - 1)H + 2(26- 1)H +.. (7 2 4 valid for I z < - cot = 2 H — 2 H 2-2 -.. (8 z 2T2 74 6 r validfor 0 < I zI < r. INFINITE PRODUCTS "289 3. Comparing 7) with the development of tan z given in 116, 4), we get I I 7r2 1 2w2 T 2 2 H2= + ++ =- — +=T 22 32 66 2! 2! 1 1 wr 1 23 4 23 (9 -ff4= 1 + 2+ 34 += - = T3 ' (9 " T241 34 + 390 30 4! 4! 6 1 1 7w6 1 257_ 257__ -U6 =I 6! 6! H - $ ~ -=-~ —=l,-26 36 945 42 6 6 etc. Here 1 W ' T3=TO 5 42 ' ' T7 9 66 In general we may set 22n1 T2n (10 (2 n) - Then 7) gives tan z - 2222 -29 1) 24(24 - 1 z3 + 26(26 - 1) 72 + (11 2! 4! 6! valid for I z 2 From 8) and 10) we get I 00: - 22n cot Z= — 7 2n-11 (12 Z 1 (2 n)! 2 The numbers 72, 2.,.. are called the Bernoullian numbers. 140. Weierstrass' Factor Theorem. 1. We have seen in 136 how the integral transcendental functions sin z, cos z may be developed as infinite products whose factors vanish at the zeros of these functions. Weierstrass by generalizing these developments arrived at the following theorem of great theoretical value: Let the one-valued integral transcendental function F(z) have al, a2,... as zeros which we suppose arranged so that Ia, aa as n oc, an m-tuple zero being repeated m times, and the a's being: 0. Then F(z)= eT(ZTJ (1) H z)-) ea an) +-n (1 where 2 i in where TP is an undetermined integral fun2ction.~ 290 FUNCTIONS OF A COMPLEX VARIABLE Before proving this theorem we wish to make a few explanatory remarks. We note that corresponding to each zero an there is a factor 1 - in 1) which vanishes at this point. Since the an exponential function never vanishes, the right side of 1) will not vanish except one of the factors vanishes, provided the product on the right of 1) is convergent. The infinite product (2 will not converge in general. For example, the zeros of -- as r (s;) we saw in 133, are 0, -1, - 2, *... The product corresponding to 2) is here z (1 + Z n This does not converge, since Elog(l + Z) is divergent. To make the product 2) converge, Weierstrass has added the factor ea 7n a, This introduces no zero into the product, but does make it converge, as we shall see. Finally the factor eT(z) has to be added, since, however the integral function T is chosen, the resulting function 1) has the assigned zeros an. 2. To prove 1) we begin by showing that (z)=1-(1 z — +... (3 n=i\ a> is an integral transcendental function which vanishes only at the a,. For take z large at pleasure and fix it. About the origin we describe a circle 9 of radius R including z. We next take m so large that m m I > R 0 ~m~~~~a= I |l> R. INFINITE PRODUCTS 291 Let H(z) denote the product 3) after deleting the first m factors, that is, the product 3) when n takes on the values n=m+1, m + 2, We now show that the corresponding logarithmic series L(z)= {log l — )+ +...+l _ n( = l (4 n=m+l ^ W n a\ n/ J converges steadily in R. For log (l- =-{ a+a-n +3( + } "V a^ 2 Cn2an 3 \an If we set Izll = R s = n I = - = P < 1, am we have 1 - n+2 Xn< + +'+ 1 /nn2 an ( n+1( (R)2 _R...} < p(l + p + p2 +...) or __ ~~~or -~Xn < 1-p Thus each term of the adjoint of 4) is < the corresponding term of the geometric series 1 Pml for any z in M. Thus H converges steadily in St and is, by 131, 2, an analytic function of z which vanishes only when one of its factors vanishes. Returning now to 3), we see this differs from H only by m factors which are analytic and vanish only at al, a2... am respectively. 3. The function G(z) defined by 3) is an integral function which has the same zeros as F(z). Let us find the most general 292 FUNCTIONS OF A COMPLEX VARIABLE integral function k (z) which has these zeros; we shall see that it will have the form given in 1). For the quotient q(z) Q (5 has no singular points in the finite part of the plane and does not vanish for any z. Thus by 118, 10, log Q is one-valued and has no singular points in the finite part of the plane. Hence T(z) =log Q is an integral function of z. This with 5) gives (z) = eT(z) G(z), which is therefore the most general expression of a one-valued integral function having the assigned zeros an. 4. The exponent in the nth factor in 1) is a polynomial of degree n, and this n increases indefinitely. Weierstrass has shown that: When the zeros an are such that I a n= | an | (6 converges, we may replace the exponential factor in 1) by Z 1 Z i \2 Z i \p-1 ean +. - + where the polynomial is of fixed degree p - 1. To establish this we need only to show that the corresponding logarithmic series L(z)= {log ) 1- + +... + 1 (z\- l (7 n=m+l a/ an.P-i al converges steadily in R. INFINITE PRODUCTS 293 This is indeed so, for here -< ) P + 1(~, < -p { +.. + 2+ Hence 1 RP C n < * am Each term of the adjoint of 7) is thus < the corresponding term of the convergent series 1 a" for any z in f. Thus 7) converges steadily in R. 5. Let us apply Weierstrass' theorem to the sine function. Here we set al=7r, a2= -, a= 2r, a4= - 2 ~ The series 6) becomes here 2 1 which converges for p > 1. Thus nor sin z = ze()H (1 - ze n = ~ 1, ~ 2,... (8 =aze rI(l;E )* (9 Thus Weierstrass' theorem enables us to write down the product expression at once aside from the unknown exponential T. The determination of T is attended with grave difficulties. To avoid this, we have developed sin z in 136 by another method. 6. From Weierstrass' theorem 1 we may write down the most general rational transcendental function with assigned zeros a1, a2, *. (10 and assigned poles b 62 a,. (11 where a zero (pole) of order m is repeated m times. 294 FUNCTIONS OF A COMPLEX VARIABLE Let us suppose that the points 10), 11) are arranged as the an in 1. Then P(z)=(1 - )ea + (an) (12 G(z) I 1(-z )e +' n+ (13 will vanish at the points 10), 11), respectively, and nowhere else. Their quotient (z) will thus have the assigned zeros and poles. Let K(z) be the most general one-valued analytic function having these zeros and poles. Then Q =K(z) H(z) behaves as the quotient in 5). We have therefore as before Q- eT Hence the most general function of the kind sought is a(z) eT(Z) () (14 where T is an integral function. 7. Let us note that the zeros 10) and the poles 11) of a rational transcendental function considered in 6 must both 00, on being properly arranged. For if in 10), for example, we could pick out a sequence a, a, a *.. which - some point a', this point would be an essentially singular point of our function K. Thus K having an essentially singular point in the finite part of the plane is not a rational transcendental function by the definition in 123, 4. CHAPTER IX THE B AND r FUNCTIONS. ASYMPTOTIC EXPANSIONS 141. Introduction. 1. In advanced integral calculus one treats of two functions called the Beta and Gamma functions which are defined by the definite integrals 41 1"jo UX-ldu B(x y) U I'-(1-u)V 1 )du= 1 ) x, y>0 (1 r(x) = e-uu-ldu, x> 0. (2 These functions enter many parts of mathematics and on account of their great importance we shall devote a short chapter to their more important properties. The B and F functions are not independent functions; in fact, as is shown in the calculus, and as we shall see in the next ~, r(x + y) Thus of the two functions we shall devote most of our attention to the r function, which is a function of a single variable, whereas B is a function of two variables. Instead of employing the definition of the B and F function as a definite integral, we can define the F function as an infinite product 1 e-_e r(x) = e - n =1, 2,... (4 ][l 1 q- )e-: n where C =.5772157... is Euler's constant. We shall see directly that the integral 2) and the product 4) have the same values for x > 0. For the function theory, however, the product definition 295 296 FUNCTIONS OF A COMPLEX VARIABLE 4) is vastly to be preferred. In fact, if we allow the variable x to take on complex values, the right side of 4) defines, as we saw in 133, an analytic function of z for the whole z-plane except at the points z = 0, -1, -2... where it has poles of the first order. We may thus regard this function as giving the analytic continuation of an analytic function which for real x > 0 has values given by the integral 2). Instead of the integral definition 1) of the B function we may now take 3) as a definition where F is now defined by 4), the variables x, y being now of course complex. From these product definitions of the B and r functions many of their properties follow very simply, as we shall show. If we prefer, however, to start with the definitions 1), 2) it is merely to preserve the continuity of the reader's knowledge. The justification of the steps we shall take in dealing with the integrals 1), 2) in the next two articles we must leave to the reader for lack of space. 2. The integrals 1), 2) may be given other forms on changing the variable. Thus in 1) let us set V 1 -v We get B(x, y) = vx-l( - v)1-ldv. (5 If we set here v = 1 - w we get B(x, y) = wy-l( - w) —ldw (6 In 5) let us set = si2 we get 7r B(x, y) = 2 sin2-1 0 cos2-1 6d6. (7 Finally, if we set 1 u = log - V in 2), -we get (x) lo dv (8 ' ~ r(^) = / lo^ -i( -)\V( B AND P FUNCTIONS. ASYMPTOTIC EXPANSIONS 297 142. B(x, y) expressed inr1'Functions. We wish to establish the relation 3.) of the last ~, the variables being real. If in 141, 2 we replace u by au, we get From this follows that 1 1 C ~~~e-(l+V)uux+Ywdu. (1+ V) x+Y F(x + yi Hence by 141,1 B (x y) = f (1~+v)x+y - F(x + y ~d V ufu-vXe-()du, or inverting the order of integration, F~x + Y)Jouxy-le-udu J vx-leuvdv. (2 In the v integral let us set uv = w;it becomes u-Xfwx-le-wdw = -XF(x). Thus 2) becomes B (x, y) = F yJ2uy-le-udu =I' (x) y) I XI y>0 (3 which establishes 141, 3. 143. r(x) expressed as a Product. From the calculus we know that e-u= lim (i - Putting this in 141, 2), gives PIx) = m in ux-( 1 -7 cdu. 298 FUNCTIONS OF A COMPLEX VARIABLE Setting u = nv, this gives r(x) -= li n (1 - v)nx-ldv. (1 n=oo t/0 Integrating by parts, we get (1 - v)nvx-dv = - - f ( v)n-lvx-dv n n-1 1 rvxd -- ---- I.. --- V)n-2vx-ldv x+n x+n-1 x +l/o n n 1 n 1 1 x+n x + 1 x We may thus write 1) ^..1 1..2... (n- 1) r(x) = lim 1* ( 1) 2.. (n 1) (2 Now 2 3 4 n., nx= -(. -...-... - (+ ( )... l+ ) Also (x+ 1)... (x+n-)= (n-l). (1+ +.)(+x - l+ x Putting these in 2) gives 1 + - X1 l+X n x log(l+-) x x 1 +n xlog (1+) -x 1 ie ne n (.+X)e e n x II + e-n ( nx B AND r FUNCTIONS. ASYMPTOTIC EXPANSIONS 299 Thus 1 er(x) (4 u(l+~)e~ where C= X nlog 1( ) ( 1 n n is Euler's constant. 144. Properties of r(z). In the foregoing articles we have seen that the B and F functions may be expressed as infinite products, the variables being real and positive. We propose now to start afresh and define these functions for complex values by these same products. Thus we say e-CZ r()= Z n= 1,2,... (1 ZHI + e ne \ n where C is Euler's constant C= { -log 1 + )}.=5772157.. (2 1 n]~ iand r/ and B(u, v)= r(u)r(v) (3 r(u+ v) Then the foregoing shows that these functions reduce to the B and r functions of the calculus when the variables are real and > 0. From the definition 1) we may obtain two other expressions on using the transformations employed in 143, viz.: F~z)=lim 1.2... (n-i) r(z)=lim G,7(z)= lim 1. (n 1) * nZ (4 n=o z (z + l)(z + 2)... (z + n-1) due to Gauss, and (1 ) r(z)=-T n =1,2,... (5 z z 1 +n where in 4), 5) we take that determination of n' and 1 + which is real and positive for z = 1. 300 FUNCTIONS OF A COMPLEX VARIABLE The expression 1) shows, as already seen in 133, that: 1~ r(z) is a one-valued analytic function of z whose domain of definition is the whole z-plane except the points z = 0, -1, -2,... which are poles of the first order. The point z = o is an essentially singular point. Since the factors in 4) are positive for z real and positive, we have: 2~ r(z) is real and positive for z real and positive. A very characteristic property of F is: 3~ r(z + i)= zr(z). (6 For using the product G~(z) in 4) we have r(Z + 1>) =nz G (z) (7 z +- n As nz lim --- = Z,?=oo n + z we get 6) on letting n - oo. By repeated applications of 6) we get r(z + n) = z(z + ).. (z - n- 1)r(z). (8 From 5) we have r(t (9 Let us set z = 1 in 8), we get r(n+ 1)=1 2.33... n=n! (10 This relation gave rise to the F function. In fact Euler proposed to himself the problem: Determine a continuous function which when x is an integer x = n shall have the value 1 2.3... n = n The relation 141, 2) shows that such a function is n(x) = e-uudu = r(x + 1). (1i The relation between r and II may be extended to complex values by defining T1(z) by T _() = r(- - 1 -(. v_\-,/ _ I \ - I -' x,-~ B AND r FUNCTIONS. ASYMPTOTIC EXPANSIONS 301 The functions II and r are of course essentially the same function. The II notation was used by Gauss, the r notation by Legendre. Both notations are currently used to-day. The fact that HI(n) = n T instead of -(1 + n) will often make it convenient to use 1 instead of r. Another important relation is: 40 r(z)r(1-z)= i. (13 sin 7rz For 1 1 n (z) aG,( 1 - z) i= 1n-z. 2 n(nAs. n As lim = 1, n=o n - we have, letting n- o, r(z)r(l z = 1 = zn(1- -sin 7rz by 136, 1). In the calculus the relation 13) is established by using the formula a u1-du - - ' Ja q -du '7r (14 1 + u sin rx whose proof is not simple. If we set z = in 13), we get r2(1)=7 or r(1)= + V/7. (15 The + sign of the radical must be taken by 2~. From 8) and 15) we get 50 tiP(n+ -)i 3 ' 5" (2ne n -1l) / (16 2s 2n Since the exponential function vanishes for no value of z, the expression 1) enables us to state: 6~ The r function vanishes for no value of z. 302 FUNCTIONS OF A COMPLEX VARIABLE 145. Expression for log r(z) and its Derivatives. From 144, 1) we have L(z) = log 17(z) -z - logz~ + {n log(1+t). ( Differentiating, we get L'(z) P1(z) C (2 11'(Z) 1 [n z+nJ 1. z+n-1} In general we find 1 (z + n - )m ( Thus LIJ(1) =- C. (5 ~~~~~~~!! 1i=(_1)m(rn_1)!Hm (6 146. Development of log r7(z) in a Power Series. We saw in 144, 60 that 17(z) has no zeros, thus log 17(z) is a one-valued analytic function about z = 1, whose nearest singular point is z = 0. Thus Taylor's development is valid, and we have log lP(z) = L(z) = L(1) 1! 2!LG 1 L 1 Replacing z -1 by z and using 145, this gives Legendre has shown how we can make the series 1) converge more rapidly. We have 2 n This when added to and subtracted from 1) gives log 17'(1 +z)=-log (1~z) + (1 - C)z~ -1 nI4 2 nl B AND r FUNCTIONS. ASYMPTOTIC EXPANSIONS 303 Changing here z into - z gives log F(I -z) =-log (- z) - (1- C)z + (H, -1)-. 2 Subtracting this from the foregoing gives log r(l + z) - log r(1 - ) = - log+ + 2(1 - C)z z2n+l - ( - 2n+1 1). From 144, 13) we have log r(l + z) + log r(l - )= log 7r sinl 7rz This in the preceding relation gives I 1+z 1i rz log r( + )= (1 -C)z- log + log z 2 1 - z 2 sin+ 1 a), z2n+l X(2+l - ) (2 valid for I z i < 1. This series converges rapidly for 0< x < 1 and thus enables us to compute log F(x) in the interval 1 < x < 3. 147. Graph of (r)x for Real x. By virtue of 144, 8) the value of r(x) for any positive x is known when its value is known for values of x in the interval (0, 1). By virtue of 144, 13) the value of r is known for x < 0 when it is known for x > 0. This relation also shows that the value of r is known in (0, 1) if it is known either in (0, -1) or indeed in any interval of length 2. Gauss has given a table of log H(x) for 0 < x < 1 calculated to 20 decimals. This gives us the value of log r(x) for 1 <x < 2. A four-place table is given in the Tables of B. O. Peirce* for < x<2. Since r(l) = r(2), the curve has a minimum between x = 1 and x = 2. This point is found to be x =1.46163.. * See reference, p. 91. 304 FUNCTIONS OF A COMPLEX VARIABLE From 145, 4) we see that L"(x) > 0, for x > 0. Hence the graph of r(x) is concave for x > 0. The adjoining figure will give the reader an idea of the graph for real x. The vertical lines x = n, n = 0, - 1, - 2... are asymptotes to the curve, and the maxima and minima of the curve lie on opposite sides of the x-axis. The distance of the elbows from the x-axis increases as we go to the left. 148. The r Integral for Complex z. 1. For real x > 0, we have r(x) =J e-U-x-ldu Jo (1 as stated in 141, 2). Let us consider the integral G(z) == euz-ldu, Jo (2 where We have z=x+iy, x>0. Uz —1 = -,txly = x --- 1iy logu = u - {cos (y log ) + i sin (y log u) f. B AND r FUNCTIONS. ASYMPTOTIC EXPANSIONS 305 Thus 00 GT = j e-uxV-1 cos (y log u)du +i e-ux-1 sin (y log u)du = H+iK. Now H and K are convergent since x > 0, hence the integral 2) is convergent for all z for which x > 0. 2. Let us show that G(z) is an analytic function of z. To this end we use 86, I). Now = e- -lux log u cos (y log u)du =a ax, o y H1= _ foe —u- 1 sin (y log u) ~ log udu = - OK ay aJ Ox As these derivatives are also continuous functions of x, y for x >0, we see that G is an analytic function of z for all z lying to the right of the imaginary axis. Since G(z) as defined by the integral 2) and F(z) as defined by an infinite product 144, 1), are analytic functions which have the same values along the positive half of the real axis, they also have the same values for any z = x + iy for which x > 0. 149. r(z) expressed as a Loop Integral. 1. In the foregoing article we have expressed the r function as an integral which converges for all z = x + iy for which x > 0. Let us now show that it may be defined as a loop integral which is valid for all values of z for which r(z) is defined, that is, for all z = 0, - 1, -2,.e. To this end let us consider the integral G(z) = e-Utz-ldu, (1 the path of integration being the L loop L extending to oo as in Fig. 1. -( -o Let =rei~, 0 < <2r. As value of uz-1, we take UZ-1 = e(z-1) log u = e(Z —l){logr+i}. 306 36 FUNCTIONS OF A COMPLEX VARIABLE The integral 1) is thus defined for all values of z and is a onevalued analytic f unction of z by 104, since its derivative dG= e-Uuz- log udu is a continuous function of z. 2. We now show that -rz (e 27riz -1)]P(z). To this end we need only.1- _ prove 2) for z = x> 0, by vir- -P tue of the principle of analyticY continuation, 113. Let us replace the loop L by the loop V as in Fig. 2. circle ac43y converges to 0. Thus (2 -0 co FIG. 2. The radius r of the CT(x) =j = + I J (3 Now on the segment (ce, a), p = 0, hence feuux-ldu -= - P(x) as r 0. On the segment (ry, o),o 2 7r. For when u passes over the circle, alry, 4 increases f rom 0 to 27r. Thus u-1 has on the segment (y, oo) the value e(x1){logr~27ri} = UX-1. e7iXD Thus 0 fe-uu 1du =e2~z e-uuxdu e - x as r 0. Finally p ie Hence ja=r uisd < 2 7rr -0, as r 0. B AND r FUNCTIONS. ASYMPTOTIC EXPANSIONS 307 Thus passing to the limit r = 0 in 3) we get 2) for real x > 0. But as the two sides of 2) are analytic functions of z and as the relation 2) holds for real x > 0, it holds for every z. 3. Let us show that e Z-1 =arrd212 7rie 'i G(z) e-u du (4 z) ( For 2) may be written e7riz e-riz G-r - e 2 ie'. r(z) 2i = 2 ieiz sin vnrz. r(z), by 58, 8). (5 But by 144, 13) sin rz* r(z)= This in 5) gives 4). Since eriz is an integral transcendental function having no zeros, and since r(1 - z) has poles of the first order at z = 1, 2, 3,... and no other singular points, we see that the function G(z) defined by 2) is an integral transcendental function whose zeros are z = 1, 2, 3,... each of order 1. From the standpoint of the function theory, the G function is simpler than the r function. 150. The B Function as a Double Loop Integral. 1. In a similar manner we can show that ZU-1(1 _-Z)V-ldz =-(1 - e2'riu)(1 - e2"iv)B(u, v), (1 where B^ I ) (2v where B(, v)= 1 (u)v) (2 r(u + v)' u, v being any complex numbers for which the quotient on the right of 2) is defined. The path of integration L is so chosen that the many-valued integrand in 1) returns at the end of the circuit to its original value. Such a path is L = llo7l -1-1,a where 10, 11 are loops about z= 0, z = 1 in the positive direction as in Fig. 1. G. It is easy to see that L may be replaced 308 FUNCTIONS OF A COMPLEX VARIABLE by the loop 8 in Fig. 2, without changing the value of the inte- 4 < gral 1). The loop 8 is a double / loop. Finally we must specify which of the many values of FIG. 2. u- e(u-l)logz (1 - )v-l = e(v-l)log(l-z) we start with at the point z = c. We take log c = log r + ia, log (1 - c)= log s + i3, as indicated in Fig. 1. The values of log z = log p + i0, log(1 - z) = log o- + i, at any point of L depend only on 0 and q, since log p, log a are the arithmetical logarithms of the positive real numbers p, a-. To prove 1) we shall first show that 1) is true when u = x, v = y are real and positive. Then reasoning as in 114, 6 we see that 1) holds for complex values of u and v. 2. Let now z run over the loop 10. When z first reaches a, 0 = a; on reaching a after the circuit about z = 0, the value of 0 is a + 2 r. Thus when z returns to z = c, the value of 0 is a + 2 rt. On the other hand, the value of ~ is unchanged. Similarly when z passes over the loop 11, the value of 0 is unchanged, while q goes over into / + 2 7r, etc. We may thus write =(=f(a, 8) + ( +27r, 3)+ + 2r, 3 +2 7) ~ / lo 1l lo. 1 + f(, + 27r) (3 where the numbers in the parentheses indicate the values of 0, f at the beginning of the corresponding circuit. As e(x-l){logp+ia+27ri} = e(x-l){logp-ia} e27ri we see that f ~+ 27r, ) = e ( '(a). *^i, ~A ASYMPTOTIC EXPANSIONS 309 Similarly Simi (a + 2 r, / + 2 7r) = e2rix e2' riy o), lo 0-1 tJ /, (3 + 2 w)-= e27ri f(a /3). Now 0 ( f 0, /3) =f+ e27rix (e27r-i tlo ^c t/0 " t/o since the integral over the little circle about z = 0 is 0. Similarly, Asymptotic Expae2sions f(a, 1) = (1 _ e-2Tri) 1 Putting these values in 3) we get 1) for real positive u, v. Asymptotic Expansions 151. Introduction. In various parts of mathematics it is important to have the approximate value of a function for large values of the argument. For example, when x is a large positive integer n, we shall see in 157, 9 that P(1 + x) or n! is nearly equal to V2 nre-nn', (1 a result of great value in the theory of probabilities and the kinetic theory of gases. Another approximate expression of this type is the following. Letting J,(x) denote the Bessel function of order n, its value for large positive values of x is approximately, as we shall see in 253, 2, ~ X/ cos x - (2 V-~- ~ ~ 4 310 FUNCTIONS OF A COMPLEX VARIABLE This asymptotic expression shows at once that J,(x) has an infinity of real roots, a result of utmost importance in the mathematical theory of heat, etc. Connected with these approximate values of a function for large values of the argument is a class of divergent series U= u1 + 2 + U2 + *. (3 which have this remarkable property:The sum of the first n terms U. gives the value of f(x) with great accuracy for values of n not too large, although the series U itself is divergent. For example, we shall see in 156, 6) that Euler's constant C=lim l1+ ++*+...+ -log n =.57721566... is given rapidly and with great accuracy by using the divergent series 1 1 1 1 - 2T n T4- - T +... (4 2n 22n2+ 4n4 66 n6 where i 4 T2 T2 6 14 = 31 T6 =' are the Bernoullian numbers introduced in 139, 3. Divergent expansions of this type have long been used with utmost advantage in astronomy. On account of their growing importance even in pure mathematics we shall give a brief sketch of them as far as they relate to the P and Bessel functions. We begin by developing a few properties of the Bernoullian numbers and a class of polynomials also named after Bernoulli. 152. Bernoullian Numbers. 1. In 139, 12) we saw that 1 0o 22n cot = (2 ) n T-2nwhere T=,, T, T3= IF T5,=.. are the Bernoullian numbers. Now cos z.eiz + e-iz cot z - = z — sin z ei - e-iz and coth z = i cot iz. ASYMPTOTIC EXPANSIONS 311 Thus coth z - ( )(-1)ZnZ 1 (2n)! 2neZ + e- 2 =-1+ (1 ez - e-z e2z -1 If we set 2 z = u, this gives for | uz < 2 7r U 2 U3 eui-1 +BlU+B2!+ 3B!+ (2 where B1= —1, B3=B5=B7=...=0 (3 and Band B2 (- 1)"+l T2,-1 n > 0. (4 Thus hB2, =(- l)n+ 2(2 n1)! (5 where as usual 1 1 H,=l+ +,++.. (6 2n 22n 32n 42n Instead of T2 we may with Lucas regard the B, as Bernoullian numbers. They may be defined therefore as the successive derivatives of U eu- 1 at u= 0. 2. Let us introduce with de la Vallee-Poussin the symbol e Bu by the relation u u2 u3 u l j =l+ B, +" U " u *=^ -T(7 {e IBu= 1 + BI.! + B2~?!+ B3 ~.+ 7... e 1. We observe that the series in the middle is obtained from eB= 1+ B + B2 +B3 +... 1! 2! 3! by replacing in it, B" by B,. This new symbol has an addition theorem analogous to the exponential function. It is expressed by the relation ueta Bu = e _ (sB+t)u = 1 + (B + t) +(B +0t)2 2 + (8 If ~~2! 312 FUNCTIONS OF A COMPLEX VARIABLE From 7) we have u- =(eu — 1)e IeBu= eBu{eBU — selBU or using 8) u= I e B(B+1) _ e}B. (9 Expanding and equating coefficients of the different powers of u on both sides, we get the symbolic relations: (B + 1)-B = 1 (B + 1)2- B2 0 (10 (B+ 1)3 -B3_ 0, etc., where (B+ 1)" stands for the expression obtained by replacing B"a by B, in the development of (B + 1)n by the Binomial formula. Thus: 1=1 2 B +1=0 (11 3B+ 3B1+ 1=0, etc. From these recursion relations we find readily the values of the T2nl given in 139. The relations 11) show that: The Bernoullian numbers are all rational. 153. Polynomials of Bernoulli. 1. Instead of the function on the left of 152, 2) let us develop t ez- _ 1(ez - 1) e}li ( JF= e^ =^eu- ~ —;B (1 in a power series about u = 0. Since euZ=l+u+( 2 2+... 1! 2! we have P=+ l+B<~C.+B~,+...}{j +C++,+... F= l+B l+B2 + 1 +! 2 3! + = 2 + (Z) +(! + Z2)! + *.. (2 ASYMPTOTIC EXPANSIONS 313 On the other hand 1) gives F-= {e(B+z) —_ IefBu U u -(B+z)- B + (B + z)2 - B2 (B+ z)3-B3. 2! 3! Comparing this with 2) gives the symbolic equation /n(Xz) _ (B + Z)n+l _ Bn+l A(Z) ~ (3 n+1 These /3n which enter as coefficients in power series 2) are the polynomials of Bernouilli. From 3) we find )(z= ) 2 (z - 1)1, /3( Z)_ _ -Z,(z- ), /3(Z) = z - 4(Z) =5 Z (Z2 )(Z- 1)(Z2 _ Z _1) 5(Z)= I Z (Z- 1)2(2 - Z - I), etc. 2. Let us set z= m a positive integer in 1). We get on performing the division indicated in the middle member F= 1 + eU + e2u +... + e(m-1)u + (i + + +... +(-1 + 2u + (2 )2 (2 U)3 ) +(~.u ~(3 U)2~ ~ ). ~..) + +(m-1+)u+.(m..)2U (m-) ) Comparing the coefficients of u" in this expansion and 2) gives 1 + 2n + 3n +... + (m - 1),n= P/(m)), (4 314 FUNCTIONS OF A COMPLEX VARIABLE which connects 3n with the sum of the nth powers of the integers 1, 2, 3,.... 3. The polynomials 13(z) have z = 0 and z = 1 as roots. That 3(0) = 0 follows at once from 3). To show that /,3(l) = 0 we set z = 1 in 3) and get (n + 1) L (1) = (B + l)n+1 - B,+i =0 by 152, 10). 4. If we differentiate 3), we get the derivative of /3,: /3' (z) = (B+ z)n = n/l(Z,_) + Bn. (5 We recall that B2,+l = 0 when n > 0. 5. We now show that: 2 ( 1 (6 82 i — 1- (1 n- )n+l(6 For (1 {e}( 2) z + e}B -=(1 + e2) e ^ -1 Z -Bz e2 - 1 Now the coefficient of gn+l in the development of the first member is in symbolic notation (B + 1)n+l + Bn+1 = (n + 1) An (1) + 2 B,+ (7 using 3), while that in the development of the last member is 2(-B\+ = 2. n+ (8 \21 2n (+l1 Equating 7), 8) gives 6). Since B2,n+1 = 0 for m > 0, we have from 6) that 2n ( 1) = 0. (9 6. The polynomial f2+1, does not take on the same value at more than two different points in / = (0, 1). For then its derivative would vanish at least at two different points within 2. But by 5) 3^+1(x) = (2 s + 1)/828(x). ASYMPTOTIC EXPANSIONS 315 Thus 128(x) vanishes at least at two points within 21; hence by 3, /,2 vanishes at least at three points within W. But by 5) /2s(x) = 2 82s-,(x) + B2. (10 Hence /2-, takes on the same value at least three times in 21; hence hence 2s-1 fP2s-3,...'P P1 each take on some value at least three times in 2. But (x) = (X2 - X) is of the second degree, and can take on the same value but twice. 7. No Bernoullian number with even index B28 can equal 0. For by 6) /32s_(x) would vanish at x= 1. This is impossible by 6. 8. The polynomial 128+l(x) does not change its sign in 1 = (0, 1). For then it would = 0, at three different points in 2, which contradicts 6. 9. The polynomial /32(x) vanishes at x = 0,, 1 and at no other point in 1 = (0, 1). For suppose /2, = 0 at two points within 21. Then /3,(x) = 0 at least at three points within W. Then 10) shows that 32,-l takes on the same value at least three times in 21 and this is impossible by 6. 154. Development of P.(x) in Fourier Series. 1. Let us develop the polynomials /3,(x) in a Fourier series, valid in the interval 2 = (0, 1). We begin by showing that x)2 =( - l)= 2 2 (l - cos 2 ntnx) 1 f -Zcos 2n7rx (1 2vr2 2 n2 I 1 where H2=1 ++ 4-. as usual.3 as usual. 316 FUNCTIONS OF A COMPLEX VARIABLE Let us denote the right side of 1) by F(x). Since,8(0) =PF(O), we need only show that 31 and F have the same derivative in W. The left side of 1) gives at once 3) = x -2. On the other hand, from 110, 8) we have - 1 sii 2 nT ( 7r 1 n But differentiating the series F, that is the last member of 1), we get precisely the series on the right of 2). This establishes 1). 2. From 1) we can express all the other /'s as Fourier series by using the relations 153, 5). Thus for n = 2, we have 32(x) = 2 /3(x) + B2. Integrating, we get, using 1), ^ HfI 1 sin 2 nwrx )2(ZX= 2 A - 3 + B2x. The constant of integration is 0 as /2(0) = 0. Using 152, 5), the last relation becomes,(x) = l X1 sin 2 n7rx 3 L32 (X) = 27 (3 3. Let us now set GS(X) = cos n 2 rx s an even integer, n=lns = sin n2 r s an odd integer. (4 n=l w Also as before let 8H,= 1+ + 3-+.. ASYMPTOTIC EXPANSIONS 317 Then, by using 153, 5) and reasoning from n to n + 1, we have for any integer s > 0 (x) =(_-1)+ (2 s)! 29(X)=(- ) 22 9) 2s+(X), 22s7r2+l ~(), 32s-1(X) = ( - 1)S+- (2 s - 1 -2 valid for 0 x <. We notice that for m an integer or 0, 92s- 7r2s G2s(m)= s=(_)sl(2 )! B2S s = _ 1, =12,.. (6 (2 s)! dxs() -)+ (7 and (1 j 8(x)dx= 2 = Gw +(x) + const. 4. From 5) we see that in the interval W = (0, 1) the signs of 131 13, 3 /, 5.. are respectively. 5. Also we see from 5) that at any point x in A, the sign of 132 is the opposite of that of /2s+2. 6. In passing let us prove a formula we shall need later. Let m < n be positive integers. Then by partial integration we have, using 7), l x) dx _ - rG2 1 G (Icx) dx 1+ 2L1 +x x27 + 2 7r, (1 + X)2 Now 7r2 Now2() = G2(n) = n = 7 by 6) and 139, 9). Thus _ I+(< - -.(8 nj,r 1 m As this 0 as mn o we see that [r a(x)dx 1+* conve g+(9 s converge+ is convergent. 318 FUNCTIONS OF A COMPLEX VARIABLE 155. Euler's Summation Formula. 1. To derive this important formula we employ an elegant method due to Wirtinger.* Let f(x) be a real or a complex function of the real variable x, having a continuous derivative in the interval I = (a, a + nb), where b > 0 and n is a positive integer. Let m be a positive integer <n. Then as in 93, 1 we have fan f(a+nb)-f(a) =b f'(a +bx)dx f(a+nb)-f(a+b) =b f'(a+ bx)dx f(a + nb)- f (a + 2 b) = bJ f' (a + bx)dx t2 In f(a+nb)-f(a+ nb)= b f (a + bx)dx where the last equation is added for symmetry. Adding these n+ 1 equations gives (n + 1)f(a + nb) - f (a + sb) = b f '(a + bx)dx. (1 s=0 s=oas Now n f s+l f s-2 n \" + +. +***+/ ] f +l t/s+1 */n-1 If this is put in 1), we see that every integral of the type occurs just s + 1 times. Thus the sum on the right of 1) may be written naf (s + 1l)f' (a + bx)dx (2 s=0 Now within the interval (s, s + 1) s + 1 = E(x) + 1 L*Acta Matematica, vol. 26, p. 255. ASYMPTOTIC EXPANSIONS 319 where -E(x) is the greatest integer contained in x. Thus the sum 2) equals Yj- l I-E(x) + 1 f '(a +bx)dx~j [E(x)+l~f'(a+,6x)dx. (3 Now E X ~)+x-1 2 where a~)= sin 2 n7rx ~1 n=1 nr 7 Gh being given in 154, 4). Thus the right side of 3) equals Ca(x) + x+ f '(a + bx)dx C fa ~f '(a +6bxdx f f'(a+bx~dx+ - ff(a + bx) dx. (4 But by partial integration Xfxf'(a + bx)dx =nf (a +nb)-ff(a +bx)dx, Ibff'(a + bx)dx-= I~f(a + bn) - f(a). Thus 1) and 4) give (n +1) f(a +nb) - If(a + s6)= 6fa(x) f'(a +bx) dx +nf(a +nb) -ff(a+bx)dx+ 1if (a +bn) -f (a) Hence If (a +sb)=I If (a +bn) +f (a) ~+ ff(a +6x)dx -R (5 where BR = GbCr(x) f(a + bx) dx. (6 820 FUNCTIONS OF A COMPLEX VARIABLE 2. The remainder R may be transformed by partial integration. Thus j(Gf'(a + bx)dx= 1 Gl(x)f'(a + bx)dx Now )) 1 G2(0) — G2(n)= 2 -2 = 7T2B2. w==l 2 Thus b TR = -B2 f'(a + bt) - f'(a)' + R, (7 where 2 n 1we = 2 2 2J G2(x)ft(a + bx)dx. (8 3. Thus by repeated partial integration we get Euler's formula of Summation: f (a)f(a + b)+f (a + 2 b)+. +f (a + nb) = f(a+bx)dx+ 1 f(a+bn)+f(a )} + b-2 f' (a+bn) -f'(a) +ab3B4 f"'(a,,+n)-f,"(a)+ 4! + b2s+1 B2,+2 f(2s+l)(a + bn) - f(2s+l)(a) + R8+1, (9 (2 s + 2). where 2 2s+2 +2 n R+,, = (- l)+,2s+1rs+ ^ Gf+2(x)f(2s+2)(a + bx)dx. (10 If we integrate partially in 10) we get, since G28s+(X) = 0, for x = 0, x = n, - 1 (1)b2s+3 an'* R+ - (2+r) s+3a (G2s+3(x)f (2s+3)(a + bx)dx. (11 4. In 10) and 11) we have expressed the remainder RB in terms of the functions r2,+2 and G2S+; let us now express it in terms of the 83 polynomials. ASYMPTOTIC EXPANSIONS 321 From 11) we have, setting for simplicity b = 1, i M '.1..../:}+~ (12 R(8-1 - 2 { o- + '"r- ) (12 __(-1) 01 = -_ --- r1 G-lz)_ X/ 2 (2s-1)(a + x) + / 2-1) (a + + 1) + * x. 22- 2 ~2s 021()/(2-( + m)x Now (- IS+fs1). 2212s- [G2() f (X2-1) (a + ++ m)]d ( - 1)8+L l A B1 ^^^ (x')f s v(1a+x+m] - 22- -) G28(x)f (28(a + x + m)dx= A - B. (3 But by 154, 5) (- 1<)s+ /-1(x) L (- '1)+ (2-1 ), G2(X) = (2 + 2 ) sv + 226 ffr 22-17_ 2 (x) = s- 1)! 2- s 2' Thus = (- - 1 H / f(2-sl) (a + m + 1) f (2S-1 (a + m), 22s —l7.2s Bm = ( 1)!1 2,l(x)f(28) (a +x+ m)dx (- 1)+ Jo + )f (2 ) (a + x + m)dx. Here the last integral is A,,. Thus 12), 13) give -1.o 1 Rs-l = (2 18- — _: 2s —l(X) f (2s) (a + X) +f(2s)(a+x+l)+ *.. +f('2)(a+x+n- 1) dx. (14 5. Let us return to Euler's formula 9). We may write it, setting b = 1, = f(a)+ f(a+l)+ (a+2+.. +f(a + n) = -f(a+n)+f(a)+ f/(a+ x)dx + D + R, (15 22c ~ v, 322 FUNCTIONS OF A COMPLEX VARIABLE where D, is the sum of the first s terms of the series -D = 2 Jf' (a + n~) - fl(a) l + B4 Iff"'(a n)- f Iff (a) (1 2! 41 (16 = d1 + d2 ~ d3 + As the Bernoullian numbers BN, increase very rapidly as n oo, the series 16) is in general divergent. Suppose, however, that f (2s+2)(X), f(2s+4)(x) have tie same sign in (a, a+ n). As P2s+1(x), P28s+3(X) have opposite signs in (0, 1), the relation 14) shows that R. and R,+1 have opposite signs. Thus -R, I < I R18- R~+1I = 1d8~11. We have therefore in this ease the remarkable result: Alt Iwugh the series 16) is divergent, the sum of its first s terms, D, enables us to calculate the sum P in 15) with an error numerically less than the (s + 1)st term in 16). 111 156. Asymptotic Expansion of 1+ + + +. + In 155, 15) let us set 2 3 n f(x)= - a=l n = m -1. x Then i 1 1 1 + + 1 + + 1 + log~ m + -DS - RS, (1 2 3 mn 2m ~ where B + B (2 + Bs 2 A1 1 118 (2s + 2)+ I )i3'28~1l (X + 1)2s+3 (x + 2)2~+3 + 1)2~}dx (3 Now the Eulerian constant C =lim l+ + +... + 1 - log m. m=00 1 2 m Let us therefore keep s fixed and let m c' o in 1); we get 1i~ B2 B4.. B28 1_ B,.23, B28 i ~ ~(4 2= L2 4 + _sj8' 2+ L2 4 2s ) I( ASYMPTOTIC EXPANSIONS 323 where e = (2 s + 2)f + 2+()+ dx. (5 P=1j Returning to 1) we have, using 4), 1 1o1 (BimB 1+ +... + =C + logm- { - +2 m 2m 2 m2 4 m4 + B 1 + (6 2s m2,J where U1 2 2s+i(x)d( U8 = (2 s + 2) ( + p)2+3 (7 p/- p=m+l P)+3 Since the derivatives f(2s+2)(x), f(2s+4)(x) have the same sign, we are in the case considered in 155, 5. The formula 6) is thus an 1 1 asymptotic development of 1 + + -. + 2 m From 7) we see that U8 has the form U8= E(m) (8 where E- 0 as m=. For in the interval (0, 1) (2 s + 2) j32s+1s(x) < G some constant. (x + p)2s+3 p2s Hence u< a '0dx by 22, 1) m X2s+3 < a 1 2 s + 2 m2s+2 Thus 8) holds since e(r<)< 2 2 m (9 2 s + 2 324 FUNCTIONS OF A COMPLEX VARIABLE 157. Stirling's Formula for n I. In 155, 5), 6) let us set a=b =1, f(a+x)=log(l+x), n=m-1. Then m-1 m-l-1 dx)1 m log (1 + s) = -logm + / log(l+x)dx- <(X)- * (I s=0 0o +xo By partial integration log (1 + x)dx = m log m - m + 1. Thus 1) gives log (1 2.2 3...n)=1(2m +1)logm-m+l-f G(x)_d-. (2 I I +x We now transform 2) by using Wallis' formula, 127, 5) 7r 2 2 4 4 6 2 1 3 3 5 5 7' (3 From 2) we have 2 log (2. 4... 2 ) = log( 22 42 62... (2 m)2) M dx (4 =(2m+l)logm+2mlog2 2m+2-2f G(x)- (4l Also if we replace m by 2 m + 1 in 2) we get log (1.2 3... 2 m + 1) = (2 m +1 + ~) log(2 m +1) 2m dx -2m-^ G(z) l +. (5 From 4), 5) we get, on subtracting, log 2.4.6...2m 11og 2.2.4 4... 2 2m l1 3 5..2m+ 1+ 2 1..3 3 5.5... -+ 2m+ 1 =-(2 + 1) log 2 m+ - log (2 m + 1) m 2 4 M-1 f~2m +2m log 2+ 2- 2m + =-2ml1o \ 1 Nm-1 -+2m =-(2m+l)logl+ 2) —og(2om+l)+2-log2 — f/. 2 Am_ 2 Jo Jm-l ASYMPTOTIC EXPANSIONS 325 Thus 1log 2.2.4.4...2n. 2m — (2,+l)log1 1 2 1 *.3.3.5.2 m-1.2 m + 1 (2 n+ mlog Mm-l >'2m -log 2+2- +f (6 *~o.JM-1 Now Now lim (2n + 1) log 1+- =1 fim-1 /a / '2m lim =, lin = 0. 7n=o0xo J m=0,Thus letting m - oo in 6) and using 3) we get log-V27/2 = 1 -r G(z)dx( J +x This in 2) gives log m! = -(2 m + 1) log m - m log 2 /7r -+ ( + ) If we use this relation in 154, 8) we get Stirling's Formula: log m! = (2 m + 1) log - m + logV/2 r + - (8 where where 0 < 0 < 7. < This may also be written i! = V2 Tm(jf)) e12 (9 e 158. Asymptotic Development of r(x). 1. In Euler's summation formula, 155, 5), 6) let us set b=1, a=x, f(a-x)= log(u+x), then we get, as in 156, 1), taking the principal branch of the logarithm, m S log (x + s) = (x + m- 1+) log (x + m) - (x - -) log ax s=0 -m n G((u)du valid f -or any complex x 0 -1, - 2 (1 valid for any complex x #f= 0, - 1, - 2 **** 326 FUNCTIONS OF A COMPLEX VARIABLE From 157, 2), we have log 1 -2.3..e m+l= mn+ log(m+)-m- (2 \ 2/ (2 Subtracting 1) from 2) and then adding (x- l)log (m + 1) to both sides, we get 1. 2. 3. m+1 ( - ( + I 1 I + 1 xx+l...x+ m 2 m+xa, log X. +1... X + 1(~1) M+~ ) lrogn+ + -.)log x i2i +( fG4- * (3 Now E lin m +z + x + - log M + = - x. w=o \ 2 m M2 m+x Thus letting m - oo in 3) and using 144, 4) and 157, 7), we get log r(x)= - +(- ) log + log 27r + y(x), (4 where ' G(u)du 7(X) -o- (5,o x + u In this relation let us set x=y + n u=v- n. Then since G(u) admits the period 1, Y(y + n)= G(v)v _- O as n oo. (6 We may thus write i41 S2 tJ73 A G G(uf)du 7() = + + +...=s j. (7 o f 2 s ~Js-l X + U In these integrals let us change the variable setting u=v+(s-1), s=1, 2,... Then the limits of integration become 0, 1. But in this interval G(v)= 2-V. ASYMPTOTIC EXPANSIONS 327 Thus G(u)dOuU r1 G(v)dv ( r1 2-v)dv J-1 x+U Jo v+x+s —1 J v+x+s —1 =(x +s-) log(1 +8 + s + 8 (+8-|)log)-l. (8 This in 7) gives a development of y(x) due to Gudermann. 2. In 5) let us integrate by parts. We get, using the functions G1, G2... of 154, 4) and the relations 6), 7) of that article, ( )1 =I GG (U) du 1 rG2(Ul C1 [ 2(u)du rJo x+u ~ 2= 2Lx + j u=O 2 7Jo ( + t)2 As G2(0)=I = 7 =2B2, we have Bi 1 r Ga(u)du( 2(.) = '. 1 u +du) (9 2 x 2 2 (X + )2 Integrating again by parts, we have ( aG2())du 1 r 3(u) T +1 r 3(u)adu J ( + (. )2 2 7L(x + )2j)o.rJ0 (x + )3 = r+r 3(u)du T rJo (x + U)3 as G3(0) = 0. Integrating again by parts, [ G3(u)du, 1 r 4(uQ)a_ 3 / G4(u)du (10 o (x+ U)3 27rL(x+ u)3j=o 2 TrJ (x + U)4 As 237r4 A (4(0) = H4 -4! 9) and 10) become B- 1 2 B 1 3 r u)du (11 1l.2 x3.4x3 +4 (Jx+u)4 3. If we continue integrating by parts we get Stirling's Series'log r(x) = logV2 r - x + ( - ) log x +B~ 1,+ B 1 5B. 1 ~+ 2.~+ -+- - -5++"' R2n + (12 1.2 x 3.4 x3 5 6 x5 328 FUNCTIONS OF A COMPLEX VARIABLE wvhere (- 1I)(2 n - 1) 92 a~n,(u)du B J0 22n -17rn (x ~ U)21 Since here f2n(x + u), f(2n+2) (x + u) have the same sign, we are under the case considered in 155, 5, and the series 12) is in fact an asymptotic development of 1(x). From 13) we see 12. has the form R In - (X), (14 2n-1 where en 0 as x - oI. oo For if we integrate hy parts we have 02n(u)du 1 GT2.+1(u) G2n+1(u)du G~~~JU ~ - 2 L (xL (X + u) 2n 2 71-1X + U r)2j wJ (x + u)2n~L n + +- J. 71 Now 0 NoJ7 1 4 (du 1 1 P-1 p2n+' (X + U) 2n+1 2 nz x p"n1 < Cr5 where a is a constant. xn Hence en(X) < 111 a constant. (15 X 159. Asymptotic Series. 1. In the foregoing articles we have heen led to divergent series D(z)=a= ao + Ki a.. ( z such that the sum of the first n terms Dn(z) gives a very good approximation of some f unction f(z) for large values of z, provided n is not taken too large. More specifically we may say that the series 1) is so related to the function f (z) that f (Z) = -Dn+J(X) + Zn - en (2 as z _o along the positive real axis. Such a series is called an asymptotic series, and we write f (z)- a, + El + z 2 (3 ASYMPTOTIC EXPANSIONS 329 Hereby we will not restrict z to move along the z real axis, but permit it to = o along any radius vector so that z=rei, r * CO, = constant. (4 This we call the asymptotic vector. Asymptotic series figure quite prominently in astronomy and also in some parts of the theory of linear differential equations. We shall meet them in this latter connection when we come to study the Bessel functions. We wish now to see how the ordinary operations on convergent series may be extended to these divergent series. We shall suppose that z oc along the same asymptotic vector unless the contrary is stated. Let us first show that: f (z) does not admit two different asymptotic developments along the same vector. For from,. For fron f(z)= a+l+..+ an+ n Z Zn Zn 6 bo + rn Z Zn Zn wehave -=(at0 - b) bl... h an - bn + e- 77n Z Zn zn Letting z = o0 we get ao = bo. Thus aal- + 2 +... + -- o. Z Zn-1 From this we get as before a1 = bl, etc. 2. Addition and Subtraction. Suppose f(Z)' ao+ + -... z (5 g () ) bo + +... Tr z t d + ez Then( 330 FUNCTIONS OF A COMPLEX VARIABLE For 5) stand f or a1 a - E bo + N + + bn 71n?In ~ (7 where y6 q. 0 as z oc1. Thus f~gq ao~ho~al~hl~..~ Zn 0 z where 19n=E6n~y'qn O as zc. Hence 6) holds. '3. Miultiplication. The functions f (z), g (z) admitting the asymptotic developments 7'), let us show that -P. el C 2 co + - ~~~~~(8 z z2 where co0=a0b0 el c=a0b1 +a100 2 c= a062+ albl+ a260... (9 as in the multiplication of series. For 7) gives 1 I~ But O n where Q. is a polynomial of order < n - 1. Thus 8) is valid. 4. -Division. Since gq g the problem of dividing one asymptotic development by another may be reduced -to finding the reciprocal of an asymptotic development. Let us suppose in 5) that ao * 0. We will write =n a a( +an +n ASYMPTOTIC EXPANSIONS 331 Thus 1 1 1 en 1 f S n f zn ffn Zn IA ao 1I + h, -ao 1+ h. Hence 1= + l 2 + n 1 ( f o+ + +. — o (10 f 2 oa0 where as z On= 0 as z oc. 5. Integration. We show that: An asymptotic development of f(z) may be integrated termwise along the asymptotic vector when the function f (z) is integrable along this vector as indicated in 12). For from 7) we have Xao 0aco ofen+l f (z)dz= +z f z + n+l dz. (11 Now along the asymptotic vector fXn+j CZ < n On ' asz ac. Thus 11) may be written = 2a dz + dzdz +... a z ( where j 0n 0 as z -o. Thus from al a2 f (z) - ao + 1 + 2+. we can infer that _ (z)-z 2 dz + 3 dz + '~ (12 ^ ~(Z)^-a,-a~~~d^a+ 0x... It/ y J,7 3 332 FUNCTIONS OF A COMPLEX VARIABLE (3. Differentiation. Suppose that wve knowt that f(z) has an asymptotic development fz) - +*,(1 z z also that f' (z) has an asymvptotic development of the form f /(Z) b2 + N + ~~(14) Then the asymptotic development off1 (z) may be obtained from that of f(z) by termwise differentiation. For from 14) we have, using 5 ff1(z~z~,fb~zHA Jk-dz+ (15 Since f (:C) II-m f (z) = 0, by 13), we have, from 15),I bfl l +(16 Since by 1, a f unction admits but one asymptotic development along the same vector, the comparison of 13), 16) gives b2 - ~ b N= - 2a2.. These in 14) establish the theorem. CHAPTER X THE FUNCTIONS OF WEIERSTRASS 160. Limiting Points. 1. At this point it is convenient to introduce a notion which is fundamental in many parts of mathematics, that of a limiting point. Let 21 be a point set. If in any domain D(b) of the point b, there lie an infinite number of points of 21, we say b is a limiting point of 21. The point b may or may not lie in 1. Example 1. Let 21 = 1, 1, 1, *. Then the origin 0 is a limit2' 3' "" ing point of 1. It does not lie in W1. Example 2. Let 91 denote all the points within a circle S. Then any point of 1 is a limiting point of W. Also any point k of the circumference R is a limiting point of X2, although k does not lie in 21. 2. A set of points W which lie in some square S is called limited, otherwise unlimited. A set of points which embraces an infinity of points is called an infinite point set, otherwise a finite set. Thus the point set formed of the points corresponding to the positive integers = 1, 2, 3,... is an infinite unlimited set. For obviously no square contains them all. The set of points on an ellipse form a limited point set. We now prove the fundamental theorem: Every infinite limited point set 21 has at least one limiting point. For a2 being limited lies in some square (. Let us divide ~ into 4 equal squares. Since 32 contains an infinite number of points, at least one of these squares contains an infinite number of points belonging to W. Call this square '1. This we divide,333 334 FUNCTIONS OF A COMPLEX VARIABLE into 4 equal squares. At least one of these must contain an infinite number of points of 21. Call this (2. Continuing in this way we get a sequence of squares 51, 2 e 3 - (' each contained in the foregoing. As the sides of these squares =0, the squares 1) shut down to a point a which lies in all of them. Since each square contains an infinite number of points of S1, any circle about a, however small, will contain an infinity of points of 2I. Thus a is a limiting point of 1. 3. Suppose the point set W is not limited. Then there are an infinite number of points of 1 without any circle S about z = 0; that is, there are an infinity of points of W2 in any domain of the point z = oo. It is convenient to say that z = oo is a limiting point of 21. We may thus say that: Every unlimited point set admits z = oo as a limiting point. Putting this in connection with the theorem in 2 gives: Every infinite point set has at least one limiting point. This may be the ideal point z = oo. 4. Let the one-valued analytic function f(z) take on the value c for the points of some set W. Any limiting point of 2t is any essentially singular point of f(z), provided f is not a constant. 161. Periodicity. 1. Let the one-valued analytic functionf(z) satisfy the relation f(z + co) =f(z), co constant T= 0, (1 for every z for which f is defined. We call o a period of f and sayf admits r as a period. We shall of course exclude the case thatf (z) is a constant. Thus ez, sin, tan z (2 admit respectively 2 r 2 (3 as periods. Obviously if w is a period off(z) so are ~*, -3 sw, -2 w, - s, co, 2 w, 3 Wt, *. (4 THE FUNCTIONS OF WEIERSTRASS 335 Thus if f admits one period, it admits an infinite number of periods. Sometimes these lie on a right line as in the case of the functions 2); sometimes they are spread over the plane as we saw in the case of the functions 4X(zm 1w+m p = 3, 4,. (5 ' 2 (z + mn1)l + MT2(2)P considered in 123, 6. It will be convenient to say that two points a, b are congruent when their difference a - b is any period of f(z). This we write a-b and read a is congruent b. If we write more specifically, a-b, mod o, read a is congruent b with respect to the modulus o, we mean that a - b = mo, m some integer or 0. If we write a=b, mod ol o2', we mean a - b 6= m11 + m2c), i, m2 integers or 0. If Cl0, c)2 are any two periods of f(z), so are obviously o1 + 6)2, and col - W2 periods and still more generally mlo1 + m2po2 are periods, ml, m2 being integers or 0. If a, b are not congruent we say a is incongruent b and write a b. 2. Let $ denote the totality of all the periods of f(z). The point set $3 must have z = o as a limiting point as it always contains a set of points as 4), and no oo as n =' oo. On the other hand, we now prove the important theorem: The point set $ has no limiting point in the finite part of the plane. For suppose V were a limiting point. Then within D1(D) there are an infinity of points of S, however small 8 is taken. If a, / are two of these, ry = a - 3 is a period and 1 71 < 2. As 8 is small at pleasure, this shows that f(z) has periods which are numerically < any given e > 0. Butf(z) is an analytic function and cannot have such periods. For if z = a is a regular point and f(a) =c, 336 FUNCTIONS OF A COMPLEX VARIABLE we know that f(z) cannot = c in some D,*(a). Butf(z) having periods v numerically < a, we have f(a+ ~r) =f(a) = and a + Xr lies in D,*(a), which is a contradiction. This shows that every point z is a singular point of f(z), and f(z) is not analytic. 3. Let w be any period of f(z). On the line I passing through the origin and o will lie an infinity of periods, for at least the periods 4) will lie on 1. Since the origin is not a limiting point, there are two periods ~ X on I nearer z = 0 than the others. All periods on I can be expressed as multiples of X. For let o be any period off(z) lying on 1. We can write o = nX + rX and take the integer n so large that 1 1 < I X. If now 0, it is a period on 1 which is nearer 0 than X, which is contrary to hypothesis. We call X a primitive period. Thus the periods 3) are primitive periods of their respective functions 2). 162. Jacobi's Theorem. 1. If the one-valued analytic function f (z) has more than one primitive period co, there exists a primitive period w2 such that every period off has the form m101) + m202, (1 where ml, m2 are integers or 0. For let to be any primitive period other than e1. In the parallelogram Q whose sides are Owo and Oo there are but a finite number of periods. None of these can fall on the edge of Q. For if D were such a period l =- r would fall in ( 0, wo) and as 1 is a period, 01, cannot be a primitive period. Let now c2 be that period in Q for which the angle w2 Owo1 is least. Then every period of f / has the form 1). / / For let P be the parallelogram whose sides are Oo1, Owc2 If THE FUNCTIONS OF WEIERSTRASS 337 there is a period w of f not included in 1), let w' be that point of P which is = —. Then w' is a period and Angle co' Ow1 < Angle W2 001 which is contrary to hypothesis. 2. An analytic function f(z) which has more than one primitive period is called a double periodic function. Two primitive periods 01, 02 such that all other periods of the double periodic function f(z) can be expressed linearly in terms of them, as in 1), form a primitive pair of periods. The functions ez, sin z, etc., are simply periodic. All their periods are multiples of a primitive period. As examples of double periodic functions we may take the functions 1 '(z) = (z + mlW + m202)P' (2 p an integer > 2 considered in 123, 6. These functions have mwl + (202 (3 as poles of order p. All other points in the finite part of the plane are regular. The point z = oo is of course an essentially singular point. 3. Let f(z) be a double periodic function having o1, W2 as a pair of primitive periods. Let a be another point. The parallelogram P whose four vertices are a a a+ 01 ' a+ o2 a+ l -1+ 02 is called a primitive parallelogram of periods. By drawing parallels to the sides of P through the points a + m10)1 + m202 we may divide the whole plane into a set of parallelograms similar to P. Any parallelogram Q built up on two periods 1l r2 not necessarily a pair of primitive periods will be called a parallelogram of periods. We shall often have occasion to integrate over the edge of such parallelograms, and in such cases we shall suppose the point a chosen so that the edge of the parallelogram does not 338 FUNCTIONS OF A COMPLEX VARIABLE pass through a singular point of the integrand. To indicate what periods 71, 2 are used we may denote Q by Q(q1, 72). The fact that f(z) admits wol o2 as a primitive pair of periods we may indicate by the notation f (, 0) ) 4. From one primitive pair of periods o1, eo2 it is possible to form an infinity of other pairs. For let 1= mll0 + m22, 2 (4 2 = nlo + n2~02 where the m, n are integers. Obviously 1l, q2 are also periods. For them to form a primitive pair it is necessary that the determinant i = mln2 - mr2n is ~ 1. In fact, solving, we get _ n271 - m22 D1 -c m1'72- nl,?l' 0.)2 D Hence when D = ~ 1, Wo, oW are linear functions of ll r2 with integral coefficients. Let us call the set of points 11(1 + 122 1, 12 =, ~ 1, ~ 2... (5 a network. We may denote it by (ct, wc2). We now see that the (vr, 72) network is the same as (1, 02) when and only when D = ~ 1. 5. Let P be a parallelogram formed by the points o0, 1 a2 1,2 + o02. (1 = al a- ibl D2 = a + ib2, we know from analytic geometry that Area P(ao1, co2)= al bi (6 a2 b2 THE FUNCTIONS OF WEIERSTRASS 339 Thus the area of P(771, r2) is mlal + m2a2 mlbl + m262 = ml m2 ajbl. nal + n2 a2 nlbl + n262 nI n2 a2b2 Hence Area P(711, 12) = A Area P(o1, o2), (7 where A= I D. 163. Various Periodic Functions. 1. From a periodic function f (z) admitting o as period we can form an infinity of others admitting this period. For example g(z) = cfm(z) m an integer, (1 admits the period w. For g(z + o) = cfm(z + O) = cfm(z) = g(Z). If o is a primitive period of f, it does not need to be a primitive period of 1). For example 2 wr is a primitive period of sin z, but it is not a primitive period of sin2 z, whose primitive period is rr. 2. If f(z),.(z) admit the period o, their sum, difference, product and quotient will also admit this period. For example let h(z) = f(z)g (z). Then h(z + o) =f(z + w)g(z + Wo)= f(z)g(z) = h(z). We must, however, guard against the case that h reduces to a constant. Thus f= sin2, g = cos2 z admit the period 7t. Their sum = sin2 Z + cos2 z = 1 is not properly periodic at all. From the above it follows that any rational function h of'the periodic functionf(z) is also periodic, guarding against the case of course that h is a constant. 3. Let f(zf(z), f.z) *.f(z) be one-valued analytic functions admitting co as a period. Then the analytic function w satisfying the equation n +fl(z)wn-' -l+f(zw) W- + ~- +.,fn(z)= 0 (2 will admit co as period. 340 FUNCTIONS OF A COMPLEX VARIABLE For let w (z) be a value of w corresponding to a value of w at z. The value of w at the point z + 4o will be w (z + co). As the coefficients of 2) have the same value at z + t as at z, we see w (z + W) = w (). 4. If the one-valued analytic function f(z) admits w as a period, so does its derivative f' (z). Forf'(z) is the limit of )_f(z + h) -f(z) h But g (z + c) f(z + co + h)-f(z + o) (3 h Passing to the limit h = 0 in 3) gives f'(z + c) =f'(Z). 5. If f(z) admits the period o, we cannot say that the primitive function F(z) admits this period, as the following example shows. Let f(z) = cos z + 2. Then (z) = (cos z + 2)dz = sin z + 2 z + is not periodic although f(z) admits the period 2 7r. There is, however, an important case when the primitive function F(z) does admit the period ao, viz.: Let the derivative f(z) of the one-valued function F(z) admit the period A. If F is an even function, F admits the period co. For from f(z + )) =f() f(z + t) =/(z) we have, on integrating, F(z + ))= F(z) + C. In this relation set z =-, then 2 A(st (-v + C. As B() =F( —), this gives C=0; thus 2^ 2/ and F admits the period co. THE FUNCTIONS OF WEIERSTRASS 341 6. From a one-valued periodic functionf(z) having no essentially singular points in the finite part of the plane let us show how to construct a periodic function having z = a as an essentially singular point. To fix the ideas let us take f (z)= cot. This has the period 7r and the poles pm ='m7r m=O, ~1, ~2, '* Let us set I fn(z) = cot (z- a - a,), n I where al, a2... is a properly chosen sequence which -O. For example we may take here i n n Consider now g(z) = Ifn(z)- =! cot z-a-. (4 The poles of fn(z) are,m = a + +pm m=0, ~1,... (5 No two terms f,, f have a pole in common. Let $ be the set of points formed of the sets 5) and their limiting points m = a +pm If z= b is not in $3, we can describe about it a circle R which contains no point of $3. Then each and every fn in 4) is numerically < some fixed a for any z in R. Thus each term of 4) is numerically < the corresponding term in the convergent series Hence the series 4) converges steadily in and as each term of Hence the series 4) converges steadily in 9 and as each term of 4) is analytic in l, the function g(z) is regular at z = b. On the other hand, each pole q of any term f, of 4) is a pole of g. For we have (z) =f(z) + h(z), 342 FUNCTIONS OF A COMPLEX VARIABLE where h is the series obtained from 4) by omitting the term f8. fFrom the foregoing reasoning h is regular at q. Thus g has a pole at q and of the same order as s. The point z = a is an essentially singular point, since it is the limiting point of the set of poles n, o= a + n Finally g(z) admits the period 7r since each term of 4) does. This shows that not only a but also Im = a + mwr are essentially singular points. This is as it should be, since the lm are limiting points of the poles 5), and g(z) has the period r. 7. Instead of the function cot z we can take a double periodic function as f (z) — (Z -~- m1co1 + m202 )3 With this we can construct a series of the type 4) which will define a double periodic function having a given point z = a as an essentially singular point. Of course all points -a will also be essentially singular points. 164. Elliptic Functions. 1. Having now an idea of some of the singularities a double periodic function may possess, let us pick out a class of great importance called the elliptic functions. These are defined as one-valued analytic double periodic functions which have no essentially singular point in the finite part of the plane. The reader will recall that, as we saw in 123, every periodic function must have z = co as an essentially singular point. Thus the elliptic functions are the simplest double periodic functions, in that they are one-valued and the number of their essentially singular points is the least possible. Such functions are p1(z)=-23 1 1() 2 (z -( 2 mlo1 -22 m2w)8 pn(Z) (Z _ 1 )n(n + 1)! ( _ 1 22)n+2 (2 (z - 2 m1w1 - 2 m2o2)~;2 (2 where 2 o1, 2 o2 are any two complex numbers not collinear with the origin and mi, m2 range over all positive and negative integers THE FUNCTIONS OF WEIERSTRASS 343 and 0. We notice that P2 is the derivative of pi, p3 the derivative of P2, etc. These functions are essentially the functions considered in 123, 6; we have replaced w w02 by 2 or, 2 02 to avoid writing the fraction 1, as we shall see. 2. From pl(z) we can get by integration another elliptic function of fundamental importance. In fact let us write 1(Z) =-2 +- (Z), (3 Z3 where the first term on the right corresponds to the values m1=0, m2 =0 in 1). The function g(z) is regular in any part of the plane which does not contain one of the points 2 m1a)1 + 2 m2o2, m1 = m2 = 0 excluded, In particular it is regular about z= 0. Thus g(z)dz = -2 ( 2 2 d oJo (z - 2 m^ o -2 m2w)3 } =h(z), (4 (z - 2 mlw - 2 m2o)2 (2 mo1l + 2 m2o2)2J where the dash indicates that in effecting the summation the combination ml = m2 = 0 is excluded. This dash we shall often employ in this sense. Iet us now set p(z) = + h(z). (5 zThen Thus 5) is th prm). Lt us nw sw tt (6 Thus 5) is the primitive of 6). Let us now show that h(z) is even. For to the term indicated in 4) there corresponds another term in which mi m2 have the same values but with opposite signs. Thus h(-z)= h(z) and h is an even function. Hence by 163, 5 the function 5) is double periodic admitting the same periods 2 wo, 2 2 as 21(z). 344 FUNCTIONS OF A COMPLEX VARIABLE Thus the function I1 f( 1 1 } P(=^ 2 (+ X - 2 m1c1 - 2 mwz)2 (2 mlol + 2 m22)J is an elliptic function admitting 2 wl, 2 o2 as periods. It is the fundamental elliptic function in Weierstrass' theory. To denote it, he has invented a modified p, viz. the symbol p, and this has been generally adopted. We shall, however, retain the ordinary p. By virtue of 6) we see that the functions defined in 1), 2) are the derivatives of p(u). 165. General Properties of Elliptic Functions. 1. Every elliptic function has at least one pole in any parallelogram of periods P. For having no singular point in P, it has no singular point anywhere in the infinite plane. It is thus a constant by 121, 2. 2. Let f(z) be an elliptic function admitting co) and %2 as periods. Then fdz = 0, (1 ~P P being a parallelogram of periods not passing through a pole off. For FJo=frff+, + + (2 Now 12 2 e4 t' Now,of dz= fdz * 43 12 since f has the same value at z' = z+ w2 as it has at z, by virtue of its periodicity. Hence 3z Similarly \ 2 J41 = 23 1C w Thus the right side of 2) vanishes. THE FUNCTIONS OF WEIERSTRASS 345 3. The sum of the residues off(z) in any parallelogram of periods P, not passing through a pole off, is 0. For this sum is by 124, 1 2 ri.f(z dz, which = 0 by 2. 4. The sum of the orders of the poles of an elliptic function in any parallelogram of periods not passing through a pole is at least 2. For if the sum is 1, f can have but a single pole z = a in P and its development must have the form f= + co +c (z- a)+... (3 z - a Here c = Resf(z). As the sum of all the residues in P is 0 by 3 and as there is but a single pole, we must have c = 0. But then 3) shows that f has no pole at a, which is contrary to hypothesis. 5. Definition. The sum of the orders of the poles in a primitive parallelogram of periods not passing through a pole is called the order of an elliptic function. From 4 we have: There is no elliptic function of order less than 2. By means of this theorem we can often show that a pair of periods of an elliptic function form a primitive pair, as the following theorem shows: 6. Let 01, c2 be a pair of periods of the elliptic function f(z). This is a primitive pair if the sum of the orders of the poles off in a parallelogram P (wI, 02) not passing through a pole is 2. For if not, let?l, 2 be a primitive pair. Then 1 = m1l- + mM2v2 02 = nll7 + n272 ~~~and A = -imln2- m2n 1, is > 1 by 162, 4. Now by 162, 5 the area of P(w1, 02) is A times that of P(Ql7, 72). From this it follows geometrically that there 346 FUNCTIONS OF A COMPLEX VARIABLE are X parallelograms which either contain no pole or a pole of order 1. Asf(z) behaves in all parallelograms of periods just as it does in any one parallelogram, we see that f violates the theorem 4. Hence 0ol 0)2 must form a primitive pair of periods. 7. From this we see that 2 ol, 2 o2 form a primitive pair of periods of the function p(u) defined in 164, 7. For as we have seen, its poles are the points of the network (2 op 2 w2) and each is of order 2. From 118, 4 and 163, 4 we also see that 2 w1, 2 e02 form a primitive pair for the derivatives p'(u), p "(u).. 8. On account of periodicity an elliptic function takes on the same values at a and b = a + where o is a period. Thus in counting up the points where an elliptic function takes on the same value in a primitive parallelogram of periods we agree to consider only one of the two opposite sides. Also if f(z) = c at z = a the function g(z)=f(z)- c will have a zero at a. If this zero is of order s, we will say thatf(z) takes on the value c at a, s times. This being agreed upon we now prove: An elliptic function f(z) of order n takes on any given value c just n times in a primitive parallelogram P. For choosing P so that no zero or pole of f(z) lies on its edge, we have, by 124, 4, 1 f(Z)-dz= mO- m, (4 2 7riJpf(z) where m0 is the sum of the orders of the zeros and m, the sum of the orders of the poles of f(z) in P. Now P being a parallelogram of periods of f(z), it is also for the function f(z)* Thus the integral in 4) vanishes by 2. Hence mo = mo. But mo = n by definition. Thusf vanishes n times in P. Consider now ) = )g(z) =f(z)- e. This vanishes whenf= c. On the other hand, P is a primitive parallelogram for g as it is forf. Finally, g having the same poles THE FUNCTIONS OF WEIERSTRASS 347 asf, and to the same orders, the order of g is n. Hence g vanishes n times in P. 9. A theorem of great use in the elliptic functions is the following: Too elliptic functions having the same periods, the same zeros, and poles to the same orders, can differ only by a constant factor. For let f(z), g(z) be two such functions. In the vicinity of a zero or a pole z = a we have f= (z - a)'(z) y = (z - a)m*(z) where a, r are regular at a and do not vanish. Hence in the vicinity of a zero or pole f(z) *(z) is an analytic function. If we give to q at a the value A(a) q is regular at a. Thus q has no singular points in the finite part of the plane. It is therefore a constant. Thus f(z) = C(z). 10. A similar theorem but not so often used is the following: If the elliptic functions f(z) g(z) have the same periods and at each pole the same characteristic, they differ only by an additive constant. For at a pole z = a, let f(4) = +(Z) + F (z), g(z) = 4(z) + (z), where 4 is the common characteristic at a. The functions F, G are regular at a by 118, 2. Thus f(z) - g(z) = h is regular at a as it is the difference of two regular functions. Thus the function is regular everywhere, and is therefore a constant. Hence f(z)= (z) + C. 348 FUNCTIONS OF A COMPLEX VARIABLE 11. Abel's Relation. Let f(z) be an elliptic function of order n. Let P be a primitive parallelogram of periods not passing through a zero or pole of f. If al, a2.** a are the zeros and pi, P2 ". pn the poles which fall in P, then (al + a2 + "' + an)- (p1 + P2 + '" + Pn) = a period. (5 Before proving this theorem let us see its significance in the function theory. It is often convenient to construct functions having assigned properties, and it is therefore necessary for us to know which such functions are possible. For example we know it is possible to construct a one-valued analytic function which vanishes at al, a2... am, which has poles at Pi' P2 "'Pn and which has no essential singularity even at oo. Such a function is (z- a).. (z- am) (Z - P1) ** (z - Zp) Now if we were asked to construct an elliptic function having these zeros and poles in a primitive parallelogram of periods P we would say at once that this is impossible unless in the first place m = n by 8. This is the first restriction. Abel's relation 5) is another restriction. It says that having chosen 2n - 1 of the zeros and poles in P, the last one is no longer free to choose; it is, in fact, completely determined by 5). Are there any other conditions to inpose? We shall see in 166, 4 that there are not. Let us note that we may write 5) am-m = o, mod w1 w2 (6 We turn now to the proof of this relation. From 124, 2 we have 1 izd logf(z)= am - ipm. (7 A ~ J. /+J +J + (8 eP '12 23 1234 41 Also C+ Le34 /us chang the vabl P Let us change the variable setting \ Z = u + co2. THE FUNCTIONS OF WEIERSTRASS 349 Then zd log f( Now if the reader will sum, he will see that influence on its value. () =I (u + o 2)d logf(u) -= uqd logf(u) + C2[logf(u). * (9 remember that an integral is the limit of a the letter chosen for the variable has no Thus u d logf(u) =j zd logf(z) = J. Also if logf(u) has the value log f(c) at u = e, its value at c + o1 is one of the many values log f(c + o1) has at this point. But f(C + Cl) = f (c); thus the value logf has at c + o1 is log f(e) - 2 m2wri, m2 an integer. Thus 9) gives f + j = 2 m2ri2' Similarly j + = 2 mnlri)l 23 ' 41 Thus 7) gives lam -p -m = -= m Ic + m2w02 = a period. 12. From Abel's relation we have: Let the elliptic function f (z) = c at the points zl, z2... z in P. Then.. For ZZm - ZPm. g(z) =f(z)-c (1i has the same poles as f(z), and its zeros are z1 *. z,. We thus need only to apply 6) to the function g. Remark. In Abel's relation 5) the a's and p's lie in one and the same primitive parallelogram. We can give this relation a slightly more general form as follows. Let us say that any set of points form an incongruent set when no two of them are congruent. Let then... al a *... a' (11 350 FUNCTIONS OF A COMPLEX VARIABLE be any incongruent set of zeros, and f I... 0 (12 P\ P2 * n (12 any incongruent set of poles off(z). Then 5) may be written a + *.- + atp +.-. +p. (13 For each a' must be congruent to some am, and no two of the points 11) are congruent to the same am, since then they would be congruent to each other, in which case 11) would not be an incongruent set. Thus a + * +- at = a1 4- + - + an + a period. Similarly + imilarly P + ** +p=pi -+ - * P +p a period. Thus 13) is a consequence of 5). A similar remark holds for the relation 10). Here it is not necessary that the zl... z all lie in the same primitive parallelogram; they can be any set of incongruent points for whichf(z) = c. 13. In case of an elliptic functionf(z) of order 2 Abel's relation enables us to solve the problem of finding all the values of z for whichf(z) takes on a given value as follows: Let Pl, P2 be incongruent poles of an elliptic function f (z, 1, cw2) of order 2. Iff takes on the value c at z = zo, then all the roots of f(z)= e (14 are given by + 20 - + ml + m202 (15 P1 + 22 -- O + "1'01 + m2~2, where mwhere mlm=O, 2= 0 1, ~ 2... For if z1 is the other value of z for whichf = c in the primitive parallelogram P(w1, w2) in which zo lies, we have, by Abel's relation, 0o + 1 - (P1 + P2) = a period. Thus -P 1 + P2 -ZO as stated in 15). Remark. In case f(z) has a double pole p we replace Pi +P2 in 15) by 2p. THE FUNCTIONS OF WEIERSTRASS 351 14. Between the poles of an elliptic function of order 2 and the zeros of its derivative there exists a remarkable relation which is expressed in the following theorem: If the elliptic function f(z, o1, 02) of order 2 has 1, P2 as incongruent simple poles, its derivative f' (z) is of order 4 and admits the incongruent points P1 - 2 li + 9 2 Z1i- 2 Z2= Z1 + 2 Z3= + 2 4= Z1 + 2 (16 as zeros. That ft(z) is of order 4 follows from 118, 4. That the points 16) are incongruent is easily seen. For suppose c)2 2 Z 3 = z1 + 2 Then i c 2 -- Z3= 2 2 is a period, which is not so, since 0, 02 form a primitive pair. From 13 we have f(P1 +P2 - ) =f(Z). Hence f'(2 z - z)= -f'(z). (17 As z1 is incongruent to Pi or P2, it is not a pole of f (z). Let us therefore set z = z1 in 17). We get f (Zi1) — f (), or, 2f' (1) = 0. This shows that z1 is a zero off'(z). Again, set z = z2 in 17); we get f' (2 z1 - Z2) = -f' (Z) (18 Now W Ai 2 z1 - - - -- = + - = Z2 -Thus 18) shows thatf'(z2) = -f'(z2) or 2f'(z) = 0. Hence z2 is a zero of f'(z). Similarly we show the other points of 16) are zeros. 352 FUNCTIONS OF A COMPLEX VARIABLE 15. Similar reasoning applied to 17) gives: If the elliptic function f (z, al, w2) of order 2 has the double pole p, its derivative is of order 3, and it admits the incongruent points AW9 Cl+ t)l "+ 6>,02 /(19 Zi- + 2 Z2=P-+ 2 Z83=P+ 2 ' (19 as zeros. Remark. The reader may ask: Why does not the same reasoning prove that p is also a zero off'(z)? As we know that a pole off is also a pole of f'(z), our reasoning would then be quite fallacious, since p cannot be at once a zero and a pole of an analytic function. The fault in such reasoning on p would lie in setting z=p in 17). Since we know that z=p is a pole of f'(z), this latter is not defined at this point. The relation 17) holds for values of z near p but not at p. 16. An elliptic function of the second order having simple poles satisfies a very simple differential equation, as the following theorem shows: If f(z) is as in 14, it satisfies (d)2=C= f- - el)(f- e2)(f -e3)(f —e4) (20 where f(zm) = em, m = 1, 2, 3, 4, and zm are the numbers 16). Let us first show that the e's are all different. For if e = e2, for example, then f(Z1) =f(Z2), and either z2 z1 or, 1 + Z2 -1 + p2 by 13. Neither is true. Let us now set gm(Z) =f(z) - em, m- = 1, 2, 3, 4, and g(z) = '192g3g4. We show that g admits c), A2 as periods and has the same zeros and poles, and to the same order as (f'(z))2 = h. Thus h and g differ only by a constant factor by 9. THE FUNCTIONS OF WVEIERSTRASS 35 358 For in the first place f and g obviously have the same periods. Next g being the product of four factors of order 2, is of order 8. As gq = 0 for z = Zm, we see that g vanishes at the four points z,. Each of these points is a zero of order 2 for g(z). In fact g'(Z) = ggyg4 ~ g+ Ag91g3g4 Y~ g 9j1g2g4 + 4 g g1g2g93 Let us set z = z1 in this relation. The first term on the right vanishes, since the factor g(zl)f(zl)= 0 by 14. The other three terms= 0, since each contains the factor gl. Thus g'(zl)= 0 and hence z1 is a zero of g(z) of order 2 at least. Hence y(z) and h(z) have the same zeros to the same order. The poles of g(z) are p1, p2 each of order 4. The same is true of h(z). Hence by -, h= C g. 17. When the elliptic function of order 2 has double poles, we have: If f(z) is as in 15, it satisfies the different;al equation (df2 Kd) = C(f - el)(f- e2) (f - e3), (21 where f(zm) = em, m = 1, 2, 8, and Zm are the points 19). The proof is entirely analogous to that in 16. 18. Application to the p function. This function is defined by p (z) 2 (z~2 2} (22 where 6= 2 m a1 + 2 n2o2. Here p = 0 is a double pole, and the periods are 2 ro, 2 o22 Thus 19) becomes z1 W 1 Z X2 = XQ2 3= 61+c02 (23 and ap(ol)=el P(3W2)=e2 P(01+'02)=e3. (24 Hence 21) shows that p(z) satisfies the differential equation (dp2= C(- ep-e1)(p - e2)(P —e3). (25 ~dzJ 354 FUNCTIONS OF A COMPLEX VARIABLE From this follows that!Z = dp. (26 J C((p -'ei)(p - e2) ( - e3) Remark. The reader can now see why we have denoted the periods of p(z) by 2 c, 2 %2 instead of o), 02. It is the half periods which enter in the definition of the e1, e2, e3, and these quantities are of fundamental importance. Also in many other relations the half period figures. If we call the periods 2 w1, 2 c2, we avoid the fraction -1 when using the half periods. The reader will also note that the period of sinz is denoted by 2 r. 19. It will greatly simplify our equations, as the reader will see later, if we introduce a half period w, by means of the relation co)1 + + 0 + = 0. (27 Then p being an even function we see that P(P1 + +02) = p(OW3)= e3s Thus the three equations 24) can be written p(coi) = e, s =1, 2, 3. (28 Also the zeros of pt(z) are _ h, 02, %w. (29 166. Elliptic Functions expressed by o(z). 1. Let the elliptic function f(z) of order n have P(2 ao, 2 W2) as a primitive parallelogram of periods. Let its zeros be a, a2, a3 *- (1 arranged so that a an+ \ > | a 0; let its poles be b, b2, 3 *.. (2 arranged so that \ b,,, l 0. Then by 140, 4, and 6, x Z 1a- Z 2 I~ -- ^)^ea "n. f (z) = e iT() ) 21a ( b (3 n~l- Z\ +Z~+ 2 THE FUNCTIONS OF WEIERSTRASS 355 since 13 I 1 converge. To determine e let us observe that d2 log f(z) g(Z) d (4 dZ2 also admits 2 Col, 2 w2 as periods. Thus 1 _x- n~ n 1n) (-X{ z - a)2 an J t (z-6 b,)2 b2 j is double periodic. Now each E here admits 2 ol, 2 co2 as periods. Hence T "(z) = g(z)+ z - a b admits 2 co, 2 c2 as periods. As T is an integral function, so is T11. But then T. (z) = 2 c, a constant. Hence = a + bz + z2. The infinite products entering 3) can be expressed as the product of m simpler products as follows: Let e1, e2 *' m (5 P1 ' P2 *Pm (5 be the zeros and poles which fall in the parallelogram P. Let e11,' 12, e13 *' (6 be all points 1) which are _ c. Let P11 P12 ' Ps13 (7 be all the points 2) which are — P. If we treat the other points in 5) in a similar manner, all the zeros 1) will be thrown into m classes, the points in each class being -some zero in 5). A similar remark applies to the poles 2). Let us therefore set t(z, el) =I ( — Cln i 21 ci (8 eln/ 356 FUNCTIONS OF A COMPLEX VARIABLE Then the numerator and denominator in 3) are each the product of m factors of the type 8). We have, in fact, f(z) = ea+bz+cz2t(Z' C)t(z, C2) *t(, m) (9 t(z, pl)t(z, P2) t(z, pm) 2. The simplest t function is obtained by taking cl at the origin. It is denoted by a(z) and is called Weierstrass' sigma function. Thus 2 -(z) = z1 (1 - -e + ), (10 where w = 2 milo + 2 m2o2 m-=mn = 0 excluded. By 140, the zeros of o(z) are z = 0 and the points co. They are of order 1. By using the a function the formula 9) can be much simplified. To show this we make use of the fact that d2log0() _1 a1 dz-2 Z X {(Z - W)2 a22 (11 From 10) we have log r(z) = log z + { log - +) + +2 (12 2 (12 The derivative of this function is so important that it has a special symbol; we set with Weierstrass ( d log o(z) ('(z) (13 dz o (z) Thus h(z)=!+f + + 2} (14 Z [ Z - GO GO Hence finally ' (z) = - () Let us note that a(z) is an odd function. For replacing z by - z in the II in 10) it becomes II 1 + e- +(Z)2 (15 As ao and - o give the same network of points, we can replace co by - in 15); but then 15) goes back to II in 10). Thus this product II is an even function. As a = zl we see a is odd. THE FUNCTIONS OF WEIERSTRASS 357 As o-'(z) is now seen to be even, the definition of '(z) given in 13) shows that ' is also an odd function. We have introduced these relations at this point in order to see how a-(z) behaves when z is replaced by z +2 wor z+2(2o We start with the relation p(z + 2 w,) =p(z). Integrating gives (z + 2 wo)='(Z) + C. To determine the constant CY, we set z = - we get 5(01 = C oil0) + 0 (1 + 619 since ~ is an odd function. Hence C= 2 2(wl). Let us set for brevity ~II 6) 1), '12 = 0)W2) (16 These two constants are of constant occurrence. Then we have '(z + 2 wl)= +(z)+ 2 771, (17 T(z + 2 w2) = rz) + 2% -2 Integrating the first eqquation of 17), we get log o-(z + 2 ol) = log a(z)~+ 2 qlz + C or o(z + 2 ol) = ce2"1Z-o-z). To determine c we set z = - wo, and remember that o-(z) is an odd function; we get -(0)) = cec2w1r1 a(- w1) = - ce2tll?' 0Cwl~. Hence E~~~ence ~~~C = - e2w1lt. Thus 2 () Thus a-~~(z + 2 w1) = - e2?1(zfw1)a(z),(1 u(z + 2 t02) = e2q2(-+c2) a Z). 3. Using the relations 18), we can now simplify 9) as follows. We saw from Abel's relation that Ce + C2...~ + cm- (pi + p2 +... + pm) = a period. (19 FUNCTIONS OF A COMPLEX VARIABLE Let us therefore pick out a set of incongruent zeros al, a2.. am and a set of incongruent poles 6b,.*. b so that a, a2 +. +am=bl + b2 + + bm. (20 From 19) this can be done in an infinite variety of ways. Let us now form the function o-(z - l) -.. (z -a b) We show that g admits 2 o1, 2 o2 as periods. For from 18) ge2"s(z-an,+w) IHo(z - a,), 2 e2e1i(z-b+X.) oII-(z - b,) e2 6lbn =e2an g(z). But by 20) = b, Thus (z + 2 o) =g(z). Similarly g(z + 2 o2) = g(z). On the other hand, the zeros and poles of 21) are the same as thdse off(z), and to the same order. Thus by 165, 9 f and g differ only by a constant factor. Hence the theorem: Let f(z) be an elliptic function of order m having 2 wol, 2 92 as a primitive pair of periods. Let al,.. am; bl, b* b e a set of incongruent zeros and poles such that la,, = lb. Then f(z) = -(z - o... -(-am) o(z- b)... -(z- b,, (22 4. From this we conclude that elliptic functions exist having assigned zeros and poles provided: 1~ the sum of the orders of the zeros in any parallelogram of periods equals the sum of the orders of its poles, and '2~ the zeros and poles satisfy Abel's relation. In fact these functions are all given by 22). 5. The relation 22) shows that every elliptic function can be expressed by means of the a function, which thus dominates the theory of elliptic functions. THE FUNCTIONS OF WEIERSTRASS 359 It is interesting to note how naturally we have been led to consider this function. By Weierstrass' factor theorem, 140, 4, 6, every elliptic function must have the form 3). The products in 3) can be decomposed into simpler products, each vanishing for one of the m classes of zeros or poles of the given function. Of all these simple products 9) the simplest is the product 10). It is an integral transcendental function like sin z, and as the circular functions can be built up on sin z as a fundamental function, so the elliptic functions can be expressed by means of this new transcendent. It is natural to denote it by or(z) where a- reminds one of the first letter s of sine. The first logarithmic derivative of sin z gives cot z which has, as poles of order 1, the zeros of sin z. The first logarithmic derivative of o(z) gives a function which Weierstrass has denoted by C(z). This also has, as poles of order 1, the zeros of a(z). It is not periodic since (z + 2 i)= =(z)+2 q i -=1, 2. Its first derivative is periodic, and this leads to the p-function dz P (z) d' 1. The minus sign is inserted so that the term - in the expression 164, 7) has a positive sign. The letter p reminds one that the most essential characteristic of this function is its double periodicity. 6. If in 10), 11), 14), defining the functions a,, p, we replace Z, (01 C2 by Hz, 1 ' W25' we see that e (stZa, a o1, /0-2) = 1A (Z', 1, 02), r(k, /c01, C02)= (z, (0:, W'2) (23 p(CZ, 1', /C02) = -2p (Z, 01' 02), which shows that ar, f, p are homogeneous functions of z, to, o2 of degrees 1, - 1, - 2 respectively. This property is useful at times. 360 FUNCTIONS OF A COMPLEX VARIABLE The relations 16) show that 7r ((1o0, oW2) = -,(f1, 602) r = 1, 2. (24 Since es =p (o,) we see that e(1uOl, 02)= es(w1, w 2) s= 1, 2, 3. (25 167. Elliptic Functions expressed by p(z), p'(z). 1. We suppose first that the elliptic function f(z) is an even function of order 2 s. Then if z= a is a zero or a pole of f(z), so is z=-a. Let a set of incongruent zeros and poles be ~a, a2, *a; b, ~b2,. of orders M, M2 *.*; l, n2,.. so that 2 ml + 2 m2+...=2n1+2 2 + *.. =2s We consider first the case that none of these zeros and poles is 0. Let p(z) have the same periods as f(z), we consider the function unc(Z) t- (pz - pal)ml(pz- pa2)m2 * (pz -pbl) z (pZ-2) * It has the same zeros and poles and to the same orders as f(z). As g admits the same periods asf, we see that it can differ from f(z) only by a constant factor. Next let us suppose that z = 0 is a zero of f(z). Since f is an even function by hypothesis, the order of this zero must be an even integer, say 2 m. Suppose now 0, a,: a * form a set of incongruent zeros of orders 2 m, m1, n2 respectively. Then as before 2m + 2m1 + 2m2 + = 2 s. Let us now form the same function g as before, where no factor, however, corresponds to z = 0. The numerator is of degree s- 1 in p and the denominator of degree s. As z = 0 is a pole of order THE FUNCTIONS OF WEIERSTRASS 361 2 for p(z), it follows that z = 0 is a zero of order 2 m for g(z). Thus as before g has the same zeros and poles and to the same orders as f(z). It can differ from f only by a constant factor. We get the same result if z = 0 is a pole of f(z). Thus in all cases when f(z) is an even function, f(z = C(pz-pal)'n1(pz-pa2)m2 (1. (f - (pz- l)pn( pz - pb 2)2... where we use a set of incongruent zeros and poles, always omitting that one which may be — 0. Case 2. f(z) is not even. Let us form (z) f(z)+ f(- z) 2 7, z) =/(z)- (- z) 2p'(z) which give f(z)= (z) + h(z)p'(z). As g and h are even functions, they may be expressed as in Case 1. We have thus proved the theorem: Any elliptic function is a rational function of p(z), p' (z). 168. Elliptic Functions expressed by t(z). 1. In 166, 167 we have learned two ways of expressing an elliptic function. Both require a knowledge of the zeros and poles of the functionf (z). When these are not readily found, it is convenient to have another representation. Such is the following, which depends on the knowledge of the characteristic at each of the poles. We will suppose, therefore, that a, b, *.. are the poles of f(z) in a primitive parallelogram of periods, and that its characteristics at these points are Ax +...+ A for z = a (z - a) - -a Bu +... b for Z=b (1 (z -b forz= (1 -(z- 6y) z-6 362 FUNCTIONS OF A COMPLEX VARIABLE We now construct a r function on the periods 2 cot, 2 2 of f(z) and then the function A3, g(z) - AjTizz - a) - A 2~-~'(z - ) f - ( - ) 2!. ly-1"~ + ~ l (z - a) + B1 ~(z - b) - B2r' (Z - b) + b) K1 B ~~'(z6+...+ z-b+ (2 2 (k - 1)! ' As f'(Z)=-p(z), (Z)=-P,(X) all the terms in the 2d, 3d... columns on the right of 2) are periodic. On the other hand, '(z + 2 o,) = 5(z) + ~2yj etc. Thus g(z + 2 wl) = 2?1(Al + B, +.)~g(z). But Al, B1... are the residues of f(z) in a parallelogram of periods. Their sum is 0 by 165, 3. Thus g(z + 2 wo)= g(z). A similar relation holds for 2 a)2 Hence g also admits 2 (ol, 2 (02 as periods. Let us now show that g has at each pole as z = a the same characteristic asf. For from 166, 14), we have obviously 1 ~(z - a) = ~+ h(z), z- a where I is regular at z = a. Hence (z ( - a) (Z a2+ 7110), (z - a)2~ (z - a ~ ) Thus the characteristic of g(z) at z = a is given by the first row in 2). Thus f and g have the same characteristic at z = a. The same is true at the other poles. Thus by 165, io f(z) = g(z) + constant. (3 THE FUNCTION'S OF WEIERSTRASS 3633 2. From the foregoing we have: Any elliptic function can be expressed in terms of C(z) and its derivatives. 169. Development of a, ', p in Power Series. i. We have now seen that the three functions 1(Z) ++1~4 (2 z Lz-o. 0 (02J I ( P Xz) + - o) wJ (3 where oW= 2 m 1 j +-2 m2ao2 MI = m 2= 0 excluded (4 may be taken as the basis of a theory of the elliptic functions. We propose in the articles which imnmediately follow to develop some of the properties of these three functions. We begin by developing them in a power series about z = 0. Since z2 is regular at z = 0 it can be developed in Taylor's series z2 which is valid within a circle R which passes through the nearest point w in 4), Now ~~p~n)~(1),n(n 1)?! (Z n Hence 10 (I)n(O)-(n+ 1) 1 n! n+2(X On+ Let us therefore set: (5 n n+2 ( iVe note that when n is odd, s8 = 0. For to each 2 m1w1 + 2 m2o2 in s, there corresponds a - 2 mio(1 - 2 mco2. 364 FUNCTIONS OF A COMPLEX VARIABLE When n is odd, the two corresponding terms in s will have opposite signs and cancel each other. We thus have p(Z) =I + 3 s2Z2 + 5 s4z4 + 7 s6Z6 + 6 z2 Integrating, we have 5(z) = - -s3z3 - s4z5 - *.. (7 Integrating again gives logo-(z) = log z - 1 s z4 - 1 s z6 - (8 Hence (z)= - z- sz- s~z7 - (986Z9 (9 2. Differential Equation satisfied by p (z). Further coefficients in the developments 6), 7), 9) can be obtained by a recurrent relation which we deduce from a differential equation. In fact, we saw in 165, 25) that p satisfies a very simple differential equation which we now proceed to find. From 6) we obtain, on differentiation and slightly changing the notation, p'(z) 2- + 6 c2z ~ 20 c33 z+ This squared gives 2 4 I. p (Z)2 24 c2 80 c3~... Also cubing the series 6) gives 1 1 p(Z)3 =-1 + 9 C2 -~15 C3 +. Let us now set 1 g=6O 2=6O=0, g3=14 Oc 140 -'. (10 These are called the invariants. From the foregoing equations we get on adding p'(z)2 - 4p(z)3g+ 2p(Z) +g83 = Z(a1 +anz+ a THE FUNCTIONS OF WEIERSTRASS 365 It thus admits 2 oi, 2 w2 and yet admits no pole in a parallelogram of periods. It is therefore a constant. Since it vanishes for z = 0, we have the desired differential equation PI (Z)2 = 4p()3 - g2p(z) - g3. (11 From this we get on differentiating p"l = 6p2 - g2 (12 3. We can now get the desired recursion formula. Let us write p(z) = 2 + alz2 + a2z4 + a3z6 + and put this in 12). Equating the coefficients of z2n-2 on each side of the resulting equation gives 2 n(2 n - 1) a, = 6 (a, + alan_2 + ** + an-_al + a,n). Hence aH = n(2n - 1) - 6 ala-2 + + an2al}. (13 This shows that a3, a4, a5 *. can be expressed as integral rational functions of al, a3, that is of g2, g3, since al = A 2, a2= -8 g3 For n = 3 we get from 13) a3-3a al =-,0o g2 For n = 4,= 3(a 4 = - (a 1^2 + 62 l) = 1 6 06 0 2,3. In this way we may continue. Thus we find (z- z -*- g - --- -— 7... (14 24 3 5 23 3 5 7 29 32. 5. 7 ( (Z) 1 — _ 9 __ gza.... (15 z 22. 3 5 22 5- 7 24 3. 52. 7 P(Z) =- +*+ g2 sz4+ g76 + (16 ) z2 + + 22. 5 22. 7 24 3* 52 P'(z) =-++.-*- +z +. (17 Z3 2. 5 7 23. 52 366 FUNCTIONS OF A COMPLEX VARIABLE From the definition of g2, g3 we see that g2 ('1' /'2) = g,2 (0i1' 2)' 1 (18 g3 (i1 /1'02) = -g3 3( l, 02)> 170. Addition Formule. 1. We have seen how important are the addition theorems eu+v = euev sin (u + v) = sin u cos v + cos u sin v, etc. for the elementary transcendental functions. We wish to establish analogous formulae for the new funictions, viz.: aC(u + v) (u - )p(u), 2 p(v)-p(zl)( (l o-2U. * a2 (u + V)= (^)) +()+ (P ----), (2 2 \pu- pv ) p(u + v) =p - _ 1 d ('u- (3 2 du\pu-pv These relations are fundamental and of constant service. We begin by proving 1). Regarding u as a constant let us look at the zeros and poles of f(v) = p(v) - p() in the parallelogram of periods P(2 ol, 2 )2). Obviously, f = 0 for v = u, and hence for v u. As p( —u) = p(u), it follows that f= 0 at v =- u, and hence at v - u. As f is of order 2, it can vanish only twice in P. Thus all the zeros of f(v) are -~ u. The poles of f(v) are v = 0, and these are of order 2. Thus the two functions of v, on the two sides of 1), have the same zeros, poles, and periods. They can only differ by a constant factor C. To determine this we develop both sides about v = 0 and compare the coefficient of. Now by 169, 14), V2 o(v) = v + av5 +... Hence 2 2 a'2v = v2 +-... THE FUNCTIONS OF WEIERSTRASS 367 Thus 1 1 I -2uC 0-2V c-2U V2 To develop o-(u + v), o(u - v) about v = 0, we set (v) = (-(u + v) = g() + vg' (0) +... = -(U) + vo-' () + *. Similarly -(- v) = (t) - va (u) + * ** Thus (c +v) )-(u -v)= o2u + Thus the left side of 1) has 1 as coefficient of -. The same is v true of the right side. Hence C= 1 and 1) is established. 2. To prove 2) we take the logarithmic derivative of 1) with respect to v, and get?(u + v)- (u - v)- 2 ~(v)= p — pv - pu But this relation holds for u as well as for v. Thus interchanging U, v gives?(u + v)+ ~(u- a)- 2 (M)= - _ p( P pv - p Adding and dividing by 2 gives 2). 3. To prove 3) we need only take the derivative of 2) with respect to u. 4. Another form of the addition theorem for the p function is the following: p(u + v) +p(u) + p(v) = P= — V ) (4 To prove this we square 2), getting?(u + V) - ~(U) _- (W)2= - V. (5,\ pu -pv / Let us denote the left side of this relation by g(u), regarding v as a constant. The right side of 5) shows that g(u) is an elliptic function. We propose now to express g by means of r and its derivatives, using 168. To this end we must find the characteristics of g(u) about its poles. These are = 0 and - v, each pole being of order 2. 368 FUNCTIONS OF A COMPLEX VARIABLE Developing about u =0 we have Hence I(~)~u-()=- v+. U and thus ~( Hence in this case the coefficients A in 168, 1) are A1=0, 2=1 Let us now develop about the point u=- v. We have ~(U+V) + a(u +v)3 +. Hence ~('u + ) - ~(U __ -V)(U + v) ~"(v) + Thus 1 A)=(U +V)2 -2~v The coefficients B in 168, 1) are here B1=0,B2 =1. Putting, these values of Al, An2, B1, B2, in 168, 2) gives g(u)= -'(u) - '(m —V) + C, or To determine the constant 61, let us equate the absolute terms of the developments of both sides about u = 0. From 6) this term on the left side of 7) is 2 p(v). On the right side of 7) it is p(v) + C. Equating these, we get U,=p(v). This in 7) gives 4). THE FUNCTIONS OF WEIERSTRASS 369 171. The co 9ij, ej and g2, g3. 1. We saw in 169, 11) that pI(Z)2 4 p3(z) -g2P (Z) -g. (1 On the other hand we saw in 165, 18, that p'(z) = 0 for z =l, (25 o3. The three roots of the cubic 4p3-g~ p-g3=0 (2 are therefore ej = p(co) i = 1, 2, 3. Thus we can write 1) T(Z)2= 4(pz - el) (pz - e)(p - e which shows that the constant 0= 4 in 165, 25). Since the coefficient of p2 in 2) is 0, we have el + e2 + e3 = 0. The other coefficients of 4) give ele2 + ele3 + e2e3 = -4 q, ele2e3 = 4 g3' To complete the symmetry let us introduce 'q3 defined by '91 + 172 + 1q3 = 0. (3 (4 (5 (6 We show that also 173 = 5(03) (7 Since f is an odd function, - - l O) = - +(Ohi ~2) — ~(Ow) - (O2) by 170, 2) 771 - 172 73, by 6), and this establishes 7). 2. Between the w1,?7j exists a relation due to Legendre. Let us suppose that we pass from a + wl to a + W2 by a positive rotation of angle < w as in the figure. Then Legendre's relation states that '71'02 - '72'01 = _' Let us take -the parallelogram P so that z= 0 lies within it. Then 27Tp)dz = I Res 2,7rie. P 370 FUNCTIONS OF A COMPLEX VARIABLE But has only one pole in P, viz. z = 0, and by 169, 2)' its residue is 1. Thus a+2(9 f'z = 27ri. Now 12 f~(z)dA =f '(z + 2 0o2)dz ~ aw =j~(z)+2%2~dz,by 166,17), =Jdz ~ 4'?12lP Similarly f(~i Jd Hence f~z= J~ i = - 4 '72'01 + 4 7(12 Putting this in 9) gives 8). By using 6) and &wl + &)2 + (O)3 = 0 we have for any two indices r, s = 1, 2, 3, Vi~ (10 where e = 1 when we pass from co, to co. by a positive rotation of angle < 7r; otherwise E=- 1. 3. The relations 166, 17), 18) may be at once 'extended to &O3 and give for r = 1, 2, 3, a-(z + 2 wot) = - e2-0r(z+wr)a(z), (11 ~(z +2 O)r)=~(Z) + 2?7. (12 172. The Co-sigmas 0-r(Z).- 1. We introduce now three new sigma functions aT(z) =e-llrzG(Z + (o) r = 1, 2, 3. (1 This we can transform as follows. Fromi o-(u ~ 2 w~r) =-e2-qr(U+wr)C(u) THE FUNCTIONS OF WEIERSTRASS 371 we have, setting u- = -,Z(- + f),) = e-2rzoz ( + Or). This in 1) gives e,() ea(or - ) ^ "rZi. (2 Replacing z by - z in 1) and using 2), we get (- Z) = r,(Z),,(O) =1. (3 We find without trouble or,( + 2 t)) = - e2Tr(Z+Wr)or(z), (4 or(z + 2 ),) = e2s(z+(s)o-r(z), (5 o-(Z ~ ~or) = ~ e7Vrzo-(fr)r(Z), (6 '(z)( o<(Z ~ o)) = T e"r(rfz) ) (7 o( -Ot -r (8 o(z ~ CO) _= -O ef~7sz-7rwsot(z), (8 where r, s, t are the integers 1, 2, 3 in any order. For example let us prove 7) for the + sign. From 1) we have a-(z +- 2 (Or,) O-r(Z + o)r) = e-r(z+wr) -Z + 2 - r) or using 171, 11), = -- e-1(z+wr)e2,7-(z+%.) -(Z) CCTO = - _ e.(z+or) - (, which is 7). 2. In 1) let us set z = w,, then o'r)O8 = - e-71rw (9 o-for since o-(f, + os) = o-(- ft)= - ot). Here as usual r, s, t are 1, 2, 3 in any order. Setting z = or in 2) gives a-,(ot,)= 0. (10 Let us put 9) in 8), we get o'r(z ~ o8) = e+szo-ra-tz. (1 372 FUNCTIONS OF A COMPLEX VARIABLE In this formula with the lower sign, set z = o,. As a,(O) = 1 by 3), it gives 0l"o = a- tO.. (12 Let us now make use of this in 7), it gives ar(Z CDO)=Fe~F eqrOa-8 0t(r)Ta-(Z) (13 From the definition 1) we see a-r(cUZ JW11 ~ P2) = r-(Z, W1, W2). (14 3. In 170, 1) let us set v = (8; we get, using 1), 2), a,2(z) a-(Z) z)e8 S (15 This shows that the square root of the left side is a one-valued function of z. We set V'p(Z)-e8= ~ a-8(z) (16 a-(z) which determines the sign of the radical. Let us set z= -o_ in 16), we get Ver-e8= 8 (17 and the sign of the radical on the left is determined. Putting 9) in 17) gives Vi_ —e, e e-wrn3~-COt Ve - 8~ a-Oaco8 Interchanging r and s, we get, dividing, Ve, - e= Ve8 eT Hence using Legendre's relation 171, 10) Veli- e2 = ie2 - el Ve - e =-i-e3-e 1 Ve2 e3 i\e3 (18 Here we suppose ail a2 such that we pass from o0 to o2 by a positive rotation < 7r. The relations 18) enable us to replace in our formula a radical Ve, - e8 by Ve8 - e~, a substitution which is often useful to make reductions. THE FUNCTIONS OF WEIERSTRASS 373 4. In 15) let us replace z by z + %w, we get o-2 + W,) p(z + o,)+ S- 2(Z + ') or using 6), 13), -= e cr2s2. 1. 2(z) (T2w e2,sZ-2os a2(z) Using 15), 17) this gives p(z + o) = e, + (e er)( - et) (19 p(z) - e, 5. To find the development of oar() in a power series about the origin we have, from 16), or(Z) = c(z)Vp(Z) - er Now from 169, 16) p(Z) - e - er r ~ g2Z2 +- ~2 z 20 Thus 1 z_2,20+ + ~ r(Z) = () { - er + 2 2Z... + 1 ( = 1 -e,2 48 (6 e - g2)z4 -... (20 6. We have obviously pu = -2 l (u)-2()u) a-u) (21 a- u 173. The Inverse p Function. Case 11. 1. We have P(Z) = #2z2 + (1 z 20 pf(z) 2= -+2 +. (2 z3 10 g2= 60, ' g3= 140 6 o = 2 mlo1 + 2 m22. (3 The relation 1) defines p as a function of z. We wish now to consider the inverse function z of p. Case 1. col real and positive, o2 = 2, O2 > 0. We note first that the invariants 2, g3 are real. For in o to each positive m2 corresponds a negative value - m2. Then the two values of o, 2 mol1 + 2 m2i0)2, 2 mlo1 - 2 m2iw2, 374 FUNCTIONS OF A COMPLEX VARIABLE are conjugate imaginary. Thus the terms in g2 g3 enter in pairs which are conjugate imaginary numbers. As the sum of two conjugate imaginaries is real, g2, g3 are real. From 1) we see that p is real for real values of z, say for z = x. As p has the real period 2 ol, p(x) is periodic. The relation 1) shows that the p-axis in Fig. 1 is an asymptote. From 2) we see that p' (x) is negative for small values of x. Thus p decreases until p'(x) vanishes. The roots of p (z) = 0 are Cl, W2), W)O of which only the first is real. Thus p decreases from x = O to x = o, at which last point p((o1) = el. Since p(x) is an even p axis I Ce C1 201 -_ O WI x axis FIG. 1. function, p is symmetric with respect to the p-axis. Thus p decreases as x ranges from 0 to - w. As p has 2 co as period, the graph of p in the interval (cl, 2 ol) is the same as in (- wl, 0). 2. The graph of p = p(x) shows that the relation 1) defines a many-valued inverse function x= (p), (4 one of whose branches may be characterized by the conditions 1~ x 0 asp T +oo, 2~ x is positive for p > e1. This branch is shown in Fig. 2. We show now how this inverse function may be represented by an integral. The derivative of p(x) is (5 dPi=~ V4p3 -2p —yg3 dx To determine the sign of the radical we observe that for 0< x < Lp, d dx I el 1 e, (6 0 FIG. 2. THE FUNCTIONS OF WEIERSTRASS 375 is negative, while the polynomial under the radical is large and positive. We must therefore take the minus sign in 6). Then the derivative of the branch of 4) as determined by 5) is dx _ 1 1 dp p -V/4 p3-g2p-g ( dx From this follows that the inverse function x defined by 4), 5) is n dp. (8 x /V= ar g (8.J V4 p3 - g2P - 3 In fact the quantity under the radical is pl(x)2; it is therefore positive for p > e. Also as p - + oo the integral 8) converges to 0. Thus the conditions 5) are satisfied. From.p(otl) = e1 follows now that x = wc when p = el. Putting this in 8) gives dp 01i= / ----.(9 V+ /4 p3 -g2p - gP This expresses the period 2 cl as a real integral. 3. We show now how 02 can be expressed as an integral. To this end we note that 1) shows that p(z) is real and negative for small values of z of the form z = iv. We have in fact q =p(iv) =v -21 22 (10 v 2 20 By analytic continuation the series 10) will give the values of q for all values of z from 0 to w2, that is, for values 0<v< w2 As the terms of 10) are all real, the values of q obtained by this process will be real. When v= = p2, q= (iW2) = p(w) = e2 which is therefore real. Asp'(z) does not vanish as z moves from 0 to W2 until it reaches %2, A does not vanish as v moves from 0 to W2, until it reaches o2. dv The graph of q considered as a function of v is given in Fig. 3. It shows that 10) defines a many-valued inverse function. v = A(q), (11 376 FUNCTIONS OF A COMPLEX VARIABLE entirely analogous to 4). One of its branches is given in Fig. 4 and is characterized by v 0 as q -oc; v > 0 forq e2 (12 -'M 0 -m~ 2WD~ v axis e2 e2 V axis e2 0 axis axis FIG. 3. FIG. 4. To represent this inverse function by an integral we observe that dqdp(z) dz dv p d ip'(z) =~iV4 p - g2p-g3. dv c -z dv 3 As the derivative is real and positive for this branch, and as p(z) = p(iv) = q, we may write this dv 1 dq +V-((4 q'q-g2q-g3) the radical being real and positive for q < e2. Let us now consider v qf dg V-(4q -g=q-g3) (13 We see this is positive for q<- e2 and that v O as q oo It is therefore the function defined by 11), 12). Setting q = e2, we have 602= (14 From this, we have expressedw 2 = w2, as an integral. THE FUNCTIONS OF WEIERSTRASS 377 4. We have seen that el, e2 are real. As g2, g3 are real, all the roots of 4 4p3-2p -3 = 0= (15 are real. Since + e2 + e= (16 eI + e2 - e - = 0, (16 it follows that at least one root e must be negative. As the first root of 15) which we meet as p moves from - oo toward the origin is e2, this root is certainly negative. The sign of the root e2 is given by 16). 174. Case 2. Periods Conjugate Imaginary. 1. Let us suppose now that C=o - io', 2= Co + iw CO > 10. Then 2 = 2 f0 =- 09 + W2 = - 03 - 03, 2 it' = 2 - 1 - (D - c3, (2 2 o =- 2 i 2 f+2io' =2)02 =0. As in Case 1, the invariants g2, g3 are real. For 2 m11- + 2 m22o = 2(ml + m,2)o + 2iow(m2 - m). On interchanging ml, m2 this period goes over into one which is conjugate imaginary. From the series 173, 1), 2), we see that p(z), p'(z) are real for real z, and that p is real for purely imaginary z. 378 FUNCTIONS OF A COMPLEX VARIABLE As in Case 1, we see that as x moves from 0 to 2 o, p(x) decreases. At x= 2 w p(x)= p()3)= e3, at which point p has a minimum, and pf (C3) = 0. Let us note that pi(x) has only one incongruent real root, as 0, %2 are complex. The inverse function defined by p =p(x) is many-valued. One of its branches is given by x=f dp = + V/4p3 - g2P-3 which is positive for p > e3, and which 0 as p -0. When p = e3, x = 2 o. Thus 2= dp (4 J + V/4 3-g2p - g3 2. Let us now express o' as an integral. Knowing co, ot, we can express the periods 2 wl, 2 C2 as integrals by 1). As z = iv moves from 0 to 2 iwt, p is real and moves from - o to p(2 ioW) = p(o3) = e3 We have v2 20 (i)=- 2-ji)v. (5 dq_ 2 dv v3 Thus q is increasing for small values of v. As dq does not dv vanish unless p'(z) = 0, and as the first root of this on the imaginary axis is o)3 = 2 i0y, we see that q increases steadily from - oo to e3. Thus the relation 5) defines an 2 inverse function v of q, one of whose branches is characterized by the condition that v-0 as q —oo, and v>O > 0 for q <e3. (6 qaxis THE FUNCTIONS OF WEIERSTRASS 379 Thus, as in Case 1, this branch is represented by v2 dq ( -~- (4 q3- 2q - 3) As v = 2 o' for q = e3, this gives 2I/ fe3 dq (8 2 W = p -- ^. (8 J- /+V-(4q3- 2q -3) 175. The p Function defined by g2, g3. 1. Up to the present we have considered p(u) as defined by means of the periods 2 co, 2 w2. These numbers being taken at pleasure but not collinear with the origin, we constructed the sum 2(z, 2 w1, 2 2)=- + 2 ( 2 (1 as in 169, 3), and showed that -P(Z) =- + | + Y^ Z2 + + Z4 + *h (2 z2 20 28 all of whose coefficients are rational integral functions of the invariants 1 2=60 4, g3 140 - 6 (3 We ask now: Can we start with two numbers g2, g3 taken at pleasure, and find the periods 2 ), 2 w2 with which to construct the p function 1)? Let us consider the roots el, e2, e3 of the cubic 4 t3 -2t -3 = 0 o= r 4 (t- el)(t- e2) (t- e3)= 0. (4 We must in the first place suppose that two of them are not equal. For we have seen, 169, 11), that p(z) satisfies the equation d = 4 p3 - 2P - Ya - -V4 (p - el)(p - e2)(p - e3). (5 dz If now el = e this gives dP 2 (3p-e~)/~P (dz Hence. Z (7 (p-e) p-e3(7 _p2(~ - e,0I e.p 380 FUNCTIONS OF A COMPLEX VARIABLE Thus z can be expressed by means of the elementary functions, and p cannot be a double periodic function. 2. The roots e8 being unequal, let us suppose they are real. We define 2 wo, 2 o2 by the equations 2\ 2 /4 t,3 -Y2t - g3 2 C = 2 dt, (8 - (4 t3 - 2t - g 3 where we suppose the e's so numbered that e2 < e3 < e1. (10 Then 2 o0 is real and 2 o2 is purely imaginary, since the radicals in both 8) and 9) are positive. As 2 w1, 2 r) are not collinear with the origin, the series 1) constructed with these two numbers defines an elliptic function p(z, 2 ol, 2 2) which we have seen satisfies 5). The reasoning of 169, 2, 3 shows that this function p will have the development 2) about z= 0 and that g2, g3 will satisfy 3). 3. Let us next suppose one root e3 of the cubic 4) is real, while the other two are conjugate imaginary. We define 2 co, 2 co' by the equations 2o= dt (11 Je V/4 t3 - Yg2t- g3 2c=fe dt= (12 2.. /J —(4 t -t - g) (12 These are real and positive, since the radicals in 11), 12) are both positive. We now set 2c=w=2a)-2io, 202=2c)+2io/. (13 Since these are not collinear with the origin, the series 1) converges and defines an elliptic function p(z, 2 o,, 2 e2). As before, we see that this function satisfies the differential equation 5); its development about z =0 is given by 2) and g2, g3 satisfy the relations 3). THE FUNCTIONS OF WEIERSTRASS 381 4. Suppose finally that g2, g3 are any complex numbers, such however that the cubic 4) does not have two equal roots. From algebra we know that a = (ei - e2(e - e3)%e2 - e3)2 = JL(gl -27 g). (14 Obviously 4) will have equal roots when and only when a = 0. For this reason G is called the discriminant of the cubic. We shall not treat the general case but merely state that: If we set fa~ dt dl(t "1e.~~I 3 - ~t -93 V4 0 - t -,t_ q3t-Y) d V4t3-g2t-g YJ3~m-2tg the series 1) constructed on these numbers is convergent and defines an elli3ptic function p(z, 2 (ol, 2 w2) having 2 ol, 2 12 as a primitive pair of periods. Thisfunction satisfies 5), its development about the origin is 2), and g2, g3 satisfy 3). 176. The Radicals Vpz) - e,,. These are factors in dp -Vp3 g2p -3 = 2 Vp —e Vp-c2_ Vp -e. In 172, 15 We saw that they are one-valued functions of z, viz. VP em Om(Z) m= 1,2,3. (1 Let us set in general 0m(Z) _Y(Z) ____ They are homogeneous functions of z, (1, w2, of orders -1,1, 0 respectively. Of these 12 functions, 6 are reciprocals of the other 6. Let us consider one of them as (z) 1 (3 q= %2(Z)= _ _ 2 -2 V/p(z) - e From 172, 1 we have q(z~2oi)=- q(z), q(z+2w2)=q(Z). Thus q admits 4 ),o, 2 W2 as periods. 382 FUNCTIONS OF A COMPLEX VARIABLE As a(z)= 0 for z_ 0 mod 2 WI, 2 w2, we see that the zeros of q are = 0, mod 2 wl, 2 C2O As o-2(z) = 0 for Z w2, we see that the poles of q are o,2 mod 2 WI, 2 W2, and simple. Thus q is an elliptic function of order 2 for which 4 WI, 2wo2 form a primitive pair of periods. By 165, 14 the zeros of q'(z) are 1 3W1, w3, 03 ~ 2 w1 mod 4 w, 2W2. At these points q has respectively the values 1 -1~ 1 - 1 Vei - e2 Vei - e2 Ve3 -e2 V\/e3 - e2 Thus by 165, 16 q satisfies the differential equation C e ~~~~~~~~(4 Uz CI el - e2, 3 - e2, ( To determine the constant C we observe that q = p~~z) - e 1 - e ~' = x% - e 2 + I = Zjl + CtZ2 + ~~) Hence a + 3 aZ2 +-~~ dz Putting these developments in 4) and equating the absolute terms, we get C (el - e2)(e3 - e2) Thus 4) becomes __ 2~- =1 -(el - e2) (e. - e2) Uzx CHAPTER XI THE FUNCTIONS OF LEGENDRE AND JACOBI 177. Rectification. 1. In the previous chapter we have studied the elliptic functions from the point of view of their most characteristic property, viz. as double periodic functions. Historically they presented themselves from quite another standpoint, and this we wish now to develop. The integral calculus enables us to find the lengths of a great variety of curves found by effecting the integration in the formula dy2 ~ dx; ~s~tt V~~ dx; =(1 for example, the circle, parabola, catenary, cycloid, cissoid of Diodes, the cardioid, etc. When the contemporaries of Newton and Leibnitz attempted to rectify the ellipse, hyperbola, and the lemniscate by means of 1) they met a most unexpected difficulty. In spite of every effort they could not effect the integration. Let us see how these integrals look. 2. The Ellipse. The equation being a2y2 + 62x2 = a2b2, (2 we have 4 e hav a4 _c2x2 dx, 2 = a2 b2. (3 aJ V (a2 _ 2)(a - c2X4) Instead of the equation 2) we may use the parameter equations of the ellipse, y=. x = asin, y= b cos. (4 Then s a fVa2 cos2 + b2 sin2 0. d+, (5 or setting a2 _ b2 2 (6 a2 a2 s=a f k2 sin2 0. do. (7 383 384 FUNCTIONS OF A COMPLEX VARIABLE As we now know, the integrals 3), 7) cannot be expressed in terms of the elementary functions; the efforts of the mathematicians of the seventeenth and eighteenth centuries in this direction were doomed to fail. And yet only failure in a narrow sense, for from their apparently fruitless efforts has sprung a whole new branch of mathematics, the elliptic functions. 3. The Lemniscate. If we take the equation in polar form, it is p = a cos2 2 (. Then 1) gives= a (8 o /1 - 2 sine 0 If we set x = sin 0, we get also s= a r dx (9 J /(1I-2)(1-22) 178. Elliptic Integrals. 1. Many problems of pure and applied mathematics lead to integrals whose integrands are rational functions of x and the square root of a polynomial P of the third or fourth degree, that is, to integrals of the type f(x, V/P)dx. (1 Such integrals are called elliptic integrals; they include the integrals 3) and 9) of 177, and cannot be expressed in general in terms of the elementary functions. They therefore define new functions in the same way that rdx r dx J x ' J- vx2 define transcendental functions although their integrands are algebraic. The question arises, how many different types of integrals are included in 1). We propose to show that all these integrals may be reduced to three, viz.: THE FUNCTIONS OF LEGENDRE AND JACOBI 385 r dx___ (2 (1- X2)(1 - k2x2) P _ x2dx (3 V/(1 - 2) ( - k2x2) ~r _dx, (4 J (x- a)/V(l - X2)(1- k2x2) which are called elliptic integrals of the 1~, 2~, and 3~ species, respectively. The number k is called the modulus, the number a which enters 4) is called the parameter. 2. To reduce the integrals 1), let us note that if the polynomial P is of the third degree, ax3 + bx2 + x + d, (5 the integral 1) may be replaced by one in which the polynomial under the radical is of the fourth degree. For let a be a root of 5), then P has the form P = (x - a)(px2 + qx + r). (6 Let us set.X - then p= yVp(y2 + )2 + q(y2 + a) + r and the polynomial under the radical is of the fourth degree if P =0. But if p = 0, the polynomial P is of the second degree, as 6) shows, and this is contrary to hypothesis. 3. Let us suppose then that P=pOx+p1x3+p2x2+p3X+p4, Po-0. (7 Since p is a rational function of x and y =V/, 386 FUNCTIONS OF A COMPLEX VARIABLE let us arrange its numerator and denominator according to y. Then + +b. O,0 =?_i6~l~_ll~(8 Co + Cly + C2y2 +... As y2=P, y3=yP, Y=P2... we see that 8) has the form A= + By (9 C+ Dy' where A, B, C, D are polynomials in x. If we multiply numerator and denominator in 9) by - Dy, we get = ~P=E+ 2-+_ a ~get = -E + Fy = E +? = _ + Y VP where E, F7, are rational functions of x. Thus fdx= fdx + r dx. (10 Here the first term on the right may be integrated by means of the elementary functions as shown in the calculus. We are thus led to considerGd f ax. (11 J VP 4. As a is a rational function of x, it may be broken up into partial fractions as shown in 122, 4). Thus G is the sum of a polynomial which may reduce to a constant and a number of terms of the type a a am \ a^-! 4... +. (x - a)m ( -- a Thus the integral 11) reduces to integrals fxx and -(12 JnVP (x- a)m-VP Both of these may be represented by Q=Sf^-aY^l V (13 (Qm = (x - a)m /- (,13 if we let m be a positive or negative integer or 0, and a any number including 0. There are now two cases. THE FUNCTIONS OF LEGENDRE AND JACOBI 387 5. Case 1. a is not a root of P. Then we may write P = Ao(x- a)4+ 4 A(x- a)3+ 6 A(x- a)2 +4A3(x- a)+A4 (14 and A4 0. For setting x=a in 14), it reduces to A4, and if this were 0, P would = 0 for x= a, which is contrary to our hypothesis. Let us note now that for any integer d- _ = -- _na) —lp + 2-(X -a)nP' dx ^ ( - a)n-\P = 2 Hence, integrating, ( _ ), -= N.( -- a) n-PdX r(D -a) 'dx If we put in the value of P and P as given by 14), we fi If we put in the value of P and P' as given by 14), we find (x - a)nV = (n +2)Ao Q+3+ 2(2 n + 3)A1Qn+2 6 (n+ 1)A2 Qn+ + 2(2 n + 1)A3Q, + nA4Qn_1. (15 If we take n= - 1, this relation enables us to express Q-2 in terms of Q-1, Q1, Q2 and an algebraic function. If we take n= - 2, we see Q-_ can be expressed by means of Q-2, Q-1, Q0. But we have just seen that Q-2 can be expressed in terms of Q-1, Q1, Q2. Thus Q-2 can be expressed in terms of Q-, Q0, Q1, Q2. In the same miianner we may reason for higher negative values of n. This shows that the integrals 13) when m is negative may be reduced to t _ 'd J d _, (16 VPJ V (X - a) VP and to Qn with positive indices. Case 2. Suppose a is a root of P. It cannot be a double root of P. For then P would have the form P = (x- a)2(px2+ qx + r), and hence VP = (x - a)/pX2 + qx +r, 388 FUNCTIONS OF A COMPLEX VARIABLE that is, the polynomial P under the radical can be replaced by one of degree 2. But in this case the integral 1) leads only to the elementary functions, as is shown in the calculus. In the present case therefore A4 = 0 but As f 0. Thus the last term in 14) disappears, but not the next to the last term in 15). Hence if we set n = - 1 in 15), this relation enables us to express Q-1 in terms of Q1 and Q2. If we set n = - 2, it gives Q-2 in terms of Q-1 and Q0, etc. We are thus led to the first integral in 16) and integrals of the type 13) for which m > 0. 6. When m > 0, the integrals 13) give rise to integrals of the form mfxMdx (17 In the relation 15) we may set a = 0, then the Q's will go over into the integrals 17). The relation 14) shows that the A's are the coefficients p in 7). In 15) let us take n = 0; this enables us to express Rs in terms of R2, R1, and R. If we set n= 1 in 15), it shows that R4 may be expressed in terms of R3, R2, and R, and hence in terms of R2, R1, and Ro, as just seen. Thus all the integrals 13) reduce to f dx f xdx [x2dx (18 J VP ' J ~^P AJ P when m > 0. 179. Linear Transformation. 1. To complete the reduction let us show how to determine the linear transformation a +bx (1 1 +ex so that dy _ dy V/Y- )/(Y - Y1)(Y - Y2)(Y - Y3)(Y - 4) dx dx ( M V(1 - x)(1 - c2x) lM i2 X THE FUNCTIONS OF LEGENDRE AND JACOBI 389 Let us determine a, b, c, k so that when Y= Y1 Y2 Y3 Y4 we have respectively 1 1 ==1, -, - (3 Then, ) /^l +- X Y-Y (=l +ex) (4 3(1 - kx') q4(1 + kx) - Y3 = +1 + Y-Y4- 1 + cx Set x =- 1, = Y2 in the first equation of 4), then 2g1 Y2-Y9= 1 _ Similarly if we put the other pairs of values of 3) in the remaining equations of 4) we have 2 g2 2- -y 22,'4 Y1-Y2 = +e Y4-Y3= Y3-4- Y1 - z1- 1~- ic k These relations give ql -- ' 22 Y — _ 1-C)(2-y ' _ (Y1-Y2)(1C), g3-2(k)(Y4-Y3), = From 4) we have y- Y1 _ (1 -x) 1 -c x- 1 Y-Y2 92( +x) l+ x+l Set here 1 = Yy =, we get Y3-/ 1-c_ 1=-. (6 Y3-y2 1 +e 1+k 390 FUNCTIONS OF A COMPLEX VARIABLE Also in 5) set 1 = Y4, x =-, we get l4 1_ - 1~+ (7 Y4-Y2 1 + 1+ (From 6), 7) we find l-)2 Y1 - Y3' Y2-IY4 = p2, say. (8 + k 2 y- y Y1 -Y Y4 This gives - k=C. (9 1+p From 6), 7) we have also - c2_ Y1-3. Y1-4= -, say. (10 + / Y2 —y Y2 —Y4 This gives - (11 1 + To get a, b we start with 1) or y + cxy- a- bx= 0. For x = 1, y = yl, this gives Y1 + cy -a - b = 0. For x = - 1, y = y/ it gives /Y2- c2 - + b = 0. Adding and subtracting these two relations give = (1 i+ 2 + ~(1 -- 2)) (12 b = 1(1 - y2 + ~(y1 + Y2)). To find Mwe differentiate the second equation in 4), which gives (1- c)dx dy= 12 (1 + cx)2 The fourth equation of 4) gives (k - ( )d& (1 + CX)2 THE FUNCTIONS OF LEGENDRE AND JACOBI 391 Hence dy22- (1 - c)(k - ) 2 dy2 24 C d(1+ cx)4 ~or qdy = Vg (l - c)(k - c) dx. (13 (1 + CX)2 Thus dy _ /2g4(1 -c)(k - c) (1 + cx)2 dx VTy (1 + c)2 Vg1g2g3g4 V Hence He- (Y1-Y2)(Y3- (14 2. Students familiar with analytical geometry will recognize that p in 8) is the cross ratio of the four roots Yl, y2, y, Y4. These four roots can be permuted in 4! = 24 different ways, to which correspond 6 values of p2. This the reader can verify without trouble. We find these 6 values are 1 P2 P2 - 1 1 p2, P'" - 1 — p2, 1 1 (15 I2 - p p2 l — p We observe that three are reciprocals of the other three. To illustrate our meaning, let us interchange yl, Y2 in 8). The middle term becomes Y2-Y3. Y1- YZ4 (16 Y1 -Y3 Y2 - 4 p2 Corresponding to this, the value of k in 9) is + +p (17 1-p We observe that 9) and 17) are reciprocals. We have therefore established this important result: By means of the linear transformation 1) we can reduce Y to V dx (18 n s a way tat t 2) (1 l 2) in such a way that the modulus k is numerically less than 1. 392 FUNCTIONS OF A COMPLEX VARIABLE 3. Let us return now to our general elliptic integral C(x, /VP)dx. (19 We had P = Pox4 + PX3 + P2X2 + P3X + p4 =P(Y - Yi) (Y - 2) (Y - Y3)(Y - Y4) = Po Y' The relations 4) show that the linear transformation 1) converts VP into c/(1-x2)(1- k22) = VX_ (1 + Cx)2 (1 + cx)2 Thus this transformation converts the integral 19) into an integral of the same form.f ((,V VX)dx (20 except the radical VP has been replaced by V/X, which is the form used by Legendre. If we had made this transformation at the start and had reasoned on the integral 20), the middle integral in 178, 18) would be IS _ xxdx (1 - 2)(1 - k22) If we set x2 = u, this becomes 1 r ~ _ du 2 fV/(1 - )(1- Cu)' which can be expressed by elementary functions. The final result of our investigation may be summed up thus: The general elliptic integral may be expressed in terms of the elementary functions and the integrals r dx rx2dx dx J f X J vX J (x-a)VX( where X= (1 - 2) (1 - k22). THE FUNCTIONS OF LEGENDRE AND JACOBI 393 We have therefore established the statement made in 178, 1). The integrals 21) may be regarded as standard or normalforms of the three species of elliptic integrals. 180. Legendre's Normal Integrals. 1. Instead of the three integrals 21) of the last article, Legendre employed as normal integrals I dx/1 V1/ - Ik2x2d V(1-x2)(1- k22) ' _l- x2 rdx( J (1 + nx2) V(1 - x2) (1 _ l122) Let us show how the former integrals may be expressed in terms of these latter. The integral of the 1~ species is the same in both cases. To express the second integral of 179, 21) in terms of Legendre's integrals we observe that 1 12 _ x2dx k2 k2 r dX 1 _V_ _ k2f,2 Jx — J vf q -- -k2J + dx. Thus the normal integral of the the 2~ species adopted in 179 is the sum of an integral of the 1~ and of the 20 species as adopted by Legendre. Turning to the third integral of 179, 21), we have.f cx x + a a dx '(x- a)x/X - - VX f= r xdx J dx + a j(2 a2) VX (2 - a2) X2 The first integral in the last member can be expressed by elementary functions, as we saw in 179, 3. 394 FUNCTIONS OF A COMPLEX VARIABLE The last integral becomes, on setting n= —, 1__ r i ___ _ dx_ aJ (1 + nx2) /(1 - x2)(1 - k2x2) which aside from a constant factor is Legendre's integral of the 3~ species. 2. Let us set with Legendre x = sin (. Then dx 7 /1 - x2 The integrals 1) become, on putting in the limits 0, Pb, F'(, kj)= f do -, E(, k)= f Vl - k2sin2. d, /1V - k2 i2 sin22 d, n(l, n)=1 do- (2 Jo +nsin2 V1-_k2sin2 ( The radical which enters in these expressions and which is of constant occurrence in this theory is denoted by Legendre by A(0), thus A() = -/1 - k2sin2 0. When (p = 7, the first two integrals in 2) are denoted by K fjr^ E=f )d~. (3r K=,, E= 4(0)d. (3 J a(o ) Jo They are called the complete integrals of the 1~ and 2~ species. Legendre denoted the integrals 3) by the letters F', E', but we shall follow the modern usage. As we shall see, they play the same role in the theory of Legendre and Jacobi as ol, *1 do in Weierstrass' theory. In practice the modulus k2 is usually real and < 1. Legendre sets k = sin 0 (4 and calls 0 the modular angle. THE FUNCTIONS OF LEGENDRE AND JACOBI 395 To make the elliptic integrals useful for numerical purposes Legendre calculated at great labor tables for the integrals F(c), E(o) for values of q and 0 for every degree from 0~ to 90~. They are to be found in Vol. 2, p. 292 seq. of his great work: Traite des Eonctions Elliptiques, Paris, 1826. Shorter tables are to be found in various works which treat of these integrals, for example in the Tables of B. O. Peirce referred to on p. 91. The reader will note that the functions F(b, k), E(Q, k) are unlike the functions log x, sin x, etc., in that they depend on two variables q, k and so require tables of double entry. This makes their tabulation extremely laborious. Turning to the elliptic integral of the 3~ species, we see that this depends on the argument p, the modulus k, and a number n which we call the parameter; in all on three variables. Its tabulation would thus require a table of triple entry, which is quite out of the question. Legendre, who had the numerical side of these integrals close to his heart, was delighted when Jacobi showed how they may be computed by means of the e functions. 3. Example. To illustrate the use of the tables let us compute the arc of an ellipse. We saw, 177, 7), that the length of an arc starting from the major axis is for a = 1, s = Vl - c2 sin2 4 do = E(o, k), where k is the eccentricity of the ellipse =/a2 - b2 a2 Suppose k= sin 0= then 0 = 30~. For - = 45~ we have from the tables s =.76719. For 6 = 600, s= 1.00755. Thus the length I of an arc between q = 60~ and 4 =45~ is 1=.24036. 396 FUNCTIONS OF A COMPLEX VARIABLE 181. Real Linear Transformations. In the foregoing reduction we have not been concerned whether the transformations employed were real or complex. In many of the applications our elliptic integrals are real, and it is often desirable to use only real transformations. With this in view let us show that: If we set x= P + y (1 l+y we may reduce dx dx V~ P V~ (- a))(X -/)(X -7)(- e) to the form (q- p) dy Va(y2- vl)(Y2- 2) where where a = ~ (q -) (q- 3)(q - ) (q - ), (4 =- _(- )(p-) (5 -- (5 (q - Y)( -) provided a ~- y-8 (7 is # 0. In case D= O, we set x=y + -(a+ 3)= y+ '(y + ) (8 and then 2, 2(y Q 2. (9 a = i -, 71. = (o _ )2,12 = ~(/ —)2. (9 Moreover if the coefficients of t/ie polynomial P are real, the coefficients p, q of the transformation 1) will be real. Also the transformation 8), which is to be employed when D = 0, is real when a, f3 or 7, 8 are real or conjugate imaginaries. Finally, r1 and 72 are real. The verification of these statements is purely algebraic and aside from its length involves no difficulty. We therefore sketch it only. Let us consider first the transformation 1). By direct calculation we find that dx (q- p)dy V~P V'~Q ' THE FUNCTIONS OF LEGENDRE AND JACOBI 397 where Q = P - + (- ) S lP- + ( -!)3M 1 -7 + (q-7)Y - 8 +( - 8)y Let us now choose p and q so that the odd powers of y drop out of Q. This requires thatt (p - a)(q - O) + (p - 3)(q - a) = 0. (p - y)(q- 8) + (p - 8)(q - y) = 0. Let us set = ) - = (P+ ), )= 2 (p-q. We findl 2 _ (a- 7 )(- - )( - )(/ - ) _ M D2 D2 But But = X+,, q =X-FL. To show that p and q are real, we observe that the coefficients of P being by hypothesis real, there are three cases to consider: 1~ Roots all real. Take a < 3 < 7 < s. 2~ Two roots real. Take = a + ib, / = a - ib. 3~ All roots imaginary. Take a, /3 as in 2~ and y = c + id, 8= c- id. Obviously in each case X is real. Thus p, q are real if /u is real and > 0. We now consider the three cases separately. Case 1~ All the factors of M are < 0; hence M> 0, D2 > 0, and thus / is > 0. Case 2~ Here M= (a - y + ib)(a- 8 + ib)(a - y - ib)(a - - ib). Now the product of two conjugate numbers is real and positive, as we saw in 2, 6. Thus 31, being the product of two such pairs, is real and > 0. As D is real, u is real and > 0. Case 3~ This is treated in a precisely similar manner. Thus in every case u is real and > 0. 398 FUNCTIONS OF A COMPLEX VARIABLE Finally, p, q being real, we see that 71, v2 are real. Thus our theorem is proved for the case that - Z 0. The case that D - 0 is at once disposed of, and needs no comment. 182. Real Quadratic Transformations. Let us now show how by a quadratic transformation we can convert dy dy dy -=?2 = -2 dy, ri n2 real (1 Va(y2 - r,)(Y2- V2) /Y into Mdz Mdz (2 V/(1 - 2)(1 - 22) (2 in such a way that if the variable y ranges over some interval I1 for which Y is real and positive, the variable z will range in the interval 3 = (0, 1). Moreover M and k2 will be real and 0 < k2 < 1. There are six cases, as indicated in the table. CASE a 171 i 12 1 + + + 2 4- + 3 + - 4 - + + 5 - - + Case 1. Suppose l< 7. As Y is to be > 0, we must have either 2 < y2 < 7 or y2> If y2 <?7, we take (3 Then 1 MT= k2 =71. (4 If y2 > 72, we take y= h/z. (5 Then M = 1 k2 =1. (6 VaG2 j2 THE FUNCTIONS OF LEGENDRE AND JACOBI 399 Case 2. As Yis to be > 0, we must have y2 > V2. We set y= V2. (7 A 1- Z2 Then M= 1 - k2= -. (8 V\/a(v2 - i) i2 - 71 Case 3. Here Y > 0 for all values of y. Let <. We set (9 y =- - T71 /1 _ (9 Then M=-, k2 = 1-. (10 V/-a?2 72 Case 4. As Yis to be > 0, we must have 1qi < y2 < 2-' We set V/77 47 - ( )z2 (11 r2/ Then i M=- 1, k2=72-1-l. (12 V-av2 172 Case 5. For Yto be >0 we must have 7l,< y2 <' 2~ We set y = VA V1- z2. (13 Then - 1 2, = 2 (14 V —a(]2- 1i) 2- 71 Case 6. As Yis to be >0, this case is impossible. 183. Rectification of the Hyperbola. As an example let us find the length of an arc of the hyperbola 2x2 - a2y2 = a22. -400 FUNCTIONS OF A COMPLEX VARIABLE We have sJ = ( dy ) Y= \ y2 + 2 ) = a2 + b2, 1 r (c2y2 + b4)dy +y2 + b2)(y2 + If we set I 1H. = _ff.. be 82 (1 )4 l =- - '2 =2- we have IlK1<I21 - The integral 1) falls under Case 3 of 182. b2 z c /1 V_ z2 and get We set therefore (2 dy (y2 + 2) 2(y+ b4 dz 1 dz a2 bV(1 - 2)((1 -k2Z) b ' V 2 We note that k is the reciprocal of the eccentricity of the hyperbola. Thus putting 2) in H we get H=b3 dzf = b3J. (I -z2) Z Let us set then z = sin 0b, J= ta b2 ) - ) =It n 4", +(O) - _E(q) 1. Hence finally b2 8 = tan 4. aO +- P(b) - E() ~ (3 in which the modulus k has the value determined by 2 k2 = C2 THE FUNCTIONS OF LEGENDRE AND JACOBI 401 184. Rectification of the Lemniscate. We saw in 177, 9) that the arc s of the lemniscate is given by aCX dx a x dxz a x_____d_____ /V2JoA/ (2 - )(x2- 1) Here 0 < x2 < 1 if the arc falls in the first quadrant. We take '1 = 2, '2 = 1 and use the transformation x = yV2 of Case 1 in 182. We get s= af Y dy, k2_ 1 2J (1 - y2) (1 - k2y2) a For c= r, the arc is just one quadrant of the lemniscate. If we 2 set k = sin 0, we see 0 = 45~. The value of the complete integral Kfor this modular angle is given by Legendre's Tables Traite, vol. 2, p. 327, as K = 1.85407. 185. Elliptic Integral of the First Species. 1. Let us now consider the function u of z defined by z dz (1 o/(1 - Z2)(1- k22) When the modulus k = 0, this function degenerates to z dz Jo 1 - z2 Now the many-valued arc sin function is of much less service to us than the inverse function z = sin u which is one-valued and periodic. So we shall find also here that it is not the manyvalued function u of z defined by 1) which interests us most, but the inverse function z of u. This we shall see presently is one 402 FUNCTIONS OF A COMPLEX VARIABLE valued and doubly periodic. In analogy to the sine function we denote this function by z= sn(u, k) or more shortly by z = sn u. (2 and read s, n of u. In analogy to the cosine function we write VI - z2 = - - sn2u = Cu u. (read c, n of u). If we set z = sin b, as we did in 178, V1 - k2z2 = VI1- k2 sin2 O = A in Legendre's notation. In memory of the A we set /1 - k2z2 = dn u (read d, n of u). Thus with the integral 1) are connected three elliptic functions sn u, en u, dn u. These are the functions of Legendre and more especially of Jacobi. It is these functions which occur in all the older literature of elliptic functions and which to-day are still used by many mathematicians. In the same way as the basis of the Weierstrassian theory of the elliptic functions is the a- function, so the base of the Jacobian theory are the 0 functions, which are integral transcendental functions differing from the a function only by exponential factors. After this slight outlook, let us return to the function u of z defined by 1). The integrand f (Z) = 1 (3 /(1 - z2)(1 - k2z2) is one-valued and analytic except at z=~1, ~j. k Obviously each of these points is a branch point and the two values off permute when z describes a small circle f about one of them. Thus in any region acyclic relative to each of these points u is a one-valued function of z. THE FUNCTIONS OF LEGENDRE AND JACOBI 403 One of the values of u for z = 0 is u = 0; also one of the values off for z = 0 is + 1. Let us start with these value that u acquires, call it uz, when z describes a circuit d about the branch point z= 1. Without changing the value of u we may replace G by the loop o C= (0 a, f, y, 0) as in Fig. 1. The circle C has a radius r as small as we please. Thus ~=J^=J~f+fp. values and find the C 'y 1 a FIG. 1. (4 The first integral on the right differs from K r dz J + V/(1 - z2) (1- k22) (5 by as little as we choose. Here the + sign indicates that the radical has the value + 1 for z = 0. Let us look at the last integral in 4). When z leaves z= 0, f has the value + 1. On reaching a it has a large positive value p. After z describes the circle C and arrives at 7, the two values which f has at this point have interchanged and the value of f is now - p. Thus as z describes the segment 7, 0, the integrand f has the same values as f had in describing the segment 0, a except that its sign is changed. Hence as a and y are really the same point, o dz a dz _ 0 I = f -- = I i — ^== K as r 0. Turning now to the middle integral, we show this is 0. For let us set z - 1 = re:, dz = rieikdc/, r being a constant on C. Then V(1 - z2)(1 - k2z2) = - 1. i(z + 1)(1 - 2z2) = Vz. (Z), where I g(z) > some G, for all z on C. Thus r ~ r2 ir? '. fdz = do, g 404 FUNCTIONS OF A COMPLEX VARIABLE and this is numerically < S7. r2 which 0 asr = O. Thus j= o and we have finally He u = 2 K. Hence' If we start with the value u = 0, f = + 1 at z = 0 and let z describe a circuit about z = 1, u will acquire the value 2 K on returning to z= O. 2. Let us set 1 dz L J +V(1 -z) (1 - C2z2) where the + sign indicates that on starting out from z = 0 the radical has the value + 1. Precisely the same considerations show that if we start out from z = 0 with the value u = 0 and make a circuit 8 about the branch point z =, u acquires the value u=2L on reaching z = 0 again. 3. We now suppose that z starts from z = 0 and makes a cir1 cuit ( which includes both branch points 1, 1. We start with the values u = 0,f= + 1 and ask what value u has acquired on reaching z = 0 again. If we denote it by u, we have fdzfdz +jr d(7 u -=+ -- (7 J+V a + V -/ - For in the first place Q may be replaced by - } *.? as in Fig. 2, since the region lying between these two curves contains no singular point of f(z). Secondly, in the last integral in 7) we have the - sign because when z reaches z = 0 after the circuit 9 about z = 1, FIG. 2. the radical has the value - 1, and it is with this value, therefore, 1 that we start out to make the circuit V about z =. Thus k u= 2 K-2 L. (8 THE FUNCTIONS OF LEGENDRE AND JACOBI 405 Since we can deform any path of integration Ckr, without changing the value of the integral u,, ) provided that in so doing we do not pass a sin- gular point of the integrandf, we see that the circuit t ~ can be replaced by that in Fig. 3, c*C) which we will call W9N. Thus FIG. 3. urf b dZ f +rJ ra dz C Ja +k - \/ Now as before S=0, r=0, pa dz +dz% Jb - / Ja + / If we let the radius r of the two circles C1, CQ, converge to 0, the last integral converges to a value denoted by Jacobi by dz iK' = r dz (10 + A/(1 - Z2)(1 - k2z2) These results put in 9) give u = 2 iK. (11 To explain why the integral 10) is denoted by iK' instead of by K' we remark that in practice the modulus k is such that k2 is real and < 1. Then (1 - z2)(1 - k2Z2) is real and negative in the segment (1, ) and thus the integral 10) is purely imaginary. The notation iK' in 10) is therefore quite appropriate. Equating the two values of u in 8), 11) shows that 2 K-2 L = 2 iK'. (12 4. Suppose next that z makes -1, a circuit @ as in Fig. 4, which (o0 includes the two branch points FIG. 4 z = ~ 1. If we start with u = 0, f= + 1, let; be the value which 406 46 FUNCTIONS OF A COMPLEX VARIABLE u acquires after (ST. Then ~~==+~~1 fa ffO0 dz ~ p > + f+ j = d$ As bef ore =- 0 =0 A/ = -\- 'K,as v -I-1. Moreover since f (- z) =f (z), J 8 dz a~ dz 0 dz-v —i0 =K as 8 % - 1. Thus, U=4K. 5. Let us now ask -what values u can acquire when beginning with the initial values z = 0, z =+ 1, 'a = 0, z describes any path ~3not passing through a branch point, and ending at some point z. Let U be its value along the path Oz a's in Fig. 5. If $3 is a loop2 about z = 1 followed by Oz, we have FiG. 5. f z dz J*'>+fzd In the last integral the radical has the - sign as indicated, since on returning to z = 0 after 2, f has acquired the value - 1, instead of + 1. Thus,,I )V i 41 T /-1 9 'MW -j LA -C L. 1, _Lj If $3 is a ioop 2 about + 1 and - 1, followed by the path Oz, u has the value ~ ~ d THE FUNCTIONS OF LEGENDRE AND JACOBI 407 For on returning to z = 0 after the loop ~, the radical has changed its sign once in going about z = 1 and once about z = - 1. It therefore reaches z = 0 at the end of ~ with the value + 1. Thus u=4K+ U. If 3 is a loop ~ about z = 1 and z= -, followed by Oz, we have A= r dz+ +r d JU+A / Jo +v/ = 2iK + U. If $ is a loop ~ about z = 1, followed by a loop 9 about both z = 1 and z =- 1 and finishing with Oz, we have J = Jd- Jo -dz =2 K- 4K- U. =2K-U-4K. We see this differs by - 4 K from the value of u in 13). By choosing various paths we find that all the values which u can acquire at a point z are given by u = U+ m 4 K+mn2 iK and lu =2K- U+m4K+m 2 iK', (14 where m, m' are positive or negative integers or 0. Thus we may state: The analytic function u defined by 1) is an infinite many-valued z function of z. If U is one value which u has at a point z, all the other values which u has at this point are represented by 14). 6. We have seen that u remains finite for finite z. Let us see what value u has as z' oo along some line I as in Fig. 6. The calculation is easily effected FIG. 6. 408 FUNCTIONS OF A COMPLEX VARIABLE if we note that Oz0zl0 is a circuit S which embraces the two branch points z = 1, -. As this is the same circuit as considered in 3 we have =2iK'. (15 But But f cl.dz + +fi (16 The second integral J *0 as R oo. For let us set Re z = Rei so that on the circle dz Rieid = izdM. Hence rt i _z e o V/( 1- (Z2 1 - k2z2) If R is taken sufficiently large, z 1/(1 - 2)(1 -_ k) <e for all z on (. Thus <e <e7r0 as R=o'o. Now, when z leaves z= 0, the integrand f(z) has the value + 1. After it describes the circuit R it returns to z = 0 with the same value, since as far as the end value off is concerned, the circuit S is equivalent to a circuit about each branch point - and 1. For each of these, f changes its sign. Thus it changes sign twice and so returns to z = 0 with the value + 1. Thus f()= I V/(1 - 2) (1 - k2z2) has the same value along Oz as along Oz1. Hence 2=o. == -- _ _dz -x) -,) as zoo. Jo o V(1- z2) (1 -c22) THE FUNCTIONS OF LEGENDRE AND JACOBI 409 This with 15), 16) gives iK +i ~V/(1-Z2) (1- k2Z2) (17 Thus the different values which u has for z = oo are iK' ~ 2 mK+ 2 m'iK'. (18 186. Inversion. 1. In the foregoing section we have considered the function u of z defined by u Z - dz p ( U, = dr ~ llf(z)dz. ( V(1 - Z2)(1 - k2z2) Let us now look at the inverse function z of u., This we said is denoted by z = sn(u). (2 Since the integrand f(z) is a one-valued analytic function about z = 0, u can be developed in a power series. It may be obtained thus: Since 1 1 f= (1 - Z2)-i(1 - 2CZ2)-z2f we may develop each factor (1 - z2)4 and (1 - k22)by the Binomial Theorem and get two power series in z. These two series when multiplied together give f= 1(1 + k-2)Z2 + -(3 + 2 2 k 3 kZ4 +... Integrating, we get U = z~ 1 (I + k2)Zl+ 3~2(3 + 2 2 + 3 k4)5 +.. (3 which is valid for IzI < either 1 or If we invert this series by 125, we get Z = u i (I + k2) U3 + (I+14k2 + k4)u5-... (4 which is valid in some circle c vhose center is u = 0. 410 FUNCTIONS OF A COMPLEX VARIABLE By analytic continuation we may extend the function so as to define it for values of u outside c. It can be shown in various ways that the analytic function defined by 1) is an elliptic function, having 4 K, 2 iK' as a primitive pail of periods. Its zeros are 0 and its poles are - iK mod 2 K, 2 iK', each of order 1. It was by inverting the integral 1) that Legendre, Abel, and Jacobi were led to consider the elliptic functions. We have seen that Legendre set z = sin (. From this point of view it was at once evident that z admits the period 4 K. But Legendre, using for the most part real variables, failed to notice that z possesses also an imaginary period, that in fact z is a double periodic function. The discovery of this property by Abel about 1825 enabled him and Jacobi to develop the theory of elliptic functions in the next few years far more than their predecessors had done in half a century. Because of the fundamental nature of the double periodicity of the elliptic functions, we began our treatment of these functions in the preceding chapter from this point of view. At the same time the older theory is so interwoven in the modern that we have not hesitated in treating the functions of Legendre and Jacobi to start from the elliptic integrals and work up to the elliptic functions. This point of view is also useful in those applications which lead at once to an an elliptic integral. 2. To show that the inverse function z = sn u is in fact an elliptic function, let us actually write down such a function which satisfies the differential equation dz = V(1 - Z2)(1 -k22), (5 du the radical having the plus sign for z = 0. The constant of integration we will determine by taking z = 0 for u = 0. There cannot be two such analytic functions. For both having the same derivative 5) about u = 0, they can differ only by an additive constant and as z = 0 for u = 0, this constant must be 0. THE FUNCTIONS OF LEGENDRE AND JACOBI 411 3. To this end we consider the function q of 176. We saw that q is an elliptic function which satisfies the differential equation dq - V(1 - (e - e2)2)(1 - (e- e2)q2). (6 du Let us choose e,, e., e3 so that k2 = (7 el - e and set =Ve Z 1 - e2 Q~ Then z satisfies the equation d Vei - e2V(1 z2)(1 1c2z2). due To remove the factor on the right we set V U, = l Vei - e2 which gives dz _ ~V(1 -XZ2)(1I-k2Z2). (8 dvv Thus Z= Ve, - e -O2(,, (9 2 0021 \Ve1 - e2 is a solution of 5). As z = 0 for v = 0, the function 9) must be 4), replacing u by v, or sn v. We saw in 176 that a-02(u) is a homogeneous function of z, o1, 02 of degree 1. We can thus write 9) Z = o02(V, 0-V1 - e2, A2Vei- e2), (10 whose periods are 4K= 4olVe1-e, 2 2 iK'-= 2 w2Ve1 - e2, (11 whose zeros are = 0, and whose poles are = iK', mod 2 K, 2 iii'. 4. Having assured ourselves that the inverse function z = sn u is a one-valued function throughout the whole plane, z= o ineluded, let us see how the study of the integral u dz (12.o -\(1 - Z2)(1i - 1ez22) 412 FUNCTIONS OF A COMPLEX VARIABLE shows us: 1~ z is a double periodic function with the periods 4 4j1 - dz 0 /(1 -z2)(l1- _ k2z2) 2iK' =2 dz (13 - ( - z2)(1 - kz2) 2~ z has as simple zeros the points = 0, 3~ and as simple poles the points = iK' both mod 2 K, 2 iK'. This will be an a posteriori verification, but it is very instructive. Precisely this path was followed by Cauchy and his successors. 5. To begin, the relations 185, 14) show that snu admits 4 K and 2 iK' as periods, while the reasoning of 185, 6 shows that sn u remains finite except in the vicinity of the points c =iK' + 2 mKK+ 2 m'iK'. (14 Let us show that these are poles of the first order for sn u. We write 12) u + =c + w, (15 where c has the value given in 14). We wish to see how w behaves about z = oo. If we set = -, it becomes t Wdt dt - lt ~ t2+ Ct4+ d Jo V(C2 t2)(1 -t)) J k =1t+ llt3 0+ k 3 k This shows, using 125, that t considered as a function of w is a one-valued analytic function. t=w(- k+ b + b2w2...). (16 Hence 1 1 1, z= -e k + dl(u-e)+d(u-oc)+..., (17 at - order. and u = c is a pole of the first order. THE FUNCTIONS OF LEGENDRE AND JACOBI 413 6. Let us show that the points u 0 mod 2 K, 2 iK' at which Sfl U = 0 are simple zeros. In the vicinity of z = 0 we have < + aiz + a2Z2 + -2 kC2z2) Hence * 1( U C +- ciz + C2X2 + el1 ~ (18 where c is given in 14). The relation 18) shows that Z = (u - C) eo ) el(,u - c) + e2(m - 0)2+ (19 and thus U = c is a simple zero. 7. Let us show that z is a one-valued analytic function about I the point u for which z = ~ 1, ~ - These latter points are points in whose vicinity the integrand of 12) becomes infinite. ILet nsa set Le sst ZI = t2 then dz = 2 tdt, Also A/ ( - Z) ( - lr2Z2) Z \ - i \/(Z 1) (/02X2 - 1) t ~ a. + alt + ~~i Thus 1 1 vt du= 2(o +bit+..)dt, u = co +eClt2 + C2t3 + el, C1#, (20 where co is one of the values which u has for t = 0, that is, for z = 1. These values are all ~= K. The relation 20) defines t as a one-valued function of U in the vicinity of u = co. As Z = 1 + t2 this gives z - 1 = d(U - c) + d2(U-C)2. (21 Thus z is a one-valued function of U about ut = co, 187. The sn, cn and dn Functions. 1. We have now two definitions of z = sn u; one as the inverse of U, fZ - clz, (1 A/ 1_ Z2 (1- k2Z2) 414 FUNCTIONS OF A COMPLEX VARIABLE the other as z==snu=V\e -e - (vw1,o2),(2 Vei - e2 Vei - e -\VP(v)- e2( xvhere k2 - (4 i~~2 = ~3L-~~4~. (4 el - e2 The radicals A/I - Z2 VI - k22X"j although two-valued functions of z, are one-valued functions of u. For from 3) we have i, 2 = _P - el 0 o-2 = ( V) p3-c o~ a(v) 1\/ Hence en U VI - 2 o1v, n(O)= 1. ( 0,2(V) Similarly _Z 0, dn u - VI I_2 k (v), dn(O)=1. (6 J2(V) If we set = sin the integral 1) becomes do do. U ~~~~~~~~~~~~(7 U Vj I -k2 Sifl2 Then snu,cnu dnu (8 become in Legendre's notation Sill cos,Cos (9 as already remarked. 2. We wish now to deduce a number of properties of the functions 8) from the definition of sn u as the inverse of the integral 1), and from the definition of c u, dn u by the relations Cnu=Vi-\S/2-, 2 U dnu= V l _ -k2 sn 2 U. (10 This is the older point of view and will accustom the reader to the use of integrals as an instrument of research. Later we propose to study these functions from the standpoint of the,P functions. THE FUNCTIONS OF LEGENDRE AND JACOBI 415 Let us first find the periods of en u. We take up the considerations of 185. Allowing z to describe a circuit about z = 1, u goes over into u + 2 K, while /l - z2 changes its sign. Thus cn(u + 2 K) = - en u. Hence en u has 4 K as a period. If we let z describe a circuit which contains both z = 1 and -, u goes over into u + 2 iK', while Vl - Z2 changes sign. Hence cn(u + 2 iK') = - en u. Thus enu has the period 2K+ 2 iK'. In a similar manner we may reason on dn u. Thus we can draw up the table: PERIODS sn u 4K, 2 iK cn uz 4 K, 2 K+ 2 iK' dntu 2 I, 4 iK' The poles of cn u, dn u are obviously the same as those of sn u, as seen from 10), and they are of order 1. From 10) we see that en u =0 when sn2 u =. But sn K= 1. Hence by 185, 10) all the points u at which snu = 1 are given by K~ 4mK+- 2m'iK'. As sn2u is even, the points at which sn2u = 1 are K+ 2 mK+ 2 m'iK'. In a similar manner we may reason on dn u. We may thus draw up the table: ZEROS POLES sn u 2 mK + 2 m'iK 2 mK + (2 ml + 1)iK' cn u (2 m + 1)K + 2 'iK' 2 mK + (2 m' + l)iK' dn u (2 m + 1)K + (2 n' + l)iK' 2 mK + (2 m' + 1)iK' Since sn u, en u, dn u have each two poles only in a parallelogram formed of the periods given in the first table, these periods are primitive pairs of periods. 416 FUNCTIONS OF A COMPLEX VARIABLE 3. Let z move from 0 to 1, at which point u = K. Hence snK=l, enK= O dn K= /- k2= k, (11 if with Legendre we set k' = 1i - 2(12 taking the positive determination of the radical for k= 0. This is called the complementary modulus. The introduction of ki is quite natural from Legendre's point of view. For as we have seen, he set k = sin 0; hence cos 0 would be k'. Next let z move from 0 to 1, at which point u =K+iK'. Then sn(K+ iK') = I T 1 ik cn(K+ iK')= \1 - = ik' (13 k2 k dn(K+ iK') = 0. To determine the sign in the second equation let us take k real positive and < 1. We let z move over the path 0 aze- in the figure. When z = 0, /1- z2 has the value + 1. Thus at z = a, V/1- z2 ____a c is real and positive. We set 1 1 - z re=( —). Then V — z=re and take /1 + z with positive sign. Then V/1- z2 is real and positive at a, as it must be; but at c, V1 - z becomes 7ri r2e = - ir, which is negative imaginary. Thus at z = -, the radical is negative imaginary. Hence we must take the - sign in 13). Thus from the foregoing we may write down the following table: THE FUNCTIONS OF LEGENDRE AND JACOBI 417 0 K 2K 3K 0 1 0 -1 0 1 0 -1 0 1 kt 1 k 1//k- 1/c iK' co - ik'/lk o ik'/k 0 0 0 1 0 -1 2iiK -1 0 1 0 - 1 - k' - 1 - 1/k - 1/k 3 iK oo ik'/k oO - ik//k/ 0 0 In each square the values of sn u, cn u, dnu are given in order for K = mK+ m'iK', m' = 0, 1, 2, 3. In the table m refers to the columns and m' to the rows. Thus for u = 3 K+ iKt' snu, e u, dn dU 1 il' - k ' ' 188. The Addition Forue 1. If we set 188. The Addition Formulae. 1. If we set x dx U= I x rY dy xy dz V= y, W= J, J y z the addition theorem log = log + log y states that w = U + 'v, or that ax ~ y 7 xy Jl.Jl Jl (1 2. Let us set u x dx J1o I - X2 (2 418 4 FUNCTIONS OF A COMPLEX VARIABLE, The inverse function defined by this relation is x = sin u. The addition theorem for the sine function is sin w = sin (u + v) = sin u cos v + cos u sin v. (3 Let ns therefore set Y dy z dz VI - 9 2 -\/1 Z2 The relation 3) states that z = xV/ - 9y2 + yV -X2, (4 and hence + X +\/J-y2+y\/i-X2 3. Let us set u X dx xG dxax u ff (6 For k = 0 this integral reduces to 2). It occurred to Euler that the integrals 6) might have an addition theorem, that is, the sum of two integrals might be expressed as a single integral fZ in which z is some algebraic function of x, y as in 4). This relation he found to exist; it is in fact A/ - 2) (1 - /e~2) +,J \(1 l -X2) (I - k222). Z= ( -' (7 I - k2X2y2 Thus if we set, similar to 2 ~=~~ " i~dy -\/( (- y2)(l -k12y2) w = U + fVd_ 1~2 Z2) (1 22 THE FUNCTIONS OF LEGENDRE AND JACOBI 419 we may write 7) sn(u + v) = snu cn v dn v + sn v cnu dn u 1 - k2sn2 2u sW V s~n u. sni'v + sn, v snf'u (8 1- lc2sn2 U sn2 V To obtain the relation 7) we notice that the differential equation dx dy - + 0~ (9 Ax Ay admits as integral x dx ~ (10 f x Ay If this integral is determined by the condition that when x = 0, y shall have the value z, the relation 10) shows that the constant C of integration must have the value (11 The value of C may also be obtained, as Darboux has shown, as follows: Let us introduce a new variable defined by ds = dx(12 Ax Then 9) becomes dy 0 ds ~ ---- I Ay or dx Ax, ds A (13 ds ds Hence d~x - 2 k2x3 - (1 + k2)x, ds2 d2y = 2 k22y3 - (1 + 1c2)y, ds2 d2x-x d2x I'- -X — 2 c2Xy (x2 - y2)1 dS2 dSP 9' 2~s X2ds/ - (1- k2x2y2) (X2-_y2). 420 FUNCTIONS OF A COMPLEX VARIABLE Thus d2x d2y ds2 - ds2 2dX\2:2dy\2 "dsJ \dsj If we multiply this by Y, 2 k2xy 1 - 2X2y2 dx dy ds ds we get or d2X d2y Yd 2 - xd 2s ds2 dS2 dx dy ds ds 2 k2Xy dx + dy \- ds ds 1 - k2x2y2 d log(yd -xzdj= d) log(l - 22y2). ds dsl "4~ Integrating, we get log -d x-d = log( - k2x2y2)+ log e or using 13), we have yAx + xAy = c(1 -- k2x2y2). (14 This is an integral of 9). To determine c we must have y = z for x = 0. This in 14) gives c = z. Hence by 14) yAx + xAy 1 - k2Xy2 ' which is 7). 4. By means of simple but rather lengthy reductions we get en u en v - sn u dn u sn v dn v en(u + v) - - 1 - k2sn2 sn2 v dn u dn v - 2sn u en u sn v en v dn(u + v) = - u 1 - k2sn2 u sn' v (15 (16 THE FUNCTIONS OF LEGENDRE AND JACOBI 421 From 8), 15), 16) we may also derive: sn(u +-v)sn(u-v) = -- sn2n v (17 1n2s-n2v cn(u + v)sn(u - v) = n v - sn2 u dn2 v (18 1 - k2sn2 u sn2 v' den2 v - 2sn2 u dn2 v dn(u + v)dn(u - v) = 1 - k2sn2 u mn (19 1 - k2sn2 u sn2 v 189. Differential Equation for K and K'. 1. We saw that K=J df b ^ -.(1 eo /Vl - k2sin220 When I k2 < 1 we can develop the integrand, getting (1 - k2 sin2 )-i = 1 + l 3 '5 (2n - 1)k2n sin2n". (2 = 2. 4.6 ~ 2 n k (2 But from the calculus we know that C2sin~2 ~fdcf 1.3.5... (2 n-1) r7 Jo 2.4.6...2n 2 Thus integrating 2) termwise gives K=v k2 + + k+.- + (3 2 2\' 2.4} valid for I k2 < 1. Comparing this with 103, 5) we see that IK-2 I(?J1 1' 2) (4 This relation, holding for I k21 < 1, must hold throughout its analytic continuation. 2. The other period 2 iK' we saw in 186, 13) is defined by 1. 1 2 iK = 2: d = 2 c dx J V(1 - x2)(1 - 22) J (2 -1)(1- 22) 422 FUNCTIONS OF A COMPLEX VARIABLE Let us set 1 X __ k'2-k1-12. / - k '22U2 Then to x = 1, x = I correspond u = 0 and u = 1. We thus get k K ' du (5 -\/(I u2)(1 -leC12u2) Comparing with 4) we see that Al =7r 1 1, 112) (6 K'=I2 2( 2 3. We saw in 103, 4 that P(ac3,yz) satisfies the differential equation z(z-1)P"+~ Qa~/3+1)z-y~F'~cLI3F=0. 7 If we take 8 = the relation 4) shows that K satisfies the equation. z(zl~d~y~d(2zYl)dY~yO z-1=2 (8 dz2 dz 4 If we replace z by 1 - Z = k'2 this equation is not changed. This shows that not only K but K' as given in 6) is a solution of 8). Thus both K and K' are integrals of 8). We shall return to this subject later. CHAPTER XII THE THETA FUNCTIONS 190. Historical. These functions were first considered by Abel. We have seen, 186, 4), that the inversion of -a'd dz U -= - I _ _ _ _ _ _ _ _ _ _ _ _ _ Jo V (- z2)(1 - /2Z2) leads to an infinite series = - (1 + k2)u3 (1 which gives the value of sn u within a circle passing through iK', which is the nearest pole of z. In a similar manner the inversion of z dz u i1 + 2 leads to a series z = tan u = u + I- U3 + -5 5 + which gives the value of tan u within a circle passing through r, its nearest pole. But we know that there exists an analytic expression 42 \z ( 4z2 -- = 1,2,... (2 n (K - (2 n - 1)27r which is valid for the entire domain of definition of the tangent function. It occurred to Abel that a similar expression might be obtained for the sn function. This is the way he found it. From the addition theorem we may express snnu in terms of sn u, cnu, dnu just as in trigonometry the addition theoreml of the sine function enables us to express sin nu in terms of sin u, 423 424 FUNCTIONS OF A COMPLEX VARIABLE cos u. If in the expression for sn nu, n being an odd integer, we replace u by - we get n f X sn - II I- n _ n 2mK+ 2 m'iK') u [ n J ~ sn u = nsn- (3 n ( u ] sn - 2 mK+ (2 m'+ 1iK) I[ ^ n n This expression is entirely analogous to that obtained for the sine function in 136, 6). From 1) we have u Sn - rn __ nt (2 mK+ 2 m'iK) 2 mK+ 2 rmniK' etc. Abel therefore concluded that I1 2 mK + 2 miK' (4 snu=u i - 2 mK+ (2 mr + 1)iK' This passage to the limit is not rigorous and the infinite products which enter 4) are not even convergent in the sense that we have given this term in 128, 2. Let us therefore regard the relation 4) merely as a stepping stone to get infinite products which do converge; these will be the 8's. To this end we write the products which figure in 4) as double iterated products, and for brevity we will set iK! u =2Kv, 7= K (5 K Then T sn 2Kv = (v) (6 where T1=2 2 Kv I(- 1W (1 — (7 nn a nn ' i \ n + m v THE THETA FUNCTIONS 425 and a similar expression for To. Here n=~l, ~ 2,.. m=0, I1, ~ 2...; rm'=~l, ~2,... Now n( i) sin7v II - - =sin rr, rnf] _ v ) Sil 7r(n'7- - v) n mm +- m' r sin mrn-rr Hence T (v) = 2 Kv sin 7v sin 7r(mT - v) 7rV i' sin m'Trr -= 2 Ksill 7rv ~f sin 7r(nT - v) sin 7r(nr + v) 7r,I Sill nZ7T sin n7rr - 2 Ksin 7rv? r (T -i (eiT-t (nr- v) - e -i(+(nv) _- e -7i(n7T+?) '7 (eirinT _ e-7rinT)2 = K Wi H e2inr (eriv e-27iv) + e4inT 7rV (e - ~~"+ ' + T (1 -_ e2rinT)2 Thus finally ( = - n (1 - 2 qcr- os 2 W+V~ q4I ) T1((v)- qsin,rrv 1(, 2. (8 w-^ - 11^ ^ ^^-"(1 -- q"")2 where _ ' q=einir e K (9 Here the infinite products in the numerator and denominator of 8) are absolutely convergent when q I < 1. This last is no essential restriction because we would merely need to interchange K, K' in 9) to get a q which is numerically < 1. A similar reduction of To(v) leads to T0(V) = - 2 q 2Sn+los 2 V ~ q4n+2), n = 1,2... (10 H1(1 - q2n+ )2 Similarly, we can express cn 2 Kv, dn 2 Kv as the quotient products whose numerators Ts, T3 are slightly different from T1, and whose denominators are T as before. Thus the development of sn, en, dn into infinite products leads us to four integral transcendental functions To. T1, T2, T3, which play the same role in the theory of Jacobi that the four functions Ga, r cr2, 0-3 play in the theory of Weierstrass. It is one of the 426 FUNCTIONS OF A COMPLEX VARIABLE immortal achievements of Abel to have introduced these functions into analysis. Although his method of deduction is not rigorous, we can and will in fact show that when T1, T as given in 8), 10) are put in 6), the resulting function satisfies the differential equation dz (- Z2)(l- 22), _z = V(1 - z2 ) (1 - k2Z2), du which defines the sn function. Here, as often happens, a method which is not rigorous, and cannot perhaps be made rigorous, leads to results whose correctness can be established a posteriori very simply. Such methods have a great heuristic value and are not to be despised by the pioneer. 191. The it's as Infinite Products. 1. It is convenient to replace the T's introduced in the last article by four new functions which differ from them by constant factors. With Jacobi we set q = ec'i. (1 #1(v) = 2 q sin 7rv 1l(1 - q2)(l- 2 q2n cos 2 7rv + q4n), 1 2(v) = 2 q cos 7rv II(1 - q2n)(1 + 2 q2n cos 2 rv + q9n), 1 (2 3((v) = 1(1 - q2n)(l + 2 q2l-1 cos 2 rv + q4n-2), #o0(V)= II(1 - q2n)(1 - 2 q2n-1 cos 2 7rv + q4n-2). We do not need to regard them in any way related to the elliptic functions, but simply as integral transcendental functions whose properties are to be investigated. The only restriction we make is that I q < 1 (3 in order that the products 2) converge absolutely. We first show that these four functions are very closely related to each other by the formulae: 3(v + 1) =9o(v), 3( + 2)= e-vq-;a2(v) (4 3(v + _ ) ie-_ivq-l(v). THE THETA FUNCTIONS 427 To prove the first relation we have, setting Q=1n(1 - q'), 3(v + -) = Qn( + 2 q2n-1 cos 2 w+(v + 1) + q4n2) = QH(1 - 2 q2n-1 cos 2 w7v + q4"-2) = 0o(VU) To prove the second relation in 4) we have 3()-== QJ((1 + q2n-le2iv)(l + q2n-le-2iriv). Hence 2)= 2Q (1 + qjnle2i(v'))(l + q-le-2e7(v)) = Q(1 + e-27riv)I(l + q2ne2riv')(1 + q'2ne-2riv) = Qe-ivq -4q. 2 cos 7rvll(1 + 2 q2n cos 2 7rv + q4n) -i =q 4e-rivta2(v). 2. In a similar manner we may prove the following relations: 1(v + 2) = 42(),?Y v + = iq-e-.ritO(v), (5 ( +i) = q-e-_ 3( #2( V + 21-)_ ~2(v + 2) = q4e(v), 2 V + -1 iq-4e-7riv 0(v) 2.1 (V + 21) =3((v), o v + tqe-r= iV?(v), (7 O(v 1 +o) -= q4eiv2(v). 428 FUNCTIONS OF A COMPLEX VARIABLE 3. By repeated application of the foregoing or directly we find: Y1(v + 1) =- - (v), z?2(v + 1) = -82(), 3(V + 1) = 3(V), (8 o0(V + 1) = Wo0(v). 1i(V + co)= - q-le-27rivl(V), 2(V + wO) = q-le-27riv2(v), s3(V + o) = q-le-iv (v), #0(v + o))=- q-le-27riVoo(v). 192. The t's as Infinite Series. The relation 191, 8) X(v + 1) = 03(V) shows that #3(v) admits the period 1, and can therefore be developed in Fourier's Series by 126. Thus 3(v) = E ane.inv (1 -oo To determine the coefficients an we use 191, 9) s3(V + o) = q-e-27riVs3(v). (2 For putting 1) in 2) gives 3(v + c) = Sane27rin(v+) = a nq2ne27rinv - q- 2-2rivane2rinv = Yanq-le27ri(n-l)v Comparing the coefficients of e2"inv gives aq2n = anlq-1 or 2 ~gor -a qn2a o Thus Th3() = ao, q2 e2,nv — o = a0l + qn2(e2&inv +e-2,rinv) = ao(1 + 2; q' cos 2 7rnv). 1 THE THETA FUNCTIONS 429 To determine ao set q = 0. Then p3(v) =1. Hence a =1. In this way we get YW(v)= 1 + 2~q cos 27nnv, 1?2(V)= 2, q (2+i)' cos (2 n + l)wrv, Y1,(v)= 21 i(- 1)'q( +2' sin (2 n + 1)Trv, (3 0 NSL(v)= 1 + 2 (- 1)nqn' cos 27nv. The representation of the Y's as infinite series is due to Jacobi (1825). As they converge with extreme rapidity for the values of q ordinarily employed, they are of immense value in all questions of numerical calculations of the elliptic functions, as we shall see. 193. The Zeros of the Y's. 1. The infinite products 191, 2), being convergent, cannot vanish unless one of their factors vanish. From this we have at once that the zeros of the Y's are as follows: 61(v); V =rn + no, 2(v); v=(m+ ')no, t~~2(0 2 O (1 r9~(v>; V = (m + 1) + (n + B/w, 'o(vU); v=m + (n + '-)w, where m,n=O, ~1, ~ 2,... To illustrate the proof, let us find the zeros of NW(v). We have 1 + 2 q2nl cos 2 7rv + (4"n- = (1 + q2n-le2iiv)(1 + q2n-e1-27riV Let us find the values of v for which 1 + q2n-le2wTiv - 0. This gives, since q = e~ii 2iv 1 1 e27iv q~n-1 esii(2n-1l)w Hence 2 7riv = log (- 1)-7 i(2 n - 1)o. As log (-l1)=wri + 2 sri, v -(mi + ')+(n + -)c m, n any integers, as required by 1). 430 FUNCTIONS OF A COMPLEX VARIABLE 2. Another method of getting the zeros of the O's is the following, it uses the infinite series 192, 3). From the series for?i1(v) we see that it vanishes at the points m+nw, since each term of the series vanishes. To show that the 4's vanish at no a/+other points let us consider /1(v) for example. We take a parallelogram P P as il the figure not passing through one of the points m + n. Then by a, +2 124, 8), 1 jf d log,(v) = M- N. (2 2 7riz As Al1(v) is nowhere infinite in the finite part of the plane, N= O. Now f=ff+f++/. (3 P 12 723 34 41 From 191, 8) d log l(v + 1) _ d log A1(v) dv dv while from 191, 9) d log a(v + c) = _ 2 ri d log (v dv dv Thus f f JUx~ / ^4a+l pa+l I =- 2 7ri dv + d log Vl(v) = -27ri+ +' These in 3) give v J = 2 ~i. Putting this in 2) gives M= 1. Thus 81(v) has just one zero and this of order 1 in the parallelogram P. The first equations of 191, 8), 9) show that?l(v) has a zero of order 1 in each of the congruent parallelograms, mod 1, o. From 191, 5) we may now find the zeros of the other thetas. THE THETA FUNCTIONS 431 194. The Y's with Zero Argument. 1. If in the infinite series and products of the i's given in 191, 2) and 192, 3) we set v = 0, we get 0 = 1 + 2 (- )nqn2 = ( I(l- qn)I(l -- q2n-1)2. (1 2 = 22 q(+) =-2 q~(1 - q2n)HII( + q2n)2. (2 0 3 = 1 + 2qn2 = n(1 - q2n)(1 + q2n-1)2. (3 1 Here, as is customary, we set 0o instead of 0o (0), etc. As 1 = 0, let us take the derivative of?l(v) = 2 q4 sin 7rvTI(1 - q2)II(l - 2 q2n cos 2 rv + q4f) = sin 7rv ~ +(v). Then S1{(v) = r cos 7rv. (v) + sin 7rv * /(v). Hence (0) = r since sin 7rv = 0 for v =0. Thus = 2r(- 1)n(2 n + 1)q(+i)2' 2 7rqIH(1 - _q2)3. (4 0 These four functions 03, 0, 82, t3 considered as functions of the complex variable q are of extraordinary interest from the standpoint of the theory of functions. As we have already seen, for these series or products to converge absolutely it is necessary that I q < 1. There is nothing remarkable about this. The series 2 3 log (1 + Z) = - + - does not converge absolutely unless I z i < 1; but by analytic continuation it is possible to extend the function represented by this series beyond the unit circle. The thing which is so remarkable about the series 1), 2), 3), 4) is that it is quite impossible to extend them by analytic continuation beyond the unit circle in the q-plane for the reason that the functions defined by them become infinite in the vicinity of every point on this circle. Here, then, is a class of functions utterly different from all the functions of elementary analysis. These latter are defined for the whole plane, isolated points excepted. The four functions 432 FUNCTIONS OF A COMPLEX VARIABLE #0', 1' 2', O' are, on the other hand, defined only for points which lie within the unit circle. If we replace q by its value q = e^iu and consider these functions as functions of w, we find that they behave in a most remarkable manner when co is replaced by aco + b (5 c) + d' a, b, c, d being integers such that ad - be = 1. The 0o3 transformations 5) form a group G, and by the aid of the functions o0, 1, 2, 81 we can construct functions which remain unaltered for G or for some subgroup of G. One of the simplest of these functions is the modulus k2 of the sn function. We shall show directly that k2 = 2 3 This function remains unaltered by the subgroup of G characterized by a _, b6=0, c=-O, d=l, mod 2. (6 For this reason all functions of this class, that is, one-valued analytic functions which remain unaltered for the substitutions of G or for one of its subgroups, have received the name of elliptic modular functions. Their theory has already reached imposing proportions, yet the modular functions forn only a small part of a still vaster class of functions called the automorphic functions. For these functions the coefficients in 5) are not restricted to integers. 2. Returning now to our Y's with zero arguments, let us prove a relation of great importance: 1 = M7Y023. (7 From 1), 2), 3), we have 0#a2#3 = 2 qHn (1 - 2n)3(1 + q2)2(l + q2n-1)2(1 - 22-1)2 Now from 135, 1), I (1 + q2n)(1 + 2n-1) ( - q2n-1) = 1. THE THETA FUNCTIONS 433 Thus 20 2 q- H (1 - q1 ) 6,8,cYJ - ^2 qH (1 - q1 ( ^ Comparing this with 4) gives 7). 195. Definition of sn, cn, dn as a Quotients. 1. With the four a functions we can form 12 quotients #7,(v) r, s = 0, 1, 2, 3, six of which are reciprocals of the other six. Of the remaining six, three are quotients of the other three. Let us therefore consider 1(v) 2() 3() (1 ~l(V) fo (V) fos(V) 0(v) ' w0(v) ' 0(v) These admit as periods, respectively, 2,co; 2, 1 +o; 1, 2o. (2 The poles of these functions are the zeros of the denominators. In a parallelogram constructed on the above periods, each of the above quotients has two simple poles respectively. They are thus elliptic functions of order 2, and the above parallelograms are primitive. Without assuming any knowledge whatever of the functions sn, en, dn, let us study the three elliptic functions 1). We will begin with qL0(V) and find the differential equation which this satisfies. To this end we use 165, 14. Here the poles are (0) f pi=-, P2 — +' Hence the zeros of dq are dv -1+) 1 1 + 1 +o)+-, mod 2,. 2 2 21^ ' 2~l ' 2 ) ' 2 mod 2' To find the values of q(v) at these points let us set C = ~/(3 ^3 J(3 v9.a 434 FUNCTIONS OF A COMPLEX VARIABLE and use the relations in 191. Then 2 =o, A-k (/1+O) _=- 1( '2 J2 - l -/ ( + 2 )q 1 ( 12 = /./ q(2+n') 2( )= - /k. q(1 + + -)= - q(1)= -. Thus 2 T (dq)= C(1= - kq2)(k- q2). Let us set 1 1 ( s = —= 1 =A 1)(4 V/k VA o0(v) We get ds = c/(l- s2) (1- k2s2). (5 dv To determine c we have from 4) ds 1 Qo(V) (v)- ) l(V)Q (v) dv Vk(v) For v = 0, this gives (ds\ = 1 Y \dvjv-o /k o The right side of 5) reduces to e for v = 0. Hence 1 ' C --- — Vrk u0 or using 3) and 194, 7), we get = 3.7 20a3 = 7.2 2 0 Putting this value of c in 5) and setting u = 7rv (6 we get ds -( ds = (- s2)(1- k2s2). (7 du THE THETA FUNCTIONS 435 Let us therefore define three functions of u by the relations =9y3 *)lV (8 sn u 3.1 (8 = 0(v), ~enu = ~-. *( (9 dnu=-0. 9 (v) (10 dn u =? 03(V) (10 ^03 o(v), where v is related to u by 6). Then if k is defined by 3), we have shown that z = snu satisfies the differential equation 7). As z = 0 for u = 0 we see that u considered as a function of z satisfies the relation f dz Jo/(1 - Z2)(l - k22) The function 8) is therefore indeed the old sn function studied in Chapter XI. 2. Before showing that 9), 10) are our old en, dn studied in Chapter XI let us note the periods, zeros, and poles of the functions 8), 9), 10). The periods may be read off from 2). If we set 2 K= 7r-j, 2 iK' = 7rcj, (11 we have as primitive pairs of periods: PERIODS snu 4K, 2iK' cn 4, 2 K + 2iK' dn 2 K, 4 iK' As zeros and poles we have ZEROS POLES sn u m 2 m 'iK' 2 mK + (2 m' + 1)iK' cn u (2 m + 1)K + 2 m'iK' 2 nmK + (2 m' + l)iK' dn u (2 mn 1)K (2 ' + 1)iK 2 mK + (2 m' + 1)iK' 436 46 FUNCTIONS OF A COMPLEX VARIABLE 3. We also note that from 191 we have Sn (u 4- 2 K) = - sn u, en(u +2 K) =-cenu, (12 dn(u +2 K) = dnu. sn (u +2iK') = sn u, en(u +2 iK') - -enu, (13 dn (u 2 iK') = -dnAu. en U' en(u +K) = d nu u' 0 1 en~~u+K)= dnu #3 dn(u +K)=d.1 sn(u +iK') = Ics dn u en(u +iK') =- U' (15 ik sn u' sn~~u-IiK~ en(u +K+ iK')-= _- (1 Ic en u dln(u + K+ iK') = il'sn u en u 4. We now show that the functions 9), 10) are the en, dn functions considered in Chapter XI, that is, we show that the functions 9), 10) satisfy the relations en2 u =1 sn2 U, (17 dn2u1 - k2sn2 U. (18 Let us prove 18); the proof of 17) is similar. The function dnuha2K,2i as a primitive pair of periods as seen f rom 12), 13). THE THETA FUNCTIONS 437 The zeros of dn2u are -K iK7. Its poles - iK'. Both zeros and poles are of order 2. On the other hand, - k2 sn2 u has the same periods, zeros, and poles and to the same order. Thus the two members of 18) can differ only by a constant factor C. To determine this, we set u = 0. This gives at once C=1. 5. Finally let us show that k2, kl2 as defined in 3), 14) satisfy the relation k2 + 2 =. (19 In fact, setting u = Kin 18), we get dn2 K= -2 sn2 K. (20 But from 14) dn2K= k2, sn2K= 1. This in 20) gives 19). 196. Numerical Calculation. 1. Let us now show how to calculate K, K1, snu, etc., by means of the 8's when the modulus k is given. We have Vp = -0 = - 12q+2q4-2q9+ (1 3 1 + 2q+ 2q4+2q9+... When q is small, we have approximately le' =1- 2 q (2 1+ 2 or 1-V (3 = 2 +V/k' To get a closer approximation, we note that 21 1- V/k_-,- 1 02 q+q9+ q25+. (4 1 + v k' 3 + o 1 +2q4+2q16+.. If we develop the right side of 4) in a power series in q and invert this series, we get q= l+ 25+ 1519+ 15013+... (5 This series converges with great rapidity. For example let k2=. We find that 432139 On the other hand q=.0432139... On the other hand =02 t =. 04'3213b... 438 FUNCTIONS OF A COMPLEX VARIABLE This shows that the first term of 5) gives q correctly to 6 decimals for this value of k. Having found q, we get Kby the relation 2Kj~=itY l +2q-2 q4~ 2 q9+. (6 As 1+-\/k =1+ o0 we can write 6) I2 K '0 ~ (0 21~+2 q4+ ~2 q6 ~..( 7r I,9 + Vk gl+ I + Vk' ~~ ( which converges more rapidly than 6). To get K' we nse 191, 1) and 195, 11), which give qrc K or K1~K /1,=lo ) 2.log(!)l (8 The values of sn u, cn u, dn u for a given u and Ic are now found from 195, 8), 9), 10). 2. Suppose on the contrary the value of z in z = Sn (u, k) (9 is given, and we wish to find the corresponding values of u. We start from the relation 195, 10) dnu Vk,"'3(v), = 2 K. (10 a0o(v) From this we get VPc _,~_O(v) 1 - 2 q cos 2 rrv ~ 2q4cos 4rv - ( V1 - kz22 '3(0v) 1~ 2 q cos 2wrv +2q4cos 47wv +. As a first approximation, this gives, I I - 2 q cos 27rv 1 + 2 q cos 2 wv cos 2wv2 = ~ (12 2 q21+ V~ THE THETA FUNCTIONS 439 To get a formula which converges still more rapidly, we set W='~8(v)- &?') = V- k2z2-,/k',3(v) + I0(v) \/I - k2z2 + /' 2(q cos 2 7rv + q9 COs 6 7rv +..) (13 1 + (2 q4 cos 4 rv + 2 q16 co.s 8 7rv +.) As a first approximation, this gives cos 2 7rv = 1 W. (14 2q 197. The O and Z Functions. 1. The a functions depend upon two variables u and q. Let us set (,r(U, K, K)= r(2 ) r = 0, 1,2, 3, (1 where as usual, Ord >0. (2 q=e K Ord > 0. (2 The properties of the O's may be read off at once from those of the Y's; in particular: The Or(u, K, K') are homogenous functions of 0 degree in u, K, K'. Jacobi denoted O0(u) by ((u) and 1l(u) by H((u); H is Greek eta. We shall not use the H notation, but shall at times write H for (0. 2. By means of any one of these O's we may express an elliptic function by virtue of the following theorem: Let f(u) be of order m and have 2 K, 2 iK' as a primitive pair of periods. Let C1e, c2 m PI; P 12, '.pm be a system of incongruent zeros and poles. Then f (u) = Ce - I Q(-Ci) 'IE(-Cm), (3 (~l(g-P1).-.~l(U —pn)m where,p is determined by Abel's relation Ecn - Epo n = 2 XK+ 2 iK '. (4 The proof of this follows along the same lines as 166, 3. 440 FUNCTIONS OF A COMPLEX VARIABLE Example 1. Let us prove the important relation nu1 - (u ~ v)e1(u - v) sn1~u-sn2 in (5 which for brevity we will write L(u) = C-(u) v being regarded as a constant. For L(u) having the periods ol = 2 If (02= 2iK', vanishes at the points 1= VI c " V. The poles of L are iK' =being double. Thns 4) becomes 0-2 2iK'= A 2 K+ ~u iKl.. Hence p = -- 1. Thns 3) gives 7ri?11 V - v)B Cu, + v) L= CI EK L C-K 21 _iK') Replacing 6E1(u) by its value expressed in terms of 6F(u), or 1 7TU 61(u) = iq4e 2KOH (U +iK1) found from 191, we get 5). The constant is found by setting u= 0. Example 2. In a similar manner we may establish another important formula: 1 -le2sn2u sn2v- 00(u + v)0(u - v) (6 3. With the O's we can define four new functions d 1og Or (U~) Or'(U) Z(u) g, r=0, 1, 2, 3. (7 du Or (U) For Z0(u) Jacobi wrote Z(u), and we shall adopt this notation at times. In developing his theory of elliptic functions Weierstrass defined the z(u) function analogous to Jacobi's Z(u). THE THETA FUNCTIONS41 441 The properties of the Z's may be read off from the corresponding t~ relations. Thus we have: Z(u +2 K) =zZ(u)j (8 These show that dZ is an elliptic function having 2 K, 2 iK' as du periods. This f unction is analogous to Weierstrass' p f unction. In a moment we shall show that Z'Qt) differs from dn2 u only by a constant. The addition theorem of Z'u) is obtained from 6) by logarithmic differentiation. Thus Z(u + v)~Z(u- v) - 2 Z(u> 7- 2k ~vnucudn (9 1 -lc2s3n2 Usn2 V Interchanging u and v gives, on noting that Z(- U) = Z~~u), (10 Z~u~v)-Z~u-v)-2Z~v)2 k2/c3sn2 u sn v en v din v I-ksn2 U S2 V Adding, we get Z(u +v) -Z(u) -Z(v) = k2sn usn vsn(u + v). (IiBy logarithmic differentiation of the '6 relations in 191 we get, without trouble: Z(u ~K) =Z(u) -2Sin ufU 4()(1 cin udn u 7rZ Z(u iK') =Z(u) + 2'(13 sn u 2K = Z1(u>:F 2 K We have also 442 FUNCTIONS OF A COMPLEX VARIABLE 198. Hermite's Formula. The reasoning used to establish 168, 2) may be applied at once to Jacobi's Z and gives, using the same notation as before, a celebrated formula due to Hermite: g~u) =AZl~u- a) A2Z'( - a) + )-(u-a) (X - 1)! I ~ BZ1(u - 6) -B2Z'(U - b)~ +. + (1k1BJ( ) )( ~.+ -— +constant. -Example 1. Let us make use of 1) to show that dn2 u= Z' (u) -Z' (K+ iK'). (2 For the poles of dn2 u are = iK' and are double. By 195, 15), dn(u +iK1)- cnu -- isnu U Thus the characteristic of din2 u at the point u - iK1 is Hence in 1), A1 = 0, A2 = I and thus dn2 u=Z'(u -iK')~ C =Z' (u) -I U, using 197, 13). (3 To determine C set u = K+ iK' and recall that dn(K~ iK') =0. This gives C= - Z'(K~ iK'), which in 3) gives 2). Example 2. In the same manner we show that 1 -Z I(0) -Z'(u +iK'). (4 sin2 u -Example 3. Let us show that 1 - 1 - ~Z0(V)~21Zji~i- v) -21Z1(u + v> (5 si2u- sin2 v -sn v en v din v To this end we note that the poles of the function of un on the left, call it g(u), are u n: ~ v. To find the characteristic of g(u) at the point un = v we have,q (11u1 snu-snv snu~snzv snu-snv h(u) Let us set u = v + w. Then, sn u sn(v ~ w) = sn v + w sn' v ~ h(u) =h(v +w) =lh(v)~+wh' (v) + -2 Sn v +. THE THETA FUNCTIONS 443 Thus 1 1 g(u) w.sn' v +.. 2snvv + - 1 1 n= + n nv. ~1 + higher powers of u - v. 2sn v cn v u -v Thus 1 Char g(u) = ru=v 2 snv cnvdnv u- v The characteristic of g at the point u =- v is the same term with a minus sign. Hence 1) becomes 1 1 L-U S — V= 2 1 s vn + ( Z(u - v) - Zl(U + v). (6 sn2 iu - sn2 v 2 sn v cn v dnv In this let us replace u by u + iK', we get k2 snU u k2sn= c2u 1+ 1Z(u - v)- Z(u + v). (7 1 - k2 sn2 u sn v 2 sn v cn v dn v If we set here u = 0, we get = 2 Z(v). This in 6) gives 5). From 7) we get k2sn v cn v dn v sn2 Z(v) + -_ Z( - v) - Z(U + v). (8 1 - k2sn2 U snv2 199. Elliptic Integrals expressed by ( Functions. Let = ff(x, y)dx (1 wheref is a rational function of x and y and = V/(1 - 2)(1k2X2). (2 In 178, 10) we saw that we can write F= Edx + dx (3 J JC Y where E and G are rational functions of x. The first integral on the right can be expressed by the elementary functions. Let us look at the second. We have G = xH(x2) + L(x2), 444 4 FUNCTIONS OF A COMPLEX VARIABLE where H,.L are rational functions of X2. Hlence f dxG fiI xdxH fLdx (4 In the first integral on the right let us set t = x2, it becomes JH(t) 2 -v/(l - t) (I - k2t) which can be expressed by elementary functions. Thus the general elliptic integral 1) reduces to the elementary functions and the integral LdX (5 where L is a rational function of x2. Let us change the variable by setting x = sn(u, IC). Th~en_ dx =V\(1 - X2)(l - k2X2)du - ydu, and 5) becomes =fR(x2du. (6 Here, B being a rational function of X2, is an elliptic fnnction admitting 2 K, 2 iK' as periods. Henlce by 198, 1), B=AZl(u - a) - A2Zl'(u - a) + - ~B1Z1(u - b) - BZ(U - b) ~ +.., + C1, This in 6) gives =Al A log 6,(,u - a) -A 2z,(, - a)++~~ + B1 log H(u - b) - BU2Zl(U - 1) + +...C + 0u, -D. 200. The Elliptic Integral of the 2' Species. This we saw in 180, 1) is r\i1 k2- -jdX. (1 THE THETA FUNCTIONS 445 If we set x = sn u it becomes, putting in the limits 0, u, 11(u) =f dn2 u dui = Z(u) - uZ'(K+ iK') (2 on using 198, 2). We have already called E= dn 2u du the complete integral of the 2' species. This corresponds to K, the complete integral of the 1V species. Let us set in analogy to iK', fKH I dn2 u du. If we set u = Kin 2), we get Z'(K4-iK')- K Thus 1(qt) = Z(u) + U * (3 K Replacing u by u + 2 K, and u + 2 iK' and using 197, 8), we find _E(u ~ 2 K) = E(u) + 211, 11(u + 2 iK')-11(u) + 2 if. (4 Thus 1E(K + iK') 11-p ilL (5 In 3) let us set u = K- iK', we get, using 197, 14), 11K' - HK=, (6 which is the Legendrian Relation, whose analogue in Weierstrass' theory is 171, 8). From 197, 11) we have the addition theorem of the elliptic 'integral of the 20 species, 1(u + v) = E(u) + ~ (v) - k2 sn u sn v sn (u + v). (7 To calculate 11 we differentiate 3), getting dn 2zc= ZI(u) +Kke 446 46 FUNCT1IONTS OF A COMPLEX VARtIABLE Setting u = 0 gives K O(0) K2 1-2q+2q4 - 2q9~* 201. The Elliptic Integral of the '30 Species. 1. We saw in 180, 1) that the elliptic integral of the 30 species used by Legendre is (~nx)ij-_dx _ + nX) ( - 2)(1-k2X2) This differs only by a constant from 1 dx (1d J(x2 - a2) <\(1 - X2) (I - k2X2) Let us setx sn, a=snv. Then we can write 1), putting in the limits 0, u, Udu Making use of 198, 5) gives fU du 1 fu ) 0!lg1(V u)}, 0sn2 u- sn2v si svcenv dn v 2 01( ur taking that branch of the log, which = 1 when u = 0. The relation 2) shows how the calculation of the integral 1) may be effected by means of the 0 series, which, as we remember, converge in general with great rapidity. 2. Instead of the integral 2), Jacobi took as normal integral of the 30 species H(U' V Puk snvcen v dn v sn du.( Using 198, 8), we find that fl(u,v) = uZ(v) + 1log O(v'- u) 2 OH( + v)' ( taking that branch of the log which = 1 when u = 0. THE THETA FUNCTIONS 447 From 4) we have a remarkable theorem due to Jacobi, which states that when the argument u is interchanged with the parameter v in 3) we have I(u, v) - zZ(v) = n(v, u) - vZ(u). (5 It follows at once from 4). From 4) we have at once the addition theorem of the integrals of the 3~ species. Denoting the parameter now by a, we get I (u + v, a) - IT(u, a) - I (v, a) =1 lo ( 4z + v - a)o(ue + ae))(v + a) (6 2 O (u + v + a)O(u - a)(v - a) Relation between the Functions of Jacobi and Weierstrass 202. Relation between sn and p. 1. We have now developed the two kinds of elliptic functions which are in current use to-day. To have two parallel theories may seem an embarras de richesse; it might seem better to choose one theory and discard the other. Perhaps one day that will be done, but at present we stand too near the time when only the functions of Jacobi were employed to neglect these latter. Besides, each set of functions has its good points, and each suffers from the defects of its very virtues. Since then we have these two classes of functions, sn u, (u), Z(u),* k2, * pU, -(U), '(U),... g2, 93, we must ask what is their relation to one another. Let us begin with pu, sn u. In 186, 9) we saw that sn(u, k) = Ve-e2, 2( - 1, ll2) (1 where e1, e2, e3 are such that k2 es- e2 (2 el - e2 as given in 186, 7). From 172, 16) we can also write sn(u, k)= Ve1 e2 (3 V 1'4 2 - e2 4483 FUNCTIONS OF A COMPLEX VARIABLE The periods of sn U are 4 K= 4 w1Vel - e, 2 iK' -2w2 Vel - e2. (4 From 3) we have ___ _ (5 AUI (1 011 W2) = e2 + e2 -e ( sn2(u-\ee - e2, k) 2. Let sn u have the periods 4 K, 2 iK', then sn2u has the periods 2 K, 2 iK'. Let p(u) be a p function constructed on these periods. Then 1 p(U) 2 (6 sn/ U have a double pole in their common parallelogram of periods. Their characteristics at u = 0 are both 1. Hence the two functions 6) differ only by an additive constant. To find this we develop both functions 6) about the point u = 0. We have p (U)= 12+ * + au2 + U2 snu~u 1-1 + k2 U2 +~.) sn2u u2(1 1 + k2u2 +) 1 1 +1+~k2/ Sn2 U U2 3 Thus by 165, 10 1 1 AU(z) = (~- ( + k2). (7 srn2 z( 3 From 7) we can get the relation between p (u) and sn u when p has the periods 2 ai, 2 w2 and sn the periods 4 K, 2 iK'. For let us suppose, as we always do, that the indices 1, 2 are so chosen that if we set T= 2 (8 then Ord i-> 0. We then set e7TiT 2 K = 7rt 2 2 iK' = 2 KT, V 2(0, T) (92 -V,2 ) sn _ (9 TP(O ) -\I-k U THE THETA FUNCTIONS 449 Then 7) gives L~2 ( 1 + 2 p(u) = l _ Ku 01 Differentiating 10) gives (u) = 2 K-2K3cn v dnv Ku PW s()l V (1 -6o3 sn3v 601 ~~~~~~~~01 (10 (11 203. The el, e2, e3 in Terms of the i' U = 61 ' 6)2, (03. Then - becomes K, iK', 01 pu becomes el, e2 s. In 202, 10) let us set - (K+ iK'). e3. k2. 1 becomes 1, 0 sn2K 01 9 Hence K21 + l'2 e= - - 01 2 3 K21 + k2 e = --, e1 + e2 + 3 = el + e2 + e3 = 0. (1 and From 1) we get K2 el -e2= -2 C<"! Hence! - w K2, el - e3 = k12 2 ' e3 -, 12 k2 _= 3 - e2 12_ e-e3 el - e2 el - e2 -e =ek2. (01 (2 (3 If now we put in the values of k2, k'2, K in terms of the?'s, we find el: ~,,o x {- 2 4 }, ea = a ) i {04 + 4}, e2= 32 oJ?2+ 3 e3= 3(2 a), 2 0 (4 450 FUNCTIONS OF A COMPLEX VARIABLE Their differences give e3- e3 = (2 - = (2 04)2 ~ el - e = Y 204. Relation between cr's and?9's. The two functions o-(u, 2 a1, 2 to2),(V, T), where V T 2 W are integral transcendental functions having the same zeros. Thus by Weierstrass' theorem, 140, 1,?~l (v) = e 9(U) a (u), where g is an integral function. If we take the second logarithmic derivative, we find as in 166, 1 that gy = constant. Hence..L (v)) - ea-bu+cu20(.u) - Aebu+CU2o(U). (1 As 01(v), o-(u) are both odd functions, e-bu+cu2 ebu+cu2 or e2bu - 1 hence h = 0. To determine A and c let us develop both sides of 1) about u = 0. We have l(v) = 0,vO + V 0' ~ eCU2 = 1 + cu2 +... eCU u(u) = U + cu3 + These in 1) give 0' ~Ef u3 A A"'U3 + 1 1l THE THETA FUNCTIONS 451 Hence 2 21 24 o) Thus 1) becomes 2.1 (, =2 ( (2 We can give this relation another form. We have O(( + 2 o ) = - e2'l((+1)"a(U, 1(u + 1, T)= -Ol(U). Thus (+1 ~~~~or ^ ^ *~,<~) ^ ^~~~'1?) --- __ 1u+ ~2 a)' 6- 2 01" * ( _(u_+ 2 ___, 2 wo 2we This gives l^ '[ +1] or 9 2 ~2 rio =-6I 1 (13 o; a (~) ~00 = - -1 e e 0 2 o0 1 2 a~l This in 2) gives This in 2) gives O ~ 'I21(- -) (U) (4 As,(u) - (u ~ w,) we have at once 2 ~l(GG)2 _,_ )_l '1(u,) = e 3 0-2(U) qe (2w (zq_,'_' ~s1 -3(U) e = e - 3, 452 FUNCTIONS OF A COMPLEX VARIABLE From these relations we get )1 (Ku )' en (u,, (6 -X/'ji -- ~r 2) cn (u, T)=(/J- —, e2 VC 1u 0 V/pu -e2 Vel - Also we find Sn ( V, T) V(e - e2 Ha ) - e dn (v, T) = u -. A/pu - e2 cn(v, A)_= ---- 2 =-, (7 \/pu - e, ~\/e, - e dn (v, 7) = -/ ---3. V/u - e% CHAPTER XIII LINEAR DIFFERENTIAL EQUATIONS 205. Introductory. 1. In the last three chapters we have given a brief account of the elliptic functions. These functions are of great importance in themselves; they also furnish a striking and brilliant example of the great power and usefulness of the theory of functions of a complex variable. As usually happens, these two theories have mutually aided each other. The function theory has furnished the viewpoint and instruments of research; the elliptic functions in return have furnished fresh problems which have given rise to a broadening and deepening of the function theory. Without the notion of a complex variable, the imaginary period of the elliptic functions would never have been discovered, and without this period, there would be no theory of double periodic functions. Yet the double periodicity is their most important and characteristic property. 2. Another theory which has been revolutionized and put on an entirely new basis by the advent of the theory of functions is the theory of differential equations. In the old days, a differential equation meant merely this: Find some combination of the elementary functions which satisfies it. The simplest type of differential equation has the form dy =f(x)dx, whose integral is formally given by y = f(x)dx. But already the simple differential equation dy -d (1 -(1 - x2)(1 - k2x2) could not be integrated in terms of the elementary functions. The problem of integrating a differential equation was a kind of 453 454 FUNCTIONS OF A COMPLEX VARIABLE game of hide and seek, the solution being usually so well hidden that no amount of seeking could discover it. We owe to Cauchy an entirely new point of view. He first taught us to regard a differential equation as defining a function whose properties are to be unfolded by a study of the equation itself. This method we have already illustrated in studying the differential equation 1). We propose now to apply it to a broad class of equations called linear homogeneous differential equations. They have the form dy pdy +i d +"Pn = 0, (2 dPo dxnthe coefficients p being analytic functions of the complex variable x. Such an equation is said to be of order n. We shall restrict ourselves to n = 2; moreover we shall generally suppose the coefficients to be rational. A number of important functions in analysis satisfy such equations, and we have chosen these equations for the same twofold reason that induced us to choose the elliptic functions, viz. to illustrate the general principles of the function theory, and to develop the properties of certain functions of great importance. Examples of this type of equations are the following: Example 1. The polynomials of Legendre satisfy (1 - x2) dy -2 x + n(n + 1)y = 0. (3 dx2 dx Example 2. The associated Legendrean functions satisfy d2y m (F+L1xdy (1 - x2) 2Y 2(m + 1)x + nn 1 + 1n(n q 1- m(nm J 1) y = O. (4 Example 3. Bessel's functions satisfy d2y dx Example 4. The functions of Lame satisfy d2y 1 1 1 1 1dy d x2- ' + ~dy dx22 x- el x- e2 x- e J dx 1 ax+b _ n r 4(x-e) (x -- e) (x - e3) kV LINEAR DIFFERENTIAL EQUATIONS 455 Example 5. The hypergeometric function satisfies x(x)- 1)d2Y+ X(V + )- d + 0. (7 We notice that all these equations have rational coefficients. We notice that all these equations have rational coefficients. 3. The general theory of linear homogeneous differential equations was first studied by Riemann. It owes, however, its present perfection largely to L. Fuchs, who began his researches in this field about 1866, and to a stately array of mathematicians who have followed in his wake. Prior to Riemann we may mention as especially important the early investigations of Gauss and Kummer of the hypergeometric differential equation 7). 206. Existence Theorem. 1. Instead of the general equation 2) of the last article let us consider one of the second order " = PlY' + P2Y, (1 the coefficients Pi, P2 being analytic about the point x = a. We propose to show that 1) admits an analytic solution y= b0 + bl(x- a)+ b2( - a)2 + *. (2 which is uniquely determined by the initial conditions that y and y' shall have assigned values y = a, y' = 3 at x = a. The reasoning we shall employ can be easily generalized so as to apply to the general case of order n. By using an equation 1) of second order we simplify our calculations without sacrificing the general method. Suppose for the moment that 1) admits the analytic solution 2). The coefficients b, are determined as follows. From 2) we have y(a)= b0, y(a)=-b,... y(n(a)=n! b. This gives at once bo= a, =. From 1) we have y"(a) = p(a)y'(a) + p(a)y(a). Let us set A1 =pl(a), A2= (a), then the last relation gives 2! b2= blA + boA = B3A, + aA2. (3 Differentiating 1) we get y"' in terms of y, y', y"f or y' f =ply" + p1y' p2y' +Y +py (4 456 FUNCTIONS OF A COMPLEX VARIABLE If we set here x = a, we get b3. In this way we may continue and so determine one coefficient b after another. This shows that only one analytic solution 2) with a given set of initial conditions is possible. The form of these b's is important. To determine it let us set oo co Pi = Pln( - a)n ' P2 = P2n(X- a)n. (5 n=O n=0 These are simply the development of P1, P2 in power series, since by hypothesis they are regular at x = a. The relation 3) shows that b2 is an integral rational function of a, 3 and pl0, P20. The relation 4) shows that b3 is an integral rational function of a, /3, p1, P20' Pi1 P21' Thus in general bn = -En(bo0 bl, Po P11' P20 P21 * ) (6 is an integral rational function of the enclosed letters with positive coefficients. 2. Having shown how to determine a solution 2) which formally satisfies 1), let us show that this solution converges for all x which lie within a circle R about x = a, and which reaches to the nearest singular point of the coefficients p (x), p2(x) of 1). To this end we seek a simple differential equation of the same type as 1), which we know admits an integral z = co + c(z - a)+ c2(z - a)2 +.. (7 converging within 9, such that 13n < yn. (8 Here, as we have so often done before when dealing with series, we denote the absolute values by the corresponding Greek or German letters. Thus in particular /, = Ib,, = l c Let this auxiliary equation be z = qlz + q2z, (9 where l - qn(x - a) ' q2 -= q2,n(X -a), (10 0 0 are the development of qj, q2 in power series about x = a. Now whatever the ql(x), q2(x) are, the coefficients ce must satisfy the relations n- = En(c0, c 1 q0l qll.* q20, q21 "'. ) (11 LINEAR DIFFERENTIAL EQUATIONS 457 where Fi is the same function as in 6), only with different arguments. As the coefficients are positive in 11), the ce will be real and positive when the arguments in 11) are real and positive. Now co, cl being arbitrary, we take them real and positive and such that > If now q > p, > (12 then < -n(/3o0 1 P10 'o P20 "') < n(7/0,' r1 q10o q20 )= yn, or An < 7n. Let us now try to choose the coefficients q1, q2 in 9) so that 12) holds. By Cauchy's inequality mn < m, m = 1, 2, where Pm > Max pm(x) on t whose radius, say, is R. But then if we only take Pm qmn —Rn the condition 12) is satisfied. In this case m- Pm + X (x-a )2 M= P-mI + -, -a<R. R Thus our auxiliary equation 9) becomes! x-a P1Z + P2z. (1 fl2-^V==p~a'+p^+P (13 We need only to show that 7) is convergent. The ratio of two successive terms of its adjoint series is 7Yn+l2 x-a. Yn+l 458 FUNCTIONS OF A COMPLEX VARIABLE Thus 7) will converge within Sg if we show that lim R/Yn+ = 1 (14 n= o n n+l To do this we derive a recursion formula to determine the y's, or what is the same, the c's. Differentiating 13) gives R R (1 x R~ )V 1 ytR = Plz" + P2z' Differentiating again, we get (1 x — a)zIV — _2 l + P2z". In general we see that (I - x-a)z(+2) _ n (n+l) = P1(n+l) + P2zn Setting x = a and noting that 1 ) en-n we get (n + 2)!c+2 - n(n 1)! n+ = (n + 1)!Plcen+ + n!P2,. Tlus -R(n + 2) Cn2 = (n + 1)! 'n + RP1 i c,, + n RP2cn, or i n RP 1 orn n + 2 1n+ (n + 1)(n + 2 (15 Let us now take P1 so that RP> >2. As the last term in 15) is positive, this shows that ReCn+2 > en l or 1 n. < 1. (16 en+l Let us now write 15) R Cn+2= n + RP1 - R2P2 1 cn.... l.. en+l n + 2 (n+ l)(n + 2) R en+1 Letting n- c and using 16) we get 14). LINEAR DIFFERENTIAL EQUATIONS 459 3. The form of proof here given is entirely general, and holds for any n. We have thus proved the Existence Theorem. The differential equation dny d'-ly0 (1~ in+ P1 + ' +PnY = (17 dxn d'n-1 admits one and only one analytic solution which together with its first n - 1 derivatives takes on assigned values at x = a. This solution is valid within a circle R which extends to the nearest singular point of the coefficients pi... pn which are regular about the point x = a. 4. Let y= 0 + 1(x - a) + 2(x - a)2+.. (18 be a solution valid in R. Let a1 be a point within M. We can write 18) y=b+ ( - a,) + b(x - a)2 + (19 which is convergent within some circle R1 about x= al which certainly extends up to the edge of R and may go beyond. If we develop the coefficients pm(x) about x = a1 and put 19) in 17), we get a power series about x=a1. Since 19) satisfies 17) within Q, it will continue to satisfy it for all points within,; moreover A1 will reach to the nearest singular point of the coefficients p. \ In this way we may extend the solution 18) by analytic continuation. Thus we have the theorem: If the function 18) is a solution of the differential equation 17), all its analytical continuations are still solutions of 17). 207. Fundamental System. To each particular set of initial conditions y = a, y' = f for x = a will correspond a particular solution of Y"b + p1Y' +P2Y = 0. Let Y1, Y2 be two such particular solutions. Then y == ely + c2Y2 (2 is obviously a solution of 1) also. Let us show that we may pick out particular solutions yl, Y2 such that every solution of 1) has the form 2). 460 40 FUNCTIONS OF A COMPLEX VARIABLE For suppose that y is to satisfy the initial conditions On the other hand, suppose the initial conditions of yl are of s~'2 are 2(a) =a2, 4(a) = /82' Then 2) shows that we must have These two relations determine el, c2 we nVI a1 2 #0. ( Let us set YiX 11/2. (4 We note that -D (a) is nothing but the left side of 3). Let us show that if -D 0 at x = a, it is also #f 0 at any point x which is not a singular point of the coefficients p in 1). For since y91 1/2 are solutions of 1), wve have 91/ ~ ]PlYl + P291/ = 0, Then if -D =t- 0, these give 11 2 D1(x) Pi - 92 Y2 ~~~ - - (5 Pi D (x) _D(x) Now we note that d dx-D (x) - -DJfX). dx ~ ~ d Ths5)gver -f$-lg (x ), or ~~~~~~~D (x) = eJP(~ (6 At x = a, -D is =,- 0, hence(J 0. Thus -D is,always 4- 0 since -D cannot van~ish unless p,(x) =o. But this pointw\o~ild be a singular point of p,. The reasoning being entirely general we see that: -If,P U.. 2 7 UP 72 - On k' I LINEAR I)IFFERENTIAL EQUATIONS 41 461 are particular solutions of dny d-'y ( dxnJlx 1 +PI( f or which y y. I I I Yi'x =91 Y2 I..fl (9 (ni) (n-i).. (-1 is =# 0 at a point x = a at which all the coefficients p,... p,,, in 8) are regular, then -D =# 0 at all such points in a connected region. Such a system 7) is called a fundamental system, and we have the theorem: Every analytic solution, qf the diffrential equation 8) is a linear function q~f any fundamental system 7 ) with constant coefficients. 208. Linear Independence. We have just seen that a fundamental system 'y1, Y21 '. Yn is characterized by the fact that D 9~1 A ( is =# 0. We now prove the theorem: Por a linear relation with constant coefficients c1y1 + c2 Y2~...~cne-Yn=O- (2 to hold, it is necessary and sufficient that _D = 0 identically. It is necessary. For if 2) holds, we get on differentiating Clyn-11 + c2y,(n11 + ~ cny~n-) =0. From these equations and 2) we have necessarily YD = 0. It is sufficient. For let _D = 0. Now y1, Iy2... y,, are all solutions of the differential equation of order n - 1, viz.: u~n1, y2n1, y3n1 yn1 u' 2~ y3 *.? -dbu d2 462 FUNCTIONS OF A COMPLEX VARIABLE For setting u = Y1, this determinant reduces to D which = 0 by hypothesis. If we set u = y2, for example, two columns of this determinant are the same; it therefore vanishes in this case. Let then u1, u2,... un_ form a fundamental system of 3). Then Yl *. yn being solutions of 3) are linear functions of the u's. Thus y1= a11l + + al, n-lUn-1 n = anllu1+ *** an,n-lUnIf we eliminate the u's from these equations, we get a relation of the type 2). In the exceptional case that the coefficients 01, 02... in 3) vanish, it reduces to an identity. Then by using a smaller number of the y's we would still get a linear relation between them; but we shall not urge this point here. Thus we may state that: Any set yl, YI2 *"'y of linearly independent solutions of a differential equation of order n form a fundamental solution. 209. Simple Singular Points. 1. Having seen that d2y +pid +P2= (1 admits a solution taking on assigned initial conditions at any nonsingular point of the coefficients P1r P2, we now turn to these singular points and ask how the solution of y behaves about one of them as x = a. We shall restrict ourselves to the case that p1, P2 have at most poles at x = a whose orders are not greater than one and two respectively. Then we can write 1) in a normal form, (x- a)2 q d + ( - a)l d + q = (2 dX2 dx Here we suppose that the new coefficients q, q1, q2 are regular at x = a and that q0 does not vanish at this point. Then we have, developing about x = a, o and at a m = 0,, 2 (3 0 and at least q00 =- 0. LINEAR DIFFERENTIAL EQUATIONS 46 463 The equation 2) may be written 2 1 (X - a) M q2-mY 0. (4 Let us try to satisfy 4) by setting y =(x -a)r Ick(X -a)k,o =/- O. (5 Our problem is to determine the unknown exponent r which in general is not a positive integer, and the coefficients ce.. From 5) we get, on differentiating, YI= (r + k)Yek(x - ark1 Y f= (r + k)(r + k - l) lCk(x - a)r+k;2. These in 4) give (x - a)2Xqon(X - a)nl(r + lc)(r + ic - 1)ck,(x - a) r+k-2 n k ~ (x - a)lql1,(x - a)nX(r + k)ck,(x -a)r+Akn Ak +:~q2n (X a) nYCk7,(x - a) r~k = 0. (6?b k The coefficient of (x - a ~~ can be written, as Frobenius remarked, as follows. Let us set fAx, r) =q2+ rq + r(r -1)qo 00 0 Then 6) can be written Ifn-k(r +k-)c7(x -a)r~n =O n=0, 1,2..co, (8 c-0o 1, 2 *n. As this power series = 0 identically, the coefficients of the different powers (x - a)r+n must all = 0. Hence cAf(r) = 0 c0f1(r) + cf0(r +1)= 0 *cf2jr)~+ clf1(r~+1) + c2f0(r +2) = 0 (9 cJf3(r) + clf2(r + 1) + c2fj~r + 2) + c3f0(r + 3) =0 Thus when 2) admits a solution of the type- 5), the coefficients co, el..and the exponent r must satisfy 9). 464 FUNCTIONS OF A COMPLEX VARIABLE As co0 0, the first equation requires r to satisfy f(r) = 0; or using its definition in 7), o(r) = q2(a) + rql(a) + r(r - l)q0(a) = 0. (10 This equation for determining r is of fundamental importance; it is called the indicial equation. It is a quadratic in r. Let now r be a root of 10). The coefficients e, c2... may be obtained in succession provided fo(r + 1), f(r +2), f(r+3)... (11 are all =/ 0. But for that root r1 of 10) whose abscissa is greatest, none of the coefficients 11) can vanish. Neither can they vanish when the two roots of 10) do not differ by an integer. Thus when the indicial equation admits two distinct roots rl, r2 which do not differ by an integer, there exist two series Y1 = (x - a)rlii + c 11(- a)+ 12(X - a)2 +..., l ~ 1 Y2 = (x - a)r2 C20-t e2((x - a)+ 22( - a)2 + *.. } 20 # 0(1 which formally satisfy the given differential equation. These series converge within a circle whose center is x = a and which passes through the nearest singular point of the coefficients P1 P2 in 1). Thiis may be shown by the method employed in 206. As the reader has been through one existence proof it is not worth while here to repeat the proof. 2. The foregoing results can be extended to the general case. Let the coefficients of dy d-y(13 dx --- +Pldx7 -l + "' +PnY= ~ 0 ( have at x = a at most poles of orders not greater than 1, 2,... n respectively. Then we can write 13) in the normal form (x - a)no + (x - a)-lqd +... + qy = 0 (14 where the q's are regular at x = a. They therefore have the form given in 3), where now m = 0, 1,... n, and as before we suppose 00 = q0(a) O 0. If we now try to satisfy 14) by a series of the LINEAR DIFFERENTIAL EQUATIONS 465 form 5), we are led to a system of equations of the form 9). The indicial equation which determines the exponent r is here fo(r) = q.(a) + rql(a) + r(r- )q2(a)+.. +r(r- 1)... (r - n + )q(a) = 0, (15 which we see is entirely similar to 10). Since by hypothesis qoo(a) is = 0, the indicial equation is of degree n. Let us arrange its roots in groups r' r,rf... r*l r2 2 1 i( r2, r, r... (16 Here rI is the root whose abscissa is greatest and the first row embraces all the roots of 15) which differ from rl by an integer. Of all the remaining roots let r2 have the greatest abscissa; then the second row embraces all the roots which differ from r2 by an integer, and so on. The roots r1, r2... (17 which head their respective rows are called prime roots. And now the existence theorem states that: To each prime root 17) corresponds an integral of 13) ym = ( - a)r cmO + c,(x - a), + cm2(- a)2 + * *, (18 cmO H= 0, whose circle of convergence reaches up to the nearest singular point of the coefficients p. 3. In case that each group in 16) contains but a single root, all the roots of 15) are prime roots. As to each prime root corresponds an integral 18), the foregoing method gives us n integrals of our differential equation 13). Let us now show that: When the roots of indicial equation 15) are all prime, the n integrals 18) form a fundamental system. For suppose there exists a linear relation alyl + a2Y2 + * + anyn = 0 (19 between them. If we put the values yl, y2... as given by 18) in 19), we get a power series; the exponents of course are not in 466 FUNCTIONS OF A COMPLEX VARIABLE tegers in general. If am, 0 in 19), our power series contains the term amtcmo (x - a)"m, and this is the only term with the exponent rm. Thus a,,eo = 0, and hence am = 0, or Cm0 = 0. Both of these are contrary to hypothesis. Hence a relation of the type 19) is impossible. The case we have just treated is the simplest case that can arise at a singular point. We therefore call such points simple singular points. 210. The Hypergeometric Equation. 1. This is, as remarked in 205, 7), x(x - 1) d2,y ~ + p+ ry - d + c)3y = 0. ( dX2 dx Its singular points in the finite part of the plane are x = 0, x =1. Let us find the indicial equation for these points. The point x = 0. Bringing 1) to the normal form (X - 1)x2y" + x(+e /+la1) -yxy'+ 3xy = 0, we have q0(x) = x - 1, (x) = (a + P + 1)x -r, q2(x) = ax. (2 The indicial equation for x= 0 is, therefore, r2 + y- 1)r- 0 or r~r-(1 y)~ 0,(3 whose roots are rl = 0, r2 = 1 - ry. The point x = 1. The normal form of 1) at this point is X(X I -)2y lf + 0 +3 + 1 )x - rl (X - 1)y' + ale(x - 1)y = 0. Here qo(x)=x, q1(X)=(a+/3+l)X-Y q2(X)=Cfl(x-l). The indicial equation is, therefore, r I r - (Y - a -/3)1 = 0, (4 whose roots are.r =O r2=7y-a-/3. LINEAR:DIFFERENTIAL EQUATIONS 467 2. Let us investigate the nature of the point x = oc. To this end we set1 U and 1) becomes + 0. ~~~~(5 dU2 u u(1 U) u u2(1l-u)0 Obviously u =0 is a singular point. The normal form of 5) at u, = 0 is (1 - U)U Here Thus the indicial equation for u = 0 is (6 r-2 - (cc -F-$r + /3 = 0, whose roots are ri = a r2/3 3. Let ns now calculate the coefficients of our solution by the formxuh~e of 209, 9). We consider first the point x = 0. Now by definition fn(r) = r(r - 1)q l()+ A i~(0) + -qi (2). (7 As the q's are linear functions as shown by 2), all derivatives beyon'd the first vanish. Thus f2(r)= 0,f3(r)= 0.. Hence the equations 209, 9) are all two-termi equations and they give fl(r~ n -i en e- (8 -ifo(r +n) 'Let us now use the root r = 0 of 3). Then 468 FUNCTIONS OF A COMPLEX VARIABLE Hence Cn =- n(- + - en-1 n=l, 2,... (9 n(y + n- 1) We thus get a-/3 (a(+1)(/3+1) a + 1./3./3+1 c1 = a 2 2(- +1) 1 1 2. 2. + 0lO etc. Hence taking c = 1, r a,.,~ a. a+l./3. /3+,2+ Yi = 1 -I + -I - 2.y ry + I ( I 17r 1.2~7.7+1~ J (10 = E(a,,,, x). Let us now use the other root r = 1- of 3). As f(r) = r(r - l) + r(a oc+ + 1) + a/ = r2 + (a +/S)r + a/, (r) = - r(r - 1 + ), we have, from 8), on taking r = 1 -, ( - y)2 + (n - r)(ae + /) + a~n ' —~ n — n(1 — y +n) (n + a- y)(n + /3 - y7) n(1 - y + n) Let us compare 9) and 11). We see that 9) goes over into 11) on replacing a, by + l-y, /3+1-y, 2-,y. Thus the integral corresponding to r = 1- 7 is y x = 1-YEY(a + 1 - r, 8 + 1 - y, 2 - ry, x). (12 4. Let us now turn to the point x = 1. The recursion formula is found to be for the root r = 0 of 4) - (n+ - 1)(n+/- 1) (13 n(n + a + / - r) We see that 9) goes over into 13) on replacing a, /3, Y by a, /3, a+/-r+l, aside from the sign which can be made right by replacing x - 1 by 1 - x. Thus the solution corresponding to the root r = 0 is y1 = i(, A, - + / - y + 1, - x). (14 LINEAR DIFFERENTIAL EQUATIONS 469 The solution corresponding to the other root r = 8y- - / of 4) is found to be Y2 = (1 - x)7-a —F(ry - p, ry - cc, 3/ - a - 8 + 1, 1 xz). (15 5. Finally we consider the point x = oo. The recursion formula for the coefficients corresponding to the root r = a of 6) is = (a + n - 1)( + n - ry) n n(a + n/ - 8) n-1 'We see that 9) goes over into this on replacing /3, y by a, a-ry l, a-/3+1. Thus the solution corresponding to the root r = a is Y= IP(Fc, a- y~1, a -3+1). (16 The solution corresponding to the other root r = / of 6) is similarly 1 1(/3 /-y~1~/3-+ ~ I)( = F8,0-y~, -a l,- 1~(17 92 x 2 2 211. Bessel's Equation. This is, as remarked in 205, 5), x2y" + Xy' + (X2 M2)y = 0. (1 The only singular point in the finite part of the plane is x = 0. Let us consider the integrals of 1) for this point. The equation is already in the normal form. Here qOa(x) l,q1(x)=l, 2(X) = 2 M2 The indicial equation for x = 0 is therefore fo(r) = - m2+ r + r(r - 1)= 0, or fo(r)= r2 - m2 = 0. (2 Also here fO(ro, f2(r)=1, f.(r)= 0, >2. Thus the equations 209, 9) become C1=0, CO~c2f2(r+2)=0, c3=0, and in general, c2f0(r 2 n) + c=0, c21 = 0. (3 470 FUNCTIONS OF A COMPLEX VARIABLE One root of the indicial equation 2) is r= m. For this root 3) gives ~~n(2 rn + n)ec, +en-2 = 0,n even. Hence c2=-~~c 2 2(2 m+ 2)' C2 -co 4(2m+4) 2..4(2rn~2)(2m+4)' etc. Thus the integral corresponding to r =mr is C X~~~n a2 a;4 Y 0X 2(2m+ 2) +2 4(2mn~2j)(2m~4) a;6 2.4. 6(2m+2)(2mn+4)(2m+6) +( In case rn is not an integer, the other root r = gn of 2) also furnishes a solution Y2 since the coefficient f0(r ~ 2 n) of CD,~ does not vanish for any n. Let us take the constant co so that 1 co = 2mH-I(m)' Then as solutions of 1) we have =Jm1(x) = (n) H (m + n) (5 and ~ 92 (a;) = (~-1) - (N2rn(6 1/2 = noH (n)HI (n-rn) 2/ They are called Bessel functions of order rn and - rn respectively. 212. The Logarithmic Case. 1. We have seen that when the indicial equation F(r) =r(r -1)q0(a)~+rql(a)~+q2ja) =0 ( has its first coefficient q0(a) =t- 0, our differential equation, which we wvrite in the normal form L(y) = (x - a)%0(X)y"1 ~ (x; - a)q1(x),y' + q2(X)y = 0, (2 has one solution of the form.yj = (x- ct~rico + clx - a) C2(X(3 LINEAR DIFFERENTIAL EQUATIONS 471 where r1 is that root of 1) whose abscissa is greatest. Suppose now the roots of 1) are equal, or at least differ by an integer. The method developed in 209 gives in general only one integral 2), viz. the integral 3). To obtain another linearly independent solution Fuchs proceeds as follows. We set y =yl zdx (4 in 2). This leads to a linear homogeneous equation of order 1. Let z be a particular solution of this equation, and let Y2 be the value of 4) for this value of z. Then yl, y2form a fundamental system of our original equation. For if Cyl + c2Y2 = 0,(5 we have, on using 4), + = 0, lYl — + f zd= O. or +C Cf zdx = 0. Differentiating this, we get Z = 0, and this requires that c2 = 0. Putting this in 5), we see that c = 0. Thus yl, Y2 are linearly independent as stated. 2. Let us now set 4) in 2) and find the resulting equation which z satisfies. We have, differentiating 4) and setting for brevity z1= J zdx, 2 1 I + MI y' = y1z1 + y1Z, y-! - Yl/1 + 2 y~Z ~ ylz'. These in 2) give z1L(yl) + (x - a) {qlyl + 2(x - a)q0yUIz + (x - a)2q0yz' =0. (6 But L(yl) = 0 since y1 is a solution of 2). Writing 6) in the normal form, we get (x - a)qoz + { ql + 2(x - a)qY z = 0. (7 If we write 3) Y1= (x a)rg, we have log y1 = rl log (x - a) + log n. 472 FUNCTIONS OF A COMPLEX VARIABLE HenceI (x -a)Y1 = r, + (x - a)k(x), Y1 where *r(a) =t 0. Thus. we may write 7) JX~(z) =(x - a) s/z + s1z = 0, (8 where s()=q() 81(x) =ql(x) + 2 %o(x) r, ~ (x - a)#(x). Thus the iudicial equation of 8) is rs0(a) ~ s1(a) = 0, or G(r) = rqo(a) + q1(a) ~ 2 q0(a)rj = 0. (9 Then (r ~ 1) G(r) = q0(a) - rl(r1 - 1) ~ (r ~ r1 + 1)(r ~ rj) as is seen by actually multiplying out. This we may write, (r + 1) a(r) -, - r - 1) q0 (a) ~ rlq1(a)~ +(r + r1 + 1)(r + rl)q0(a) + (r + r1 + 1)q,(a) ~ IBut the first term. on the right is q2(a), since r1 is a root of P(r) = 0. Thus the last equation becomes (r+l)a(r)=(r+r-1~1)(r+rl)q0(a)+(r+r1+1)q1(a)+q2(a) = F(r +r1 + 1) Hence the root of G(r) = 0 is r2 -r1 -1I=- M, an integer, since by hypothesis r1 and r2 differ by an integer, which maybe 0. From this we have as result that the differential equation 7) admits a solution, LINEAR DIFFERENTIAL EQUATIONS 473 whose coefficients may be obtained as before. Then fzdx- hk Czdx -(x )m-l +... +- + h log (x~- a) + kl(x - a) - k2(x - a)2 + *+. (10 But we have seen that Y2 = l zdx is a second solution of 2). Putting in the value of Yl as given by 3), we get Y2 = (x - a)t2o(x) + (x - a)ril1(x) log (x - a), (11 which may also be written = (x - a)r2< +(x) + h(X - a)m-l(x) log (x - a), (12 where +(x), r(x) are regular at x = a and do not vanish at this point. 3. Thus when the indicial equation at the point x = a has two roots which differ by an integer, there exist always two linearly independent solutions of the form 2) and 11) or 2) and 12). Let us note that the logarithmic term in Y2 may not be present. This takes place, as 12) shows, when h = 0. That the two roots of the indicial equation may differ by an integer without Y2 containing a logarithmic term, is illustrated by Bessel's equation 211. For let m = 1 + l in 1) of that article, I being an integer. Then the two roots of the indicial equation are 1 1 - whose difference is 2 1 + 1, an integer. However, the recursion formula 211, 3) for determining the coefficients cn is such that the cn of odd index vanish, and thus c, for even index are uniquely determined if only m is not an integer. 4. There is no difficulty of generalizing the foregoing result. We may therefore state the theorem: At the point x = a let the indicial equation of y() + ply"-1 +..." + py = 0 (13 be of degree n. Let r, r', r... r(s) (14 474 FUNCTIONS OF A COMPLEX VARIABLE be the group of roots belonging to a prime root r, arranged according to diminishing abscissae. Then y = (x - a)r>(x), Y1 = (x - a)r' {10(x) + -11 log (x - a), Y2 = (x - a)r" 2 0(X) + +21 log (x - a) + 022 log2 (x - a) }, (15 s = (x - a)(S) 8o0(x) + sl (x) log (x- a) +... +,s log8 (x- a) f are solutions of 13). The functions c are one-valued analytic functions within a circle about the point x = a, and passing through the nearest singular point of the coefficients p of 13). Each group of roots as 14) of the indicial equation furnishes a group of integrals as 15). The total number of integrals obtained in this manner is n. They form a fundamental system. 5. When the degree of the indicial equation at a singular point x = a is n, the same as the order of the differential equation, we say x = a is a regular point. They include the simple singular points of 209. When the indicial equation at the singular point x= a is of degree less than n, the foregoing method does not give us all the integrals of 13). Such singular points are called irregular, and their theory is too difficult to treat in this work. We shall soon see that Bessel's equation has x = oo as an irregular point. 213. Method of Frobenius. 1. In the foregoing article we have established the existence of a fundamental system when the roots of the indicial equation differ by an integer, using a method due to Fuchs. Knowing the form of the solution, the coefficients may be obtained in any given case by the method of undetermined coefficients. Frobenius has given a method which leads more quickly to the desired result. Let us take the singular point x = a at the origin; we write our equation in the form d) 2 dy (1 L(y) = x2 y+ xp(x) - + q(x)y = 0. dX2 dx LINEAR DIFFERENTIAL EQUATIONS 45 475 Using still the notation of 209, 9) let us set cof1(s) + e1Jfos + 1) = 0, cof2(s) ~ clf1(S -F 1) + c2f0(s + 2.) = 0, (2 where s is not necessarily a root of the inclicial equation, but an arbitrary parameter. Then en~will have the form e"'(8) = Vs) c(s) (3 f0(s + 1)f0(s ~ 2)...f0(s ~ n) Let us now set Y =Xs e~nXn=g(x, s) (4 in 1). It becomes L[g(x, s)] = v ~c1fo(s -i-n) + c.-if,(s -F n - 1)q-...+ cofll(s) x n = cofo (s) x, (5 since all the terms on the right vanish except that which corresponds to n = 0, by reason of the relations 2). Thus when s is a root of the indicial equation f0(r) =r(r - 1) + p(0)r + q(O)= 0, (6 we see thatX1CX( 0 satisfies the equation 1). Suppose that the twvo roots rl, r2 of the, indicial equation differ by an integer, say r1 = r2 -I m, m> 0. Then 6) has the form I j~r =(r- j)(r- r + m). For co let us take co0= CJ~ (s + l)fo (s +2)..fo(s-j-m). (8 Then the en% in 3) will have the form 476 FUNCTIONS OF A COMPLEX VARIABLE in which the denominator does not vanish. Also the coefficient of x8 in 5) has the form ofo(S) (s - rl)(s - r + m)2S. Hence in this case L[g(x, s)] = (s - r1)(s - rl + m)2Sxs. (10 Now a a d2w a dg g a L(g) =-X2 -_ + Xp-. -* +q- * = L( J, as as d2 Os dx As as a - (S - rl)(s - r1 + m)2S (s - r1+ m)2S + 2 (s - r1)(s - r1+ m)S Os + (s-r)(s - r + m)2 d. ds Hence differentiating 10) with respect to s and then setting s=r2, we see that -Las Os=r2 is a solution. Thus, provided the series 4) can be differentiated termwise, we have as a second solution of 1) Y = z= Xr2 log x Xn 4- X2 o () Xn. (11 2 o =r2 2. When the coefficients of 4) are determined by 3) and s = r, the first prime root of the indicial equation, the series 4) is a solution. But if we give the cs values as determined by 9) and take s = r2 the second root of the indicial equation, we see that the series 4) will also be a solution in the case that rl, r2 differ by an integer. 214. Logarithmic Case of the Hypergeometric Equation. 1. We saw in 210 that the two roots of the indicial equation at x = 0 are O and 1-y. Thus when is an integer, we have the logarithmic case. To fix the ideas let us suppose that 7> 1. Then our two integrals have the form y1 = P(E-,, 3, x), (1 Y2 = (a, /, y, x) log x + x —( G(x), LINEAR DIFFERENTIAL EQUATIONS 477 where G is regular at x = 0. We proceed to apply the method of Frobenius given in 213 to find G. We have here fo(s = - s(s - 1) - ys = - s(s + g), fl(s)= (s - )+ s( + + )+ 3, (2 f2(s) =f3(s)..' = 0 Thus the relations 213, 2) become nfjo(s + n) + elfl(s + n - 1) = 0, or (s +n+ a- l)(s + n + -1) (3 (n + s)(s + n + 7- 1) The coefficient co is by 213, 8) o(s).= Cfo(s + 1) *..fo0( + g) = (- 1)YC(s + 1). ( + g)(s + + 1) (s + 2 g). (4 As C is arbitrary, let us take, in order to get simple formule, (- 1) C (5 (s + )... (s + + g - )(s + ) ( p + g - 1) Then 3), 4), 5) give c(s) = CO(s) CU(s), ( where (s + ) (... (s + g)(s + g + 1)... (s + 2 g) o ( + a)... (s + a + -1-)(s+,p)... (8 + + -1)' C,(s) = (s + ~a) (s + a + n- 1)(s + )..** (s + + n- 1) (8 (s + 1)... (s + n)(s + ry)... (s + +n-) ' C0(s)= 1. Thus y =o(S) )xS C,(s)xn, C= 1 (9 n=O is a solution for s = 1 - ry. We call this yi. From 8) we have Co(S) Cn.+(S) (s +1) *. (s+2g)(s + a)... (s+a +g + - 1)(s + 3... (8 + a)... (s + a +y - 1)(s +/3)... (s + /3 + g - 1)(s + 1).. (s + + n)(s +y). (s +g+ + n-1) (s+g+ a)... (s++g+ +n-1)(s+g+ ^)... (s+3+g+n- 1) (s++ 1)... (s+++n)(s+g+y)... (s+g+7y+n-1) = C(s + ). 478 48 FUNCTIONS OF A COMPLEX VARIABLE Thus we can write 9), y = c0(n) O n= Cj0x +x nC(S + g)Xn~g. (10 This series satisfies formally,the hypergeometric equation for 8 =1 -7C = - g. As c0(s) contains the factor (s + g), co = 0 for s = 1 - y/. Thius 10) becomes 0 n=,,O~x=Fa 8 1 ) since the recursion formula 3) goes over into 210, 11) for s =1 - ry. In order to apply 213, 11), let us show that 10) may be differentiated termwise wvith respect to s at the point s = 1 - y.To this end we show that the series n=O is steadily convergent in a small circle c about the point s =1 - y In cwe will have < <IS+YI<T Thus if we set I al a, 1,31 = b, we have Let us now consider the series, PE= e0 + e1RBe282 +.. 0 <ft< 1. This series is convergent since the ratio of two successive terms is (Tr+a +n)(,r+6+ n) I (ar+ n+ 1)(o +7~ +n) and this ~Ras n=o. Thus 10) converges steadily and we may differentiate it termwise. Tjhe new series so obtained is a solution of our differential equation for s = 1-7q by 21. We get thus = b 1gx8)Xnx X+ C0( n - + X8 I Cl(sg g)Zn+g. (12 n=O LINEAR DIFFERENTIAL EQUATIONS 479 Let us now set s = 1- = -g. Then cO(s) = ( y = (, 17, x). To find Ct(O) we take the logarithm of 8) and then differentiate with respect to s. This gives C71(s)o c s ~ C+K s13+ s+y/ C + s + P s +1 s + ry Setting in this s = 0 gives Cr(O) =Cn() {-+ + + 1- - + a+n-1 + 3+n-lJ Cn(o) + +...+ ++-...+ }+. 1 n 7 7Y +r n-1 To find CL( - g) we note that e (-g) = lim c0(8) = co(- ) = lim ~c8) s =- s +g s==-g s ^_ _____ - (1)(y — 2)! (y -1)! (-1)... (- + 1)(3-1)... ( - +1) We have thus a second solution. Y2 = F(',, /, x) log x + F1(a,, 3, y, x), where Fj(_._ A __ X- 1)_(_- 2)!(-)!______ _ Fl~(,,, x)- (, _)... (_ - + )(/- l).. (- + 1) x+-' + ~~- )-1 + + - - +a --- (13 a( a+1)( + 1) 1<~ 1 +1 1 1 1 1 x 1.2.y(y+ 1) a ac+1 3 1+1 1 2 y y+1J +... 2. In the foregoing we supposed y > 1. If we suppose y is 0 or a negative integer, we have a fundamental system y = xl'-y (ac + 1 - y, / + 1 - 7, 2 - y, x), (14 Y2 = Yl log x + xl~-yEl( i + 1 -- 7,y + 1 - 7y, 2 - y, x). 480 FUNCTIONS OF A COMPLEX VARIABLE 3. Let us now consider the point x = 1. ITf y - a - /3 is an integer, we have the logarithmic case. If ry - a 3 -30, a f undamental system is Y = F(a, /3, a +/3-ry +1, 1 -x), If ry - a - /3> 0, a fundamental system is y1=1x 'P~-3Y Cay 0 /+ L11- X), (15 4. Finally let us consider the point x = cia. If a. - /3 is a positive integer, we have Y, =x-F a, a -ry +lc a/3 + 11), (16 1 a( If a / is 0 or a negative integer, we have yi~x~I7/3~ /3y+1, /3a+1~!) (17 YY1log /3 -P, /3,8- y+l /31- cc+i~) 215. Logarithmic Case of Bessel's Equation. 1. The indicial equation of Bessel's equation x2d~Y + X-t+ (X2-mW2)y=0 ( dX2 dx has, at x = 0, the two roots ~ m, as we saw in 211. When m is an integer, we have the logarithmic case. As in most applications m is an integer, we wish to find a fundamental system in this case. Applying Frobenins' method given in 213, we have here fo~) =82 M f1 (s) = 0, f2(s)= 1 and fA(s) =0.for n >2. The equations 213, 2) have the form e~f0(sg + n) + C.- = 0, or c(s+n2-m2~ + eC,2 = 0. LINEAR DIFFERENTIAL EQUATIONS 481 Thus - Co 2 (s + 2)2 _ -2' 4 (s + 4)2 _m2 (s + 2)2 - m' etc. We notice thatfo(s + n) occurs in these denominators only for even n. We may therefore modify the formula for e0 in 213, 9) and take eo = f7o(S + 2)f0(s + 4)...f( + 2 m). (2 Let us set Let us set P(s) = (- )m ( + 2)2 -m2... (s + 2 m - 2)2 - m2 so that so that 0 = (- )m" (s + 2 m)2 - M2 P(s). (3 Then the series y = Xs C (Yexn n=O becomes here x2 X4 y = e-oX -_(s + 2)2 - m2 +(s + 2)2 - m2 { (s + 4)2 - m2} X.... + Cxs+2m 1- m - 2_2-m2 pP- (s +2m +2)2- T2 (4 4 - + ( + 2 m + 2)2- 2 (S + 2 m + 4)2 _ m2 = COXu + CXs+2mv = U+ V. (5 Here U embraces only a finite number of terms. The series v is steadily convergent for every x and for any s > - (m+ 1). For let x I < B. Then (s + 2 m + 2)2 - m2 > > 0. Hence each term in v is numerically < the corresponding term in R2 R22 R2.3 1+ - + + +... o- (a- + 12) o-(o- + 12)( + 22) The ratio of two successive terms is here R2 + n2 and this 0 as n = c. o til n P C1 CD CD,, 1 o. o 0 I2 I 0 110 0c CD ~~II~I C cF O?o + m 0c 00 c 0~s 0 -r CD 0c (1 II 10 '-i II + fca I ~1 -ct + a, P-' 11 cr + 1rc I5, ^i Or 0-? CD CD 0 11 II ^-4 ImD 10 I ltt ^ I I 12 a f ~ N ^s N ^r^ r? -4 Q3 I - + -c (-N S I S 10^ + I 0 CD 00 (7I Co CR o_ + 0 CO o0 -1 - NID 10 CD ct-. F -o CD 0 cP -. 0 CD,. c0 3 -- o0 C I - Ss o II. 0 N w — ^? CD 0 CO 0 ~ 0 hI o I i~ r 11 17i I I IC ~lel co + 0 CD O-4 CD 0 c: CD ~ct0 -CD 0 O CD c-t CD 0 CD 0 CDi 0 t4 -q t-. t, — LINEAR DIFFERENTIAL EQUATIONS 483 leads us to take as second independent integral mm-1 ( 1 k) 2k y = - ~ H1(k) Jm.) ) log x _ x M 00 ~ ~ o(k)+t(k +m)l (7 -:2 n1 (m)n1 (m + k) 216. The Differential Equation for K, K'. 1. In 1.89 we saw that 1 a K=v( 12' 1 k) K' w )(l, i, 1-k. 2' ~' ' 2 2 Thus.K and K' satisfy a special case of the hypergeometric equation for which a = / =}, Y = 1, viz.: dyl 1=k2. x(z- 1)d2 +(2 -1) + y X,=2. (1 d+2 c- 4 Referring now to 214, we see that a fundamental system of integrals of 1) for x = 0 is _= '1 1 (2 Yz y= slog x + Fj(1, il, ). 3 Here ~( 22 )2(1+2-4.4 5 > + + + x (4 2. Let us find the development of K' about x = 0. Since K' is a solution of 1), we must have K' = Ay1 + By2, (5 or since 1 _ 7rK' = 2 AK+ 2 BKlog k2 + 7rBF1. (6 From 196, 2) we find k2 1 - 16q+... Hence K'...+... x = k2=216q+....16e +... 484 FUNCTIONS OF A COMPLEX VARIABLE or ~ 71TK '=log 16 - log k2~.. (7 Also F 1. k2 + Also B-1 =,7B 2'k + - Bi%2 + ~~ ~ (8 In 6) let us divide by Kand put in 7), 8); we get (4 log 2 - 2 A) - (I+Blogl +.=O. Thus A = 2 log 2, B= —. Hence 5) gives K'- (2 log 2 - log k 2)F(}!,, 1, k12)- IP1(~, 1,, k2). (9 217. Criterion for a Regular Point. We saw in 212 that x = a is a regular point of d2y dy d2+p(X) +~q(x)Y =O0 (1 dX2 dx 0 if p, q have the form g s) h(x) x-a (x - a)2'( where g, h are regular at x = a. When x = a is a regular point, 1) admits a fundamental system of integrals, Yj = (x- a)?'i01fx) ( 92= (x - a)"2 '021(x) ~ 022(x) log (x - a) where r1, r2 are roots of the indicial equation at this point. We wish now to establish conversely: If 1) admits 3) as a fundanental system of integrals at the point x = a, it is necessary thatp, q have the form 2). For we saw in 212 that. if we set =2 =Y1fzdx, (4 then z satisfies the equation dz + gz = 0, whereg=p~2 1. (5 dx 91 LINEAR DIFFERENTIAL EQUATIONS 48 485 From 4) we have d Thus from 3) we see that z must have the form z = (x - a)s (x)~ # (x) log (x - a)h where, J' are one-valued about x = a. Let now x makce a circuit about x = a. If z acquires the value, Z, this must be a solution of 5). Hence = CZ. (6 But 27i -e e2isz~+ 2 wie2,,is(x - a)~-r Putting this in -6) gives z(c - e2,ri) -1 277wie 2lr!is - a)s'4r 0. This requires that -f 0. Hence z= (x - a)so(x). Thus S ~~+f(x), (7 zdx x-a where f is regular at x = a. On the other band, 5) gives 1 dz _ 2r1 a+k(x), (8 where k is regular at a. Thus 7), 8) show that p(x) has at most a pole of order 1) at x = a. From 1) we have, setting y = l q(x)- -p y1 Yi Now -, 1(x) where I is regular at a. Hence q has at most a pole of order 2. 486C FUNCTIONS OF A COMPLEX VARIABLE 218. Differential Equations of the Fuchsian Class. 1. When all the singular points of a linear homogeneous differential equation are regular, it is said to belong to the Puchsian class. Now in order that x = a is a regular point of d2y + P3) dy, (1 the coefficients p, q being one-valued, it is necessary that x = a is at most a pole of p and q. Hence p, q having only poles, even at x = 00, must be rational functions of x. As the poles of p cannot be of order > 1, and those of q of order > 2, we can write h(x) g2(X) wheref, g are polynomials and h = xX` + cix +1... + cm = (x- al)(x - a2)... (x - a,). To find the degrees of these polynomials we use the fact that x = oo must be a regular point. Let f (x) = aoxr + axr-1 + + a7, g(x) = boxs 8 61x s-+ ~6s. We set nowX= in 1). Since U dy 2d d2y 4 2 d dx dul dX2 Udu2 du' we find as transformed equation d2y~{ 2U3a-u2rd q dU2 zc4 du As ao + a-a + ~u + ar ar ur i~+l+..CO. Umr, q 1 + d1U +... + d2~mu bo + blze +... + byus 2m-s 4=l~zcf-~-~J~,2m. U LINEAR ]DIFFERENTIAL EQUATIONS 48 487 we see that 21 u3 - u2p- P(u) Pi = -4- u r-m~2' q Q(u) 4 s-2im+4' U U where P, Q are regular at m = 0. Asp, cannot have a pole of order > 1, andl q1 one of order > 2, we have r-m~2<1,s-2m~+4<2, or r<m-l s<2mi-2. Thus m-r=1+Jc 2m-s=2+1, le1>0. Also1 U2 Hence P(u)= 2 - aoUk +.;Q(u)-boUt+. (3 Let us set X =lim-P=lim xp, (4 u=:OU Xx= ~k= lilinI 9=lim X2 q. (5 U=O0U2 X0 Then we see f rom. 3) that P(0) =2 -, Q(0)=~ (6 2. At the singular point x = ai let ri, pi Ibe the roots of the indicial equation. The roots at x = oo we will denote by r., p~. Fuchs showed that these roots must satisfy the relation This is called Fuchs' relation. Let us find the indicial equation at x = ai. We bring 1) to the normal f ormn (x-a,) 21 (X) dY+ (X - a5) hi (x)f (x) d-Y ~ g (x) y = 0. 71 dx2 dx The indicial equation is r(r - 1)14(a5) ~ rh5(aj)f(aj)~+g(a5) = 0. (8 Now h (x) = (x - aj)1hi (x). 488 FUNCTIONS OF A COMPLEX VARIABLE Hence h' () = h (x) + (x - ai)(x). Thus Th~us hi(a) = h'(ai). We may thus write 5) r(r- 1) + r+ (a)= o. A' ( aj) hi(ai)2 Hence r p - f(ai) ri + pi = 1. (ai) /' (ai) Let us now write 1) in the normal form for x = -, we saw that it takes the form u u2dY~,updg~ QY=0. du2 du Its indicial equation is therefore r(r - 1) + rP() + Q(O) =0, or, using 4), r2(,. r2 + (1- X)r +/~ = 0. Thus roo + p = X- 1. From elementary algebra we have f(x) = f(a~) - Hence from 4), f(a) x=oo h'(ai) Thus 1 TrGO + pO = C f(ai) __ 1 rFr + p~ =o h'(ai) From 9), 10) we have 7). (9 x= oo. Setting (10 219. Expression of F (a,, /, x) as an Integral. We leave now the general theory of linear differential equations and return to the hypergeometric function. Let us show that when x <1, O</ <r, (1 we may express F (a, /, ^y, x) as a definite integral, viz.: F (Y1 /,. x) =B(/ - 1 ). UP-1(1 - u),-P-(l - XU)-adu (2 J( - B(8, y-/3)' LINEAIR DIFFERENTIAL EQUATIONS 48 489 where B(p, q) is the, Beta function B(p, q)f-Upl(.1 _ U)q-ldu. (3 For by the binomial theorem (t~~xa~a=1~ftxu~~ xa+ l x2a2 ~.. 1 1.2 whenI xuI < 1. Hence the integral J In 2) may be written J=fu0AI( - u)'Y-P1du ~ uPX(I - u)~'Y —du 1. 1 x2fX ul3+(1- u)/Y4-1du + - + cc (-~ a x2B (, 213~2y-I) +. (4 Now ~~B (fl~ 1,y -/3) B (I3 y -fl). ry Hence etc. Putting these values in 4), we get 2). 220. Loop Integrals of the Hypergeometric Equation. 1. In the last article we have shown that the hyp~ergeornetric equation X~l- ) d Y+ y -Qxc,+I3~)x a-c4y =O (1 dxl ~~~~dx admits as solution the integral 2) when the conditions 1) of that article are satisfied. Let us replace the path of integration (0, 1) by a more general path L, properly chosen; we proceed to show that 1) admits a solution of the f orm y =f -zX) - au (z) dz. (2 In fact, putting 2) in 1), we get O dF A __ d ~( dz,Idz 490 FUNCTIONS OF A COMPLEX VARIABLE where v = ( - x)-a. (4 F= z(1 -z)du _ -- + (/3 - z}, (5 dz a=z(l-z)U -V-vdI +~ -Y+(f3-c+1)zuv. (6 ( dz dz J To prove this we may proceed as follows. From 5) we have vdF =(l - z) vu1+ 1 - a + ry + z(a - - 3) }vu - (/3-+ 1)uv. dz From 6) we have dG = - z(1 - z)vuw' + a-vy - 1 + (/3-a + 3)z}vu' da dz + a(~C+ l)Z(l- Z) (z - x)-a-2 + (y7- ~- 1 + (a-/3 + l)z) (Z- x)-a-l +(3- a + 1)v}u. Thus V + dG = ( (a + ) (l-Z) (z X)-a-2 z dz [- a- 1 + (c -/3 + 1) ] (z- x) a~- } = Hu. On the other hand we have from 2) dr == ac (z - x)-a-udz, dx d2y + ( 1)f(z - x)a-2 udz. dx L Thus 1) becomes 5(e+l) X (I - ) (Z- X)-a-2+ (7_- (+ 4+ )X) (Z - )-a-l - (, (z - x)-a dz. Now we have identically x(1 -x) =z (1 -z) + (2 z- 1) (z-) - (z-)2, 7 - (a + /+ 1) x = y - (a + 3 + 1)z + (c + 3 + 1) (z - X). Thus the brace in the foregoing integral reduces to the function H above, and this establishes 3). An integral of= o u= e = za-y(z _ 1)yiP-1 (7 LINEAR DIFFERENTIAL EQUATIONS 491 'This in 6) gives G = GCc-Yl, _ YB(Z X) —1 (8 Thus when u is chosen as in 7), F= 0 and hence the first integral in 3) vanishes for any path. Also if L is so chosen that a in 8) takes on the same value at the end of L that it had at the start, the second integral in 3) vanishes. In this case 2), or what is the same, y =f z-Y'(z - 1)/'P1(z - x)-dz =fw(z)dz, (9 is an integral of 1). Here W - zo —Y(z - 1)Y"'P'1(z - Xj a. (10 2. Let 10, 11, I, l denote loops about the points z = 0, 1, x, xo, respectively, each circuit being described about the corresponding point in the positive sense. Let Q,, iw be the end values of G0, wo after describing I., etc. After a circuit about z = 0, Za ---+l = ea-y-YI-) logz goes over into e(a-y+1-)(1ogz+27ri) e27i(a-,y+l)Za- 'Y+l 3 = e27i(a-Ty~y1. a Thus -0= e2,irica- Ca0. Similarly = e27i(a —w0) In the same manner we find 2-i (-y - P) 2 7ri(-y-p)W-B)W G = e2 CT' a1 W, = 6%, = e-28 a, Wt = e-2 awx -O e 27P ao, wit,0=e2743w Co. Let a, 6 be any two of the four points 0, 1, x, oo. Let Lab be a path about a, 6 as in Fig. 2, ~ 150. Obviously, as far as the values of a and the integral 9) are concerned, this path is equivalent to 1a1b1a_1b16 As a returns to its original value, &IabJ wdz 492 FUNCTIONS OF A COMPLEX VARIABLE is a solution of 1). Since we can choose the points a, b in 4.3 -=6 1.2 ways, we get in this manner six solutions of 1). They must of course be linear functions of a fundamental system, as shown in 210. 3. As an illustration let us consider yo. For simplicity let us take Ix > 1 and suppose that a, / do not differ by an integer. As the loop Lo1 let us take a double loop 2 running over two little circles about z = 0, z= 1 and the segment of the real axis joining them. Then on 2, < 1 and x (z - X)-a = aox-(1 - X =x-{ao+al+a2+.... = X aX + X2 + a +I Now the two fundamental integrals at x = o are, as we saw in 210, 16), 17), 771 =a X a, X- 7 + 1 3 ~- + 1 ), = x-^jP(/3, -y)/3 i + 1 a a -,!I). Hence y must have the fo Y =_no at [z,+- (z - 1)Y-P-ldz y = C0 + + 22 As y does not coundanta in any powegrals of x in common with as we saw in 210, 16), 17), z771i X-aF(, 1 —,, --— 1,r-1 -X Hence y must have the fornm As y does not contain= 0. Hence 11) differs fr in om monly by a constant factor.must =. Hence 1) differs from l only by a constant factor. CHAPTER XIV FUNCTIONS OF LEGENDRE AND LAPLACE Functions of Legendre 221. The Potential. 1. We wish in the present chapter to develop some of the more important properties of these functions which are of great importance in mathematical physics. We begin with the polynomials introduced by Legendre, who was led to study them while treating of the attraction exerted by the earth on a mass exterior to it. Such questions arise in celestial mechanics and in geodesy. Let us find the attraction exerted by a body B on a unit mass p situated at the point A. The force exerted by an element of mass dm situated at P on lu is, by Newton's law, dm f= --- 52 = ( - )2 (y - )2 + (Z - c)2. Z If AP makes the angles a, 3/, y with A ) the x, y, z axes, we have abe \ X x-a, y-b o cosa = —, cos/3= — ' z-c ~^\ ^ \x cos 7 = 8 ' The x, y, z components of f are therefore dm x-a dm y-b dm z - ~ - - 5 e - ~ ~C —. 82 8 ' 82 8 82 If we denote the total force of attraction exerted by B on tL by F and the x, y, z components of F by X, Y, Z, we have v rX -- a X= - dm Y = dm Z -dm -3 8 83-~- dm ' 493 494 FUNCTIONS OF A COMPLEX VARIABLE Let us consider the function V= of. (1 We have ay m, r. Q-t fl\ ci sa dFiidm x-CL a (o a=cjd. = _ X. (2 ax Ox=) 8 J Similarly aTr_ v=- Z By IOz Thus the function 1) has the remarkable property that its first partial derivatives are, aside from sign, the components of the force exerted by the body B on a unit mass / situated at A. This function V is called the potential of the body B with respect to the point A. It is of extraordinary importance in many parts of applied mathematics. For simplicity we shall set c = 1. 2. Let us now show that V satisfies the partial differential equation 2+ + V 0. (3 ax2 0y2 az2 This is known as Laplace's equation and is often written av= 0. (4 We have from 2) 02 fs3 (x -a)2 1 m v_ f ( Jl - - dm, OX2 85 83 J and similar expressions for the two other derivatives in 3). Thus adding, 32 3 A-'d 8m3 dm =0. 3. As a special case we see that 1 V-7=,18 (5 is a solution of 3). 4. As an exercise in the calculus the student may transform 3) to polar coordinates, x=rcosOcoso, y=rsinOsino, z=rcosO. (6 FUNCTIONS OF LEGENDRE AND LAPLACE 495 It is convenient to call 0 the altitude and q the azimuth of the point x, y, z. After a lengthy calculation we find that the left side of 3) becomes,AT /_ a r\2 + 1 0 aV\ 1 Sl 0(7 a ar ar )sin 0ao (s a sin2 a02 When the attracting masses are symmetric with respect to an axis, we may take this to be the z-axis. Then V cannot change when b changes. Hence a0 and in this case 7) becomes /xV=A a r2a +- jl (sin 0 a (8 ar ar sill 90 ) ao 222. Definition of Legendre's Coefficients. 1. In many investiga1 tions it is useful to develop the quantity - in a series. In doing this we are led directly to Legendre's coefficients. Z Let 0 be the angle between a and p. Then Then =a2 p2- 2 ap os.. P xyz Let a a Xyz r=, when a <p P - P, when a> p. a Then 8=a2(1-2rcos0+r2), a>p =p2(1- 2 r cos0 + r2), a < p. In either case the development of - leads us to develop V=-, O<r<l. VI - 2 r cos 0+ r2 This we now do, using the binomial series 1- 123.24 1 3. 54 3+ (1 u) + - 1 + U2 + U3 -2 =1 2 —4 2 4 6 (1 (2 (3 496 IiUNCTIONS OF A COMPLEX VARIABLE which is valid when I u < 1. Let us therefore set x = co 0, = 2rx - r2, whence m Ibm = ( — I) m- )! (2 x)m-srm+. s=O s! (m - s) This in 2), 3) gives 7=^ n(- 1 ) (2m) i ) m+'~) (2 3 rmTsxm-8 n==o s==o m+8s. s (m - s), ( )8) (2 m - 2,.! m-2s, < 2 s < s1 _rm (-1)s (2m-2)!.Xm-2s 0 <2s<m.,in=O s= 2m(m - s)!(m - 2 s)!s! Thus o V= Pm(x)rm = Po + Plr + P2r2 + ' (4 m=t where 1. 3. 5.- (2 m -1) PMm(x) = m! f x m m- m-2 m- m-l m —2.m-3 xm.. (5 t 2(2m ) 2.4.2-1.2m-3 These are Legendre's coefficients or polynomials, for on the one hand they are polynomials in x, and on the other they are the coefficients in the expansion 4). We have Po=, P=x, p2=-x 2- (6 P3 - x3- - x, 4 - -2+8, etc. 2. From 5) we see that P(- x) =(-1)mPm(x). (7 Thus Pm(x) is an odd or even function as m is odd or even. 3. When 0=0, x= cos = 1. Then V= = 1 +r+r2+.. 1 - T 1-r Comparing with 4), we see Pm(1)=1, m=l,2,.. (8 FUNCTIONS OF LEGENDRE AND LAPLACE 497 4. From 5), 6), we have P2m(O) = ( —1)m.5... 2 m (9 2.4.6...2 m P2,+1 (0) = 0- (10 5. The equations 6) enable us to express x, x2, x3..* in terms of Po, P, P2,." Thus we find x= Pl), X2 = P2(x) + 3 Po(x), X3= P3(x) + 3 P1(), etc. In general we see xn has the form xn = aoPo(x) + alPl(x) +.. + a,P,n(), (11 the coefficients being constants. 223. Development of Pm in Multiple Angles. 1. We have 1 - 2 r cos 0 + r2 = (1 - rei0) (1 - re-i). But (1 - rei0) — = a 0 + a1rei0 + a2r2e2 + e... where 1. 3 1 3.5 ao=1, a1=~, a==, a3, 2 a2=24 a3a2. 4.6 Hence V= = (a0 + a1lei0 + -.. ) (a + alre-iO +..) V1 - 2 rcos 0 + r2 = 1 + Plr+ P2r2 +* * Thus P,(cos ) = 2 aoan cos nO + alan_1 cos(n- 2)0 +a2an_2 cos(n-4)0+. -1 35.5 2os nO- + n. cos (n - 2)0 (1 2.4.6.. 2n 1 2n-1 _2.3.. 2n-n-1 + ^ *2^ Cos (-4)0 + 1. 2'2 n- 1 ~ 2 n - 3cs (n-4)0 +... From this we have P0=1, P1=co s, P2=-(3cos2+1)... (2 0 5 1 "VU " 2 I( 498 48 FUNCTIONS OF A COMPLEX VARIABLE 2. We note that all the coefficients in 1) are positive. Thus P.,(cos 0) has its greatest value when 6 = 0, for then cosnO, 'cos(n-2)0,... (3 all take on their maximnun positive value 1. Thus P,,(x) has its maximnm value for x = 1. On the other hand P.~ is certainly greater than the right side of 1) when we replace the quantities 3) by - 1. Thus - P"(1) 7~ P,,(cos 6) P"1) or using 222, 8), 1 < P,,(cos 0) < 1. (4 224. Differential Equation for P,, (x). Let A be on the z-axis. Then 1 ~ ~~~~~~,5 V"1- 2r-cos-0+ a is independent of p.Now V satisfies Laplace's'eqnation A V= 0 as we saw in 221, 3. This we saw in 221, 8) is here (r2~ + si6Osin VO ) O. (1 Let us set A o,o, a x = COS 0. (2 Then 1) becomes Now by 222, 4) Putting this in 3) gives dx dxJ' Hence P,. satisfies ddn 0 dx dx or (1 x2) C2 -2xAV+ n(n +1)y 0. (5 FUNCTIONS OF LEGENDRE AND LAPLACE 499 If in this we set U = x it becomes U)d2y +, d u(1 3 u y + n(n_ - l)y =0. (6 uCd2 ~ ( d~ This is a special ease of the hypergeometric differential equation. Comparing with 210, 1) we get n = n1+ 2 la 2 A fundamental system of integrals of 6) is, as we saw, 210, 10), 12), F R- n, n+22 2 ' '. 2 2 2 x2> = n - 1 n + 2 3 ) Y2 = ~2 2 ' 2. Now when aI or 1 is a negative integer, PQa, 3, y, x) reduces to a polynomial. Hence when n is an even integer, y, is a polynomial and Y2 is an infinite series; while when n is odd, &2 is a polynomial and Yi is an infinite series. This shows that P,(x) = cly,, n even c=y2, n odd. Comparing with 222, 5), we get I1nl 3 -... n -1 +1 1 ( P2+ (X) =( —) ( — n(, n 28 x2 2.4.6.. 2n 225. Integral Properties of Pn(x). 1. In 224, 4) let us set y = P,mand then y =Pn; we get m(m + 1)Pmi(lx2)d. 0O dx dx n(n 1)P, +(ld (XdP2 = 0. dx dx 500 FUNCTIONS OF A COMPLEX VARIABLE Multiply the first by P,,, the second by Pm and subtracting, we, get on integrating (m - n)(,m + fl + l)fPMPndx = 0. Thus JPmPndx =0, mn. (1 From this f ollows the thieorem: Let Pm(x) be a polynomial of degree m < n. Then tfmx)Pnx)dx = 0. (2 For by 222, 5, 11) Fm= c0P0 + c1P1 +-. + CmPm. Thus the left side of 2) s=( C~'P X "()d =0,byl1). 2. We have _________ =1 + rP1(x) + r2P 2 (X) +...= rm'P,,(x). VI - 2 xr +r2 m=Q0 Squaring, we get 1 1-2 xr +r 2 r~m )P (X),m, n =0, 19,2.b (3 Now Hence, integrating 3) and using 1), 4), we have 2 ir Z 2nf P (X) dX. Hence, equating the coefficients of like powers of r, we have f ~1P (x)dx 92 (5 n 2 n + FUNCTIONS OF LEGENDRE AND LAPLACE 501 3. We have 1= = 1 + zPl(x) +2P2(x) +.. -V/1 - 2 xz + z2 Hence, C denoting a small circle about the origin in the z-plane, fdz~1V1 dzc + Pl(x) J z; ~ + Pn(X)f d. z"+I-/l - 2 xz + z2 +.. P( = 2 7riP,(x). Thus n);I n1 (6 ~~~Thus P,(x) = 2 +/i Y -\1V - 2 xz + Z2' (6 the radical having the value + 1 for z = 0. Let us set z =- in 6), we get U P.(x) = 2 wil'7 Pn 7ri2 r VI - 2 xu + u2' where D is a large circle about u = 0, which u describes in the positive direction. 4. In 7) let us set - 2 xu+ u2 = w - u, or 1-2xu+u2=(w-u)2. Then w2- 1 w- u U =, du = dw. 2(w- x) w - While u describes the large circle D, w will describe a curve t which is approximately a circle of radius 2 R. Thus 7) gives 1 ~ (w 2 -)n Pn(X) 2 1if;2n8 - ~x)nj: dw. (8 Since the integrand has no singular points in the distant part of the w-plane, R can be regarded as a large circle whose center is x. The relation 8) is due to Schlafli. 502 FUNCTIONS OF A COMPLEX VARIABLE 226. Rodrigue's Formula. 1. Let f(x) = (x2 - 1)n. Then by Cauchy's integral formula f(x)= I ( - Jn dw, where R is a circle about the point w = x. Hence fn(x (W2- 2 j)n - 21n!P,(x) by 225, 8). Thus 1 dn P-(x) 2n d (X2 - l ), (1 2 n,! dXn a relation due to -Rodrigue. 2. From this relation we can prove the theorem: The n roots of P,(x) = 0 are all real, and lie in the interval We start with f (X) = (X2 - J)n = (X - I)n(X + l)n. rhis shows that x = 1 is an n-tuple root, and the same is true of x = - 1. As f is of degree 2 n, f(x) has no other roots. By Rolle's theorem f(1)-f( —1)=0=2f'(aj) -1<al<l. Hencef'(x) vanishes at x = a,, a point within 2f. But f' (x) = 2 n (X2 - l)n-IX has x= ~1 as roots of order n - i. Thus f'(x) = 0 atx= ~ land at x = a1, and only at these points. We may reason in the same way onf"(x). We have f" (x) = 4 n (n - 1)(X2 - 1)n-2X2 + 2 n( X2 - )n-1 This has x = ~ 1 as roots of order n - 2. Rolle's theorem again shows thatf"(x) must = 0 at some point h, within (- I, a,), and at some point 62 within (a,, 1). We have thus found 2 n - 2 FUNCTIONS OF LEGENDRE AND LAPLACE 503 roots off"(x). Since the degree of f"(x) is 2 n - 2, there are no other roots. Thusf"(x) vanishes at just two points bi, b2 within 1. Continuing in this way, we see that f(n)(x) vanishes at n and only n points within W. By Rodrigue's relation 1), P,(x) and f (n)(x) differ only by a constant factor. Hence Pn(x) vanishes n times within W. As P, is of degree n, these are all the roots of Pn (x). 227. Development of f(x) in Terms of P (x). 1. Let f(x) be a one-valued continuous function of x having only a finite number of oscillations in the interval 2 = (- 1, 1). Then it can be shown that f(x) can be developed in a series of Legendrian functions f(x) = Co + P1 () 2P2 (X) + *** (1 which is valid for any x in W. Moreover this series can be integrated termwise in 1. Admitting this, let us show how the coefficients c, may be found. Multiplying both sides by Pn (x) and integrating, we get 1 1 I f(x)P, (x) dx = cnPPndx. d-l Jn-1 All the terms on the right vanish by 225, 1), 5) except that corresponding to c,. Thus f (X)Pndx = cnf P (X)dx= 2X1 1 -Hence2 n+ f( ( enn2 ff(x)P.(x)dx. (2 Thus we have the theorem: Let f(x) be a one-valued continuous function having only a finite number of oscillations in the interval (- 1, 1). Then = 2 +n~ ( (3 f W = S 2 2+Pn( Xf(W)Pn(x)dx. (3 2. Since Pn(x) satisfies the condition of this theorem, we have dP,(x)- Pn ()- an-,P,- + a-2P 2 + * dx 504 FUNCTIONS OF A COMPLEX VARIABLE Since PI is odd or even with n - 1, we must have P'(x) = an1P,1 + an-3Pn-3 + Here by 2) am =2 j P IdX. a, P P'dx Integrating by parts gives 2 m + = 2 m + 1{ since the integral vanishes as PI (x) is the sum of P's whose index is < n. Thus PI (x) = (2 n - 1 Pn-,+ (2 n - 5)Pn-3 + (2 n - 9)Pn5 + (4 3. Let us show that f(x) can be developed in a series of Legendrian functions f(x) = ao + al1P(X) + a2P2 (x) + (5 in but one way. For suppose that f(X) = bo + blPl(X) + b9P2(x) ~ + (6 were a second development valid in (- 1, 1). Subtracting we get 0=co + cP(x) + c2P(x) + (7 where C-= a_ - b. Let us multiply 7) by P,(x) and integrate between - 1 and 1. Granting we can integrate the resulting series termwise, we get 0= f 1 0 = e Pndx + e P,I dX + e P2Pndx +~ (8 Here each term is 0 by 225 except the term Thus 8) reduces to 1 2 ce 0 = ef Pn2dX = 2n+ Hence C1 = 02 and thus an=bn n= corresponding to c,. 1, 2... FUNCTIONS OF LEGENDRE AND LAPLACE 505 228. Recurrent Relations. 1. (n +1)P,,,1- (2 n +1)xP,,~+nP1,= 0. (1 (1 - x2)P' ~ nxP,, - nP,,,1 0. (2 (1 - 12p,_ ~ nPn - nxPn-i = 0. (3 These may be proved by putting in the values of P., -P'. as given by 222, 5). There results a polynomial in x whose coefficients are all zero. A more expeditious method is the following. Let V= (1 - 2xz + Z2)-1 - P(X)~ +zP1 (X)+ 2ZP2x F..( Thus 07 V 1Z 2x + Z or (l2xz +Z2) a0 ~(Z -x)V=0. ( On the other hand, we get from 6) a K P,~+2P2+32P3 ~ (8 Putting 6) and 8) in 7) gives J~n (n +I 1)P,~1 - 2 xnP,, + (n - 1)Pn-l + Pn1j xPn.j 0. As all the coefficients are 0, the coefficient of Zn here gives 1). 2. To get 5) we use 227, 4). Thus PI (2 m + 1)P, + (2 n - 3)Pn-2 ~ Subtracting gives 5). 3. To get 4) we have only to differentiate 1) with respect to x and use 5). 4. To get 2) we multiply 4) by x, getting x2PI = XP1_1 ~ nxP,,. 506 FUNCTIONS OF A COMPLEX VARIABLE Hence (1 X2)PI - P - XpP or (1 - x2)P' nxP, = PI - xP, on using 4), 5). = nPn-1i 229. Legendre's Functions of the Second Kind. We saw in 224, 6) that Pn(x) satisfies the equation d2y (3 IN dy U = X u(1 - u) ~ -u - + (~n ~1) )y = u du 2 2 2 dL This equation admits, by 210, 16), 17), two integrals about U, = C, viz.:?I n-1l 2n-1 I\ Y2=f1 (2 n~2 2n~3 1) 92 = Z Lt I I (3 X111 2 2 2 xY Since P(a, /, y, x) is a polynomial when a or 3 is a negative integer, we see that whether n is odd or even nz n - I 2n - 1 ~ 2' 2 2 is a polynomial in w, and thus 2) is aside fronx a constant factor nothihg but P,(x). The other integral 3) multiplied by a constant factor gives rise to Legendre's Functions of the second kind, viz.: Q() = l.2.3...n 1 jn ln+2 2n~3 1> 3. 5.7....2n + 1 X 2'2 ' 2 'x2' 1xx>1. (4 230. Recurrent Relations for Q,. 1. If in 229, 4) we set n = 0, we get x 2 21al or Q0(x) = 0gX+ (4 2 x-(1 FUNCTIONS OF LEGENDRE AND LAPLACE 507 Using 229, 4) we prove at once that Q, -xQ01=0 (2 and (n + 1) Q~6+1 - (2 n + 1)xQ. + nQ,.- = 0. (3 These show that Qn(X) = n(X>8log - 1 + T,(x) (4 x - when 5, 1 are polynomials. We can go fnrther by observing that the recursion formula 3) for Q, is the same as that for Pn in 228, 1). Let us set g Z L log Z1=1 x-1 in 2), 3),weget Q, xQ0 1- 1 'PL -Z,. T'hen 2 Q2 =3xQ,-Q0 2P2==3xP1-P0. Hence = P2L2 2 - 2 if weset 2=Ip=2 X This is perfectly general. For let us admit that Qn - _' Pn-L - Zn (5 is true for n and shov that it holds for n + 1. Here Z, is a polynomial of degree n - 1. For by 3), (n + 1) Qn+1==(2 n I)x Q-nn Qn-1, or using 5), (2 n, + 1)xHI — PnL - Zj) - nn 1 Pn_,L - Znj3 - 'L (2 n ~ 1)XPn -- nP,, - (2 n ~ I)xZn + nZn1 - -'(n + 1)LPn+1+ InZnZj - (2 n + 1)xZ&, which goes over into 5) on setting -(n+ 1)Zn~j= lnZ -1-(2 n + 1)xZ,. (6 This is a recursion formula for Zn and shows that Z, is odd or even according as n - 1 is odd or even. 2. Since L is a logarithm and Pn, Zn are polynomials, we see that Legendre's equation 224, 6) does not define any new class of functions, that is, its general integral is a combination of polynomials and logarithbms. 508 FUNCTIONS OF A COMPLEX VARIABLE 231. Development of Z, in Terms of the P,. Since Z, is a polynomial, we can develop it in terms of P0, Pp P2 *.. by 227, 1, or by 222, 5. Thus Zn = alP,_ + aP,,-3 + *~ (1 the a2m being all 0 on account of the parity of Zn. To determine the coefficients in 1) we use the fact that Qn is a solution of Legendre's equation 224, 4) d (L1 - X2) d + n(n + 1)y = 0. dxL TX^] 4-1^0 This gives d- d 1 g- x2) ] Z + n(n + 1)Zn-2 P =. (2 But by 227, 4) P = (2 n- 1)Pn,_ + (2 n - 5)P,_3 +. (3 Thus 1) and 3) in 2) give 2n- 4m- 1 a2m+1= (2m+ )(n-m) = 1.. Hence 2n- 1 2n - 5 2n-9 p (4 Zn= - Pn-l + -- P 3 + Pt n-5 * * (4 1. n 3(n - 1) 5(n - 2) 232. Laplace's Equation. 1. One of the most important equations in mathematical physics is Laplace's equation 02u 02u 0% Au = + ~ + 0. (1 9x2 aY2 a v Example 1. Suppose heat is passing into a body at certain points of its surface S, and leaving at other points. It is easy to show that the temperature u at any interior point P of the body satisfies the partial differential equation u = eau, (2 ot where c is a constant. In many cases a stationary state sets in; as much heat leaves an elementary cube described about the point P as enters it. In this case the temperature u is constant, FUNCTIONS OF LEGENDRE AND LAPLACE 509 and hence a- =0. Thus u satisfies in such a case Laplace's 9t equation. It can be shown that when u is known on the surface S in this case, the value of u can be found at any point P within the body; in other words, the solution of 1) is uniquely determined when u is given on the boundary S. Any function u satisfying 1) is called a harmonic function. Example 2. Suppose a fluid, that is, a liquid or a gas, is in motion. At the time t, the particle at the point P is moving with a certain velocity U whose components call u, v, w. Let V be the volume of an element of the fluid at the time t; at the time t+-dt, this volume has changed to V +dV. Thus the rate at which V is changing is dV-; it is called the divergence of the dt' vector u. It is denoted by div. div a. One finds easily that Ou av aw divu = + +- (3 Ox ay az If the fluid is incompressible, as it is sensibly for liquids like water div u = 0. (4 In an important class of problems the velocity t is such that its components are the derivatives of some function )(x, y, z), that is U, v=, W = -. (5 ax ay az We call h the velocity potential. If the fluid is incompressible and its velocity has a velocity potential q, then 4 satisfies Laplace's equation 1) as is seen at once by putting 5) in 4). The surfaces +(x, y, z)= C (6 are called equal potential surfaces. A curve in space such that the tangent at each point of it has the direction of the vector u at that point is called a stream line. 510 FUNCTIONS OF A COMPLEX VARIABLE These cut the surfaces 6) orthogonally. For the normal at a a point of 6) has direction cosines which are proportional to ax a y ' z but these by 5) are proportional to the direction cosines of the vector u. 2. When the particles of the fluid are all moving parallel to a plane, which we take as the x, y plane, we may neglect the component w of the motion since it is 0. Let f(z)= U+iV (7 be an analytic function not necessarily one-valued. Then, as we have seen, the Cauchy-Riemann relations hold, or aU 1 aU V (8 Ox oy y ax From these follow that 02 U 2U 02 V 02 =~U dU 0 8 + =0. (9 ax2 by2 ax2 ay2 Thus U, V satisfy Laplace's equation for two variables. Let us take one of the functions U, V (to fix the ideas, say U) as a velocity potential. Then by definition the components of the velocity U are aU AU =-, = - - - Ox Oy The relation 9) shows that div u = 0, thus the fluid is incompressible. From 8) we now have oUoV oUov d+ - =0. ax ax Oy Oy Thus the two families of curves U = const, V= const (10 FUNCTIONS OF LEGENDRE AND LAPLACE 511 cut each other orthogonally. This gives the theorem: The two components of an analytic function 7) may be used to define two families of curves 10), such that one family represents the stream lines of the motion, the other the curves of equal potential. 3. To illustrate this theorem let us take as analytic function f(z) = z2 = 2 y2 + i. 2 xy. Then U=x2-y2, V=2xy. These give rise to two families of equilateral hyperbolas whose asymptotes are the lines y=x, y=-x and y=O x=0. 233. Theorems of Gauss and Green. 1. In studying the solution of Laplace's equation we shall find it extremely useful to use some theorems relating to surface and volume integrals due to Gauss l yv and Green. Let S be an ordinary ~dy -_.-7 closed surface. Let us deffect a rectangular division of the yz-plane. Each rectangle dydz x -- may be used as the base of a cylinder which cuts out elements of surface dr', darf, da-"'... on S whose normals call n', n', n.. Then as the figure shows dydz =- da' cos (n'x) = da" cos (n"x).. Let d = dxdydz be an element of volume. Let u be a vector whose components are u, v, w. Then if u is one-valued and continuous, ax - dT= dydza dx= udydz = ucos (nx)d-. (1 JsOx J J Ox js 512 FUNCTIONS OF A COMPLEX VARIABLE We get similar relations for v, w. Thus adding, ay Oz 'fmu. In ale w. au + -v a+ W d= I div u d aJsx ay az Js = (u cos (nx) + v cos (ny) + w cos (nz))dc. (2 Now the component of the vector u normal to S is Un = u cos (nx) + v cos (ny) + w cos (nz). (3 If u denoted the velocity of a fluid, UtdS would denote the amount of fluid which passes across the element of surface dS per unit of time. For this reason we call quite in general UndS the flux of u across dS. Thus 2) may be written fdiv d =u flux u. do. (4 This is Gauss' theorem. 2. Let us now deduce Green's theorems. To this end we consider the integral taken over a volume bounded by S, UJ= aavdT i =1,2,3 (5 OJ xi Oxi where x1, X2, X3 are simply x, y, z. We use the subscript notation in order to use the E sign on account of brevity. We have now aUO= A. ua -U- u 0 Oxii Ox i Oxi aOx Let g be the vector whose three components are U-V. Then 5) becomes,' xi J-= |div g dT - UA VdT. By Gauss' relation 4) div g dT = flux g do= - U- d ns FUNCTIONS OF LEGENDRE AND LAPLACE 513 if we reckon the normal n inward. Thus J= - <U vdo- -fUAr dr, (6 or interchanging U, V - a Ud- -C VAUdTr. Equating these two values of Jgives pU vFF TaU~do-=J(VA U UAF)dT. (7 an y- an /d=(a ~d~( If we set U= 1 in 7), we get an = Fdo VdT. (8 If we set U= Fin 6) it becomes a V2 a Tra~ a dT= FaV- 61C - FA dd (9 a? an is If Vis harmonic, that is, if A V=z 0, this gives aF2dT Y a Vdo- (10 ax2 ~ an ITf U and Fare both harmonic, 7) becomes Ua do C aUdo. (11 Jan jan If Vis harmonic, the relation 11) gives for U= 1 O aF I - doa = 0. (12 js an, These relations are due to Green. 234. Potential Expressed in Terms of Boundary Values. 1. Let a be a point within the surface S. The distance from a to any point x in S is r = V~r1a1)2 + ( )2- a2 )+ (X3-a3)2. 514 FUNCTIONS OF A COMPLEX VARIABLE Then 1 U=r is not continuous in S. Let us therefore describe a small sphere Kof radius k about a and let T denote the remaining volume, as well as the surface bounding this volume. The normals n we will reckon inward as in the figure. / From 233, 7) we have, denoting an element of surface by do, / Jrr a V d-F a - VdT (1 | since A U= 0. Let us suppose Vsatisfies the relation v< some G, as r 0. (2 r Then \ A-I-Vdr < G dr=, as k-0. J r JK Hence the integral on the right of 1) converges to - AFVdr, as k 0. j Sr Let us turn to the integral on the left of 1). We have, taking account of the sign of the normals,.( )d~-( )d — ( ) da. Now relative to K, a 1 Or 0 1 1 On r an Or r r2 Let now V1, V2 be the minimum and maximum of Von K. Then 4 7rk2 1 f Vda < 4 tk21T7, JK or Vd = 4 7k2 V JK where Vm is a mean value of Von K. FUNCTIONS OF LEGENDRE AND LAPLACE 515 Let now k 0, then Vm- Va, the value of V at a. Thus. Orr Let us now look at r V X - da. JKr an Since the first partial derivatives of V are continuous, so is Hence,V. Heuncen O < some Hfon K. an Thus i1 a 4w7rk2 I - doa < HH =0, assk=0.;r an k We thus get, the point a being within S, 4 7r Va ( V 1 1 ro - A Vdr. (3 AY n r r An r In case Vis a harmonic function, this gives a = r; V idn r dn. (4 4 7r an r r An, 2. In the foregoing, the point a was taken inside the surface; let us now take a without S. Let.9 be a sphere of radius K=oo. About a as a center let us describe a sphere t of radius k 0. The three surfaces 9, S, t limit a region T whose boundary may be denoted by the same letter. Let x be any point in T. Then U=1 r is continuous in T and we have again f(! KV;a )d= f1A/Vdr. \r an anr JT r 516 FUNCTIONS OF A COMPLEX VARIABLE Then if we reckon n inward as in Fig. 2 we have J( )~+daf( )da, + ~ ( d= ) i A VdT. (5 We have already seen that d 1 t-dr r_ 0,/ Jr An y f,l - da -= 4 rV,. an r Let us suppose now that V is such that 2) still holds and that also V-=O,and R2 an <some G as R-o. (6 Then / 0 iI V ' V- da - d O as K-too. J r do dOn r Thus 5) gives, the point a being without S& O4 7r V — l do -— lA VdT, (7 Js\ nd r rr adn Jr where E denotes all space outside of S. If Vis harmonic, this gives 4 r Va =f(V 1 )do-. (8 s On r r an We notice that 4) and 8) are the same in form. 235. Outline of a Solution of Laplace's Equation. The following method is applicable to the sphere, the cylinder, and the ellipsoid. It depends upon the fact that each of these three surfaces belongs to a family of triply orthogonal surfaces, viz.: 1~ Sphere, cone, meridian plane. 2~ Cylinder, meridian plane, plane perpendicular to the axis. 3~ Confocal ellipsoid, hyperboloid of one and two sheets. FUNCTIONS OF LEGENDRE AND LAPLACE 517 In passing let us note that the rectangular xyz co6rdinates are also defined by a system of triply orthogonal surfaces, viz.: planes. To solve 02u 02u 0% 0z2 + A-+ 0= (1 aX2 ay " z2 by the method we have here in view, we first pass from the rectangular coordinates x, y, z to a system of coordinates determined by the triply orthogonal surfaces. To illustrate this let us consider the case of the sphere. Here the family of surfaces are given by x2 + y2 + z2 r = 0, 2 + y2 _ 2 tan2 =0, (2 y - xtan 0 = 0. If we solve these, we get f1(x,,, )-= /2 + y2 + z2 = r, f2(x, y, z) = arctg v2 + = -, (3 f3(, y, z) = arctg Y =. x Giving r, 0, 0 definite values, we can solve 2) for x, y, z, getting X = gl(r, 0, ),y = g2(r, 0, Zb), = g3(r, 0, O). In the case of the sphere these are x= rsin0cos, y =rsin0sin, z = rcos0. The new coordinates are polar coirdinates. In general the family of orthogonal surfaces corresponding to 3) may be written fl(x, Yz) = %, f2(x, Y, z)= r, fa(x, y, z)-=. To each triplet,, 7, 4 will correspond one surface in each family. Their intersection, taking account of the octant, will be the required point. The next step is to take ~, V, C as new independent variables and transform Laplace's equation 1) to this set of variables. For polar coordinates this has been done already. 518 FUNCTIONS OF A COMPLEX VARIABLE We saw, 221, 7), that 1) becomes a 2 q I 1 u (4 2 Au == Af.2 + l-i Sin 0 + n_ b = 0. (4 Or arr sin 0 aO9 0- sin2 0 02 -Having transformed 1) to the new coSrdinates ~, V, f we try to find solutions of the very special form u = F() G(-)H(), (5 where F,, H, depend respectively on a single variable as indicated. With this end in view we set 5) in the transformed Laplace equation Au = 0 and find that it is possible to break it into three ordinary linear differential equations of the second order, of the type d2F dP dJ2_ dF tp+ + qF + =0, (6 and similar equations for X and C. Let F1(4) be a particular solution of 6), while Gl(Q), Hl(0) may denote particular solutions of the equations analogous to 6). Then Ul-= E,1 GlH (7 is a solution of Au = 0. As we shall see, it is possible to get an infinity of solutions U1 u 2 u3... of Au = 0 of the type 7). Then M = Cl'cl + C222 + C3U3 + *" (8 is found to be a solution and it is possible to determine the constants c which enter so as to satisfy the given boundary values. All this will be made clear in the following. 236. Solution of Au = 0 for the Sphere. Axial Symmetry. 1. Let us apply the method outlined in the last article to find a solution of Au = 0 for the case that u must assume given values on a sphere S of radius R, which are the same on all meridians having the same axis. This axis we call the axis of symmetry, and we say the boundary conditions have axial symmetry. FUNCTIONS OF LEGENDRE AND LAPLACE 519 Let us take this axis as the z-axis. Since the boundary values are symmetrical, we take u as independent of p. Then Ou = 0 and Laplace's equation becomes 9 9 u\ 1. d / a Qc\ p 0(1 a 2(r a+ 1 -a(sin 0 s )=0. (1 ar ar) + sin99 a6 o According to the general scheme we now set = F (r) G(0) (2 where P depends only on r, and G only on 9. We find 1) becomes 1 d fr d}\ = 1 1 d d( d&) (3 FdrA drrr G sin8d\ d 'dO Here the left side is a function of r alone, the right side is a function of 0 alone. Suppose then that we determine F so that X= d (r2d _ O a (4 P dr\ d-r - and G so that a1 - d-(sin 0-d + a= 0. (5 aGsin deo d) ( The corresponding values of F, G put in 2) will obviously satisfy 1). 2. Let us look at 4). This may be written r2 d2+ 2 r - aE=, (6 dr2 dr which is a linear homogeneous differential equation and so belongs to the class of equations treated in the previous chapter. Its only singular point in the finite part of the plane is r = 0. At this point the indicial equation is s2+ s-a=. (7 Let s be one of its roots. We then set y = r8(eo + elf + c2r2 +...). (8 520 FUNCTIONS OF A COMPLEX VARIABLE Here the e's are determined by 209, 5), viz.: cofl(s) + clfo(s + 1) = 0 Cof2(s) + elf l(s + 1) + Cfo(8+ 2)= 0, etc. Now in the present case f,f2'* are all zero. Thus c1, c2... are all zero. Thus 8) reduces to y = r9. Since we are seeking only particular solutions of 1), let us choose a in 7) so that its roots are integers. Then if n is one root, the other must be - (n + 1). Hence a = n(n+1). (9 For this value of a, 6) admits the two integrals r and, n 1, 2,... (10 r n+ Both types of solution are useful, as we shall see. 3. Let us now turn to 5), which becomes on giving a its value in 9), and setting X = COS 0, (1-)Xd2 2 x +- n(n + 1) = 0. (11 But this is Legendre's equation for which Pn (x) Qn(X) form a fundamental system. Thus by 2) = rnP,(cos 0) ~1 n=n = 1, 2,... (12 un,+1 Pn(cos 0) r~l are solutions of Laplace's equation Au = 0. The boundary values being symmetrical with respect to the zaxis, tie value of u is known on S when it is known on a meridian. Call this value v, it is a function of 0. If continuous and having only a finite number of oscillations in the interval (0, 7r), it can be FUNCTIONS OF LAPLACE 521 developed in a series v = a0 + alPl(cos 0) + a2P2(cos 0) + *. (13 where n 2 - vPn(cos 0) sin OdO (14 as we saw in 227, 1. According to the general scheme we now set U = C1Ul + C2U2 + 3t3 - "' and try to determine the coefficients c so that u reduces to v when the point P = x, y, z is on the sphere S. There are two cases. If P is within S, we take u = cO + e1 P(cos 0) + (- P2P(COS)+ 0) (15 If P is without S, we take u = c, + c(R Pi(cos 0) + c2 P2(cos 0) +... (16 To determine the c's we take r = R. Then 15), 16) give, since u = v now, v = co + clPl(c 0) P(cos 0) + ( ) +. (17 Comparing this with 13), we see that the boundary condition is satisfied if we take en = an where an is given in 14). Functions of Laplace 237. Spherical Harmonics. 1. We have just seen how to solve Au = 0 when the values assigned to u on the surface of a sphere S are symmetrical with respect to an axis. We wish now to consider the case that the values assigned to u on S have no such symmetry. This general case was considered first by Laplace, and the functions he introduced to effect the solution have been named after him. We begin by proving a number of theorems. 2. The equation 02 v= + + =0, Ox2 Qd2 0z2 522 FUNCTIONS OF A COMPLEX VARIABLE admits as a solution the homogeneous polynomial U = _aijkiyizk (2 of degree n = i +-j + k containing 2 n + 1 parameters. For 2) contains (n n+ (+ l)(n +2) 2 terms. Putting 2) in 1) we get a homogeneous integral rational function of degree n - 2 which contains n(n- 1) 2 terms. As AU must = 0 identically, the coefficients of all its terms must = 0. The number of independent parameters is therefore (n + 1)(n +2)_ n(n - 1) 2 2 We call such polynomials harmonic polynomials of order n. 3. We have at once the following theorem: There exist 2 n + 1 linearly independent harmonic polynomials of order n. As examples of such linearly independent harmonic polynomials, we add the following table: n = 0 a constant. n = 1 x, y, z. n = 2 X2 - y, y2 - z2, xy, yz, xz. n = 3 x2y- y3, 3 X2z - Z3, 3 yx2 - y3, 3 y2z - 3, 3 z2x- X3, 3 z2y - y3, xyz. 4. Let us pass to polar coordinates, x=rsinOcosb, y=rsinOsino, z=rcosO. (3 Then the harmonic polynomial U of order n becomes U=r Yn, (4 where Y. is a homogeneous polynomial of degree n in sin cos, sin sin, cos 0. (5 FUNCTIONS OF LAPLACE 523 We call Yn a spherical harmonic of order n, and have the theorem: There exist 2 n 1 linearly independent spherical harmonics of order n. If we transform 1) to polar coordinates 3), we get, as already seen, seen, 02V 1 02V 1 O2V 2~ OeV cotOa V + _ 4 + O. (6 ar2 r2 002 r2sin2 0 a02 r ar r2 a If we put 4) in 6), we see that Y, satisfies 02Y 1 a2 Y Y a2 + + + n(n + 1 Y= O. (7 Let us also note in passing that a U irn-1. (8 dr 5. If U, V are two harmonic polynomials or two spherical harmonics of orders m = n, then fUVd = 0, (9 the integration extended over the sphere S. Let us first suppose that U, V are polynomials. By Green's relation 233, 7) U V PO U do-= O. a s -n An / Using 4) and 8) this gives f (n YR-m Y n - mRm-Ym Y,,) da = 0, or f, Y,,,do = 0. (10 If we multiply this by Rm+n, it goes over into 9). If we suppose, on the other hand, that U, V are spherical harmonics, say U= Ym, V= Y,, they may be converted into harmonic polynomials by multiplying by rm, rn respectively. Then we are led to 10) again. 524 FUNCTIONS OF A COMPLEX VARIABLE 238. Integral Relations between Ym and P,. Let v= pmr (1 be a harmonic polynomial. Let z P(p', O', 0') be a point inside the sphere S, and Q(p, q, 0) a point on its surface. Let r=Dist(P, Q). Then by 234, 4) 4^7r Vp F-I l --- 6ad. (2, / Js\ dnar r on j Let Fu = cos (p, p') = cos o = COS 0 cos 0' + sin 0 sin 0' cos (s - 0'). (3 Then by 222, 4) 1 (p). (4 rP o Hence a 1 1. On r Op r - ps+ Also, m a V__- pm-1 Ymt dn da = p2 sin OdOdq. These in 1), 2) give 4 7r VI, 4 7pm Y(0, ) pm Ym Y (n +- 1) p+ Pn(/) + pm —Y X n() I p2 sin 0OdOdd A ~ ~ 2 p/ pm = pl P J A-n + A- +1) YIn(Fl) sin Odcl f j n=n }P The right side is a power series in p'. Equating coefficients of like powers gives fdc f Y,(0,,)Pn(v) si OdO = 0, m - n (5 Ym, ) = n + l f2 d f Ym(0O, >)P,(/i)sin OdO (6 4i 7r Jo o where Lt is given by 3). FUNCTIONS OF LAPLACE 525 239. Development of f(0, 4) in Terms of Ym. Suppose the values of a function f are given at all the points 0, Q of a sphere S. Iff is continuous and has only a finite number of oscillations along any great circle, it can be proved that f admits a development of the form f= Yo(0,,) + Yi(0, ) + Y 2(0 ) +,, (1 where the Ym are spherical harmonics of order m. Moreover this series may be integrated termwise. Admitting this, we can easily show how to determine the terms in 1). Let P (0, 0) be an arbitrary but fixed point on S; let Q(a, 3) be a variable point on; let =COS O O ==os cos COS 0 + ill a sin 0 cos (3 - ). (2 We multiply 1) by P(,() sin adcd,8 and integrate over S. Then by 238, 5) every term on the right will drop out except that with the index n. Thus p27r P7Tr P27r ~7r f d/f f. P,. sin ad = d/3 YnPnp sin ad = - 4 - Y,(0, ), by 238, 6). (3 Hence 1 ~2n+1 Hence ( ) 2 + ifT d f ~ Pn(cos 0) sin ada, (4 where cos o is given by 2). For later reference we note that 3) gives YT (0, <) = 2n + d3 ((, /)Pn(cos o)sin ada. (5 4 vr 240. Fundamental System of Harmonics of Order n. 1. We saw in 237, 2 that there are 2 n + 1 linearly independent spherical harmonics of order n. Such a system we call fundamental. We show how to form them. We saw in 237, 4 that any spherical harmonic Y, of order n is homogeneous in sin cos 0, sinsin, cos 0. (1 526 FUNCTIONS OF A COMPLEX VARIABLE Thus Yn = 2Ap, q(si 0 cos )'()1(si 0 sinll )qn cos -P-q 0 = qAp, q cosp sinC q sinlP+ 0 cos-P -q 0. (2 Now cosp < sinq k can be expressed as a linear function of sines and cosines of the angles (p+ q), (p+q- 2),... In fact cosP sin0 e P(e -i CosPgsinq=(ei ~ - e-i~)(e-e-i Expanding this gives 2p+q iq cos sinq = ei(p+q) + (p - q)ei(p+q-2) +-... + ( 1)l)-i(p+Q). Hence, when q is even, cosP < sinq = a0 cos(p + q)> + a1 cos(p + q - 2) +... = aj cos (p + q- 2j). When q is odd, we get similarly cosP 4 sinl f =:j sin (p + q - 2j)>. If we set these in 2), we get Yn,= C,., sinm 0 cos-m cos - (3 lsin (m- 2j)0' Nsinm 0 cosn-m 0 = sinm-2i 0(1 - cos2 0)i cosn-m 0. This in 3) gives, on setting n - m = k, Y, = F, cos kf + G sin k }, (4 k=O where Fk =Lk sink 0, G= k sink 0 (5 and Lk, Mk are polynomials in cos 0. Now we saw in 237, 7) that Yn satisfies 02Y 1 2t Y a Y si -cot0- + n(n + 1)Y=O. (6 002 sin2 042 o0 FUNCTIONS OF LAPLACE 521 Putting 3) in 6), we get sin2Od~k~sinCos dk+ [n(n +1) sin29 ]kkl~jcos kc n=O dO2 dO ~ X similar expression in Gh sin ko = 0. n=o This relation holding f or any 0 requires that Fk, ak are solutions of Sin2 O d- +SinO cos O~Y ~ n(n+1)i2 12~ y = 0. If now we set Y=usn we get by 5) the equation that L,., M. ~satisfy, viz.: Sin2O0d2U +(21c~+ 1) sin OcsO du + ~n (n~+1) - k c+ 1) sin2O. U=0. If we set x = Cos 0, this becomes dx2 dx This is closely related to Legendre's equation (1 X2) dV_2 x j+ n(n + 1)v = 0. (8 For if we diff erentiate 8) A- times, we get 7). Thns one solution of 7) is Since now every solution of 7) is the lcth derivative of a solution of 8), it follows that Lk., 11~ are. But L, Al are polynomials in x. Now we have seen that 8) admits no solution besides cP, (x), which is a polynomial. Hence 19k, 3~4 are aside from constant factors the function given in 9). Thus by 5), 1',, a,. have the form k. sink OP~k (COS 0) =I (- X2)2Pll (X) = P., k(X). (10 They are called associated Legendrian Punctions. 528 FUNCTIONS OF A COMPLEX VARIABLE Thus we have: P,1= Vi - xP 2,1=3XV\1x2, P22 3 (1 - X2), PP =3(5:2l)/lx2 - )\,X P,=15x(1- x2), P3, =315(1 - X2) -1_x2. 2. Returning now to 4) we see that Y,, has the form n k0 f l,kPn,k(cos O)sinA ~o 6P,,,kP(cos O)coskOl. (11 Since sin k= O for le = 0, Y, is the sum of 2 n + 1 terms of the type sin c-PO,, k(CO), coskOPf,k(cos60). (12 It is easy to show that the functions 12) are homogeneous in the quantities 1) of degree n. To do this we reverse the process used in 1. Thus each of the 2 n + 1 terms 12) which enter 11) being homogeneous and also satisfying 6), is by definition a spherical harmonic. Since any spherical harmonic Y, of -order n can he expressed linearly in terms of the 2 n + 1 harmonics 12), these latter form a fundamental system of order n. We have thus the theorem: The 2 n + 1 harm.onics Pn(cos 0), (13 cos Pn,l(cos0), cos2cP, 2,(eos0), *.. cosn0P.,,(eos6), sin cP,,1(cos 0), sin 2Pn2(cos 0), *..sin n0bP,,((eos6), form a fundamental system of order n. 241. Integral Relations between Pn, k'n m We now prove the important relations Pmk(X)Pnk(x)dx = 0 m f (1 (n-k)! 2n+1 FUNCTIONS OF LAPLACE To this end we consider J= PmkPnkdx -1 529 dkPm dkP. -1 d-k dxk To fix the ideas suppose n > m. We integrate by parts, using the formula u udv = [Uv] 1 — vdUL. We take u = (1- X2)() = P-l) Then Te [v]LI = 0 and hence - P d J= - P(k- - (1 - X2)kP(!dx. J-i dx Repeating this process k times gives I dk J= (- 1)k p (I - x2)kPk)dx =(- 1)kfPnFmddx, (3. -1 where Fm is a polynomial of degree m. Thus when n > m, J= 0 by 225, 2), which gives 1). Suppose n = m. We have from 222, 5) 1.23... 2 n - 1 Pn(X) = 1*3 -2- 1 ~ 2... Axn + ' " Hence P (k)() = An(n - 1)... (n- k + l)x'"- +. Also (1 - x)1 = (- 1) ^x+... Hence Gd a= P (\X)(X(1 - X2) =(- 1),An(n -- l)...+(n ( - k + 1)xn+ +.dk F, = da+ = (- l) An(n, - 1)...(n - /c + 1)(n + k) (,, + k - 1) n dxlin (~-iA.. (+.)Xn +.. ) (n + )! Axn.. (,, - k).! 530 FUNCTIONS OF A COMPLEX VARIABLE Now F. is a polynomial of degree n; it can therefore by 227 be expressed in terms of P0, P1 '* Pn. We get F ( ) (n +- k)! Pn + AP_1 + A2P,_2 +.. Thus 3) becomes (n + k)! /'P (n - Ic)! J since each of the other terms = 0 by 225, 1). If we now use 225, 5), we get 2) at once. 242. Development of f(O, 4) in Terms of Pnk. In 239 we saw that when f.(0, c) is a one-valued continuous function of 0, ( having but a finite number of oscillations along any great circle, it could be developed in a series of spherical harmonics f(O, )= Y0,+ Y1 + Y2,+' (1 But in 240, 10) we saw that each Yn is a linear combination of certain fundamental harmonics. Thus f(0, 4) = E: Ank cos k>B + Bn, sin k tP,k(x), (2 n=0 k-=0 where X = cos 0. To determine the A's and B's we note that 27r f cos mq sin nodf = 0, always I cos mnf cos ncfdc = 0, m = n, q27-r /2V a27r f cos2 ncdo =, cos nofd = 0, f sin nodo = 0, n> 0. o o Jo Let us multiply 1) by dc and integrate, we get j fds = 2 7rEA,,oP,(x). Jo n-0O FUNCTIONS OF LAPLACE '531 This we multiply by Pmdx and integrate, getting Jo n J-i Jldz foPmd5= 2 7E;Anofi PmP,~dx 47r 2n+ 1 A since all the terms on the right = 0 except when m = n. Hence A o2+-1 = fr fp sil ada n = 0, 1, 2... (3 4 i O Jo Let us now multiply 1) by cos kodo, k > 0, and integrate, we get 2r I f cos k1db = wEAnkPnk. J/o n=O We now multiply this by Pnkdx and integrate. We get, using 241, 1), 2), 1 r, 7 j D + 2 7r (n + k)!.I-i! 2nw. (n-k)! nk Thus 2n+ 1(n -) 2, rT (IQ $- 7/) I ^ * /7 / AT k = ~l(n- k)T f df f cos kP,,,;(cos a) sin ada. (4 2 r (n + k). Similarly we get B^ -= 2 + 1 (n + k dC f sin klP,,k(cos a) sin ada. (5 2-2 7( +,k). J There are no coefficients Bn since the factor sin k 0 = 0 for k = 0. Putting in these values of the A's and B's in Y, we get Y., =fdafi 1kP, k(oo s ) 2 n + l (n - k) f(c'. / 0 0 27r, (n+kk)! cos kf cos k,/ + sin kf sin kf3Pnk(cos a) sin ad/3, where 7k-= =2 when k=0 =1 when k > 0. Thus we have oYn= o "] ( - da t k)! 2 n+-.kf (a, )P, k)-.(cos a)P,,kncos 0) cos k (( - /3) sin ad,8/3. (6, 532 FUNCTIONS OF A COMPLEX VARIABLE 243. Expression of Pn(cos co) in Terms of P,, Pnk. Let P(O, 13), Q(O, ) be two points on a sphere whose center is 0. If w is the angle between OP, OQ we have cos o = cos a cos 6 + sin a sin 0 cos (8 - ). Now in 239, 5) we saw Yr,(0, q) = 2 n + sil ad f (adlf(, )P(cos o)d. 4 r ~0 o ao0 If we compare this with 242, 6) we get P,(cos co)= Pn(cos a)P,(cos 0) + 2 (n - ) Pnk(cos )Pn,(cos ) cos k(3- ). (1 k=l ' 244. Solution of Dirichlet's Problem for the Sphere. 1. This problem is to find a solution V of Laplace's equation AT = 0 which takes on assigned values f(0, )) on a sphere S of radius R. If the point P is in X, we set V= Y+ Y R+.2 + ' r <R. (1 Each term is a solution of A V= 0 and hence 1) is. For r = R it reduces to = y0 + Y + 2 +.. (2 Thus V8 must =f(0, q) if 1) is to satisfy the boundary conditions. Thus the coefficients Yo, Y1 *. in 1) are the terms of the development of f(0, qb) given in 242, 6). If the point P is outside S, we set V=Yo +y + y2()) +... r>R (3 and reason as before. 2. By the above we have solved the problems: 1~ Determine the temperature at any point in a sphere S which is in a stationarystate, the temperature being given on the surface of S. 2~ Determine the potential for any point outside a sphere S, knowing its value on the surface of S. 3~ Determine the motion of an incompressible fluid having a velocity potential F, knowing V on the surface of S. CHAPTER XV BESSEL AND LAME FUNCTIONS Bessel Functions 245. The Integrals of Bessel's Equation. 1. This equation was studied in Chapter XTIJ; it is __ d2d+xLy +(2-M)y=0 dX2 dx When 2 m is not an integer, we saw in 211 that 1) has two linearly independent integrals Jm(x) = U(n,)I(~n) (X~ m+2n (2 and Lm~~(X) I(n)LI (n - m)(2) The first converges for every x, the second for every x -— L 0. In the applications m is usually an integer. For m= 0, 1, we have ~ J~(x)=z1-22~2 242- 2 —4f2 -.~ (4 The function defined by the series 2) is called a Bessel's function of order m. 2. When m is an integer, we have Jm(x>&=( 1)mj-m(X). (6 For k being an integer or 0, ll(-lk) 533 534 54 FUNCTIONS OF A COMPLEX VARIABLE He-nce the first m terms in 3) vanish. Let us therefore change the index of summation in 3), setting v = n - m. Then 3) becomes (1)M+v (x\\v+?n V= (m ~7) H')UQ) 2?) 3. Since 2) is a power series, we may differentiate it termwise, which gives Jj) _ A0 2 )n(M ~ 2 n) m~02nh(n) JJ(m ~ n)Xm2l.( 4. When x is complex, x==r(cos0+ i sinO0), (8 we may write Jr()=(mmx) Now n 2 (XN~ = K~(cos 2 nO + i sin 2 nO). Thus When m is real, this enables us to write Jrn(x) in the form Jm(X) = Urn+ ~ ~Vrn where U, V are real. 5. From 2), 3) we have J (X) =\I2 sill x, (10 J1(X)=-\ —Cos X. 246. Relations between the J.. and the Jn'. 1. The following recursion relation exists betwveen three consecutive Bessel's Funetions. Jn+1(x) = 2 n~h(X) - JA1 (X).( X BESSEL FUNCTIONS 535 For Xn-1 X2s+n-1 Jn-1 + (2 2n- =l(n - 1) + ) 2-+2S+,-1lI(s)H(n - 1 + )' X2s+n-I =+1 = ( _ )2z+2s(s - 1)JII(n +s) Hence _ n-1 Jn-, + Jn 2n-1 (n- 1) 2xn_1> oX2s+n-1~ 21) s2s-I 1) +(s) - 1(n-l+s) H(s-I)n ( +8s) =- n-l 2n(-l)+n-l =2 l (n - 1) 8- z 2" f"-(s)H (~n f+ ) 'M *^\yx2s+n x s= 2+2n2s-1 T1 (s) n (n + s) = Jn (X) X 2. We show next that 2 J(z) = Jn_, (x)- J+,(x). (4 For subtracting 3) from 2) gives T _- =________ + 1 2s.n- n + 2s -1 1 2n-1H(n - 1) _ 2n21-) 1 l(s)f(n + s) _0 ( C ( n~ + 2 -1s)'n+n- s): - - 2n+2s-l- (8) 1- (n + 8) =2 J7'(). 3. From 4) we get, on replacing Jn+j by its value given by 1), Jn(x) = — J() + J l_(x). (5 From 1) we also get From 1) we also get J (z) = jT (x) -J x). x (6 536 FUNCTIONS OF A COMPLEX VARIABLE 4. From '245, 7) we have, 5. We have also d(x)- xJ.~ dx or using 5), d X j dx x"nI= I (X). (8 6. By means of 1) and 245, 10), ii.) we, have the theorem: -When n is a positive or negative odd integer, J,,(x) can b~e expressed in terms of the elementary functions. Thus in particular the recursion formula 1) gives 2 rsin x J,, (x) = —es -x sn w x - \-x/xj / 2 (Cos x 3(x) = t~~+siflx, 2~ 7tXt xJ J 5x)~\I ICos x f4_1 )~sinxT. 2 7T7XX X J 247. Integral Relations. 1. As a first such relation let us prove fxfo(x)dx = xJ,(x). (1 For n = 0 Bessel's equation becomes xd2 YI dy x-r-~ ~+ x'=O0. This is satisfied by Jo; it therefore gives Integrating this gives xJO'(x) +f(XJ~dx = 0. Using 246, 7), this goes over into I). BESSEL FUNCTIONS 537 2. To get other relations let us set in Bessel's equation x2 2 d+ + (X2_ m2)y= 0. (2 Xd2 dx u = X /. y(ax). It becomes d2 1 d2+ $- -- ( 4=0. (3 dX2 - 4 X2 0. (3 Thus u = (/xJ(ax), v= VxJ-(,3x) (4 are solutions of equations of the form d2u dx2 gu =0, d2v + hv v=0. (5 dx2 From 5) we have d2u d2 d2 d- (h -g)uv (6 v j-X2 dX2 On the other hand the left side of 6) d f du dv\ -V- - Z-. dx\ dx dx) Thus 6) gives du dv vT —u- I (h - g)uvdx +. (7 dx dx Substituting 4) in 7) gives (f32 - a2) xJ(ax)Jn(3x)dx = x cIJ (.x)J'(ax) - j3acx)J(3x). (8 If we use 246, 6), we get from 8) (/32 - 2) fxJ(ax)J~(/x)dx = XI\Jn((ax)Jn.+l(x) - aJTn(lx)Jn+Tl(x). (9 538 58 FUNCTIONS OF A COMPLEX VARIABLE Let us differentiate 8) with respect to fi and then set 3 a. \Ve, get axxJ,,( ax) - 1 ( ax)fJ,( ax) - axJ-T,,( x) J,,'(ox)' (10 Expressing J" (ax) by means of 1) this gives fxJX 2(12)~1 a2X2)nx} (1 Using 246, 6, this gives x,,(a) x= - ~J2(aX) 2-V J~(ax) - l J)( (12 Similarly if we diff erentiate 9) with respect to 3 and thenr set fi = a we get x fJ,2)dx = x Jn ax> J,,~ (ax) 248. The Roots of Jm,(x), m Real. 1. In many of the application's it is important to know that Jm,(x) has an infinite number of real roots. Let us consider the general question of the nature of the roots of J,,(X). The roots of Jm,(x)= 0 are all real when mn is real. For suppose J.m=0 for x =a +ib, b = —0. Then the series would give A~i 0 whence A=0 B-0. This shows that then the conjugate numnber x' a - ib must be a root. Let us therefore suppose that a=a+ib, /3a-ib are two roots of Jm, Then a2- /2= 4iab. Jm(ax)= P~+ iQ, Jm(/3x)= - i Q. BESSEL FUNCTIONS 539 These in 247, 9 give - 4 iab x(P2 + Q2)dx =0 (2 t o since Jm(a) = 0, Jm(1) = 0. But the integrand in 2) is positive. Hence the left side cannot vanish unless a or b = 0. Suppose a = 0, so that a = ib, b > 0 is purely imaginary. Then 1) becomes fib\m 1 J,(i) = bn=0 (_n) 1(m + n) Here im is a series all of whose terms are positive. It cannot vanish. As b > 0, Jm does not vanish. Thus J, has only real roots. 2. The development 1) shows that: Jm() = 0, when m > 0. It also shows that: If x = a > is a root, so is x= - a a root. 3. No two consecutive functions em(x)), J,+1(z) have a root in common, aside from x = 0. For if a were such a root, 247, 9) gives X Jn (ax) Jn (/3x) dx = 0. In this relation let3, a; we get j xJ2(ax) dx = 0. This is impossible as the integrand is > 0 for x > 0. 4. The roots of Jm (x) are all simple, aside from x = 0. For consider f(x)= mJm (). This does not vanish for x =- 0, unless Jm =. 540 FUNCTIONS OF A COMPLEX VARIABLE But from 246, 8), f (x) = mJm (x) As Jm, Jm-l have no root = 0 in common, f (x) does not vanish for any non-zero root of Jm. 5. Jm(x) has an nininity of roots. For we have seen in 247, 2 that u = x\/Jm (x) satisfies d2 - gu = 0, dx9 where 4 m2- 1 g= 1-.. 4 x2 The index m being fixed, let us take ~ > 0 so that 0<g<l for x>I. The equation d2v dx2 admits -d = sin x as a solution. Then by 247, 2 du~ dv- r7 V - u -T - ( -g) uvdx. dx dx jo J If we take 2 ( +1) U=2nzr, = (2n+ 1r, we get u (13) + u (a) = - (1- g) uvdx. (3 Suppose now u is positive in the interval = (ac, /3). Then the left side is positive and the right side is negative, since 1 -g > 0 in any case, and v is positive except at the end points of W. Thus the two sides of 3) have opposite signs, which is a contradiction. Similarly, if u is negative in t1, we are led to a contradiction. Thus u must vanish at least once in W. Hence in any interval (a, 6) of length rr, Jm(x) vanishes at least once, provided a>. BESSEL FUNCTIONS 541 249. Bessel Functions as Loop Integrals. We have seen that Bessel's equation d2 +XdY (X2_M2)y =O ( dx2 dx admits J, as a solution. As J, has the form Jmxm(ao + aix +***), let us set b= X"U in 1). We get ~2 m 1 ~u x ) - m+~ + xu 0. (2 dX2 + + CI This is a special case of a class of equations (aob+ o X) ~ (a, + h1x) 1+ *.. + (a~ + bn)X = 0, (3 0 dn dxn-1l whose integral may be expressed in the form u = exzw(z)dz (4 Let us suppose n = 2 in 3) and let us change the independent variable x by setting = a0 + box. If we make this substitution in 3) and then drop the prime from xi, we get an equation of the form d2a d~U x d2U (a + bx) + (c+ dx)u = O. (5 dIX2 dx Comparing 5) with Bessel's equation 2), we see that a=2rn+1, 6=0, c=0, d=-1. (6 If we divide through by x, the equation 5) becomes d2u +( a~6du~ ( du = 0. (7 dx2 x dx ~x / Here the coefficients have poles of order 1 at x = 0. Hence the integrals are regular at this point. 542 FUNCTIONS OF A COMPLEX VARIABLE Let us now consider the point x = oo. If we set x =-, z we get d2u '2-a du 7\ - ~+ + )u0 =(8 dz2 - z 2 )dz (z3 z4) As the coefficient of -d can have a pole of order at most 1, and dz the coefficient of u, a pole of order at most 2 when x = oo is regular, we see this point is an irregular point of 5). Bessel's equation 2) becomes, on setting 6) in 8), d2u 2m-ldu 1 - + - u=O. (9 dz2 z dz z4 2. Suppose we try to satisfy 5) by a power series of the form U =xiho + h 2 +~+.} (10 I x xa2 which shall be valid about x = oo. We shall find it possible to determine the coefficients h0, hi... so that 5) is formally satisfied, but we shall find that 10) is divergent. To illustrate this let us consider the equation 2) for m=0 which is satisfied by Jo(x). If we set y = exu, it becomes d2u i \ i + - +2i z) u=O. (11 dx2 \X dx x Comparing this with 7), we see that a=l b =2i, c =i d=O0. If we put 10) in 11), we find r=-1 and 2 in (2 n -)2 2 inh,, = n- 1)hn-_ BESSEL FUNCTIONS 543 The ratio of two successive terms in the adjoint of 10) is in 1 2 ~1 f The series 10) diverges therefore for every x. 3. Returning now to 5), let us try to determine w and L in 4) so that the resulting integral satisfies 5). Putting 4) in 5), we get fexzdz +J = 0, (12 where d P= (az 4c) w-.w(z2 +bz + d), (13 dz G - exz (z2 + bz - d)w. (14 Thus if we determine w so that P = 0 and choose L so that GC takes on the same value at the beginning and end of L, the integral 4) will be a. solution of 5). Let us write 13) d dzi 3W) qw. Then ( dz q. Hence logpw zJ~ldzy or cz W. ~-e P Let us now decompose g into partial fractions. We have P q az+c = X + p z2 +1)z + d z - acc where a, 3 are the roots of p = 0, which we will suppose unequal. Thus we may take 1 _ W (z ( - a) (X IP E = (z a-) h — (z - P8)-I (15 P~~~~~~~~~~~1 This in 14) gives a = exz(z - a)X(z - a) '. ( (16 544 FUNCTIONS OF A COMPLEX VARIABLE As a path of integration L we may take a double ioop about ae, fi as in 220. Hence, remembering that xz Az22 exz =1+ + ~ +. 1! 2! we get from 4) Uap = fe(z - a-)x1-(z -,9ld (17 = n! fj(z -)X(z -3<dz. 4. For Bessel's equation 5), 6) p-z2 + 1=0 gives a=~,/=i q=(2r m~+1)z. Hence q (2 m +t) -1(29m l) which gives X u= ~M~. Thus 4), 17) become ~~~xz(Z2 + 1)M —d2 (18 =, fZn(z2 + 1)mildz. (19 Now L -lklclk-ll where 1, Ic are loops about z = i, z = - i When z describes a small circle about z = i, the end value of is e(m-i)27riZ = e21imZ, =rZ. Thus J=1+ + + i~.(20 But obviously J+ f=0, +?7J=0 since the integraud has the same values, while the direction of integration is reversed in 1 and 1-1, etc. BESSEL FUNCTIONS55 545 Thus 20) can be written fZndz = (1 - Zi), Zdz -fZ~ndz. (21 Now when n is odd, when n is even, fad Jnz Thus JL, = n odd, =-2(1 '~ Z,,dz n neven. To compute the integrals we set z =i 2y. Then the ioop I will go over into a ioop 5 about y =1, and JZ2,dz (-1) nif 2(1 - )2y Now - i=(1+ 9 ')f y(I - y)mldy Thus finally 19) gives U,=(~ rm 2 [(I)[ -1)" (22 2vm by 245, 2). Thus xtmua,, aside from a constant factor is nothing but Jm(x). 250. Other Loop Integrals for x>0. 1. A second path of integration L for which the function ~~=ex(z - ac~f(z-f),(1 considered in 249, 16) takes on the same value at the beginning and end is indicated in Fig. 1. We will denote them by A and B; 546 FUNCTIONS OF A COMPLEX VARIABLE both are parallel to the real axis and pass about the points a, 3 respectively. On A for example the real and imaginary parts of = + z = u + iv -IZ0-Z - A co are such that I v < some r while u comes from -oo, moves up to a, and recedes again to -oo. If we set z - a = reie, z- = sei, we have Thus G = exurXsf ei(xv+X0+o4). I G I = exur^S - 0 as u = - oo, and G takes on the same value at the beginning and end of A or B. 2. Let us now consider the integral ua= f exzwdz= exz(z - a)'-l(z - 3)-ldz JA JA (2 where w is given by 249, 15). Similar results hold for the other integral ua for the loop B. Setting -=, -=a, z-a=y, a-,8=a, 2) becomes Ua=- ey(y+a)y-l(y + a)u-dy JW (3 where W is the new path. As y approaches indefinitely near 0 for a part have Y- <1. a of AX, call it 1i, we (4 Then (y + a)1-l = al- 1 + )-l (O a- 1 + -ly + -1- 2 -2 +} 1. ' 1 a1 1.2 a2'" = I ~yn, (5 n=0 l= - 1. - 2... -n a,-n_ ( 1 2... n I where II BESSEL FUNCTIONS 547 For the other part of?I, call it W2, the relation 4) does not hold, and we write 5) (Y + a), —1 = co + cly + ** + Csys + R. (7 This in 3) gives Ua = eax c | e^yA;+n-ldy + eax eyyk-R.l dy (8 n=0 J c =eax(U+ V). We set now xy = - t - (. in the U integral. Then S goes over into a loop? as in Fig. 2. FIG. 2. We get now U = (- 1)X-y (- 1)nCnx-n e- th- n-ld. (9 n=0 t] But, as we have seen in 149, 2), J e-tt^+-dt = (e2i - 1)P)(X + n) (10 where by 144, 8) r( + n)= X(X +1)... (X + n- )(X). (11 This in 10) gives U=(-1 )Ax-(e2iA _- 1) (- 1)n'r( + n)e,-n (12 n=0 - X —A$ where __ where 3 = ( 1)(e2i - 1) coPr(X) - lr(X + 1) + cr(X + 2). -* -.... (13 'X( 251. Relation between up and ua, up. We have now found three integrals of our differential equation d2U, c~u x -d + (a + bx) + (c + dx)u = 0, (1 dx2 dx visz.: " =(exzu= (z- )a-l(z -/3 0)-ldz (2 Li 548 FUNCTIONS OF A COMPLEX VARIABLE of 249, 17), and a,= exz(z - a)A-l(z - 3)y-ldz, t (3 up = fexz(z - a),-1(z - 3)yk-dz (4 of 250, 2). Since 1) is of the second order, a linear relation exists between them. Here the path of integral L in 2) may be taken L = ABA-1B-1. Since (z - a)A is multiplied by the factor e2"Tix after describing the loop A, and a similar result holds for B, we have ua[ = _f =f + e27riAj- + e27ri(X+'L) f + e27ri f L A B A-1 JB-1 (5 Since JAA 1 =0, we have JA+ e2iJ\ = - 0,,-i A - e-2eri;f 1,l A or Similarly Thus 5) gives = - 2 13i' f. IA I Uap = ( i = (1 - e27ril)ua_- (1 - e2i7rA)up, and this is the relation sought. (6 252. Asymptotic Solutions. 1. We show now that ua admits the asymptotic development u,,e-aXxA, x real and = + oo. the solution (1 Referring to 250, 8), 12) we have uze-0x = $3 + xA V. (2 Thus 1) holds if where by 250, 8) BESSEL FUNCTIONS lim x+A = 0 X=+-o V= exyyX-lR dy. ~72l 549 (3 (4 The path of integration 2 we, take as follows. About y= 0 we< describe a circle c of radius y. Let 0 a = (-o, - y) as in Fig. 1. Then FIG. 1. a = a c. a-. Hence V=-+ f+t- 'L (5 and Sand 4AX +A X+A X + Xs+ X = A =A+B+ C. (6 We show that A, B, C all - 0 as x + oo. 2. We consider first A. From 250, 7) we have Rs = (y + ay —1 ( + cly +.* + c s). Hence A = S+AJ (y + a) -lexyAldy - ckX+ eYy^Y+k-ldy k=0 ict = Ao - CkAk. k=O We show now that Ao, Al,... A, 0 as x +oc, beginning with Ao. For complex z = re'0 we have log z = log r + iO + 2 m7ri. Thus for large values of y on the path of integration, | logy =log I y, nearly. But from the calculus, lim -lo = 0. x=+oo X Thus for I y I > some Y, I logy < IYI, Ilog(a +Y) l< I y 5;50 FUNCTIONS OF A COMIPLEX VA RIABLE Hence for some q > 0, log y"-l(y + a),,-, I I< Thus Thus X1(y + a)"-' < ey, y K som1e yg < 0. Hence JA0 I x ey(x-?)dy 0, as x + co. le now show that Ak 0. Changing the variable by setting xy = - we have Ak = (- 1)X+kxs-k e-ttxkldt =0 as x + o. 3. Having shown that A 0, it is easy to see that C~ 0 also. For C differs from A only in two respects. The path of integration is reversed and the integrand has another value at y = - y, due to the fact that y has described the circle c. After this circuit y,1' is multiplied by the factor elffiX, while (y ~ a)E'-' is multiplied by e2liL. Thus the C integral behaves essentially as the A integral, and we see at once that lim C = 0. 4. We show now that B 0. We have B = xs+x fex~ y Ay 8dy where Rt8 is the remainder of the series t=ax (Y + a)~1-e 0cC + c1y eCy2~ (.. begrinning with the exponent s +. If we set xy = - t, we get t=Xr t =0o B=(-I)')Axsfe-tt'x4BRdt (7 ti where S is the circle corresponding to c in Fig. 1. FIG. 2. BESSEL FUNCTIONS 551 (551 As the singular point of R,1 is t = ax which lies t=o t=6rIX' outside of G, we can replace G by the loop V in Fig. 3. This loop is made up of a segment 1 and FiG. 3. two small circles about the poiuts t = 0, t = xry. Thus B will be the sum of three integrals B = BO ~ B, 1+ B,, (8 corresponding to these three parts of V. Since the integraud in 7) is one-valued about t = xry, the integral x = 0. The integral Bo = 0 also. For Rs= cs+Ysf+l + cs+2Yf2 +... ~1ts+lf t (9 (1)s 1- ~ s+1Cs+2e + *2- * (9 Xs~l( X Thus R I 0 as the radius of the circle about t = 0 converges to 0. On the other hand, the reasoning often employed shows that B1 = (-1)x- x(e27iX- 1)-ettALR~dt. To estimate the numerical value of R. we use 112, 3). here a= 0, ItI=p~ Ixyl, r=laxl. Thus a denoting a sufficiently large constant, aI S — Xs+ ' Xli Hence B1 e, fiXB,-1 =HIfe-IttKdt, (10 taking the real positive value of the integrand. Let now x ~ o. Then the integral in 10) converges to P(X), and hence B, _'0. Thus lim. B = 0. We have now shown that each term of 8) is 0 or 0. Hence B -_ 0 as asserted. Thus the proof of 1) is finished. 552 FUNCTIONS OF A COMPLEX VARIABLE 253. Asymptotic Development of Jm(x). 1. Let us apply the results of the last articles to the equation 249, 2) which results from Bessel's equation 249, 1) on setting - = XMu. As in 249, 4 a-=i, f3=-i, \=,-u=m+ —, a=a- /=2i. The coefficients c are given by (y + a)y-' = + cly y2+.. =(y+2i)= — (2i)P{1+ YP p 1 2.. 2 =(2 i)P(1+ (l) y + p) y2 I+ ' + ((2i)2 Thus (2 m - 1)(2 m -3)... (2 m -2 r + 1) 2nm-2r-|i2r — 1 2... r Substituting in 250, 13) and 252, 1), we get 1 1 n'i/ 1 u ie-ixm+2 2m-2e —2 - (1 + e2rrim)(m + D, (1 where -1 + X -1)! 4m2-1 4M2 - 9 4m2-(2s- -1)2 1 (2 + V.4"-. 1(2 s=1 2sis 4 4 4 82 In a similar manner we have for the integral up = u_i 1 1 iri/' 1' uiexxm+~2 212m-eKm-,2 (l + e2)rim) F (n +)E (3 where __1 4m2 -1 44 2- 4m42-(2s-1-)2 I _E= I +:~ ~~ i-. - (4 = + is 8! 4 4 4 X- (4 2. Another integral of Bessel's equation is xmuea as given in 249, 23). Now in 251 we have expressed u, in terms of u,, us. Thus the asymptotic expressions 1), 3) just found enable us to express Jm(x) asymptotically. In fact 251, 6) gives here Uag =(1 + e 27rim) (u-_ -,). BESSEL FUNCTIONS 553 Hence if we take only the first terms in D, E of 2), 4), we have Jm,)~ - X iF { ( ei(x- (m-)) - ei(-(m-) ) } or, since F -= V/r by 144, 15), 2 { r 2 J"() cos {-j(in +)}. (5 254. An Expansion in a Series of Bessel Functions. Let us show that u-u-1 e" 2 - = unJn(x) (1 -oo for aly x and for u q 0. For u —u-1 1 X e 2 =e e 2u I~XU XU2x X2 = ~ + "' -+222 -..4..I2 I {1 2 22. 2! 2 22 2 Now for any x, and for any u = 0, the series in the braces are absolutely convergent. Their product may therefore be written in the form x2 (x)4 1 1-_+ _ _... 22 +\2 2! 2! (x 1 (x 3 1 (x\5 } + 2* *2* 3! 2!2 ***) +. + J= o(X) + uJ(x) + 2J2(X) +.._JlU _. 2. Uf U2 554 FUNCTIONS OF A COMPLEX VARIABLE 255. Jn(x) expressed as an Integral. 1. Let us show that J-(x) = nx+_/r l + cos (x cos ) sin2 ndf (1 2,nV7F(n~+ )Jo where n is a positive number. For 00 u28 cos U= E (-1) (2 " - (2s )! Hence cos (x cos () = (-S! x2S cos28. o (2s)! Thus 1 s cos (x Cos <) sin28q = ( -)X2S COS2S < sin2( ). o (2 s)! co2s2~. As this series converges steadily in the interval (0, 7r) for any value of x, we may integrate termwise, getting J cos (x cos 4)sin2n cd =d (- 1)x2s cos2* b sin2 4df o (2 s)!.0 ( -1)8 (2s~ 2 n~1'\ by 141, 7) o (2 s) 2 2 2 8+ 1r(2n + 1 V(-;Z^ r. 2 J -2 I by 142, 3). o (2s)! r(s + n1 +4, ' But r (2 + l) 1- 3 5... (2 s- ) r b14416) B t ^ 2 ) l 3 5 *28. v, by 144, 16). Thus the last series above -= r f(2n + 1 ( — 1) 3.5... (2s- 1) x2 = "'^ 2 J (2 s)! 2.sI1(n~+ s) Thus 2nVr (n + 1) os (x cos c) sin2f df: '2 (_ + =n 2 (-1)x28+!n = J"(X). s=O 22s+n,! fl(n + s) BESSEL FUNCTIONS 555 2. Another integral expression is 1 / 7r Jm(x) = I cos(m - x sin cd)do, (2 rrJo where m is a positive integer. For from 254, 1) setting u = eiO, exisin = E-emi Jm(x) = Jo(x) + 2 cos 2 m J2m(x) + 2 i I sin (2 m - 1) ) J2m_1(x). 1 Since Since eisinf = cos(x sin ) + i sin (x sin 4), we have,cos (x sin ) =Jo(x) + 2 i cos 2 m) J2m(x), (3 sin (x sin q5) = 2 I sin (2 m - 1)( ~m-1(x). (4 Let us multiply 3) by cos 2 m4 and integrate, then fcos 2 m) cos (x sin (f) = 7rJ2m(X), (5 since all the other terms = 0, by virtue of the relation cos mx cos nxdx = 0, m - n 7r m=n. If we multiply 3) by cos (2 m + 1)4 and integrate, we get f cos (2 m + 1)) cos (x sin )d = 0. (6 Similarly we get n 7r f sin (x si ) in ) sin 2 md = 0, (7 sin(x sin 4) sin (2 m + 1))d4 = 7TJ2m+l(). (8 Adding 5) and 7), or 6) and 8), we get 2). 556 FUNCTIONS OF A COMPLEX VARIABLE 256. Bessel's Solution of Kepler's Equation. This equation is 2 7rt - = u -e sin u = 7, (1 where T is the period of the planet, e the excentricity of its orbit, u the eccentric anomaly, and t the time. As u is a periodic function of t or of r, it can thus be developed in Fourier's series, u= E an sin nT, (2 n=-1 where 2 7r an, = u sin nrdr, or by partial integration =2 {( tr77- + J cos nTddu } 77r n n j This in 2) gives U= 2 ( -)f..+sin n7 +2 sin n- cos n(u- e sin u)du. 1 fln r 1 n o But the first series on the right = T, while the integral in the second series is rJn(ne). Thus 3) gives u= = T +2 J,(ne ) sin n n=l T7 which is Bessel's solution. 257. Development of f(x) in Terms of J,. It can be shown that if f(x) is continuous and oscillates but a finite number of times in l = (0, a) then f(x) admits a development of the form f(x) = cln(alx) + c2Jn(a2x) + (1 where a1 < a2 <.. (2 are the positive roots of Jn(ax). To determine the coefficients c we make use of the relations I xJ(aCx)Jnx)dx = 0 r # 8, (3 J x)dx = a2 (4 obtained from 247, 9) and 12). obtained from 247, 9) and 12). BESSEL FUNCTIONS55 557 L-et us therefore multiply 1) by xJn(a,,x). Granting that the resulting series can be integrated terrnwise, we have Xf ( J U)dx= c'fJLm X dX (5 since all the terms on the right = 0 except the one written down, by 3). Thus 4), 5) give cm = 2 a fxJ,(xd.(6 a'J2J,+(ama) Hence 1) gives f W = JQ rn ) f (x n(~)d.(7 am=1J2,_(ama) J 258. Development of f (r, ~) in Terms of the Jn. Let f (r, 0) be a one-valued continuous function in a circle W about the origin, of radius a. Then, f admitting the period 27w, we have, for a given r, by Fourier's theorem, f (r, 0) = a0 + a, cos o + a2 cos 20~+ +61 sink+ b 2sin 2 P+. ( where 1 27f =. 2Jf (r5 b)dc, (2 an ~ f (r, b) cos nfdo, (3 b~= I f f(r, c0) sin nodco. (4 But these coefficients a, 6 are functions of r and may be developed by 257. Let therefore ank, k = 1, 2, 3. be the positive roots of Jh(ar) =0. (5 Then b 257,a0 = A0tJo(a01jr) + A02J0(a02r) ~ (6 an = A n1J,(an1r) + An2Jn (a,2r) + (7 bn= Bnljnfctnlr) + B,2-jQ",2r) ~ (8. (8 558 FUNCTIONS OF A COMPLEX VARIABLE where 1 12 -f01 Aok ~e. (oa dojrf(r, ck)JO,(CtOkr)dr, (9 Amnk a2W-J 2( ka Cos mkdo frf (r, P) Jm amkr)dr, (10 Bmk a2wJ~cm ka) fsin modkf rf (r, ck)JmQ~mkr)dr. (11 259. Solution of Au = 0 for the Cylinder. 1. Let us apply the method outlined in 2~35 to find a solution u of Laplace's equation which takes on assigned values on the surface of a given cylinder C. Here the triply orthogonal surfaces are a family of cylinders, meridian planes, and planes perpendicular to the axis of C,~ which we will take to be the z-axis. Our new coordinates are therefore r, 4, z, where x==rcoso y=rsino. Transforming to the new codrdinates, we find that Laplac6's equation Au = 0 becomes 02U lu+ ~1 02U a2u ~. ( O__2 r-r r2 a02 aZ2 ( According to the general scheme, we now set u =IPRPZ, (2 w~here 1? is a f unction of r alone, 1D of c,, and Z of z. If we set 2) in 1), we find it gives rise to three equations: d2B ld (2 (3 dr2 Ird r2 ' d2- + n2CF 0, (4 d0b2 d2Z 2Z (5 dZ2 BESSEL FUNCTIONS 559 The general solution of 4) is = A cos nf + B sin no. (6 As 4 must admit the period 2 vr, we take n a positive integer. The general solution of 5) is Z = C cosh az + D sinh az. (7 The equation 3) is a form of Bessel's equation. To reduce it to the standard form we have only to set r= -. (8 Thus a special solution of 3) is R = J,(r). (9 2. Problem 1. It is now time to specify the boundary conditions on the cylinder C, which we will suppose is of length I and radius a. Let us suppose that on the lower base and on the convex surface of C, u has the value u0, a constant. On the upper base z = I let u = f(r). (10 The boundary values being symmetrical with respect to the z-axis, 4) is independent of tb and is hence a constant. Thus we take n = 0. A special solution of 1) is therefore u = RZ = A sinh (az)Jo(ar). With special solutions of this form we now construct the series ~Z /- \~ +, sinh (anO T z) u = = (z, r)= -uo + en h ( Jo (anr) (11 n=l sinh (al) where 0 > a1 > a2 > ~. are roots of Jo(ar). Since each term of 11) is a solution of Au = 0, < is a solution. Let us see if *- satisfies the boundary conditions. For z = 0, = u0 since sinh(aMz)= 0. For since ( ) r=a, = u0 since J0(aa)= 0. 560 FUNCTIONS OF A COMPLEX VARIABLE Let us now look at the boundary condition 10). If it is satisfied, we must have, setting z = I in 11), Uo + icnJo(anr)= f(r); (12 n=l or setting or setting g(r) = f(r)- uo, we must have g()= n=l Using 257, 7), we see 12) is satisfied if we take -- == a- fr(r)o(nr)dr. (13 en a2j2(ana) Thus the en being taken in this way, the solution of our boundary value problem is 11). 3. Problem 2. Let us keep the boundary values as in 2 except for z=l, =f(r,( ). (14 In this case u is no longer symmetrical with respect to the axis of U and hence u now depends on (. A special solution of 1) is therefore (An cos no + Bn sin n4) sinh (az)Jn(ar). With special solutions of this form we now construct the series u = r(z, d, r) = uo + (Ak cos ncb + Bnk sin nqb) sinh k) JQ~kr) (a15 n=o -= sin.-1) where ank, Ic = 1, 2, 3 *.. are the roots of J7(ar) = 0. Since each term of 15) is a solution of Au = 0, we see u is a solution. Let us see if 15) satisfies the boundary conditions. If we set z = 0 in 15), we see that u = u0, since sinh (akz) = 0. If we set r = a in 15), we get u = uo, since Jn(anka)= O. LAME FUNCTIONS 561 Thus two of the boundary conditions are satisfied. If the condition 14) is satisfied, we must have g(r, ) = A ~(A,, cos no + Bk sin n0)J((cnkr), (16 n=O k=l where we have set = f g = f(r, p)-.o (17 Referring to 258, we see the condition 16) is satisfied if we choose the coefficients Ak, Bfk as in 258, 9), 10), 11), where, however, we should replacef (r, b) by g(r, 4) in 17). Lamen Functions 260. Confocal Quadrics. 1. We wish now to consider very briefly a class of functions introduced by Lame which play the same role for the ellipsoid as Laplace's functions for the sphere. Suppose we wish to find a solution of Laplace's equation 02U a2U % 01 + 02+ - (1 Oz2 Oy2 aZ2 which takes on assigned values on the surface of an ellipsoid e whose equation is 2 2 -+ +-=1, a<b<c. (2 According to the general scheme outlined in 235, our first step is to replace the x, y, z coordinates by a set of coordinates defined by a family of triply orthogonal surfaces, one of which is the given ellipsoid (. This family is the family of confocal quadrics defined by x2 z2 2 F += + Iy + -1= 0, (3 a-X b-X c-X the parameter X ranging from -oo to +.o We note that for X = 0, 3) reduces to 2). We observe that F is an ellipsoid for X < a, a hyperboloid of one sheet for a < X < b, a hyperboloid of two sheets for b < X < c. Let us give to X values X1, X2, X3 lying respectively in these three intervals. The corresponding surfaces 3) will cut in 8 points 562 FUNCTIONS OF A COMPLEX VARIABLE symmetric with respect to the origin, one in each of the 8 octants. Thus, if we state in which octant the point lies, the three numbers X1, X2, 3 determine the position of the point in that octant uniquely. Let us now show that through a given point P = x, y, z there passes one and only one of each of these three kinds of surfaces. To this end we have only to show that the cubic equation in X x2 /y2 z2 (X) = + + -1=0 (4 a-X b -X c~-X has a root in each of the above intervals. Let e be a small positive number. Then X2 y2 z2 F(a -e)= + + -- 1>0, e b-a+e c-a+e x2 since the first term - + oo as = 0. Let X = X, a large negative number. Then (X0) < 0 since F(X) -1 as X —oo. Thus F (X), having opposite signs at X0 and a-e and being continuous in the interval (X0, a-e), must vanish somewhere in this interval. Similarly we see that F = 0 for some point within the interval (a, b) and within (b, c). As F (X)= 0 is a cubic, it has no other roots. These considerations show that we may take Xl, X2, X8 as coordinates of a point. They are called ellipsoidal coordinates. When we do not wish to use subscripts, we may denote these coordinates by any three letters as X, pz, v. 2. Let us show that the three surfaces X, j/, v meeting at a point x, y, z cut orthogonally. Since the X and p surfaces pass through the point xyz, we have x2 z 2 z x2 2 z2 aV+ --- b - X +k 1=0, + q-+ - 1= 0. (5 a-X b -X c-X a —p b - c - The direction cosines of the normals to these surfaces at xyz are proportional to x y z a- ' b-X ' c-X and to __ y z a —u b- c —p, LAME FUNCTIONS 563 These angles are at right angles if X2 2 z2 + __ + =0. (6 (a - X) (a - ) (-) - X) (a ) This is indeed so, for if we subtract the two equations 5) and discard the factor X -, we get 6). Thus any two of the three quadrics X, /, v meeting at the point x, y, z cut at right angles. 3. Let us now express x, y, z in terms of X, I, v. To this end let us establish an identity in u which will also be useful later, viz.: X2 Y2 2 x + + Z _ 1 (u - X) (u - A) (.- v) (7 + +1= ~. (7 a- u b- u c —u (a - u)(b - u)(e - u) To prove 7) let us consider =(u) = + 2 + + } (a - ) (b - ) ( _- ), (8 I - b - u c - uz which is a polynomial of third degree in u. The coefficient of u3 is 1. Since the X surface goes through the point xyz, the first factor of G vanishes for u = X. Hence u = X is a root of G. Similarly u = a, u = v are roots. Thus G(u) = (u - )(U - A)(u - V). Putting this in 8), we get 7). Having established 7), let us multiply it by a - u and then set u = a. We get (b- a)(e - a) Similarly, y2 (b - X)(b - at)(b - v) (c- b)(a-b) ' Z2 ( -) (e - p)(s - v) (a - c)(b - c) These determine 8 points ~ x, ~ y, ~ z, one in each octant. 4. For later use, let us find an element of arc, or the value of ds2= dx2 + dy2 + dz2 (10 in terms of ellipsoidal coordinates. 564 FUNCTIONS OF A COMPLEX VARIABLE Taking the logarithmic derivatives of 9), we get 2d= d + dj + dv x X - a -- a v-a 2 dy _ dX ds dv 2dz= dX + du + dv y X- b p -b v-c dz dA d)J dv Z X - c /A - c V - C Putting these in 10), we get ds2 = Adx2 + Bdp2 + Cd2, (11 where 2i f 2 z A + + 4l(a -X)2 (b X) (C-_x)2 -and similar expressions for B, C. The other terms which result from this substitution vanish by virtue of the relations of the type 6), which express the orthogonality of the X, pa, v surfaces. To eliminate the x, y, z in the coefficients of 11), let us differentiate the identity 7) with respect to u and then set u = X, u, v. This shows that A (X - 0-)(X- v) 4 (a -X)(b- X)( - X)' B = 1 ( - v) ( - () (12 4 (a- I) (b-/a) C - A) C= 1 (v - X) (- ) 4 (a - v)(b - )(c - v) 261. Elliptic Coordinates. 1. The equations 9) show that x, y, z are not determined as one-valued functions of the coordinates X, u, v, or using subscripts X1, X\, X3. We may remove this ambiguity by introducing three other quantities u, v, w or using the subscripts uz u2, u3 defined by pua =X, a= 1,2, 3 (1 as follows. LAM~~ FUNCTIONS 565 The ellipsoid c on which the boundary values are given is, by 260, 2), x2,2 z2 -+ - -+ 1 =0. a b c Let us set a=X0-e1, b=X0-e, c=Xo-e, (2 and determine X0, el, e2, e3 so that el +e2 +e3; e2 <e3,< el. (3 Then the equation of @ is 1I- + + 1= 0, (4 (4 k0 - e X0 - e2 0 - e3 and the equation of our confocal quadrics is L + 2 + 1= 0. (5 X-e1 X-e2 X-e3 We see that 5) is an ellipsoid when X > el, a hyperboloid of one sheet when e3 < X < el, a hyperboloid of two sheets when e2 < X < e3. Let us now set 4 p3- P - Y3 = 4(p- el)(p - e2)(p - e) (6 and suppose the p function introduced in 1) to be constructed on the invariants g2, C3, in 6). The periods 2 o, 2 o2 of this p function are given by 1.fV4p3-9P - f3-. - (f4 p3-g9P - g3) 0,0 a/41>8L~tz79=6la 2 t(< =( y p(7 as we saw in 173, 9), 14). Putting 1) il 260, 9), we get )x(p = u3-ea) )(8 (a - p) (ea - e,) where a, /, y is a permutation of 1, 2, 3. Now from 172, 16), 17), we have V/pu- e (u) Ae - e- =-. (9 0-(U) O() 566i FUNCTIONS OF A COMPLEX VARIABLE These in 8) give X,= ~,,CrW1, auIG"u a-l, 2, 3. (10 The quantities ul, uT, u3 are called elliptic coYrdinates. Having once chosen the ~ sign in 10) they determine x, y, z as onevalued functions of uI, u2, u23. Let us agree to take the + sign in 10). From 172 we have o.( - u) a-,,(u). o,(u~-:20U)),,) ( 0ju2wB) o-ju) o(-u) G0(u)' o0(u ~ 2 w) 0(u)' 0(u~2w,) o-(U) (11 These relations show that the x's are periodic functions of the u's admitting 4w1, 4W2 as periods. They also show that if we restrict u1 to range in the interval (0, w1),= -11 u2 to range in (w2 - 019 2 c+ W) = U2 u3 to range in (co - 2 w021 6 + 2C02) = U3 the point x1, x2, x3passes over every point in space once and only once. Such restricted u's we shall call normal elliptic coordinates. Let u2 be the value of u1 lying in Ul such that PU? = Xo. Then the point xj, x2, x3 describes the given ellipsoid Q once and only once when u2, u3 range in the normal intervals U2, U3. 2. The expression for ds2 = dX2 + dy2 ~ dZ2 is extremely simple in elliptical codrdinates. We saw in 260, 4 that ds2=A dX2 + A dX2 + A dX2 (12 where A (13 4(X1 - el)(X- - e2)(X01- e3) ( and similar expressions for A2, A3. Making use of 1) and remembering that du dp V4(p - el)(p - e2)(p - e,) LAMI] FUNCTIONS 56T we see that ds2 = (pu1 - 2) ()pU1 - PU3) du2 + (pit2 - PU3) ( uP2 - p 21) dU (PU3 - PU1) ( Pu3 -PU2) d42 (14 262. Transformation of Laplace's Equation. 1. The next step in the solution of Laplace's equation 02V a2 _ a2_ a~2 ay2 az2 as outlined in 235 is to transform the equation to the new coirdinates. This is a lengthy process even in the polar coordinates; for the new coOrdinates X1, XT X3, or ul1, ut2, U3 it is far longer. In order to avoid this we will make use of a theorem due to Jacobi: * Let \avp be any system of triply orthogonal coirdinates in terms of which an element of arc is given by ds2 = AdX2 ~ Bd)a2 + Cdv2. (2 Then Laplace's equation 1) becomes in the new coordinates 0 D0V\+ 0 /D0V_\ 0 (D0TJ\ _ X aVA OX / 0 aB + a\ 01 (3 ax axj a p B auj ap Cev where D=VABC. 2. To illustrate this theorem on a simple case, let us transform 1) to polar coOrdinates: x=rcos9, y=rsin~cos, z=rsin sino. As ds2-dX2+dy2 +dz2= dr2 + r2d92 ~ r2 sin2 Odf2, wehaveontaking X=r, p, A=1 B-r2 Cr= 2sin2 9 D = r2sin90. Thus 3) becomes a 2aV +\ 1 a in O a V + 1 a2 0Y _ _ _ _ 1 02 V arQ ar2O)in 0 (sino9{) sin2 9 02 which agrees with 235, 4). * For a proof of this theorem the reader may consult: C. Jordan, "Cours d'Analyse," vol. 3, p. 640; H. Weber, "Differential-Gleichungen," vol. 1, p. 94. 568 FUNCTIONS OF A COMPLEX VARIABLE 3. Let us now transform 1) to ellipsoidal coordinates ' X, X2, By 261, 12), 13), we have, on setting q = 4(\ X -el)(X-e2)(Xa- e3), a=l, 2, 3, D=i(X3-X2) -. A q2q3 Hence a (D ) a) i(X2- ) ( f) Oa1 A axi q2q 3 a1 V Thus Laplace's equation becomes (X2-X3)qlaa1 l -) + (X3- Xl)q22 (2( a2 + (X —X2)qa (a )=0. (4 3. Let us pass to elliptic co6rdinates. We have a_ _ =au __a = _ etc. a\1 X8\1 aU ql aul Thus 4) becomes 02 V. a2 V a2 V (PU2 - pU3) -2 + (pU3 - PU1) ~- + (pU1 pu2) -2 = 0, (5 whose form is extremely simple. 263. Lame's Equation. Having reduced Iaplace's equations to the new coordinates, the next step in the solution as outlined in 235 is to set V=f (u1)g(2)h(u3), (1 where f, g, h are each functions of one of the variables ul, I2, U3. If we put 1) in the transformed Laplace equation 262, 5), we get PU2 - d2 p f + pU pU d2 pu - pu2 d2h. ( f(u1) du2 g(u2) du+2 h(u3) du_ LAMIV FUNCTIONS 569 Suppose f, g, h satisfy the equations, d2f = (apel + b)f(ul), d2_ d2g = (apu2 + )g(2), (3 d2h = (aP3 + b)h(u3). du2 Putting these in 2), we see that the left side-O identically, howPutting these in 2), we see that the left side = 0 identically, however a, b are chosen. We are thus led to consider differential equations of the type d2y d2 p(x) + bjy = 0. (4 As in the case of the sphere, we are not looking for the general solution of 4) for arbitrary a and b; rather we seek to determine a and b so as to get an infinite number of particular solutions 1) with which to construct a series which will satisfy the conditions imposed on Von the surface of the given ellipsoid A. Let us apply the general method developed in Chapter XIII to the equation 4). Its singular points are those of p(x) or the points = 0 mod 2 ol, 2 02. Let us consider the point x = 0. Writing 4) in the normal form, f2yo" + 0. yl -x2Iap(x)+ b6y = 0, we have q(x)=1, ql(x) = 0, q2(x) = - ax2p(x) - 2. Hene (O) = 1 q(O) =, q2() = - a. The indicial equation for x = 0 is therefore r2-r-a=0. (5 This shows that if we take a=n(n +), n an integer 5) will have - n, and n + 1 570 FUNCTIONS OF A COMPLEX VARIABLE as roots. This choice of a gives us especially simple particular solutions, infinite in number. Putting this value in 4) it becomes d2y f-n(n +l)p(x) + bly = 0. (6 dx2 These equations are called Lame equations, in honor of Lame, who first studied them. 264. Lame Functions. 1. In Lame's equation, 263, 6), b is still undetermined. Let us see if we cannot choose it so as to get particularly simple solutions of the form y = L(p), y = L(p) -(x ) y= L(p).(-p) *. (P) (1 () (x() (x(' where L is a polynomial in p =p(x). Since the sigma quotients admit 4 wl, 4 w2 as periods, as noted in 261, 1, the expression 1) will admit these as periods also. We set then y= L (2 where Mis the product of v = 0, 1, 2 or 3 factors O'I(x) 0'2(X) O3(2) (3 a(xX) ' v(x) v (x) and L = ampm(x) + amlpm-l(x) +. + aO. (4 If then we set 2) in Lame's equation d2Y (5 d2-y -n(n + 1)p(x) + b y = 0, ( tlhl result should be an identity in x. The coefficients of the (ifferent powers of this result developed about x = 0 will thus all l) 0. These will then give us a system of equations which if cotlsistent will enable us to find the quantities we seek. LAME FUNCTIONS 571 To set 2) in 5) we need to calculate y". Let N be the product of the factors 3) which do not enter M, we call it the cofactor of M. Since a (x) we observe that N = 2 is a polynomial in p. Also let us note that MN= - p'(x). We have now = L3t + MLyp(x). But M' = l p'(x) _ _ IN 23M Hence y'=-N(2 L' + LN') = NQ. (6 Thus the first derivative of LMis the product of a polynomial in p and the cofactor of M. Hence at once 6) shows that y" = MR, where R is a polynomial in p and MJis the cofactor of N. Thus setting 2) in 5), its left side becomes y"'- in(n + 1)p + b6LM= R - n(n + 1)p(x) + bL[M= D$M, (7 where $ is a polynomial in p. As 7) must= 0 identically and as MO=0 we see $ must=0 identically. Now y has a pole of order 2 m + v at x = 0, hence the left side of 7) has a pole of order 2 m + v + 2. Hence $ has a pole of order 2 (m + 1). Thus $ considered as a polynomial in p is of degree m + 1, or 3 = am+-lPm+ + ampm + "- + aO. (8 As $ is to vanish identically, all the a's = 0, and conversely if the a's = 0, $ = 0 identically. 2. Suppose then we develop 7) about x = 0. Equating the first m + 2 coefficients of the development to 0 will give us a system of relations between the a's and the a's. These we shall see are linear in these letters. If now we set the a's = 0, we have 572 FUNCTIONS OF A COMPLEX VARIABLE a set of equations to determine the a's in 4). Let us now carry out this scheme. We have L=am a+... X2m x2m-2 where a', a~... are linear and homogeneous in the a's. Also 1 21-| 2+.., k(X) =1 + 2 ak(X) = + + G-(x) k oa(x) x Xv Xv — Hence a y = LM= am + f... x 2m+v x2m+v-2 2 where a', a'... are linear and homogeneous in the a's. Hence y,, (2m + v)(2 m. + v +1 ) +A X2m+v+2 2m+v where A1, A2... are linear and homogeneous in the ao, a,... Also Also-i^ r ^ T i\/r n l)a, A\ n(n + 1)p(x) LMJ= n( 2 + A + x2m+v2 X 2m+v where the A', A' *.. are linear and homogeneous in a0, a * Finally ba ba bLIM= sm~ + 1 + xlm+V x2/)n+v2 These set in the left side of 7) give x-(2'm+v+2) (2 m + 2)(2 m + v + 1)a a - n(n + 1)a(m + x-(2m+)Ai -A A -bam } + x-(2m+v-2) A2 - Al - bat (9 + X-(2nz+v-4) A3- A3 - ba'- +... We turn now to the right side of 7). We have am+l + +...(10 x2m+2 x 2m X2m —2 LAME FUNCTIONS 573 where a', a *... are linear and homogeneous in ag, ac... Hence aM+1 + ~a a 2 ++ X2m+2+v X2mfv + -2m-2+ where all, al.. are linear ani homogeneous in al, a... We now equate the coefficients of like powers of x in 9), 11), getting a =V2m,~v)(2m + l)-n(n+l)~am (12 and the system al = A - A' - bam, all =A Al - ball a2 A2 - A2 - b( (S alf +I AM+1- A~ +- baf. As the a's are to be all 0, set am+, = 0, then 12) gives a relation between m, n, and v, viz.: (2m + v)(2 m + p + 1) - ~(n + 1)= 0, or (2 m + v)2 + (2m + v)=n(n 1) Hence 2 m + v= n, or -(n+ As m and v are not negative, this gives 2 m + v = n. (13 Consider now the system 8). Since a a... are linear and 1 ne, 2,~~aelna n homogeneous in ao, a *... a,1,+1 they all vanish when the a0, al... vanish. Hence S) goes over into the system of m + 1 equations: Al - A' - ha, = 0, - -A a = 0, (St "2 "2 1" Am+i - A$,~ - ba" - 0. These equations are linear and homogeneous in the unknowns a0, al... am. In order that the system S') has a solution different frIom a0 = 0, a1 = 0... it is necessary that its determinant A~(b) = 0. (14 574 FUNCTIONS OF A COMPLEX VARIABLE Thus b satisfies an equation of degree m + 1. Let us put a root of 14) in S'). This system of equations allows us now to determine the ratios a0: al: a2**.: am Suppose the a's are determined thus. Then 4) and S) show that am+l a, a,..** C+1= ~ From this follows that al a1i a2, m= 0 For the development 11) shows that acl contains besides am at most am+l; that ait contains besides am-1 at most am, am+,, etc. Thus 2 = (2(am-, am, a), (S+) -2 = 2(am —1 am m+l) (s O31 = 3(arn-2 am-1, am, am +)l, where the O's are homogeneous functions. Now aot and a+,, being = 0, the first relation in S") shows that am = 0, as (1 is homogeneous. Putting this in the second equation of S"), we see that am_l = 0, and so on. Thus 3 = 0 identically. 3. Let us see in how many ways we can satisfy the relations 12) and S). There are two cases. Case 1. n = 2 s. Then 13) gives 2m+v=2s (15 and v is even. Hence or 2 v = 0, or 2. For v= 0, m=s. For each of the m+ 1 =s+ 1 roots of A(b) = 0, the system S') gives us a set of coefficients for the polynomial L in 2). We thus get s + 1 polynomials L which satisfy Lame's equation 5). For v= 2 we get, from 15), m= s-1. Each of the m + 1 = s roots of A(b) gives us a polynomial L of degree s —1. As here v = 2, y=L () c(X) = L. (16 v(x) r)(x) LAME FUNCTIONS 575 Thus we can take the factor Min three ways corresponding to the indices 1,2; 1,3; 2, 3. Each of these determinations gives us a new system of equations S) and hence a new equation A(b) = 0. Hence v being = 2, there are 3 s values of b, each of which leads to a solution 16). We thus get in all s + 1 + 3 s = 4 s + 1=2n+1 solutions of the desired type. Case 2. n= 2 s + 1. Reasoning in exactly the same way, we see that also in this case there are 2 n + 1 solutions. As it can be shown that these solutions are distinct we may state the fundamental theorem due to Lame: The constant b in 'i Y =n(n+ 1)p(x) + b y, dx2 may be determined in 2 n + 1 different ways such that this equation admits a solution of the type L(p), Vp-e- L, /p -eap-e L, V(p - el)(p - e2)(p- )3) * L (17 Moreover the b's belonging to different n's are also different. The functions 17) are called Lame functions or polynomials. 265. Integral Properties. 1. In order to develop an arbitrary functionf(O, 0) in terms of Laplace's functions we made use of certain integral relations established in 238. We wish to establish analogous relations for Lame's functions. We saw in 264 that for a given n, the constant b can be taken in 2 n + 1 ways, bnl, bn2, * bn, 2n+1, 1 so that y" = In(n + 1)p(u) + b,,y (2 admits a Lame function, denote it by Lnk(u), as solution. We saw in 261, 1, that if u1 has the value u~ lying in Ul = (0, ol) such that P= Pu1l = X0 576 FUNCTIONS OF A COMPLEX VARIABLE and if u2 ranges in U2 = (2 - ol, w1 2 + l) and u3 in U3 = (ot1l-2 c2, Clo + 2 Co2), then the point x x2 X3 corresponding to u~ u2 us describes just once the ellipsoid @ on which the boundary values are given. For brevity let us set a = - 2 o, b = co + 2 o1, c=o1-2 )2, = c1 + 2co2. Thus the interval (a, b) is twice as long as 12 while the interval (c, d) is the same as U1. Obviously if u2 ranges over (a, b) and u3 over (c, d), the point x1 x2 x3 corresponding to u~, u2, u3 will now describe the ellipsoid ( twice. We wish now to prove: J= fbduJ LI- mr(U2)Lmr(U3)Lns(U2)Lns(u(3) ( P2- u3)du3= 0 (3 a c when Lmr f Lns, For, let us set A = Lmr(U2)Lns(U2)pu2 du2, d B = dLmr (Us)Ln (u)ps3du3' (4 C = J Lmr(u2)Ln,(U2)du2, a D = JdLmr (U) L (u3) dus. Then J= AD-BO. (5 Now 2) gives, u being either u2 or u3, d2Lmr( - {m(m + 1)pu + bmr}Lmr(U), dU2 d2L(u) =- n(n + 1)pu + bns Lns(U). dU2 Let us multiply the first equation by Ls and the second by Lm, and subtract. We get on the left side Lns"r - L "mrnsL =- d L -L mrL's I d F(u). s du du LAME~ FUNCTIONS 77 577 On the right side we get pum~ + 1) - n(n + 1) +-LmrLns (bmr - 8,,) = G(1u) ~ H(u). Thus we have dF(u) (()+Hu.( du rU)+Uu) ( Set here u = u2 and integrate over (a, 6). We get (U) m(m + 1)- n(n + 1>~A + (bmr -.C As P(u'2) admits the period 4co,, the lef t side= 0. Thus ~m(m~+1)- n(n~+1)'A +(bmrhs)=. (7 Similarly integrating with respect to u3 over (c, d) gives ~m(m +1)- n(n +1)~B +Qbmr - bns)-D=0. (8 Suppose now m #- n. We multiply 7) by _D and 8) by C, then subtract. We, get J~m(m ~ 1) - n(n + 1)~ 0. Hence J= 0. Suppose r =t s. We multiply 2) by B and 3) by A and subtract, getting (bm - i) = 0. Hence again J= 0. 2. We next wish to calculate K= fdu2f L2r(ut2)L2(us) p2-p3d which is what J becomes for m = n. Let us set P = L'( B = ]~ir~2)du2dU SR =fdL2nau (9 (10 578 FUNCTIONS OF A COMPLEX VARIABLE Then K= PS- Q.R. (11 We begin by observing that V= Lr(U)p(u), W= Lr(U), (12 admit 2 ow, 2 w2 as periods. They are even functions and have but a single pole in the parallelogram of periods Q2 constructed on 2 o1, 2 o2. Thus Q2 is a primitive parallelogram. Thus by 168, V(u) = Vo + Vlp(u) + v2p"(u) + v3piv(u) + * (13 W(u) = wo + + W1P(u ) + W"(u) + W3pi(u) +... (14 Here we have used the fact that V, TW being even functions of u, cannot contain derivatives of even order of '(u); also that 1(u) = -p(u), 1 "'(u) = -pl(u), etc. We have now rb /~b Pb P= V(u)du = v du + v, p(u)du, the other terms dropping out since they admit the period 4 w1. Thus P = V0o2 01 + W02- (O2 - 2 Wl) - v l ( 12 o + 02) - (0- 2 l1) Now '(u +2 01) ='(u) +21, (u - 2 w1) = (u) - 2 1. Thus P = 4 v - 4 vl1. Similarly Q = 4 v002 -4 vl7 R = 4 w0ol - 4 wlV1. - = 4 wo02 -4 wl.2 S=4W0eo2 —4wly2. Thus K= 16(v0w- VlWo) (W27l - W)12) = 8 i(vowl - v1w), by 171, 8). Hence finally 7b d f du j L2r(U2)L2s(U) (pu2 - opu3)du = 8 r7i(vo1 - vlw), (15 where v0, w, v1, w1 are given by the developments 13), 14). LAMI FUNCTIONS 579 266. Solution of AV= 0 for an Ellipsoid. We are now in a position to obtain a solution V of Laplace's equation O2 V. a2. at2 V (PU2 - pu) du2 + (pu3 - P8) 02 + (Pm1 - Pm2) duT 0 (1 which will take on assigned values f(u2, u3) on the surface of a given ellipsoid x 12 X0 - e X0- e2 X0- e3 For let Lns(um), m= 1, 2, 3 be Lame polynomials satisfying du2 Thenn 1by 263 t,, he du, Then by 263, the product L,,(U1)L,((u u2) Lns(u3) is a solution of 1), and hence the series, f (u1,U2,U3) = EEans,,L(ul) Ln,(u2) Ln(u3), (2 n s if convergent, will satisfy 1). For the boundary values to be satisfied it is necessary for 2) to take on the value f(u2,uw) for u1 u~. If we set anLns(?u?)= An,, (3 the boundary condition is satisfied if f (u2,u) = 2 ALns(u2) Ln,(u3). (4 n, s To determine the unknown coefficients am, or what is the same, Amr, we multiply 4) by Lmr(U2) Lr(U3)(PU2 -pu3) and integrate. Granting that termwise integration is allowed, we get Pb fd f du f (f(u,v) Lr(U) L mm(v) (pu - pv) dv J~a Jd~c sb Pd == A,, du Ld ( u) L.(v) (pu- pv) dv, (5 AmrJ duMJL~ (in) LMr(V) (PUPV)dV, (5 a c since all the other terms = 0 by 265, 3). Let L2p =o l +a * Let L(U) (u) = ao + alPU + C2p,"(u) +... Lmr(U) = 13' + lpu + 02 P"(u) + * Then by 5) and 265, 15) 1 b Amr 8 vi(ao1 - a0) (u V)Lmr(U)Lmr(v) (pu-pv)dv. (6 Thus the coefficients amr in 2) are given by 3) and 6). INDEX (Numbers refer to pages) Abel's relation, 348 Abscissa of z, 4 Absolute value, 11 Acyclic, 139 Analytic continuation, 224 Argument, 11 Asymptotic development, 309, 328 series, 328 vector, 329 Axial symmetry, 518 Axis, real, 4 imaginary, 4 Bernoullian numbers, 289 polynomials, 312 Binomial coefficients, 17 Branch, of a function, 95 point, 95 Casus irreducibilis, 22 Calculation of 7r, 205 Cauchy, theorems, 211, 214 Cauchy-Riemann equations, 175 Characteristic, 238 Circle of convergence, 72 Confocal quadrics, 561 Conformal representation, 184 Congruent, 335 Connex, 139 Continuity, 134, 167 uniform, 139 Convergence, absolute, 37, 271 steady, 197, 271 Cross cut, 214 Cyclic, 139 Definitely infinite, 141 Differential equation, Bessel's, 454, 469 Differential equation, hypergeometric, 455 Lame's, 454 Laplace's, 494 Legendre's, 454 linear homogeneous, 454, 466 Dirichlet's problem, 532 Domain of a point, 133 of definition, 130 e notation, 32 Ellipsoidal coordinates, 562 Elliptic coordinates, 564 integrals, 385, 444, 446 integrals complete, 394 modular functions, 432 Essentially singular point, 244 Euler's formula, 106 summation formula, 318 Flux, 512 Fourier's development, 262 Frobenius' method, 474 Fuchs, 455, 486 relation, 487 Functions, of z, 164 B, r, 295, 299 p, 344 a-,, 356 07', n, 381 O, 426, 428 0, Z, 439 su, cu, du, 402 algebraic, 89 algebraic explicit, 90 analytic, 210 Bessel's, 533 circular, 111 581 582 INDEX Functions, elliptic, 342 even, 115 exponential, 102 harmonic, 509 hyperbolic, 118 integral rational, 87, 249 inverse, 178 inverse circular, 125 Lam6, 575 logarithmic, 62 odd, 115 periodic, 334 double, 337 rational, 88, 251 regular, 234 transcendental, 253 Fundamental system, of differential equations, 459 spherical harmonics, 525 Gauss' theorem, 511 Green's theorem, 511 Gudermann, 327 Harmonic polynomial, 522 Hermite's formula, 442 Images, 167 Incongruent, 335 Indicial equation, 464 Integrals, 186 curvilinear, 150 surface, 149 Invariants, 364 Irregular points, 474 Jacobi's theorem, 336, 447 normal integral, 446 Kepler's equation, 556 Lame's equation, 570 Laplace's equation, 494 Laurent's development, 234 Legendre's normal integrals, 393 relation, 369, 445 Limiting points, 333 Limits, 26, 131, 133, 167 Modular angle, 394 functions, 432 Modulus, 11 of elliptic integrals, 385 Moivre's formula, 17 Network, 338 Normal form, of differential equation, 462 of product, 276 Order, of elliptic function, 345 of pole, 239 of zero, 237 Ordinary point, 234 Ordinate of z, 4 Paradox of Cauchy, 62 Parallelogram of periods, 337 primitive, 337 Parameter of elliptic integrals, 385 Period, 334 primitive, 336 pair, 337 Point at infinity, 247 Polar form of z, 11 Poles, 237 Postulate, fundamental, 29 Potential, 154, 493 Prime roots, 465 Primitive unit roots, 21 Rectangular form of z, 11 Region, 138 connected, 138 simple, 138 Regular point of function, 117 of differential equations, 474 Residue, 238, 256 Rodrigue's formula, 502 Scale, logarithmic, 46 Schlafli's formula, 501 Series, adjoint, 35 alternate, 30 associate logarithmic, 271 asymptotic, 328 binomial, 52 INDEX 583 Series, deleted, 26 double, 77 geometric, 27 harmonic, 28 hypergeometric, 54 hyperharmonic, 29 inversion of, 259 niormal, -al power, 71 remainder, 26 row and column, 80 Stirling's, 327 two-way, 75 Set of points, 333 limited, 333 Singular point, 234 simple, 466 Spherical harmonics, 523 Stokes' theorem, 158 Taylor's development, 218 remainder, 223 Tests of convergence, Cahen, 40 Cauchy, 44 ID'Alemnbert, 42 Gauss, 51 Raabe, 40 Unit roots, 21 primitive, 21 Vall6e-Poussiii, de la, 311 Wirtinger, 318 Zeros, 237 LIST OF SYMBOLS (Numbers refer to pages) Ill, 11 D oo), 248 P.,, 527 21 -, 167 div, 509 q, 425 Abs a, 4 A (O), 394 Q,-, 506 Arg a, 4 A (u), 508 11(z), 300 a =- b 335 E,,k).p )0 zl 439 B (u, v), P (z), 295, 299 11 (p, a), 394 sn, en, din, 402 Char f(z), 238 El, K, 394 y,,, 523 Res f (z), 238 flux, 512 Z&, 508 Ds (a), -D8*(a), 133, 134 Pm, 496 17 The symbol, is used throughout the book and means converges to, or has as limit.