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University of Illinois Library L161—H41 eee a if A ALGEBRA AN ELEMENTARY TEXT-BOOK _ Digitized by the Internet Archive in 2021 with funding from University of Illinois Urbana-Champaign httos://archive.org/details/algebraelementarO2chry ALGEBRA AN ELEMENTARY TEXT-BOOK FOR THE HIGHER CLASSES OF SECONDARY SCHOOLS AND FOR COLLEGES © BY G: CHRYSTAL, M.A., LL.D. HONORARY FELLOW OF CORPUS CHRISTI COLLEGE, CAMBRIDGE ; PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF EDINBURGH Leeds ad EDINBURGH ADAM AND CHARLES BLACK MDCCCLXXXIX WATHE MAINE i 5 a - / \s \ 4 7 PREFACE TO PART II. Tue delay in the appearance of this volume finds an apology partly in circumstances of a private character, partly in public engagements that could not be declined, but most of all in the growth of the work itself as it progressed in my hands. JI have not, as some one prophesied, reached ten - volumes; but the present concluding volume is somewhat larger and has cost me infinitely more trouble than I expected. The main object of Part II. is to deal as thoroughly as possible with those parts of Algebra which form, to use Euler’s title, an Jntroductio in Analysin Infimtorum. A practice has sprung up of late (encouraged by demands for premature knowledge in certain examinations) of hurrying young students into the manipulation of the machinery of the Differential and Integral Calculus before they have grasped the preliminary notions of a Limit and of an Infinite Series, on which all the meaning and all the uses of the Infinitesimal Calculus are based. Besides being to a large extent an educational sham, this course is a sin against the spirit of mathematical progress. The methods of the Differential and Integral Calculus which were once oI ; EMATICS LIBRARY vl ; PREFACE an outwork in the progress of pure mathematics threatened for a time to become its grave. Mathematicians had fallen into a habit of covering their inability to solve many particular problems by a vague wave of the hand towards some generality, like Taylor’s Theorem, which was sup- posed to give “an account of all such things,” subject only to the awkwardness of practical inapplicability. Much has happened to remove this danger and to reduce d/dz and /dz to their proper place as servants of the pure mathematician. In particular, the brilliant progress on the continent of Function-Theory in the hands of Cauchy, Riemann, Weierstrass, and their followers has opened for us a prospect in which the symbolism of the Differential and Integral Calculus is but a minor object. For the proper understanding of this important branch of modern mathe- matics a firm grasp of the Doctrine of Limits and of the Convergence and Continuity of an Infinite Series is of much greater moment. than familiarity with the symbols in which these ideas may be clothed. It is hoped that the chapters on Inequalities, Limits, and Convergence of Series will help to give the student all that is required both for entering on the study of the Theory of Functions and for rapidly acquiring intelligent command of the Infinitesimal Calculus. In the chapters in question, I have avoided trenching on the ground already occupied by standard treatises: the subjects taken up, although they are all important, are either not treated at all or else treated very perfunctorily in other English text-books. Chapters xxix. and xxx. may be regarded as an PREFACE j Vll elementary illustration of the application of the modern Theory of Functions. They are intended to pave the way for the study of the recent works of continental mathe- maticians on the same subject. Incidentally they contain all that is usually given in English works under the title of Analytical Trigonometry. If any one should be scandalised at this traversing of the boundaries of English examination subjects, [ must ask him to recollect that the boundaries in question were never traced in accordance with the principles of modern science, and sometimes break the canon of common sense. One of the results of the old arrangement has been that treatises on Trigonometry, which is a geometri- cal application of Algebra, have been gradually growing into fragments more or less extensive of Algebra itself: so that Algebra has been disorganised to the detriment of Trigono- metry ; and a consecutive theory of the elementary functions has been impossible. The timid way, oscillating between ill- founded trust and unreasonable fear, in which functions of a complex variable have been treated in some of these manuals is a little discreditable to our intellectual culture. Some expounders of the theory of the exponential function of an imaginary argument seem even to have forgotten the obvious truism that one can prove no property of a function which has not been defined. I have concluded chapter xxx. with a careful discussion of the Reversion of Series and of the Expansion in Power-Series of an Algebraic Function— subjects which have never been fully treated before in an English text-book, although we have in Frost’s Curve Tracing an admirable collection of examples of their use. Vill PREFACE The other innovations call for little explanation, as they aim merely at greater completeness on the old lines. In the chapter on Probability, for instance, I have omitted certain matter of doubtful soundness and of questionable utility ; and filled its place by what I hope will prove a useful exposition of the principles of actuarial calculation. I may here give a word of advice to young students reading my second volume. The matter is arranged to facilitate reference and to secure brevity and _ logical sequence; but it by no means follows that the volume should be read straight through at a first reading. Such an’ attempt would probably sicken the reader both of the author and of the subject. Every mathematical book that is worth anything must be read “backwards and forwards,” if I may use the expression. I would modify Lagrange’s advice a little and say, “Go on, but often return to strengthen your faith.” When you come on a hard or dreary passage, pass it over; and come back to it after you have seen its importance or found the need for it further on. To facilitate this skimming process, I have given, after the table of contents, a suggestion for the course of a_ first reading. The index of proper names at the end of the work will show at a glance the main sources from which I have drawn my materials for Part II. Wherever I have consciously borrowed the actual words or the ideas of another writer I have given a reference. There are, however, several works to which I am more indebted than appears in the bond. Among these I may mention, besides Cauchy’s PREFACE 1X Analyse Algébrique, Serret’s Alyébre Supérieure, and Schlo- milch’s Algebraische Analysis, which have become classical, the more recent work of Stolz, to which I owe many indica- tions of the sources of original information—a kind of help that cannot be acknowledged in footnotes. I am under personal obligations for useful criticism, for proof-reading, and for help in working exercises, to my assistant, Mr. R. E. ALLARDICE, to Mr. G. A. GiBson, to Mr. A. Y. FRASER, and to my present or former pupils— Messrs.’ B. B. P. BRANDFoRD, J. W. Butters, J. CROCKETT, J. GOODWILLIE, C. TWEEDIE. In taking leave of this work, which has occupied most of the spare time of five somewhat busy years, I may be allowed to express the hope that it will do a little in a cause that I have much at heart, namely, the advancement of mathematical learning among English-speaking students of the rising generation. It is for them that I have worked, remembering the scarcity of aids when I was myself a student ; and it is in their profit that I shall look for my reward. G. CHRYSTAL. EDINBURGH, lst November 1889. VOL. II b , > e ; } Fae i a ae i ler =. _ - P , ¢ 7 ? ‘ ’ ie » @ - ee ee cL a . ; 7 * , Pitta ee is : F ‘ ; : he } by | Se ‘ Fiak Ae a ao ti : ri 7) = eee fj 44 ad Pi | We} . 4 beg a s . as 7 . ee. * j Lf Se ; r ’ i] ; = ! S . . i r ‘ ' f , i ~ ; A er ; : a = “ } 4 Oe a CONTENTS. The principal. technical terms are printed in italics in the following table. CHAPTER XXIII. PERMUTATIONS AND COMBINATIONS. Definition of r-permutation and r-combination Methods of Demonstration Permutations Number of r-per olen 03 n hs Kramp’s Notation for Factorial-n (7!) Linear and Circular Permutations : Number of 7-permutations with repetition . Permutations of letters having groups alike Examples Combinations Combinations from Sete Number of 7-combinations of » ieee Various properties of ,C,—Vandermonde’s THestett Combinations when certain letters are alike Combinations with repetition Properties of ,H >—Number of 7-ary Prodne Exercises I. : : : Binomial and Multinomial iietae Examples Exercises II. Examples of the sapike te of the List of Distribution Distributions and Derangements Distribution Problem Derangement Problem Subfactorial n (7j) defined Theory of Substitutions Notation for Substitutions PAGE = ae AnNnAOr RKP DY OD WH 21-22 22-25 22 24 25 25-32 26 Xll CONTENTS Order and Group Cyclic Substitutions and Transpostiiane Cycles of a Substitution : Decomposition into Transpositions . Odd and even Substitutions Exercises III. Exercises IV. CHAPTER XXIV. GENERAL THEORY OF INEQUALITIES. Definition of Algebraic Inequality Elementary Theorems . : Examples Derived Theorems A Mean-Theorem for rachions (x? - 1) [eee (a —1)/q mom—-Ve — 1) _ = (am -1) = m(@ —1) ma"-Wa — Daa am — bm —mb"-(a, —b) Inequality of Arithmetic and Geometric Means Lpa"/Xp> <(Zpa/zZp)” . Exercises V. Applications to Maxine and Minima- Tiesrens Fundamental Theorem Reciprocity Theorem Ten Theorems deduced Grillet’s Method Method of Increments . Purkiss’s Theorem Exercises VI. CHAPTER XXV. LIMITS. Definition of a Limiting Value and Corollaries Enumeration of Elementary Indeterminate Forms Extension of Fundamental Operations to Limiting Values Limit of a Sum Limit of a Product Limit of a Quotient . Limit of a Function of Limits Limiting forms for Rational Functions Forms 0/0 and « /oo PAGE 27 27 27 28 29 32 33 35 36 88 41-50 41 42 43 CONTENTS Fundamental Algebraic Limit L(w—-1)/(~-1) when x=1 Bxarhples—L7"( e+1)/ie, Ladx/itx, when c=0, &e. Exponential Limits : L(1 +1/x)* ae =o, Nenierian eee : : L (+a, L (+y/2), L (tay), L (a*—-1)2 2=0 %= 00 0 “=0 Exponential and Logarithmic Inequalities . Euler’s Constant General Limit Theorems L{ fe) ie rae : Lif(e+1)-—f(x)} =Lf(x)/a when x= Lf«e+1yfix)=Li f(x ‘ Vz when x= 00 Exponential Limits Resumed . Daeye, L log.ve, L vloge L=0 t= x=0 Examples— L a*/n!, L m(m-1)...(m—n+1)/n! NM= co nNn=n fe 1 2=--0 General Theorem regarding the form 0° Cases where 0°+1 Forms «° and 1% Trigonometrical Limits Fundamental Inequalities Lsinaz/z, Ltanaz/x, when x=0 . 1( sin 1 . LU nti / Un< > iL 101 102 103 105 106 107 107-118 107 107 X1V CONTENTS Examples—Integro-Geometric, Logarithmic, Exponential, nomial Series ; Cauchy’s Condensation Test . Logarithmic Criteria, first form Logarithmic Scale of Convergency Logarithmic Criteria, second form Examples—Hypergeometric and Binomial Ser les Historical Note Semi-Convergent Series Example of Direct Discussion Uy — Ug + Ug — . Tr ptomercial aye Abel’s Inequality Convergence of a Series of Complies Tors Necessary and Sufficient Condition for Conver ele Convergence of the Series of Moduli sufficient Examples—Exponential and Logarithmic Series, &c. Application of the Fundamental Laws of Algebra to Infinite Series Law of Association Law of Commutation Addition of Infinite Series Law of Distribution . Theorem of Cauchy and Mertens Special Discussion of the Power-Series Condition for Convergency of Power-Series . Circle and Radius of Convergence Discontinuity and Infinitely Slow diene Uniform and Non- Uniform Convergency Du Bois-Reymond’s Theorem Abel’s Theorem Principle of Indeterminate Goefliclents ‘Infinite Products ; Convergent,” Divergent, and Oscillatae Products Discussion by means of 2 log (1+ 2p) Criteria from Du, Independent Criteria Convergence of Complex Products General Properties of Infinite Products Estimation of the Residue of a Series or Product Convergence of Double Series . ~ Four ways of Summation Double Series of Positive Terms Cauchy’s Test for Absolute Convergency Examples of Exceptional Cases Imaginary Double Series General Theorem regarding Double Paver Series Exercises VIII. PAGE 108-109 109 112 113 115 116-118 118 119-123 120 121 122 122 123 124 124 125 125-128 125 125 ee Be 127, 127 129 129 129 129 130 131 133 135 135-144 136 136 137 139 139 7AU 145 147-157 148 ~ 150 153 155 157 158 158 CONTENTS CHAPTER XXVII. BINOMIAL AND MULTINOMIAL SERIES FOR ANY INDEX. Binomial Series Determination of Detrieces medic fe assumed Euler’s Proof S Addition Theorem for the Bacned Bere: Examples Ultimate Sign of We Tecnee Integro-Binomial Series Examples Exercises IX. Series deduced by Expansion of Ratanel Hanciiong Expression of a”+ 6" and (a”tl— 6T1)/(q—) in terms of a8 and a+, and connected series Sum and Number of 7-ary Products Examples Exercises X. Multinomial Series Numerical Approximation by iavrgp eel tees, Numerically Greatest Term . Limits for the Residue Exercises XI. CHAPTER XXVIII. EXPONENTIAL AND LOGARITHMIC SERIES. Exponential Series Determination of inetedate attatia, iste mead Deduction from Binomial Theorem . Calculation of ¢ Cauchy’s Proof Addition Theorem for the inane cope Bernoulli's Numbers : : Expansions of #/(1-e-*), n(e# + o-*)|(c i ed Mant aoe Bernoulli’s Expression for 17+ 27+... +n” Summations by means of Exponential Theorem Integro-Exponential Series . Examples Exercises XII. . Logarithmic Series Expansion of log (1+ 2) Derived Expansions Calculation of log 2, log 3, ae XV PAGE 162-175 162 164 165 168 169 170 172 175 176-186 177 181 184 186 189-191 191-194 19] 192 195 197-204 197 198 200 202 203 204-209 205, 208 208 209-211 209 210 212 213-226 214 215 216 Xvi CONTENTS PAGE Factor Method for calculating Logarithms . ; ‘ : 219 First Difference of log x : P ; : , 221 Summations by Geisha Series. ; ; 221-226 Zp(n)u"/(n+a)(n+). ; : : : 222 Examples—Certain Si Conver tot) Series, Aa, : ; ; 224 . Inequality and Limit Theorems ‘ : : ; : 226 Exercises XIII. : ; : . ; ‘ 227 CHAPTER XXIX. SUMMATION OF THE FUNDAMENTAL POWER-SERIES FOR COMPLEX VALUES OF THE VARIABLE. Preliminary Matter . : . 230-248 Definition and Properties of re Gaadlar Rumouens : A. 230-238 Evenness, Oddness, Periodicity : , : - . 231 Graphs of the Circular Functions . , . ‘ 232 Addition Formulex for the Circular Fainetions : ; ; 234 Inverse Circular Functions . 7 : ; , , 234 Multiple-valuedness . ; a a ; : 236 Principal and other rie ah Z 3 , 236 Inversion of w=z2" and wP=24 : : : ; 238- 247 Circulo-Spiral Graphs for w=% : : ; : 240 Multiplicity and Continuity of aay w , : : ; 241 Riemann’s Surface . : : : : 241 Principal and other Bonen of a w : ; 243 Circulo-Spiral Graphs for w=24 : : 245 Principal and other Values of £/w? : 246 ‘Exercises XIV. ‘ 247 Geometric and Integro- sCaomounte Barias ‘ : ; : 248 Zr" cos(a+né), &e. . ? , 249 Formule connected with ero S Theoret sf the Binortak Theorem for an Integral Index . : : 250-255 Generalisation of the Addition Theorems for the Gr cular Functions 251 Expansions of cos 70, sin 76/sin 0, &c., in powers of sin @ or cos@ . 252 Expression of cos”@ sin”@ in the form 2a, cos pé or Za, sin pO ; 253 Exercises XV. : : ; : ‘ 255 Expansion of cos 13 0 and sin @ in ‘powers sof Pare. : : 256-259 Exercises XVI. e. : i ; ; 260 Binomial Theorem for a Gone: Vanighle : : ; 261-264 Most general case of all (Abel) : : : 263 Exponential and Logarithmic Series for a Contples Vanable : 264-273 Definition of Expz . ; : : ‘ : 264 Exp (@+ yt) =e*(cosy+7sin Ay ; , : ; F 266 Graphic Discussion of w=Expz . : : : 3 266 Imaginary Period of Expz . : . : : : 268 Log w=log (mod w) +7 amp w , ‘ . ‘ é 269 CONTENTS XVil PAGE Principal and other Branches of Log w : ; : P11 p.269 Definition of Exp,2 . : : ; : } 270 Addition Theorem for Log w : : sl Reet : 271 Expansion of ;Log(1+z) . ; ; ; ; 272 Generalisation of the Circular F SpwabRE : : i 273-289 General Definitions of Cosz, Sinz, &c. : : : ; 278 Euler’s Exponential Formule for Cosz and Sinz . ; : 274 Properties of the Generalised Circular Functions. . : . 275 Introduction of the Hyperbolic Functions F : : 276-289 Expressions for the Hyperbolic Functions . : : ‘ 276 Graphs of the Hyperbolic Functions : ; : ; 277 Inverse Hyperbolic Functions ; é i : 279 Properties of the General Hyperbolic Wane ‘ ; 279-283 Inequality and Limit Theorems ; 283 Geometrical Analogies between the vias and ae i epeebelie Functions d ‘ ‘ : : : : 284 Gudermannian Function. : : . ‘ ; 287 Historical Note : : : ; : < j 288 Exercises XVII. : : , 289 Graphical Discussion of the Generaleed @renian Tnactione : 292-301 Cos(@+yt),.° «. 2 : : : ' wheal ae 2 Sin(#+yt) . : B : : : : : 295 Tan (a+yi) . ; : , : : 297 Graphs of f(# + yt) and Tyee) : : ; ; : 298 General Theorem regarding Orthomorphosis ‘ Hp ipwi F 299 Exercises XVIII. : : ; ; 301 Special Applications to the Cireulas mena : ‘ 302-310 Series derived from the Binomial Theorem . ‘ : ; 303 Series for cos m@ and sin m@ (m not integral) . ; : 303 Expansion of sin—'w, Quadrature of the Circle ; : ‘ 305 Examples—Series from Abel, &c. . : : : ; 306 Series derived from the Exponential Series. . ; : 307 Series derived from the Logarithmic Series. : ; : 307 Sin d-—4sin 20+4sin30—...=40 ; , : : 308 Remarkable Discontinuity of this last Series 5 hee : 308 Series for tan—!z, Gregory’s Quadrature of the Circle : ; 309 Note on the Arithmetical Quadrature of the Circle : é 309 Exercises XIX., XX. . : ; : ; A 310, 311 CHAPTER XXX. GENERAL THEOREMS REGARDING THE EXPANSION OF FUNCTIONS IN INFINITE FORMS. Expansion in Infinite Series. me ; - 313-320 Expansion of a Function of a eneton : : : : 313 XVill CONTENTS PAGE Expansion of an Infinite Product in the form of an Infinite Series 313 Examples—Theorems of Euler and Cauchy . 5 : : 315 Expansion of Sech w and Sec : : : : : 317 Euler’s Numbers. A 318 Expansion of Tanh a, Coth a, Cosech on Tan ©, 2 Cot 2, Gosee a 319 Exercises X XI. : : 320 Expression of Certain Haneuons as Takis Procucee : 322- 333 General Theorem regarding the Limit of an Infinite Product —_ . 322 Products for sinh pu, sinh w; sind, siné . 2 : : 324 Wallis’s Theorem. 7 : ; .- | ead Products for cosh pu, cosh wu ; cos nO, cosé . : 327 Products for cos # +sin ¢ cot 8, cos ¢—sin ¢ tan 8, 1 + cosec 0 sing. 330 Product for cos ¢ — cos 0 4 i : : : 331 | Remark regarding a Certain Waller : : : - 332 Exercises XXII. : ; : 333 Expansion of Circular and Hyperbolic Wacie Hoven in an ata Series of Partial Fractions : : : 335-338 Expressions for tan 6, 6 cot 0, 6 cosec 8, sec 6 : ; : 336 Expressions for tanh u, wcoth wu, wcosech wu, sech wv ; ; 338 Expressions for the Numbers of Bernoulli and Euler . , 338-343 Series for B,, : ; F : ’ L : 339 Product for By» ; . orl ; P ! , 340 Certain Properties of B, . : ; 340 Radii of Convergence of the Power-Series for tan 0, 0 cot 6, 0 cosec 0 340 Series and Edner for En: ; A : : ; 341 Certain Properties of E,,, : ; ; 342 Radius of Convergence of the Foc Series for sec 7) pr ; 342 Sums of Certain Series involving Powers of Integers é : 343 Power-Series for log sin @, &c. . ; : ; : : 343 Stirling’s Theorem : : ie y : ; : 344 Exercises XXIII. ‘ iy : i 348 Reversion of peer ashe of an Bie Function. 349-370 General Expansion-Theorem regarding 2(m, n)v”y"=0 : : 350 Reversion of Series . : 354 Expansion of the various Bethe: 3 an Alchente Wanewse : 355 Irreducibility, Ordinary and Singular Points, Multiple Points, Zero Points, Poles, Zeros, and Jnfinities of an Algebraic Function ; : : \ : ; 356 Expansion at an Ordinary Bont : ‘ : : : 358 Expansion at a Multiple Point : 359 Newton's. eich Degree-Points, Effective Beau g jateere: Points . : 362 All the Branches fe an ange iede Hanetien ecpanae : : 365 Algebraic Zeros and Infinities and their Order , : y 367 Method of Successive Approximation ; : : ; 367 Historical Note : : ; 5 : : ‘ 370 Exercises XXIV. ; : ; : : 3‘ . 370 " CONTENTS XIX CHAPTER XXXI1. SUMMATION AND TRANSFORMATION OF SERIES IN GENERAL. iy PAGE The Method of Finite Differences : ; te: : 372-383 Difference Notation , ; , : ; 372 Two Fundamental Difference FEfieors ems , ‘ : : 375 Summation by Differences . ae7: A , ; 376 eee eel Series, = sin es +nB) . ; : : 377 Expression of Se in terms of the Differences of ‘ : 379 1 : Montmort’s Theorem . : : ‘ jie 381 Euler’s Theorem : ; i R : ; : 382 Exercises XXV. : . 4 ; : : : 383 Recurring Series : ; : ; : ‘ 385-389 Scale of Relation ; ‘ : ; ; , : 385 Generating Function ; ; . : j : 386 To find the General Term . : 387 Solution of Linear Difference Teton wait Goetant Coenen 388 Summation of Recurring Series ; : , : : 388 Exercises XX VI. i ; 389 Waring’s Method for shines ed taking every “kth fre of a Series whose sum is known ; : , 390-392 Miscellaneous Methods : : . ; ; 392-394 Use of Partial Fractions ; : é ; » 392 Euler’s Identity ‘ : ; ; . ; 392 Exercises XX VII. ; ; ; : : : : 394 CHAPTER XXXII. SIMPLE CONTINUED FRACTIONS, Nature and Origin of Continued Fractions. ) 396-402 Terminating, Non- Terminating, Recurring or Perini, Simple Continued Fractions . : : ; : 396 Component Fractions and Partial Guokes : ; : 396, 397 Every Number convertible intoaS.C.F.. : ; 397 Every Commensurable Number convertible into a Terminating S.C.F. 399 Conversion of a Surd into aS.C.F. . ‘ : : ‘ 401 Exercises XXVIII. 5 ; : ; ee: 403 Properties of the Convergents to a S. c. F. ; : ‘ 404-414 Complete Quotients sh Convergents . : : : : 404 Recurrence-Formulz for Convergents : : : ; 405 Properties of pp, and gn ; , : : . ; 406 XxX , CONTENTS PAGE Fundamental Properties of the Convergents : - : 408 Approximation toS.C.F.. ‘ s ; ; 410 Condition that Pn/dn be a Gonvercent to 2, : : ‘ 412 Arithmetical Utility of S.C.FF. . : . . alee Convergence of 8.C.F. ; : : : ; : 414 Exercises XXIX. : : : 415 Closest Rational Leagenesetteon of Given Comes , 417-424 Closeness of Approximation of py/qdn ; ; ; : 418 Principal and Intermediate Convergents ; a - : 419 Historical Note : ; h ‘ ; 421 Examples—Calendar, Belipses, oes a, : : : : 423 Exercises XXX. . : : . - : 424 CHAPTER XXXIII. ON RECURRING CONTINUED FRACTIONS. Every Simple Quadratic Surd Number convertible into a Recurring S.C.F. ; “ ; : , : 425-430 Recurrence-Formule for P, Stal \Jerte : ; : 4 426 Expressions for P, and Qn . ; : : 5 428 Cycies 01 Py, Qu, Gn. 2 , 429 Every Recurring 8.C.F. equal tb a Srnite Gnstietic Sard Nuwiber 430-432 On the S.C.F. oN represents 4/(C/D) : ‘ : 432-441 Acyclic Quotient of \/N/M . : : ; ; 434 Cycle of the Partial Quotients of /N/M ; : ; 435 Cycles of the Rational Dividends and Divisors of ./N/M ‘ . 436 Tests for the Middle of the Cycles . P s 439 Examples—Rapid Calculation of High Pres Re &G. ; 440 Exercises XX XI. : : ; 44] Applications to the Solution of Tian Probie: : 445-460 ax—by=c . : ‘ : i ; ; : 446 (et Oy = 6 | : é : : : 5 : 447 ax+byt+c=d, wutby+cz=d' . : : : , 449 Solutions of x? — Cy?=+H and 2?- Cy?=+1 ; : ; 450 General Solution of z?- Cy?=2+H when H./C : : 454 Remaining Cases of the Binomial Equation : : 458 General Equation of the Second Degree : é : ; 458 Exercises XX XII. ‘ : ; : : sie CO 461 CONTENTS CHAPTER XXXIV. GENERAL CONTINUED FRACTIONS. Fundamental Formule Meaning of G.C.F. . : G.C.FF. of First and Second Class Properties of the Convergents Continuants Continuant arene sane Continuant Functional Nature of a Continuant . Euler’s Construction Euler’s Continuant-Theorem : Henry Smith’s Proof of Fermat’s Theorem fiat a Erin: of ite fort 4X+1 is the Sum of Two Integral Squares Every Continuant reducible to a Simple Continuant _ C.F. in terms of Continuants Equivalent Continued Fractions Reduction of G.C.F. to a form having Unit N umer tore Exercises XX XIII. : Convergence of Infinite C.FF. . : Convergence, Divergence, Oscillation of C.F. Partial Criterion for C.F. of First Class Complete Criterion for C.F. of First Class Partial Criterion for C.F. of Second Class Incommensurability of certain C.FF. Legendre’s Propositions Conversion of Series and Continued Byoddts Hie C. F li - Euler’s Transformation of a Series into an equivalent C.F. Examples—Brouncker’ s Quadrature of the Circle, &c. C.F. equivalent to a given Continued Product Lambert’s Transformation of an Infinite Series into an Infi ite 0, Fr. Example—after Legendre C.FF. for tana and tanh x Incommensurability of w and e Gauss’s Conversion of the Hypergeometric Series Si aC. F. Exercises XXXIV. CHAPTER XXXV. GENERAL PROPERTIES OF INTEGRAL NUMBERS. Numbers which are congruent with respect to a given Modulus Modulus and Congruence Periodicity of Integers Examples of Properties deduced ean ered feityetinteprality, of a(e+1)...(%+p-1)/p!, Pythagorean Problem, &c. Broa! PAGE 463-466 464 464 464 466-47 4 466, 467 467 468 470 471 472 473 473 474 474 477 477 478 479 482 “484 486 484 486-496 486 488 489 489 492 494 495 495 497 500-506 500 501 50] Xxil . CONTENTS PAGE Property of an Integral Function . : : : a 504 Test for Divisibility of f(~) . : : : $ : 504 J(x) represents an Infinity of Composites . : : : 504 Difference Test of Divisibility : hal: iti ‘ ; 505 Exercises XX XY. ; 4 ee ee : ; 506 On the Divisors of a given Tnteper : : : ‘ 508-518 Limit for the Least Factor of N : ; $ ; ; 508 Sum and Number of the Divisors of a Composite . : . ;, 09 Examples—Perfect Number, &. . : : : 510 Number of Integers - = + pO GOsoy epee = p+qUs (6). If we divide p +q letters into two groups of p and gq respect- ively, the »4,C; s-combinations of the p+q letters may be classified exhaustively and simplexly as follows :— XXII VANDERMONDE’'S THEOREM 9 Ist. All the s-combinations of the p letters. The number of these is (Aer 2nd. All the combinations found by taking every one of the (s—1)-combinations of the p things with every one of the 1- combinations of the g things. The number of these is AO shee 3rd. All the combinations found by taking every one of the (s— 2)-combinations of the p things with every one of the 2- combinations of the g things. ‘The number of these is Wi ee oie ge And so on. ‘Thus the theorem follows. It should be noticed that Cor. 4 and Cor. 5 furnish proposi- tions in the summation of series. For example, we may write Cor. 5 thus— pp-1)... (p—s+1), pw). 9 Se: . ae Ss eee... (s— 1) 41 _P(p—3).. » @-st+3) ag-}) I ome (874) ae iP Wg-1) G84) LpPearp earner? (si) Ug=1).. . g-st)) di he peinee Aceh mag rg— ly) pt gosel) (7) | Pa lar eS It is obvious that (7) is an algebraical identity which could be proved by actually transforming the left-hand side into the right (see chap. v., § 16). If we take this view, it is clear that the only restriction upon 9, q, s is that s shall be a positive integer. Thus generalised, (7) becomes of importance in the establishment of the Binomial Theorem for fractional and negative indices. Cor. 6. If we multiply both sides of (7) by 1.2... 5, and denote p(p—1)... (p—s+1) by ps, we. deduce (Pe d)p=— 0s + sO, Ds. Gi FCs Dada + area Gp (8), which is often called Vandermonde’s theorem, although the result was known before Vandermonde’s day. 10 p LETTERS ALIKE CHAP. §$ 9.] To find the number of r-combinations of p+q letters p of which are alike. Ist. With the g unlike letters we can form Cr 7-combina- tions. 2nd. Taking one of the p letters, and 7-1 of the g, we can form ,C,_, 7-combinations. 3rd. Taking two of the p, and 7-2 of the g, we can form gOr-2 7-combinations ; and so on, till at last we take r of the p (supposing p>r); an form one 7-combination. We thus find for the number required OSM ES oc Ung) * GIrls FTG al Cor. The number of r Pe of p+q things p of which are alike is gir} her) 4H Ree ke oY! A Bes 4. “trl -r)! UWr-I)q-rt+ 1)! 2'r- 2)Mq—7r + 2)! ae ee (r—1)!1(q-1)! rig!” For, with the ,C, combinations of the 1st class above we can form gOr 7! permutations ; With the ,C,_, combinations of the 2nd class, 7C,_, 7! permu- tations ; With the ,C,-, combinations of the 3rd class (in each of which two letters are alike), ,C,_, 7!/2! permutations: and so on. Hence the whole number of permutations is gor?! hgQCr-ari/li+ oC,7!/21 +... pC ynl ee whence the result follows. A similar process will give the number of 7-combinations, or of 7-permutations, wheh we have more than one group of like letters ; but the general formula is very complicated. §$ 10.] The number of r-combinations of n letters (,H,), when each letter may be repeated any number of times up to 1, is mn+1)(n+2)... (ntr—1)/1.2.3... 297 (0). XXIII COMBINATIONS WITH REPETITION i In the first place, we remark that the number of (7+ 1)-com- binations, in each of which the letter a, occurs at least once, is the same as the number of 7-combinations not subject to this restriction. This is obvious if we reflect that every (7+ 1)- combination of the kind described leaves an r-combination when a, is removed, and, conversely, every 7-combination of the n letters gives, when a, is added to it, an (7 + 1)-combination of the kind described. It follows, then, that if we add to each of the 7-combina- tions of the theorem all the n letters, we get all the (n + 7)-com- binations of the n letters, in each of which each letter appears at least once, and not more than 7+1 times. We may therefore enumerate the latter instead of the former. This new problem may be reduced to a question of permuta- tions as follows. Instead of writing down all the repeated letters, we may write down each letter once, and write after it the letter s (initial of same) as often as the letter is repeated. Thus, we write asssbsscs . . . instead of aaaabbbcc . .. With this notation there will occur in each of the (n +7)-combinations the n letters a,, a, . . ., @ along with r s’s. The problem now is to find in how many ways we can arrange these n + 7 letters. It must be remembered that there is no meaning in the occur- rence of s at the beginning of the series ; hence, since the order of the letters a,, a,, . . ., d is indifferent, we may fix a, in the first place. We have now to consider the different arrange- ments of the n — 1 letters a,, a,, . . ., d, along with r s’s._ In so doing we must observe that nothing depends on the order of Us, Uz, » » +» Gn imler se; so that in counting the permutations they must be regarded as all alike. We have, therefore, to find | the number of permutations of n-1+7 things, n-1 of which are alike, and r of which are alike. Hence we have ae Ue ls (n—1)!7! | (2) ii Vere (keh) =~ 7 Be cue at 12 THEOREMS REGARDING ,,H, CHAP, Corset Help = pepe This follows at once from (2). Cor. 2. nuly =o ee For the r-combinations consist, lst, of those in which a, occurs at least once, the number of which we have seen to be ,H,-, ; 2nd, of those in which a, does not occur at all, the number of nici ies ey clas Cor. 3. ,Hy =azjHe tae Heist pe illpee This follows from the consideration that we may classify the 7-combinations into Ist. Those in which a, does not occur at all, ,_,H, m number ; | 2nd. Those in which a, occurs once, ,-,H,_, in number ; 3rd. Those in which a, occurs twice, ,-,H,_-, in number: and so on. Cor. 4. The number of different r-ary products that can be made with n different letters is n(nt+1)... (mt+r—-1)/1.2...7; and the number of terms in a complete integral function of the rth degree in n variables is (n+1)(n+2)... (m+r)/1.2...7%. The first part of the corollary is of course obvious. The second follows from the consideration that the complete in- tegral function is the sum of all possible terms of the degrees 0,1, 2, .. ., 7 respectively. Hence the number of its terms is 1 +H, +e ee But, by Cor. 3, this sum is ,,,H,. We have thus obtained a general solution of the problems suggested in chap. iv., §§ 17, 19. As a verification, if we put n=2, we have for the number of terms in the general integral function of the rth degree in two ‘ variables 3.4... (r+2)/1.2... 7, which reduces to (r+1)(r+2)/2, in agreement with our former result. EXERcIsEs I. Combinations and Permutations. (1.) How many different numbers can be made with the digits 11122333450 ? (2.) How many different permutations can be made of the letters of the sentence Ut tensio sic vis ? XXIII EXERCISES I 13 (3.) How many different numbers of 4 digits can be formed with 0123456? (4.) How many odd numbers can be formed with the digits 3694 ? (5.) If SUnatlan=a0n= 132/35, find n. (6. ) If m= ,Co, show that noon Ce. (7.) In any set of n letters, if the number of 7-permutations which con- tain @ be equal to the number of those that do not contain a, prove that the same holds of v-combinations. (8.) In how many ways can the major pieces of a set of chess-men be arranged in a line on the board ? If the pawns be included, in how many ways can the pieces be arranged in two lines ? (9.) Out of 13 men, in how many ways may a guard of 6 be formed in line, the order of the men to be attended to? (10.) In how many ways can 12 men be selected out of 17—1st, if there be no restriction on the choice; 2nd, if 2 particular men be always included ; 3rd, if 2 particular men never be chosen together ? (11.) In how many ways can a bracelet be made by stringing together 5 like pearls, 6 like rubies, and 7 like diamonds ? How many different settings of 3 stones for a ring could be selected from the above ? What modification of the solution of the first seh of the above problem is necessary when two, or all three, of the given numbers are even ? (12.) In how many ways can an eight-oared boat be manned out of 31 men, 10 of whom can row on the stroke side only, 12 on the bow side only, and the rest on either side ? (13.) In a regiment there are 10 captains, 20 lieutenants, 30 sergeants, and 60 corporals. In how many ways can a party be selected, consisting of 2 captains, 5 lieutenants, 10 sergeants, and 20 corporals ? (14.) Three persons have 4 coats, 5 vests, and 6 hats between them; in how many different ways can they dress ? (15.) A man has 12 relations, 7 ladies and 5 gentlemen ; his wife has 12 relations, 5 ladies and 7 gentlemen. In how many ways can they invite a _ dinner party of 6 ladies and 6 gentlemen so that there may be 6 of the man’s relations and 6 of the wife’s ? (16.) In how many ways can 7 ladies and 7 gentlemen be seated at a round table so that no 2 ladies sit together ? (17.) At a dinner-table the host and hostess sit opposite each other. In how many ways can 2n guests be arranged so that 2 particular guests do not sit together ? (18.) In how many ways can a team of 6 horses be selected out of a stud of 16, so that there shall always be 3 out of the 6 ABCA’B’C’, but never AA’, BB’, or CC’ together ? (19.) With 9 consonants and 7 vowels, how many words can be made, each containing 4 consonants and 3 vowels—lst, when there is no restriction on the arrangement of the letters; 2nd, when two consonants are never allowed to come together ? (20.) In how many ways can 52 cards, all different, be dealt into 4 equal 14 BINOMIAL THEOREM . CHAP, hands, the order of the hands, but not of the cards in the hands, to be attended to ? _ In how many cases will 18 particular cards fall in one hand ? (21.) In how many ways can a set of 12 black and 12 white draught-men be placed on the black squares of a draught-board ? (22.) In how many ways can a set of chess-men be placed on a chess-board ? (23.) How many 8-combinations and how many 8-permutations can be made with the letters of parabola? (24.) With an unlimited number of red, white, blue, and black balls at disposal, in how many ways can a bagful of 10 be selected ? In how many of these selections will all the colours be represented ? (25.) In an election under the cumulative system there were p candidates for q seats ; (1) in how many ways can an elector give his votes; (2) if there be v voters, how many different states of the poll are there ? If there be 15 candidates and 10 seats, and a voter give one minute to the consideration of each way of giving his vote, how long would it take him to make up his mind how to vote? BINOMIAL AND MULTINOMIAL THEOREMS. § 11.] It has already been shown, in chap. iv., § 11, that (a+ b)® =a" + ,C,a"-10 4... Cp O90 ee ee where »C,, nCe, - + -> Cy - . - denote the numbers of I-, 2-, . .,7-combinations of » things. Using the expressions just found for ,C,, ,C., &¢., we now have -1 (a+ dr =at + nar-¥ 4 OD gmap ‘i n(n—1)...(m-r+1),_2. + iO Valet: dad Cale site ST (1). This is the Binomial Theorem as Newton discovered it, proved, of course, as yet for positive integral indices only. § 12.] We may establish the Binomial Theorem by a some- what different process of reasoning, which has the advantage of being applicable to the expansion of an integral power of any multinomial. Consider (@, + dg +. . «+ Gm)” (2). We have to distribute the product of n factors, namely, — (hy + ig +... + Amn) (dy + Oy +o oe + im) + (01 + Mg +. + + Om) (3). XXIII MULTINOMIAL THEOREM 15 and the problem is to find the coefficient of any given term, say Gia. ee dt cee Ona (4), where of course a, +a, ... +a ,=%. In other words, we have to find how often the partial product (4) occurs in the distribu- tion of (3). We may write out (4) in a variety of ways, such as Clr 0, 0,0 ae (5), there being always a, @,’s, %, a,’s, &e. Written as in (5) we may regard the partial product as formed by taking a, from the lst and 2nd brackets in (3); a, from the 3rd, 4th, and 5th; a, from the 6th; and so on. It appears, therefore, that the partial product (4) will occur just as often as we can make different permutations of the n letters, such as (5). Now, since a, of the letters are all alike, a, all alike, &c., the number of different permutations is, by § 6, m!/a,!a,.!.. . am!. Hence we have | (G+ 0.+...+4,)"= Seer eet 7 a Peden oc Canton Opes O)-Ag: «2 » Am: wherein a, 0, . . . &» assume all positive integral values con- sistent with the relation Opt Og iets < tani Ib (7). This is the Multinonial Theorem for a positive integral index. The Binomial Theorem is merely the particular case where m=2. We then have, since a, +a,=n, and therefore a,="—-a,, n! a,!(v—a,))"" - stn — Ly alr le —a,+ 1) a aig, a,! ay a4 : (a, + A)" = 2 which agrees with (1). Cor. Yo find the coefficient of a” in the. expansion of POPU tes.) 40, ate te (8) we have simply to pick out all the terms which contain 2”. The general term is n! tect 0,4 ec ee Dy tater zest + (m-1)Om, Qa) Ag. . ° . Am: 16 | EXAMPLES CHAP. Hence we have to take all the terms which are such that dg+2a,+... +(m—1)ay=7 (9): The coefficient of 2” in the expansion of (8) is therefore ype i Ps Rn (10), ae ent 0; ag eee where a, a2; . - -, Gm have all positive integral values subject to the restrictions (7) and (9). Example 1. The coefficient of ab? in the expansion of (a+b+c+d)’ is 5! 3121010! 1° Example 2. To find the coefficient of # in (1+ 2x+.2?)*. Here we must have aj +a,+a3=4, a2+2a3=5. Hence a=a3—1, ag=5—2az. Since a; and ag must both be positive, the only two admissible values of ag are land 2. We have therefore the following table of values :— ay a2 | a3 0 abe aclemeert 1 1 2, The required coefficient is therefore 4! 4! on trai | The correctness of the result may be easily verified in the present case for (1+2x+2?)*=(1+42)8, the coefficient of x in which is gC;=56. Example 3. To find the greatest coefficient, or coefficients, in the expansion of (ay +det+. . . +Mm)”. . This amounts to determining a, y, 2, .. . sothatn!/a!y!z!... shall be a maximum, where e+y+z2+...=%n. This, again, amounts to deter- mining x, y, 2, . . . so that UW=Ligiel eee (1) 1121 ee shall be a minimum, subject to the condition Ct+yte+... =n (2). Let us first consider the case where there are only two variables, x and y. We obtain all possible values of z!y! by giving y successively the values 0, 1,2, ..., ”, « taking in consequence the values n, n—1,m-2,..., 0. The consecutive value to w!y! is (2z—1)!(y4-1)!, and the ratio of the latter to the former is (y+1)/x; that is (since x+y=n), (n+1-~2)/x, that is, XXIII MAXIMUM COEFFICIENT 1g (n+1)/e-1. This ratio is less than unity so long as (n+1)/e<2, that is, So long as x>(n+1)/2. Until x falls below this value the terms in the series above mentioned will decrease ; and after falls below this limit they will begin to increase. Ifm be odd, =2h+1 say, then (n+1)/2=4+1. Hence, if we makex=k+1, the ratio (n+1)/a—1=1, and two consecutive values of x ly!, viz. (K+1)!k! and &!(%+1)!, are equal and less than any of the others. If m be even, = 2k say, then (n+1)/2=k+4. Hence, if we make x=h, we obtain a single term of the series, viz. k!k!, which is less than any of the others. , Returning now to the general case, we see that, if « be a minimum for all values of x, y, 2, . . . subject to the restriction (2), it will also be a minimum for values such that x and y alone are variable, z, . . . being all constant. In other words, the values of x and y for which xw!y!z!... is a minimum must be such as render #!y! a minimum. Hence, by what has just been proved, « and y must either be equal or differ only by unity. The like follows for every pair of the variables x, y,z, ... Let us therefore suppose that p of these are each equal to £; then the remaining m-—p must each be equal to +1. Further, let g be the quotient and 7 the remainder when 7 is divided by m; so that n=mq+7. We thus have pét+(m—p)(—E+1)=mqtr. Hence mé+(m—p)=mg+r ; so that E+(m—p)/m=q+r]/m. Now (m-p)/m and 7/m are proper fractions ; hence we must have f=q, mM-—p=r. It follows, therefore, that 7 of the variables are each equal to g+1, and the rest are each equal to g. The maximum coefficient is therefore mi(gym—"i(q-+1) 1}; that is, to n'i(q!)"(q+1)" (8). This coefficient is, of course, common to all terms of the type mag... ieee Pepe tS. So Oy Tt}, As a special case, consider (a; +a2+a3)*. Here 4=3x1+4+1; g=1, r=1. Hence the terms that have the greatest coefficient are those of the type ayaa”, and the coefficient in question is 4!/(1!)°2'=12. This is right; for we find by distributing that ; (a +de+ a3)4 =Dayt+ 43 a;2aq+ 6Day2ae? + 122a;7a203. Example 4. Show that 1-2 1+e% n(n—-1) 142% n(n-1)(n-2) 1+8¢ ai Mira 1.2) (liane 1.2.8 anepo ( Wolstenholme.) The left-hand side may be written nm 1 n(n —1) i n(n —1)(n—- 2) 1 si “I itnat 122) ena 2 8 ney uw mMn-1) mn-1)(n-2) 82 f Biipar 1.2 (+n 1.2.30 dene 8%: VOL. II C 18 PROPERTIES OF ,,C, CHAP. Pt» ¥ n(n —1) 1 _ n(n —1) (n= 2) 1 ‘ Lil+ne" 1.2 (lena)? 1.2,8)) 10 442) 9) ae na {1 (n=1) 1 (n-1)(n—-2) 1 \ aad ee Soe een, . ° . 5 ~ L+nx ¥ =2) T (Gtne)' = ir2 ” (ener oT —1 Sh er realik aoe eeeee a l+naz 1+nx L+nzx J _f ne \" _ ne NL pr 7 Ul+ne l+nell+nzx ? 2 Nx \ { ne \" > AL 4+ na 1l+naeJ ’ ==(), § 13.] The Binomial Theorem can be used in its turn to establish identities in the theory of combinations; as the two following examples will show :— Example 1. We have 1=(1+e-2) =(1+a)"— 0, 2(1+2)r14,Coxr(1+e)"4-... (- WC a. On the right-hand side of this identity the coefficient of every power of must vanish. Hence, s being any positive integer less than 7, we have Cs Xie r-10.-1 x rC1 D5 r—20s—2 x C2 9s ait ( ix ate 0 x BOP 4s ( = ),C,= 0. Example 2. To find the sum of the squares of the binomial coefficients. We have (1+a)*"=(1+a)" x (w+1)" =(1+nCivt+nCov? +... + Cpa”) X (20% + Ope} + Cor —2 +. 2 e+ wear If we imagine the product on the right to be distributed, we see that the coefficient of x is 12+ ,Ci2+,Co?+ . - - +nCn?; the coefficient of a” on the left is onCy. Hence 124+ ,0r2 + nO2+ . . » +nOn?=onCn=2n!/n!n!. Since Qn !=2n(2n—-1)(2n—2)... 4.38.2.1=2%1.2 os mx1.3... (2n-1), we have 12+,Ci2+,0e2+ .. - tnCn?=2".1.3... (2n—-1)/n!. A great variety of results can be obtained by the above process of equating coefficients in identities derived from the binomial theorem ; some specimens are given among the exercises below. ExERrcIssEs II. (1.) Find the third term in the expansion of (2 +82)”. (2.) Find the coefficient of x” in the expansion of (1+a+27)(1-2)”. (3.) Find the term which is independent of in the expansion of (a+ 1/x)", XXIII EXERCISES II 19 '(4.) Find the coefficient of x?" in the expansion of (a —- 1/x)2”, (5.) Find the ratio of the coefficients of x2” in (1+a)@ and (1+a)%, (6.) Find the middle term in the expansion of (2+ 42)", (7.) The product of the coefficients in (1+2)"+1: the product of the co- efficients in (1+a)"=(n+1)":n!. (8.) The coefficient of @” in {(7-2)a?+na — r\(a+1)" is 2,Cp—9. (9.) If I denote the integral part and F the proper fractional part of (3+4/5)", and if p denote the rational part and o the irrational part of the same, show that T= 218" + 028%, 5+ nCy8"-4, 524...) 1, F=1 ho /5)", p=4(1+), o=43(1+2F-1). (10.) If (\/2+1)"H=1+F, where Fisa positive proper fraction and I is integral, show that F(I+F)=1. 5, (11.) Find the integral parts of (24/3+3)2”, and of (24/3 + 3)2m+1, (12.) Show that the greatest term in the expansion of (a+2)" ig the (7+1)th, where 7 is the integral part of (n+ 1)/(a/z+1). Exemplify with (2+3) and with (2+4), (13.) Find the condition that the greatest term in (a+) shall have the greatest coefficient. Find the limits for x in order that this may be so in ei 7), (14.) If the pth term be the greatest in (4+a)™, and the gth the greatest in (a+2)”", then either the (p+q)th or the (p+q-1)th is the greatest in (a+ax)mtm, (15.) ‘Sum the series nV nC2 nC3 ar Shi Ag ESE ay gerN ae a 5 oat (16.) Sum the series Hepes Ort GeO, Paka a (17.) If p, denote the coefficient of a” in (1 +)", prove the following rela- tions :-— 1°. 71-224 8p3-. . -+2( —1)"-1y,=0, ° (-1)»-1 n- Aeed Ti GPa et, 2° Fe eel ee We ! Pi, Pe Dn eae) 3. 1t+o+3 cha ate 2 Le SRY 7 ras , (18.) If p, have the same meaning as in last question, show that ro +4 fee estoy eee | M-sMt+tps—.. . nm wn=ltgtat. .. a: (19.) Show that grey pa s-1 X C1 +2059 X Co +. +e by—sp1 Cr X Cy +1 x C= 0,28. (20.) Show that (L—nCot+nCy—.. .)?+ (nCi-—nCst+. . poate Ce Hen Garkik “> 20 EXERCISES II CHAP. (21.) Show that 1x pCotnCiXnCst. + «+nCn-2X nCn=(2n)!/(m+2)!(n—2)Ke = 2 e att 2 (22.) Show that 1-8 + (“4=2)) ~ (Menge) +... S0if n be odd, and =(-1)"?(n+2)(n+4)... 2n/2.4... nif n be even. (23.) Show that n(n +1) n(n +1) (n+ 2) 2! 3! +... .=2(2n+1)!/(n+2)!(n-1)!. L.n(n+1)+7(n-1)n+ (n —2)(n—1)+ (n — 3) (n-2) (24.) If w, stand for x +1/x", show that Uta + pCi Up—a + rp1C2 Urge. - = Ua(Urt pO Up—g + pCo Upnats « + )s (25.) If a, denote the coefficient of #? in (1 + )2"-7)(1 — x)", show that iy — nC Gi +nCed2—-. . . =0 for all values of p except p=n, in which case the right-hand side of the equation is 4”. (26.) Show that a ot nC2 ( aa La v7 ! ira + =. hee eee OE hn OLE T Gan Pas es ¢+1 2+2 x+n tal): 2 Gee (27.) Find the coefficient of a in(l+a+a°+...) (28.) Find the coefficient of a'$ in (1+2°+a%+a)*. (29.) Find the coefficient of a in (l+a+20°+3a%+.. .)?, (30.) If ao, a1, - + +, Gon be the coefficients of the powers of x in (1+20+2a?)", show that dod@en—- ident. . - +Aento = 0 if n be odd, =2"n!/{(4n)!\? if m be even. (31.) If a, be the coefficient of x in (1 +a+a2+. ..+a?)", show that Ay — nO bp—1 + nCoGr-g—-'. « - = 0, unless 2 be a multiple of y+1. What does the equation become in the latter case ? (32.) Find the coefficient of a in (1+ 2a+ 3a? + 42%)”, (33.) Write out the expansion of (a+b+c+d)’. (34.) Show that 5 12 vie OT nad) 1 plein ee eee 2 “ where 7, s, . . -, & have all values between 0 and p, both inclusive, subject to the restriction r+s+...+k=p. (35.) If ,»H, have the meaning of § 10 above, prove that 1. m4nHy=mHy + mH y-1 X nH + mH y-2 X nHo t+. ~ « +mHi X ny. 2°) 1-101. X nHitnCe XnHe-nC3XnHst-. - .+(—-1)nCr nbn =0. | (36.) Ifa,=2(a@+1)... (e+r-—1), show that (20 + Y)p= Ly Oy Cp—1 Yt + Co Vyp2Yat. » -+Yrn 3 (37.) Find the largest coefficient in the expansion of (a+b+ce+d+e)*. XXIII LAW OF DISTRIBUTION USED a EXAMPLES OF THE APPLICATION OF THE LAW OF DISTRIBUTION. § 14.] Lf we have r sets, consisting of my, ny .. ., Ny different letters respectively, the whole number of different ways of making com- binations by taking 1, 2, 3, .. . up tor of the letters at a time, but never more than one from each set, is (m4 +1)(m,+1)... (n,-+1)-1. Consider the product (l+a,+0,+. . ., letters) x(1+a,+0,+ . . .%, letters) x(1+a,+6,+ . . .m, letters). In the distributed product there will occur every possible com- bination of the letters taken 1, 2, 3, . . ., 7 at a time, with the term 1 in addition. If we replace each letter by unity, each term in the distributed product will become unity, and the sum of these terms will exceed the whdle number of combinations by unity. Hence the number required is (l+)(1+n)...(1L+m,)-1 = 2, + 2N,Ne +. . e+ MMe sss Ny This result might have been obtained by repeated use of § 7. § 15.] Lf we have r sets of counters, marked with the following numbers— O15 B,, siunrese INi5 Oo, Ps, s+ 49 Kay Any Bry soe ty Kyy the number of counters not being necessarily the same for each set, and the inscribed numbers not necessarily all different, then the number of different ways in which r counters can be drawn, one from each set, so that the sum of the inscribed numbers shall be n, is the coefficient of x” m the distribution of the product 22 DISTRIBUTION PROBLEM CHAP. (a 4+ oPh +... + aX) x (2 + oP +, yh Hot?) x (a + oPr4+ yy. . + ake), This theorem is an obvious result of the principles laid down in chap. iv. | Cor. 1. Jf in the first set there be a; counters marked with the number a,, b, marked with B,, &c., in the second a, marked with as, b, marked with B,, &c., the number of ways in which r counters can be drawn so that the sum of the numbers on them is n, is the coefficient of x” in the distribution of (a, 0% + bh hee) x (4,07? + bt te, A ak) x (au + barr 4... Eke). Cor. 2. In a box there are a counters marked a, b marked B, ke. A counter is drawn r times, and each time replaced. The number of ways in which the sum of the drawings can amount to n is the co- efficient of a” in the distribution of (ax* + ba +...) DISTRIBUTIONS AND DERANGEMENTS. § 16.] The variety of problems that arise in connection with the subject of the present chapter is endless, and it would be difficult within the limits of a text-book to indicate all the methods that have been used in solving such of these problems as mathematicians have already discussed. The following have been selected as types of problems which are not, very readily at least, reducible to the elementary cases above discussed.* $17.] To find the number of ways in which n different letters can be distributed among r pigeon-holes, attention being paid to the order of the pigeon-holes, but not to the order of the letters im any one pigeon-hole, and no hole to contain less than one letter. Let D, denote the number in question. * For further information see Whitworth’s Choice and Chance. XXIII DISTRIBUTION PROBLEM 23 If we leave s specified holes vacant and distribute the letters among the remaining 7—s holes under the conditions of the question, we should thus get D,_, distributions. Hence, if ,C, have its usual meaning, the number of distributions when s of the holes are blank is ,C, D,._;. | Again, the whole number of distributions when none, one, two, &c., of the holes may be blank is evidently 7”, for we can distribute the n letters separately among the r holes in 7” ways. Hence ee ee CL te he Get Cnn iD part ar CA}. The equation (A) contains the solution of our problem, for, by putting r= 2, r= 3, &c., successively, we could calculate D,, D,, &e., and D, is known, being simply 1. We can, however, deduce an expression for D, in terms of n and 7, as follows. Writing 7-1 in place of r we have Dey a, poate Bee ee Ol els rel , = (B): From (A) and (B), by subtraction, remembering (§ 8, Cor. 3) that Cg — paiCg-1 as ris, we derive ye oe cosh Op 1D Pek aE oak OSs De 2S Soe 5 rary D, =" — (r — 1)” (1). From (1), putting r - 1 in place of r, we derive 1B a ce AAs DP. absent ats RAL ey D; ocean — 2)" (1’). From (1) and (1’), by subtraction, we derive | D, Ly “216s 19 a5 vases Des Ae ig Bal (dak, peers IBS =r" — 2(r — 1)" + (r — 2)" (2). Treating now (2) exactly as we treated (1) we derive D; ny pas Das ay AeKOE Ds5 eed Sia AD D, = 7" — 3(r — 1)" + 3(r — 2)" — (r — 3)” (3). The law of formation of the right-hand side is obvious, the coefficients being formed by the addition rule peculiar to the binomial coefficients (see chap. iv., § 14). We shall therefore finally obtain | 24 DERANGEMENTS CHAP. D,= 1" — ,0,(r — 1)" + ,0,(r — 2)" -—. 2. (-—)?4,C,-11”, — 9n — ir 1)" + a — oe bls 2)" aS ( “I race (4). 2) realy rf the order of the pigeon-holes be indifferent, the number of distributions is D,/r!. In other words, the number of partitions of n different letters into r lots, no vacant lots being allowed, is D,/r!. We shall discuss the closely-allied problem to find the number of r-partitions of n—that is, to find the number of ways in which 1 letters, all alike, may be distributed among r pigeon-holes, the order of the holes being indifferent, and no hole to be empty—when we take up the Theory of the Partition of Numbers. § 18.] Given a series of n letters, to find in how many ways the order may be deranged so that no one out of + assigned letters shall occupy its original position. Let ,A, denote the number in question. The number of different derangements in which the 7 assigned letters do all occupy their original places is (n—7)!. Hence the number of derangements in which the r assigned letters do not all occupy their original places is n!-(n-r)!. Now, this last number is made up of— lst. The number of derangements in which no one of the r letters occupies its original place ; that is, ,A,. 2nd. The number of derangements in which any one of the r letters occupies its original place, and no one of the remaining r—1 does so; that is, ,C,’,-,A,-1. 3rd. The number of derangements in which any two of the r letters occupy their original places, and no one of the remaining 7 — 2 does so; that is, ® agp peas eon Hence Wi — (n— 1) = yA oO; noieoy Fore neaorseet ee ee 1 pr n-r+1O1 (A). If we write in this equation n—1 for n, and r—1 for 7, and subtract the new equation thus derived from (A), we deduce mi — (nm — 1) $= Ap t+ p10) nepAp-a + ri Gon aOe-s +. = 72 prey n-r+1Oi (1). XXIII SUBFACTORIAL 7” 25 We can now treat this equation exactly as we treated equa- tion (1) of § 16. We thus deduce fA, =n! =F (n- 1)! ge 2)!-...(-)(m—-r)! (2). If we remember that (n-—1)!, above, stands for the number of derangements in which the 7 letters all occupy their original positions, we see that, when r =n, (n—7)! must be replaced by 1. Hence Cor. The number of derangements of a series of n letters in which no one of the original n occupies its original position is ni{l-ptge HSL (3). The expression (3) may be written m.. .{4(8(211 -1)+1)-1)+1).. .-(-1)) +(-1)% Hence it may be formed as follows :—Set down 1, subtract 1; multiply by 2 and add 1; multiply by 3 and subtract 1; and soon. The function thus formed is of considerable importance in the present branch of mathematics, and has been called by Whitworth swbfactorial n. He denotes it by ||n. A more con- venient notation would be nj. SUBSTITUTIONS. § 19.] Hitherto we have merely counted the permutations of a group of letters. If we direct our attention to the actual per- mutations, and in particular to the process by which these per- mutations are derived from each other, we are led to an order of ideas which forms the foundation of that important branch of modern algebra which is called the Theory of Substitutions. ’ Consider any two permutations, becda, beade, of the five letters a, b,c, d,¢ The latter is derived from the former by replacing a by ¢, b by b, ¢ by a, d by d, e by c. This process. may be represented by the operator (ee) ; and we may write (Fee) becda = beade : 26 THE SUBSTITUTION OPERATOR CHAP. ° or, omitting the letters that are unaltered, and thus reducing the - operator to its simplest form, (ea becda = beade. ace The operator (! and the operation which it effects, are called a Substitution; and the operator is often denoted by a single capital letter, 8, T, &c. Since the number of different permutations of a group of n letters is n!, it is obvious that the number of different substitu- tions is also !, if we include among them the identical substi- tution (ey (denoted by 8’ or by 1), in which no letter is altered. We may effect two substitutions in succession upon the same permutation, and represent the result by writing the two symbols representing the substitutions before the permutation in order from right to left. Thus, if S= Gah T= Ugh STaebed = ecabd. We may also effect the same substitution twice or three times over, and denote SS by 8’, SSS by 8S’, &. Thus, S being as before, S*aebed = Sceabd = becad. It should be observed that the multiplication of substitution symbols is not in general commutative. For example, S and T being as above, STaebcd = ecabd, but TSaebcd =caebd. If, when reduced to their simplest form, the symbols S and T have no letter in common, they are obviously commutative. This condi- tion, although sufficient, is not necessary ; for we have dcab\ /badc badc\ (deab (3) es abode = cdbae = ay (owe abede. § 20.] Since the number of permutations of n letters is limited, it is obvious that if we repeat the same substitution, §, sufficiently often we shall ultimately reproduce the permutation that we started with. The smallest number, », of repetitions for which this happens is called the order of the substitution S. XXIII ORDER AND GROUP 27 Hence we have S“=1, and S?#=1, where p is any positive integer, : We may define a negative index in the theory of substitu- tions by means of the equation SY = S?!-7, p being the order of S, and p such that pu>q. From this definition we see that SIS 4 = SISP4-7 = SP" = 1. In other words, 8% and S~ are inverse to each other ; in particular, if dabe abed beda tes (vied) ee = (Gate) ~ (ced): A set of substitutions which are such that the product of any number of them is always one of the set is called a group ; and the number of distinct substitutions in the group is called the order of the group. The number of letters operated on is called the degree of the group. It is obvious from what has been shown that all the powers of a single substitution, S, form a group whose order is the order of 8. § 21.] A substitution such as ae ih where each letter abcdef. is replaced by the one that follows it, and the last by the first, is called a Cyclic Substitution, and is usually denoted by the symbol (abedef).* The cyclic substitution (a), consisting of one letter, is an identical substitution ; it may be held to mean that a passes into itself, The cyclic substitution of two letters (ab), or what is the same thing (0a), is spoken of as a Transposition. The effect of a cyclic substitution may be represented by writing the n letters at equal intervals round the circumference of a circle, and shifting each through 1/nth of the circumference. Thus, or otherwise, it is obvious that the order of a cyclic sub- stitution is equal to the number of the letters which it involves. § 22.] Every substitution either is cyclic or is the product of a number of independent cyclic substitutions (cycles). Consider, for example, the substitution * Or, of course, by (dcdefa), (cdefab), &c. 28 CYCLES CHAP. alone appears. The modifications necessary when the other sign appears are in all cases obvious. I. If P>Q, Q>R, R> S, then P>S. Proof —(P — Q) + (Q-R)+(R- S) =P —S,hence,since P — Q, | Q-R, R-S are all positive, P —S is positive, that is, P>S. IL If P>Q, then PER>QER For (P + R) -(Q+R)=P-Q; hence the sign of the former quantity is the same as the sign of the latter. Cor. 1. If P+Q>R+S, then ‘P4Q-Rs8, -R28>—-P-Q =Po Osa It thus appears that we may transfer a term from one side of an inequality to another, provided we change its sign; and we may change the signs of all the terms on both sides of an imequality, pro- vided we reverse the symbol of inequality. Cor. 2. Every inequality may be reduced to one or other of the forms P>0 or P<0. In other words, every problem of inequality may be reduced to the determination of the sign of a certain quantity. Ill. If P,>Q, PrSQ, - - +» Pn>Qns then P+ Ppt... +Pn = Qt Qt. + - Qn; for (P,+P, +... +Pn)-(Q,+Q+ ~~ - +Qn) » =(P, — Q,) + (P; — Q:) Oe ry cae + (Pa —- Qn); whence the theorem follows. It should be noticed that it does not follow that, Tht ees, P, > Q,, then P, - P.> Q, — Q,. * See, for example, the proof that every equation has a root. XXIV ELEMENTARY THEOREMS OW IV. If P>Q, then PR>QR, and P/R>Q/R, provided R. be positive ; but PRQR, and R>S8, then P>QS, provided Q be positwe. Cor 2. Every fractional inequality can be integralised. For example, if P/Q>R/S, then, provided QS be positive, we have, after multiplying by QS, PS> QR; but, if QS be nega- tive, PS < QR. If there be any doubt about the sign of QS, then we may multiply by Q’S’, which is certainly positive, and we have QPS’ > Q’RS. V. Tf P,>Q,, P2>Q,, . . ., Pu >Qn, and all the quantities be positive, then EC bahar: Pa> CC Fe atm Qn For ete ee ete Pak oc Pe since P,>Q, and P,P, . . . P, is positive ; OLN HA a i ey ieee since P,>Q, and Q,P, . . . P,, is positive; and so on. Hence, finally, we have lam eerste ma). ()o ue Ge Cor. 1. Jf P>Q, and both be positive, then P > Q”, n being any positive integer. Cor. 2. If P>Q, and both be positive, then PU > QU n being any positive integer, and the real positive value of the nth root being ‘taken on both sides. For, if P¥"2Q'”, then, since both are real and_ positive, (PUn)nz(QU”)”, by Cor. 1; that is, PZQ, which contradicts our hypothesis. Cor. 3. If P>Q, both being positive, and n be any positive quan- tity, then P-™ —3, and —3> —A4, yet it is not true that (— 2)(-—3)>(-3)(- 4). These restrictions might be removed in certain cases ; for example, it follows from —3> —4 that (-3)’>(-— 4), in other words, that — 27> —64: but the importance of such particular cases does not justify their statement at length. Cor. 4. An inequality may be rationalised if due attention be paid to the above-mentioned restrictions regarding sign. § 4.] By means of the theorems just stated and the help of the fundamental principle that the product of two real quantities is positive or negative according as these quantities have the same or opposite sign, and, in particular, that the square of any real quantity is positive, we can solve a great many questions regarding inequalities. The following are some examples of the direct investigation of inequalities ; the first four are chosen to illustrate the paral- lelism and mutual connection between inequalities and equa- tions :— Example 1. Under what circumstances is (8a —1)/(a—2)+(2x-3)/(e@—5)> or < 5? 1st. Let us suppose that # does not lie between 2 and 5, and is not equal to either of these values. Then (x — 2) (x—5) is positive, and we may a by this factor an reversing the signs of inequality. Hence = (8a —1)/(~@— 2) + (2a -3)/(x-5)> <5, according as (8a —1)(a—5)+(2xe- 38) (@—2)> <5(w— 2) (x- 5), according as 5a? — 28a -+11> <5x*- 385x450, according as 124 > <89, according as e> <3}. Under our present supposition, « cannot have the value 3}; but we con- clude from the above that if e>5, F>5, and if <2, F<5. Qnd. Suppose 25, and if 34<2<5, then F <1? Multiplying by the positive quantity (x - 2), we have (3% —4)/(«—2)> <1, according as (8a — 4) (w-2)> <(%—- 2°, according as {(8a — 4) —(a—2)}(a%-2)> <0, according as 2(2—1) (w-2)> <0. Hence Resist gore 2 s eel base a Example 3. Under what circumstances is a+ 25a> <8x?+ 26 ? a+ 25> < 8x? + 26, according as 2° — 8x7 + 25% -26> <0, according as (x — 2) (a? -6x%+13)> <0, according as (2-2) {(w-3)?+4}> <0. Now («-—3)?+4 is positive for all real values of x; hence ve +25a> <8x?+4+ 26, according as Gon 8, Example 4. If the positive values of the square roots be taken in all cases, is A/(2e@+1)+a/(@-1)> <8z, according as 2/{(2e+1)(e#-1)} > <0. Now, provided w is such that the value of a/{(2%+1)(#-1)} is real, that is, provided «>1, 2a/{(2e+1)(w@-1)} =0, therefore A/(2a+1)+4/(e@-1)>/(82), if w>1.. Negative values of x less than —4 would also make «/{(2%+1)(x-1)} real ; but such values would make 4/(2%+1), ,/(w-1), and 4/(3a) imaginary, and, in that case, the original inequality would be meaningless. Example 5. Ifa, y, 2... bem real quantities (n — 1)? + 22axy. Since all the quantities are real, D(a —-y)? +0. Hence, since # will appear once along with each of the remaining n-1 letters, and the same is true of y, z, . . ., we have (n — 1) 2a? — 2zay +0, that is, (n —1)2a? + 2Daxy. * The graphical study of inequalities involving only one variable will be found to be a good exercise. 40 EXAMPLES CHAP, In ‘the case where e=y=z=... we have Ya%=nz?, 22ey=2,C,27 =n(n—1)z?, so that the inequality just becomes an equality. When n=2, we have the theorem a? +y? + 2ay ; or, if we put e=/a, y=~/), a and b being real and positive, a+b+2/(ad), a theorem already established, of which the preceding may be regarded as a generalisation. .A more important generalisation of another kind will be given presently. Example 6. Ifa, y, z, . . . be” real positive quantities, and_p and q any two real quantities having the same sign, then LPTI + PTI ¢ ePyd + atyP, NILP+I + DaePLao, We have seen that x?-y? and xw%—y? will both have the same sign as x—y, or both opposite signs, according as p and g are both positive or both negative. Hence, in either case, (x? —y?) (x?—y%) has the positive sign. Therefore (x? — y?) (at — y2) +0, whence CPTI 4 YPTT <3xyz, accord- ing as Zr~> <0. For ra? — 8xyz = Tx( Ta? - Tay), =42rr(x-y). Hence the theorem, since =(#—-y)? is essentially positive. Example 8. To show that 1 _ 1.3 ioe (2-1) _ V(r +1) AM (Qm4+ 1) | 2s4 seen Qn+1 ’ where 7 is any positive integer. XXIV EXAMPLES 4] From the inequality a+ b>2,/(ab) we deduce (2m — 1) + (20+1)>2a/{(2n—-1) (Qn+1)} ; whence (22 —1)/2n2/{n(n+1)}, that is, 2n+1>2r/{n(n+1)}. Hence we have the following inequalities— (2n +1)/2n>/{(n+1)/n} (1)’, (21 — 1)/2(n —1)>A/{n/(n-1)} (2)’, 7/2.8>/ {4/3} (n- 2), 5/2.2>/{3/2' (n-1), 3/2.1>A/ {2/1} (n)'. Multiplying these n inequalities together, we get 1.8.5... (2n+1) DcdaiewocOnunee Won oi noe 1.3.50... » (22-1). afin+1) (B). NOS a ae Te (A) and (B) together establish the theorem in question. Since 4/(n + 1)/(2n+1)>a/(n+1)/(2n+2)>1/2./(n+1), we may state the above theorem more succinctly thus, a Hous ive (270— 1.) 1 > > : A/(2n +1) BiAven. «on 2/(n +1) DERIVED THEOREMS. § 5.] We now proceed to prove several theorems regarding inequality which are important for their own sake, and will be of use to us in following chapters. Ff by, bs, . « «, bn be all positive, the fraction (a, +dg+.. .+ An) (0,+0,+...+0n) is not less than the least, and not greater than the greatest, of the n fractions a,/b,, a/b.» «, On| On. Let f be the least, and f’ the greatest of the n fractions, then afi, tf, dlbotf, .- + dnldntfe 42 MEANS AMONG RATIOS CHAP. Hence, since 0,, 0, . . ., J, are all positive, iy <(at-1)/q according as p> <4. | Since p and g are positive, (2? — 1)/p> <(#t- 1/4, according as g(a? —1)> <0. If p>q, we have X= (x —1){o(wP-1 + eP—2 +. £1) — (ett at 7+... + ID}, Bee gee + P62 +... + at) —(p— gat + at +... + 1). Now, if «> 1, P14 gP-24. . + at>(p—g)rt; ee Se a a a qui! ; therefore, X > (w—1) {a(p — get — (p— gant 5 > Wp — gata — 1)”, =U; Again, if «<1, X>(w— 1) @(p— ga? — (p— 9a} » > op — 4) (e— 1) (a? — 1), > 0. ASE AMO A A Giore VED) Hence, in both cases, (a? — 1)/p > (a4 - 1)/¢. By the same reasoning, if g>, (a4 — 1)/q > (a? — 1)/p, (a? — 1)/p < (a4 - 1)/¢. § 7.] If x be positive, and +1, then ma—(¢—-1)>a2"—-1>m(a-1), unless m lie between 0 and +1, in which case mala — 1) <2 -1 <1, which is part of the theorem to be established. In (2) we may replace « by 1/z, where x is any positive quantity +1, and the inequality will still hold. ~ Hence (1/z)™—1> ma™—l(¢ — 1), that is, ma™—Y¢—1)> <1. We have thus established the theorem for positive values of m. Next, let m= —n where n is any positive commensurable quantity. Then a-"-1> <(-n)(«-1), according as l-—a™> < —na(x—- 1), according as a” —1< >nx(x—- 1), north —ngt> <(-n)(a- 1), according as (n+ 1)e(a—-1)> 1, therefore, by what we have already proved, (n+ 1)a®(a—-1)>a"tl- 1. Hence a—"—1>(-n)(#-1) (4). In (4) we may write 1/x for z; and then we have (1/x)-" —1>(-n) (1/x- 1). If we multiply by — a”, this last inequality becomes a™—1<(-n)a-™ (x - 1), that is, (—n)a-"-We —-1)>a-™- 1. Hence, if m be negative, mea —-1)>a™—1>m(a-1); which completes the demonstration. XXIV ma 1(o Ae? y) eS y= my “Va Wi y) AB Cor. If x and y be any two unequal positive quantities, we may replace x in the above theorem by 2/y. On multiplying throughout by 7”, we thus deduce the following— Tf x and y be posite and unequal, then mala — y) > am — y™> mya — y), unless m lie between 0 and +1, in which case marl — y) m2z(1+z2)"/(1+2), {1-ma/(1+2)}(1+a)">1. If mz<1+a, 1—me/(1+2) is positive, and we deduce (l+a)">1/{1-ma/(1+2)}, >(1+a)/{1+(1-m)z}. The other cases may be established in like manner. Remark.—lt should be observed that (lba)"> <1+ma, according as m does not or does lie between 0 and +1. Example 2. Show that, if 2, wo. . ., Un be all positive, then (1+) (1+u2) .. © (L4+Un)>14+umtuet...+Un; also that; if uw, #2. . ., Up be all positive and each less than 1, then (l—m)(1—we) . . . (1—-Un)>1-M—-te-...-U,. The first part of the theorem is obvious from the identity (1+) (1+)... (L4+upn)=14+ Duy t+ Duy ete + Duy Ugugt...+UUg... Un. The latter part may be proved, step by step, thus— 1l-w=1-%U. (1 — wy) (1 — Ue) = 1 — Wy — Ug t+ W1U2, >1-Uy,- Us. * Several mathematical writers have noticed the unity introduced into the elements of algebraical analysis by the use of this inequality. See especi- ally Schlémilch’s Handbuch der Algebraischen Analysis. The secret of its power lies in the fact that it contains as a particular case the fundamental limit theorem upon which depends the differentiation of an algebraic function. The use of the theorem has been considerably extended in the present volume. 46 ARITHMETIC AND GEOMETRIC MEANS CHAP. Hence, since 1 — ws is positive, (1 — wy) (1 — a) (1 — wg) > (1 — wg) (1 — 41 — ua), > 1 — my — Ug — Ug + Us(U + U2), >1—-Uy- Ug— Us. And so on. These inequalities are a generalisation of (l4x)">l+tnzx(~7<1 andna positive integer), They are useful in the theory of infinite products. § 8.] Lhe arithmetic mean of n positive quantities is not less than their geometric mean. Let us suppose this theorem to hold for m quantities a, b,c, ..., k, and let 7 be one more positive quantity. By hypothesis, (a+b+er+. . ie) tt (abe. mere that is, Or O+ 0+... | +a 00ers ge i) een Therefore atbtet+r...tk+istniabe .. . b+]. Now, mabe... ky +t (n+1) (abe. . . kl)Wm+, provided m {abe . . kim +1 (n+ 1) {abe . . . kb fir+ Het), (n+ 1) {abe . . , k/imp ety), that is, provided nertt 1 t(n + 1)E%, where CENT) HOC eee that is, provided (LEM Ee 1) fee bea, which is true by § 7. Hence, if our theorem hold for n quantities, it will hold for n+1. Now we have seen that (a+b)/24(ab)?, that is, the theorem holds for 2 quantities; therefore it holds for 3 ; there- fore for 4’; and so on. Hence we have in general (at+b+e+.. .+kh)/nt(abe .. . k)in, It is, of course, obvious that the inequality becomes an equality ‘when =0=¢c=. =k. XXIV ARITHMETIC AND GEOMETRIC MEANS 47 There is another proof of this theorem so interesting and fundamental in its character that it deserves mention here.” Consider the geometric mean (abe... k)/™ Tha, bc, .. . be not all equal, replace the greatest and least of them, say a and h, by (a+ k)/2; then, since {(a+k)/2}*> ak, the result has been to increase the geometric mean, while the arithmetic mean of the n quantities (a+ k)/2, b,c. . ., (a+k)/2 is evidently the same as the arithmetic mean of a, 0, c,. . ., & If the new set of n quantities be not all equal, replace the greatest and least as before ; and so on. By repeating this process sufficiently often, we can make all the quantities as nearly equal as we please; and then the geometric mean becomes equal to the arithmetic mean. But, since the latter has remained unaltered throughout, and the former has been increased at each step, it follows that the first géometric mean, namely, (abe . . . k)"", is less than the arithmetic mean, namely, (a+b+ce+...+h)/n. As an illustration of this reasoning, we have (1.3.5.9)"4 <(5.3.5.5)§<(5.4.4.5)§<(45.45.4:5.45)§<45<(14+3 + 5+ 9)/4. Cor. If a, b,. . ., & be n positive quantities, and p,q, . . ., t be n positive commensurable quantities, then paotgqo+.. SE + (orl BMH. +0. pr+gt...t It is obvious that we are only concerned with the ratios p:q:...:t. Hence we may replace p,q,.. ., ¢ by positive integral numbers proportional to them. It is, therefore, suffi- cient to prove the theorem on the hypothesis that p,q,.. ., ¢ are positive integers. It then becomes a mere particular case of the theorem of the present paragraph, namely, that the arithmetic mean of p+q+. . .+# positive quantities, p of which are equal to a,qtob,...,¢ to &, is not less than their geometric mean. * See also the ingenious proof of the theorem given by Cauchy (Analyse Algébrique, p. 457), who seems to have been the first to state the theorem in its most general form. 48 Zpa"/Zp= (Zpal/=p)™ CHAP. Example 1. Show that, ifa,b,..., Eben positive quantities, (Age . - +k*\a+b+. . at yeas oes ¢(s : Se » otk, n The first part of the proposition follows from the above corollary. by taking VU ie ay ea To prove the second part let us assume that abe... ht {(atbt. . .+k)fnpotor. - tk. Then abe... REP 4S atbt. © .+h)/nyotot. . tk. But Bi(a+b+. . .+h)[nprtot.- ths (asb4. . .+h4+D/(nt1)yetor.. +k+1, provided 1+ {n(1+2)/(n+1)} {(1+1/x)/(n+1)}, that is, {(1+1/x)/(n+1)}*+ (n+1)/n(1+2), where x=//(a+b+. . .+k), which we may suppose +1, since there is no loss of generality in supposing a, b,. .., &, l arranged in descending order of magnitude. Now, by § 7, since x+1, (1+ 1/x)/((n+1)}*$142{(14+1/z)/(n4+1)-1}, > (n+2-—nzx)/(n+1). Also, (n+2—nzx)/(n+1)> (n+1)/n(14+2), provided n(n +2 —nexe) (1+x2)> (n+1)?, Nn? + Qn + Wnw — nx? pn? +2n+1, 0+ (nx —1)?, which is true. Hence, if the proposition hold for quantities, it will hold for n+1. But, obviously, a*<+(a/1)*, hence &c. Example 2. Prove that 1.38... (2n-1) {1.8 . . . (Qn-1)}1m, that is, vin> {1.38 ... (2m —1)}1/m, Hence neo 1.3 °s (Set) § 9.] Hf a,b,...,k ben positive quantities, and ‘0, ees be n positive quantities, then par + qb +... ay (Me - - tth\™ (1) prat...t+t prat...tt i according as m does not or does lie between 0 and + 1. If we denote Pipt+qt...+8, g/(ptqt...+2), &e, DYAY jG 2 27, and alAatpb+...t+ch), b/Aatpbt+. . + tk), &¢., DY2, te 7,200; sol that NOP eS ea (2), D+ pyt.. .+Tw=1 (3), XXIV Sam (Za/n)™ 49 then, dividing both sides of (1) by {(patqb+...+th)(p+qt. ..+h}™ we have to prove that da” + py +... + Tw™t pl (4), according as m does not or does lie between 0 and + 1. Now, by § 7, if m does not lie between 0 and +1, 2”-1 ma —1), ¥”-14¢m(y-1), &. Therefore, since d, p, &c., are positive, =e" — 1) ZAm(a - 1), tm {zAx — ZA}, <+m(1 - 1), by (2) and (3), that is, LAxv™ — TA 0. Hence De te, In like manner, we show that, if m lies between 0 and + 1, Age ial Ne Cor. If we make p = 7 == = te Ove am + OM +. + km ‘at+b+,..+kh\™ ; ee eh" bs that is to say, the arithmetical mean of the mth powers of n positive quantities is not less or not greater than the mth power of their arith- metical mean, according as m does not or does lie between 0 and +1. Ltemark.—lIt is obvious that each of the inequalities (1), (4), (5) becomes an equality ifa=b=. =k if tee.0; Or itm = Le Example. Show that =dx”, considered as a function of m, increases as m increases when m>+1, and decreases as m increases when m <-—1, eres), %, /, 2, . . . being as-above. Ist. Let m>1. We have to show that =\w™t"> Zw”, where r is very small and positive, that is, . Dx" (a” —1)>0. Now, LAw™a" —1)> Trw™ra"-l(a2 -1), >rzramtr-l(¢ — 1), * The earliest notice of this theorem with which we are acquainted is in Reynaud and Duhamel’s Problemes et Dévelopmens sur Diverses Parties des Mathématiques (1823), p. 155. Its surroundings seem to indicate that it was suggested by Cauchy’s theorem of § 8. The original proof rests on a maximum or minimum theorem, established by means of the Differential - Calculus ; and the elementary Sade hitherto given have usually involved the use of infinite series. VOL, II KE 50 EXERCISES V CHAP. Since m>1, m+r>1, therefore (m+r)x™"-l(a-1)>(m+r)(#-1), that is, etr-l(¢—-1)>(a-1). Hence Dru" (a —1)>r=BrX(a- 1), >r(ZAx— ZA), >, Therefore Drwmtr > Drv”, 2nd. Let m< -1. ZDAxL™(a” —1)(m+1)(a-1), since m+1 is negative. Hence, dividing by the negative quantity m+1, we have w™(x-1)<(x—1). Hence ZAx™(x" -1)y>z2, then aty+yiz+24e> cyt + yx + zat, (3.) If x, y, 2 be any real quantities, then 2(y—2)(z-x)+0 and Zyz/ 2a? > 1. (4.) If 2+ y?+22+4+2vyz=1, then will all or none of the quantities x, y, z lie between —1 and +1. (5.) If # and m be positive integers, show that 2m < g(a +1) (2a +1) (8x? + 8e+1)"/2.38"%<(~+1)mts, (6.) (a2/b)} + (B2/a)? tat +82. (7.) If x, %,..., % all have the samosigon, and 1--2,, 1g swat ee be all positive, then Il(1+a,)>1+ 22. (8.) Prove that 8xyz+ II (y+2) > 322°. (9.) If x, y, % ..., a 6 ¢... be two sets, each containing 7 real : quantities positive or negative, show that Za*zau* + (Zaza)? ; also that, if all the quantities be positive, 2X(x/a)/Da+ Dx/Zax ; and, if Zv=1, Zl1/e<+n*. (10.) If a, v2, ...«, % and also yy, Yo, . +» +, Yn be positive and in ascending or in descending order of magnitude, then Lay /Zayi > Day?/Zax}. (Laplace. ) * Unless the contrary is stated, all letters in this set of exercises stand for real positive quantities. XXIV EXERCISES V | 51 (11.):Ifa, b, . ..., 2be in A.P., show that A Geleucee wact = (0eL: SS —— (12.) For what values of x is (w—3)/(aw?+a+1)>(a—-4)/(a®-a+1)? (13.) Find the limits of # and y in order that c>ax+by>d, a>cx#+dy>b; where ad — be +0. (14.) x® — ay + 4aty? — 2a8y3 + 4a?y4 — xy’ + y8 > 0, for all real values of x and y. (15.) Is 10x? + 5y? +132? > = <8yz4+ Qry+182x ? (16.) If p+ 2-—A/2, then s/(a?+ y?)+pr/(ay)>at+y. (17.) Is \/(a?+ab +b?) — \/(a? -ab+ 0?) > = <2a/(ab)? (18.) If x and a@ be positive, between what limits must « lie in order that wh a>r/{3a?-+20-+0")} +/ {302 — 2a-+0)}? (19.) Ifa<1, then {a+a/(2®-1)}3 + {a—/(a? -1)}d<2, (20.) If all the three quantities ./{a(b+c—a)}, \/{b(c+a—b), /$e(at b-—c)} be real, then the sum of any two is greater than the third. (21.) If the sum of any two of the three x, y, z be greater than the third, then 22a2a?> Zax? + xyz. (22.) Z1/x+ Da8/aFy3e8, (23.) If p, denote the sum of the products 7 at a time of a, b, c, d (each positive and <1), then p.+ 2p4> 2p. (24.) Zat+ xyzda. (25.) Ifs=a+b+ce+.. .n terms, then 3s/(s—a)+n?/(n—1). (26.) If m>1, x<1, and mxv<1+za, then 1/(l=mzx)>(lba)">1+ime. If m<1, x<1, me<1+a, then (1+2)/{14(1-m)x} <(lta)m< lime. (27.) Ifz"=a"+y", then 2™> 2a™, m being a positive integer. (29.) n{(m+1)—1} <141/2+... +1/n Zp, (33.) If 2 be integral, and « and n each >1, then xe” —1L>n(alrt)i2 — a(n- 12), (34.) Prove for x, y, 2 that (QDyz— Za)” + (Zx)**Tl( Ta — 2x). (35.) Ifs=a+a2+.. . +a,, then II(s/a,—- 1) +(n—-1)*. 52 INEQUALITIES AND TURNING VALUES CHAP. (36.) 3m(38m+1)?>4(8m!)U™, (37.) If sm be the sum of the nth powers of a, d2,.. +, Gn, and Pm» the sum of their products m at a time, then (7—-1)!8m+ (1 —m)!m!pm. (SS. )eliwiyets>. . 5 any ten (1 — Gn)"-1 > (nm — 1)"—1(ay — Ae) (42-3). . « (Gn-1— Gn) Hence, or otherwise, show that {(7-1)!}?>n"-*. (39.) Which is the greatest of the numbers 4/2, 4/3, 4/4, ..- ? (40.) If there be m positive quantities 2, w2,.. ., %m, each>1, and if £1, £2,. . +, &, be the arithmetic means, or the geometric means, of all but 1, aut DUL Re, +1. +> all DUE ayy Loen Ta*! >} TE"), (41.) If a, 6, c be such that the sum of any two is greater than the third, and a, y, z such that Zz is positive, then, if Za?/x=0, show that xyz is negative. (42.) If A=a+det+...+M, B=b, + bo+...+bn, then =(a,/A — b,/B) (ay/b,)” has the same sign as 7 for all finite values of n. (Math. Trip., 1870.) APPLICATIONS TO THE THEORY OF MAXIMA AND MINIMA. § 10.] The general nature of the connection between the theory of maxima and minima and the theory of inequalities may be illustrated as follows :—Let ¢(2, y, 2), {(@ y, %) be any two functions. of 2, y, z, and suppose that for all values con- sistent with the condition T(t y 2)=A (1), we have the inequality LY, DELG % #) (2). If we can find values of 2, y, 2, say a, 0, c, which satisfy the equation (1) and at the same time make the inequality (2) an equality, then (a, 0, ¢) is a maximum value of ¢(2, y, 2). For, by hypothesis, ¢(a, 6, c)=A and ¢(2, y, 2)A; therefore (2, y, 2) cannot, for the values of a, y, 2 considered, be greater than A, that is, than ¢(a, 0, ¢). Again, if we consider all values of a, y, 2 for which ble, y, 2) =A (1), if we have K&, 2) G(2, Y, 2) +A (2'), it follows in like manner that, if a, b, c be such that (a, 0, c) =A, fla, 6, c)=A, then f(a, 8, ¢) is a minimum value of /(@, ¥, 2). XXIV RECIPROCITY THEOREM 53 The reasoning is, of course, not restricted to the case of three variables, although for the sake of brevity we have spoken of only three. The nature of this method for finding turning values may be described by saying that such values arise from exceptional or limiting cases of an inequality. § 11.] The reader cannot fail to be struck by the reciprocal character of the two theorems deduced in last section from the same inequality. The general character of this reciprocity will be made clear by the following useful general theorem :— Tf for all values of x, y, 2, consistent with the condition it (2, Y; 2) a A, f(z, Y, %) have a maximum value (a, b, c)=B say (where B depends, of course, upon A), and if when A increases B also increases, and _ vice versa, then for all values of x, y, 2, consistent with the condition (a, 4; 2) = B, K(@, y, 2) will have a minimum value f(a, b, c) = A. Proof—Let A’B’ where B’a eee Whence, putting 7 = II(a/X)*(k/Za)*, we see that, if Ix? =j, the minimum value of SAz! is Sa{j/II(a/d)*}**, corresponding to %=[a {j/T(a/A)*}P/Ape . . Cor, If we put /=>m=n=...=1, p=ger=...=1, we obtain the following particular cases, which are of frequent occurrence :— If dAux =k, Ux is @ maximum when de=py=.. .; If Wx=k, DAx is a minimum when de =py=.. . Example 1. The cube is the rectangular parallelopiped of maximum volume for given surface, and of minimum surface for given volume. If we denote the lengths of three adjacent edges of a rectangular parallelo- piped by x, y, 2, its surface is 2(yz+z2e+ay) and its volume is xyz. If we put E=yz, n=2x, f=xy, the surface is 2(£++ ) and the volume «/(én$). Hence, analytically considered, the problem is to make én¢ a maximum when &+n+¢ is given, and to make £+y+¢a minimum when é7¢ is given. This, by Th. I., is done in either case by making £=y=§, that is, yz=2v=ay; whence x=y =z. Example 2. The equilateral triangle has maximum area for given peri- meter, and minimum perimeter for given area.. The area is A=Vs(s—a) (s—b) (s—c).. Let s=s—a, y=s—b, 2=s—c; then «+y+2=s; and the area is \/sayz. Since, in the first place, s is given, we have merely to make xyz a maximum subject to the condition x+y+z=s. This leads to x=y=z (by Th. I.) Next, let A be given. Then (e+ y+2)eyz= A? (1); s= A2/eyz (2). If we put £=27yz, n=ayz, f=axy2, we have E+q+f=2? (1'); s= A®/(Eng)4 2 56 DEDUCTIONS FROM § 9 CHAP, Hence, to make sa minimum when A is given, we have to make énf a mawi- mum, subject to the condition (1’). This leads to £=n=¢, that is, ayz= cyr2e= xyz? ; whence r=y=2. Example 3. To construct a right circular cylinder of given volume and minimum total surface. Let « be the radius of the ends, and y the height of the cylinder. The total surface is 2r(z?+ ay), and the volume is 7a’y. We have, therefore, to make w=xz?+-xy a minimum, subject to the con- dition «*y=c. We have U=2? +ary=cly +e]x (1); ey =C (2). Let lja=2, l/y=n; then w=c(2é+7) he ie yn =1/4c (2'). We have now to make 2+ (that is, €+&+7) a minimum, subject to the condition £7= constant. This, by Th. II, leads to —=£=y, which gives 2e=y. Hence the height of the cylinder is tal to its diameter. By the reciprocity theorem (applied to the problem as originally stated in terms of zand y), it is obvious that a cylinder of this shape also has maximum volume for given total surface. 13.] From the inequality of § 9 we infer the following :— VIL. Lf m do not lie between 0 and +1, and if p,q,7, . . . be all constant and positive, then, for all positive values of a, M2 oy a ae eur that Lpx = k, Lp” (m unchanged) has a mimum value when 2=y=ez=.. . If m lie between 0 and +1, instead of a minumum we have a maximum. In stating the reciprocal theorem it is necessary to notice that, in the inequality, =pz occurs raised to the mth power ; so that, if m be negative, a maximum of Ypx corresponds to a mini- mum of (2px). Attending to this point, we see that— VIII. Ifm> +1, andif pg, 7, .. . be all constant and posi- tive, then, for all positive values of x, y, 2, . . . such that Lpu” = k (m unchanged), Dpx has a maximum value when v=y=ez=... If m< +1, we have a minimum instead of a maximum. Theorem VIII. might also be deduced from Theorem VII. by the substitution £=2", »=y™, C=2™, &e ... XXIV DEDUCTIONS FROM § 9 57 § 14.] Theorem VII. may be generalised ah a slight trans- formation into the following :— IX. Jf m/n do not lie between 0 and +1, and if p,qr,.. ., A, pv, » + . be all constant and positive, then, for all positive values eye, 2. . such that LAa" = k (n unchanged), zp” (m unchanged) has a minimum value when pa/dx” = qy" py =... Tf mn lie between 0 and +1, instead of a minimum we have a maximum. The transformation in question is as follows :— Let Ax” = p&, py” =on (1), pe” = pl, qy™ = ont (2). From the first two equations in (1) and (2) we deduce EF} = pam-n/d, pt-1 = Mafn-™/p, &e. Hence, if we take fn =m, that is, f=m/n, p, o, . . . will be all constant and obviously all positive ; we have, in fact, eae eae As Maree tee (Gara fA) 12 ay) cite =D), p = (M/py"G-D, o = (pi{gyug-D, . .. (4); and we have now to make =pé/ a maximum or minimum, subject to the condition Dpé =k. Now, by Th. VIL, >pé/ is a minimum or maximum, according as f does not or does lie between 0 and +1, when ==. . . Thus the conditions for a turning value are (pa) X)UF-D = (gy 2p U-D= . . ,, which lead at once to FB OM PMA TAT I Se mae Cor. A very common case is that where n = 1A=p=.. 48 We then have, subject to the condition 27 =k, Spa™, a mini- mum or maximum when p2”-1=qy"-1= .. ., according as m ‘does not or does lie between 0 and + 1. § 15.] We have hitherto restricted p,q, 7, .. . in the in- 58 EXAMPLES CHAP. equality of § 9 to be constant.. This is unnecessary ; they may be functions of the variables, provided they be such that they remain positive for all positive values of 2, y, 2. We therefore have the following theorem and its reciprocal (the last omitted for brevity) :— X. If p,q 7, . . . be functions of x, y, z, . . . which are real and positive for all real and positive values of x, y, %, . . ., then, for all positive values of x, y, 2, . . . which satisfy 2px = k, (Spa) (Sp)"-1 (m unchanged) has a minimum or maximum value when w=y=..., according as m does not or does lie between Oand +1. For example, we may obviously put p=d2, q=py?,.. . We thus deduce that if m>+1 or <0, then, for all positive values of X,Y, % .. . consistent with Drrt!=h, (Zrw™+4) (Lda)"-1 is a minimum When t= Yi, Theorem X. may again be transformed into others in appear- ance more general, by methods which the student will readily divine after the illustrations already given. . Also the inequalities of § 8 may be used to deduce maxima and minima theorems in the same way as those of § 9 were used in the proof of Theorem X. Example 1. To find the minimum value of w=a+y+z2, subject to the conditions a/a+b/y+c¢/z=1, «>0, y>0, z>0, a, b, c being positive constants. Let w=ptl, y=onl, z=7h7; aje=pé, bly=on, clz=cs. Hence p/1=a//z/H1. If we take f= —1, we therefore get w=rfatt, y=rfby, 2=r/eg-; aju=/ak, by=a/in, ce=A/eg, The problem now is to make w= Za/aé-! a minimum subject to the con- dition Z\/aé=1. By Th. VII. this is accomplished by making £=n=f Hence =n=(=1/Za/a. The minimum value required is therefore (Z/a)?; the corresponding values of x, y, z are \/aZa/a, \/bEa/a, v/cZa/a respectively. x Example 2. To find a point within a triangle such that the sum of the mth powers of its distances from the sides shall be a minimum (m>1). Let a, b,c be the sides, x, y, z the three distances ; then we have to make u=Zxe" a minimum, subject to the condition Dax=2A, where A is the area of the triangle. . | XXIV GRILLET’S METHOD 59 If p&e™ == a”, pé = ax, then pm-t = ae. p =qm I(m-1)_ Hence, if we put aw=a™{ "DE, by=b" (MN y, cz = c™/(™-1)&, we have U= Lam|(m—l)Em, ZA= Danes, The solution is therefore given by £=n=[=2A/Za™/(™-}), Whence a= 2Aal/m-liSam{m-l), yx=ke., z=&e. Example 3. Show that, if 2° +y4+2°=3, then (at+y°+ 2°) (2?+43+24) has a minimum value for all positive values of x, y, z when x=y=z=1. This follows from Th. X., if we put m=2, p=2*, q=y?, r=‘, which is legitimate since x, y, 2 are all positive. Example 4. If x, y, z, . . . be m positive quantities, and m do not lie between 0 and 1, show that the least possible value of (Zx™"-1) (21/x)™-1 is n™. This follows at once from the inequality of § 9, if we put p=1/2, J eee § 16.] The field of application of some of the foregoing theorems can be greatly extended by the use of undetermined multipliers in a manner indicated by Grillet.* Suppose, for example, it were required to discuss the turning values of the function u= (ax + p)'(ba + q)™(cx + 7)” (1), where /, m, n are all positive. We may write w= (daz + Ap)'(pba + pq) (vex + vr)? /Npmy” (2), where A, », v are three arbitrary quantities, which we may sub- ject to any three conditions we please. Let the first condition be lAa + mpb + nve = 0 (3); then we have I(Aaa + Ap) + m(pba + wg) + n(vex + vr) =Ihp + mpg + wr =k (4), where / is an arbitrary positive constant. This being so, we see by Th. III. that I(Aaw+ Ap)! is a maximum when at + Ap = phe + py = vow + vr = k/Xt (5). * Nouvelles Annales de Math., ser. i., tt. 9, 16. 60 EXAMPLES CHAP. The four equations (3) and (5) are not more than sufficient to exhaust the three conditions on A, p, v, and to determine 2. We can easily determine by itself. In fact, from (3) and (5) we deduce at once la[(aa + p) + mb/(bu + q) + ne/(ca + r) = 0 (6). This quadratic gives two values for a, say z, and z,; and the equations (5) give two corresponding sets of values for d, p, v, in terms of /, say A,, py, v, and d,, as Vat If, then, ,’n,"1," be positive, 2, will correspond to a maxi- mum value of w; if A,r," be negative, x, will correspond to a minimum value of w; and the like for 2,. Example 1. To discuss w=(x+3)?(a— 8). We have W= (w+ BA)?(uH — Bu)/A2 un. Now 2(Av+ 8A) + (ux —-38u)=h, provided 2\ + w=0 (1), 6A — 3u=k (2). Therefore (\v-+3))(ua — 8u) will be a maximum, provided ; Aw + BA= par — 38y (3). Hence, by (1), 2/(e+38)+1/(x-3)=0; which gives e=1. From (2) and (3) we deduce A=k/12, w=.—k/6; so that \7u is negative. We therefore conclude that w is a minimum when v=1. The student should trace the graph of the function w; he will thus find that it has also a maximum yalue, corresponding to z= — 8, of which this method gives no account. Example 2, For what values of « and y is U= (aye + byy + c1)? + (ae + doy + C2)? + 6. © + (Ane + day ten)? a minimum ? Let i, Ax, . . . » An be undetermined multipliers. Then we may write U=ZNj? {(a@ + dy + ¢1)/M}? (1); and p= Dy" { (aya + byy +¢1)/ra} (2), where & is an arbitrary positive constant, that is, independent of # and y, provided 2a\1=0, 2011 =0, 2CyA\y =k (3). This being so, by Th. VII., w is a minimum when (ay + bry + ¢1)/M = (ax + boy + ¢2)/g=. . . =k/DAP (4). The n+2 equations, (3) and (4), just suffice for the determination of M1, da, Ge Ate ns x, Y. From the first two of (3), and from (4), we deduce XXIV METHOD OF INCREMENTS 61 Day (Mx + by +c) =0, Dbi (ax af by i ¢) —ii Hence the values of # and y corresponding to the minimum value of n are given by the system Zay"a+ Daybyy + Za,c,=0, Da,byx + Vdy2y + Dye, =0. This is the solution of a well-known problem in the Theory of Errors of Observation. § 17. Method of Increments.—Following the method already exemplified in the case of a function of one variable, we may define | T= (ath, y+k, 2+1)- g(a, y, 2) as the increment of ¢(,y,2). If, when z=a, y=), z=¢, the value of I be negative for all small values of h, i, J, then f(a, b,c) is a maximum value of (2, y,2); and if, under like circumstances, I be positive, 4(a, b, c) is a minimum value of f(x, ¥, 2). Owing to the greater manifoldness of the variation, the ex- amination of the sign of the increment when there are more variables than one is often a matter of considerable difficulty ; and any general theory of the subject can scarcely be established without the use of the infinitesimal calculus. We may, however, illustrate the method by establishing a case of the following general theorem, which includes some of those stated above as particular cases. Purkiss’s Theorem.*—J/f (a, y, 2. . .) f(a, 4%» « .) be sym- metric functions of x, y, 2, .. ., andif a, y, 2... be subject to an equation of the form SB %» . )=0 (1), then p(% y,2, . . .) has im general a turning value when w=y=2 =..., provided these conditions be not inconsistent with the equation (1). In our proof we shall suppose that there are only three ‘variables ; and so far as that is concerned it will be obvious that there is no loss of generality. But we shall also suppose both “Given with inadequate demonstration in the Oxford, Cambridge, and Dublin Messenger of Mathematics, vol. i. (1862). 62 PURKISS’S THEOREM CHAP. (2, y, 2) and f(a, y, 2) to be integral functions, and this supposi- tion, although it restricts the generality of the proof, renders it amenable to elementary treatment. We remark, in the first place, that the conditions a=y=z2 and f(a, y, 2)=0 are in general just sufficient to determine a set of values for 2, y, 2. In fact, if the common value of a, y, z be a, then a will be a root of the equation f(a, a, a) = 0. - Consider the functions T=$¢(a+h, a+k, a+1)—- $(a, a, a), and fath,a+k, a+l). Each of them is evidently a symmetric function of h, k, J, and can therefore be expanded as an integral function of the element- ary symmetric functions th, Zhk, hkl. We observe also that, since each of the functions vanishes when h=0, k=0, 1=0, there will be no term independent of h, &, J. Let us now suppose h, k, / to be finite multiples of the same very small quantity 7, say h=ar,k=Br, l= yr. Then Sh=rda =ru say, thk=r°SaB =r'v, hkl=7*w. Expanding as above in- dicated, and remembering that by the conditions of our problem fia+h, a+k, a+l)=0, we have, if we arrange according to powers of 7, I= Aur + (Bu* + Cv)r* + &e. (1%; 0 = Pur + (Qu’ + Roy’ + &e. (2), where the &c. stands for terms involving 7° and higher powers. From (2) we have ur = — (Qu’ + Ro)r’/P + &e., ur =0+ &e, 22087" = — Dar’ + &e., &c. as before including powers of 7 not under the 3rd. Hence, substituting in (1) and writing out only such terms as contain no higher power of r than 7°, we have I=(C— AR/P)er’ + &., = — $7°(C — AR/P) 30’ + &c. Now (see chap. xv., § 10), by taking 7 sufficiently small, we may cause the first term on the right to dominate the sign of I. XXIV EXERCISES VI 63 Hence I will be negative or positive according as (CP — AR) /P is positive or negative; that is, ¢(a, a, a) will be a maximum or minimum according as (CP — AR)/P is positive or negative. Example. Discuss the turning values of ¢(z, y, 2) =ay2z + b(y2z+ 2a + xy), subject to the condition 2?+y?+22=3a, The system L=Y=Z, e+ y+ 2° — 3a7=0 has the two solutions Cay = aa 0, If we take c=y=z= +a, we find, after expanding as above indicated, [=(a? + 2ab)ur + (a+ b)vr? + &e., 0=2aur + (wu? — 2v)r?. In this case, therefore, A=a?+2ab, C=a+b, P=2a, R= -2 ; and (CP - AR)/ P=2a+ 3d. Hence, when e=y=z= +a, ¢is a maximum or a minimum according as 2a + 3b is positive or negative. . In like manner, we see that, when z=y=z=-a, ¢ is a maximum ora minimum according as — 2a +30 is positive or negative. EXERCISES VI.* (1.) Find the minimum value of bex+cay+abz when xyz=abe. (2.) Find the maximum value of xyz when a°/a?+ y?/b? + 2%/c2=1. (3.) If 22?=c, Dla isa maximum when w:y:2:... =limins. (4.) Find the turning values of \w"*+ wy + v2, subject to the condition px + qy +re=d, (5.) Find the turning values of ax? + by? +cez" when ayz=d?, (6.) If ayz=a*(a+y+z), then yz+2x+ay is a minimum when C=y=2= A/3a. (7.) Find the turning values of (w+) (y+m) (z+n) where atbve=d., (8.) Find the minimum value of av” + b/x. (9.) Find the turning values of (3x - 2) (a — 2)?(a — 3), (10.) If ca(b-y)=ay(c — z) =b2(a — 2), find the maximum value of each. (11.) Find the turning values of a”/y” (m>n), subject to the condition “-y=c. (Bonnet, Nowv. Ann., ser. i, t. 2.) (12.) If apy? + atyr=a, then x?t7+ yP?t7 has a minimum value when Rayos (a/2)'(P+9) ; and, in general, if Sxv?y?=a, DxPt2 has a minimum value, a/(n — 1), when w=y=2=.. . = {a/(n-1)n}1/+), Discuss specially the case where p and q have opposite signs. (13.) If wryt+arys=c, then ay” is a maximum when «?-"/(ru — st) = ered! (g¢—pu), the denominators, ru -st and gt—pu, being assumed to have the same sign. (Desboves, Questions d’ Algébre, p. 455. Paris, 1878. ) Sau * Here, unless the contrary is indicated, all letters denote positive quantities. 64 EXERCISES VI. CHAP. XXIV (14.) If p>q, and #?+yP=a?, then x%+y% is a minimum when w=y= a/2'/p, State the reciprocal theorem. (15.) Find the turning values of (ax? + by?)/\/(a?z + by”) when 2?+y?=1. (16.) If a1, %, . . .,%» be each >a, and such that (a—-a@)(a-a)... (ap, —@)=b", the least value of av... & is (a+b)", a and b being both positive. (17.) If,(m) denote the greatest product that can be formed with m integers whose sum is m, show that f(m+1)//(m)=1+1/q¢ where q is the integral part of m/n. (18.) ABCD is a rectangle, APQ meets BC in P, and DC produced in Q. Find the position of APQ when the sum of the areas ABP, PCQ is a minimum. (19.) O is a given point within a circle, and POQ and ROS are two per- pendicular chords. Find the position of the chords when the area of the quadrilateral PRQS is a maximum or a minimum. (20.) Two given circles meet orthogonally at A. PAQ meets the circles in P and Q respectively. Find the position of PAQ when PA,AQ is a maximum. or minimum. (21.) To inscribe in a given sphere the right circular cone of maximum volume. (22.) To circumscribe about a given sphere the right circular cone of minimum volume. (23.) Given one of the parallel sides and also the non-parallel sides of an isosceles trapezium, to find the fourth side in order that its area may be a maximum. (24.) To draw a line through the vertex of a given triangle, such that the sum of the projections upon it of the two sides which meet in that vertex shall be a maximum. a «thi in = OH A PARE RO RX XC Ve Limits. § 1.] In laying down the fundamental principles of algebra, it was necessary, at the very beginning, to admit certain limiting cases of the operations. Other cases of a similar kind appeared in the development of the science; and several of them were discussed in chap. xv. In most of these cases, however, there was little difficulty in arriving at an appropriate interpretation ; others, in which a difficulty did arise, were postponed for future consideration. In the present chapter we propose to deal specially with these critical cases of algebraical operation, to which the generic name of “Indeterminate Forms” has been given. ‘The subject is one of the highest importance, inasmuch as it forms the basis of two of the most extensive branches of modern mathematics—namely, the Differential Calculus and the Theory of Infinite Series (including from one point of view the Integral Calculus). It is too much the habit in English courses to postpone the thorough discussion of in- determinate forms until the student has mastered the notation of the differential calculus. This, for several reasons, is a mistake. In the first place, the definition of a differential coefficient involves the evaluation of an indeterminate form : and no one can make intelligent applications of the differential calculus who is not familiar beforehand with the notion of a limit. Again, the methods of the differential calculus for evalu- ating indeterminate forms are often less effective than the more elementary methods which we shall discuss below, and are always more powerful in combination with them. VOL. II F 66 MEANING OF A LIMITING VALUE CHAP. § 2.] The characteristic difficulty and the way of meeting it will be best explained by discussing a simple example. If in the function (2° —1)/(@-—1) we put x= 2, there is no difficulty in carrying out successively all the operations indicated by the synthesis of the function ; the case is otherwise if we put «=1, for we have 1*—-1=0, 1-1=0, so that the last operation in- dicated is 0/0—a case specially excluded from the fundamental laws ; not included even under the case a/0 (w+ 0) already dis- cussed in chap. xv., § 6. The first impulse of the learner is to assume that 0/0=1, in analogy with a/a=1,; but for this he has no warrant in the laws of algebra. Strictly speaking, the function (x — 1)/(x— 1) has no definite value when +=1; that is to say, it has no value that can be deduced from the principles hitherto laid down. This being so, and it being obviously desirable to make as general as possible the law that a function has a definite value corresponding to every value of its argument, we proceed to define the value of (7° — 1)/(@—1) when a= 1. In so doing we are naturally guided by the principle of continuity, which leads us to define the value of (# —1)/(e7-1) when x=1, so that it shall differ in- finitely little from values of (a*-1)/(e-—1), corresponding to values of « that differ infinitely little from 1. Now, so long as «+1, no matter how little it differs from 1, we can perform the indicated division; and we have the identity (2° -1)/(e-—1)= «+1. The evaluation of «+1 presents no difficulty ; and we now see that for values of « differing infinitely little from 1, the value of (a — 1)/(a - 1) differs infinitely little from 2. We there- fore define the value of (a? 1)/(w— 1) when «=1 to be 2; and we see that its value 7s 2 in the useful and perfectly intelligible sense that, by bringing « sufficiently near to 1, we can cause (x*—1)/(%7-1) to differ from 2 by as little as we please.* The value of (2° — 1)/(« — 1) thus specially defined is spoken of as the linuting value, or the limit of (u*-1)/(a—-1) for a=1; and it is symbolised by writing * The reader should observe that the definition of the critical value just given has another advantage, namely, it enables us to assert the truth of the identity (a”—1)/(@-1)=a+1 without exception in the case where z=1. s- - —~ + a Se Sv FORMAL DEFINITION OF A LIMIT 67 ib v— 1 ao aay — 1 where L is the initial of the word “limit.” The subscript «= 1 may be omitted when the value of the argument for which the limiting value is to be taken is otherwise sufficiently indicated. We are thus led to construct the following definition of the value of a function, so as to cover the cases where the value indicated by its synthesis is indeterminate :— When, by causing « to differ sufficiently little from a, we can make the value of f(x) approach as near as we please to l, then 1 is said to be the limiting value, or limit, of f(a) when 2=a,; and we write L f(x) =1 LG Cor. 1. A function is in general continuous in the neighbourhood of a limiting value , and, therefore, in obtaining that value we may subject the function to any transformation which is admissible on the hypothesis that the argument « has any value in the neighbourhood of the critical value a. , We say “in general,” because the statement will not be strictly true unless the phrase “ differ infinitely little from” mean “differ either in eacess or in defect infinitely little from.” It may happen that we can only approach the limit from one side; or that we obtain two different limiting values according as we in- crease w wp to the critical value, or diminish it down to the critical value. In this last case, the graph of the function in the neighbour- hood of «=a would have the peculiarity figured in chap. xv., Fig. 5; and the function would be discontinuous. The latter part of the corollary still applies, however, provided the proper restriction on the variation of a be attended to. When it is necessary to distinguish the process of taking a limit by increasing « up to a from the process of taking a limit by decreasing « down to a, we may use the symbol L_ for the x=a—-0 former, and the eee es Bo the latter. Cor. 2. ts Li fF () Hy a f(a+h)=14+d, where dis a function =64 of aand h, ibis, value may be made as small as we please by suffi- ciently diminishing h. 68 CONSEQUENCES OF THE DEFINITION | CHAP. This is simply a re-statement of the definition of a limit from another point of view. Cor. 3. Any ordinary value of a function satisfies the definition of a limiting value. For example, L (#* — 1)/(a@- 1) =(2°- 1)/(2-1)=3. This re- L=2 mark would be superfluous, were it not that attention to the point enables us to abbreviate demonstrations of limit theorems, by using the symbol L where there is no peculiarity in the evaluation of the function to which it is prefixed. § 3.] It may happen that the critical value a, instead of being a definite finite quantity, is merely a quantity greater than any finite quantity, however great. We symbolise the process of taking the limit in this case by writing L is (2), or Tigeayitae. n= - 0 e=+ according as the quantity in question is positive or negative. For example, L (w@+1)/a= L(1+1/2)=1. In this case, we can, strictly speaking, approach the limit from one side only ; and the question of continuity on both sides of the limit does not arise. If, however, we, as it were, join the series of algebraical quantity POD Son Lin 2 6 Oe Ss), aed es ee 00. thronoh Inia considering +o and —o as consecutive values ; then we say that /(z) is, or is not, con- tinuous for the critical value z=, eb as L f(x)and L f(z) have, Y= 0 “= - 0 or have not, the same value. For example, (v+1)/x is continuous for z=, for we have L (@+1)/fe=1= L (a#+1)/x; but (z?+1)/x is not continuous woo v=-0c LOY 20500 § 4.]| The value 0 may of course occur as a limiting value ; for example, L a(z~1)’/(«°—1)=0. It may also happen, even il | for a finite value of a, that f(v) can be made greater than any finite quantity, however great, by bringing @ sufficiently near to a, In this case we write L f(z)= 20. In thus admitting 0 and « v= as limiting values, the student must not forget that the general rules for evaluating limits are, as will be shown presently, sub- ject in certain cases to exception when these particular limits occur. xv CLASSIFICATION OF INDETERMINATE FORMS 69 ENUMERATION OF THE ELEMENTARY INDETERMINATE FORMS. § 5.] Let w and v be any two functions of 2 We have already seen, in chap. xv., that «+v becomes indetermin- ate when w and v are ‘infinite but of opposite sign; that wx v becomes indeterminate if one of the factors become zero and the other infinite; and that w+v becomes indeterminate if x and v become both zero, or both infinite. We thus have the indeterminate forms—(I.) «© — 0, (II.) 0x 0, (III.) 0 +0, (LY.) co +0. ' It is interesting to observe that all these really reduce to (III.). Take coo —o for example. Since w+0=(1+/u)/(1/w), and L1/%=1/00 =0, this function will not be really indeterminate unless Lv/uw=-1. The evaluation of the form o — oo therefore reduces to a consideration of cases (IV.) and (IIT.) at most. Now, since w+v=(1/v)+(1/w), case (IV.) can be reduced to CEDTe and finally, since wx v=w-+(1/v), case (II.) can be reduced to (III.) To exhaust the category of elementary algebraical operations we have to discuss the critical values of uw’. This is most simply done by writing wu” =a"°%“ where a is positive and >1. We thus see that w” is determinate so long as rlog,u is determinate. The only cases where vlog,w ceases to be determinate are those where—(V.) v = 0, loggu = + o,thatisvo=0,u=0; (VL) v=0, loggw= —, that is v=0, uw=0; (VIL) veto, loggu = 0, that is v= +0, uw=1. There thus arise the indeterminate forms—(V.) «0°, (VI) 0°, (VII.) 1=%.* All these depend on a@°* ; or, if we choose, upon @°/9; so that it may be said that there is really only one fundamental case of indetermination, namely, 0+0. EXTENSION OF THE FUNDAMENTAL OPERATIONS TO. LIMITING ; VALUES. § 6.] We now proceed to show that limiting values as above defined may, under some restrictions, be dealt with in algebraical * The reader is already aware that 1° gives 1; and he may easily convince himself that 0+, 0-©, ot, a -o give 0, ao, to, 0 respectively, no matter what their origin. 70 FUNDAMENTAL OPERATIONS WITH LIMITS CHAP, operations exactly like ordinary operands. ‘This is established by means of the following theorems :— I. The limit of a sum of functions of a is the sum of their limits, provided the latter does not take the indeterminate form 2 — a. Consider the sum f(a) — $(z)+ x(x) for the critical value a=a; and let Lf(#)=/’, Ld(z) = ¢', Ly(z)= x’. Then, by § 2, Cor. 2, f(a) =f +4, Pm=$ +P, x(@)=x' +7, where o, 8, y can each be made as small as we please by bringing x sufficiently near to a. Now, f(z) - da) +x@)=f-o +x +-B+y). But, obviously, « — 8 + y can be made as small as we please by bringing z sufficiently near to a. Hence L{f@) - $2) + x@pal-$ +X; that is, = Lf(v)— L(x) + Ly(~) (1). This reasoning supposes f’, ¢’, x’ to be each finite ; but it is obvious that if one or more of them, all having the same sign, become infinite, then f’— ¢’+y’ and L {f(z) — $(a) + x(#)} are both infinite, and the theorem will still be true in the peculiar sense, at least, that both sides of the equality are infinite. If, however, some of the infinities have one sign and some the opposite, f’ — 6’ +x’ ceases to be interpretable in any definite sense ; and the proposition becomes meaningless. Il. Lhe limit of a product of functions of « is the product of their limits, provided the latter does not take the indeterminate form 200 © Using the same notation as before, we have F(a) $@) x) =F + ah! + BY + y) = f'd'y' + Tad'y’ + ZaBy’ + aPy. Now, provided none of the limits /’, ¢, x’ be infinite, since a, f, y can all be made as small as we please by bringing « sufficiently near to a, the same is true of Yad’y’, Lay’, and aby. Hence Lf(x) $(2) x(a) =f'¢'x' = Lf) L(x) Lx@) (2). If one or more of the limits /’, ¢’, y’ be infinite, provided none _ of the rest be zero, the two sides of (2) will still be equal in the xxv —- LF{ f(a), $@),---} = FLF(@), Loa), -- +} 71 sense that both are infinite ; but, if there occur at the same time a zero and an infinite value, then the right-hand side assumes the indeterminate form 0 x ©; and the equation (2) ceases to have any meaning. III. Lhe limit of the quotient of two functions of a is the quotient of their limits, provided the latter does not take one of the indeterminate forms 0/0 or «@ /«. We have fe) _ fire _f fire fi_f', ob'- Bf oa) $B $ F+B 6 SF $+) From this equatjon, reasoning as above, we see at once that, if neither f’ nor ¢’ be infinite, and ¢’ be not zero, f@)_f _Lfle) Ha) Lda) CG} It is further obvious that if f’ = «©, ¢’+ , both sides of (3) will be infinite ; if ¢’= 0, f+, both sides will be zero; and if ¢' =0, f’+0, both sides will be infinite. In all these cases, therefore, the theorem may be asserted in a definite sense. If, however, we have simultaneously /’ = 0, ¢’ = 0, the right hand of (3) takes the form 0/0; if f’=a, ¢'= 0, the form o/c ; and then the theorem becomes meaningless. § 7.] If the reader will compare the demonstrations of last paragraph with those of § 8, chap. xv., he will see that (except - in the cases where infinities are involved) the conclusions rest merely on the continuity of the sum, product, and quotient. This remark immediately suggests the following general theorem, which includes those of last paragraph as particular cases :— Tf F(u;-v, w,.. .) be any function of u, v, w,. . ., which ts determinate, and finite in value, and also continuous when eae) = ihe) wi Ly (2), 6. then tit), pe), x(a). . -} = FLL, Ld)Ly(a), 2) ..} The reader will easily prove this theorem by combining § 2, Cor. 2, with the definition of a continuous function given: in chap. xv., §§ 5, 14. 72 LIMITS OF RATIONAL FUNCTIONS CHAP, The most important case of this proposition which we shall have occasion to use is that where we have a function of a single function. For example, L {(a2 -1)/(@-)}2={ L (@?- 1l@-1)j2=4. t= vik L log {(a? —1)/(# —1)} =log aul & —1)/(a@- 1)} = log a, v=1 ent ON THE FORMS 0/0 AND «/ IN CONNECTION WITH RATIONAL FUNCTIONS. § 8.] The form 0/0 will occur with a rational function for the value x=0 if the absolute terms in the numerator and denominator vanish. The rule for evaluating in this case is to arrange the terms in the numerator and denominator in order of ascending degree, divide by the lowest power of « that occurs in numerator or denominator, and then put «=0. The limit will be finite, and +0, if the lowest terms in numerator and denominator be of the same degree; 0 if the term of lowest degree come from the denominator; % if the term of lowest degree come from the numerator. All this will be best seen from the following examples :— Example 1. 2x? + 80% + a4 2+3xe+27 _ 2 4 eno O02 +38 + 08 yap ota +e as ow Example 2. L 28 + Bato L [n+ sete 0 0 pa De te +O. ny oa ae) Oe ' Example 3. Doty gh 2 ae eee ————_ = ————. = =) 100). go at a pay eee UO § 9.] The form « / © can arise from a rational function when, and only when, 7=«. The limit can be found by dividing numerator and denominator by the highest power of 2 that occurs in either. If this highest power occur in both, the limit is finite ; if it come from the denominator alone, the limit is 0 ; if from the numerator alone, the limit is «2. Example 1. ae a Ca 3/e+1 O+1 1 ga 0 2+ oP + 8a pwerije+8 0+0438 8 XXV USE OF THE REMAINDER-THEOREM io Example 2. a+ 8aF+ det | Lett 8/a+4/a®? 0 0 pao 2 tar + bas Wa +1/ee+6 6 Example 3. w+ 8a+ 4o8 Afxt+d3fae+4 14. ¥ gan 2U+ 8x7 + a7 B/a + 8/at+1/x 3 0 § 10.] If the rational function f(z)/(7) take the form 0/0 for a finite value of a, +0, say for « =a, then, since f(a) = 0, 6(a) = 0, it follows from the remainder-theorem that x—-a is a common factor in f(z) and (7). If we transform the function by remoy- ing this factor, the result of putting x=a in the transformed function will in general be determinate ; if not, it must be of the form 0/0, and # — a will again be a common factor, and must be removed. By proceeding in this way, we shall obviously in the end arrive at a determinate value, which will be the limit of F(x)/$(z) when z = a. Example. Evaluate (3x4 — 10a + 3a? + 12a - 4)/(a4 + 2a? — 2227 + 32” - 8) when x=2. The value is, in the first instance, indeterminate, and of the form 0/0; hence x—2 is a common factor. If we divide out this factor, we find that the value is still of the form 0/0; hence we must divide again. We then have a determinate result. The work may be arranged thus (see chap. Hey S13) :— 3-10+ 3412 -4 1+2 —22 +32 -8 2/0+ 6- 8-10+4 2/0+2+ 8 -28+8 8- 4- 5+ 240 1+4-14+4+ 4/+0 0+ 6+ 4- Q| 0+2+4+12- 4| Seo 10 Le6 = 9 arp (0+ 6 +16 0+2 +16 3+ 8|+15 1+8|/+14 The process of division is to be continued until we have two remainders which are not both zero. The quotient of these, 15/14 in the present case, is the limit required. The evaluation of the limit in the present case may also be effected by changing the variable, an artifice which is frequently of use in the theory of limits. If we put e=a+ 4, then we have to evaluate Lf(a+2z)/f(a+2) when z=0. Since f(a+z) and $(4 +z) are obviously integral functions of z, we can now apply the rule of § 8. It will save trouble in applying this method if it be remembered—1st, that in arranging f(a +2) and (ua +2) according to powers of z we need not calculate the absolute 74 CHANGE OF VARIABLE CHAP, Pas pare terms, since they must, if the form to be evaluated be 0/0, be zero in each case; 2nd, that we are only concerned with the lowest powers of z that occur in the numerator and denominator respectively. L Bart — 10a? +8a?+ 120-4 _ L 3(2+2z)4-10(2+2)8 +3(2+2)?+12(2+2)— 4 ney @+ 208 — 22a? +B2QH—8 ~ yg (2 +28 + 2Q+z2)9 —22(2+2)P + 82A(2+z)—8 an 1527+ P2+&e. * " 7a0l 42? + Qe + Xe.’ ‘7 _ , 15+P2+&e. ; ~ ga 0l4 t+ Qet &e.’ _ 165 3 uals i This method is of course at bottom identical with the fone for, since 4 z=a-—-a, the division by z* corresponds to the rejection of the factor (c-a?, = § 11.] The methods which are applicable to the quotient of two integral functions apply to the quotient of two algebraic sums of constant multiples of fractional powers of x Each of the two sums might, in fact, be transformed into an integral function of y by putting x=y%, where d is the L.C.M. of the denominators of all the fractional indices. It is, however, in general simpler to operate directly. Example. Evaluate ty, ! ve P+ a8+3xt 2=0 as + 2x8 +0 7 ae 1 If we divide by x, the lowest power of x that occurs, we have Ue at +a + Sart n=0 14+2at+a8 ra = § 12.] The following theorem, although partly a special case under the present head, is of great importance, because it gives the fundamental limit on which depends the “differentiation” of — algebraic functions :— If m be any real commensurable quantity, positive or negative, Ae: -1)/@-1)=m (1). XXV L(a™ — 1)/(a-1)=m 75 First, let m be a positive integer. Then we have ee Wiel yaa nt. et 1. Bees L (@@™—1)/(@-1)=1+1+...+1+1 (m terms), “=1 =. Next, let m be a positive fraction, say p/q, where p and q are positive integers. Then the limit to be evaluated is L i: PIG — )/ (a—1).* If we put «=2%, and observe that to «= 1 Pages anak z= 1, the limit to be evaluated becomes L (z? — 1)/(24-1). This i may be evaluated by removing the common factor ¢—1,; or thus gress | eS _ De of = = - — = Ba Ie ) ule l ) 2—1/” erly: eS Ear) heer) plq=m. Finally, suppose m to have any negative value, say — 1, where II n is positive. ‘Then DL @*- 1-1) =L0-@)/ne- 2), Fy = LG" Vile - 1) = et 1 eles Lae aye bah Now, by the last two cases, since n is positive, L (2” —1)/ T=! (7-1)=n. Also L Aah =1. Hence 5 Ga See) = that is, in this case as L (w™ —1)/(a-1) =m oe 1 gi Second Demonstration.—The above theorem might also be deduced at once from the inequality of chap. xxiv., § 7, as follows :—For all positive values of z, and all positive or negative values of m, w”-1 lies between ma”—\(a— 1) -and m(a-1). Hence (a%”—1)/(z-1) lies between mx” and m. Now, by * Theye is here of course the usual understanding (see chap. x., § 2) as to the meaning of «?/2, ® > 76 EXAMPLES CHAP, bringing a sufficiently near to 1, ma™-! can be made to differ as little from m as we please. The same is therefore true of (w”—1)/(w-1); that is to say, L(a” —1)/(a@-1)=m for all real values of m. Example 1. Find the limit of (#? — a?)/(a2 - oe when x=a. We have L (w? —a?)/(at-at)= L a?-*}(x/a)P —1\/{(x/a)2-1}, x= c= ee i ae (f=), ras Gres Ares where y=2/a. Hence we have, by is theorem of the present paragraph L (a? —a?)/(at — at) =a?~tp/q. x= Example 2. Evaluate log (x? — 1) — log (a ?—1) when v=1. L{ log (a! - 1) — log (at -1)} =Llog {(#? -1)/(a4-1)}, =log {L(«! - 1)/(e#-1)}, by §7, =toe{ (5 =1)/1(=2)}, =1 tly lo Example 3. If dx, [%, ... denote logz, log(logx), . . . respectively, then, when v=o, Li(v+1)/la=1. In the first place, we have Uae+1)/le= Uet+1)-let+lx} /lx, =1(1+1/x)/le+1. Now, when x=, U(1+1/z)=l1=0 and la=w. Hence Li(x+1)/le=1. If we assume that Li7(a#+1)/l™x=1, we have ret ll) les {er o(est 1) —- Het reo eee, =U{i(ae+1)/irx} firAe+1. O relco ~ II OB 6 [e) Go Hence LY e+ D/P heal, alls that is, the theorem holds for 7+1 if it holds for 7. But it holds for 7=1, as we have seen, therefore for r=2, &e. It is obvious that this theorem holds — for any logarithmic base for which Joo =o. ieole 4. If 7 have the same meaning as before, and » have a similar | meaning for the base a, then L \a/lx=1/loga. w— co Let p=I1/loga. Since \¥x=wlx, the theorem clearly holds when r=1. It is therefore sufficient to show that, if it is true for 7, it is true for r+1. Now Athy Tre (Xe) Pe = pl ( Nx arte, = wy a) — Ie + Ig} [re =m UN alla) [He + 1}, Hence, if we assume that L\"2/l’e=p, we have LA Ha/l +12 = w {lpn =e 1} ; » ’ | GE dint ° Pod . XXV L(1 + 1/2)" =e 77 EXPONENTIAL LIMITS. § 13.] The most important theorem in this part of the sub- » ject is the following, on which is founded the differentiation of exponential functions generally :— The limit of (1 + 1/x)* when a is increased without limit either positively or negatively is a finite number (denoted by e) lying between 2 and 3. The following proof is due to Fort.* We have seen (chap. xxiv., § 7) that, if a and b be positive quantities, and m any positive quantity numerically greater than 1, then maa — b)>a™ — b™>mb™-VWa — 5) (1). In this inequality we may put a =(y+1)/y, b=1, m=y/z, where | y>a>1. » We thus have 4 y/x =). “yh y ny LY poten Hence 9 (1 + = >1+-, y a Tie, 1\¥ iy that is, (1 +) -(1 + ) (2), y vt where y>z«. Again, if in (1) we put a=1, b=(y—-1)/yp (m, y, @ being as | before), we have 1 Ge ae —>l]-—(' : © y . y|% Hence (1 - “| >] - y Y, Me y ce (aay y © 1\-¥ 1 and therefore (1 - - | <(1 ~ =) (3); y D where y>z. We see from (2) and (3) that, if we give a series of in- » * Zeitschrift fiir Mathematik, vii., p. 46 (1862). 78 L( Sir 1ye)* =6€ OHAP. creasing positive values to x, the function (1 + 1/x)* continually increases, and the function (1 -1/z)-* continually decreases. Moreover, since 2°>2° — 1, we have x r+] > ? y—- 1 y ‘ 1\=! 1 that is, (1 a =) oh pea 4 O, 4 ke Hence (1 - ay = € + 2) (4). The values of (1 —1/z)-* and (1+ 1/z)* cannot, therefore, pass. each other. Hence, when « is increased without limit, (1—1/z)-* must diminish down to a finite limit A, and (1 + 1/x)* must increase up to a finite limit B. The two limits A and B must be equal, for the difference (1 — 1/x)-* —- (1 + 1/z)* may be written {a/(x - 1)}* — {(v@ + 1)/2}*; and by (1) we have a x fi ( x i ea) 1 er 5 — > | —— — {|—} >——— > | ——_ ty b o\n—1 “—1 7 e(L—1 jary Se ©) But, since, as has already been shown, {z/(a#—1)}* and {(@+1)/e}* remain finite when «=o, the upper and lower limits in (5) approach zero when « is increased without limit ; the same is therefore true of the middle term of the inequality. It has therefore been shown that L(1+1/z)? and L=P L (1 —1/x)-* have a common finite limit, which we may denote iy the letter e¢. | ; | Since (1+ 1/6)°=2°521.. . and (1-1/6) *= 2,03 —ueue e lies between 2°5 and 2°9. A closer approximation might be — obtained by using a larger value of x; but a better method of | calculating this important constant will be given hereafter, by — which it is found that é e= 2*7 1828182386 van The constant e is usually called Napier’s Base *; and it is the — logarithmic or exponential base used in most analytical calcula- tions. In future, when no base is indicated, and mere arith- * In honour of Napier, and not because he explicitly used this or indeed ' any other base. = L(a* — 1)/x = loga 79 metical computations are not in question, the base of a logarithmic or exponential function is understood to be e; thus logz and expx are in general understood to mean log,2 and expt (that is, e”) respectively. i Breeds (1 '+.2)b% = ¢, f t= () Bement +tiz’=¢; and if we put ¢=1/2,:s0 that #=0 Z=0 corresponds to z= 0, we have L at Pay ya gs “x=0 Cor. 2. ph loge {(1+1/2)*} = nae lo8a {(1 +x)¥*} = logge. For, since ie y is a continuous ties of y for finite values of y, we have, by § 7, TilGe, (1+ La) log, { L (1+ 1/2}, II = logge. . The other part of the corollary follows in like manner. Cor. af L (1 + y/x)* = i + ay)ue = ey, If we put | Re = i then io “= 00 corresponds z= @ ; hence ae Abad? z)PY, = L(+ 1}, = {L141}, by § 7, —— ey, Cor. 4. L (a* - 1)/x = loga. “z=0 If we put y=a*—1, so that «=log,(1 +y), and to «=0 corre- sponds y = 0, we have L (a@—1)/z=L 4 Y/08a (1 +4), v=0 y=0 = . ifloga( + yyy, | = =1/log a ae + yy}, | | = 1/log Pa fee | It will be an excellent exercise for the student to deduce directly from the fundamental inequality (1) above, the important result that L (a*-1)/a is x7=0 80 EXPONENTIAL AND LOGARITHMIC INEQUALITIES CHAP. finite ; and thence, by transformation, to prove the leading theorem of this : paragraph. * _ Cor. 5. If « be any positive quantity, 7 e>1+4, log(1 +a)1-—a, -—log(l—2)>za. 4 Since ¢>(1 + 1/n)", when nm may be as great as we please, 4 e— 1>(1 + 1/n)* -1, i >nz{(1+1/n)—1}>a, , by chap. xxiv., § 7, for, however small 2, we can by sufficiently increasing » make nz>l. Hence et=>1 +2. It follows at once that loge*>log(1 + «), that is, e>log(1 + 2). ] Again, since oe —1/n)-” and e71>(1 — 1/n)", “© —1>{(n—- 1)/n} ™—-1, >nxz{(n-1)/n-1}, > a Hence e-*>1 — 2, and therefore 1/(1 — 2) >&. § It follows at once that log {1/(1 — z) } , thatis, —log(1 - 2a # Cor. 6.¢ If Ix, Pa, . . . denote loga, log(logu), . . . respect wely, « be positive and >1, and r any positive integer, then Valea... a> +e+1)-i teal (e+ lat LE (w +1). Pee ea) For, by Cor. 5, U(ae+1)-—iz=U(1 + 1/2), 7 <1 /z. This proves the first inequality when 7=0. It remains to show that, if the mequality holds for 7, it holds also for 7 + 1. We have We + 1) Pte l {ita t Lyrae, =1+ {tet 1)-— Pte Pg) <{P e+] ig ire by Cor. 5. Hence, if we assume ("+ Wa+1)—itie<1/ale ... Va, it follows that’ Peal) — ithe align re * See Schlomilch, Zeitschrift fiir Mathematik, vol. iii., p. 8387 (1858). + Malmsten, Grunert’s Archiv., viii. (1846). XXv EULER'S CONSTANT 81 Again, by Cor. 5, we have lx —ex—1)= —K1—-1/2), > 1/2. Therefore Ua +1) —le>1/(x + 1). This proves the second inequality when r=0. If we suppose it to hold for 7, we have PY%e+1)—-7 +2 = —Virt ies t+ Wa + 1)), = iL {Pee l)— tle irl + 1)), af ita t+ 1)— tle} irtle+1), by Cor. 5, eifer i iat ly... Mee ii tas 1. Hence the induction is complete. Cor. 7. From the inequality of Cor. 5, combined with the result of Example 3, § 12, we deduce at once the following im- portant limits :— L {?(<+1)-fe}=0, i fe a ae Hol) — i tet alee. ON Pee Example : Show that the limit when » is infinite of 1+1/2+ : +1/n—logn is a finite quantity, usually denoted by y, lying between 0 anal i (Euler, Comm. Ac. Pet. (1734-5). ) Since, by Cor. 5, —log (1 ae >log (1+ 1/n). We have log {n/(m—1)' >1/n >log {(n+1)/n', log {(n— aes \ ve n—1)>log fale —1)}, log {8/2} >1/3>log {4/3}, log {2/1} >1/2>log {3/2}, Ll log 2ii;. Hence 1+log n> Z1/n>log (n +1). Therefore ; 1>21/n -log n> log (1+1/n). Now, when n=, log(1+1/n)=0. Thus, for all values of 2, however great, 21/n —logn lies between 0 and 1. The important constant y was first introduced into analysis by Euler, and is therefore usually called Euler’s Constant. Its/value was given by Euler himself to 16 places, tamaely, = '5677215664901532(5). (See Jnst. Cale. Diff., chap. vi.)* . * Euler’s Constant was calculated to 32 places by Mascheroni in his Adnotationes ad Euleri Caleulum Integralem. It is therefore sometimes called Mascheroni’s Constant. His calculation, which was erroneous in the 20th place, was verified and corrected by Gauss and Nicolai. See Gauss, Werke, Bd. iii., p. 154. For an interesting historical account of the whole matter, see Glaisher, Mess. Math., vol. i. (1872). VOL. II | G SY CAUCHY ’S THEOREMS CHAP. Example 2. Show that L {1/1+1/2+.. .+1/n'/logn=1. n= nN This follows at once from the inequality of last example. From this result, or from Example 1, we see that L {1/1+1/2+...+1/n} N=P =o; andalsothat L {1/k+1/(k+1)+. ..+1/n} =o, where £ is any finite N=n > _ = positive integer. cy GENERAL THEOREMS. N § 14.] Before proceeding further with the theory of the limits of exponential forms, it will be convenient to introduce a few general theorems, chiefly due to Cauchy. Although these theorems are not indispensable in an elementary treatment of limits, the student will find that occasional reference to them will tend to introduce brevity and coherence into the subject. I. For any critical value of x, Lf f(a) }% = {Lfla)}™, provided the latter form be not indeterminate. This is in reality a particular case of the general theorem of § 7. The only question that arises is as to the continuity of the functions of the limits. We may write | { lx) 1964) — pt)logs @®. : ; Now w = logw is a continuous function of w, so long, at least, as u lies between +1 and +o; and e” is a continuous function of vand w. Hence, so long as L¢(x) and L log f(x) are neither of them infinite, we have L{ f(a) yee) — Te? Io8 Ka) Lata) log fe) __ plala)log L/() Henee Li fla) }* = {Lflay (1) An examination of the special cases where either L¢(x) or Llog f(z), or both, become infinite, shows that, so long as {Lf(z)}'* does not assume one of the indeterminate forms 0°, «°, 1+”, both sides of (1) become 0, or both «©; so that the theorem may be stated as true for all cases where its sense is determinate, ¢ a XXV CAUCHY’S THEOREMS 83 Il. Lf{f(x+1)-f(@)} =L f(x)/a, provided L { fla + 1)—f(a)} be not indeterminate.* (Cauchy’s Theorem.) Since 2 is ultimately to be made as large as we please, we may put «=h+n, where i is a number not necessarily an integer, but as large as we please, and 7 is an integer as large as we please. First, suppose that L {f(« + 1) —f(z)} is not infinite, = say. Since L{ f(z+1)—j/(w)}=%, we can always choose for 4 a definite value, so large that for «=h and all greater values fiw +1)-f(v) —& is numerically less than a given quantity a, no matter how small « may be. Hence we have numerically f(b +1) — fh) — k1, f(ih +n) —fiht+n—1)>1, where / is a definite quantity as large as we please. * Theorems II. and III. are given by Cauchy in his Analyse Algébrique (which is Part I. of his Cowrs d Analyse de ? Ecole Ror yale Polytechnique). Paris, 1821. 84 CAUCHY’S THEOREMS CHAP. Hence fh +n) —f(h)>nl, that is Ja) —f(h)> (a — hp. Hence He) >] HC) — ue ns Set Wane Since f(h), h, 7 are all definite, we can, by sufficiently in- creasing z, render f(h)/x—hi/x as small as we please, therefore f(z)/z>l. Now, by properly choosing h, 1 can be made as large as we please ; hence Lf(«)/x = © : The case where L { f(z + 1) —f(x)} ='— @ can be included in the last by observing that (— f(x + 1)) — ( —f(x)) has in this case + # for its limiting value. I. L f(x+1)/f(a) =L { f(x) }*, provided L f(x +1)/fla be ws = : — not indeterminate. This theorem can be deduced from the last by transformation, as follows :—* We have A { War +1)-¢(@)} =L awl where y¥(x) is any function such that L {¥(x + 1) — Y(x)} is not indeterminate. Let now y(z) = log ae 3 so that ¥(a# + 1) - Y(a) = log f (¢ + 1) — log f(x) = log { 1)/f(a)}; and y(a)/a = {log f(a) }/x =log { f(x) }1™. ha we rae § fle+1) Bes SSL 0g eae 0$ l tli r) if 0 iS( )} ? peeic 108 { een | We - rs [ Die) "); provided Lf(z + 1)/f(@) be not indeterminate. Hence, finally, p ED = Li} n=o f(a) t=0 | Cauchy makes the important remark that the demonstrations of his two theorems evidently apply to functions of an integral variable such as «!, where only positive integral values of z are admissible. * The reader a na it a good exercise to establish this theorem direetly from first principles, as Cauchy does, i xxv Lat®/z, Lloggz/a, Lalogyx 85 For example, we have L (w#+1)!/a#!=L (w@+1)=0. Hence L @!)*=0, rL=D v=” r=0 and consequently L (1/ax !)'/*=0. L= OP ro | 9 hy So EXPONENTIAL LIMITS RESUMED. too v=+0 The first of these follows at once from Cauchy’s Theorem (§ 14, IL.) for we have | ee L(a**! — a”) = La*(a-1)= 0. Hence Lat/z = 0. ; | _ As the theorem is fundamental, it may be well to give an | §15.] Uf a=1, then L at/x= 0; L log,z/z=0; L alog,x=0. } B= CO 1 ’ | | independent proof from first principles. | First, we observe that it is sufficient to prove it for integral values of a alone, for, however large x may be, we can always put w= f+2 where f is a positive proper fraction and z a positive integer. Then we have ae ad +2 lips ee a= g=0 & z he = L as. 5 2'== 00 Te Z Z lie a = as I, ; 4 ORG TE ts (| = CO z | m deg*y Ae | =atL—, (1), g=n© where we have to deal merely with La/z, z being a positive integer. em Let u,=07/z,then u,4,/u,=az/(2+1)=a/(1+1/2). Now, since La/(1+1/z)=a>1, we can always assign an integral 2=2 value of 2, say z=7, such that, for that and all greater values of 2, Uz4,/U,>b, where b>1. We therefore have | De Se Uy+1/Uy > 0, Ur+o/Ur+1 = Ugh tee iar Ob = 86 1 20/01 ORC, CHAP, Hence, by multiplying all these inequalities together, we deduce Ug OP tty > POG Now w,/" is finite, and, since b> 1, l* can be made as great as we please by sufficiently increasing z. Hence L.w,= ©, on the c= supposition that 2 is always integral. But, since a is finite, it follows at once from (1) that L a*/a= ©, when # is unrestricted. w= The latter parts of the theorem follow by transformation. If we put a*=y, so that #=logay, and to x= © corresponds y= a0, we have o = Lat/z=L y/logay w= y=n Hence Ee loayy feel fame): 4, ='00 If we put a*=1/y, so that x= —log,y, and to “=o corre- sponds y= 0, we have = wje= — Li Liv log L=o y=+0 Hence L logy =— 1/0 =: y=+0 Example 1. Show that, if a>1 and x be positive, then L a*/a"=0 t=O” L SOB e we) ;, L. oPlogar=0, = s=-+0 ghee 6 ine (te V n, L=P ee = (ati) mae ; a v=A =o n— So) for, since a>1 and x is positive, we have aln>1, so that L(a!/")*/x= 00 and COPECO, , The two remaining results can be established in like manner, if we put y =logax in the one case, and y= — log,x in the other. It should be noticed that if n be negative we see at once that L a*/a”=0 ; ~ ; '‘2=2 menlog ria" —o; L eloggz= —a, L=n x=0 Example 2. If x be any fixed finite quantity, L a”/n!=0. N=PD Since 2 is to be made infinite, and x is finite, we may select some finite positive integer & such that x or <-1. . First, let m>—-1, then m+1 is positive. We can always find a finite positive integer / such that m+1(m+1jk+(m+1)/(R+1)+. . .+(m+1)/n, | by § 13, Cor. 5. Also, by $138, eeney 2, the limit of (m+1)/k+(m+1)/(K+1) +...+(m-+1)/n is infinite when n=. It follows, therefore, that LP=0, and therefore that L,,C,=0. Next, let m< —1, ‘say m= —(1+a), where a is a positive finite quantity. We may now write “Co yilteGte) (abe en Lee Or 29), =(-)"P, say. Now a a a log P= — log (1-; 5.) -log (1-5--)-. . .-log (---), >af(l+a)+a/(2+a)+...+a/(n+a), >afl+p)t+a/(2+p)+. . .+a/(n+p), where p is the least integer which exceeds a. But the limit of a/(1+ +a/(2+p)+...+a/(n+p) is infinite. Hence LP=o. When m= —1, -mOn=(-—1)", and the question regarding the limiting value does not arise. $ 16.] The fundamental theorem for the form 0° 1s that L at =1. r=+0 This follows at once from last paragraph ; for we have es re Lé logy _ out loge _ 0 2} aN = = =e =], a Example 1. L (a#)?=1. 2=-+0 For Lian * = Lane = L(at)" = (La*)" =1"=1. 4 Example 2. L at"=1 (n positive). ee 19 For La” = Let” logw — La” loge — 0], by § 15, Example 1. N.B.—Ifn be negative, L a*”=0" =0. c=-+0 88 THE FORM 0° CHAP. $17.| If wu and v be functions of x, both of which vanish when v=«a, and, are such that L v/u" =1, where n is positive and neither 0 v= nor ©, and 1 is not infinite, then Lu’ =.1, provided the limit be so z=a approached that u is positive.* For Ly" - L(u Te & (La pelea ls: Now, by § 16, Exam le 2, since 7 is positive L uw“”=1. Hence y § I I U=--0 Lu® = 1!=1. lf L v/u” =o, this transformation leads to the form 1° x= = and therefore becomes illusory. The above theorem includes a very large number of parti- cular cases. We see, for example, that, if Lv/w be determinate and not infinite, then Lu” =1. Again, since, as we shall prove in chapter xxx., every algebraic function vanishes in a finite ratio to a positive finite power of z—a, it follows that every such function vanishes in a finite ratio to a positive finite power of every other such function. Hence Lu®=1 whenever w and v are algebraic functions of 2. Example. Evaluate L{«-1+4/(a-1)} V@-)) when a=1. Here w=/(a-1){(V(a@-1)+V(2+a+))}, v= nf (ae 1) wPly= iA/(a@—1) +A/( ++ 1)} 7/8, 2/: 2); ye Hence La?*/y= i/3. Therefore Lu? = Lue ole pee § 18.] In cases where the last theorem does not apply, the evaluation of the limit can very often be effected by writing w” in the form ¢”'°s“, and then seeking by transformation to deduce the limit of vlogw from some combination of standard cases.t “Example. Evaluate a!/!08 @*-)) when 2=0. a SAM Sas suggested to attempt to make this depend on Sah 1)/x}=1. This may be effected as follows. We have cpl log (e* - 1) — ¢ log a/ log (e*- 1), * See Franklin, American Journal of Mathematics, 1878. t See Sprague, Proc. Hdinb. Math. Soc., vol. iii., p- 71 (1885). + At one time an erroneous impression prevailed that the indeterminate form 0° has always the value 1.—See Crelle’s Jowr., Bd. xii. - ————— ee XXxV THE FORMS «0°? 1” 89 Now log a Bee Bi log (e*-1) log {(@-1)/x} + log a 1 ~ log {(@—l/jx}floga+1 Since L log 1)/x} =0, by § 13, rie 4, and L log a= —o, we see that a a a iat Hence La log (e*—1) — ¢, § 19.] Since w= 1/(1/u)’, indeterminates of the form a ° can always be made to depend on others of the form 0°, and treated by the methods already explained. Example. Evaluate (1+2)/* when z=o., Let 1+2z=1/y, so that y=0 when x=; then we ae L( cS eae as AT ny =1/L(y”) V0 rT=02 Now Ly’=1 and L1/(1 aie ; hence L, +a)**#=1. , T=100) § 20.] The fundamental case for the form 1° is L (1 + 1/2)" an + L=0” . =L (1 +2)" =e, already discussed in § 13. A great variety of x=0 yt other cases can be reduced to this by means of the following theorem. 7 If wu and v be functions of « such that uw=1 and v= @ when a=a, then Lu” =e“), provided Lv(u — 1) be determinate. We have in fact ) aes {(1 oh Ea aes oe a) Hence, by § 7, Tie Ss L{ (1 a G1) i -1) ehh hs te ga 1) provided Lv(u— 1) be doemrinates Example 1. gh em Y= TL (1+e2-1e-Y=e. diff Example 2. Saas (1+log x)V@-) when w=1. We have 1=1(1 + log w)¥@-D=L{ (1 +log a) Vlog ® } log a(a—1), — gl log x((w-1). Now Llog x/(z-1)=L log 2¥@-) =log La@-D=loge=1. Hence l=e. TRIGONOMETRICAL LIMITS. § 21.] We deal with this part of the subject only in so far as it is necessary for the analytical treatment of the Trigono- metrical Functions in the following chapters. 90 TRIGONOMETRICAL INEQUALITIES _. GHAP. We shall require the following inequality theorems :— If x be the number of radians (circular units) in any positive angle less than a right angle, then ifs tan > 27> sin @ ; IL. _ a>sing>a—a’; inal 1>cosa>1-— 4a". If PQ be the are of a circle of radius 7, which subtends the central angle 22, and if PT QT be the tangents at P and Q, then we assume as an axiom that PT + TQ>are PQ > chord PQ. Hence, as the reader will easily see from the geometric defini- tion of the trigonometrical functions, we have or tan 2> Qra>2Qr sin ze; that. 1s, tang> GS S102, which is I. To prove IIL, we remark that sina = 2sin}zcos 92 = 2 tan dacos*4a = 2 tande(1—sin*}2). Hence, since, by L, tan $2 > 4a and sin $a < 42, we have sinw>2.da {1 —(42)}, >a — 42°. The first part of IIL is obvious from the geometric definition of cosz. To prove the latter part, we notice that cosa=1-2sin*4x; hence, by L, cos a> 1 — 2(42)’, >1— 42% § 22.) The fundamental theorem regarding trigonometrical limits is as follows :— If « be the radian measure * of an angle, then L (sin a/x) = 1. x=0 This follows at once from the first inequality of last para- graph. For, if e<47, we have tan > 2>sinZ; therefore secxz>2z/sing> 1. * In all that follows, and, in fact, in all analytical treatment of the trigono- metrical functions, the argument is assumed to denote radian measure. = Lesina/z, L tan w/a aa If we diminish « sufficiently, sec can be made to differ from 1 by as little as we please. Hence, by making ~@ sufficiently small, we can make «/sinz lie between 1 and a quantity differing from 1 as little as we please. Therefore Lz/sin a=. Hence also isin = 2 i Ltan 2/2 = 1. i) For L tan a/x = L(sin x/x)/cos x = L sin a/a x L1/eosa=1 x 1=1. Cor. 2. L sin : / = = L tan = i : =1 provided « is either a con- L=o L=a oy stant, or a function of « which does not become infinite when w= oe. This is merely a transformation of the preceding theorems. It should also be remarked that L (sin af =a0 (tan af)" = 1, ay ian] Po ae wl o provided «a and f are constants, or else functions of « which do not become infinite when z= a. If, however, a were constant, and £ a function of « which becomes infinite when z= «, then each of the two limits would take the form 1%, and would require further examination. § 23.| Many of the cases excepted at the end of last para- graph can be dealt with by means of the following results, which we shall have occasion to use later on :— If a be constant, or a function of x which is not infinite when a= ao, then x L (si 2 aa ly a sin or <1. If in (2) we put for p in succession 1, 2, 3, .. .,, m and add all the resulting inequalities we deduce eet or A) (17 + Oe. ee Hence | {(1 + 1/n)"2 — 1/nP*1}/(7 + 1)E (17 + 27 +e. tM) [nth ~1f/@ +1). That is to say, (17 + 27+... + ")/n"*1 always lies between 1 /(r He) and {(1+1/n)'t1—1/n*1}/(r+1). But L (1+1/n)jtt=1; and L 1/n’+1=0, since r+1 1s positwe. Hence the second of N=D the two enclosing values ultimately coincides with the first, and our theorem follows. It may be observed that, if r+ 1 were negative, the proof would fail, simply because in this case L 1/n7t!= a, Cor. 1. If s be any finite integer, and r + 1 be positive, Deer 2+... + (ns) hint = 1 /(r + 1). N=nN This is obvious, since L{17+ 27+. : . + (n—s)"}/n"+! differs from L(17+ 27+. . .+n”)/n"*+1 by a finite number of infinitely small terms. Cor. 2. If a be any constant, and r + 1 be positive, Pe ty + (at 2) +... t (etn) h/t =1/(r + 1). n=n This may be proved by a slight generalisation of the method ‘used in the proof of the original theorem. 94 DIRICHLET’S LIMIT CHAP, Cor. 3. If aand c be constants, and r + 1 + 0, L {(na +c)" + (na + 2c)"+. . 2+ (na + ne)"}/n | N= A = {(a +c) ti — attic gl), This also may be proved in the same way, the only fresh point being the inclusion of cases where 7 + 1 is negative. . § 25.] Closely connected with the results of the foregoing | paragraph is the following Limit Theorem, to which attention has been drawn by the researches of Dirichlet :— Tf a, 6, p be all positive, the limit, when n= ~% Be the sum of n terms of the series : + : + : + + : + (1) at? (a+ byt? (+o? . (a4+nby™ ta is finite for all finite values of p, however small; and, if Fl /(a+nb)'*? denote this limit, then n=1 L p S 1/(a + nb)'*? = 1/d (2). PHO 1 By means of the inequality (1) of last paragraph, we readily establish that fat+(p- 1b} a panes eee a+pb}~?"*> {a+ pb}? ae ea graf. | Putting, in (3), 0, 1, 2,..., successively in place of p,7| adding the resulting inequalities, and dividing by bp, we deduce (in, < saan ee 1 ia 1 bp ( fa—b}? fa+mnb}’} p=ola + phyte bp(% {a+(n+1)b}? (4). Since Li/{a+nb}’=0, and Ll/{a+(n+1)b}’=0, when n= a, we deduce from (4), 1 2 1 1 Rel Sieh D). ph(a—b)’ p=0(a+ pby'*? if pba? (5) From (5) the first part of the above theorem follows at once; and we see that 1/ph(a—b)’ and 1/pb°** are finite upper _| and lower limits for the sum in question. XXV GEOMETRICAL APPLICATIONS 95 We also have a 0 i 1 ——_>p } ———__ > — b(a — b)? BES (a+ pby'*? ba’ whence it follows, since L1/b(a— 0)’ = L1/ba’ = 1/b, when p= 0, that ] 1 ee (a +pby't? fe v 5 From the theorem thus proved it is not difficult to deduce the following more general one, also given by Dirichlet :— Tf ky, hoy. ~~ kn, . . . be @ series of positive quantities, no one of which is less than any following one, and if they be such that L T/t=a, where T is the number of the k’s that do not exceed t, t= 00 io/9) 1 : ; ae : then 21 /Ky *? 4s finite for all positive finite values of p, however 1 small ; and L oll [ent *? maeg p=0 1 Cor. It follows from (5) that 1 1 1 1 1 = > |, 1a + = es arenes fae (6), p(a—1)P n=ao la"? (a+1) (atn) '? J pa? an inequality which we shall have occasion to use hereafter. - GEOMETRICAL APPLICATIONS OF THE THEORY OF LIMITS. § 26.] The reader will find that there is no better way of strengthening his grasp of the Analytical Theory of Limits than by applying it to the solution of geometrical problems. We may point out that the problem of drawing a tangent at any point of the graph of the function y =/(x) can be solved by evaluating the limit when h=0 of {f(v+h)—f(x)}/h; for, as will readily be seen by drawing a figure, the expression just written is the tangent of the inclination to the axis of # of the secant drawn through the two points on the graph whose abscissae are a and a+h; and the tangent at the former point is the limit of the * See Dirichlet, Crelle’s Jour., Bd. 19 (1839) and 53 (1857) ; also Heine, - ibid., Ba. 31, 96 GEOMETRICAL APPLICATIONS CHAP, secant when the latter point is made to approach infinitely close to the former.* Example. To find the inclination of the tangent to the graph’ of y=e* at the point where this graph crosses the axis of . If 6 be the inclination of the tangent to the a-axis, we have tan 6 = L(e°+ — ¢°)/h, = L(e*-1)/A, — Alyy aa Hence 6=4rn. § 27.] The limit investigated in § 24 enables us to solve a problem in quadratures ; and thus to illustrate in an elementary way the fundamental idea of the Calculus of Definite Integrals. We may in fact deduce from it an expression for the area in- cluded between the graph of the function y=27/I’-1, the axis of v, and any two ordinates. _ Let A and B be the feet of the two ordinates, a, b the corresponding ~ abscissae, and b—a=c.t Divide AB into m equal parts ; draw the ordinates through A, B, and the 2-1 points of division ; and construct—Ist, the series of rectangles whose bases are the m parts, and whose altitudes are the Ist, 2nd, ..., th ordinates respectively ; 2nd, the series of rectangles whose — bases are as before, but whose altitudes are the 2nd, 3rd, ..., (n+1)th— ordinates. IfI,, and J, be the sums of the areas of the first and second series A of rectangles, and A the area enclosed between the curve, the axis of « and — the ordinates through A and B, then obviously I,— ("1h /(@—1)P—@— 1), a= 1. (am — 1)? — (am —1) (a —1)+(a"-1)? % 0.) Gem e em —1)@ 1+ era1e 7" (10.) {a-a/(a?—2")' /a®, «2=0. (Kuler, Diff. Calc.) RL 1S ) { A/(a+2) — V (a- a) } 1M a2) - J (a- 2) 9 ee 0. (12.) {(a2 + aa +02) ~ (a? — ae +22)2} | {(a +2)? — (a -x)t}, x=0. (Euler, Diff. Cale.) 4 (18.) {(2a%x— x) — a(a2x)3} | {a — (ax*\t}, aw=a. (Gregory, Examples in Diff. Calc.) - (14.) fa+/(2a?- 2am) — /(2a0—«*)}/{a-a+r/(a?-2*)}, x=a. (Kuler, Diff. “A (15.) e—a/(a?—y?), when z=, y=oo, but y*/a finite =2p. (16.) Da(y.— nity —#), v=Yy=z. ; (17.) Samy — 2)/DaeP(y? — a LY se, (18.) na”—3/(a"—a")-1f(w-a), x2=a. + (19.) 2%(qHl2” -1), w=0 (202) 2a. nao, l(t )*,; a=, (22. joa] (1 aa a 4 (23.) ee x= 0. (24.) (141/x)", w=0. (25.) av@-l?, o2=1. (26; ern) s ee 1 (27) /aF le; w=m. (28-) (loz a)", ..a==00 (29°). (lop afa)"*, 2—00. (30.) loge™a/log"x, x=. (31.) a*f(x), z=, where f(x) is a rational function of x, and a@ a con- Stant. (32.) (aa+bor-14+...)¥*, x=. (Cauchy.) (38.) gi/@t2logz) y=, (34.) {(a@?+a4+1)/(a?-x2+1)}*, w=. (35.) {4(@*+07)}¥*, x2=0. 3 (36.) {1-+2/x/(a?+1)} rl edge (Lanochadthal) Apt ayer. . -+a,U"\A0 +A w : : x =o. (Math. Trip., 1886. ie aoe oe) pA IB Dd VOL. II H > 98 EXERCISES VII CHAP. (38.) {1(et-D}¥=, x=. (39.) {log (1+a)}log +=) »—9, (40.) log (1+ax)/log(1+bx), wx=0. (41.) (e*-e-*)/log(1+z), w=0. (Euler, Diff. Calc. ) (42.) (44 - x) tan 2, n=5. (43.) tan—z/x, 2=0. (44.) (1-sin #+cosx)/(sinz+cos7—1), w=4n, (Euler, Diff. Cale.) (45.) sin a/(1—2?/n?), v=. (46.) a {cos(a/x)-1}, x=. (47.) (sinw-sina)/(w—a), x=a. (48.) secu—tanga, x=4n. (49.) (sin* — tan4x)/(1+cosx)(1—cosz), 20. (50.)* sinha/x, w«=0. (51.) (cosha-1)/x?2, x=0. (52.) tanh—z/a, a«=0, (53.) sinaw/log(1+a), w=0. (54.) snawloga, wx#=0. (55.) coswlogtana, w=4n. (56.) log tan mz/logtannx, «2=0. (57.) (log sin max — log «)/(log sinna -logx), «=0. (58.) singSim® 9, (59.) sina tan” y—9, (60.) (sinha) »—9, (61.) {(x/a)sin(a/x)}*"(m <2), w=o., (62.) (cos ma)"/=", x=0, (63.) (cos ma) COSC? mg), (64.) (2—ax/a)n72a vig (65.) log, (log .)/cos a tae: (66.) Show that sin x cot (a/x) log (1+tan (a/x)) has no determinate limit when =o, (67.) If 7,2a stand for loga (logax), 12x for logs (log, (logy2)), &c., show that L [L— {a?a/la?(w@+1)}™Jalgarlgee. . Uq?2=mM(lae)”. (Schlémilch, Alge- r=n braische Analysis, chap. ii.) s=N (68.) Show that L = (a+s)'"/n=1., t=o s=] s=Nn (69.) Show that L & {(a+s)/n}” lies between e* and eth, N= s=] s=nN » (70.) Show that L = {(a+1, and when a<1. (72.) Trace the graph of y=«'/* for positive values of «; and find the direction in which the graph approaches the origin. ee eee * For the definition and elementary properties of the hyperbolic functions cosh, sinha, tanha, &c., see chap. xxix. All that is really wanted here is coshz = 4(e* + e-*), sinhv=4(e* — e-*), XXV EXERCISES VII 99 (73.) Trace the graph of y=(1+1/x)*;.and find the angle at which it crosses the axis of y. (74.) Find the orders of the zero and infinity values of y when determined as a function of « by the following equations :—* (a) 2(x?-ay)?-y’=0. (Frost’s Curve Tracing, § 155, Ex. 3). (B) 20746 + a8y® — xy? + aby —a’e?=0. (Zb., Ex. 7.) (-y) (w= 1)yP+ (a2 — Ly? — (w— 2)°y +a(@- 2) =0. (75.) If « and v be functions of the integral variable n determined by the equations %p,=Un—1+VUn—1, Un=Un-1, Show that L wp/v,=(1£4/5)/2. How N=0 ought the ambiguous sign to be settled when 2% and 2% are both positive ? (76.) Show that : n\” (n-1\"-1 Qo NALINE n(n+1) nt+if — ae =a ae ae) (7) ( 2 ) Cea) (;)- 1 (77.) Show that L (m+1)(m+2).. . (m+n) Un N= 1 Fi 2, 2 Calica 7) * For a general method for dealing with such problems, see chap. xxx. CHAPTER XXVI. Convergence of Infinite Series and of Infinite ; Products. § 1.] The notion of the repetition of an algebraical operation upon a series of operands formed according to a given law presents two fundamental difficulties when the frequency of the repetition may exceed any number, however great, or, as it is shortly expressed, become infinite. Since the mind cannot over- look the totality of an infinite series of operations, some defini- tion must be given of what is to be understood as the result of such a series of operations; and there also arises the further question whether the series of operations, even when its meaning -is defined, can, consistently with its definition, be subjected to the laws of algebra which are in the first instance laid down for chains of operations wherein the number of links is finite. That the two difficulties thus raised are not imaginary the student will presently see, by studying actual instances in the theory of. sums and products involving an infinite number of summands and multiplicands. § 2.] One very simple case of an infinite series, namely, a geometric series, has already been discussed in chap. xx., $15. The fact that the geometric series can be summed con- siderably simplifies the first of the two difficulties just. men- tioned ;* nevertheless the leading features of the problem of infinite series are all present in the geometric series ; and it will be found that most questions regarding the convergence of infinite series are ultimately referred to this standard case. * The second was not considered, CHAP. XxvI CONVERGENCY, DIVERGENCY, OSCILLATION 101 The consideration of the infinite geometric series suggests the following definitions. Consider a succession of finite real summands 2,, 2, Us. - «; Uy « » 4 unlimited in number, formed according to a given law, so that the nth term w,, is a finite one-valued function of m; and consider the successive sums Soy, ee, ti, Dg ht et YU, Ae Uy umes at Un When » is increased more and more, one of three things must happen :— | | Ist. S,, may approach a fixed finite quantity S im such a way that by increasing n sufficiently we can make 8, differ from S by as little as we please; that is, in the notation of last chapter, L S,=8. In N =A this case the series Uk tle Fl OS ea is said to be CONVERGENT, and to converge to the value 8, which as spoken of as the sum to infinity. | Leet a Bering Le Pot os oe ces Here S =a a = 2. ee 2” iO and. S,, may increase with n in such a way that by increasing n sufficiently we can make the numerical value of S, exceed any quantity, however large ; that is, L S,= 0. In this case the series is said to be DIVERGENT. eta) Example. 1+2+3+... Here L S,=0. N=0 3rd. S,, may neither become infinite nor approach a definite laimat, but oscillate between a number of finite values the selection among which is determined by the integral character of n, that is, by such considerations as whether n is odd or even; of the form 3m, 3m + 1, 3m+2, &c. In this case the series is said to OSCILLATE. Example. 3—1-2+3-1-2+38-1-2+... Here L 8,=0, 3, or 2, according as 7 is of the form 3m, 3m+1, or 8m+2. prebes In cases 2 and 3 the series Ur Us tsb... runt’. is also said to be non-convergent. In many important senses 102 CRITERIA FOR CONVERGENCY CHAP. non-convergent series cannot be said to have a sum; and it is obvious that infinite series of this description cannot, except in special cases, and under special precautions, be employed in mathematical reasoning. Series are said to be more or less rapidly convergent according as the number of terms which it is necessary to take in order to get a given degree of approximation to the sum is smaller or larger. Thus a geometric series is more rapidly convergent the smaller its common ratio. Rapid convergency is obviously a valuable quality in a series from the arithmetical point of view. It should be carefully noticed that the definition of the con- vergency of the series Uy\+ Ugh Ug Ff ba Ry involves the supposition that the terms are taken successively in a given order. In other words, the sum to infinity of a conver- gent series may be, so far as the definition is concerned, dependent upon the order in which the terms are written. As a matter of fact there is, as was first pointed out by Dirichlet, a class of series which may converge to one value, or to any other, or even become divergent, according to the order in which oes terms are written. § 3.] Two essential conditions are involved in the definition of a convergent series—Ist, that S, shall not become infinite for any value of n, however great; 2nd, that, as n increases, there shall be continual approach to a definite limit S. If we introduce the symbol ,,R, to denote Up4.+Untot. . tees that is, the sum of m terms following the nth, we may state the following criterion :—- The necessary and sufficient conditions for the convergency of a series are that Sy, be finite for all values of n ¢ and that, by taking n sufficiently great, it be possible to make Ry as small as we please no matter what the value of m may be. That S, must be finite is obvious from the definition of convergency. Since L §,=S (where S is finite), therefore XXVI RESIDUE AND PARTIAL RESIDUE 103 Lo Shim=S. Hence L (Srim—Sp)=S=-8=0; that 1s, n=co Li Ry = 0. n=0 The two conditions are sufficient. For, since §,, is finite for all values of , the limit of S, cannot be infinite. Also the limit of S,, cannot have one finite value when nm has any particular integral character, and another value when n has a different integral character ; for any such result would involve that LS, N=0 and LS,+m should have different values; but such cannot be n=0 the case, since L(Sp4m— Sn) = LmRn = 0. The above criterion is often stated with the omission of the first part regarding the finiteness of S,, it being implied that the second condition L ,,R,, = 0 involves the first. Cauchy, the N=0 originator of the modern theory of convergence, states the matter in this way. The discussion of this subtle point need not be taken up here, because in most cases a slight alteration of the demonstration which proves that L ,,R,, = 0 shows that n=n S, is always finite. Cor. 1. In any convergent series L Uy, = 0. N=P For y,=Sn—Sn-1=1Ryn-1, and, by the criterion- for con- vergency, we must have L ‘R,-1=9. This condition, although N= 0 necessary, is not of itself sufficient, as will presently appear in many examples. Cor. 2. If Ry= Li Ry, and S and S,, have the meanings above m=n assigned to them, then S,=S — Ry. For Snim=Sn+ mRn, therefore L Sp4m=Sn + ete vette Ma=n” M=on L Sy+m=S, hence the theorem. R,, is usually called the residue m=n of the series, and ,,R, a partial residue. Obviously, the smaller R,/S» is for a given value of n, the more convergent is the series ; for R, is the difference between 8,, and the limit of 8, when n is infinitely great. R, is, of course, the sum of the infinite series U4; +Untot+ Unpo+..+; and it Is an obvious 1} Le ; : =* 5 4 i . 104 EXAMPLES CHAP, remark that the residue of a convergent series is itself a convergent Series. Cor. 3. The convergency or divergency of a series is not affected by neglecting a finite number of its terms. For the sum of a finite number of terms is finite and definite ; and the neglect of that sum alters L S, merely by a finite N=” determinate quantity ; so that, if the series was originally con- vergent, it will remain so ; if originally oscillating or divergent, it will remain so. Example 1. Consider the series 1/1+1/2+1/3+... + Ijan+... Here mRy=1/(n+1)+1/(n+2)+ ... +1/(n+m), >1f(m+m)+1f(nt+m)+ ... +1/(n+m), >m/(n+my), >1/(n/m +1). Now, however great n may be, we can always choose m so much greater that , n/m shall be less than any quantity, however small. Hence we cannot cause mln to vanish for all values of m by sufficiently increasing x. We therefore conclude that the series is not convergent, notwithstanding the fact that the terms ultimately become infinitely small. We shall give below a direct proof that LS,=0o. Example 2. 1 2 ih of fi f 1, AE SLOG OO ite a Helo 1 °87.3 7 2°89 4 n?n(n +2) Since (7 +1)?/n(n +2) =(1+1/n)/{1+1/(n+1)}, we have Ryo jopttimtt), 1 1 1+I1/(m+2) cee can ee | °81 + 1/(n +2) n+2 81 +1/(n +3) 1 1+1/(n+m) wre °8] +1/(m+m+1y fe Ao Led 1) 1+1/(n+2) 1+1/(n+m) - nly 81+ I/(n +2)” 81 + I/(n +3) vin USES CRE | i 1 1+1/(n+1) (1). n+l FF +1/(n+m+1) Now, whatever m may be, by making 7 large enough we can make 1/(n+1), and, @ fortiori, 1/(n+m-+1), as small as we please, therefore L miig=0 tora all values of m. Seat If in (1) we put 0 in place of n, and n in place of m, and observe that Snr=nRo, we see that Jl ae S,, < lost Tin +1) ? so that S,, can never exceed log 2 whatever n may be. - Both conditions of convergency are therefore satisfied. XXVI ELEMENTARY COMPARISON THEOREMS 105 Putting m= in (1), we find for the residue of the series Ry <[log {1+ 1/(n+1)} I/(m+1); a result which would enable us to estimate the rapidity of the convergency, and to settle how many terms of the series we ought to take to get an approxi- mation to its limit accurate to a given place of decimals. § 4.] The following theorems follow at once from the criterion for convergency given in last paragraph. Some of them will be found very useful in discussing questions regarding convergence. We shall use Xv, as an abbreviation for wu, + u, +. ..+Un+.. ., that is, “the series whose nth term is Up,” I. Tf Uy and vy, be positive, Uy, V, for all values of n, and Xp divergent, then Xu, is divergent, For, under the first set of conditions, the values of 8, and mk, belonging to Lv, are less than the values of the correspond- . . X/ / . . ing functions 8’, and ,,h’, belonging to =v, Hence we have 0S’, Hence, since L S’, = 0, we must also have L S,= 0; that is, Dw, is N= 0 N= divergent. Il. Jf, for all values of n, %,>90, and u,/v, as finite, then Lu, is convergent if Sv, is convergent, and divergent if Xv, 2s divergent. By chap. xxiv., § 5, if A be the least, and B the greatest of the fractions, Un+1/ Unt Jd Dae mes bre Einel Ua ins then U, == Wh ara ty elas A = n+1 n+2 n+m xz B. Now, since u,/v, is finite for all values of nm, A and B are finite. Hence we must have in all cases ,,R, =C,,R’,, where C is a finite quantity whatever values we assign to m and 2. 106 RATIO OF CONVERGENCE, ABSOLUTE CONVERGENCE cmap. Hence §,, (that is, ,R,) will be finite or infinite according as S’, 1s finite or infinite; and if L ,,R’,=0, we must also N=0 L min = 0. ao have v7 Ill. Jf u, and vy, be positive, and if, for all values of n, Un+i/Un Un+s/Uny and Wp is divergent, then Vy, is divergent. We have, if Un11/Un Un+i/ Un, and xv, be divergent, then Sw, is divergent. N.B.—In Theorems -I., II., III. we have, for simplicity, stated that the conditions must hold for all values of n; but we see from § 3, Cor. 3, that it is sufficient if they hold-for all values of n exceeding a certain finite value r; for all the terms up to the 7th in both series may be neglected. 7 It is convenient to speak of tp, [Un as the Ratio of Converg- ence of Yu, Thus we might express Theorem III. as fol- lows :—Any series is convergent (divergent) if its ratio of con- vergence is always less (greater) than the ratio of convergence of a convergent (divergent) series. IV. If a series which contains negative terms be convergent when all the negative terms have their signs changed, it will be convergent as. at stood originally. For the effect of restoring the negative signs will be to diminish the numerical value both of S,, and of ,,R,. XXVI GEOMETRIC STANDARD 107 Definition A series which is convergent when all its terms are taken positively is said to be ABSOLUTELY CONVERGENT. It will be seen immediately that there are series whose convergency depends on the presence of negative signs, and which become divergent when all the terms are taken posi- tively. Such series are said to be semi-convergent. In §§ 5 and 6, unless the contrary is indicated, we suppose any series of real terms to consist of positive terms only, and convergence to mean absolute convergence. SPECIAL TESTS OF CONVERGENCY FOR SERIES WHOSE TERMS ARE ULTIMATELY ALL POSITIVE. § 5.] If we take for standard series a geometric progression, say 2”, which will be convergent or divergent according as r< or >1, and apply § 4, Th. IL, we see that 2w,, will be con- vergent if, on and after a certain finite value of n, Up <71", where + <1; divergent if, on and after a certain finite value of N, Upn>7”", where r>1. Hence © I. Sup, is convergent or divergent according as U,'" is ultimately less or greater than unity. This test settles nothing in the case where u,\ is ultimately wnity. Example. 21/(1+ 1/n)”” is a convergent series ; for Ltt, P= 1/L(1 + 1/2)“ =e, N=0 by chap. xxv., § 13, where e>2, and therefore 1/e<1. If, with the series >” for standard of comparison, we apply § 4, Th. III., we see that Sw, is convergent or divergent accord- IN aS Upn+,/%y is, on and after a certain finite value of n, always 1. Hence II. Su» is convergent or divergent according as its ratio of convergency is ultimately < or > 1. Nothing is settled in the case where the ratio of convergency ts ultimately equal to 1. 108 EXAMPLES, INTEGRO-GEOMETRIC SERIES CHAP, The examination of the ratio w4+,/U, is the most useful of all the tests of convergence.* It is sufficient for all the series that occur in elementary mathematics, except in certain extreme cases where these series are rarely used. In fact, this test, along with the Condensation Test of § 6, will suffice for the reader who is not concerned with more than the simpler applications of infinite series. Notwithstanding their outward difference, Tests I. and II. are fundamentally the same. This will be readily seen by recalling the theorem of Cauchy, given in chap. xxv., § 14, which shows that Le un4,/u,= L up”. It is useful to have the two forms of Nn=e0 N=O0 test, because in certain cases J. is more easily applied than II. Example 1. To test the convergence of Sw", where 7 and z are constants. We have in this case Unti/Un=(n+1)'a"F/n7an, =(1+1/n)"x. Hence Lv,41/%,=2. The series is therefore convergent if a<1, and diverg- ent if x>1. Ifw=1, we cannot settle the question by means of the present test. Example 2. If ¢(n) be any algebraical function of n, Do(n)xz”™ is con- vergent if x<1, divergent if #>1. This hardly needs proof if L ¢(7) be finite. If L ¢(n) be infinite, we N=nN Nn=n know (see chap. xxx.) that we can always find a positive value of 7, such that L ¢(n)/n” is finite, =A say. We therefore have n=PD ‘ ante (n+1)/P(n), oe (Leet vig pee oe (n+1)? = ari al Se =v, This very general theorem includes, among other important cases, the integro-geometric series (1) t+ P(2)u*+. . .4+h(n)ar+ where ¢(7) is an integral function of n; and the series ee ihe ~ we etot... 1 rth Beer (1), which, as we shall see in chap. xxviii., represents (when it is convergent) * We here use (as is often convenient) ‘* convergence ” to mean “the quality of the series as regards convergency or divergency.” ———e eee. eee XXVI LOGARITHMIC AND BINOMIAL SERIES 109 -log(1—a). It follows, by § 4, Th. IV., that since the series (1) is con- vergent when z<1, the series u he a —_ \n-l a” Reps es! paar (2) is also convergent when 7<1. When (2) is convergent, it represents log (14+ 2). Example 3. Dx”/n! (the Exponential Series) is convergent for all values of x. Unsa/Un= {aH Yf(n +1) / far/n}} =2a/(n +1). Hence, however great x may be, since it is independent of n, we may always choose 7 so great that, for all values of n>r, #/(n+1)<1. Since the limit of the ratio of convergence is zero in this case, we should expect the converg- ency for moderate values of « to be very rapid ; and this is so, as we shall show by examining the residue in a later chapter. Example 4. =(-)"m(m-1)... (m—n+1)a"/n! (@ positive), where m has any real value,* is convergent if #<1, divergent if a>1. m—-N w+1 min —1 1+1/n For Littnya/U,= — cL =-a«L = Hence the theorem. The series just examined is the expansion of (l-a)™ when x<1. It follows, by § 4, Th. IV., that the series Dm(m—1)... (m—-n+1)a/n}, whose terms are ultimately alternately positive and negative, is convergent if <1; this series is, as we shall see hereafter, the expansion of (1+2)™ when 2=1. § 6.] Cauchy’s Condensation Test.—The general principle of this method, upon which many of the more delicate tests of convergence are founded, will be easily understood from the following considerations :— Let Sw, be.a series of positive terms which constantly de- crease in value from the first onwards. Without altering the order of these, we may associate them in groups according to some law. If 2,, U2, .-+ Um)-- + be the Ist, DATS ice CUIAB SPE oa 811 these groups, the series 2vyp, will: contain all the terms of Lwp,; and it is obvious from the definition of convergency that Lun is convergent or divergent according as vm 1s convergent or * If m were a positive integer, the series would terminate, and the ques- tion of convergency would not arise. 110 CAUCHY’S CONDENSATION TEST ; CHAP, divergent ; we have in fact L Su,= L Xv, Itis clear that the N=0 mMm=n convergency or divergency of >v,, will be more apparent than that of Sw,, because in >”, we proceed by longer steps towards the limit, the sum of n terms of Sv, being nearer the common limit than the sum of n terms of Sw». Finally, if Sv’, be a new : : . convergent . series such that v’,, 2 v,, then obviously Sw, is if Dv’, divergent Be overeent * divergent ; We shall first apply this process of reasoning to the following case :— Example. The series 1/1+1/2+...+1/n+.. . is divergent. Arrange the given series in groups, the initial terms in which are of the following orders, 1, 2, 27, ... 2™, 2m+1, . . . The numbers of terms in the successive groups will be 2—1, 27~—2, 23-22,.. . gmtl_Qm, Omt2_ Omir, respectively. Since the terms constantly decrease in value, if 2”+! be the ees oan of 2 ees does not exceed n, then g i es eee a I 1 4 1 ee =f oy TETse at 5) te - Ml Gar pect one Fees) a : "Gta 0 2 + (241 — Qm) ——_ eee 1 vee #3 > 15 2 aieter's 2? m Fice Hence, by making n sufficiently great, we can make S,, as large as we please. The series 1/1+1/2+1/3+. .. is therefore divergent. This might also be deduced from the inequality (6) of chap. xxv., § 25. Cauchy’s Condensation Test, of which the example net dis- cussed is a particular case, is as follows :— If f(n) be positive for all values of n, and constantly decrease as n increases, then =f(n) is convergent or divergent according as Sa" f(a”). . as convergent or divergent, where a is any positive integer + 2. The series =f(n) may be arranged as follows :— [fl)+...+fla-1)] + {fla)+fat1)+...+fl(a-1)} | + {f@)+f@+1)+...4+f(-1)} : «Flam) ++ fa +1)+ | - + a1 — 1) XXVI CAUCHY’S CONDENSATION TEST RE? Hence, neglecting the finite number of terms in the square brackets, we see that =/(n) is convergent or divergent accord- ing as 2 {fam + flam+ 1+... +f(amt—1)} (1) is convergent or divergent. Now, since f(a”)>f(a"+1)>... >f(a"* — 1)>f(a™+), we have ee Se an) > fam) ef +1). ot flat! 21) (er? as hua | Cr hikaad that is, (a —1)a™fla™)>fla™) + fam +1)+... +f(a™t1-1) > 4(a—ljia} ata). ) Hence, by § 4, Th. I., the series (1) is convergent if S(a—- 1): af(a”) is convergent, divergent if Y{(a—1)/a}a™+ f(a) ig divergent. Now, by § 4, Th. IL, (a- 1)a™f(a™) is convergent if Za™fa™) is convergent, and Y{(a-1)/a}a™*f(a"+1) is divergent if Ya™*1f(a™*1) is divergent; and for our present purpose zanfia™) and xa™ttf(am™+t) are practically the same series, say 2a"f(a"). Hence Cauchy’s Theorem is established. N.B.—Ii is obviously sufficient that the function f(n) be positive and constantly decrease for all values of n greater than a certain finite value 1. Cor 1. The theorem will still hold if a have any positive value not less than 2. ; Let a lie between the positive integers b and b+1, (b<2). If 2a"f(a") be convergent, then L a”f(a”) = 0, that is, L af(x) = 0. n= L=O Hence, on and after some finite value of 2, the function f(z) will begin to decrease constantly* as z increases. We must therefore have (b+ 1)"f{ (b+ 1)"}a”/(a") is convergent, a fortiori, will X(b + 1)” Ji (> +1)" } be convergent, and therefore, by Cauchy’s Theorem, >f(n) will be convergent. If 2a" f(a”) be divergent, xf(x) 1° may, or 2° may not decrease as # increases. * This assumes that #f(x) has not an infinite number of turning values ; -so that we can take x so great that we are past the last turning value, which must be a maximum. 112 CRITERIA OF DE MORGAN AND BERTRAND CHAP. In case 1°, f(b") > a” f(a"). Hence the divergence of 2a”/(a”) involves the divergence of Lb"f(b") ; and the divergence of >f(n) follows by the main theorem. In case 2°, the divergence of =f(n) is at once obvious ; for, if L af(z) +0, then ultimately 2f(x)>A, where A>0. Hence r= f(a)>A/x. Now ZA/n is divergent, since 21/n is divergent ; therefore >f(n) is divergent. In what follows we shall use ev, ¢7, ... to denote a”, a”. . ., a being any positive quantity 2; and Az, A’a,. . lx, Pa, .. . to denote log,2, log, (log,2),. . . log.a, log,(log.2), . . ., where ¢ is Napier’s Base. Cor. 2. Sf(n) is convergent or divergent according as Yene'n. . . enfle"n) is convergent or divergent. This follows, for integral values of the base a, by repeated application of Cauchy’s Condensation Test ; and, for non-integral values of a, by repeated applications of Cor. 1. Thus =/(n) is convergent or divergent according as Yenf(en) is convergent or — divergent. Again, Yenf(en) is convergent or divergent according as Dene(en)f {e(en)}, that is Lene nf(e'n), is convergent or divergent ; and so on. Cor. 3. Sf(n) is convergent or divergent according as the first of the functions T= Af T; = M af(x) }/Aa, ToS Al cAeyfle) yn Tp Ad PAGAN so eR aie Ae which does not vanish when x= «©, has a negative or a positive limit. By Cor. 2, =/(n) is convergent or divergent according as Senen ... &nf(e"n) is convergent or divergent. Now the latter series is (by § 5, Th. I.) convergent or divergent according as Lf{enén... &nf(en)}™1; n=” that is, according as L log, {enen . . . enf(e'n) }"<>0; N=0 | XXVI DE MORGAN’S LOGARITHMIC SCALE 9%: that is, L: logy {ene'n . . . e'nf(en) }/n<>0. N=0 If we put «=e"n, so that Av =e-In, We=e-™, . -ltr=en, Me=n, and 7=0 when n=, the condition for convergency or divergency becomes : LA {arwra .. . dN -laf(x)}/Ma<>0 (1). rw=0 If, on the strength of Cor. 1, we take ¢e for the exponential base, the condition may be written Evipiers . . U-tafle) ties 0 (2), w=o0 where all the logs involved are Napierian logs. We could establish the criterion (2) without the intervention of Cor. 1 by first establishing (1) for integral values of a, and then using the theorem of chap. xxv., § 12, Example 4, that L "e/I"x = 1/la. L=0 Cor. 4. Hach of the series D1 /nit~ (1), D1 /n {in} i+ (2), 21/nin { Pn jr (a) Zl /niniPn . . . la {in ite (r + 1), is convergent if a>0, and divergent if a =or <0. As the function nlni’n . . . I’m frequently occurs in what follows, we shall denote it by P,(n); so that P,(n) =n, P,(n) = nin, &c. Ist Proof—Apply the criterion that Yf(n) is convergent or divergent according as Li { P,(z)f(x)}/’+1x<>0. In the pre- sent case, f(x) = 1/P,(x) (Ix). Hence 1 {P,(alf(a) }/ x = 1A f(Uay fH, ee It follows that (r+1) is convergent if a>0, and divergent ifa<0. Ifa=0, the question is not decided. In this case, we must use the test function one order higher, namely, 1 P4(a) f(a) }/'+?x. Since f(x) = 1/P,(z), we have VOL, II I 114 DE MORGAN'S LOGARITHMIC SCALE CHAP. L{ Phas (n)fta) yee = 1 ey =i Faby Hence, when a= 0, (7 + 1) is divergent. 2nd Proof.—By the direct application of Cauchy’s Condensa- pus Test, the convergence of (1) is the same as the convergence Sa"/(an)'+, that is, S(1/a*)”. Now the last series is a geo- ore progression whose common ratio is 1/a%; it is therefore convergent if a>0, and divergent if a= or <0. Hence (1) is convergent if a>0, and divergent if a= or <9. Again, the convergence of (2) is by Cauchy’s rule the same as the convergence of a”/a” { la” }4+«, that is, D1/(Ja) tent? ; and the convergence of this last the same as that of 21 jaite Hence our theorem is proved for (2). Let us now assume that the theorem holds up to the series (r). We can then show that it holds for (r+ 1). In fact, the ~ convergence of (7 + 1) is the ‘same as that of La" alae", oe U1" {ran }+, that is, D1/(nla)l(nda). . .0-?(nla) (-*(nla)} OF, First suppose a>0, and a>e. Then la>1, nla>n. Hence 1/(nla)U(nda) . . . I~ ?(nla) {°-1(nla) jo <1 iting, th ens ey. But, since a> 0, 21/P,_,(n) {l”~‘n }* 1s convergent, a fortwo, ~1/P,(n) {in }* is convergent. Next suppose a} 0, and 2 0, and a’ is very large. | xxvI DE MORGAN AND BERTRAND’S SECOND GRITERION 115 If we put «=I1"-'n, we may write L alta Lief get e)f N=0 w=c lx}**«, Hence, however small a, so long as it is greater than 0, f / and however large a’, Li'y/t,= 0. If we suppose the character of the logarithmic scale estab- lished by means of the second demonstration given above, we may, by comparing 2wv, with the various series in the scale, and using § 4, Th. I., obtain a fresh demonstration of the criterion of Cor. 3. We leave the details as an exercise for the student, This is perhaps the best demonstration, because, apart from the criterion itself, nothing is presupposed regarding f(z), except that it is positive when a is greater than a certain finite value. By following the same course, and using § 4, Th. IIL, we can establish a new criterion for series whose ratio of con- vergence is ultimately unity, as follows, where pz =f(x + 1)/f(a). Cor. 5. If f(x) be always positive when a exceeds a certain Jimte value, Xf(n) is convergent or divergent according as the first of the Jollowing functions— To Px ly 7, = P,(@ + 1)pe — P,(z) ; oni Pix + 1)p, - lei : ty = ere Ge “t 1) pz = Epa). which does not vanish when % = o has a negative or a positive limit. Comparing =f(n) with 51/P,(n){i'n}*, we see that > I} (n) will be convergent if, for all values of x greater than a certain finite value, Pa 0. Now (1) is equivalent to : P,(@ + 1) pa — P,(x) < P,(2)[{ Ua/i"(a + 1) }* — Dy Also LP,(x)[{ l"a/I"(@ + 1)}*- 1] i s Ve {lala +1)}*-1 — LP,_,(a){0"(a + 1) — Fa} - a4), none Tae =-lxlxa=-—a, by chap. xxyv., §§ 12 and 13. af 116 EXAMPLES CHAP. Hence a sufficient condition for the convergency of =f(m) is L {P,(% + 1)pz — P,(a#)} < - (a positive), ant <0. In like manner, the condition for divergency is shown to be L {P,(@ + 1)px—- P(x} > —« (a negative), eet ==(); Example 1. Discuss the convergence of cel hs ss stele Here To=1{f(n)} /n, laa) 2a -+1fn+riln n Now, by chap. xxv., § 13, Example 1, 1+(rt1)ma14+1/2+.°. . +1/n+rin>rint+ln+1). Hence LT)=0. We must therefore examine T;. Now T,=L{nf(n)} /ln, = — {141/2+. . .+1/n+(r—1)in}/in, = — {141/2+. . .4+1/n}/in-(r-1). By chap. xxv., § 13, Example 2, L(1+1/2+...+1/n)/in=1. Hence LT,=—-1-r+1l=-7. The given series is therefore convergent or divergent according as 7> or <0. If r=0, LT)=0, and LT;=0. But we have T,=L{nlnf(n)\ /Pn, =1-{141/2+. . .+1/n—In}/Pn. Now, when 2 is very large, the value of 1+1/2+. . .+1/n-In approaches Euler’s Constant. Hence LT,;=1>0. In this case, therefore, the series under discussion is divergent. Example 2. To discuss the convergence of the hypergeometric series, a6, aat1).A(8+1) y.o | y+). 5(5+1) The general term of this series is ae (a+n—1).B(B+1). . .(B+n-1),, ~a(ytl). . . (ytn—-1).4(5+1). . . (6+0-1) ‘ . The form of f(n) renders the application of the first form of criterion somewhat troublesome. We shall therefore use the second. We have | _(a+n)(B+n),, Pa= (en) (8+) _(a+n) (B+) a (y +n) (6+) Lr=x- Ly M+... 1+ b] TO ’ Hence the series is convergent if «<1, divergent if #>1. If z=1, Lro=0, and we have XXVI HYPERGEOMETRIC AND BINOMIAL SERIES 117 h _(n+1)(a+n)(B+n) = (y+n)(6-+n) : _(a+B—y-5+1)n?+An+B. iA, n+On+D P Lr =a+B-y-6+1. If, therefore, ~=1, the hypergeometric series is convergent or divergent according as a+B-—y—5+1< or >0. Ifa+B-y-6+1=0, L7;=0. But we have (a+n) (B+) (y+) (d+n) =[n(n+1)— ln} +(a +841) {Un+1)-In} + {Al(n +1) + Bln} /n + Cl(m +1)/n?]/[1 + E/n + F/n?], Hence, since Ln{i(n+1)-im}=1, L{Un+1)-In '=0, Li(n+1)/ns=0, Lin/ns=0(s>0), &c., we have T2=(n+1)l(n+1) —nin, lre=T> 0: In this case, therefore, the series is divergent. Example 3. Consider the series m m(m—1) p. Sea enon tT: rete 1) This may be written eee Vt) (—m +1) | ,(=m)(~m+1). : gle CORE yi 1 9 - he eee m(m—-1).. SUAS D hee Fie a oe It is therefore a hypergeometric series, in which a= — Ms Bee Oe #=1. It follows from last article that the series in question is convergent or divergent according as -m<>0, that is, according as m is positive or negative. ‘ “re This series is the expansion of (1-—x)”, when 2=1. Example 4. Consider the series m m(m-—1) m(m—1).. .(m-—n+1) 7) ab pe ta Sas tr seen mmmemnde PUR Ta A) In this series the terms are ultimately alternatively positive and negative in sign. Hence the rules we have been using are not directly applicable. 1st. Let m be positive ; and let m-~, be the first negative quantity among m, m-1,m-2,.... &c., then, neglecting all the terms of the series before the (r+1)th, we have to consider mm—1)...(m—-r-+1) m-7r (m-7r)(m-r-1) Oe toe [eer (r+ 1)(r+9) } @. If we change the signs of the alternate terms of the series within brackets, it becomes r—-m (r—m)(r—m+1) Pe lsb gre) 2 (3) Now (3) is a hypergeometric series, in which a=r— m, B=y, d=r+l1, *=1. Hence a+B-y- 6+1=r-—m-(r+1)+1=-m<0. Therefore (3) is convergent. Hence (2), and therefore (1), is absolutely convergent, 1+ 3 ee 118 HISTORICAL NOTE CHAP, ond. Let m be negative, =-—p say. The series (1) then becomes Bw, petl) | avn Me td)... (4 +01) Lele 1c Be .. - +(-1) ao kee ar) res Since « is positive, the hypergeometric series wh eMuet1) Awl)... (uta 1) 1+5t+-19 SEC NINE Bo RTI + ae (5), is divergent. Hence (4) cannot be absolutely convergent in the present case. Since pyp= —(u+n)/(n+1), the terms will constantly increase in numerical value if ~>1. Hence the series cannot be even semi-convergent unless w<1. . If « be less than 1, pa<1, and the series will be semi-convergent provided U,n=0. Now log Un= = log ce n p-1l Tey = 2 los { ae \. Since Llog {1+(u—1)/(n+1)}/{(H—-1)/(n+1)} =1 (see chap. xxv., § 18), | the series Dlog {14+(u—1)/(n+1)} and 2(u—-1)/(r+1) both diverge to an . infinity of the same sign. But the latter series diverges to —# or +o, according as #< or >1. Hence Lw,=0 or , according as “< or ols Hence the series (1) is divergent if u>1, semi-convergent if «<1. It obviously oscillates if w= 1. Hence, to sum up, the series (1) is absolutely convergent, if 0-m< +0; semi-convergent, if —-l yahoo vencd =an ee (1). Compare this with fhe ie : - ; 43 1 i 1 = +5) +7 Z - (545 titeia-(gotm)t-- (2), that is, mh series whose (27 — oe term is 1/(32-2), and whose (2n)th term is — (1) (32 ~1)+1/3n). If S, S,’ denote the sums of 2 Wes of (1) and (2) respectively, then Sn 2a = Son 23; San— 1=Son -1- 1 7t (82 —1),Ssn =Seon’. Since L1/(8n —1)=0,; we have in all cases LS,=LS,’. Hence (1) is convergent or divergent according as (2) is convergent or divergent. That (1) is really divergent may be shown by comparing it with the series = {1/(3n — 2) —1/(8n - 1) — 1/8n} (3). If 8,” denote the sums of ” terms of this last series, we can show as before that LS,”=LS,. But the nth term of (3) can be written in the form (-9+12/n —2/n?)/(3 — 2/n) (8 —1/n)3n ; and therefore bears to the nth term of 1/n a ratio which is never infinite. But 21/n is divergent. By § 4, II., (3) is therefore also divergent. Hence (1) is divergent. It should be noticed that in the case of an oscillating series, where Lu, +0, the grouping of terms may convert a non-convergent into a convergent series; so that we cannot in this case infer the convergency of the original from the convergency of the derived series.* * This remark is all the more important because the converse process of splitting up the nth term of a series into a group of terms with alternating signs, and using the rules of § 8, often gives a simple means of deciding as to its convergency. The series 1/1.2+1/3.4+1/5.6+1/7.8+ ... may be tested in this way. XXV1 Uy — Ug t+ Us—Ut+ - - 121 Example. 1\2 1\2 1\2 Ley? (145) - (145) sk +(1455) -(145-55) eo is obviously a non-convergent oscillating series. But { (45) e3)} + (042) -(24))} of (4d) 1 acl Cr) Noe aes _ whose nth term is (877+ 87 + 1)/(4n2+2n)?, that is (8+ 8/n + 1/n?)/16(1 + /2n)?n2, { ) is convergent, being comparable in the scale of convergency with 21/n?. § 8.] The following rule is frequently of use in the discus- | . . . ° sion of semi-converging series :— 1 FFU, >U,>U,> .. . >U,>... and all be positive, then ome —. . . (— oa, (iy te ee (1) | converges or oscillates according as L u, = or +0. N=0 Using the notation of § 3, we have min = + (Un —Untet. . = Un+-m)s + {Unti- Unte— Unts)—+ » «}, + {(Un4i— Unts) + (Unts— Unt) t. . .} Il ' Hence we have Un+1 = Pant > Un+ti— Un+te (2), numerical values being alone in question. — If, therefore, Lu, = 0, _we have Lu,4, = Luy.,=0; and it follows that L nivis Ulorall N=0 ‘values of m. Also U, > PAK = Sn > U, — Us, so that S,, is finite for all values of n. The series (1) is there- fore convergent if Lu, = 0. itd, —a+0, then L ,,R,=« or =0 according as m is odd N= or even, Hence the series is not convergent. We have, in fact, LSsn41 — Son) = Litton; = 0, which shows that the sum of the series oscillates between S and S +a, where S=LS 2n° Cor. The series | (a, —U,) + (Us — U,) +. . . + (en; — Won) +. where U,, Ua, . . . are as before, is convergent. | | | PS 122 ABEL’S INEQUALITY CHAP. Example 1. The series =(-1)"-1/n is convergent, notwithstanding the fact, already proved, that 21/n is divergent. Example 2. >(—1)"""(n+1)/n is an oscillating series ; but 2(-1)-? {(n+1)/n —(n+2)/(v+1)} is convergent. § 9.] The most important case of periodic series 1s 2a,C0s (nO +p), where dy is a function of n, and ¢ is independent of 2, commonly spoken of as a Z’rigonometrical or Fourier’s Series. The question of the convergence of this kind of series is one of great importance owing to their constant application in mathematical physics. We observe in the first place that L Jf Sa, be an absolutely converging series then 2p cos(nO + ) is convergent. 3 This follows from § 4, I. Il. If 06=0 or Qkr (k being an integer), Vay cos (nO + p) 428 _ convergent or divergent according as Za, 1s convergent or divergent. This is obvious, since the series reduces to 2dp cos ¢. Ill. If 6+0 or Qher, then Say, cos (nO + ) is convergent if for all values of n greater than a certain finite value, Up constantly decreases as n increases, in such a way that L a, = 9. n=” This is a particular case of the following general theorem, which is founded on an inequality given by Abel :— IV. If Sup be convergent or oscillatory, and ay, M2) + + +> ny + + + be w series of positive quantities, constantly decreasing as n wmereases, so that L ay, = 0, then Yann is convergent. N=P Abel’s Inequality is as follows :—If, for all values of n, A>U, + Ugt. . o+Un> B, where %,, Us; ., Un are any real quantities whatever, and ., dy be a series of positive quantities constantly decreasing as 7 increases, then a, A > Ay + Agllg +. « + Ann > 4,B. This may be proved as follows -—Let §8,=u,+U@+. ..+ Um and S! = Ay, + Ag+... + Onn Then %=8, %= S.-8, &e; eS” -_- | XXVI TRIGONOMETRICAL SERIES 123 Ba =4,9, + 0,(9,-9,) +... 4 an(Sn—- Sn.) | = §\(@; — a2) + S,(dg— dy) +. 2 + Sn -1(Gn-1 — Un) + Spdn. _ Hence, since §,, 8,, . . ., S, are each B and (a, — dt), (2-5), » » +5 (An-1 — Mm), Mm are all positive, {(@, — Gy) + (@,— 4,3) +. . .+ (ny —- An) + dy A =n > {(@, - ad.) + (dg— G3) +. . 4 (Qn-1 — Gn) + On} B; that is, aA>S8,/>a,B (1). ‘Theorem IV. follows at once, for, since Sw, is not divergent, |S, 1s not infinite for any value of n. Hence, by (1), S,’ is not | infinite. Also, by Abel’s Inequality, / An+1C a min = Ont iUn+y Zs An+tolnte Peters aaa Soke Cn+muntm / =8,) +m — ae An+.D (2), where C and D are the greatest and least of the values of | ain ( = Un 4-1 if Un+te AF ta eee + Un+tm — Sn ca Sn) for all different positive values of m. Now, since Sw, is convergent or oscillatory, Snim—Sn is either zero or finite, and L Qn4,=0, by hypo- n=n thesis. Therefore, it follows from (2), that L ,,R,’=0 for all values of m. Hence Sap2p is convergent. We shall prove in a later chapter that, when Un = cos (nO + ¢), Sn = sin $70 cos {h(n + 1)0 + '/sin £0. ‘If therefore, we exclude the cases where 6=0 or Qh cir, We see that S, cannot be infinite. Theorem III. is thus seen to be a particular case of Theorem IV. Cor. If ay be as above, S( - 1)" la, cos(nO + ), San sin (nO + ), and X(—1)"-1a, sin (nO + ) are all convergent. CONVERGENCE OF A SERIES OF COMPLEX TERMS. § 10.| If the nth term of a series be of the form In + Ynt, where 7 is the i imaginary unit, and a, and y, are functions of n, we may write the sum of n terms in the form S,, + T,2, where i ' ee 124 CONVERGENCE OF COMPLEX SERIES CHAP. Sn Hm + y+. . 2+ hn, = Ua eee anne By the sum of the infinite series X(, + Yi) is meant the limit when n= 00 of S,,+T,t; that is, (LS,) + (LT,)z. The necessary and sufficient condition for the convergency of Z(Xp + Yni) ts therefore that Yay, and Xyp, be both convergent. For, if the series Sz, and Sy», converge to the values S and T respectively, 2(a+Ynt) will converge to the value 8+ 71; and, if either of the series Sw», Ly, diverge or oscillate, then (LS,,) + (LT,,)¢ will not have a finite definite value. | $11.] Let z, denote %,+Yni; and let pp be the modulus and 6, the amplitude of z,;* so that 2, =pn(cos 6, +7 sin On), In = pn COS On, Yn = pn Sin On. We have the following theorem, which is sufficient for most elementary purposes :— The complea series Szp is convergent if the real series > mod zy, 1 convergent. For, since Sp, is convergent, and py, by its definition is always positive, it follows from § 4, I, that Zp,cos6, and Spnsin 0, are both convergent; that is, 2x, and Ly, are both convergent. Hence, by § 10, =z, is convergent. It should be noticed that the condition thus established, although sufficient, is not necessary. For example, the series (1 -1)/1-(1 -1)/2+ (1 -1)/3—. . . is convergent since 1/1 — 1/2 +1/8-...and —1/1+1/2-1/3+. . . are both convergent; but the series of moduli, namely, /2/1+ /2/2+ /2/3+.. 4, is divergent. When Sz, is such that = mod zn is convergent, Xz, 1s said to be absolutely convergent. Since the modulus of a real quantity wp is simply wu, with its sign made positive, if need be, we see that the present definition of absolute convergency includes that formerly given, and that the theorem just proved includes § 4, IV., as a particular case. Cor. If Ay be real, and za complea number whose modulus is not infinite for any value of n, however great, then =(XAn%m) will be absolutely convergent if XA, is absolutely convergent. * See chap. xii, § 13. XXVI LAW OF ASSOCIATION FOR SERIES 125 For mod (A,,2,) = mod A, mod zp ; and, since SA,,"is absolutely convergent, YmodA,,_ is convergent. Hence, since mod z,, is cys finite, =(mod A,, mod z,,) is convergent by § 4, II. ; that is, Ymod(A,zp) is convergent. Hence X(Anzn) 18 ateclitely | convergent, Example 1. The series =z"/n! is absolutely convergent for all finite values of z. Example 2. The series 2z”/n is absolutely convergent provided mod z<1. | Example 3. The series (cos 0+ sin 0)"/n, is:convergent if 0+0 or 2kr. For the series = cos n0/n and > sin nO/n are convergent be §-9 LUE: | Example 4. The series (cos +7 sin 0)"/n? is absolutely convergent. For the series of moduli is 21/n?, which is convergent. | APPLICATION OF THE FUNDAMENTAL LAWS OF ALGEBRA | TO INFINITE SERIES. ! § 12.] Law of Association.—We have already had occasion to observe that the law of association cannot be applied without imitation to an infinite series. It can, however, be applied without limitation provided the series is conver gent. For let Sm denote the sum of m terms of the new series obtained by associating the terms of the original series into groups in any way Re aver. Then, if S, denote the sum of » terms of the original series, we can always assume m so great that S,, includes at least all the terms in S,. Hence §,,’-—S, = pp, where p is ‘a certain positive integer. Now, since the original series is con- vergent, by taking n sufficiently large we can make pp as small as we please. It follows therefore that L S,,’=L Sn. Hence m= N=0 the association of terms produces no effect on the sum of the mfinite ‘convergent serves. § 13.] Law of Commutation.—The law of commutation is even more restricted in its application than the law of association. In fact, the law of commutation can be appled only to absolutely convergent serves. __ We shall consider here merely the case where each term of the series is displaced a finite number of steps.* Let Su, be * See below, § 33, Cor. 2 126 LAW OF COMMUTATION FOR SERIES CHAP. the original series, w,,/ the new series obtained by commuta- tion of the terms of Suv,. Since each term is only displaced by a finite number of steps, we can, whatever » may be, by taking m sufficiently great always secure that §,,’ contains all the terms of S, at least. Under these circumstances §,,’ —S, con- tains fewer terms than ,R,, where p is finite, since m is finite. Now, since Sw, is absolutely convergent, even if we take the most unfavourable case and suppose all the terms of the same sign, we shall have L pk, = 0; and, a fortiori, L Sm’ -L S,=0. M=OA N=0 N=P Hence L S,,.’=L 8,; which establishes our theorem. M=A N=0 The above reasoning would not apply to a semi-convergent series, because the vanishing of L,R, does not depend solely on the individual magnitude of the terms, but partially on the alterna- . tion of positive and negative signs. Riemann has shown that the series =(-—1)"~!u,, where Lu, = 0, and Luyn+, and Yu, are both divergent, can, by proper commutation of its terms, be made to converge to any sum we please ; and Dirichlet has shown that commutation may render a semi-convergent series divergent. Example 1. The series dia 15 21 CU Gale oie ea Jl Af2° AB n/h 6 taf Qutb) -~/Gae is convergent by § 8; but the series ( he aa Nae ee Nil AUB. APL) NAIOP IN Ovo eta 1 1 1 (4m +1) /(4m+38) (m+ y te (2), which is evidently derivable from (1) by commutation (and an association which is permissible since the terms ultimately vanish), is divergent. For, if m= 1//(4m + 1) + 1/r/(4m + 8) - L//(2m+2), and %m=1/r/m, then Litm/Um = L41//(4 + 1/m) + 1/r/(4+38/m) - 1/a/(2 + 2/m) } =1/2 + 1/2 — 1/n/2= 1-4/2. Hence %»/vm is always finite ; and Zv, is divergent, by § 6, Cor. 4. Hence Zw» is divergent. (Dirichlet.) Example 2. The series joi ee (al 1.25.3 4). 5s eee nd 20 ee (1), PSY caer 6 Spee i! 1 it ($+5)-5+ (547) “at Bn +(Gesitaes) mat co's = A2F are both convergent ; but they converge to different sums. For, by taking | . j | | XXVI LAW OF DISTRIBUTION FOR SERIES 127 ] | successively three and four terms of each series, we see that the sum of (1) lies between *583 and *833 ; whereas the sum of (2) lies between ‘926 and 1°176. { | Addition of two infinite series. If Sm and Lvy be both con- | vergent, and converge to the values S and T respectively, then X(tn + Up) _ ts convergent and converges to the value S + T. | We may, to secure complete generality, suppose w, and v,, to be complex quantities. Let S,,, T,,, U, represent the sums of | terms of Yun, Loy, X(Un + Vp) ‘respectively ; then we have, how- ever great n may be, U,=S,+T7T,. Hence, when n = om | LU,, =LS,, + LT,,, which proves the proposition. | § 14.) Law of Distribution.—The application of the law of | distribution will be indicated by the following theorems :— Tf a be any finite quantity, and Su, converge to the value S, then | 2dUn, converges to as. The proof of this is so simple that it may be left to the “reader. Tf Xun and Xv, converge to the values S and T respectwely, and at least one of the two series be absolutely convergent, then the series UV, + (UV, + Ugd,) +... + (thtn + Ube + Un) ats EEL) | converges to the value ST.* | Let S,, T,, U, denote the sums of » terms of 2, ais NMA, Ugty 1 +... + Unv,) respectively ; and let us suppose that 2p is absolutely convergent. We have | Pas = Us ay Ly where Lin = UVp + Uy t . . . + Unde | + UsUn +... + Und, sr UnVn = Uy + Us(Up + Up)... + Un(Un +... +%,) (2). “ The original demonstration of this theorem given by Cauchy in his Analyse Algébrique required that both the series 2Un, Zn be absolutely con- vergent. Abel’s demonstration is subject to the same restriction. The more | general form was given by Mertens, Crelle’s Jour., lxxix, (1875). Abel had, however, proved a more general theorem (see § 20, Cor.), which partly in- cludes the result in question. 128 THEOREM OF CAUCHY AND MERTENS CHAP. If therefore n be even, = 2m say, i Fi — [ UeVom a8 ts Vem. + Vom - :) pegver i. Un (Yom + : oe Um-+s) | + [Qenei(Yem toa Uma) b-\ - + om Vent +2 40s) Am If.n be odd, =2m+ 1 say, LL, “ [ tYem-+, + Ug Verna 2 Ven) See hae Urn (Gene thee Um-+-s) | Fl ina Oana hares st Omepalito aa emi amit > + a) Now, since Sv, is convergent, it is possible, by making m sufficiently great, to make each of the quantities modmy mod (Yn—1 + Yem)) » + +» mod Cr Vem), 000 Yom MOG: (Vom +Vem4i)) - > + mod (Omagt + + + + Vem-+1) 28 small as we please. Also, since mod T,, mod T,, mod T,,... mod T,, . am are all finite, and mod (T,.— T,) + tam) > + +3 mod (v,+ . . - +Vem)s THO (Vypy/ng ee ce amia)y) > es mod (v,+ . . » +%sm-+1); are all finite. Hence, if «,, be a quantity which can be made as small as we please by sufficiently increasing m, and 2 a certain | finite quantity, we have, from (3) and (4), by chap. xu, § 11, @ mod L, <¢,(mod wu, +modw,+.. . +mod Um) | + B(mod U4) + MOd Unie t+ . + + + mod Up). If, therefore, we make x infinite, and observe that, since Duyp, is absolutely convergent, mod w, + modu,+ .. . + mod Uy, is” finite, and L(modw,»4, + Mod Unte+ . - + + mod w,) = 0, we have (seeing that Le, =0) L mod L,=0. Hence LS,T, = LUp, that is, LU, =ST. , 4 Cauchy has shown that, if both the series involved be semi- convergent, the multiplication rule does not necessarily apply. | Suppose, for example, u,=Un= (-—1)°/4/n.. Then both Yu, and Lv, are semi-convergent series. The general term of (1) is wa =( Lasers ss Gs + = + d ) (5) w= t( Siar tyme} 11 m= 1)} Em} 3 Now, since 7(n—r+1)=}(0+1)?- (a(n +1)- r\?, therefore, for all values of 7, 7(v—7r+1)<}(n+1)*, except in the case where 7=3(n+1), and then there is equality. It follows that mod Wn>n/3(n+1)>2/(1+1/n). The terms of Xw, are therefore ultimately numerically greater than a quantity which is infinitely nearly equal to 2. Hence 2w, cannot be a convergent series. XXVI RADIUS AND CIRCLE OF CONVERGENCY 129 SPECIAL DISCUSSION OF THE POWER SERIES Dayz”. § 15.] As the series 2a,2" is of great importance in algebra- ical analysis we shall give a special discussion of its properties as regards both convergence and continuity. We may speak of it for shortness as the Power Series, and we shall consider both ay and « to be complex numbers; say dy = Tn(COS an +7 S1N ap), “= p(cos 0 +7sin 6), where 7, and a, are functions of the integral variable , but p and @ are independent of n. § 16.] Lana” is convergent if mod x {f(n,v+h)—f(n,x)} is - convergent, and has a definite finite value; but this value is not necessarily zero for all values’ of x. Suppose, for example,* that f(n,v) =a/(nx+1)(na—a2+1). Since f(n,x) = na/(nx + 1)- (n —1)z/{n-1a+1}, we have, in this case, S, =na/(na + 1). Hence, provided «+0, LS,=1. If, however, x=0 then 8, =0, however great n may be. The function ¢(2) is, therefore, in this case, discontinuous when z= 0. The discontinuity of the above series is accompanied by another ‘peculiarity which is often, although not always, asso- ciated with discontinuity. The Residue of the series, when «+0, is given by Rn = 1-8, = 1/(na + 1). Now, when z has any given value, we can by making m large enough make 1/(n# +1) smaller than any given positive quantity a. But, on the other hand, the smaller z is, the larger must we take m in order that 1/(na +1) may fall under a; and, in general, when 2 is variable, there is no finite upper limit for n, independent of x, say v, such that if n>v then R,vmod R, {x/n(na + 1) (na — @ + 1) — a?/(na? + 1) 1 (na? — a2 + 1)} that infinitely slow convergence may not involve discontinuity. In point of fact, the sum of this last series is _ always zero, even when z=0 ; and yet, when a=0, the con- _ vergence is infinitely slow. The object of the present paragraph has been, not to intro- _ duce the student to a discussion of exceptional cases in the _ functionality of infinite series, but to lead him to see the neces- sity of the demonstrations now to be given of the continuity of the Power Series in certain cases. § 19.] As regards the power series 2a,2" there are two cases of great practical importance—I1st, when G 18 independent of # and we regard Sa,” as a function of a, say (a); 2Qnd, when a, is a function of n and y, say f(ny), and x is’ regarded as constant, so that (n,y)e” is a function of Y, say V(y). The points involved were first raised and discussed by Abel ; but the following theorem, together with its elegant demonstra- tion,+ will give us at once all that is here required, Let pn be independent of z, and W,(2) be a single valued function of om and 2, finite for all values of n, however great, and finite and con- _ tinuous as regards z from z=a to z= b, then, if Zum be absolutely convergent, ZpnW,(z)t is a continuous function of 2 from z=a to z=, | / Let S,(2) = wyw,(z) + pyw(z)+. . . + Prvr(2), and assume py, _to be positive for all values of », which will not limit the “ The distinction here involved was first pointed out by Seidel, Abhi. d. Bayerischen Akad. d. Wiss., Bd. v. (1850). It has assumed great import- ance in the Theory of Functions developed by Weierstrass and his followers. + Both due to Du Bois-Reymond. See Math. Ann., iv. (1871). We have | presented the original notation and phraseology as closely as possible. _ + Under the circumstances supposed, Zu,,w,(z) is, of course, convergent _by § 4. 132 DU BOIS-REYMOND’S THEOREM CHAP. generality of the demonstration, since Sp, is absolutely con- vergent. Let Aw, = w,(2 + h) — wp(2), so that L Aw, =0 for all values h=0 of p. Then we have S,.(2 + h) — Sn(2) = pr Am, + poAWe +. - » + Pm AWm > + Pane Want? © I) + Pansies? + A) ee + pin Wy(2 + fh) oe Pan-+1Wm-+i(2) is Pm-+2Wm-+a(2) Tt ae = pin Wp (2) Let AW,, be a mean among Aw,, Aw, . « »; Aw» (that is, be greater than the least and less than the greatest) ; W' nn & Mean among Wns +h), Wm+a(% +h), - + + Wy(z +h) 5 Winn @ Mean among Wmri(2)> Wmt2) - - + Wy(2). Then Sn(z + h) — Sn(2) = AWS’ + CW Gages W win) nina where 9’, and n-mR’m have the usual meanings as regards 2pm. WLS, , If, now, we make n infinite, Wm. and Wm. become, by virtue of our hypotheses, finite determinate quantities for every value of m and z; and we have S..(2 + h) — S..(2) = AW S'm + (W' ince — Wie) R'm: Our object is to prove that L {S,, (2 + h) — Sao (2)} (Say, Sun (# + 0) h=0 _§,,(2))=0. Now when i= 0, Aw, = 0, Aw,=0,. . ., AWm=9, since all the functions ~,(z), w.(2), . - + Wm(Z) are continuous. Therefore, since 8’, is finite for all values of m, owing to the convergency of 2pm, we have S..(z + 0) — So(2) = (LL W'mnco — Wace \R peal h=0 We cannot be sure that LW'mo = Wm; but since both are finite their difference is finite. Hence, since R’,, is the residue of the convergent series 2pm, by making m sufficiently great we can make R’,, and therefore the right-hand side of (1) as small as we please. It follows that the left-hand side of (1) must be numerically less than any assignable quantity ; that is, must be zero. We have supposed all the quantities involved to be real ; but the extension to the case where pn and w,(x) are complex is obvious after what has been said in §§ 10, 11. r a P| XXVI CONTINUITY OF POWER SERIES 133 As an example, take the first of the series discussed in§18. The nth term may be written {1/n*} {a/(a + 1/n)(@-a/n+1/n)}. Hence, if we take a= 1/n*, W(x) =2e/ (v+1/n)(e-a/n+1/n), we see that all the conditions of Du Bois-Reymond’s Theorem are fulfilled, except when 2=0 plore (ay 1) which becomes infinite when a=0. We conclude therefore that this infinite series is a continuous function of x for all positive values of x except x=0, in which case the theorem does not apply. Cor. 1. If the power series Sayx" be absolutely convergent when mod x = R, then, for all values of x such that mod x < R, Layx” is a con- tinuous function of 2. We have a,x" =a,R"(2/R)". Now Ya,R” is an absolutely convergent series by hypothesis. Hence, if we take pn =a,R®, W(x) = (x/R)”, all the conditions of Du Bois-Reymond’s Theorem will be satisfied, and the corollary follows. Cor. 2. If the power series Yf(n,y)a” be convergent when mod «= (<1), and f(n,y) be a function of y which is finite and stngle- valued for all values of n, and finite, single-valued, and continuous as regards y from y=a to y=b, then, from y=a to y=b, W(y) = >f(n,y)x" 1s a continuous function of y so long as mod > R. This follows at once from Du Bois-Reymond’s Theorem, if we take pp =a", z=y, and w,(z) = f(n,y). § 20.] We have seen that, so long as x lies within the circle of convergence, the power series Ya,x” is a continuous function of z Nothing, however, has been established regarding values of « that lie on the circumference of the circle of convergence. Hence the importance of the following theorem of Abel’s, which we prove for real series, but which can at once (see § 10) be ex- tended to imaginary series. [f the series Lay, be convergent, and if Layx” be convergent for all values of x less than 1, then Le Sanz = Say, %=1-0 This is tantamount to asserting the continuity of Sa,” up to the circumference of its circle of convergency, so far as real values are concerned.* If f(a) denote Sanz” we have to show that L f(x), say (1-0), = Day. x=1-0 * Proofs have been given by Abel, Dirichlet, Du Bois-Reymond, and others. The above is a modification of Dirichlet’s demonstration (see Liouville’s Jour. ) 134 ABEL’S CONTINUITY THEOREM CHAP. Since Sa, is convergent, if Ss) =a, $,=W +My» + +) Sn =A + G43", l Yag. . . then % 3°» +) Sm... tena enn have for their limit 8, the sum of the infinite series 2d». Also we have dy = Sp) 0, = S, — Spy Mg = Sg— 81)» + +) On =Sn— Sn-y * Hence f(%) = Sy + (8, — 8)@ + (Se — 8,)0 +.» + (Sn — Sn 1)" + =s(1—2)+sa(l—a) +... +52"1—-a2)+... This transformation will be legitimate so long as @ is less than 1, by however little. Let «= 1—&, then we have, however small € may be, F(l-§ =s€ + (1 - SE +. - «+ 8n-i(1 — €)"7E + 8 (1 — €)"§ + ata = Eye es wherein 7 ay be taken as large as we please. Let now o’, be a mean among %, (1 — &)s,, .. + (1 — €)"""8n-1 and on a Mean aMONE Sp, Sntiy Sn+a > + = Then Lo’, is finite, and Lo, =s. We have fl - 8 =ngo'n+ (1+(1-94+(1-£)'+. . FE Pom = néo'n + (1 — €)"on, Since m may be made as large as we please, we may cause € to approach the limit zero by putting €=1/n’, and then making m= 00. Hence f(1 - 0) =Lo’,/n + LQ - 1/n')"on. Now, Lo’,,/n = 0, since Lo’, is finite. Also L(1 -1/n)"=L{(1—1/n’)-™ }-¥* = Lee eae and Lo,=s. Hence L(1 -1/n’)"on=s; cae we have finally f(1-0)=s, = athens It should be observed that a, need not be absolutely con- vergent, but if it be semi-convergent the order of its terms must not be altered. By considering the series 2uw,2”, Sv,z", and the series L(Y + Un Vet.» + U%p)e"t1, which is their product when both of them are absolutely convergent, and applying the theorem just established, we easily arrive at the following theorem, also due to Abel. XXVI INDETERMINATE COEFFICIENTS ioe, Cor. If each of the series Stn and Sv, converge to the limits wu and v respectively, then, if the series 2 Utit Uy, © Vado ave U,Un) be convergent, it will converge to uv; and this will hold even uf all the three series be only semi-convergent. § 21.] Principle of Indeterminate Coefficients. Tf the real series Xanax” be convergent for all values of « such that mod z>R, and if for all the values in question Ay + Layx” = 0, then @=0,0,=0,a,=0,.. 4. a,=0,... Sin¢e a,z is convergent, it follows that L >a,v”=0. %=0 Since a+ 2a,2"=0 when z=0, we must have a,=0. There- ~ fore, by our original hypothesis, we have Sa,” = 0 for all values of « such that modz+R. Now, by § 14, Sa,z" = LUA,0"-1, where Za,v"-* is a convergent series for any value of « which renders 2a,2” convergent. Since, then, we have 2>a,v"-!=0 for values of « other than 0, it follows that Sa,,2”-1= 0. But, since Ya,av"-1 is convergent, L Sa,z"-1=a,. Thus we must have a, = 0. x=0 Proceeding in this way, we can show that all the coefficients must vanish. Cor. Lf, for all values of x such that mod xR, ay + Sane” = b+ Xbnw”, both series being convergent, then ay = he = Ue Casals te a For we must have (a, — 0,) + d(dy — bn)a” = 0 where, by § 13, 2(an — bn)u” is a convergent series. Hence, by the main theorem, a —b,=0, a,-—b,=0, &c. INFINITE PRODUCTS. § 22.] The product of an infinite number of factors formed in given order according to a definite law is called an Infinite Product. Since, as we shall presently see, it is only when the factors ultimately become unity that the most important case arises, we shall write the nth factor in the form 1 + Um. By the value of the infinite product is meant the limit of (1 +4,)(1 + uw). ... (1 + un), (which may be denoted by II(1 + w,), or simply by P,,), when n is increased without limit. 136 ZERO, CONVERGENT, DIVERGENT, CHAP. It is obvious that if Lu, were numerically greater than unity, then LP,, would be either zero or infinite. As neither of these cases is of any importance we shall, in what follows, suppose modu, to be always less than unity. Any finite number of factors at the commencement of the product for which this 1s not true, may be left out of account in discussing the converg- ency. We also suppose any factor that becomes zero to be set aside ; the question as to convergency then relates merely to the product of all the remaining factors. Four essentially distinct cases arise— 1st. LP, may be 0. ond. LP, may be a finite definite quantity, which we may denote by II(1 +»), or simply by P. olds Utaiuay be infinite. 4th. LP, may have no definite value; but assume one or other of a series of values according to the integral character of n. In cases 1 and 2 the infinite product might be said to be convergent ; it is, however, usual to confine the term convergent to the 2nd case, and to this convenient usage we shall adhere ; in case 3 divergent ; in case 4 oscillatory. . § 23.] If, instead of considering P,,, we consider its logarithm, we reduce the whole theory of infinite products (so far as real positive factors are concerned *) to the theory of infinite series ; for we have log P,, =log (1 +,) + log (1 +m) +... +log (1+ Un) N = Zlog (1 + Un); and we see at once that 117 Ist. If Slog (1 + wp) is divergent, and LE log (1 + up,)= - ©, then II(1 + w,) =0; and conversely. ond. If Slog(1+%,) be convergent, then I(1 + Un) 18 con- vergent. nN 3rd. If Slog (1 + wy) is divergent, and LY log (1 + w,)= + 2, then II(1 + w,) is divergent. , ee ee —N * The logarithm of a complex number has not yet been defined, much less discussed. ——— eee aa we Lad XXVI AND OSCILLATING INFINITE PRODUCTS 137 4th. If Slog (1 + w,) oscillates, then II(1 + u,) oscillates. § 24.] If we confine ourselves to the case where wu, has ultimately always the same sign, it is easy to deduce a simple criterion for the convergency of II(1 + wp). Tf Lu,<0, then = log(1 +-w,) = — ©, and II(1 + w,) = 0. Tf Lu, > 0, Dlog (1+ u,) = + 0, and II(1 + w,) is divergent. It is therefore a necessary condition for the convergency of I1(1 + un) that Lu, = 0. Since Lu, = 0, L(1 + Un, in =e; hence Llog(1 + Up)/u, = 1. It therefore follows from § 4 that log (1+ u,) is convergent or divergent according as uw, is convergent or divergent. More- over, if wv, be ultimately negative, the last and infinite parts of Yu, and Y log (1 +u,) will be negative; and if uw, be ultimately positive the last and infinite part of wu, and Ylog (1 +u,) will be positive. Hence the following conclusions— * If the terms of Yup, become ultimately infinitely small, and have ultimately the same sign, then Ist. II(1 +) is convergent, if Duy, be convergent ; and con- versely. 2nd. II(1 + u,)=0, if Xu, diverge to — © ; and conversely. 3rd. II(1 + u,) diverges to + ©, if Yu, diverge to +o; and conversely. Since in the case contemplated, where wu, is ultimately of invariable sign, the convergency of II(1 +.w,) does not depend on any arrangement of signs but merely on the ultimate magnitude of the factors, the infinite product, if convergent, is said to be absolutely convergent. It is obvious that any infinite product in which the sign of uy, is not ultimately invariable, but which is convergent when the signs of Up» are made all alike, will be, a fortiori, convergent im its original form, and is therefore said to be absolutely convergent ; and we have in general, for infinite products of real factors, the theorem that Il(1 + Up) ts absolutely convergent when Yup is absolutely con- vergent ; and conversely. Cor. If either of the two infinite products TI(1 + uw»), T1(1 - up) be absolutely convergent, the other is absolutely convergent. 138 _ INDEPENDENT CRITERIA CHAP. For, if Sw, is absolutely convergent, so is >( — tn) 5 ; and con- versely. Example 1. (1+1/12)(1+1/2?). . . (1+1/n?).. . is absolutely conver- gent since 21/n is absolutely convergent. Example 2. (1—1/2)(1-1/3)... (1-1/n) . . . has zero for its value since (—1/n) diverges to —©. Example 3. (1+1//2)(141//8) . . . (1+1/a/n) . . . diverges to +0 since 2(1//n) diverges to +. Example 4. (1+1/a/1) (1-1//2) (1 +1/n/8) ( )(1—1/s/4) .. . Since the sign of w» is not ultimately invariable, and since the series 3(- 1)"-1/4/n is not absolutely convergent, the rules of the oe, paragraph do not apply. We must therefore examine the series = log (1+(-—1)""1//n). The terms of this series become ultimately infinitely small; craters we may (see § 12) associate every odd term with the following even term. We thus replace the series by the equivalent series LV log {1+ 1/a/(2n - 1) — 1//(2n) — 1/a/(4n? - 2 ' It is easy to show that 1/a/(2n — 1) — 1/a/(2n) - 1/r/(4n? — 22) <0, for all values of n>1. Hence the terms of the series in question ultimately become negative. Moreover, 1/4/(20-1)-1/a/(2n) - aes (4n?—2n) is ulti- - mately comparable with -1/2n. Hence Zlog(1+(- ek” diverges to —o. The value of (1+1/a/1)(1—1/a/2)(1 + 1//8)(1-1/n/4). . « is there- fore 0. This is an example of a semi-convergent product. Example 5. eltlg-1-3,l+3,-1-4 .. . The series Zlog(1+u,) in this case becomes (1+1)-(1+4)+(1+4)-(14+2)+ which oscillates. The infinite product therefore oscillates also. Example 6. II(1—a”-1/n) is absolutely convergent if #<1, and has 0 for its value when «=1. § 25.] We have deduced the theory of the convergence of infinite products of real factors from the theory of infinite series by means of logarithms ; and this is probably the best course for the learner to follow, because the points in the new theory are suggested by the points in the old. All that is necessary 1s to be on the outlook for discrepancies that arise here and there, mainly owing to the imperfectness of the analogy between the properties of 0 (that is, + a—a) and 1 (that is, x a+a). It is quite easy, however, by means of a few simple inequality theorems,* to deduce all the above results directly from the definition of the value of II(1 + w,). * See Weierstrass, Abhandlungen aus d. Functionenlehre, p. 203 ; or Crelle’s Jour., Bd. 51. ‘ XXVI COMPLEX PRODUCTS 139 If P,, have the meaning of § 22, then we see, by exactly the same reasoning as we used in dealing with infinite series, that the necessary and sufficient conditions for the convergency of II(1 + w,) are that P,, be not infinite for any value of n, however large, and that L (Prim — Pn) = 0. Nn=0 If we exclude the exceptional case where L P,,=0, then, NnN=0o since P,, is always finite, the condition L (Paim—Pn)=0 is N= equivalent to L (Paim/Pn—1)=0, that is, LPasm/Pn = 1. If, therefore, we denote (1 + n+4,)(1 + Uns)... (1 + Unsm) by mQn, We may state the criteria as follows— The necessary and sufficient conditions for the convergency of I(1 + up) are that P,, be not infinite for any value of n, however large, and that LQ, = 1. Nn=0 Tf Un be complen, then the two conditions obviously (see chap. xii.) are that mod P,, be not infinite for any value of n, however large, and that L mod (mQn—- 1) = 0. We shall not stop to re-prove the results of § 24 by direct deduction from these criteria, but proceed at once to complete the theory by deducing conditions for the absolute convergence of an infinite product of complex factors. § 26.] (1 + u,) is convergent if T1(1 + mod un) is convergent. Let p,=mod wu,, so that p, is positive for all values of n, then, since I(1 + p,) is convergent, Li(1 + pnti)(1 + pnts). - - (1 +pnim)-1}=0 (1). Now mn —1=(1 + Un4:)(1 + Unss). . . (14+ Untim)—1, Seti ouys Ung tits. Unaaa te). 5 Unda Hence, by chap. xii., §§ 9, 11, we have OP mod (Qn — 1) 2pnti + ZPntiPnte t+ «+ + Pn+iPn-te+ ++ Patms pee Gata) (Ur nisl (LF Pam) L Hence, by (1), L mod (,,Q, — 1) = 0. = Sil 140 ASSOCIATION AND COMMUTATION CHAP, | Also mod P,, = mod (1 + ,)(1 + %,). . . (1 +n), . = mod (1 +w,)mod(1+u,).. . mod(1 + Un) $(1+p,)(1 + ps). . « (1 + pn)- Hence mod P,, is finite, since II(1 + p,) is convergent. Remark.—The converse of this theorem is not true; as may be seen at once by considering the product (1 + 1)(1—4)(1 + 3) (1-4). .., which converges to a finite limit +0; although (L+1)(1+4)(1+4)(1 +4). . . is not convergent. When (1+) is such that IL(1 + mod Un) 1s convergent, TI(1 + tp) is said to be absolutely convergent. If II(1 + Un) be con- vergent, but I1(1+modu,) non-convergent, I1(1 + Un) 1s said to be semi-convergent. The present use of these terms includes as a particular case the use formerly made in § 24. § 27.] If Xmodu, be convergent, then I1(1 + Un) is absolutely convergent ; and conversely. For, if Smodw, be convergent, it is absolutely convergent, seeing that mod wu, is by its nature positive. Hence, by § 24, II(1 + mod w,) is convergent. Therefore, by § 26, I(1 +p) is absolutely convergent. Again, if II(1 + up) be absolutely convergent, II(1 + mod w,) is convergent ; that is, since mod w, is positive, II(1 + mod w,,) is absolutely convergent. Therefore, by § 24, 2modw, is abso- lutely convergent. Cor. If Sun be absolutely convergent, T1(1 + up) is absolutely convergent, where x is either independent of n or is such a function of n that Lmoda+ «© whenn=o. | Example 1. II(1—2”/n) is absolutely convergent for all complex values such that mod a<1, but is not absolutely convergent when modw=1. Example 2. II(1-—«/n?), where « is independent of n, is absolutely con- vergent. § 28.] After what has been done for infinite series, it is not necessary to discuss in detail the application of the laws of algebra to infinite products. We can at once deduce the fol- lowing results— I. The law of association may be safely applied to the factors of II(1 + Up) provided Lu, =0 ; but not otherwise. | XXVI CONTINUITY OF INFINITE PRODUCT 141 II. The law of commutation may be safely applied to I(1 + un), provided it be absolutely convergent, but not in general otherwise. Ill. Jf both W(1 +») and Il(1+w',) be absolutely convergent, then IL {(1+wp)(1+u'n)} is absolutely convergent and has for its limit {TI(1 + wp)} x {T(1+w'n)} 3 also IL {(1 + up)/(1 + w'n)} ts absolutely convergent, and has for its limit {TI(1 + up) }/{ 00 +'n)} provided none of the factors of I1(1 + w'n) vanish. Since Slog {1 + prWn(2) } = DpnWn(2) log {1 + pnp (z) rrr. If p, and w,(z) satisfy all the conditions of Du _ Bois- _ Reymond’s theorem, given in § 19, we have Lynw,(z) = 90, t Llog {1 + ppwp(z) Hn = 1, and w,(z) log {1 + pnWn(2) } ihren satisfies all the conditions imposed upon w,(z) alone. Hence — Ylog {1+ pyw,(z)} is a continuous function of z from z=a to z=b. Hence IV. Jf p, and w,(2) satisfy the conditions of § 19, (1 + pnwn(2)) is a continuous function of 2 from z=a to z=. Cor. 1. If Sanu” be convergent when mod a = R, then H(1 + apx”) converges to h(x), where f(x) is a finite continuous function of « for all values of « such that mod x as ie Pwint1 | (5) ; LW nga z Wnts : (6). § 31.] Residue of an Infinite Product. Let us consider the infinite products, TI(1 +») and Il(1— wp), im which Un becomes ; ultimately positive and less than unity. If the series 2, converge in such a way that the limit of the convergency-ratio pp is a positive quantity p less than 1, then it is easy to obtain an estimate of the residue. Let Q,, Q’, denote the product of all the factors after the mth in II(1 +w,) and II(1 — up) respectively, so that Q,>1, and Q',<1. We suppose 2 so great that, on and after ”, %, is positive, p, less than 1, and either (a) pp» never increases, or else (b) py never decreases. In case (a), 2p falls — under case (1) (a) of last paragraph ; in case (b), Du, falls under { case (1) (0) of last paragraph. We shall, as usual, denote the residue of Su, by R,; and we shall suppose that n is so large — that mod R,, < 1. : Now (by chap. xxiv., § 7, Example 2), Qn = (1 +Ungi) (1 + tnge) © + > Le Uy ti Units bs Soe eat pt (1). Q'n a (1 WE Un+i) (1 i Un+s) a 9 ol eae (2). Also, 1/Qn = {1 - Unga/(1 + Unai)b {1 - Unto] (1th as) eee >l- Un+i/(1 a= Un-+1) = aaa | @ a Un+-3) = os Mees > L-Unti-Unte—+ + 4 >1—-R,. : Whence Q,- 11+ Unaal(l - Un+1) tT Uatel(l men Un+-2) sare rt >l]+ Untit Untet-. + sy ; Slay. ' Whence 1 Qlger Ral lena) (4). XXVI DOUBLE SERIES DEFINED - 147 From (1), (2), (3), and (4) we have Ry < Qn -1< R,/(1 oT R,,) (5) ) R,,/(1 * R,) <1-Q’, CL): Un ,1 Un, 2 Un ,3 Un,4 Ie a ea Un,n Un, n+l Pe K’ —- Li y Van | Abn, 2) en 8 | ak | ears Um,n | Um,nt a N Um-AL, 1 Um+1,2 |Um+1,3 | Um, 4 Um+i,n|Umt1,ntl| ++ > | ) It may, however, happen—lst, that both these operations lead to an infinite value; 2nd, that neither leads to a definite — value; 3rd, that one leads to a definite finite value, and the XXVI SUM OF A DOUBLE SERIES — 149 © other not ; 4th, that one leads to one definite finite value, and the other to another definite finite value.* In all these cases we say that the series is non-convergent for the first way of summing. | Second Way.—Sum to n terms each of the series formed by taking the terms in the first m horizontal rows of (1); and call OOM Sd Aa Tin,n- Define on Ap Te an + Ake Wet eas lhe. (2) as the finite sum. Then, supposing each of the horizontal series to converge to T,, T.,. . ., Tm respectively, and >T,,, to be a convergent series, define SP SE sed Det ee ad MA ea Ba (3) as the sum to infinity in the second way. Third Way.—Sum to m terms each of the series in the first m columns; and let these sums be UE ase itu tenes al yrene aes Define } Shae - U; ym + lope 4 RI Coss Uh U, ym (4) as the finite sum. Then, supposing these vertical series to converge to U,, Us, me. .U, respectively, and =U, to bea convergent series, define 87=U,+U,+...+U,+...adoo (5) as the sum to infinity in the third way. So long as m and 1 are finite, it is obvious that we have eae = an nS pee ; so that, for finite summation, the second and third ways of summing are each equivalent to the first. The case is not quite so simple when we sum to infinity. It is clear, however, that cae oL ehh: (6) ; and sos L nae Sinn} Alias N=0H M=0 ee eee ee a a ea Eee * Examples of some of these cases are given in § 85 below. 150 DOUBLE SERIES OF POSITIVE TERMS CHAP. ‘so that 9’ and 8” will be equal to each other and to S when the two ways of taking the limit of S,,, both lead to the same definite finite result.* Fourth Way—Sum the terms which lie in the successive diagonal lines of the array, namely, AA’, BB’, CCl} eis and let these sums be D,, D,, / . ., Dn4i respectively ; that is, D, = 1, Ds = the + Ue,ty - + + Das: = tin + tenes 4 Define Ses De Darna eae (8) as the finite sum; and, supposing =D, to be convergent, define SD, + Diet Data @ -adieg 8 as the swm to infinity in the fourth way. The finite sum according to this last definition includes all the terms in the triangle OKK’; it can therefore never (except for m=n=1) coincide with the finite sum according to the former definitions. Whether the sum to infinity (S’”’) according to the fourth definition will coincide with S, S’, or 8S”, depends on the nature of the series. It may, in fact, happen that the limits S, 8’, 8” exist and are all equal, and that the limit $’” is infinite. + § 33.] Double series in which the terms are all ultimately of the same sign. By far the most important kind of double series is that in which, for all values of m and m greater than certain | fixed limits, wm, has always the same sign, say always the | positive sign. Since, by adding or subtracting a finite quantity to the sum (however defined), we can always make any finite number of terms have the same sign as the ultimate terms of | the series, we may, so far as questions regarding convergency | are concerned, suppose all the terms of wm. to have the same | (say positive) sign from the beginning. Suppose now (1) to represent the array of terms under this last supposition ; and let us farther suppose that vw, is convergent in the first way. Then, since L(Snipntq—Smn)=S-S=0, when m=, n= © whatever p and g may be, it follows that the sum of all a Neh * For an illustration of the case when this is not so, see below, § 35. + See below, § 35. XXVI DOUBLE SERIES OF POSITIVE TERMS 151. the terms in the gnomon between NMK and two parallels to NM and MK below and to the right of these lines respectively, must become as small as we please when we remove NM suffi- ciently far down and MK sufficiently far to the right. From this it follows, a fortiori, seeing that all the terms of the array are positive, that, if only m and » be sufficiently great, the sum of any group of terms taken in any way from the residual terms lying outside OKMN will be as small as we please. Hence, in particular, Ist. The total or partial residue of each of the horizontal series vanishes when n=. 2nd. The same is true for each of the vertical series. 3rd. The same is true for the series >D,. The last inference holds, since S8’”, obviously lies between Se.n-¢ and Sn ijn | Hence Theorem I. Jf all the terms of tm m be positive, and if the series be convergent in the first sense, then each of the horizontal series, each of the vertical series, and the diagonal series will be convergent, and the double series will be convergent in the remaining three ways, always to the same limit. If we commutate the terms of a double series so that the terM Um» becomes the term py, Where m' =f (m,n), n' = g(m,n), F (m,n) and g(m,n) being functions of m and n, each of which has a distinct value for every distinct pair of values of m and n (say non- repeating functions), and each of which is finite for all finite values of m and n (Restriction A*), then we shall obviously leave the convergency of the series unaffected. Hence Cor. 1. If Ym» be a series of positive terms convergent in the first way, then any commutation of its terms (under Restriction A) will leave its convergency unaffected ; that is to say, it will converge im all the four ways to the same limit S as before. * No such restriction is usually mentioned by writers on this subject ; but some such restriction is obviously implied when it is said that the terms of an absolutely convergent series are commutative ; otherwise the character- istic property of a convergent series, namely, that it has a vanishing residue, would not be conserved. 152 - DOUBLE SERIES OF POSITIVE TERMS CHAP. Cor. 2. If the terms (all positive) of a convergent single serves Sun be arranged into a double series Lupin, where m’ and n' are functions of n subject to Restriction A, then Diy will converge im all four ways to the same limit as Lup. It should be noticed that this last corollary gives a further extension of the laws of commutation and association to a series of positive terms; and therefore, as we shall see presently, to any absolutely convergent series. Let us next assume that the series 2w,,. is convergent in the second way. ‘Then, since T,, is convergent, we can, by sufii- _ ciently increasing m, make the residue of this series, that is, the sum of as many as we choose of the terms below the infinite horizontal line NM, less than $<, where « is as small as we please. Also, since each of the horizontal series is, by our hypothesis, convergent, we can, by sufficiently increasing n, make the residue of each of them, less than «/2m; and therefore the sum of their residues, that is, as many as we please of the terms above NM produced and right of MK, less than $< Hence, by sufficiently increasing both mand n, we can make the sum of the terms outside OKMN, less than «, that is, as small as we please. From this it follows that =u,» is convergent in the first way, and, therefore, by Theorem I, in all the four ways. In exactly the same way, we can show that, if Dm 18 con- vergent in the third way, it is convergent in all four ways. Finally, let us assume that Zu,» 1s convergent in the fourth way. It follows that the residue of the diagonal series 2D, can, by making » large enough, be made as small as we please. Now, if only m and n be each large enough, the residue of S,, n, that is, the sum of as many as we please of the terms outside OKMN, will contain only terms outside OKK’, all of which are terms in the residue of 8”. Hence, since all the terms in the array (1) are positive, we can make the sum of as many as we please of the terms outside OKMN as small as we please, by ? XXVI CAUCHY’S TEST FOR ABSOLUTE CONVERGENCY 153 sufficiently increasing both m and n. Therefore 2m» is con- vergent in the first way, and consequently in all four ways. - Combining these results with Theorem I., we now arrive at the following :— Theorem II. [f a double series of positive terms converge in any one of the four ways to the limit S, it also converges in all the other three ways to the same limit S; and the subsidiary single’ series, horizontal, vertical, and diagonal, are all convergent. Cor. Any single series Luy consisting of terms selected from Zima (under Restriction A) will be a convergent series, if Summ be convergent. Restriction A will here take the form that n’ must be a function of m and n whose values do not repeat, and which is finite for finite values of m and n. Example. The double series Zx™y” is convergent for all values of « and 7 Y, such that 0m) be absolutely convergent ; and find its sum. The series may be arranged thus :— 1424+ + .°. Ree ae . —(n+1ljym-... Ba ee 7 : + YOAV (nt hee S* Cy ips i martin +(— Yn saCayrat + +) "inn yr at If x gaat y' 4 the aaa or Conte values, of x Be Y, ees the series Zw’ myn corresponding to the above will be L+o'tat%+.. +a 4+... ty +2y'a! +8y'e?% +... +(n+1jy’en+... In order that the horizontal series in this last may be convergent, it is necessary and sufficient that 2’ <1. Also T’m=y'™/(1-a’y"+1 ; hence the BceSeAgy. and sufficient condition that ZT’, be convergent is that y'<1-2’, which implies, of course, that Cp ae The given series will therefore satisfy Cauchy’s wonders of absolute con- vergency if moda<1, moda+mody<1, and consequently also mod y<1. These being fulfilled, we have Tn=(—)™y™/(1—ax)" ; XXVI EXCEPTIONAL CASES 155 jis 1 Yy A y m =, {1-;4+. ° ea 74) ae }, 1 l-x+y’ and the sum of the series in whatever order we take its terms is 1/(1-7+y). o” ott 2 Example 2. If UW, =a + 92"* Pash Um,n— > Um, ns n=1 n=m+1 =(m+1){1/(2m + 2) -1/(m+2)} —m{1/(2m+1)-1f(m+1)} —(m+1){0-1/(2m+2)} +m{0-1/(2m+1)}, =1—-2m/(2m +1) -(m+1)/(m+2)+m/(m+1), =(m?+m+1)/(m+1) (m+ 2) (2m +1). Now the convergence of =T’,, is of the same order as that of 21/m, that is: to say, ZT", is divergent. Hence Cauchy’s conditions are not fully satisfied ; and the anomaly pointed out above ceases to be surprising. The present case is an excellent example of the care required in dealing with double series which are wont to be used somewhat recklessly by beginners in mathematics. * “ Before Cauchy the reckless use of double series and consequent perplexity was not confined to beginners. See a curious paper by Babbage, Phil. Trans. R.S.L. (1819). XXVI | COMPLEX DOUBLE SERIES 157 Example 2. The double series =(-)”+t"1/m”, whose horizontal and verti- cal series are each semi-convergent, converges to the sum (log 2)? in the second, third, or fourth way (see chap. xxviii., § 9, and Exercises XIII. 14). But alteration in the order of the terms in the array would alter the sum (sce chap. xxviii., § 4, Example 3). Example 3. If the two series Da, and Xd,, converge to a and D respectively, and at least one of them be absolutely convergent, then it follows from § 14 that the double series a,b, converges to the same sum, namely ad, in all the four ways, although it is not absolutely convergent, and its sum is not inde- pendent of the order of its terms. The same also follows by § 20, Cor., provided Zap, Zby, U(anb1+An-1b2 +. .. +n) be all convergent, even if no one of the three be absolutely convergent. * If, however, both Za, and 2b, be semi-convergent, then the diagonal series may be divergent, although the series converges to the same limit in the second and third way. This happens with the series 2(—)”+"1/(an)* where a is a quantity lying between 0 and 3. This series obviously converges to the finite limit (1—1/2%+1/8%-. . . )? in the second and third ways. For the diagonal series we have De — r I Ma 1/r*(n- 1)”. tities Now, since 0 or $1. (18.) Show that at/™ + qlimt]/(m+)) 4 ql /mtiimt)+1 (mt?) 4+... . is conyerg- ent or divergent according as a< or or +1. (Bertrand.) (21.) Show that 21/n log {log log 2} * is convergent or divergent accord- ing as a> or <1. (22.) Show that 21/(n+1+cos nz)? is convergent. (Catalan, Zraité £1. d. Séries, p. 28.) . is convergent or Examine the convergency of the following infinite products :— (23.) IL{1+/(n)r"}, where f(x) is an integral function of x. (24.) ID}{(x?" + a)/(a?m +1). (25.) I{nt44/(n -1)*(n+2)}. (26.) If Zf(n) be convergent, show that, when n=0o “17 Lille +/(n))}2" =a, 1 (27.) If w denote one of the series of primes 2, 3, 5, 7, 11, ..., then =/(p) is convergent if Zf(p)/logpis convergent. (Bonnet, Liowville’s Jour., vili. (1848), and Tchebichef, 7b., xvii. (1852).) (28.) If <1, show that the remainder after m terms of the series 172 + 27a? + 8°93 + is <(n+1)art!/{1-(14+1/n)"z}. (29.) If wo, wi, . . ., Un be all positive, and Zw,x” be rere for all values of x?0. Hence show 160 EXERCISES VIII CHAP. that there can be no function ¢(1) such that every series Zw» is convergent or divergent, according as L $(n)u,= or +0. (Abel, Huvres, ii., p. 197.) © n=e (33.) If Du, be any convergent series whose terms are ultimately positive, we can always find another convergent series, Zv,, whose terms are ultimately positive, and such that Lv,/u,=0. If Zw», be any divergent series whose terms are ultimately positive, we can always find another divergent series whose terms are ultimately positive, and such that Lu,,/v,=0. (These theorems are due to Du Bois-Reymond and Abel respectively ; for concise demonstrations, see Thomae, Hlementare Theorie der Analytischen* Functionen. Halle, 1880.) (84.) If wn41/n=(n™+An®1+, . .)/(n™+A'n*1+...), then Zu, will be convergent or divergent according as A—A’> or +1. (Gauss, Werke, Bd. iii., p. 189.) | (35.) If wnqa/un=a—B/n+y/n7+6/n? +. .., then Du, is convergent or divergent according asa< or >1. If a=1, Zw, is convergent only if B>1. (Schlomilch, Zettschr. f. Math., x., p. 74.) (36.) Z1/u, is convergent if w+ — 241+ U» is constant or ultimately in- creases with n. (Laurent, Nowv. Ann., ser. ii., t. 8.) (37.) If the terms of Zw, are ultimately positive, then— (I.) If y(n) can be found such that y(n) is positive, Ly¥(n)w,=0, and LiW(n2)un/Unsi — Y(m+1)} +0, Zw» is convergent. (II.) If y(n) be such that Ly(n)u,=0, L{Y(n)up/Undi- Y(n+1)} =0, and LyY(2)Un/ {W(12)Un/Unsa — W(n+1)} +0, Zu, is divergent.: (III.) If 2/41 can be expanded in descending powers of n, Zp 1s con- vergent or divergent according as Linu,/un4i—(n+1)} > or +9. (IV.) If wp/%en41 can be expanded in descending powers of n, Zp is con- vergent or divergent according as Liw,= or +0. (Kummer’s Criteria, Crelle’s Jour., xiii. (1835) and xvi.) (38.) If a terms of Dw, be ultimately Hoaitice and if, on and after a certain value of 2, Qn2tp/Unti-—Anti>-, Where @ is a function ef n which is always positive for values of 2 in question, and yu is a positive constant, then Zw, is convergent. From this rule can be deduced the rules of Cauchy, De Morgan, and Bertrand. (Jensen, Comptes Rendus, c. vi., p. 729. 1888.) Discuss the convergence of the following double series :— (39, ) = Datel (40.) 2(—1)"-7™/n !. (41. > { i (n sas 1)™/nerl a vrl(n ak Be mt+1 MS (42.) Damy"/(m +2). ) 21/(m+ny. (44.) Z1/(m-+n). ) 21/(m? - n?). (46.) Under what restrictions can 1/(1+a+y) be expanded in a double series of the form 1+ ZA, ,v”y"? (47.) If Zum,» converge to S in the first way, and if its diagonal series be convergent, show that the diagonal series converges to 8 also. a XXVI _ EXERCISES VIII 161 Deduce Abel’s Theorem regarding the product of two semi-convergent series. (See Stolz, Math. Ann., xxiv.) | (48.) If 2/1 can be expanded in a series of the form 1 + @/1 + @2/n? + eey ‘show that 1°. If q=0, a2=0, .. ., dpr1=0, au + 0, then u,=UW+,/n, where u is a definite constant + 0 Ao + o, and Lv, is finite when 2=0 2°. If a +0, and the real part of a be positive, Aan eae when es. 3°. If a +0, and the real part of a,=0, then Lw, is not infinite, but is not definite. 4°, If a + 0, and the real part of a, be negative, then Lu,=0. Apply these results to the discussion of the convergency of Zw,«#", and, in particular, to the Hypergeometric Series, and to the following series :— = ofee yiCn(x+ yt)”, Larne tri, ZmCn/(m+n)?, = - YonCnr/(m+n)?. (See Weierstrass, Uber die Theorie die Analytischen Facultdét.— Crelle’s Jour., li.) (49. ) Discuss the convergence of 3 = mCn(a — 2B)""(a+nB)”. (50.) If w, and v, be positive for all values of n, never increase when 2 increases, and be such that Lu,=0, Lv,=0, when n=, find the necessary’ and sufficient condition that PAE Uaeape » » + Un) = Zn X ZV. (See Pringsheim, Math. Ann., Bd. Xxi.) WoL. 11 M CHAPTER XV LE Binomial and Multinomial Series for any Index. BINOMIAL SERIES. § 1.] We have already shown that, when m is a positive integer, (Vt zy 1 + Cie + Ce +s. F Cnt? +... tnCae™ CL where mOn=m(m—-1) 2... (m-n+1)/n! (2). When m is not a positive integer, ,,Cy, although it has still a definite analytical meaning, can no longer be taken to denote the number of n-combinations of m things; hence our former demonstration is no longer applicable. Moreover, the right-hand side of (1) then becomes an infinite series, and has, according to the principles of last chapter, no definite meaning unless the series be convergent. In cases where the series is divergent there cannot be any question, in the ordinary sense at least, regarding the equivalence of the two sides of (1). As has already been shown, the series 1 tinCit + Ce +00 Cee (3) is convergent when a has any real value between — 1 and +1; also when a = + 1, provided m> —1; and when # = — 1, pro- vided m>0. We propose now to inquire, whether in these cases the series (3) still represents (1 + ~)” in any legitimate sense. In what follows, we suppose the numerical value of m to be a commensurable number ;* also, for the present, we consider * If m be incommensurable we must suppose it replaced by a commensur- able approximation of sufficient accuracy. CHAP, XXVII FIRST PROOF . 163 only real values of a, and understand (1 +2)” to be real and positive. § 2.] If we assume that (1 +2)” can be expanded in a con- vergent series of ascending powers of a, then it is easily shown that the coefficient of 2” must be m(m—-1).. . (m—n+1)/n!. For, let | : I TG 9 ie ae I SE (1) where ET Oe. Oye SL. (2) is convergent so long as moda0. Hence, by Abel’s Theorem, chap. xxvi., § 20, (1+1—0)"= U 4 Petes aren eens ce Ont to eens x=1- that is, | | Ores [eth Oni teat scale mn onit ets (15), provided m> —1, with the condition that, when —-10, we have, by Abel’s Theorem, ed 0" = L (—nCtt Ca. it. (—)%nCna™ +: --.); “=1-0 that is, Ore Or Om. <5 Um) Cn tal (16), provided m be positive. The results of (15) and (16) complete the demonstration of 168 PARTICULAR CASES CHAP. the Binomial Theorem in all cases where its validity is in question. Cor. If x+y, it follows from the above result that we can always expand (a+) in an absolutely convergent series. We have in fact, if z>y, that is, y/x<1, (e +4 yy = ald | ,(n) is any integral function of » of the 7th degree. Such a series stands in the same relation to the simple Binomial Series as does the Integro-— Geometric to the simple Geometric Series. We may therefore speak of it as an Integro-binomial Serves, We may always, by the process of chap. v., § 22, establish an identity of the following kind, b(n) =A, +An+ Agn(n—-1)+...+A,n(n-1)...(n—r+1) (1), where Ay, Assigns of n. We can therefore write the general term of the Integro- Binomial Series in the following form :— helm Ont” = AgmOnX? + AitmCnt? +. -.. AR ee (1 —17 + 1)mOnx™, .. A, are constants, that is, are independent = AymCn x” + MA Lm 1Cn- Fa + mm —1)Ag2'm-2Cn-s0" +. » «Emm 1) (maf +1)A,2 pi -pGun pte ep XXVII SoM) Cee |( +a)(n+b)...(n+hk) 171 Hence, if the summation proceed from 0 to «, we evidently have oe ire) © = h()mCnx” Aye nte F MAW an Ay les tes a 0 0 1 : ioe) is m(m a 1) vee « (m p+ DAG Cera (By Te =Ai(l +a)" mA WT + eyo t +... + m(m =" Dy ernie 7 tL yA el se), Since all the Binomial Series are evidently complete.* Hence Dh(0)mCn2” = {A+ mA,v/(1 +) + m(m—1)A,a7/(l+2) 4... 0 +mm—-1)...-(m—r41)Aat/d +a)" }(1+a)™ (4); and the summation to infinity of the Integro-Binomial Series is effected.f ee os The formula will still apply when m is a positive integer, although in that case the series on the left of (4) has not an infinite number of terms. The only peculiarity is that a number of the terms within the crooked bracket on the right-hand side of (4) may become zero. | Cor. We can in general sum the series D$,(2) nOn/ (n+ a)( +b) ... (v+k), where a, b,.. ., k are unequal positive integers, in ascending order of magnitude. For by introducing the factors n+1, 1+2,...,n+a-—1, N+a+1,n+a+2,...,n+b-1, &., we can reduce the general term to the form | W() mr kCn+eertel(m+ 1)(m+ 2)... (m+k)at. (5); where y(n) is an integral function of n, namely, ¢,() multiplied by all the factors introduced. * Tf the lower limit of summation be not 0, then the Binomial Series on the right-hand side of (3) will not all be complete, and the sum will not be quite so simple as in (4). + It may be remarked that the series is evidently convergent when x<1. The examination of the convergence when x=1 will form a good exercise on chap. Xxvi. 72 EXAMPLES CHAP. Hence Sh/(22) mCi” (n+a)(n+b)...(n+k) | = (S(0) n4¢On4 40 TY} /(m+1)(m+2)...(m+k)a® (6). The summation of the series inside the crooked bracket may be effected; for it is'an Integro-Binomial Series. Hence the summation originally proposed is always possible. We have not indicated the lower limit of the summation, and it is immaterial what it is. Even if the lower limit of summa- tion be 0, the Binomial Series into which the right-hand side of (6) is decomposed will not all be complete (see Example 6, below). It should also be noticed that this method will not apply if m be such that any of the factors m+1, m+2,..., m+k vanish. In such cases the right-hand side of (6) would become indeterminate, and the evaluation of its limit would be trouble- some. The above method can be varied in several ways, which neéd not be specified in detail. It is sufficient to add that by virtue of Abel’s Theorem (chap. xxvi., § 20) all the above summa- tions hold when x= +1, provided the series involved remain | convergent. Example 1. To expand (w7+y)™ in a highly convergent series when x and y are an equal. From the obvious identities {(x+-y)/2u\™ = {Qa/(a+y)}-™= {1+ (@—y)/(a+y)}p™, {(x-+y)/2y}™= {Qy/(@+y)}-m™= {1 -(w@-y)/(@+y)p-™s (x+y) ym 41)( (2o)m=E1/(2y)™} = {1+(e- y Vle+y)} yom {1—-(e@-y)[(at+y)}-™ we deduce at once \mn — Mam aae ee me " (a+ y)n=2mam (142 n(— Ha EY)" f — 9M 7;m > Y med {14 min ee, } where mHy=m(m+1).. .(m+n-1)/n}, Qmtlomym ‘ene +1)/a-y\? _m(m+1)(m+2)(m+8) (x-y \4 = ————_ 4 1+ erry hoz ie ee i a i 2! L+y 4! x+y eee) meaner ay 7 (2 7 mn ne zou? \ = Seo Pe ae Gola cic em — ym L+Yy 3! x+y All these series are cae convergent, since («—y)/(v+y) is small. XXVII EXAMPLES I Ey Example 2. To sum the series 212 2 pele 2 * 2.5.8 2 ff 9-2 t\9 38!\9 POD AG! Rimes If we denote this series by %+%+uUg+. . ., we see that 2.5... {2+(n-2)3} 2” in n ! Ba 4.3.8... (-h4n-1)/2\" a -n! Dye , << {ct as Sa ae aah Ny ~ n! oie Pea LS 2) nin (So) 2 \" Se amen nm! Gir Hence 1 — (uy tugt+u3+ 3. ti .)=(1- 3), =1/ 0/3. Therefore, Ujtuetugt...=1-1/ NL ot Example 3. To sum the series m(m — 1) mm — 1)\(m- 2). m+ {= 1.2 de! whenever it is convergent. Here we have mm—1)(m—-2).. . (m-n) Uni = te > n | _mm—-1)(m-1-1)...(m-1-n+1 4 n! a ben hon Hence UtUgtugt.. . =mM{1L+m raCit+m-rCet. . «} Sse Lt tm 2. provided m-—1>-—-1, that is m>0. It should be observed that we have at once from § 2 (5) the equation M1 +2)" 1, Cy +2mCott. . . tMmCnar i+... (1), from which the above result follows by putting z=1. By repeating the process of § 2, we should deduce the equation mm—-1)...(m—-k+1)(14+a)"-*¥=1.2... bm CO, +2.8..-(K4+1)mCrpet... +(n-k+1)(n-K+2). 2. MmCnrar-Ft... (2), whence it follows that mm—1)...(m—k+1)2"-*¥=1.2...kmCe+2.8...(K4+1)mCrut-:. (8), _ provided m>k. These results might also be easily established by the method first used. Example 4. To sum the series ui m1 mae emer Sou (eh 1), 340 tO) 174 EXAMPLES CHAP. Here we have Fee monte mth (n+1)(n+2).... (w+ky mia wit pce car ar . (m+k)jak Hence (1a )nre oe (m+1)(m+2).. “Un Ebets mk Den ea) Sa ee + m4xC1@ + mi xCo” = [vee ome + ne Opa? } + { ty + Ug + Ug + ene ths Therefore ail + Me — 1 yp Cyt — myn Coe? =.» » — mtn Cn—10*t (m+1)(m+2).. .(m+k)ak If m> —k-1, this gives as a particular case LmCn]{(m+1)(n +2)... (nm ee (4). Srtias ole te =k- fomtk_y—- = “miaCeh (One L) (nD) ..(m+k) (5). The formule (1), (2), (3), (4), and (5) contain of course a considerable variety of particular cases. Example 5. Evaluate E80 Let n?= Ap + Ayn + se —1)+Ag3n(n —1)(2-2), then we have the follow- ing calculation to determine Ap, Ay, Az, Ag (see chap. v., § 22). 1 +0 +0|+0 ~ Ao=0, LOS ae 1 +1|+1 ge ah 210 +2:. “1[+3 As= 3, A3=1. Hence in C,z"=0. Enna + LinieZn10n- "1 + 38m(m - 1)2°Zma20n—220"9 +m(m-—1) “a 2)282maCn-s0"4, ; =ma(1+a)"-1+ 3m(m —-1)x me + a0)m—? + mm = 1)(m = 2)a3(1 +02), = {ma3 + m(3m —- 1)z?+ mx} (1+ a)"-8. Example 6. Evaluate Sn Cnaenr +2\(n+4). mont” LS (n + 1) (n + 3)m+4Cnp4ert4 (n+2)(n+4) x(m-+1)(m+2)(m+3)(m+4) (n+1)(n+8)=n7+4+4n+4+38, =Ay+A;(n+ 4)+ Ao(n+ 4)(2+4+ 8). 1 +4 +3 -4/]0 -4 +0 i038 Ay=3, 0-8 ies | A= -8,-As=1. XXVII EXERCISES IX 175 We therefore have ment 1 0(n+2)(n+4) amt 1 (m+ 2) (an £8 )(ar + 4) 18 | Snir sorts — 3(m ae 4\x Zn aOn4ot? az (m ay 4) (m a yee m2Cn490"*?} ’ | 0 | 1 al(m+1)...(m+4) E — mpsC3x3' — 8(m + 4)a{(1+ayet3 — 1 = 43012 - my3Cox”t + (an + 4) (m + 3)a? {(1 + 0)? — 1 — mpoCre} J, ~ a(m-+1)(m+ ae 3)(m+4) [1 (2 +1)(m + 8)a? — 3(am + 2)a+ 35 (1+ a) | set 18 lem ease tl {(1 “us Pack SN nie m4012 = m-4Con EXERCISES IX. Expand each of the following in ascending powers of « to 5 terms ; and in each case write down and simplify the coefficient of x”. (1.) (1+a)5. ; (2.) (L-a) 74. (3.). (= ne (4.) (2-42)3. (5.) (a+ 3a). 6 (6)2 Kat (7.) %/(1- na). . (8.) 1/(1 - 3a2)8. (9.) Manes. (10.) Write down the first four terms in the expansion of {(a+z)/(a-«)}3 in ascending powers of a. Determine the numerically greatest term in (11.) (8+), <3. (12.) (2-3/2)4”. (18.) (1—5/7)-235, (14.) Find the greatest term in (1+2)"", when e=$, n=4. (15.) If 2 be a positive integer, find the greatest term in (n -1/n)?"+", (16.) The sum of the middle terms of (1+) for all even values of m 1 (including 0) is (1 - 4x) 72. ax ent) 1\2 Ch7. x =14n(1-2)+ 7 (1-3) + (18.) Show that, if m exceed a certain value, then may 4 mri, (m+ 1) (m-2 ! gm 4 et Om m woes )(m — 2) | (19.) Sum the. series a—(a+b)m+ (a+ 2b) Pah eis m(m~-1)m-— 2) a ee faye = ly Bue )_ (+38) for such values of m as render the series convergent. 5.7 sig 3132 ple RO 24 3 9831 2441." O55! (20.) NOY by eas een (21.) 176 EXERCISES IX CHAP. (22.) Sum to infinity i lsd lea St @igcerorien vie (23.) Sum the series m(m—1)(m-2) , m(m-1)... (m—-rtl) SKY tee et — (7-2)! : aE i Soke for such values of m as render the series convergent. (24.) If m be even, show that nn+2)... (2n-2)//1.38... (1-1) =o" (25.) In the expansion of (1-«)-™ no coefficient can be equal to the next following unless all the coefficients are equal. (26.) Prove by induction that mm —1)+ m(m +1) m(m+1)... (m+r—1)_(m+r)! Sse 2! uci ny 7! ~ m!r! where 7 is a positive integer. Hence show that, if «<1, i otra l lat ee aa (m-1)!r! (27.) The sum of the first 7 coefficients in 1/ x/(1-«): the coefficient of the rth terrm=1+n(r-1):1. (28.) If F(a) = 14 24 MOTD on MO . « «, the series @ being absolutely convergent, then F(a)F(b)= F(a +0). What is the condition for the convergency of the series ? (29.) Show that oe od a Bani g tla zg - oe eh {(n+1)e+1}(1—a)rtt]/(m +1) (m+ 2). Sum the following series, so far as they are convergent :— (30.) Z(n-1)’m(m—-1).. - (m-n+1)a"/n!, from n=1 ton=0. (31.) 2(—)"-'n4+1)(m+2)1.3.5 ... (2n—5)a/n!, from n=0 to n=. (32.) Sm(m+1).. . (mt+n—-ljarf(n+3)n}, from n=0 ton=0. (38.) Z(n—1)71.4.7 . . . (8n—2)/(n+2)(n+3)n}, from n=1 ton=0. (34.) Why does the method of summation given in § 5 not apply to La"/(n+1)? SERIES DEDUCED BY EXPANSION OF RATIONAL FUNCTIONS OF &. — § 6.] Since every rational function of can be expressed in | the form I+F, where I is an integral function of a, and F a | proper rational fraction, and since F can, by chap. viil., § 7, be | } XXVII EXPANSION OF (2 —px)/(1 — px + qa?) | 177 expressed in the form YA(x -—a)~”, where A is constant;ait-follows that for certain values of z a rational function of x can be exe panded in a series of ascending powers of a, and for certain other values of x in a series of descending powers of 7.* We shall have occasion to dwell more on the general consequences of this result in a later chapter, where we deal with the theory of Recurring Series. There are, however, certain particular cases which may with advantage be studied here. § 7.] Series for expressing a” + B” and (a®t+1 — B"+1)/(a — B) in terms of a8 and a+ B, n being a positive integer. | If we denote the elementary symmetric functions a + B and af by p and q respectively, it follows from chap. xviii., § 2, that we can express the symmetric functions a” +B”, (a"+1 — Bt] (a — 8) as follows :— foo 5 Oop” + Op" *g +. 2. + apt *9" +. 2.4) (A), fe eee) i(a — 8) = bop” + 0p" 29 +. ~. + bp" 9" +... (2), where both series terminate. By the methods of chap. vii., § 8, or by direct verification we can establish the identity I le a! | Z 1 9 l—put+qa® (l-az)(1-—fBx) l-ax 1-fe oe) Now if z be (as it obviously always may be) taken so small that px — gx’ <1, we have by the Binomial Theorem 2 — pu 1 — pu + gx" = (2 — pax){1 — (pa — ga*)}-1 = (2 — pa){1 + (px — qx’) + (pu -quy+...+(pe-qvyr+...} (4). Now (by chap. xxvi., § 34) if w be taken between — a and +a, a being such that the numerical value of +pa+qa’<1, that arrangement of signs being taken which makes + pa + qa” greatest, then each of the terms on the right-hand side may be expanded in powers of x and the whole rearranged as a convergent series proceeding by ascending powers of z. * Strictly speaking, this is as yet established only for cases where a is real. The cases where a is imaginary will, however, be covered by the ex- tension of the Binomial Theorem given in chap. xxix. VOL. II N 178 a” + B” IN TERMS OF af, a+ f CHAP. We thus find that 2 — pu ~ 1 ieee 4 a (2 —pu){1 + 2(p" — naiCio" 99 + nae eh sn a8 ( = pene ONS Ab nities ys ya (5), =2{1+> &.} —pofl +2 &.} (6). The coefficient of 2” on the right-hand. side of (6) is 2{p" — n_ ee Cap” *q° + a +(- ye at 4 De Oe sug — pip" ‘lex a a OF len 5 oe AG ea se oe ml. YneriCip"™ a ome Now Dna = n-riOr=nn—7—1)(n—7—- 2). . (n- 2r41)frt. Hence aia : et eRe; 24 nN 7 n-2 n(n — 3) 400 ly + 9r Lal ht Le Ls 21 is ~e ~A\in are 2 —~9r+1 ‘ i ( # yr nn —f7 y(n a Ze (1 ar ) yn-2rgp Fi he (7). Again a Saas RRO | fico a ae . 2. tates. . .$4{1 4 fo to 12 fr .~ ; + Sit aaie ae RP Ue ee = 2+ (a + B”)a” . (8). All the series involved in (8) will be absolutely convergent, provided z be taken so small that mod am and mod fz are each <1. Now, by (3), the series in (7) and (8) must be identical. Hence, comparing the coefficients of #”, we must have (by oe REVI S21) rae ier ate 29 + ——— in 3 = 9) on ~ 492 _ n(n — em .. (n— 2r4+1) r! pr argt 4 Saat (9), As we have indicated (by using=), the equation (9) is an — algebraical identity, on the understanding that p stands for a+ PB Hts dui XXVII SERIES FOR a” + 2", (a™t! — B"+1)/(a — B) 179 and g for a8. The last term will or will not contain p according as n is odd or even. In like manner, from the identity a oa My a4 1 1 1 l—pe+ge 1-(a+ B+ ofa? {i-=-iez loa we deduce (an44 — Br+3)/(a — B) =p —"=* pray 4 (n—r)(n—-7-—1)...(n—-2r +1) a fence ae We is mC LO). subject to the same remarks as (9). n-2)(n—-3 . mee If we write the series (9) in the reverse order, and observe that, when 7 is even, = 2m say, only even powers of p occur, and that the term which contains ”* is Ree ee a) ar A) . (28+ 1) prgn-s, (m—s)! that is, ? (—yn-e 2m(m + s—1)(m+s—2)...(m+ 1)m(m—1)...(m—s +1) (2s) ! pag *, that is, m>(m2—12) . . . (m?—s—1?) — ym-s9 jd m-s ieee) (2s) | FEN IIA Sn then we have m m*(m? — 1 oe 2m == nN 2,m-1 m—2 a + fma(— yma} gm pm joey waar +(-)= (sy! Similarly, we have m>(m? — 1?) . . (mi? =e- 14) ign-# a (9, . gzmt+1 it Been. =f _ ym(2m 4: 1) | pg _ (an = - a P _ {m + 2)m(m? — 1?) one 5! LA fe tig: m+s—1)m(m2— 12)... (m?-—s— 2? | 2 ( i ye-1l ( (2s 7) i ] ( ) p28-1gm-s+1 AAS \ (9”). 180 SERIES FOR fa + /(a2 + y?)}" + fu -— JA (+ y?)}” CHAP. Sf Wi —_ mym-l1 3,7m-2 fein ee? \qrPt 3, 2 en wii 2 oi ay 2 eyo et he ce et Py om 4+. de (10’). (2s—1)! eee) Cala , (m+ ym, 4s, (m+ 2)m(m* — 1%) Sal Pg =} ge re rN | (m+ sym(m2 —1%). 0. (P= S31) 4~m-2 _ EPS pig We ea | NOE vt os) rr rags + are. | (Lie. Since a and B are the roots of the quadratic function 2 —pz+q, we may replace a and f in the above identities by Lint J(p—4q)}, and 4{p- V(p’- 4q)} respectively. lf this be done, and we at the same time put p= and - 4¢= yf, we deduce the following :— fat J (a2 + y2)}™ + {a — J (a? + y)}™ F n n(n — 3) = )nN nv N29 2 n-4) 4 =P) oe Toe Ui crane Tenens pt ‘- pee Ss 2).. (n — 2r + 1) oar, 7 eae rl ger oo8 ti: oJ mn n* 2,-2 4yn—-4 1 a et wale coral: MCR? Li) Mai ge ad « n(n? — 27) f n(n? — 27) (n? — .. (n2 = 25 — 22) i , ey g28yjn- 284 Or ( C a 4 best, if m be even ; 224 nay" Vy AVGah Babe mB uations Wile 3! 5 er n(n? — 17) (n? — 3?) . (n? — 28 '— 2s — 12) Jn ee aaaee 1 ; , if m be odd. : aol ) AP Et ond Pe ue xxvit' SERIES FOR {x +°4/(a? + ?)}" — fa-— /(a?- + 4)” 181 {a + J (x? +42)" — {a ac a/ (0? + y?)} , ee 2) an -3y2 2, (m eo 9) 22 219 Efi key BSG, bar gn 5yA oP, me r ie - oa : (n 27) AS atid cane I _=2 2 n(x? ce Ye 2.) rn ay” ia n(n* — 2°) fd tke ip 3 ! (Gas a be + n(n* — 2*).. (2s yi peices 2?) g2s—lyn— 2s a ; if m be even ; ; | = 2 /(2? + 7?) \y (AVze oh yr 4 (n? — 1%)(n? — 3?) 2! 4! 40-5 (n” Bk es (n? = 37) sos (i? — 2s- 05 12) os, ) de ae (Os) | alae ae a I, if n be odd. ) These series are important in connection with the theory of the circular and hyperbolic functions. § 8.] A slight extension of the method of last paragraph enables us to find expressions for the sum and for the number of r-ary products of n letters (repetition of each letter being allowed). The inverse method of partial fractions gives us the identity 1/(1 —a,t)(1—a,%) ... . (l—a,v)=2A,(1 —a,x)-1 (1), where ~ ‘Age year ( tts — a.) ee A es (as — aye Also, since (1 — at)" = 1+ a,"2", we have (by chap. xxvi., § 14), provided z be taken small enough to secure the absolute convergency of all the series involved, 7 1/1 —< a2) (1 a Gy) cer (1 a Ay) a, ae) (Lt Daa). 2 tee oan a) i te(2), =1+2,K,0" (3), where ,K, is obviously the sum of all the 7-ary products of Gy) Oo, . . . Gy. Since the coefficients of 2” on the right-hand sides of (1) and (3) must be equal, we have aie ep las fats (Ge = 0.,) (as = Gd.) si «(ag — On) (4). CHAP. SUM AND NUMBER OF 7-ARY PRODUCTS 182 If, for example, there be three letters, a,, a,, a, we have r+2 +2 r-+2 Ree pote ie rr! (a, =e ay) (a, = a3) (a, ak a) (a, ay a3) (a, a a) (a, “an dy) a,” F*( ag 03) it a” t*(a, = 011) il ag! T (oy 2 dy) (5). (a — as) (a3 — a) (a4 — tg) Il =a, =1, then each of the terms in If we put a,=a,= nK, reduces to 1, and ,K, becomes ,,H,. Hence, from (3), (l—a)*=] + 2,00 (6). Equating coefficients of a” on both sides of (6), we have nHp=nn+1)... (v+r-1)/r}, a result already found by another method in chap. xxii, § 10. § 9.] Some interesting results can be obtained by expanding 1/(y+2)(y+ut+1)... (y+#+n) in descending, and in ascend- ing powers of ¥. If we write L(y+a)(ytur+])... Yratny= S A(y +2 47)74, r=0 — then we find, by the method of chap. viii, § 6, that 12A,(~1r)(—7r 71). 22 (- 1). 2 ae A,=(—)nC,/1!. (y+ utny=X-)aC(ytur+r)* (1). . . denote respectively the sum of . ata Hence Therefore ni[(y+%)(y+%+1).. Blance it) br ea bes a+n, and of their products taken 2, 3, . UAL ged <2 ay time (without repetition), we have ! Peat =4 at eer ke % 2 3(=).0; (a ; y diene, y a i. Ree ) fe Ee : ) 1- +++ aN Sr retro). | y y y yy ) \ § = 2( — in | AS 22") I (-ynG, P14 (- REZ)" (2), XXvII EXPANSIONS OF 1/(yt+a)(y+ut+1)...(yt+u+n) 183 where we suppose y to have a value so large that all the series involved are convergent. Since there is no power of 1/y less than the nth on the left of (2), the coefficient of any such power on the right must vanish. ‘Therefore (a + )§ — ,C,(2 +0 —1)8§ + C(a + —- 2)®-...(-)"a®*=0 (3), where s is any positive integer <2. Equating coefficients of 1/y", 1/y"*, and 1/; /y"*?, we find (a+n)"—C,(a+n-—1)"+ nUs (a+n — 2)” — | (- ng = =o (4); (7 + n)°t! — ,C,(a + — 1)"*1 4+ ,0,(@ +n — 2)rt} — Caayettt in! P. = (n+ 1)!(e-+ 4n) (5); (a+ n)yet2 — O,(2 +0 — 1)? + C,(a+ 0 — 2)" +? — ( — )"gft2 = n1(P,° — P,), = 4(n + 2)! {a" + na + yyn(3n + 1)} (6); and so on. Again from (1) wejhave n! Be died Hwee \ ate r {is pew: he (7); where Q,, Q,, 0, . . . are respectively the sum of 1/z, 1/(«+ 1), .. 1/(@+n), and the sums of their products taken 2, 3, .. . atatime. From (7), by expanding and equating coefficients of |: ¥, we get ) aa + be if ¥ i n! Teanga 1 Heel)... St aets em} 1 ny nUs n “oe ey Gey If we put z=1, we get the following curious relation between the sum of the reciprocals of 1,2, . . .. 7 +1, and the reciprocals of their squares :— 184 . ' EXAMPLES E CHAP. + ieee Lid “Wy aC, AST oe el be oe th (=) Cray 3), § 10.] We have now exemplified most of the elementary processes used in the transformation of Binomial Series. The following additional examples may be useful in helping the student to thread the intricacies of this favourite field of exercise for the tyro in Mathematics. Example 1. Find the coefficient of 2” in the expansion of (1-«)?/(1 +a)? in ascending powers of x. If (1+2)-32=142a,2, then (1—a)/(1+a)? = (1—-2u+2*) (1+ Zana). Hence the coefficient required is @, — 2dn-1 + n-2. If we substitute the actual values of Gn, Gn—1, An—2, We find that 3:5. (2 2.4.62. 3 eee Example 2. If /(z)=ao+aiv+aox"+. .., then the coefficient Oh ain the expansion of /(«)/(1 +a)” in ascending powers of x is a mHy+@1 mHy-1 +o mHy-2+. . . +a, This follows at once from the equation Ka)(1 — x) = (ao + Za,2") (1 + 2 mHx"). In particular, if we put /(~)=(1-«)-” and m=1, we deduce that | Par ed seeds bout See Pan ner and, if we put f(z)=(1-«)-”, we deduce that mtnHr= mH t+ mH y-1 nHi+ mHy-2 Hat. » »+nHn results which have already appeared, in the particular case where m and 7 are integral (see chap. xxili., § 10). Example 3. Show that Gy — 2An—1 + On—-2=(— )"™(16n? — 8n - 1) Pa OY 07) + m4iCn/2? + m42Cn/23 + oy cae ad coo =1 + nC1 + mC2 +. a +mCn (1). The left-hand side of (1) is obviously the coefficient of x” in X=(1+a)"/2+(1+a)mt1/2? 4+ (1+ax)mt?/23 +... ado. Now X=4(14+a)™[1 + {(1+2)/2} + {((1+)/2}?+. . . ado, =(1+a)"/2{1-(1+2)/2}, if we suppose x<1. =(1+a)"/(1- 2), =1+ (1 a mC1 a8 mC2 rs Ba cee mn es by last example. Hence the theorem follows. Example 4. Sum the series n-3 (n—4)(n—-5) (n—5)(n—-6)(n—-7) Gime e131 £1 o being a positive integer. XXVII EXAMPLES 185 The equations (9”’) of § 7 being algebraical identities, we may substitute therein any values of 2 and y we choose, so long as no ambiguity arises in the determination of the functions involved. We may, for example, put *=-land y=2%. We thus find { =1+./8i\", aera ee =(-)"{1-nS}. Hence, if w and w? denote, as usual, the two imaginary cube roots of +1, we have S= {1+(- )"-1(w% + w)! /n. If we evaluate w”+w" for the four cases where 2 has the forms 6m, 6m+1, 6m+2, 6m+3 (remembering that w°”=1, wt=w?, w-?=w), we find that S has the values —1/n, 0, 2/n, and 3/n respectively. Example 5. Sum the series m(n—1), n(w—1)(n—2)(n-38) , n(n—1) (n—- 2) (n—- 8) (n- 4) (n—-5). S=1+ or+1)' 2.4(ar+1)(Qr+8) 1 2.4.6(2r-+1)(2r+3)(ar+5) ae n being a positive integer. If we denote the series by 1+wy+%g+ug+. . ., then = nm(n-1)... (n-—28+1) Us=O 4... 29(2r+1)(2r+3)... (2r+2s—1) _7\(2r)! (r+1)(r+2)... (r+s) 2 (2 — 2s) 1(2r + Bs) te restricting r for the present to be a positive integer. We may therefore write n\(2r)! ~ (n+ 2r) +2r)! Now 74;C, is the coefficient of a*” in the expansion of #"+°5(1 + 1/x?)"+s; that is, in the expansion of a°7t°s{,/(1+1/x")}?"+"s, Hence 2u, is one part of the coefficient of x” in the expansion of i AG hy een OP m1(2r)! 2). n-+-2 § 2} n+2r Ao eG + 1/a?)hnter + $1 — aa/(1 + 1/ac?) bet ne Hence 28 is the whole coefficient of x” in the expansion of n!(27)! {1 1 2\t n+27 ae il 22\ 1. n+27 (n+ Ory aN lace) ee AL A/C Aas) preies Ie Now, by § 7, {+ /(1+a?)peere {1 — a/(1 +2?) ree er ere s—2)...(m+2r—2s41) x =} beri. (s)! ale the coefficient of x2” in which is itary er) et) ne «(t+ 1) 7! Q?r ; 186 EXERCISES X CHAP. Hence 1 (2r) !(n+2r)(n+r—-1)! (n+2r)Irini2r ? pe TAU ess Miers Vie ee). “(1 +2%—1)(n+2r—-2)... (2r+1) The summation is thus effected for all integral values of r. So far, how- ever, as 7 is concerned, the formula arrived at might be reduced to an identity between two integral functions of 7 of finite degree. Since we have shown that this identity holds for an infinite number of particular values of r, it must (chap. v., § 16) hold for all values of 7. The summation is there- fore general so far as 7 is concerned. S= gnt2r-1 EXERCISES X. Find the coefficient of #” in the expansion of the following in ascending powers of x. (1.) a/(a%— a) (a — 0) (wc). (2.) at+3/(a-—a) (a —b) (x —C¢). (3.) w™t3/(a — a) (a—b)(x%—c¢), where m is a positive integer <7 —3. (4.) (8—@)/(2-—ax) (1-2). (5.) 2a?/(a— 1)"(ae* +1). (6.) (1—paym(1 — gx)-™. (7.) If (1—-3a)/(1— 2x)? be expanded in ascending powers of x, the co- efficient of a+7-1 is (-1)"(r — 2n)2”-1, n and r being positive integers. | (8.) Find the numerically greatest term in the expansion of (a —- a) /(b + x)? in ascending powers of x. (9.) Show that (+B) (+28). . .(e-+nB) (% — B)(%—- 2B). . .(%— nf) As rey yn —viele +1) (n? — 1) (n? — 2?). . (0? - r—1?) Bp. Fie ree (r!)? 2—7B’ and hence show that 7 rT=N 2— 17A\(72 — 92). 2p = 42 = ( af ese Maa, (n aoe a A tess yn Ay TS : (10.) If x be a positive integer, show that 1-307 +4. Cesioe s(n Ce aaa (11.) If 2 be an even positive integer, mOn—mOn—1*° mO1+ mCn—2 = mca— ss «tm Cnn eee (12.) If m and x be positive integers, show that mo + mj20n-+ mCe « (m—2)/2Cn—1 + mC4 - (m—4)12Cn—2 + » 6» tmCon © (n—2any/2Co _ mm? — 27)... (m2 —2n— 2?) | (2n) ! : mCi ° (m—1)/2Cn tr mCs C (m—3)2Cn—1 <6 PAGS ° (m—5)/ 20 n—2 es ore ankeed * (m—2n—1)/ 2Co _ m(m? — 17) (m? - 37)... (m?-2n- 1?) ‘a (Qn +1) ! : (See Schlémilch, Handb. d. Alg. Anal., § 38.) XXVII EXERCISES X 187 (13.) Show, by equating coefficients in the expansion of (1 — a~')™(1 — 2) where m is a positive integer, that 3 aki -1) tm. = 1") . (m?* —m—1?)_ 1—-m’*+—._, (Ip +... +(-1) (mi =, (14.) If m be a positive multiple of 6, then nC1 — n038 + nC53? — ec = 5 1 1 nO1 ~ nOaz + nC5g5 — ; oe =O, (15.) If (1+a)-3=1+a,x+a.77+ ..., sum the series 1-ay+a2-a3+.. to m terms. 6.) lf (+2)"=1l+ae+ag7?+..., then l-atafJ-... = (-1)"2n(2n-1). . . (n+1)/n!. r! 2° +1)! (—1)r2""(2r)!_ (-1)r oy tl = 1)13! as O!(Qr+1)! A@r+1- (18.) "ES 1/4r(r'){2n — 2r)!=(47) 1/4" {(2in) hs r=0 (19.) Sum to m terms 2(2n — 2)!/2?”-In {(n —1)!}?. (20.) Sum the series Li 14.7 1.4... .(3n-5) 1 4 m+(n—1)5+(n- 2) gt (n- 3)3-6 9 Ron tag Mune (SH eo aih (21.) Find for what ee of 7 the following series are convergent ; and show that when they are convergent their sums are as given below. vee ae n(n-1) 1 eee EL) } m 1!m+1 2! m+2 °° *° (m+1)(w+2)... (n+m)’ a om «Cel nmn-1) 1 = (m—1)t aa Meiimyl 2b mee ~ (n+1)(n +2)... (n +m) y Lmtn€ ae = minOm—22" +...» +(- ym-lgntm +. (— mi}, m in both cases being a positive integer. (22.) *S (rts)! (mtn—r—s—1)t_ (m+n)! ; sao T!s!(m—r—-1)!(n—-s)! ~~ min! - r=m S= (r+)! (m+n—-7r—s)! (m+n+1)! (23.) = 2", a0 sno OT!) Sim =7)!{n—3)! m!n! (24.) The number of the r-ary products of three letters, none of which is to be raised to a power > the nth, where n<7r<2n, is r(3n —1r) +1 -— 3n(n-1). (25.) Prove, for a, b, c, that 2a”/(a—b)(a-c) if r=2; and generalise the theorem. (26.) Show that a(b—c)(be—aa'\(a™—a'™) b(c—a)(ca— bb')(b™—b'") ca — b) (ab — cc’)(c™ — c'™) es $e a-a b-b c—C¢ = (b-—c) (c— a)(a—6) (be — aa’) (ca — bb’) (ab — ce’) Sn—s/abe, AON EY IAC) iy ON Lid 1 where aa’=bb'=cc’, and Sm-3 is the sum of the (m-—8)-ary products of mre, co, a, 0’, c. (Math. Trip., 1886.) ' 188 EXERCISES X CHAP, (27.) If S, be the sum of the r-ary products of the roots of the equation x + aa") 4 age"? +... +a4,=0, then 0=Si+q, 0=S.4+ 81a + de, 0=S,4Sn-1041+S,-2de+.. . +4n, 0=S,+5,-101+S,-o09+ . . « +8;—nGn. . | (Wronski. ) (28.) IfS, be the sum of the v-ary products of n letters, P,. the sum 1 of the products 7 at a time, =, the sum of their rth powers, then >,=78,—-—(n-1)PiS,31+ ... +(-1)"(n—-71)P,, ifrn-1. (Math. Trip., 1882.) (29.) If v=(1l—-azx)-(1-Bx)-?. . Le the number of ways of distributing n things, \ of which are of one sort, u of another sort,. . ., into p boxes placed in a row is the coefficient of #a*8" . . . in the expansion of (v—1)? in ascending powers of x, namely, Uy — pOiUle+ pCotlg—.. «5 where Us=(pt+rA—S)!(ptmu-—s)!. . . /(p—s)IAl(p-s)! ul. (Math. Trip., 1888. ) (30.) With the same data as in last question, show that the whole number of ways of distributing the things when the order in which they are arranged inside each box is attended to is n'(n—-l1)!/(n-p)!(p-l)!Atpwlyt... / (Math. Trip., 1888. ) Show that (31.) 1+1/2+. .. +1 /e=20,-$,Cot §.C3-. - (m+1)m,, 4 (m+ 2) (m+ 1)m(m — 1) es Mad 8 (32.) en 51 DF oes : Sonat DY ee as ntlan ee 2_ 92 (33.) ee, ee we dp ae a. Jobs. =(-1)™ (34.) If m and x are both positive integers, and m>n, then 2-" (m—n)(m-n-1) nas (m—n)(m—n-1)(m—n-2)(m—-—n—- rae? f n! l!(n+1)! 2!(m+ 2)! ; ce re ek i ee (m+n)! (35.) If 7 be a positive integer, 12 (72 — 12) (7? — 2?) » a (Pre = 2) Grae aa ) rae" 31 x + BI 7 ci = (a+ 2)"—) — »-2Ci(@ + 2)r-3 4. ._ gCo(a + 2)7-> — p-aC3(ae + oyt = XXVII CONVERGENCY OF MULTINOMIAL SERIES 189 ¥ y > “¢g MULTINOMIAL THEOREM FOR ANY INDEX. § 11.] Consider the integral function a,7 +a, a7 +... +.4,2", whose absolute term vanishes, the rest of the coefficients being real quantities positive or negative. Confining ourselves in the meantime to real values of x, we see, since the function vanishes when «= 0, that it will in all cases be possible to assign a posi- tive quantity p such that for all values of « between —p and +p we shall have , GL+Ae +... +a,2°<1 ey In fact, it will be sufficient if p be such that ap+ap +... .+ap"<1 where a is the numerical value of the numerically greatest among @,, 2, ..., a». That is, it will be sufficient if ap(1 — p")/(1 - p) <1; a fortiori (supposing p < 1) it will be sufficient if ap/(1—p)<1; that is, if p<1/(a+1)* (2). p is, in fact, the numerically least among the roots of the two equations One ear Oe el 0, as may be seen by considering the graph of a,a7+.. . + aa. Therefore, whether m be integral or, not, provided —p<%< +p we can always expand (1 +a,¢7+a,a°+...+a,x7)™ in the form | 14+2,,0,(a,0+ a0 +... +a,2")8 (3); and the series (3) will be absolutely convergent whether m be positive or negative. Hence, since qv+ag+...+a,2" is a terminating series and therefore has a finite value for all values of ~ positive or negative, it follows from tlie principle established in chap. xxvi., § 34, that we may arrange (3) according to powers * This is merely a lower limit for p; in any individual case it would in general be much greater. id e a 190 MULTINOMIAL COEFFICIENTS CHAP, of a, and the result will be a power series which will converge to the sum (1 +a,e+a. +... +a,2")™ so long as —p<%< +p. Since s is a positive integer, we can expand ,,C, (av + ae + . . + a,2")§ by the formula of chap. xxiii.,§ 12. The coefficient of x” in this expansion will be a “x a N 1 2 r A Os ! a, Os . . . Uy /a, ! Og ! . . . Oy He that is, Ya, ‘a, ". . . d, *m(m —1).. .. (m—s +1) ayo, eee where the summation extends over all positive integral values of G@,, Gy, . . «, ay, including 0, which are such that Gag, ios O. = 8 1 2 Tr (om Q, +.20g+.. « «t+ T= In order, therefore, to find the coefficient of 2” in (3) we have merely to extend the summation in (4) so as to include all values of s; in other words, to drop the first of the two restric- tions in (5). . Hence, whether m be integral or not, provided x be small enough, we have (l+a,0+ ag? +. . . +a,x"m mm—-1)...(m—2a,4+ 1 Deas ( ca cae ie ot Ge mi Aner ey tert) the summation to be extended over all positive integral values of a,, a2, . .y ay, including 0, such that a,+2agt+...+fa,=N. The details of the evaluation of the coefficient in any parti- cular case are much the same as in chap. xxili., § 12, Example 2, and need not be farther illustrated. It need scarcely be added that when nis very large the calculation is tedious. In some cases it can be avoided by transforming 1 + a,7 + a,2° +... +a," before applying the Binomial Expansion, but in most cases the application of the above formula is in the end both quickest and most conducive to accuracy. XXVII CONDITIONS FOR GOOD APPROXIMATION 191 Example. To find the coefficient of a” in (lt+a+a?.. .+a7)™, We have (ltata?+. .. +arym={(l-a"t)/(1-a))™, =(1—art-)m(1 —x)-™, = (1—art!)™(1] + 2,27). Hence, if n<7r+1, the coefficient of 2” is simply mHn=mm+1)...(m+n—-1)/n!; but if n¢7r+1, the coefficient of x” is mn — mCi» mH n—r—-1 + mC2 + mHn-2r-2- » NUMERICAL APPROXIMATION BY MEANS OF THE BINOMIAL THEOREM. § 12.] The Binomial Expansion may be used for the purpose of approximating to the numerical value of (1 +2)”. According as we retain the first two, the first three, . . ., the first n+ 1 terms of the series 1+,0,0+,0,2°+... ., we may be said to . take a first, a second, . . . an mth approximation to (1 +2)”. The principal points to be attended to are— Ist, To include in our approximation the terms of greatest numerical value; in other words, to take m so great that the numerically greatest term, at least, is included. 2nd, Ta take n so great that the residue of the series is cer- tainly less than half a unit in the decimal place next after that to which absolute accuracy is required. 3rd, To calculate each of the terms retained to such a degree of accuracy that the accumulated error from the neglected digits in all the terms retained is less a unit in the place next after that to which absolute accuracy is required. The last condition is easily secured by a little attention in each particular case. We proceed to discuss the other two. § 13.] The order of the numerically greatest term. In the case of the Binomial Series (1 + x)™, if € denote the numerical value of 2, so that 0 <é<1, we have for the numerical value of the convergency-ratio p+, re 192 NUMERICALLY GREATEST TERM — CHAP. mn n—m La aa oe Cie a g; (1) according as m—% is positive or negative. Hence it is obvious, in the first place, that if -1Sm<+1, that is, if m be a positive or negative proper fraction, the condi- tion o,<1 is satisfied from the very beginning, and the first term will be the greatest. If m>+ 1, the condition o,<1 is obviously satisfied for any value of ” which exceeds m; in fact, the condition will be satis- fied as soon as | (m—n)E(m& — 1)/(1 + &) (2), _ the right-hand side of which is obviously less than m. ‘This con- dition is satisfied from the beginning if €< 2/(m-— 1). If m be <—1= —p, say, where p»>1, the condition o,<1 will be satisfied as soon as (ut ny&E (pé —1)/(1 - &) (3). This condition is satisfied from the beginning if &<2/(u + 1). § 14.] Upper limit of the residue. We have seen that, ulti- mately, the terms of a Binomial Series either (1) alternate in sign or (2) are of constant sign. To the first of these classes belong the expansions of (1 + x) and (1 + x)-™, where « and m are positive. If m be greater than the order of the numerically greatest term, and in the case of (1 +2) (see § 4) also >m, then the residue may be written in the form Ra = = (Uni — Unis PUnks 3 eee) (1), where Unij, Unto Unt -- - are the numerical values of the various terms, and we have Ups, > Un+2>Unt3> Hence, in the present case, the error committed by taking an nth approximation is numerically less than w,+,. In other words, XXVII UPPER LIMIT FOR RESIDUE 193 if we stop at the term of the nth order, the following term is an upper linut for the error of the approximation. Cor. A lower limit for the error is obviously Uns) — Un+e The expansions of (1-2) and (1-—2)-™ belong to the second class of series, in which the terms are all ultimately of the’ same sign. It will be convenient to consider these two expan- sions separately. ee In the case of (1-2), if we take n>m, then we shall certainly include the numerically greatest term; and o,, the numerical value of the convergency-ratio, will be (7 — m)zx/(m + 1), that is, {1-—(m+1)/(n+1)} a This continually increases as n increases, and has for its limit 2, when n= a. Hence | OP the he aes he ee an Therefore, Uy41, Unto... having the same meaning as before, (Una gt Unig +. Une t+ = +); hee Unai(l + Ont + On+:8n+2 + Tn4:0nt2Fn+3 +: - H Therefore Mod Ry (mz — 1)/(1 — 2). We have, in this case, n= (t+ m)a/f(nt+1)={1-C- m)( m+ 1)}a, ={1+(m—-1)/(n+ le. Hence, if m <1, Onti 1, Loa ae ee ee and an upper limit of Ry will be un4,/1-—on4:), that is, mn 2" t/{1 — (n+ 1+ m)a/(n+ 2)}, a lower limit being un4,/ (1-2). The error for (1 — a)~™ is, of course, always in defect. Example 1. To calculate the cube root of 29 to 6 places of decimals. The nearest cube to 29 is 27. We therefore write 8/29 = (33 + 2)'3 = 3(1 +. 2/33), = Up- Ui— Uo tla Ila ee The first term is here the greatest ; and the terms alternate in sign after 2. Also w,, written in the most convenient form for calculating successive terms, is 6r — 8 Uy =38( sr) (rez) Cees) (ass) (Pk). - - 81r ): Therefore 4- = Ue 3°000,000,00 Uy = U2/81 = 74,074,07 Uy = Uy 4/162 = °001,828,99 Us = U210/243 = 75,27 Us = 316/324 = 3,72 3°074,149,34 ‘001, 832,71 ‘001,832,71 3°072,316,63 U5 = U42.2/405 20 Hence the error in defect, due to neglect of the residue, amounts to less than 2 in the seventh place. The error for neglect of digits does not exceed 1 in the seventh place. Therefore, the best 6-place approximation to x/29 is 3°072,317. In Barlow’s Tables we find 3°072,316,8 given as the value to 7 places. | Example 2. To calculate (1 —«)”/(1+a+22)" to a second approximation, « being small. (1—a)"(1+2-+27)-m ~ { 1- mae = Doo | x { 1 -- m(a+a?) + sedi) e ~ __ 5 XXVII EXERCISES XI 195 where we have already neglected all powers of « above the second in each of the two series ; = { 1-ma+ ee” e}{1-me+ Maa 4) m —1) et, 25 z is P =14+(-m-m)xe+ ee m2 RD) has, where higher powers of x than x have again been neglected in distributing the product ; =1-2mx+m(2m —- 1)x?. EXERCISES XI. (1.) The general term in the expansion of (l+u+y+ay)/(l+%+y) is Be hymn — 2)lary"/(m —1)\(%—1)!. Determine limits for « within which the following multinomials can be expanded in convergent series of ascending powers of x; and find the co- efficients of (2.) of in (1 — 2x +a? - 823) ~4, (3.) a in (1-38-72? +23)73 Cee ro fal ain (ar oe be ae a eT. (5.) «7 in (1-80 +a3 - x) - 3, (6.) a” in (2+ 8x +22)-2, (7.) Show that in (9a?+6ax+4a*)-! the coefficient of 2” is 2°"(3a)—3"-? ; and that the coefficient of every third term vanishes. (8.) The coefficient of w” in (1+a+«*) (m a positive integer) is m(m—1) m(m—-—1)(m— 2) (m— 3) ape Ch 5 (9.) The coefficient of 2711 in (1+%)/1l+a+a?)? is -(r+1). (10.) Evaluate *%/(100/99), and 1/(1002/998), each to 10 places of deci- mals ; and demonstrate in each case the accuracy of your approximation. Er Find a first approximation to each of the following, « being small : — (11.) {e+ a/(x? +1)} 2m — fa — a/(a?+1)}™ ; {a+ n/(x? + 1)t2mtl — fa — A/ (a? +1)} 2rd 12.) (14+a)(1+rex)(1+7%z).../—2)(1-az)(1l-a)?... 3 (18.) /(2-v/(2-/(2-... —a/(l+a) ... ))); where / is repeated 2 times. (14.) If x be small compared with N?, then ,/(N?+2)=N+2/4N+ N2w/2(2N?+.2), the error being of the order z‘/N’. For example, show that A/(101)=10<4°45, to 8 places of decimals. (15.) If p differ from N* by less than 1 per cent of either, then X/p differs from 3N + 4p/N? by less than N/90000. (Math. Trip., 1882.) 196 EXERCISES XI CHAP. XXVII (16.) If p=N*+a where z is small, then approximately Ne li 27 4 = 3 4). 4/p= 5g N+ gepINe+ 54 Nal(7p + 5N4) ; show that when N=10, x=1, this approximation is accurate to 16 places of _ decimals. (Math. Trip., 1886. ) (17.) Show that L {1/a/n?+1fa/(n?+1)+ . 2 © + 1/a/(n?+2n)} = 2. n=0 (Catalan, Mowv. Ann., sec. i., t. 17). (18.) Find an upper limit for the residue in the expansion of (1+a)" when m is a positive integer. yy CHAPTER XXVIII. Eixponential and Logarithmic Series. EXPONENTIAL SERIES. § 1.] We have already attached a definite meaning to the symbol a” when a is a positive real quantity, and « any positive or negative commensurable quantity. We propose now to discuss the possibility of expanding a” in a series of ascending powers OL a If we assume that a convergent eapansion of a® in ascending powers of x exists, then we can easily determine its coefficients. For, let P=A, tA ctAg+i. .+A wee. . a) then, proceeding exactly as in chap. xxvii., § 2, we have L(a#t* — a®)/h=A,+2Aa+. . .+nApe-l+. , A and the series on the right will be convergent so long as z lies within limits for which(1)is convergent. Now (by chap. xxv., § 13) L(at+® — a®)/h = a®XL(e — 1)/ ah, — ine where A = log,a, and e is Napier’s Base, namely, the finite quantity L (1+1/n)". Hence 3 Mee ALF OAr +. . vt NA, et. (2). Therefore, by (1), eae. ts +A, 2-14... .) . SAT AAs eA net tr. 8) (8). Since both the series in (3) are convergent, we must have te Aa DAG XA) ay) nA, = AAR. 198 DETERMINATION OF THE COEFFICIENTS CHAP. Using these equations, we find, successively, A,=A,A/1!, Ag=A,A7/2!, . . 5 An=AoA”/n! (4). Also, since, by the meaning attached to a, a’= +1, putting a = 0 on both sides of (1), we have | +1 =A, (5). Hence, finally, a® =1+Aw/1!+(Azy/2!+.. .+(Aw)M/ni+... (6). We see, a posteriori, that the expansion found is really con- vergent for all values of « (chap. xxvi, § 5), and also that the series in (2) is convergent for all values of w Our hypotheses are therefore justified. This demonstration is subject to the same objection as the’ corresponding one for the Binomial Series: it is, however, interest- ing, because it shows what the expansion of a* must be, provided it exist at all. We shall next give two other demonstrations, each of which supplies the deficiency of that just given, and each of which has an interest of its own. § 2.] Deduction of the Exponential from the Binomial Expansion. By the binomial theorem,* we have, provided z be numeric- ally greater than 1, {Ren Dizzee — 1 (1+ =) Nee ere hae Me z 2! ge ga(e—1)...(ze-n+1)1 + — n! ze 2 ne = aN = ed ye Y eo AU Le), wh 1 /ew)... (1 —m— 1/en) 2! n! +R, (1), where , _ at] — Lfea)... (1 — n/ea) ji ant2(1 — 1 fea)... (1— n+ 1/ea) it (n+1)! (n+ 2)! 4) a * In what follows we have restricted the value of the index za. Since zis to be ultimately made infinite, there is no objection to our supposing it always so chosen that zx is a positive integer. We then depend merely on the binomial expansion for positive integral indices. This will not affect the value of L(1+1/z)*, for it has been shown (chap. xxv., § 13) that this has the same value when z becomes + or — ©, and whether z increases by integral or by fractional increments. f XXVIII DEDUCTION FROM THE BINOMIAL THEOREM 199 Suppose now 2 to be a given quantity; and give to n any fixed integral value whatever. Then, no matter what positive or negative commensurable value z may have, we can always choose z as large as we please, and at the same time such that zz is a positive integer, p say, where p>n. ‘The series: (2) will then terminate; and we shall have l/ex<2/a<...2 x 2, this root lies 206 COEFFICIENTS IN EXPANSION OF 2/(1 — e~*) CHAP, between 1 and 2.* It will, therefore, certainly be possible to expand «/(1—e-*) in a convergent series of powers of 2 if —-l —— ee =(A,+Ac+ Ay +t... +A2)e ye Cor. We can in general sum the series Xp,(n)a"/n!(n + a)(n + 6) . (n+k), where a,b, . . ., kare unequal positive integers. The process is the same as that used in the corollary of chap. xxvii., § 5, only the details are a little simpler. (See Example 5, below.) | Example 1. To deduce the formule (3), (4), (5) of chap. xxvii., § 9, by means of the exponential theorem. (a+n)§—,Cy(xtn—-1)§+... (-)nO(w+n—r)ys+... (- ae is evidently the coefficient of 2° in 3! { elxtn)jz — Cy elttn—l)z4. te ( wi ¥iGy e(x+n—r)z are ae ( ey yrer*} sl.e7(e— 1)", maar n =s!{1+ae+% : Ae es ee eis ae \{ n n(s8rn+1), } s! { 102+ a1 “oct. came Lig oe ee Pee The lowest power of z in the product last written is 2”, and the coefficients of 2”, zt], 2nt? are s!, si(u+4n), 4s! {u?+ nxv+7yn(8n+1)} respectively. Hence (+n) —nCi(e+n—-1)§+. ow (=) Cher wer) eae ee gs O,ifis(1+1/n)”. (5.) Sum from 0 to » 2(1—- 8n+n7)er/n!. Sum to infinity (6.) 12/2!4+27/8!+37/4!+.. (7.) 13/214 23/3 14+39/4!+.... (8.) 1—23/114 89/2! -48/31+. . (9.) 144+ 24/2!1+34/3!+... Show that (10.) 1/(Qn)!—1/1'(2n-1)!4+1/2'(2n-2)!—. . . - 1/1(2n-1)!4+1/(2n)!=0. (11.) If m>3, m+ nCo(n—- 2) +nCa(n-4)% +... =n(n+3)2"-4, (12.) n — ,0i(n — 2)" + »Co(n-4)"-. . . =2"n!. (13.) By expanding e“(!-*), or otherwise, show that, if N=” A= 2 (ntr—1)!/n'(n—-1)!, then Apu — (27 +1)A,+7(7-1)A,a=0. n=1 , (Math. Trip., 1882.) (14.) Prove that (2 —23/38!+a°/5!-. . )(L—w2/2t+at/4!— 2. 2) = 2 — )ro%aPrtt/(2r+1)!. (15.) Solve the equation «?-w-1/n=0 ; and show that the nth power of. | its greater root has e for its limit when n=. ; | (16.) For all positive integral values of 2 armory Wey. (<=) p(n)a?/(n + a)(n +b)... (wth) 293 If «= 1, the series under consideration will not be convergent unless the degree of ¢(n) be less than the degree of (n+) (n+b)...(n+k). It will be absolutely convergent if the degree of ¢(n) be less than that of (n+a)(n+b)...(n+k) by two units. If the degree of (nm) be less than that of (n+ a) (n+b)...(n+k). by only one unit, then the series is semi- convergent if the terms ultimately alternate in sign, and divergent if they have ultimately all the same sign. In all cases, however, where the series is convergent we can, by Abel’s Theorem, find the sum for 7=1 by first summing for a<1, and then taking the limit of this sum when z= 1. In the special case where ¢(n) is lower in degree by two units than (n+a)(n+0b)...(n+k), and a, b,..., & are all positive, an elegant general form can be given for Sp(n)/(n + a) 1 (n+b)...(n+h). From the identity p(n)/(n + a) (n +b)... (n +k) =A/(n+a)+B/n+b)+. . .+K/(n+h), we have p(n) =A(n + b)(n+c)...(n+h)+Bint+a)(n+c)...(n+h+... +K(m+a)(n+6)... (+7) (5), _ and, bearing in mind the degree of (n), we have | A+B+...+K=0 (6). Also, putting in succession n= —a, n= —),..., n= —k, we have A = ¢( -a)/(b- a) (c- a)... (k-a) B= ¢(-))/(a—b) (¢- 6)... (&-8) (7). K= $(-h)/(a-k) (b-&)... (G-#) _ Reverting to the general result, we see from (4) that E p(n)" [(n +.a)(n+b)...(n +k) = —2Ax U/l +a/2+. . .+2%/a)-log(1—2).2Ax-* (8), _ where the = on the right hand indicates summation with respect me 0,4, ., ke. 224 EXAMPLES CHAP. Now, since A+B+...+K=0, ZAx~* is an algebraical function of x which vanishes when z=1. Also 1-—za is an algebraical function of x having the same property. Therefore, by chap. xxv., § 17, we have | L log (1 -— a). 2Ax-4 “=1 L log{ (1 — )2Ae*}, t=] = log 1, = 0. Hence, taking the limit on both sides of (8), we have, by Abel’s Theorem, SH(n)/(n+a)(n+d)... (wth) = -SA(V/1 41/24... + 1a) _ sho MU/l + 1/24... + 1a) 7 + Ga) (esa) ate SE the = on the right denoting summation with respect to Cs ete a eee: io.0) Example 1. Evaluate 2n3x”/(n — 1) (n+2). 2 We have = n8x”/(n — 1) (n+2)=(n—1)ar+ 40" (n—1) + sarl(n +2), { j ao Now Z(n — 1)" = 1a? 4-2a8 4 Bet+., . ., 9 > ey ‘ (1 — x)?2(n — 1)a" = 1a? + 243 4+ B8a44... i 2 | = ile 2. Sot " leas =i. 5 ra Hence (2 — 1)a=27/(1 - x), 2 loa) ie2) Also $200"/(n —1)= haeda"-l/(n - 1), 2 2 = — 3x log (1-2); a) 2 4] $2a"/(n + 2) = $a-*Zant?/(n + 2), ai 2 2 = — 3a? {a/1 +2°/2+293/3 +log (1-2)}. Hence the whole sum is / / S( )} 2*/(1 — a) ~ $1 — $ — 89— ¥(w + 82%) log (1-2). ive) Example 2. Evaluate 21/(n — 1) (7 +2). 2 By the same process as before, we find 17.2) ee —1)(m+2)= fa) + 34 4a+4 Ma — x) log (1-2). Vili EXAMPLES 225 Now, since L (l-a)*-*=1 1 (chap. xxv., § at} L (a-*- x) log (1-—w)=0. a=1 z=1 Therefore Z1/(n —1)(n+2)=4§4+34+4=H. This result might be obtained in quite another way. It happens that D1/(m—1)(2+2) can be summed to 2 terms. In fact, we have 1/(m— 1) (w+ 2)=$ {1/(n-1) -1/(n+2)}. Hence, since the series is now finite and commutation of terms therefore permissible, sha Dele di 1 1 1 age Mie 2) al gina / re nin Bene CT 1. ant. pean noha! 4 nm-4 n-3 n-2 n-1l T 1 1 m n+l n+2 me eee aie Vogl 1 rip ds sahuny ts neo Hence, taking the limit for n=, we have 1 fl il ‘alt B=3(7+5+5 “18 Example 3. ‘To sum the series az oa GR te ee int uctnd oh a fae Giaila Giro : (Lionnet, Nouv. Ann., ser. il, t. 18.) Let the (7+1)th term be wy, then, since w,, =0, association is permitted (see chapter xxvi., § 7), and we may write en % él i 1 -e4ep-)° 4n+3° 2242’ 1 il 1 Ni 1 1 =In+1 In+27 n48 dnt" dnF2 n4¥ Aa ae TS) eee) ~\4n+1 4n4+2° 4n4+3 4n4+4/ °2\2n41 - 2n+2/’ =Un+ Wn, Say. Now, as may be easily verified, v, and w, are rational functions of ., in which the denominator is higher in degree than the numerator by two units at least. Hence (chap. xxvi., § 6) Dv, and Zw, are absolutely convergent series. Therefore (chap. xxvi., § 13) iva) “t = Z(Un+ Wn), 48 i?) Unt ZW 0 SOMS8o OLA TT 226 INEQUALITY THEOREMS CHAP. Hence, again dissociating v,, and wy, (as is evidently permissible) we have 2 Ds lee lle ele anl ad l= Tesi ag eal 1 1 1a Rie 373747508 eee ) =log,2+4lo by § 9 above, = $log,2. « This example is an interesting specimen of the somewhat delicate opera- tion of evaluating a semi-convergent series. The process may be described as consisting in the conversion of the semi-convergent into one or more absolutely convergent series, whose terms can be commutated with safety. It should be observed that the terms in the given series are merely those of the series 1 - 1/2+1/3- 1/4+1/5-—. . . written in a different order. We have thus a striking instance of the truth of Abel’s remark that the sum of a semi- convergent ee may be altered by commutating its terms. APPLICATIONS TO INEQUALITY AND LIMIT THEOREMS. § 15.] The Exponential and Logarithmic Series may be applied with effect in establishing theorems regarding inequality. Thus, for example, the reader will find it a good exercise to deduce from the logarithmic expansion the theorem, already proved in chapter xxv., that, if z be positive, then e-1>loga>1—-I1/e (1). It will also be found that the use of the three funda- mental series—Binomial, Exponential, and Logarithmic—greatly facilitates the evaluation of limits. Both these remarks will be best brought home to the reader by means of examples. Example 1. Show that loo ot 1 Do ee See n+1 m-1l m m+1l m+2 nN If we put 1-1/x=1/m, that is, z=m/(m—-1), in the second part of (1) above, and then replace m by m+1, m+2, .. ., m successively, we get log m — log (m-—1)>1/m, log (m+1)-logm>1/(m+1), log n — log (~-1)>1/n. Hence, by addition, log-log (m—1)>1/m4+1/(m+1)+. . .+1/n (2). ie XXVIII LIMIT THEOREMS, EXERCISES XIII 227 Next, if we put z-1=1/m in the first part of (1), and proceed as before, we get log (m+1)-logm<1/m, log (2 +2) —log (m+1)<1/(m+1), log (x +1) -logn<1/n. Hence log (n+1) -logm<1/m+1/(m+1)+. . .+1/n (3). From (2) and (38), — . log {n/(m—1)} >1/m+1f(m+1)+. . .+1/n>log {(n+1)/m}. Example 2. If p and q be constant integers, show that L {1/m+1f(m+1)+. . .+1/(pm+q)} =logp. m= 0% F (Catalan, Zraité Hlémentaire des Séries, p. 58.) Put n=pm-+q of last example, and we find that log {((pm + q)/(m —1)} >1/m+1/(m+1)+...+1/(pm+q)>log ((pmt+qt1)/m}. Now L log {(pm+4q)/(m—1)} =log p, M>=P : and L log {(pm+q+1)/m} =log p. m= Hence the theorem. Example 8. Evaluate L(e* — 1)?/{a- log (1+z)} when w=0. Since (era (ober eee 1 aa...) S x—log(1+a)=427-Je8 +... =he7(1-3e+...). Therefore (c*—-1)/{a—-log (1+a)} =2L+4e+. . .)/(L-get.. .). Since the series with the brackets are both convergent, it follows at once that . L(e* — 1)?/{a—-log (1+a)} =2. Exercises XIII. - (1.) If P=1/81+-1/3.31°+1/5.31+.. ., Q=1/49 +1/3.493+1/5.49°+. . ., R=1/161 +1/3.1613+1/5.1619+. . then log 2=2(7P+5Q+3R), log 3=2(11P+ 8Q+5R), log 5=2(16P +12Q+7R). (See Glaisher, Art. ‘“ Logarithms,” Zncy. Brit., 9th ed.) (2.) If a=-log(1-1/10), b= —-log (1-4/100), c=log(1+1/80), d= — log (1 — 2/100), e=log (1+ 8/1000), then log 2=7a—- 26+ 3c, log3=1la-3b + 5c, log5=16a-4b+7c, log 7=4(39a-10b+17¢-d)=19a —4b+8c+e. (Prof. J. C. Adams, Proc. R.S.L. ; 1878.) (3.) Calculate the logarithms of 2, 3, 5, 7 to ten places, by means of the formule of Example 1, or of Example 2. (4.) Find the smallest integral value of x for which (1:01)*>10z. ae 228 TREX ERCIShoe Le CHAP. ~ Sum the series :— (5.) £24 /1 (a8 — Bar)! + 23/3(a3 — eas ; (6) 1+ (5+ sat +(G+ +3)at + (547 7 ne (7.) oh/1.2—2?/2.8+27/3.4—-. . eerie e ie (8.) 22/3+a4/154+.. . +2°"/( (4nd — 1) 4 (9.) wef 12+ a2 /(12 + 22) +03/(12 +274 37)4+ pan? +2 +. 0.407) 4.2 ae also 1/124 1/(127+27)+1/(174+2?+37)+. . +H (1?+2?+.. peers oy - (10.) 4/1.2.84+6/2.3.44+8/3.4.5+... (11.) If a>100, then, to seven places of decimals at least, log(#+8)= _ 2 log (a+7)— log (w+5) — log (a+ 8) + 2 log w—log (w-3)—log (xw-5) + 2 log (z—-7) —log (w- 8). (12.) Expand log (1+#+2*) in ascending powers of x. (13.) From log (+= log (x+1) +log (a-—a#+1), show that, if m be a positive integer, then 6m—2 (6m-8)(6m—4) (6m — 4) (6m-—5) (6m—-6 . * - ea! )( + )( hy Aer (Math. Trip., 1882.) (14.) {log.(1 + a)}2= 20/2 - 2(1/1+1/2)a/3+ . . (—yr2{a/14+1/2+... 1/(n-1)ta/n. . . Does this formula hold when tes) 2 ae log (1+2)!og Q-#) = — Qia?/1 — Qsxt/2-. . »- Qon-1 nl. a where . Qon-1= 1/1 -1/2+1/8-.. . .+1/(2n-1) (16.) If <1, show that xt a7 + fart + zeal. : .=log; {1/( (1—- a)} = LP; 4P5+é Pg —+P7-4Po9+75Pi0- Sie where P,=a"+a2"+ai+a84ql6r4,.., and the general term is (—)"P,/n, unless 7 is a power of 2, in which case there is no term. (Trin. Coll., Camb., 1878.) (Li vale ot x eels x e@l5 , | = Ant Aye+.. ., then Ao-=—Aoj—1.0-9 0 (Or 1)/204,6 5 soar, (18.) If vtage?+ase®+.... Hytasy®tasyi+. . .= {(et+y)/(l-ay)}i+ a3{(v+y)/(1—ay)}? +a5{(w+y)/(1—ay)}>°+.. ., for all values of x and y which - render the various series convergent, find ag, @,... ° Show that (19.) log (4/e)=1/1.2 -1/2.84+1/8.4—-1/4.5+... (20.) log 2=4(1/1.2.841/5.6.7+1/9-10.11+1/13.14.15+. . .) (Euler.) (21.) (1 — 1/2—1/4) + (1/3 - 1/6 - 1/8) + (1/5 -1/10-1/12) +. . .=glog2. (See Lionnet, Nowv. Ann., ser. ii, t. 18.) (22.) o4/1!—noe/2!4+n(2—-1)o3/3!-. .. to n+1 terms =1/(n+1), where Gp= 1/1 l/2+1/8+...4+1/r. (Math. Trip., 1888.) - (23.) e~(1+1/m)™ lies between e/(2m+1) and e/(2m+2), whatever m may be. (Nowv. Ann., ser. ii, t. 11.) (24.) L{a/(w@-1)-Ifloga} =4, when w=1. (Kuler, Jnst. Cale. Duff) (25.) Liet—-1-—log(1+2)}/z?=1, when 2=0. (Euler, /,c.) (26.) Liat—ax)/(1-a+logx)=-2, when z=1. (Euler, Zc.) Oise bo bo eo) XXVIII EXERCISES XIII (27.) L(L+1/n)¥(1+2/n)/". 2. (L+n/n)™=4/e, when n=0. (28.) L{(2n-1)!/n?"-1} n= 4/e?, when n= (29.) e*>1+2, Be all real values of a. (30.) e-1>log~>1-1/x, for all positive values of xz; to be deduced from the logarithmic expansion. (31.) e*'>(1+n)"/n!, n being any integer. (32.) If m be an integer >e, then n™+1>(n+1)”. (33.) If A, B, a, b be all positive, then (a — b)/(A — B)+(Aa-— Bd) log(B/A) (A-B)? is negative. (Tait.) (34.) Ife>y>a, then {(x+a)/(z—-a)}*< {(y+a)/(y-a)} 5.) L{1/(n+1)+1/(n+2)+...4+1/2n} =log2, when n=. (Catalan.) (36.) log {(m+4)/(m—4)} >1/m+1f(mt1)+. . .+1/n>log {(n+1)/m}. (Bourguet, Nouv. Ann., ser. ii, t. 18.) (37.) log 3=5/1.2.8+14/4.5.6+. . .+(9n-4)/(8n—-2)(38n—-1)38n+... (38.) If > —)-1g(n)/(n+a)(n+b). . . (n+), where a, b,..., kare all 1 positive integers and ¢(z) is an integral function of x, be absolutely conver- gent, its sum is sa d(—a){1/a-1f(a-1). . .(-)* 11/1} /(b-a)(c—a). . . (k-a); COs etess Ke and, if it be semi-convergent, its sum is S+log2 ° = (-)*(-a)/(bD-a)(c—a)... (k-a). thy Ooo 5 "(39.) Show that the residue in the expansion of log {1/(1-2)} lies between art {1 +(n+1)a/(n+2)'/(n+1) and arth 1+ (+ 1)e/(1— 2x) (1 +2)}/(n+1). (40.) In a table of Briggian Logarithms in which the numbers are entered to 5 significant figures, and the mantisse of the logarithms to 7 figures. Calculate the tabular difference of the logarithms when the number is near 30000 ; and find through what extent of the table it will remain constant. CHAPTER XXIX. Summation of the Fundamental Power Series for Complex Values of the Variable. GENERALISATION OF THE ELEMENTARY TRANSCENDENTAL FUNCTIONS. § 1.] One of the objects of the present chapter is to generalise certain expansion theorems established in the two chapters which precede. In doing this, we are led to extend the definitions of certain functions such as a%, log, 2, cos 2, &e., already introduced, but hitherto defined only for real values of the variable «; and to introduce certain new functions analogous to the circular functions. Seeing that the circular functions play an important part in what follows, it will be convenient here to recapitulate their leading properties. This is the more necessary, because it is not uncommon in English elementary courses so to define and discuss these functions that their general functional character is lost or greatly obscured. § 2.] Definition and Properties of the Diréct Circular Functions. Taking, as in chap. xii, Fig. 1, a system of rectangular axes, we can represent any real algebraical quantity 0, by causing a radius vector OP of length r to rotate from OX through an angle containing 6 radians, counter-clockwise if @ be a positive, clockwise if it be a | negative quantity. If (a, y) be the algebraical values of the co- ordinates of P, any point on the radius vector of @, then 2/7, y/r, y/a, zy, v/2, r/y are obviously all functions of 6, and of @ alone. The functions thus geometrically defined are called cos 0, sin 8, CHAP, XXIX EVENNESS, ODDNESS, PERIODICITY — 231 tan 6, cot @, sec0@, cosec@ respectively, and are spoken of collectively as the circular functions. All the circular functions of one and the same argument, 0, are algebraically expressible in terms of one another, for thei definition leads immediately to the equations tan 0 =sin 6/cos @, cot 6 = cos 6/sin 0 ; sec0=1/cos 0, cosec 0=1/sin 0; (abe cos°6+sin°@=1, sec’O—tan*O=1; from which it is easy to deduce an expression for any one of the six, cos 0, sin 0, tan 0, cot 0, sec 0, cosec @, in terms of any other. When F(6) is such a function of 6 that F(— 6) = F(@), it is said to be an even function of 6; and, when it is such that ° F( — 0) = — F(6), it is said to be an odd function of @ For example, 1 + @° is an even, and 6 — 46° is an odd function of @. It is easily seen from the definition of the circular functions that cos 9 and sec 9 are even, and sin 0, tan 6, cot 6, and cosec 0 odd functions of 6. When F(@) is such that for all values of 6, F(@ + nd) = F(A), where A is constant, and ” any integer positive or negative, then F(0) is said to be a periodic function of @ having the period A, It is obvious that the graph of such a function would consist of a number of parallel strips identical with one another, like the sections of a wall paper; so that, if we knew a portion of the graph corresponding to all values of @ between a and a+ A, we could get all the rest by simply placing side by side with this an infinite number of repetitions of the same. Since the addition of + 27 to @ corresponds to the addition or subtraction of a whole revolution to or from the rotation of the radius vector, it is obvious that all the circular functions are periodic and have the period 27. This is the smallest period, that is, the period par eacellence, in the case of cos @, sin 6, sec 0, cosec 8. It is easily seen, by studying the defining diagram, that tan @ and cot 0 have the smaller period 7. Thus we have 203 ZERO AND TURNING VALUES CHAP. XXIX cos (9 + 2nr) =cos 0, sin (0 + 2nm) =sin 8, sec (0 + 2nz) =sec 0, cosec (0 + 2nm) = cosec 6, \ (2) tan (0+ m7) =tan 6, cot (0 + mr) = cot 8. Besides these relations for whole periods, we have also the following for half and quarter periods :— cos(r + 6)= —cos6, sin(r+6)= ¥sin@; cos (47 + 6) = Fsin 0, sin(47r + 0)= +0080; : tan (47 + 0) = + cot 0, cot(4r+0)= = tan 0; (3), Ge all easily deducible from the definition. We have the following table of zero, infinite, and turning values :— 0 0 47 1 Sar Qr | &e. God ttt Ok0 eae eer sin@| O es eee a, —1/ 0 tan) 0 | o | 0] o'| 0 | . 77 mana cot 6 | a 0 0 0 CO j sec@| +1.) wo }—1] > oj} +1 cosée G4 co) co Sha bo which might of course be continued forwards and backwards by adding and subtracting whole periods. Hence cos @ has an infinite number of zero values correspond- ing to = 4(2n + 1)z, where n is any positive or negative integer ; no infinite values ; an infinite number of maxima and of minima values corresponding to 6=2n7r and 6=(2n+ 1) respectively ; and is susceptible of all real algebraical values lying between —land +1. Sin 6 is of like character, But tan 0 is of quite a different character. It has an infinite number of zero values corresponding to 6=n7; an infinite number of infinite values corresponding to 0=4(2n+1)m; no turning values ; and is susceptible of all real algebraical values between — © and +o, Cot 0 is of like character. 4 4 ’ \ 27 47r ee) ote { Py \ \ ry A A A A 1} A A 4 1} H 1 a ' 4 iy asaeccaere2e eee _eownsswcwon sama os os" oo? Nee he -[ A Ff \ ES \ ook t A t t t ‘ ' ‘ ‘ t ‘ é ' a ! i] a a - .. \ . XN \ 4 ‘ \ \ 1 \ 1 ‘ ‘ \ ‘ \ ry \ 1 ‘ 1 \. \ ‘ : H ‘ ‘ ‘ ‘ 1 1 t i U / NAN TT cae 30 hes d -¢ ’ A feo r oY ' \ 1 1 1 ‘ 1 i \ ’ i { \ t \ ' a ‘ \ ! a ' H i | ' H ’ ’ i] i} 2 K Fic. 1, 234. ADDITION FORMULA CHAP, Sec 6 and cosec 6 have again a distinct character. Hach of them has infinite and turning values, and is susceptible of all real algebraical values not lying between —1 and +1. The graphs of the functions y= sina, y = cos z, &c., are given in Fig, 1. The curves lying wholly between the parallels KL, K’L’, belong to cos # and sin a, the cosine graph being dotted ; all that lies wholly outside the parallels KL, K’L’, belongs either to sec a or to cosec 2, the graph of the former being dotted. The curves” that lie partly between and partly outside the parallels KL, K'L’, belong either to tanz or to cota, the graph of the latter being dotted. Again, from the geometrical definition combined with ele- mentary considerations regarding orthogonal projection are deduced the following Addition Formule :— cos (6 + ¢) = cos 6 cos } + sin Asin ¢ ; ) J sin (6 + $) =sin 6.cos d + cos Osin ¢ ; (5) tan (6 + ¢) = (tan 6 + tan ¢)/(1 + tan 6 tan ¢). i As consequences of these, we have the following :— cos @ — cos = 2 sin 3(6 + 4) sin (0 — 4) ; cos 6 + cos f= 2 cos 4(6 + ) cos $(0— 4); ; (6) sin 0+ sin ¢ = 2 sin $(0 + ¢) cos $(6 = @). cos 6 cos fd = 4 cos (4 + ¢) hit | sin Osin ¢ = } cos (6 - d) — $ cos (0+ 4); (7) sin 0 cos $= 4} sin (0+) + $sin(6 — ). i cos 20 = cos °6 — sin*6 = 2 cos 9 - 1 =1-2sin*6 | = (1 - tan °*6)/(1 + tan °6). (8). sin 26 = 2 sin 6 cos 6 = 2 tan 6/(1 + tan *6). tan 20 = 2 tan 6/(1 — tan 6). § 3.] Inverse Circular Functions. When, for a continuum — (continuous stretch) of values of y, denoted by (y), we have a relation a = Fy) (1), ve; XTX INVERSE CIRCULAR FUNCTIONS 235 which enables us to calculate a single value of « for each value of y, and the resulting values of « form a continuum (2), then the graph of F(y) is continuous; and we can use it either to find x when y is given or y when w is given. We thus see that (1) not only determines @ as a continuous function of 7, but also y as a continuous function of z The two functions are said to be inverse to each other; and it is usual to denote the latter function by F-1(z). So that the equation y =F) (2) is identically equivalent to (1). It must be noticed, however, that, although F~'(@) is con- tinuous, it will not in general be single-valued, unless the values in the continuum (a) do not recur. This condition, as the student is already aware, is not fulfilled even in some of the simplest cases. Thus, for example, if #=y", for -0 1, then tan-ta + tan-!y =a + tan! [(@ + y)/(1 — ay)];* and, in general, it is easy to show that mtan~1% +, tan~'y=(m+n-+ p)r + tan-l{(a + y)/(1 — ay)}, = m+n+P tan~*{(# + y)/( — ay)} (1 1), where p=1, 0, or — 1, according as tan-!7+tan-!y is greater than $7, lies between 47 and — 3a, or is less than — $z. ON THE INVERSION OF w=2", § 4.] When the argument, and, consequently, in general, the value of the function are not restricted to be real, the dis- cussion of the inverse function becomes more complicated, but the fundamental notions are the same. 7 For the present it will be sufficient to confine ourselves to the case of a binomial algebraical equation. Let us first consider the case pow (1), where » is a positive integer, z is a complex number, say 2=%+ yt, and, consequently, w also in general a complex number, say W=U+ UM. To attain absolute clearness in our discussion it will be * In English Text-Books equations of this kind are often loosely stated ; and the result has been some confusion in the higher branches of mathematics, such as the integral calculus, where these inverse functions play an important part. XXIX INVERSION OF w=z” . 239 necessary to pursue a little farther the graphical method of Shap. xv., § 17. It follows from what has there been laid down, and from the fact that any integral function of ~ and y is continuous’ for all finite values of « and y, that, if we form two Argand Diagrams, _ one for x + yi (the z-plane), and one for w + vi (the w-plane), then, whenever the graphic point of 2* describes a continuous curve, the _ graphie point of w also describes a continuous curve. In this sense, _ therefore, the equation (1) defines w as a continuous function of _ # for all values, real or complex, of the latter. For simplicity in what follows we shall suppose the curve described by z to be the whole or part of a circle described about the origin of the z-plane. We shall also represent z by the standard form r(cos 6 + 4 sin 8), and w by the standard form s(cos $ +i sin ¢) ; but we shall, con- _ trary to the practice followed in chap. xii., allow the amplitudes @and ¢ to assume negative values. Thus, for example, if we wish to give ¢ all values corresponding to a given modulus r, without repetition of the same value, we shall, in general, cause 6 to vary continuously from —7 to +7, and not from 0 to 27, as heretofore. In either way we get a complete single revolu- tion of the graphic radius; and it happens that the plan now _ adopted is more convenient for our present purpose. It is obvious that by varying the amplitude in this way, and then giving all different values to r from 0 to + 0, we shall get _ every possible complex value of z, once over; and thus effect a complete exploration of any one-valued function of z. Substituting in (1) the standard forms for w and 2, and _ taking, for simplicity, n = 3, we have s(cos @ +7 sin ¢) =7"(cos 6 +7 sin 6)? =1"(cos 36 +7 sin 30) (2) by Demoivre’s Theorem. Hence we deduce | s=r, 6=30+4 2Qnz; _ * For shortness, in future, instead ‘‘of graphic point of z” we shall say *“‘z” simply. 240 CIRCULO-SPIRAL GRAPHS CHAP, or, if (as will be sufficient for our purpose) we confine ourselves to a single complete revolution of the graphic radius of 2, Sn wipe (3). F i] If, therefore, we give to 7 any particular value, s has the — fixed value 7°; that is to say, w describes a circle about the origin of the w-plane (Fig. 4). Also, if we suppose z to describe its circle (Fig. 3) with uniform velocity, since A= 36, w will describe the corresponding circle with a uniform velocity three times as great. To one complete revolution of z will therefore Hie: 3. Fia. 4. correspond three complete revolutions of w. In other words, the values in the (w)-continuum which correspond to those in the (z)-continuum are each repeated thiee times over.* The actual course of w is the circle of radius 7° taken three times over. We may represent this multiple course of w by drawing round its actual circular course the spiral 0’, 1, 1’, 0, 1’, 1, 0’, which re-enters into itself at 0’ and 0’. The actual course may then be imagined to be what this spiral becomes when it is shrunk tight upon the circle. * To indicate this peculiarity of w we shall occasionally use the term ‘“Repeating Function.” A repeating function need not, however, be periodic eG as w=2 is. ee XXIX RIEMANN’S SURFACE IA) If we now letter the corresponding points on the z-circle ‘with the same symbols we have the circle 0’11’ in the w-plane, cor- responding to the circular are 011’ in the z-plane, and so on, in this sense that, when z describes the arc 0'11’, then w describes the complete circle 0'11’, and go on. It follows from this graphical discussion that the equation ° w= 2", which defines w as a one-valued continuous function of z for all Bg of 2, defines 2 as a three-valued continuous function of w for all values of w. | In other words, since, in accordance with a notation already defined, (1) may be written XR = th NAGLE: we have shown that the cube root of w is a three-valued continuous function of w for all values of w. It is obvious that there is nothing in the above reasoning peculiar to the case n= 3, except the fact that we have a triple spiral in the w-plane, and a trisected circumference in the z-plane. Hence, if we consider the equation | Dee it (4), and its equivalent inverse form = lw (4’), all the alteration necessary is to replace the triple by an n-ple spiral, returning into itself on the negative or positive part of the w-axis, according as ” is odd or even; and the trisected cir- cumference by a circumference divided into n equal parts. Thus we see that the equation (4), which defines w as a continu- ous one-valued function of 2 for all values of 2, defines z (that is, the nth root of w) as a continuous n-valued Sinha of w for all’ values of w. §5.] Riemann’s Surface. It may be useful for those who are to pursue their mathematical studies beyond the elements, to illustrate, by means of the simple case w=2", a beautiful method for representing the continuous varia- tion of a repeating function which was devised by the German mathematician Riemann, who ranks, along with Cauchy, as a founder of that branch of modern algebra whose fundamental conceptions we are now explaining. VOL, II R v 242 BRANCHES OF fw - CHAP, Instead of supposing all the spires of the w-path in Fig. 4 to lie in one E plane, we may conceive each complete spire to lie in a separate plane super- — posed on the w-plane. Instead of the single w-plane, we have thus three separate planes, Pj, Po, Pi, superposed upon each other. To secure continuity between the planes, each of them is supposed to be slit along the w-axis from 0 to —«; and the three joined together, so that the upper edge of the slit in Po is joined to the lower edge of the slit in Py; the lower edge of the slit in Py to the upper edge of the slit in P,; the lower edge of the slit in P; to the upper edge of the slit in P;, this last junction taking place across the two — intervening, now continuous, leaves. We have thus clothed the whole of the w-plane with a three-leaved continuous flat helicoidal* surface, any continu- ous path on which must, if it circulates about the origin at all, do so three times before it can return into itself. This surface is called-a Riemann’s — Surface. The origin, about which the surface winds three times before return- ing into itself, is called a Winding Point of the Third Order. Upon this three-leaved surface w will describe a continuous single path corresponding to any continuous single path of z, provided we suppose that there is no con- tinuity between the leaves except at the junctions above described. § 6.] If we confine @ to that part 1/01’ of its circle which | is bisected by OX, and ¢ to the corresponding spire 1/01’ of its path, so that # lies between —7z and +7, and @ between —a/n and +7/n, then z becomes a one-valued function of w for all values of w. We call this the principal branch of the n-valued — function %/w; and, as we have the distinct notation w/” at our — disposal, we may restrict it to denote this particular branch of — the function z. In other words, if w=s(cospt+isind), -7 tan ¢; tan 0... tan 6,. The formule (1), (2), (3) are generalisations of the familiar addition formule for the cosine, sine, and tangent. From the usual form of Demoivre’s Theorem, namely, cos nf + 7sin nO = (cos 6 + isin 8)”, we derive, by expansion of the right-hand side, cos n@ + 7sin nO = cos”6 + 1,0, cos”~16 sin 6 — ,C, cos” -26 sin24 : — inG; cos”- 76 sin?O + ,C,cos”-40 sin40 +... Hence cos 20 = cos”0—,,C,cos”~70 sin?6 + ,C,cos”- 40 sin46—... (4);* sin n6 = ,C, cos”-10 sin 6 — ,,C, cos” 36 sin3@ +nU,cos"-°@sin®F—... (5); pet ne neytan @ —,O0,tan’é+,C.tan°d—... . (6) | 1, Op tan: 070) tan*é— 4. These are generalisations of the formule (8) of § 2. The formule (4) and (5) above at once suggest that cos 70 can always be expanded in a series of descending powers of cos 6; that, when m is even, cos n@ can be expanded in a series of even powers of sin @ or of cos 6; sin 6/sin 6 in a series of odd powers of cos@; and sin n6/cos 6 in a series of odd powers of sin 0: and, when m is odd, cos ”@ in a series of odd powers of cos 6; cos 76/cos 6 in a series of even powers of sin 0; sin 6 in a series of odd powers of sin @; sinn6/sin 0 in a series of even powers of cos 6. * The formule (4), (5), (6), (8) were first given by John Bernoulli in 1701 beee Op,,.t..1., p. 387). 252 EXPANSIONS IN POWERS OF SIN @ AND COS 0 CHAP. Knowing, « priori, that these series exist, we could in various ways determine their coefficients ; or we could obtain certain of them from (1) and (2) by direct transformation; and then deduce the rest by writing $7 — @ in place of 0. (See Todhunter’s Trigonometry, $§ 286-288.) | We may, however, deduce the expansions in question from the results of chap. xxvii., § 7. If in the equations (9), (10), (9’), (9”), (10’), (10”) there given we put a =cos 0 +7sin 6, 6 = cos 6 — 7sin 6, and therefore oe = 2cos 6, g=1, we deduce Le BU Sane cos 6)"-4 — 2 cos n0 = (2 cos 0)" — a “(2 cos O)—2 + ne—r—1)(a-F=2).. (= 3 + apg gery (7) * i), r! : Bi vir < =o ~3)(n—4 sin n9/sin 6 = (2 cos 6)"-1 -— - (2 cos O)"-3 + ee) ies ue ) ie pats n—r—1)(n-—r—2)...(n—- 2r) (2.008 0)" ete (oe (2:cos'@)" 7-2 anna Re Tk BAL NORE ae (-)' n(n — 2')... (wi = 28-2 sia ; fo even) (9); (2s) ! cos 16 = Sas g hm - 1) neag , Mm — 1) (w= 8) 3! 5! 2 ae it Pz fa eee Men yen(n L*) (nm i i aie Lee : 25-1") esti Pe (n odd) (10); sin nO /sin 0 = ( — )”/2- ne os 0 — alee, cos*O+... Pi re ee, ) ee ~ cos +19 +...t(meven) (11); (Say -L@veven) (11); * The series (7), (9’), (10') were first given by James. Bernoulli in 1702 (see Op., t. Se ., p- 926). He deduced them from the formula 2sin2nd=- a1 (28 sin 6)? 2 in (2sin 0)* + at ee ae zs which he established by an induction based on the previous results of Vieta regarding the multisection of an angle. (2 sin 8)® — XXIX EXPANSIONS IN POWERS OF SIN @ AND COS @ 253 told 2 2 2 2 2 sin?6/sin 0 = (yon 1 - sae cos’0 + iia) rest) ne ee (= yd sel) — 3’)...(n — Dg ars) (2s) ! If in the above six formule we put da — 6 in place of 6, we derive six more in which all the series contain sines instead of cosines. In this way we get, inter alia, the following :— cos6+.. Jo odd) (12). he. 9, n(n —2°) . ; cos n@ = 1 — sin |S og egumee th mvp TReVel yuma as).: ] : gon he ‘ entity vay . sin nd = 7 sin 9 — MAT) ain 2g 4 aN et Fi (n odd) (10’); , ; . cate h ue ref) tes hae ae sinn@/cos 6 = {sind = By on 6 + eset) C ") Gie-even) “(11')s oe a ek? Ha OU le cos n0/cos Gel = sin 6 + eee gin “@ = (nx odd) (12’). The formule of this paragraph are generalisations of the familiar expressions for cos 26, sin 20, cos 36, and sin 30, in terms of cos 0 and sin 0. § 13.] The converse problem to express cos”6, sin "@, and, generally, sin” 6 cos”@ in a series of sines or cosines of multiples of 6, can also be readily solved by means of Demoivre’s Theorem. If, for shortness, we denote cos 6 +7 sin 6 by a, then we have, by Demoivre’s Theorem, the following results :— #=cos@+isin6, 1/z=cos 6—isin 6; a”=cosné+isinné, 1/a”=cosnO—isinnd: 1 1 | oe O= 52 + 1/a), sn@= a =i ja); (1). 1 ; ] cos n0 = 5” Sf"), sinnd= 92” a1 a"), 254 EXPANSIONS IN COSINES AND SINES OF MULTIPLES OF @ cnap. Hence 1 cos "6 = 55,,(@ + 1/2), (ae a a “i Pn ON Ee: ah gees) tam Og 0 Tt Lila!) ers {cos 2718 + omC, cos (2m — 2)0 + omC,cos(2m — 4)0 + + domOm} (2). = —— { yan a 0°m — 24m 1 Similarly, i bos “UTA O Fam t©O8 (2m + 1)6 + om4iC, cos (2m — 1)0 + omtiU, cos (2m ~— 3)O +... .+oma,0m cos Op (a)e ; ( — 1) + omO, cos (2m—4)0+.... (—)™dbenCm} (4); [ey sin 2m+19 = {sin (2m + 1)0 — om4.C, sin (2m — 1)0 92m + on-tiOg sin (2m'=3)0. . . {— ong. Cy Sit Gene These formule are generalisations of the ordinary trigonometrical formule sin?0= — }(cos 20 — 1), cos 36 =1(cos 36 + 3 cos 8), &e. In any particular case, especially when products, such as sin "4 cos”6, have to be expanded, the use of detached coefficients after the manner of the following example will be found to con- duce both to rapidity and to accuracy. Example 1. To expand sin °@ cos *@ in a series of sines of multiples of @. sin 56 cos 34 = aes ( - 1x) +1/zx). Starting with the coefficients of the highest power which happens to be remembered, say the 4th, we proceed thus— Coefficients of Multiplier. Coefficients of Product. 1-4+ 6- 441 1-1 | 1-5+10-104+5-1 141) 4 =—445 95 0b ae | 14+1/)1-34+ 14+ 5-5-143-1 | 1+1 | 1>9= 94 °640-64+290—" | The coefficients in the last line are those in the expansion of (a — 1/x)5(#+1/a)°. Hence, arranging together the terms at the beginning and end, and replacing 7 3 7s XXIX EXERCISES XV 255 — 1/x°) by sin 86 — 1/2) by sin 60, and so on, we find 1 (0 er ae 2 sin 60 — 2 sin'40+6 sin 20+4, 0}, =e isin 80 — 2 sin 60 — 2 sin 49+ 6 sin 20}. The student will see that sin™6cos”@ can be expanded in a series of sines or of cosines of multiples of 8, according as m is odd or even. The highest multiple occurring will be (m + n)0, Example 2. If @=27/n, and a any angle whatever, and mUn=cos”a+cos™(a+0)+...+c0s m™(a+n-16), mVn=sin™a+sin™(a+0)+...+sin™a+n— » — 16), where m is any positive integer which is not of the form kis coe then omUn=mVn=1.1.8... Qm—1 1)/2.4... 2m; em+1U n= am+1V n= 0. This will be found to follow from a combination of the formule of the present paragraph with the summation formule of § 11. EXERCISES XV. Sum the following series to » terms, and also, where admissible, to infinity :— (1.) cosa—cos (a+6)+ cos (+26) — (2.) sina —sin(a+6)+sin(a+20)-... (3.) Zsin 2n0. (4.) 2 cos 6+(n—1) cos 20+(n-2)cos80+... (5.) Zsin né cos (n+1)0. (6.) Zsin nO sin 2n8 sin 3n0. (7.) sina—cosa sin(a+6)+cos?asin(a+20)—... (8.) 1+cos 0/cos 8+ cos 26/cos 26 + cos 30/cos90+... ton terms, where d=nr. (9.) 1—2rcos 0+ 8r? cos 20-473 cos 80+... eee sin@6+3sin20+5sin30+7sin4d@+... (11.) Sn? cos (nO +a). (12.) En(v+1) sin (2n+1)8. (13.) sin 270 — o,0; sin (2 — 2) + o,Ce sin (2n—4)0—... (na positive inte- ger). (14.) sin (22 +1) + 2n41Cy sin (27 — 1)0 + on41Co sin (2n-—3)0+... (na posi- tive integer). (15.) Zm(m+1).. . (m+n-1)r"cos (a+76)/n! to infinity, m being a positive integer. (16.) Does the function (sin °0+sin*20+. . . +sin 2n6)/(cos20+cos 220+. . . +c0s *70) approach a definite limit when n= ? (17.) Expand 1/(1-2 cos 0.a+22) in a series of ascending powers of x. 256 FUNDAMENTAL SERIES FOR COS 8 AND. SIN 0 CHAP, (18.) Expand 1/(1—2cos 0.2 +.?)? in a series of ascending powers of «. (19.) Expand (1+ 2a)/(1-«°) in a series of ascending powers. of x; and show that 1- 37+ (8n—1)(8n-—2) (32 —2)(38n — 3) (32 — 4) a -/- . 2! 3! (20.) Show that 1/(1+a+2?)=1-—a2+2%-at+a%-a7-+a9—-al0+, . .5 and that, if the sum of the even terms of this expansion be ¢(z), and the sum of the odd terms y(x), then {¢(x)}?- {Y(x)}?= (x) + p(x”). a, pee Prove the following identities by means of Demoivre’s Theorem, or other- wise. and II refer to the letters a, 6, y:— (21.) Ysina/(1+Zcosa)= —II tan $a, where a+B+y=0. (22.) Dsin (0 — B) sin (0 — y)/sin (a — 8) sin (a —-y)=1. (23.) Dsiti d(a+) sin 3(a +7) cos a/sin 3(a — B) sin $(a —-y) =cos(a+B+7). (24.) cos o cos (o — 2a) cos (o — 28) cos (o — 2y) + sin o sin (¢ — 2a) sin (a — 28) sin (« — 2y)=cos 2a cos 28 cos 2y, where c=a+fB+y. Expand in series of cosines or sines of multiples of @:— (25.) cos 19, (26.) sin 70. (27.) sin 80, (28.) cos ®@ sin 70. (29.) cos 6 sin 40, Expand in series of powers of sines or cosines :— (30.) cos 100. (31.) sin 76é. . (32.) sin 30 cos 60. (33.) cos m0 cos né. EXPANSION OF COS 6 AND SIN O IN POWERS OF @. §$ 14.] We propose next to show that, for all finite real values of 0, cos0=1- 6/2!+ @/4!-O/61+... ado (ie sin6=0-@/3!+@/5!-@/7!+... ado (2). _ These expansions* are of fundamental importance in the part of algebraical analysis with which we are now concerned. They may be derived by the method of limits either from the formule of § 12, or from two or more of the equivalent formule of § 13. We shall here choose the former course. I¢ will appear, however, afterwards that this is by no means the only way in which these important expansions might be introduced into algebra. * First given by Newton in his tract Analysis per cequationes numero terminorum infinitas, which was shown to Barrow in 1669. The leading idea of the above demonstration was given by Euler (Introd. in Anal. Inf., t. 1., § 132), but his demonstration was not rigorous in its details. XXIX FUNDAMENTAL SERIES FOR COS @ AND SIN @ 257 From (4) and (5) of § 12, writing, as is obviously permissible, 6/m in place of 6, and ae n=m, we deduce, after a little rearrangement, con = cos m4 1 — ot af 2), m 2! ” ml m Baye uO iat mene 4 ae 1/m) (1 — 2/m) (1 21) 6 (tan oe) Heo | (3), 4! m/ m 6 = COs iam ug ttn OL saye = (3/) ; and sin 0 = cos al (tan b iz) m m/m _ (1 = 1/m) (1 - 2/m) (tan Ses | (4), a m/ m 6 = cos ™ pat Ni hg Ses eed ed Ch (4’). Here, from the nature of the original formula, m must be a positive integer; but nothing hinders our giving it as large a value as we please, and we propose in fact ultimately to increase it without limit. On the other hand, we take 6 to be a fixed finite real quantity, positive or negative. The series (3), as it stands, terminates ; and its terms alter- nate in sign. We have Uonte (1 — 2n/m) (1 — 2n + 1/m) a =f) ee en (2n + 1) (2n + 2) : aa m/ - Hence, so long as n is finite, Uv, G° 1b d en+e2 t , a ae Won (2+ 1)(2n+ 2) _Ii, therefore, we take 2n + 1 > 6,* we can always, by taking m large enough, secure that, on and after the term w,,, the numerical value of the convergency-ratio of the series (3) shall be less than unity. * Strictly speaking, it is sufficient if @<4/{(2n+1)(2n+2)}. VOL. I g 258 FUNDAMENTAL SERIES FOR COS 0 AND SIN 0 CHAP. From this it follows that, if 2n+1> 6, and m be only taken large enough, cos 6 will be intermediate in value between 0 | and 6 COS ewe a Us 2 U4 eae sO ( rs ) Mon 23 ( - lente} (6). Therefore cos 6 will always lie between the limits of (5) and (6) for m =o. Now (see chap. xxv., 23) Lcos™(0/m)=1, Luy=@/2!, Lu= 6/4}, Li tig, = OP /(20) $1, Litengg = 0 4) (2 ee Hence cos @ lies between 1-@/2!+O/4!—.. .(-)re"/(2n)! and 1 — 62/214 O/4!—. . .(—)"62"/(2n) 1+ (— +16 42/(20 + 2) 1. In other words, provided 2n +1 > 0, cos 0=1-— 67/214 O/4!1—. . .(—)6"/(2n)!4+ (— "Ren (7); where Ron < 2" +2/(2n + 2) 1. Here 2n may be made as large as we please, therefore since L 6"+2/(2n+2)!=0 (chap. xxv., § 15, Example 2), we may N=o write cos 90=1-0/2!+@/4!-. ..ad Gh ys By an identical process of reasoning, we may show that, provided 2n + 2 > 0,* then sin 0=0— @/3!+. . .(—)”O'™+1/(Qn + 1)! + (— Reng (8), where Reap < P"t3/(2n + 3)!, an therefore sin 0 = 0-— @/3!+.@/5!-. . .ad (8’). It has already been shown, in chap. xxvi., that the series (7’) and (8’) are convergent for all real finite values of 0; they are * More closely, if 0<4/{(2n+2) (2n+3)}. XXIX > EXAMPLES 259 therefore legitimately equivalent to the one-valued functions cos @ and sin @ for all real values of 6, that is, for all values of the argument for which these functions are as yet defined. From this it follows that the two series must be periodic functions of @ having the period 27. This conclusion may at first sight startle the reader; but he can readily verify it by arithmetical calculation through a couple, of periods at least. When @ is not very: large, say + 17, which is the utmost value of the argument we need use for the purposes of calcula- tion,* the series converge with great rapidity, five or six terms being amply sufficient to secure accuracy to the 7th decimal place. We shall not interrupt our exposition to dwell on the many uses of these fundamental expansions. sin 0>0- 46°; 1—1622). 0. (15.) Show, by employing the process used in chap. xxvii., § 2, that the series for sinn6/cos@ in powers of sin @ can be derived from the series for cos n@ in powers of sin #; and so on. (16.) Show, by using the process of chap. xxvii., § 2, twice over, that, if cos nO6=1+A;sin 26+ Assin40+. ..+A,sin?@+..., then —n? cos nO =2A,+(8. 4Ao— 27A;) sin 70+... + (2r+1)(2r+2)Ap41 -— (27)?A,} sin "0+. . Hence determine the coefficients Ay, Ag, &c.; and, by combining Exercise 15 with Exercise 16, deduce all the series (7). . . (12’) of § 12. (17.) Show (from § 13) that cos”@ and sin”@ can each be expanded in a convergent series of powers of @; and find an expression for the coefficient of the general term in each case. In particular, show that sin 8/3 != 03/8! — (1+37)x5/5 14+ (1438? 4 34)a7/71—- (14374 344 3%)29/9 14... XXIX BINOMIAL THEOREM 261 BINOMIAL THEOREM FOR ANY COMMENSURABLE INDEX. . § 15.] If, as in chap. xxvii:, § 3, we write F(m) =1 + DnCne” (10), where m is any commensurable number as before, but z is now a complex variable, then, so long as modz<1, 3,,0,2” will (chap. Xxvi., § 3) be an absolutely convergent series ; and f(m) will be a one-valued continuous function both of m and of z -Hence the reasoning of chap. xxvii., § 3, which established the addition theorem /(m,)f(m,) =f(m, + m,) will still hold good; and all the immediate consequences of this theorem—for example, the equations (4), (5), (6), (7), (8), (9) in the paragraph referred to— will hold for the more general case now under consideration. In particular, if p and q be any positive integers (which, for simplicity, we suppose prime to each other), then S( P/O}? = {f0)}, =(1+2)P CEL). It follows that /( p/q) represents part of the g-valued function /(1 +2)”; and it remains to determine what part. Let z=7r(cos0+isin 6), then, since we have merely to ex- plore the variation of the one-valued function I(p/@, it will be sufficient to cause 0 to vary between — 7 and +7. Also, let W=1l+z=1l+autyi, ) =1+rcos0+z'sin 6, (a), = p(cos + 7sin ¢), f so that p={(1 +a) + 9°}? = (14+ 2rcos 6 + 1°)"; ) tang=y/(1+2)=rsin6[(1+rcosd), § If we draw the Argand diagram for w=1 +2Z+ Yl, we see that when r is given w describes a circle of radius 7, whose centre Is the point (1,0). Since r<1, this circle falls short of the origin. Hence ¢, the inclination to the «z-axis of the vector drawn from the origin to the point w, is never greater than 262 EXPANSION OF (1+a@+y1)” CHAP, tan-1{7/(1 —77)¥?}, and never less than —tan-1{r/(1 —7?)42}, Hence ¢ lies in all cases between —37 and +47. Therefore, since f(p/q) is continuous, only one branch of the function x/(1 +2)” is in question. Now, if we denote the principal branch by (1 + 2)?/%, so that (1 + 2)?! = p?/%(cos ph/q + isin pd/q), we have, by § 8, V(1 + 2)? = (1 + 2) Pog?! (12), where ¢=0, +1, +2, ..., according to the branch of the function which is in question. Hence we have F(p/q) = (1 + 2)? tang”, where ¢ has to be determined. Now, when z= 0, we have /(p/q) = 1, hence we must have = w_P, Hence ¢= 0, and we have F(p/q) = (1 + 2)?!1 = p?!1(cos.pp/q + 1 sin.pd/¢), where ° —i7<¢<3r. Next consider any negative commensurable quantity, say —p/q. Then (by chap. xxvii., § 3 (9)), HK - p/d) =f) (f( 2/9); = 1/f(p/4)- If, therefore, we define (1 + z)-”/% to mean the reciprocal of the principal value of (1 + 2)?/%, we have I - pla) =(1 + 2)-P!4 = 1/(1 + aria p = ~?!2{cos( — pp/q) + tsin( — p¢/g)} (18). To sum up: We have now established the following expansion for the principal value of (1 + 2)", in all cases where m is any commen- surable number, and mod z< 1 :— (1 +g)" Ge (14). The theorem may also be written in the following forms :— 1 + mn (% + yt)” = {(1 + x)? + y?}™?[cos.m tan —{y/(1 + x)} +isin.mtan~{y/(1+2)}] (15); xxix GENERAL STATEMENT OF THE BINOMIAL THEOREM 263 1 + ,C,7" (cos n6 + isin n6) = (1+ 2r cos 6 + 77)™?(cos mg + isin m¢), where —$r0, then, as we have already proved (see chap. xxvi., § 6, Example 4), =,,C, is absolutely convergent, and, a fortiori, 1+2 nC, cosné and >,,C,, sin v6 are both absolutely convergent. It follows, therefore, that the equation (Lot ge) Me gs will hold when mod z= 1, in all cases where m>0; and also when m lies between — 1 and 0, provided that in this last case the imaginary part of z do not vanish, that ts, provided the amplitude of z is not + x. In other cases where modz=1, the theorem is not in question, owing to the non-convergency of Y,C0,2”. In all cases where modz>1, the series ¥,,C,2” is divergent, and the validity of the theorem is of course out of the question. EXPONENTIAL AND LOGARITHMIC SERIES—GENERALISATION OF THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS. § 18.] The series L+z4+2/2!4+2/31+. is absolutely convergent for all complex values of z having a finite modulus (see chap. xxvi., § 10). Hence it defines a single- valued continuous function of z for all values of 2 We may call this function the Exponential of z, or shortly Expz;* so that Exp z is defined by the equation Expz=1+2+4+2/2!+2/3!+... (1). The reasoning of chap. xxviii, § 5, presupposes, nothing but the absolute convergence of the Exponential Series, and is therefore | applicable when the variable is complex. We have therefore the following addition theorem for the function Exp z:— * When it is necessary to distinguish between the general function of a complex variable z and the ordinary exponential function of a real variable z, we shall use Exp (with a capital letter) for the former, and either e or exp x for the latter. After the student fully understands the theory, he may of course drop this distinction. It seems to be forgotten by some writers that the ¢ in ce is a mere nominis wumbra—a contraction for the name of a function, and not 2°71828 . . . Oblivion of this fact has led to some strange pieces of mathematical logic. XXIX ADDITION THEOREM FOR Exp z 265 Exp 2, Expz,... Exp gm, = Exp @ +2 +... +2m) (2), WaIOLOay 82a). <5 Sm BLO any values of z whatever. In particular, we have, if m be any positive integer, (Exp 2)” = Exp (mz) (3). Also : Exp z Exp (- z) = Exp 0, and therefore Exp (— 2) = 1/Expz (4). We have, further, Expl1=1+1+1/2!+1/3!+..., ae (5) ; and, if a be any real commensurable number, Expo=1+a+a7/2!+a7/8l+.. ., ad (6), by chap. xxviii, where ¢ denotes, of course, the principal value of any root involved if a be not integral. It appears, therefore, that Exp coincides in meaning with e*, so far as ¢* is yet defined. We may therefore, for real values of a and for the corre- sponding values of y, take the graph of y = Exp « to be identical with the graph of y=¢e*, already discussed in chap. xxi. Hence the equation y = Exp (7) defines x as a continuous one-valued function of y, for all positive real values of y greater than 0. We might, in fact, write (7) in the form ope (8); and it is obvious that Hzp-1y may, for real values of y greater than 0, be taken to be identical with log y as previously defined. If we consider the purely imaginary arguments + ay and — iy, _ we have, by the definition of Exp z, 266 MODULUS AND AMPLITUDE OF EXP (%+1y) CHAP. Exp (+ wy) = 14+ ty—y°/2!- iy'/314 y//4! 4+ iy'/5!-. . ., =(1-7'/2!+y'/4!-. ..) +Uy—y'/3!+y°/5!-. . .), =cosy+isiny Oy; Exp (- ty) =(1-y°/2! + y'/41-. . .) | -—iy-—y/3!+y7/5!-. ..), = cosy —7siny (On: by § 14. . : Finally, by the addition theorem, Exp (# + yt) = Exp (x) Exp (yi), = e"(cos y +78in ¥) (10). The General Exponential Function is therefore always expressible by means of the Elementary Transcendental Functions e*, cos y, sin y, already defined. _ Inasmuch as the function Exp z possesses all the character- istics which e? has when z is real, and is identical with ¢ in all cases where é is already defined, it is usual to employ the nota- tion ¢* for Exp in all cases. This simply amounts to defining é* in all cases by means of the equation @=lL+et+e/at+e/3i+. .., which, as we now see, will lead to no contradiction. § 19.] Graphic Discussion of the General Exponential Function — Definition of the General Logarithmic Function. Let w be defined as a function of z by the equation | w= Exp2z (1); and let z=a+y, and w=u+vi=s(cos?+isind). Then, since Exp («+ yi) = &(cosy +isin y), we have s(cos @ + isin ) = e*(cos y +7 sin y) (2). Hence Se fy KS | (3); | where we take the simplest relation between the amplitudes that will suit our purpose. q Suppose now that in the z-plane (Fig.7) we draw a straight line 2'1'1'2’ parallel to the y-axis, and at a distance # from it. Then, XXIX GRAPH OF Exp (v7+yi) 267 7 a Q| S ol Fia. 7. 268 GRAPH OF Exp (7+ 7) CHAP. if we cause z to describe this line, « will remain constant, and therefore ¢” will remain constant ; that is to say, the point w will describe a circle (K) (Fig. 8) whose radius is ¢ about the origin in the w-plane. If we draw parallels to the a-axis in the z-plane, at distances 0'l’=7, 0'2'= 37,. . ., above, and 0'1’=z, 0'2’= 37, . . - below, then, as y varies from — 7 to +7, z travels from 1’ to 1’; as y varies from +7 to + 32, z travels from 1’ to 2’, and so on; and each of these pieces of the straight line corresponds to the circumference of the circle K taken once over. To make the correspondence clearer, we may, as heretofore, replace the repeated circle K by a spiral supposed ultimately to coincide with it. Then to the infinite number of pieces, each equal to 27, on the line K corresponds an infinite number of spires of the spiral K. In like manner, to every parallel to the y-axis in the 2-plane corresponds a spiral circle in the w-plane concentric with the circle K. To the axis of y itself corresponds the spiral circle BAOAB;; to the parallel DO’D to the left of the y-axis the spiral circle DO"D ; and so on. To the whole strip between the infinite parallels DB and DB corresponds the whole of the w-plane taken once over; namely, to the right half of the infinite strip corresponds the part of the w-plane outside the circle BAOAB; to the left half of the strip the part of the w-plane inside the circle BAOAB. To each such parallel strip of the zplane corresponds the whole of the w-plane taken once over. Hence the values of w are repeated infinitely often, and we see that the equation (1) defines w as a continuous periodic function of 2 having the period Q7i. Conversely, the above graphic discussion shows that the equation (1) defines 2 as a continuous cw -ple valued function of w. Taking the latter view, we might write the equation in the form z= Exp-tw (1’). XXIX Log w=LOG(MOD w) +74 AMP (w) 269 Instead of Exp~'w we shall, for the most part, employ the more usual notation Log w, using, however, for the present at least, a capital letter to distinguish from the one-valued function log y, which arises from the inversion of y = ¢, when a and y are both restricted to be real. In accordance with the view we are now taking, we may write (3) in the form t=logs, y=d¢. Hence z= Logw / gives “+ yi= Log {s(cosp+isin d)t, mm Wy where a=logs, and y= ¢. . In other words, we have Log w = log (mod w) + i amp (w) (2°); and, if we cause ¢ (that is, amp (w)) to vary continuously through all values between — o and + , then the left-hand side of the equation (2’) will vary continuously through all values which Log w can assume for a given value of mod w. If we confine ? to lie between — 7 and + 7, then Log w be- comes one-valued ; and we have Log w= logs +i¢ (4), Where s=modw = ./(u’+¢"), and cos 6 =u//(u?+#), sind =v/J(w' +), -r$ op +7. This is called the principal branch of Logw; and we may denote it by 2. It is obvious from the graphic discussion that, of 2 or ,Logw denote the value of Logw in its t-th branch, z being the value in the principal branch corresponding to the same value of w (that is, a value of w whose amplitude differs by an integral multiple of 27), then tLog w = % = 2 + Qtri, = logs +i( + Qtr) (5), where is the amplitude (confined between the limits — 7 and + 1) of w, and t is any integer positive or negative. If w be a real positive quantity, = wu say, then s= mod w= u, p=ampw=0.; and we have, for the principal value of Log u, Log wu = log u. 270 DEFINITION OF ExP 2 CHAP, Hence, for real positive values of the argument, log u is the principal value of Logu. The other values are of course given by ,Log u = log wu + Qtari, t being the order of the branch. We have also the following particular principal values :— Log (+7) = dat, Log (-—%) = — 47%, Log (-—1)= +71: the principal value in the last case is not determinate until we know the amplitude; and the same applies to all purely real negative arguments. § 20.] Definition of Exp gz The meaning of a’, or, as it is sometimes written, Exp 42, has not as yet.been defined for values of z which are not real and commensurable. We now define it to mean Exp (z.,Log a), where ;Log a is the t-th branch of the inverse function Loga, and ¢ may have any positive or negative integral value including 0 Thus defined, a? is in general multiple-valued to an infinite extent. In fact, since ,Log a = logs + i(¢ + 2tr), where s = mod a, and ¢=ampa(—7<¢< +7), we have, if z=x+ 4, at+y' = Exp [(a + yi) {log s + i(h + Qtr) }], = Exp [{z logs — (4 + 2tr)y} + i{y logs + (fp + 2tr)x}], = exp {alogs — (4 + 2tr)y}.[cos {y log s + (fp + 2tr)x} +isin {ylogs+ (f+ 2tr)x}] (1). If we put ¢=0, that is, take the principal branch of Log a, in the defining equation, then we get what ma be called the principal branch of a*+¥’, namely, atty' = Exp (z Log a), = exp {zlogs — dy}.[cos{ylogs + 42} + isin{ylogs + 4x}] (2). The value given in (1) would then be called the ¢th branch, and might for distinction be denoted by ,a*+¥ or by ,Exp q(x + 2). It is important to notice that the above definition of a* agrees with that already given for real commensurable values of z provided we take the corresponding branches. In fact, when y= 0, (1) gives a® = exp («log s). [cos (h + 2tr)x + isin (p+ 2tr)z] ; XXIX ADDITION THEOREM FOR LOG z 271 that is, if «= P/G [s(cos @ + isin #)]2/4 = s¥/9[cos.(p + 2ér)p/q + isin.(p + 2tx)p/q] (3); the right-hand side of which is the ¢-th branch of the left as ordinarily defined. Cor. It follows from the above that when x is an incommensuradle number the function a® has an infinite number of values even when both a and x are real. The principal value of a%, however, when both a and 2 are real and a is positive, is exp (a log a), which differs infinitely little from the principal value of a®, if a’ be a commensurable. quantity differing infinitely little from 2. § 21.] Lhe Addition Theorem for Log 2. By the result of § 19 we have mLog w, + ,Log w, = log(mod w,) + log (mod w,) + vamp w, + damp w, + 2(m + n)ri. Now (chap. xii., § 15) mod w, mod w, = mod (w,w,), and, if amp (w,w,) were not restricted in any way, we should have amp w, + amp w,=amp (w,w,). Since, however, amp (w,w,) is re- stricted in the definition of Log (w,w,) to lie between — x and T, we have amp w, + amp w, = amp (w,w,) + Qpr, where p—.+ 1, 0; or — 1 according as amp w, + amp w,> +7 lies between +7 and —7z, or<—-7. Hence we have . mLog W, + nLOg W, = m4n4pLog (w,w,) (1), where p is as defined. In like manner, it may be shown that mUog W, ae nLog Wy, = m-n+plog (w, We) (2), where p= +1, 0, or —1 according as amp w,—amp w,> +7, between +7 and —7z, or <—7. Taking the definition of a*+¥’ given in § 20, and making use of equation (1) of that paragraph, we have 272 EXPANSION OF ,LOG (1 +2) CHAP. ,Log ,a®t¥' = log (mod ,a*t¥") + (amp ,a* +4! + Qhkr)i, =alogs —(p + 2tr)y + {ylog s + (fp + 2ta)a}i + 2(k + Dat, where / is an integer, positive or negative, chosen so that —ar il), Hence x =log {y+ /(y?—1)}; that is, cosh ~ly = log {y+ »/(y? -1)} (3), the upper sign corresponding to the positive or principal branch of cosh~'y, the lower sign to the negative branch. In like manner we can show that sinh~!y=log {y+ J(y’°+1)} toe tanh~1y = $ log {(1+y)/(1-y)} (5); coth “ty = 4 log {(y + 1)/(y - 1)} (6); sech-ty=log[{1+ /(1-y')}/y] (7); cosech ~1y = log [ {1 + /(1 + y)}/y] (8). § 27.] Properties of the General Hyperbolic Functions analogous to those of the Circular Functions. We have already seen that the properties of the circular functions, both for real and for complex values of the argument, might be deduced from the equations of Euler, namely, (A). In like manner, the properties of "the general hyperbolic functions spring from the defining equations Cos z= 3 {Exp (+ iz) + Exp (- éz)} ; Sin z= ={Exp (+ iz) — Exp ( — iz)} 280 PROPERTIES OF HYPERBOLIC FUNCTIONS CHAP. {Exp (+2) + Exp (—)}5) {Exp(+2)-Exp(-z)}} J ce} We should therefore expect a close analogy between the functional relations in the two cases. In what follows we state those properties of the hyperbolic functions which are analogous to the properties of the circular functions tabulated in § 2. The demonstrations are for the most part omitted ; they all depend on the use of the equations (B), combined with the properties of the general exponential function, already fully discussed. The demonstrations might also be made to depend on the relations connecting the general circular functions with the general hyperbolic functions given in § 24,* namely, Cosh z= 4 1 2 Sinh z= Cosh z = Cos 22, iSinh z = Sin zz ; +tTanhzg=Taniz, -1Cothz=Cotziz; (C). Sech z=Seciz, —127Cosech 2 = Coseciz; Algebrac Relations. Cosh *z —Sinh*z=1, Sech*z + Tanh *z=1 (1), &e. Periodicity.— All the hyperbolic functions have the period 277; and Tanh z and Cothz have the smaller period zi. Thus Cosh (z + 2n7i) = Coshz; &c. ; Tanh (2 + mt) = Tanhz; &c. (2). Also, Cosh (ri +z)= —Coshz, Sinh(zi+2)= +Sinhz; Cosh ($71 + 2) = +7¢8inhz, Sinh ($ai+z)=71Coshz; $ (3). Tanh ($77 + z) = +Cothz, Coth($7i +z) = +Tanhz; Addition Formule. Cosh (z, + 2,) = Cosh z, Cosh z, + Sinh 2, Sinh g, ; Sinh (¢, + 2,) = Sinh z, Cosh z, + Cosh z, Sinh g, ; (5), Tanh (2, = 2,) = (Tanh z, + Tanh z,)/(1 + Tanhz, Tanh z,). * This connection furnishes the simplest memoria technica for the hyper- bolic formule, XXIX GENERAL HYPERBOLIC FORMULA 281 Cosh z, + Cosh z, = 2 Cosh £(z, + 2,) Cosh $(z, — 2,) ; Cosh 2, — Cosh z, = 2 Sinh 4(z, + 2,) Sinh 4(z, — z,); (6). Sinh z, + Sinh z, = 2 Sinh $(z, + z,) Cosh Wee): Cosh z, Cosh z, = 4 Cosh (z, + z,) + $ Cosh (2, —2,)3 Sinh z, Sinh z, = 4 Cosh (z, + ,) — £ Cosh (2,-2,); + (7). Sinh z, Cosh z, = 4 Sinh (z, + z,) + 4 Sinh (2, — 2,). Cosh 2z = Cosh *z + Sinh *z = 2 Cosh 2s — ii, =1+2Sinh “z= (1 + Tanh *z)/(1 — Tanh *2). ' Sinh 27 = 2 Sinh z Cosh z= 2 Tanh z/(1 — Tanh *). Tanh 2z = 2 Tanh z/(1 + Tanh *z). Inverse Functions.—Regarding the inverse functions Cosh ~}, Sinh~], &., it is sufficient to remark that we can always express them by means of the functions Cos~1, Sin71, &. Thus, for example, if we have Cosh~1z = w, say, then z= Cosh w = Cos iw. Hence ww = Cos-}z; that is, w= —1Cos-lz, So that Cosh-1z= —i Cos~}z; and so on. | In the practical use of such formulz, however, we must attend to the multiple-valuedness of Cosh-! and Cos-l If, for example, in the above equation, the two branches are taken at random in the two inverse functions, then the equation will take the form | Cosh ~1z = 2mmi + i Cos-1z, where m is some positive or negative integer, whose value and the choice of sign in the ambiguity + both depend on circum- stances. § 28.] Formule for the Hyperbolic Functions analogous to Demowvre’s Theorem and its consequences. We have at once, from the definition of Cosh z and Sinh zg, 282 ANALOGUE TO DEMOIVRE’S THEOREM CHAP. Cosh (z,+2,+. . -+%n) £Sinh (2, +2,+. . . + %n) =Exp+(z¢,+2,+. «+. +2n); = Exp +z, Exp+z,... Exp + Zp, _ = (Cosh z, + Sinh z,) (Cosh z, + Sinh z,) . ... (Cosh z, + Sinh zp) (A); and, in particular, if 7 be any positive integer, Cosh nz + Sinh nz = (Cosh z + Sinh z)” (B). These correspond to the Demoivre-formule, with which the reader is already familiar.” 7 We can deduce from (A) and (B) a series of formulz for the hyperbolic functions analogous to those established in § 12 for the circular functions. Thus, in particular, we have Cosh (2; +.2).4..5 .4+2n) = Pat Paistiby- «ee where P, = > Cosh z, Cosh z,... Cosh z, Sinh z,~,... Sinh 2. Tanh (2, + 2+. - .42n) =(T,+7,4+7,+.>.)/(1+ 17, +7, where Ts > Tanhz, Tanke... lanhs. Cosh nz = Cosh “z + ,C, Cosh ”~ 22 Sinh 22 + ,0, Cosh™~*z Sinh 424) ate Sinh nz = ,C, Cosh ”~!z2 Sinh 2 + ,C, Cosh n-8» Sinh 32 +',0, Cosh *-5z Sinh *a-- yeaa 2! 4! ve (tm Dg 29 (2s)! (7 even) ; 2 ) 2/,2 9 n° Te Ghee a Cosh nz = ( - yn 1 — — cosh 72 + aU ta) cosh 4z-. m?(n? — 27). a) cosh #2 +5493 bo * As a matter of history, Demoivre first found (B) in the form y=43{1/ J f/(1 +07) —v} - WATVAGl +v7)—v}], where y is the ordinate of P in Fig. 10 below, and v the ordinate of Q, Q corresponding to a vector OQ such that the area AOQ is nm times AOP, and OA is taken to be 1. He then deduced the corresponding formula for the circle by an imaginary transform- ation. (See Miscellanea Analytica, Lib. I1., cap. i.) 7 a 4 XXIX HYPERBOLIC INEQUALITIES AND LIMITS 283 n(n? — 2?) 31 Cosh #4. Ak , ’ cad a Sinh nz/ sinh z = ( — )@-2? a cosh 2 — (-) n(n? — 27)... (n? — 282) (2s +1)! cosh +1 ri a fan (nm even) ; | and so on. We may also deduce formule analogous to those of § 13 such as ’ Snag — {sinh (2m + 1)2 = om4iC, sinh (Q2m—1)e+. . . (i) Cn sinlliar, § 29. Fundamental Inequality and Limit Theorems for the Hyperbolic Functions of a real argument. If wu be any positive real quantity, then tanh w < wu < sinh wu < cosh u (1). By the definitions of § 24 we have sinh uw = ${exp (w) — exp (—u)}; =u+w/3l+w/5!+... (2h: coshu=1+u'/2!+u*/4!4... whence it appears at once that sinh u > w. Again, coshu= + /(1+sinh *w), so that cosh w>sinhw. Finally, since P tanh w = sinh u/cosh wu =wl+w/3!+ui/5l+.. (+ /dleut/4te. . .), and w/[3i; L( sinh © /*) = 1k 1 (tanh © /*) =a, n n/n n/n 284 GEOMETRICAL ANALOGIES CHAP, We have ne (Ges n 1 + e72e/n n ( %41- y+ Yr)3 N=c0 1 n= and Cr Yrs — erp Vy = «{cosh.ru/n sinh. (r + 1)u/n — sinh.ru/n cosh. (r + 1)u/n}, =a sinh.u/n. Therefore ) U = sa Ln sinh. u/n, | | = }@uL(sinh.u/n)/(u/n), = ta’u, by § 29, (3’). Hence, if the area AOP =U, and w= 2U/a’, then, « and y being the co-ordinates of P, we might give the following geometric definitions of cosh uw, sinh u, &¢. :— coshu=a/a, sinhu=y/a, tanhu=y/z, cothu=za/y, &e. It will now be apparent that the hyperbolic functions are connected in the same way with one half of a rectangular hyperbola, as the circular functions are with the circle. It is _ from this relation that they get their name. We know, from elementary geometrical considerations, that the area O is the product of $a? into the number of radians in the angle AOP. It there- fore follows from (3) that the variable 6 introduced above is simply the number of radians in the angle AOP. Our demonstration did not, however, rest upon this fact, but merely on the functional equation cos26+sin 20=1, This is an interesting point, because it shows us that we might have intro- duced the functions cos@ and sin#@ by the definitions cos @=4{Exp (i) +-Exp(-78)}, sin 9=5{Exp (10) - Exp(—76)}; and then, by means of the | above reasoning, have deduced the property which is made the basis for their | geometrical definition. When this point of view is taken, the theory of the circular and hyperbolic functions attains great analytical symmetry ; for it | becomes merely a branch of the general theory of the exponential function as | defined in § 18. When we attempt to get for w a connection with the are AP, like that which subsists in the case of the circle, the parallel ceases to run on the same elementary line. To understand its nature in this respect we must resort to the theory of Elliptic Integrals. § 31.] Expression of Real Hyperbolic Functions in terms of - Real Circular Functions. 288 GUDERMANNIAN CHAP. Since the range of the variation of cosh w when w varies from —« to + is the same as the range of sec @ when @ varies from —7z to +7, it follows that, if we restrict 6 and w to have the same sign, there is always one and only one value of w between — o and + and of 6 between —7z and +7 such that cosh u = sec 6 (1). If we determine @ in this way, we have sinh w= + /(cosh*u — 1), = n/ (SCC pu al is hence, bearing in mind the understanding as to sign, we have sinh wu = tan 0 (2). From these we deduce e“ = cosh u + sinh w, = sec 0 + tan 0; wu = log (sec 6 + tan 6), = log tan (47 + $6) (3). Also, as may be easily verified, tanh $u= tan $0, (4). When @ is connected with w by any of the four equivalent equations just given, it is called the Gudermannian* of u, and we write = gd u. * This name was invented by Cayley in honour of the German mathe- matician Gudermann (1798-1852), to whom the introduction of the hyperbolic functions into modern analytical practice is largely due. The origin of the functions goes back to Mercator’s discovery of the logarithmic quadrature of the hyperbola, and Demoivre’s deduction therefrom (see p. 282). According to Houel, F. C. Mayer, a contemporary of Demoivre’s, was the first to give shape to the analogy between the hyperbolic and the circular functions. The notation cosh. sinh. seems to be a contraction of coshyp. and sinhyp., pro- posed by Lambert, who’ worked out the hyperbolic trigonometry in consider- able detail, and gave a short numerical table. Many of the hyperbolic formulz were independently deduced by William Wallace (Professor of Mathematics in Edinburgh from 1819 to 1838) from the geometrical pro- perties of the rectangular hyperbola, in a little known memoir entitled New Series for the Quadrature of Conic Sections and the Computation of Logarithms (Trans. R.S.E., vol. vi., 1812). For further historical information, see Giinther, Die Lehre von den gewihnlichen und verallgemeinerten Hyperbel- Sunktionen (Halle, 1881); also, Beitrdge zur Geschichte der Neweren Mathematik (Programmschrift, Ansbach, 1881). | XXIX EXERCISES XVII 289 It is easy to give a geometrical form to the relation between @ and wu. If, in Fig. 11, a circle be described about O with a as radius, and from M a tangent be drawn to touch this circle in Q (above or below OX according as wu is positive or negative), then, since MQ?’= OM? - 0Q?=22 - a?=7, we have acoshu=x=asecQOM. Therefore QOM=8, and we have y=MQ=atan 6. From this relation many interesting geometrical results arise which it would be out of place to pursue here. We may refer the reader who desires further information regarding this and other parts of the theory of the hyperbolic functions to the following authorities :—Greenhill, Differential and Integral Calculus (Macmillan, 1886), and also an important tract entitled _4 Chapter in the Integral Calculus (Hodgson, London, 1888) ; Laisant, ‘‘ Essai sur les Fonctions hyperboliques,” Mém. de la Soe. Phys. et Nat. de Bordeaua, 1875 ; Heis, Die Hyperbolischen Functionen (Halle, 1875). Tables of the functions have been calculated by Gudermann, Theorie der Potential- oder Cyclisch- hyperbolischen Functionen (Berlin, 1833); and by Gronau (Dantzig, 1863). See also Cayley, Quarterly Journal of Mathematics, vol. xx. Exerrcisrs XVII. (1.) Write down the values of the six hyperbolic functions corresponding to the arguments 477, i, 27i. ; Draw the graphs of the following, x and y being real: — (2.) y=sinh x/2. (3.) y= cotha. (4.) y=od a. (5.) y=sinh —{1/(2-1)}. (6.) Express Sinh ~!z, Tanh —z, Sech ~!z, Cosech -1z by means of Sin ~ly Cos —1z, &c. (7.) Show that cosh %w — sinh &=143 sinh 2x cosh 2u. (8.) Show that 4 cosh *u — 3 cosh u— cosh 3u=0 4 sinh ?w+3 sinh w—sinh 3u=0. ? (9.) Show that any cubic equation which has only one real root can be numerically solved by means of the equations of last exercise. In particular, show that the roots of a3— qzu—r=0 are /(9/3) cosh u, 2/(q/3) (cos 3r cosh w=E¢ sin $7 sinh w), w being determined by cosh 30 =387a/3/2/q3. (10.) Solve by the method of last exercise the equation 2°+ 62+7=0. Express (11.) tanh -4%+ tanh —ly in the form tanh —le, (12.) cosh -14x+ cosh —ly in the form cosh —lz, (13.) sinh ~14z- sinh -y in the form cosh —:. Expand in a series of hyperbolic sines or cosines of multiples of w:— (14.) Cosh 1%, (15.) sinh7w. (16.) cosh®w sinh*z, VOLO TT U 290 EXERCISES XVII CHAP. Expand in a series of powers of hyperbolic sines or cosines of v:— (17.) Cosh 10u. (18.) sinh 7w. (19.) cosh 6w sinh 3. (20.) sinh mw cosh nw. Establish the following identities :— (21.) tanh 4(u +) — tanh $(u—v)=2 sinh v/(cosh u + cosh 2). sinh (w— v)+sinh w+sinh (u+%) (92. \erse ee cosh (w —v)+cosh w+ cosh (w+ v) (23.) tanh w+tanh (4ri+u) +tanh (3ri+u)=8 tanh 3u, cosh 2u + cosh 2v + cosh 2w + cosh 2(w+v+w)=4lIl cosh (v+w). (24.) Tan 3(u+%)=(sin u+7 sinh v)/(cos w+ cosh 2). =tanh w. (25.) Express Cosh °(w + iv) + Sinh >(w+7iv) in terms of functions of wand v. Eliminate w and v from the following equations :— (26.) e=a cosh(w+A), y= sinh (wtp). (27.) y cosh u—a sinh w=a cosh 2u, y sinh w+ cosh w=a sinh 2u. (28.) e=tanhw+tanhe, y=cothw+cothy, wtv=c. (29.) Expand sinh (w+) in powers of h. (30.) Expand tanh—z in powers of #; and deduce the expansions of cosh —4z and sinh-!w. Discuss the limits within which your expansions are valid. (31.) Given sinh w/w=1001/1000, calculate wu. ; ice) gli2n aie 1 (32.) Show that the series 2 —~~—— Lg

1, the upper or lower sign being taken according as u is positive or negative. (8.) Ifw+vt=cot (v+yi), show that u=sin 2x/(cosh 2y—cos 2”), v= —sinh 2y/(cosh 2y — cos 22) ; uw+v?—2ucot2e-1=0, wu?+v?4+2vcoth 2y+1=0. (9.) If w+vi=cosee (7+ yi), show that w=2 sin # cosh y/(cosh 2y — cos 2x), v= —2cos a sinh y/(cosh 2y — cos 22) ; (uw? +v°)P?=w?/cos "x —v?/sin?y, (wu? +v?)2=u?/cosh 2y + v?/sinh 2y. Express the following in the form w+ vi, giving both the principal branch and the general branch when the function is multiple-valued :— (10.) om Na+ yi). (11.) Tanh —'(x+ yi). | (12.) = Log {(a+yi)|(a—yi)}. (13.) Log Sin (7+ y/). fl (14.) (cos 047 sin 6). (15.) Log w+,e(a@ + 2). (16.) Show that the general value of Sin —(cosec 6) is aE a: log cot (tm +6), where ¢ is any integer. (17.) Show that the real part of Exp; {Log (1+7)} is e-7"8 cos Gr log 2). (18.) Prove, by means of the series for Cos 6 and Sin @, that Sin 26 =2 Sin 0 Cos 0. (19.) Deduce Abel’s generalised form of the binomial theorem from §§ 20, 22. (20.) Show that 1+ pea eer OE =F m+niC2 A tad S ease io =(1+2)™[cos {n log (1+x)} +7sin {n log (1+2)}]. (21.) Show that the families of curves represented by sinw coshy=X, cosz sinhy=u are orthotomie. (22.) Find the equation to the family of curves orthogonal to r” cos N0=X., (23.) Find the condition that the two families Au? + 2Bay +Cy?=)r, Ala? +2Blry+Cly?=p be orthotomic. SPECIAL APPLICATIONS OF THE FOREGOING THEORY TO THE CIRCULAR FUNCTIONS. § 37.] In order to avoid breaking our exposition of the general theory of the elementary transcendents, we did not stop a Lx APPLICATIONS TO CIRCULAR FUNCTIONS 303 to deduce consequences from the various fundamental theorems. To this part of the subject we now proceed ; and we shall find that many of the ordinary theorems regarding series involving the circular functions are simple corollaries from what has gone before. | Let us take, in the first place, the generalised form of the binomial theorem given in § 15. So long as 1+ 3,,0,2" is convergent, we have seen that it represents the principal value of (1+2)". Hence, if z=r(cos0+i sin 4), where 7 is positive, and — 7} O+ + 7, we have 1+ 2mCn7”(cos nO + isin n 0) = (1 + 2r cos 6 + 77)"2(cos md + ¢ sin md), where —jr}=tan-1{r sin G/(1 +7 cos 0)} + 4a. Hence, equating real and imaginary parts, we must have 1+ mCnrr”™ cos n6 = (1 + 27 cos 6 + 72)™/2 cog mp (1); . mC sin nO = (1 + 27 cos 6 + rymi2 sin md (2). These formule will hold for all values of m, provided r < 1, When 7 = 1, we have o = tan~*{sin 6/(1 + cos @)} = 18, and (1) and (2) become 1 + 2nC,, cos né = 2” cos’ 26 cos 1m C1): YmOn Sin NO = 2” cog ™L6 sin m0 (2’), These formule hold for all values of 6 between — 7 and + z,* when m>-—1; and also for the limiting values—7 and +7 themselves, when m> 0. § 38.] Series for cosmd and sin md, when m is not integral. If in (1) and (2) of last paragraph we put 6=47, and r=tand, so that ¢ must lie between — qm and +17, then (1 + 2r cos 6 +77)? = sec™; and we find cos Mp = cosMA(] — C, tan" + mC, tanth —. . Pas): sin Mh = cos™ d(C, tan d — mC, tan*p+.. .) (4). * Since the left-hand sides of (1’) and (2’) are periodic, it is easy to see that, for 2or-r+O+2pr+7, the right-hand sides will be 2” cos mig cos 37(0 — 2pm) and 2” cos 46 sin (0 — 2pm) respectively, where 2” cos mG, _ being the value of a modulus, must be made real and positive. 304 SERIES FOR COS muh AND SIN mo CHAP. Whence / | Boe tan d — mC, tan“ +. — mUg tan’ + mC, tan*d — (5). These formule are the generalisations of formule +, (5), (6) of $12. They will hold even when ¢ has either of the limiting values + 17, provided m>— 1; so that we have tan md = gm/2 COS Jnr = 1 = oa Ui = eA SP es coe 3 gm/2 sin dinar = oe a AB + oe ane Since cos ake: ae (1 i; sin”) Wh Nae Le 2( a Varo sin 5, and the terms of this series are ultimately all positive, it follows that the double series deducible from (3), that is to say, from (= ino cos” 2" sin?" by substituting expansions for the cosines, satisfies Cauchy’s conditions (chap. xxvi, § 34), for there is obviously absolute convergency everywhere under our present restriction that — 47} + {r. Hence we may arrange this double series according to powers of sin ¢. The coefficient of (—- )” sin *"¢ is s=r iM ll ~ (m- ey/sUr—s lat $ mm—2)... (m—2r+t 2) — 1 aes oe (2r Sad 1) 2(m-s1)sUs Grain Now, by chap. xxiii, § 8, Cor. 5, Z(m- nals (27 - nhaes = (in-t2r—s)jsUre Hence the coefficient of (— )" sin?”¢ is m(m—2)... (m—2r+2)(m+ Qr—-2).. . (m+2)n 1.3. 5. (2r = 1) 2 a ns) on —m(m* — 2")... (m* — Ir — 2°) (27)! LMNs RG Ra ps cos mp = 1 - 9, in ot ay ——_——sin “¢ — (6). Hence XXIX fp IN POWERS OF SIN 305 In like manner, we can show that shee sp ae 1 mim — 1) ee! he 2 pos a wih + ann - ) ELA Cee EAL iia WF Also COS Md = cos aL me 7 : sin“ (me ae — 3°) es at (8), as — 2°) sin md = cos aha nop- sin “ mm — 2°)(m — 4 mle 2D 8-1) nee} The demonstration above given establishes these formule under the restriction —4r¢47r. It can, however, be shown that they hold so long as — 17+ $47; that is to say, so long as the series involved are convergent. Cauchy, from whom the above is taken, shows that by expanding both sides in powers of m and equating coefficients we obtain expansions for ¢, ¢’, ¢', &c., in powers of sin 4. Thus, for example, we deduce i lsin*? 1.3 sin*? 1.3.5 sin’ (PS OSA a a EC WY) If we put ~=sin ¢, this gives Wee OL om eae a Lc ES an el tee Shatin sae on 0631 5 kh eG ee (10). In particular, if we put z=4, we obtain ie Sed Teg | r=O5+ oe et eet: ) | (11), from which the value of + might be calculated with tolerable rapidity to a moderate number of places. The result to 10 places 1s 7 =3°1415926536 .. . | VOL. II Xx 306 EXAMPLES CHAP, The important series (10) for expanding sin -!a is here demonstrated for values of « lying between —1/a/2 and +1/a/2. It can be shown that it is valid between the limits x= —1 and «= +1. The series was discovered by Newton, who gives it along with the series for sinz and cosx in powers of # in a small tract entitled Analysis per «Liquationes Numero Terminorum Infinitas. Since this tract was shown by Newton to Barrow in 1669, the series (10) is one of the oldest examples of an infinite series applicable to the quadrature of the circle. Example 1. If m>0, and O=2-m > mOn COS (m — 2n)a, n= S=a-m > mCn Sin (mv — 2n)a, r= C’=2-" > (—)"-1,,C, cos (m — 2n)x, n=0 S’=2-m S (—)r-1,,0, sin (m—2n)e, 2 — Oo then, p being any integer, 1°. C=(cos x)” cos2mpr, S=(cosx)”sin 2mpr, from «=(2p —4)m to x=(2p+4)r. 2°, C=(-cosx)”cosm(2p+1)r, S=(-cosx)™sin m(2p+1)z, from 2=(2p+4)r to v=(294+3)z. 3°. C’=(sin xz)" cos m(2p+4)r, S’=(sin x)” sin m(2p+4)r, from «=2pmr to x=(2p+1)r. 4°, C’=(-sin x)" cos m(2p+3)mr, S'=(-sinw)” sin m(2p+)r, from x= (2p+1)m to x=(2p+2)r. These formule will also hold when m lies between —1 and 0, only that the extreme values of 2 in the various stretches must be excluded. (Abel, (uvres, t.1., p. 249.) If we multiply (1’) and (2’) above by cosa and sina respectively, and add, we obtain the formule COS A+ ZmCy COs (a — 20) = 2” cos 40 cos (a —- 4m + mpr), wherein it must be observed that cos$@ is the modulus of (1+ 27 cos 6+ 72)™2 when 7=1, and must therefore be always so adjusted as to have a real positive value. | From the equation just written, Abel’s formule can at once be deduced by a series of substitutions. Example 2. Show, by taking the limit when m=0 on both sides of (1) and (2) above, that the series (1) and (2) of § 40 can be deduced from the generalised form of the binomial theorem. Example 3. Sum to infinity the series =n*,,C, sin”0 cosn@.. This series is the real part of 2n?;,C, sin”@ (cosn0+7sin n0). Hence S=R[2n?,,C, sin "6 (cos 6+7 sin 6)", =R[{m? sin 20 (cos 6 +7 sin 6)? + m(3m — 1) sin 20 (cos 0 +7 sin 0)? +m sin 6 (cos @+7sin 6)} {1+sin @ (cos 0+7¢sin 0)}™-%], XxIX FORMULA! FROM EXPONENTIAL & LOGARITHMIC SERIES 307 | by Example 5 of chap. xxvii., § 5, =[m* sin °0 cos {80 +(m—3)p} + m(3m— 1) sin °0 cos {20 + (1 —3)} +m sin 6 cos {6+ (m—8)¢}](1+2 sin 6 cos 0 +sin 39 (8) where p=tan—" {sin 6/(1+sin 6 cos 0)}. § 39.] Formule deduced from the Exponential Series, From the equation — e*(cosy +isiny)=1+X(x+yi)™/n!, putting =, cos 6, y=rsin 6, we deduce e” 89 cos(r sin @) + isin(r sin 6)} =1 + 21" (cos nO + isin n6)/n! Hence e"°°89 cos(r sin 8) = 1 + Sr” cos n/n! ay eres? sin (r sin 8) = Xr” sin nO/n! (2) ; which hold for all values of 7 and 6. In like manner, many summations of series involving cosines and sines of multiples of 6 may be deduced from series related to the exponential series in the way explained in chap. xxviii, § 8. Thus, for instance, from the result of Example 8, in the paragraph just quoted, we deduce S13 +234, ~ » +Nn*)a"/n tae" SFr, cos(0 +7 sin 6) +47? cos(20+7 sin 0) 1 + 27° cos (30 +7 sin 0) +} cos(40-+r sin 6)}. § 40.] Formule deduced from the Logarithmic Series. Since the principal value of Log (1+2) is given by Log (1 +2) = log mod (1+2)+iamp(1+2), and since the series -— 2/2 +2°/3—.. . represents the principal value of Log(1 + eo) we put =r(cos 6+7sin @), we have log(1 + 2r cos 6 + r?)¥? +i tan -1{r sin 6/(1 +r cos 6) } = 2( — )"-4" (cos n + isin n0)/n, where — 37>} tan ~l{r sin 6/(1 +7 cos 6)} $7, that is, the prin- cipal value of the function tan -1 is to be taken. Hence we have the following :— b log(1 + 2r cos 6 + 17) = S( — )"-7” eo n/n Gls tan ~*{r sin 6/(1 +7 cos 6)} = S( — )-1y" gin nO/n (2). 308 SIN @—4 SIN 20+4SIN30-—...=40 CHAP. Although, strictly speaking, we have established these results for values of 6 between —7z and +7 both inclusive, yet, since both sides are periodic functions of 0, they will obviously hold for all values of 0, provided r<1. If + =1, (1) and (2) will still hold, provided 6+ +7; for the series in (1) and (2) are both convergent, and we have, by Abel’s useey cos 6 — 4. cos 20+ 4 cos 30 - = Ld log(1 + 27 cos 0 +7”), r=1 ; = log (2 cas: $6) (6); sin @—4sin26+4sin 30-—. . .=tan~l{sin 6/(1 + cos 6)}, = tan ~1{tan 4(0 + 2kr)}, =40+kr (4), where & must be.so chosen that $6+%m lies between — 3a and +47. Thus, if 0 lie between —7 and +7, k=0, and we have simply sinO—4sin260+4sin30-...=130 (4’). In particular, if we put 0 = 47, we oa — al 1 1 i 1 tele eter se St eee (5), which is Gregory’s quadrature ; see § 41. When @=+4(2p+1)z, the series in (3) diverges to —-o, and the right- hand side becomes log 0, that is —«, so that (3) still holds in a certain sense. The behaviour of the series in (4) when 6=+(2p+1)m is very curious. Let us take, for simplicity, the case 0=-+7. With this value of 6 we have for values of 7 as near unity as we please tan—'{r sin 6/(1+,7 cos @)} =0. Hence, by Abel’s Theorem, when 6=+7, sind-4sin20+...=0, as is otherwise sufficiently obvious. On the other hand, for any value of @ differing from +7 by however little, we have L tan-!{r sin 6/(1+7 cos 0)} = 40. once. again, by Abel’s Theorem, r=1 for 0= ir =¢, where ¢ is infinitely small, we have sind-$sin20+...=t4rz39. The series y=sin 6—} sin 26+. . . is therefore discontinuous in the neigh- bourhood of 0=+7; for, when 6=+7, y=0, and when @ differs infinitely little from +7, y differs infinitely little from 7/2. This discontinuity is accompanied by the phenomenon of infinitely slow convergence in the neighbourhood of r=1, 0=-+7; and the sudden alteration of the value of the sum is associated with the fact that the values of the double limits XXIX GREGORY’S SERIES 309 Gs tan {rsin @/(1+rcos0)} and LL tan ~* {7 sin 6/(1+7 cos 0)} r=1 0=4+-7 6=+7 r=1 are not alike. When @ lies between + and 37, we may put 6=27+ 90’, where 6’ lies be- tween —7 and +7, then, for such values of 6, we have y=sin 6’—$sin 26+. . ., =46', as we have already shown, =40-r. Hence, however small ¢ may be, we have, for 0=r+9, y=4o- 27. but, as we have just seen, for 0=x -— ¢ we have Y=—th+47. Hence, as 6 varies from r-¢@ to r+, y varies abruptly from —4¢+47 to 3¢-47. In other words, as @ passes through the value 7, y suffers an abrupt decrease amounting to 7.* We have discussed this case so fully because it is probably the first in- stance that the student has met with of a function having the kind of dis- continuity figured in chap. xv., Fig. 5. It ought to be a good lesson regarding the necessity for care in handling limiting cases in the theory of infinite series. § 41.] Gregory’s Series. If in equation (2) of last paragraph we put 6= 47, we deduce the expansion PAT Pe ered (6), where tan ~1r represents, as usual, the principal value of the in- verse function, and —-1$r}1. In particular, if r= 1, we have m=4(1-2+4-.. .), The series (6), which is famous in the history of the quadrature of the circle, was first published by James Gregory in 1670; and independently, a few years later, by Leibnitz. About the beginning of the 18th century, two English calculators, Abraham Sharp and John Machin (Professor of Astronomy at Gresham College), used the series to calculate 7 to a large number of places. Sharp, using the formule gm =tan~11//3=(1//3) {1-1/3.84+1/5.32-. . on suggested by Halley, carried the calculation to 71 places; that is, about twice as far as Ludolph van Ceulen had gone. Machin, using a formula of his own, for long the best that was known, namely, 47=4tan—11 /5 — tan-11/239, went to 100 places. Euler, apparently unaware of what the English caleu- lators had done, used the far less effective formula 47 =tan-14.+tan-!4, Gauss (Werke, Bd. ii., p- 501) found, by means of the theory of numbers, two remarkable formule of this kind, namely :— 4m =12 tan—11/18 + 8 tan—!1/57 — 5 tan—11/239, = 12 tan~'1/38 + 20 tan—11/57 +7 tan-11/239 + 24 tan —1]/268, * The reader should now draw the graph of the function y, for all real values of 0. Sho EXERCISES XIX CHAP. by means of which 7 could be calculated with great rapidity should its value ever be required beyond the 707th place, which was reached by Mr. Shanks in 1873!* EXERCISES XIX. Sum the following series to infinity, pointing out in each case the limits within which the summation is valid :— 1 18 17855 ae 1 — 5 cos 8 + 5] cos 20 — 5 1.6 me cos 30 1.3 ,cos50 gear eo et ee (3.) , 160 889 1.300850 | ' : 1 v5 3 2.4 5 Ae is? result 4 cos —1(1 — 2 sin 6). (4.) 2(2n-1)(2n-8) cosnd/n! (5.) Dsinno/(n+2)n! (6.) e~* sin 0 — 4¢~°“ sin 80 + fe“ sin 50-—.. . cos 86+. os 6 (2.) «= Ae oe (7.) in Ot ate oe ene oo ! 203 3.4 (8.) sin?6-4sin 220+4sin730-. . .; result 4 log sec @. (9.) 2 cos 2n6/n(n — 1). (10.) Zsin n6/(n? - 1). (11.) 4 sin 6 sin 0-4 sin 26 sin26+4sin 34 sin96-... (12.) cos(a+) —cos(a+8)/3!+cos(a+58)/5!-.. . (13.) cos6—4cos20+4cos380—-—...; result 4 log (2+2 cos 0), except when 0=(2p+1)r. (14.) cos8+4 cos 20+4cos80+...; result —4 log (2-2 cos 0), except when 0=2pr. (15.) sin 8+4sin 260+4sin36+.. .; result =0, if @=0; =3(r-8@), if 0 mod a, (mod x)"=0 N=0 where T,, T,,..., T, ... are all finite. Also the infinite series 1 + dT, is convergent , and converges to the same limit as the infinite product T1(1 + up). 314 INFINITE PRODUCT REDUCED TO A SERIES CHAP. After what has been laid down in chap. xxvi., it will obviously be sufficient if we prove the above theorem on the assumption that all the symbols w,, uw, . . ., U,.. . represent positive quantities. In the more general case where these are complex numbers the moduli alone would be involved in the statements of inequality, and the statements of equality would be true as under. Since w,, Uy). . +) Un... are-all positive, we see, by the Multinomial Theorem (chap. xxiii, § 12), that O < pQUh, Ug. . Up < (th FUgt. . + Un) /r! <(utUgt...+U,+...ado)/r!} =
» Let (1+az)(1+a72)...(1+amz) =1+Aye+ Aoz?+. . . +Ane™+. . .+Am2™ (2), where Aj, As, ... are functions of 2 which have to be determined. Put «xz in place of z on both sides of (2), then multiply on both sides by (1+2az)/(1+a"tz), and we get (1+az)(1+a?z)...(1+a™2) = {14+(1+ Aijwzt (Ar + Agate? +...4+(An-1+An)ure"™ +... +Amaemtignth} | x {1 —amtg 4 gmt, | (- yin) ae (3). Hence, arranging the right-hand side of (8) according to powers of z, replacing the left-hand side by its equivalent according to (2), and then equating the coefficients of 2” on the two sides, we get An= (An oe An—1)x" =e seme Ay 1 An—9)ar—t . atlm+t) (An-2 + An-3)x"~ ( = a Tyin-1) (m-+1) (Ay 4 1)x ( = \nagn(m-+1) ; whence 1—an - aT — ayn =An—1— Ayo + AR-ge ™—.., . ( 7 em St (4). Putting 2-1 in place of m in (4), we have La wi ayn = ‘Anaa — An-3 x + An—472™ sys ts ( == )n—2 gp(n—2)m (5). If we multiply (5) by w” and add (4), we derive, after an obvious reduction, él a x")An = (on = SOL Ay (6). In like manner, (1- net) Aer — (ans ae amtl) Ao (62), (l= i Agu = (ane? or gmt) Ais (63), ie AYN, oy fe ntl) (6). Ot egypt nt 7 , © t a AY / XXx _ EXPANSION OF SECHZ AND SEc x 317 Multiplying (61), (62), . . ., (6,) together, we derive > (x a gem) (x fol lage ane) (xr = gett) aaa (1—a)(1—2)... (1-2) (7), _(1=a™)(1— al), , (1 — mnt) - an(n-+1)/2 (l-x)(1-a)...(1-a) * (8); which establishes our result. If modw<1, the product (1+zz)(1 +2z).. . . will be convergent when continued to infinity, and will, by the theorem of the present paragraph, be expansible in a series of powers of z. The series in question will be obtained by putting m= in (1). We thus get nm (9), (1+) (1+22z) ado =1+ $ gn(n+1)/2 - a= —x) (1-27)... (1 — a”) an important theorem of Euler’s (Introd. in Anal. Inf., § 306). § 3.] Expansion of Sechx and See x. We have, by the definition of Exp a, 2/(Exp « + Exp -— 2) = 1/(1 + 322"/(2n)!) (‘any Hence, if y = 2x" /(2n)! (2), 2/(Exp « + Exp —) = 1/(1+4+), =1+2(-)"y (3). The expansion (3) will be valid provided mod y <1; and the series (2) is absolutely convergent for all finite values of . Hence, if €= mod, it follows from § 1 that the series (3) can be converted into a series of ascending powers of x provided & Gn/(2n)! (—)"E,2”"/(2n)! convergent. We shall determine the radius of convergency of this series presently. Meantime we observe that (5) as it stands may be written ; Sech # = 1 + 2(— )"E,2?"/(2n)! (5% and, if we put 2 in place of a, it gives Seca = 1+ SE, a?"/(2n)! (10). * See Inst. Cale. Diff., § 224: the last five digits of Ey are incorrectly given by Euler as 61671, XXX EXPANSION OF TANH x, &C. 319 Cor. Sech™x and Sec™x can each be expanded in a series of even powers of x. ' The possibility of such an expansion follows at once from the above. The coefficients may be expressed in terms of Euler’s numbers. We may also use the identity 1 =(1 + YA,,22"/(2n)!) cos “a; expand cos x first as a series of cosines of multiples of «; finally in powers of x; and thus obtain a recurrence formula for calculating A,, A,,... The convergency of any expansion thus obtained will obviously be co-extensive with the con- vergency of (10). § 4.] Expansion of Tanh x, x Coth *, Cosech a; Tanz, « Cot 2, Cosec x.* We have already shown, in chap. xxviii., § 6, for real values of x, that Bes = li dak Sf — eee ten) the expansion being valid so long as the series on the right is convergent. In exactly the same way we can show, for any value of x real or complex, that a/(1 — Exp —2)=1+40+4>(-+ eT Ba" /(2n)! (1), where Exp —z is defined as in chap. xxix., and « is such that ‘mod @ is less than the radius of convergency of the series in (1). From (1) we derive the following, all of which will be valid SO long as the series involved are convergent «(Exp x — Exp — @)/(Exp « + Exp — 2) - = 4x/(1 — Exp — 42) — 2a/(1 - Exp — 22) — a, = 2 — )*-122n(Q2n _ 1)B,2?"/(2n)! (2) ; «(Exp « + Exp - 2)/(Exp x - Exp — i} = x/(1 — Exp — 22) - 2/(1 — Exp 22), Se) e228)! (3) ; 2a/(Exp « — Exp — x) = 2e/(1 — Exp — 2) — 2x/(1 — Exp — 22), = 1+ 23( - )”(2?"-1_ 1)B,a?"/(2n)! (4), From these equations, we have at once Tanh = 3( — )—122n(92n _ 1)B,a?"—1 (Qn)! (6); Coth # = 1 + 3( — )"-122"B,, 2n/(2n)! (Glee « Cosech # = 1 + 2X( — )m(Q2n—1 _ 1)B,2?"/(2n)! (7). * Euler, J.c. 320 EXERCISES XXI CHAP. If in (2), (3), and (4) we replace a by iz, we deduce Tan @ = 222"(229 — 1)B,a?"-1/(2n)! | (8); a Cota = 1 —22?"B,a"/(2n)! ao) a Cosec & = 1 + 22(27"—1 — 1)B,2"/(2n)! (10). Cor. Each of the functions (Tanh x)”, (« Coth x)”, (a Cosech x)”, (Tan x)", (w Cot x)", (a Coseca)” can be eapanded in an ascending series of powers of x. EXERCISES XXI. (1.) If 0=gd w (see chap. xxix., § 31), show that O=ayu—azgv®taswe—.. ., U=MHO+a3;PR +a;0°+. . ., where don41= E,/(2n+1)!. (2.) Find expressions for the coefficients in the expansions of Sin”x and Cos “x. (3.) Find recurrence-formule for calculating the coefficients in the expan- sions of (x cosec x)” and (sec 2)”. In particular, show that s 8» E,+ sha En +. ..+8) | +5 En+p gn oe (2p)! (2n)! where §,. denotes the sum of the products 7 at a time of 1”, 3°, 5°, . . ., (2p- ie (Ely, American Jour. Math., 1882.) Sec ?tHy= (4.) If moda<1, show that (1 +22) (1+a4) (1+a°)... ad oo =1+4 Dar’tn/(1— 2%) (L-at)... (1-2), (5.) If moda>1, and p be a positive integer, show that gen(n+1—2p)/2 ® (a —1) (@-1)... (a1) =0. 1+ (6.) Show that the Binomial Theorem for positive integral exponents is a particular case of § 2, Example 2 (7.) Show that (1+ (Lay (Cauchy, Comptes Rendus, 1840.) enn, (8.) Show that | (1 — a) (l—at)... (lL—gmi) = ] NaN - r (1 —wz) (1-22)... (1 —a™z) ae (l—a) (1-2)... (1-2) also that, if mod «<1, mod za<1, 1/(1 — az) (1 —o2)...,ad co =14 Da"e/(1—x) (1-2)... (1-2). (Euler, Znt. in Anal. Inf., § 313.) XXX EXERCISES XXI 391 (9.) If m bea positive integer (1-2) (1-a™-1)... (1- gm—nt1) is exactly divisible by (1 — a) (1—2)...(1-a”). (Gauss, Summatio quarumdam serierum singularium, Werke, Bd. ii., p. 16.) : mn — mm-1 — ym—n+l1 (10.) Tffta, m)=1+2(- ee eee where mod z>1, show that J (x, m)=flx, m—- 2d) (1 — a1) (1 — m8). (1 qm — 22-1) a l-am-l | ~xgm-3 1 —gm-5 7 ee re eh Hence show that, if mod <1, then a Oo 1-2? 1-a* 1-28 1+ Zaminty2— 5° ot zs l-z% 1-2 1-2 .adoo, (Gauss, 70.) (11.) Show that, if m be a positive integer, : ih germ’ al ae ge2m—2 Pee (1 ew! eal) il if a) ee mm) — ‘ ——— ——_—_—~—_————". ( +2) ( +x") (l+2 ) 1+ 2x (1 — a) (1—at)... (1—a?n) (Gauss, 7b.) (12.) Show that 1 (1 — az) (1-282)... (1—a2m—1y) (1 as oy (1 i geome) Vg (1 a Holle ase s Na : es (1-2?) (1-24)... (1—-a™) Also that, if mod <1, and mod za<1, 1/(1 — az) (1 — 282)... ad co =1+4 Earren/(1 — a”) (1— a)... (1-22), (13.) Show that, if moda<1, 1/(1— a) (1-2) (1-2). ..ada =(1+2) (1+?) (1+23)...ado., . (Euler, 7.c., § 325.) (14.) If moda<1, . +a (1—a) (1-2?) (1-23)... adoo= > ( — )raelbn?-+n)/2, (Euler, Nov. Comm. Pet., 1760.) (15.) If modz<1, log (1 — x) (1-2) (1-23)... ado} =-— S/(n)a"/n, 1 where /(m) denotes the sum of all the divisors of the positive integer n ; for example, /(4)=14+2+4. Hence show that v2) ") ive) maple => f/(njan tee at (Euler, 70.) (16.) If d(m) denote the number of the different divisors of the positive integer n, and moda<1, show that 3 d x = oon ; (n)x a 1 Aa (Lambert, Essai @ Architectonique, p. 507.) VOL. II Ne 322 EXPANSION IN INFINITE PRODUCT CHAP. Also that 17/2) ioe) i mn 3 d(n)ar=S an?( —t = i i 1-2 (Clausen, Crelle’s Jowr., 1827.) (17.) If modw<1, show that x ng x Zz ee Tae lee eee “peti ia © (18.) Bett — atl} Dae] a"). D( — 1)" nat /(1 +2") = 2(— eta + a)? (19.) The sum of the products 7 at a time of x, 2, . . ., x” is ger tA)2(ger-t1 — 1) (ar t2—1)... (a —1)/(a-1) (2-1)... (a "—1). (20.) If S, be the sum of the products 7 at a time of 1, a. . ., 2”-*, then S; — Soe ge— (n—1) (n—2r)/2, (21.) Show that, if 2 lie between certain limits, and the roots of ax +ba+e be real, then (px+q)/(az?+bx+c) can be expanded in the form w+ D(Unx” + nz") ; and that, if the roots be imaginary, no expansion of this kind is possible for any value of x. ae ee ON THE EXPRESSION OF CERTAIN FUNCTIONS IN THE FORM OF FINITE AND INFINITE PRODUCTS. 5.] The following General Theorem covers a variety of cases in which it is possible to express a given function in the form of an infinite product ; and will be of use to the student because it accentuates certain points in this delicate operation which are often left obscure if not misunderstood. Let f(n, p) be aw function (with real or imaginary coefficients) of the integral variables n and p, such ie L ae p) is finite for all finite values of n, say L f(n, p) = f(n) ; aa let us suppose that for p=n all values of n and p (n modf(n) be convergent (that is, provided II{1 + f(n)} be absolutely conver ae Let us denote inf + f(n, p)} by Pp; L ir {1 +f(n, p)} by =] = p=0 n=1 P; mod f(n, p) by ho: p); and mod f(n) by f,(n). Xxx GENERAL THEOREM 320 We may write m p P,=I {1+ f(n, p)} I {1 + f(n, p)}, n=1 n=mM+1 Panldms Say, (2). Just as in chap. xxvi., § 26, we have P mod(Q,,—1)+ IL {1 +fi(n, p)}—1. n=m+1 Now, by one of our conditions, if m, and therefore p, exceed a certain finite value, we may put Fn, p)[f,(n) = An, where A,, is not infinite. If, therefore, A be an upper limit to A, and there- fore finite and positive, we have Fi(n, p) bAf(n). Hence p : mod (Qn—1) I {1+ Af,(n)}-1, n=m+1 $ IL {1+ Af(n)} —1, (3). m+1 Let us now put p=o in (2). Since m is finite, and L f(n, p) = f(n), we have p=an : m L Pp =I {1 +f(n)}. p=0 il = m Therefore P=II{1 + f(n)} Qn (4), 1 where Q,,, is subject to the restriction (3). Let us, finally, consider the effect, of increasing m. Since IT {1 + f,(7)} is absolutely convergent, II{1 + Af,(n)} is absolutely convergent. It therefore follows that, by sufficiently ‘ o increasing m, we can make II {1 + Afi(n)}-1, and, a fortiori, m+1 mod (Q;, — 1) as small as we please. Hence, by taking m suffi- ciently great, we can cause Q,, to approach 1 as nearly as we please. In other words, it follows from (4) that P=I{1 4 fn)} (5). In applying this theorem it is necessary to be very careful to see that both the conditions in the first part of the enunciation regarding the value of J(”, p) are satisfied. Thus, for example, it is not sufficient that L S(2, p) p=n have a finite definite value /(7) for all finite values of m, and that =f\(7) be 324 INFINITE PRODUCTS FOR SINH pu, SINH & CHAP. absolutely convergent. This seems to be taken for granted by many mathe- matical writers; but, as will be seen from a striking example given below, such an assumption may easily lead to fallacious results. § 6.] Factorisation of sinh pu, sinh u, sin pO, and sin 6." From the result of chap. xii, § 20, we have, p being any positive integer, p—1 Nar : 2? —1=Q?-1) 0 (2 2 2ncos™ + 1) (1). : n=1 P From this we have a0) as g2p —] -P-l nr 5 = II (a2?- 2acos— +1}; ae a) ) wy nm=1 ] . whence, putting z = 1, and remembering that L(x??— 1)/(2?-1) =p, we have | pot p= 2”-1 II (1 — cos. n/p) (2) ; 1 p-1 = 4P-1 TJ sin2.nr/2p (3) ; 1 and, since sin.7/2p, sin. 27/2p, . . all positive, ., sin. (p — 1)/2p are obviously = ee v 1 sin.na/2p (4). If we divide both sides of 4) by z?, we deduce oP — 4 P=(x—a-1)I (x + a} — 2 cos. n/p) (5), where for brevity we omit the limits for the product, which are as before. If in (5) we put 7 =e", we get at once sinh pu = 2?~1 sinh wII(cosh u — cos. n/p) (6), = 4?-lsinh wII(sin’.na/2p + sinh*.u/2) (7). Using (3), we can throw (7) into the following form :— sinh pu =p sinh uII{1 + sinh.*u/2/sin.’na/2p} (8). Finally, since (8) holds for all values of u, we may replace w by u/p, and thus derive \ * The results in §§ 6-9 were all given in one form or another by Euler in his Introductio in Analysin Infinitorum, His demonstrations of the funda- mental theorems were not satisfactory, although they are still to be found unaltered in many of our elementary text-books. Sax INFINITE PRODUCTS FOR SINH pu, SINH wu 325 p= 1 [ . 2 9 sinh w= p sinh =e Tl {1 + eed (9). 1 cur sin*. na /2p We shall next apply to (9) the general theorem of § 5. Before doing so, we must, however, satisfy ourselves that the requisite conditions are fulfilled. In the first place, so long as n is a finite integer, we have (10). This can be deduced at once, for complex values of wu, from the series for sinh.w/2p and sin.nm/2p. When uw is real it follows readily from chap. xxv., § 22. The product II (1 + w?/n?x°) is obviously absolutely convergent. We have, therefore, merely to show that, for all values of n and p exceeding a certain finite limit, . 2 9 2 sae 2 gtd . nS a . ; eee , 0 sin? —=( — )"/?2"—-2 cos — cos 2 N Y n (14.) Show that II(1 + sec 2”0)=tan 2”6/tan 0 ; 0 and evaluate iif ee \ . 0 (15.) Show that < 4 6 sin 6 Il SRR ey eee Wee earn (4 S sin ) a 2a 6 (1 ~4sin? 5 ) =cos 6; 1 By EU and write down the corresponding formule for the hyperbolic functions. ’ (Laisant. ) Prove the following results (Euler, Znt. in Anad. Inf., chap. ix.) :— ebtr 4 ectx 4(b-c)x+ 4a? =a } ete Yt ten -1 e+ (be) Gets Riper ait Qe 4(b—c)x+ 4a? e-ge 7 (3 +55) mf 1+ Gna (b CF ite (16.) cosh y + cosh ¢ hk 2Qcy+y? 17.) ——*——_- = Artec ei EBs, fF ee 1+cosh ce 1) 14 ems te ae = (1-4) 0 {1- zh 2ey +4? \; 1 —coshe o (200)?7r? + ¢? II ] j San 2 sinh y+sinh ¢ (1+4) {14 acy+y? ) | sinh ¢ nm? + ¢ ; . wae —\n-19, 2 sinh y sinh Brey, (3 -¥) n{1+& hs 2oyty }. sinh ¢ ¢ n°1* +c Write down the corresponding formule for the circular functions, and deduce them by transformation from § 9. Xxx EXERCISES XXII 335 “SORA I $s Bie, 1+cos 0 Re (on 1 eee 2h 2p (19.) cos 6+tan 46 sin d= If (1+ Geate 3) (1- cae \. cos (0—) _ 2p ME nea . (20. ) eee Ut (1+ cannot ceareathe sin (0—¢) | ~ ~ en DP sin 0 {be yn { (+525) (1 seas) \. (21.) Show that cosh 2v— cos 2u=2(u?+v?)IT m{ Cage +0 n*1* ; ((2n - 1)r$2u)?+ 40? } (22 — 1)?2r? i 4 cosh 2 — cos 2u—= 4y2 ‘I { if +o ad eae ahs cosh 2v+ cos 2u=2II {S Wee cosh 2u + cos 2u= am{ 1 +7, sim } : 2n — 1)47r4 (Schlomilch, Hanb. d. Alg. Ana. , chap. x1.) 4n?—4n+5 29. ater |= (22.) Evaluate u( aa ni) (23.) If o=log (1+4/2), show that 2 2 V2= (14ers ) (144g) (ead oo. EXPANSION OF THE CIRCULAR AND HYPERBOLIC FUNCTIONS IN AN INFINITE SERIES OF PARTIAL FRACTIONS. § 10.] By § 8 we have, provided 6+ 4(2n — 1)r, i & 2064+ ¢ cos — sin tan = IL 1 Greet (1). Now, referring to § 2, Cor. 2, we have here 6° ! a i Te d OPES) moa | ie ie ie j (mo ) ] 2 + 4 mod, ie De ie ial d) ’ 8 6’ ent L)*r* — 407}? * mod {Qn — We sage? where 6 =mod@, ¢’=mod¢. It follows, therefore, that the product in (1) may be expanded as an ascending series of powers of ¢. 336 INFINITE SERIES OF PARTIAL FRACTIONS CHAP. pene also on the left of (1), we have eaten mers ea | bo 1 =]-4(2 f = : eee Rags —1)0°- 46 +16(20¢+ ¢°)'> [Cm T= FF (Gn— Tye a | Since the two series in 2) must be identical, we have, by comparing the coefficients of ¢, 2 1 = 802 3). tan 0 et pres eel, (3) This series, which is analogous to the expansion of a rational function in partial fractions obtained in chap. vill, is absolutely convergent for all values of 0 except 47, 37, $7,... Itshould be observed, however, that when 6 lies between $(2n — 1)r and 4(2n +1), the most important terms of the series are those in the neighbourhood of the nth term, so that the rate of converg- ence diminishes as @ increases. We may, if we please, decompose 86/{(2n — 1)’x* — 46} into 2/{(2n — 1)a — 26} — 2/{(2n — 1)x + 26}, and write the series (3) in the semi-convergent form 2 A 2 2 tan =~ —=9 gad 3720) Br 12d 9 2 +50. Se 20) In exactly the same way, we deduce from (1) and (3) of § 8 the following :— G cot 8 = 1 - 273 (4), or 6 9 6p 0 Y ctf ane pi ae ee 6p i a ere XXX INFINITE SERIES OF PARTIAL FRACTIONS Oot provided 0+7, 27, 37, ...; and ai La jl 0 cosec O='1 + 2 20S ee Re (5), or 6 6 7) 0 COU NAS Sane on ary, pa ae fae eae 6 0 eae gee g ot a) provided 0+7, 27, 37, . . We might derive (4) from (3) by writing (17 - 0) for 6 on both sides, multiplying by 6, decomposing into a semi- convergent form like (3’), and then reassociating the terms in pairs; also (5) might be deduced from (3) and (4) by using the identity 2 cosec 6 = tan 40 + cot $8. When we attempt to get a corresponding result for sec 6, the method employed above ceases to work so easily ; and the result obtained is essentially different. We can reach it most readily by transformation from (5’). If we put (5’) into the form 1 1 1 1 1 OSC i aay emery ak Seer 2r +0 1 1 SY CRE Fa which we may do, provided 6+ 0, and then put 47-0 in place of 6, we get el 2 2 2 2 earl ai 2) 8n20. Bee 2 2 ; br — 20° br +20 (6); or, if we combine the terms in pairs, sec 0 = 4>( — )”-t CGae (6), Ah — 1)" where 0 +47, 2a, 52, The series (6), unlike its congeners (3), (4), and (5), is only y VOL. II Z 338 INFINITE SERIES OF PARTIAL FRACTIONS CHAP. semi-convergent ; for, when 1 is very large, its nth term is com- parable with the nth term of the series >1/(2n — 1). We might, by pairing the terms differently, obtain an abso- lutely convergent series for sec 6, namely, 2 9 S - Ly ea a\e ond es ee (7) 5 but this is essentially different in form from (3), (4), and (5). Cor. 1. The sum of all the products two and two of the terms of the series Y1/{(2n — 1)’x" — 46°} is (tan 6 — 6)/1286°; and the like swum for the series 31/{n'x° — 0} is (3 — @ — 36 cot 6)/86. This may be readily established by comparing the coefficients of ¢ in (2) above, and in the corresponding formula derived from § 8 (1). Cor. 2. The series 21/{(2n — 1)’r° — 46°}° converges to the value (0 tan’ 0 — tanO + 6)/646°; and Y1/(n'x’ — 6)’ to the value (6 cosec*O + 6 cot 0 — 2)/46". sec 6 = Since the above series have been established for all values of 6, real and imaginary, subject merely to the restriction that 0 shall not have a value which makes the function to be expanded infinite, we may, if we choose, put =u. We thus get, mter alia, tanh wu = 8w>1/{(2n — 1)" + 4u*} (8) ; wcoth u=1 + 2u’X1/{n'x* + u*} (9); u cosech w= 1 — 2u’d( — 1)"—1/{n'n* + u’} (10) ; sech wu = 42( — )"-1(2n — 1)r/{(2n —1)’r* + 4u} (1.11). EXPRESSIONS FOR THE NUMBERS OF BERNOULLI AND EULER. RADIUS OF CONVERGENCY FOR THE EXPANSIONS QF TAN 6, COT 6, COSEC 6, AND SEC 0. § 11.] If mod 6<7z, then every term of the infinite series >6’/(n'x — 6°) can be expanded in an absolutely convergent series of ascending powers of 6. Also, when all the powers of 6 are replaced by their moduli, the series arising from 1/(n’x’ — 6°) will simply become 1/{n°x°— (mod 6)"}, which is positive, since mod 6 1/2?” + 1/32” 4 1/42 +... >1/(2m—1)2?m-1, which shows that L(1/22” 41/324 Diderot ote yest when m=. t See Introd. in Anal. Inf., § 283. 340 PROPERTIES OF BERNOULLI’S NUMBERS CHAP, If we take the next prime, namely 3, and multiply (1—1/2?")o—m by 1-1 /32”, we shall deprive the series of all terms involving multiples of 3; and so on. Thus we shall at last arrive at the equation (ieee 26) (1 elon) ie L/D oa) ee aie Lye =141/g+. 2. (6); where 2, 3, 5,. . ., p are the succession of natural primes up to p, and q is the next prime to p. We may, of course, make q as large as we please, and therefore 1 /g+.. . (which is less than the residue after the g— 1th term of the convergent series 2m) as small as we please. Hence Cope Lf ya) oe) UR bess (7), where the succession of primes continues to infinity. Hence By= 2(2m) fOr)" — 1/29") — 1/82) (1 = 1/57 ee § 13.] Bernoull’s Numbers are all positwe; they increase after B,; and have © for an upper limit. That the numbers are all positive is at once apparent from § 11 (4). The latter part of the corollary may also be deduced from (4) by means of the inequality of chap. xxv., § 25. For we have 1/(2m—1)>1/2?" + 1/37 + 1/424... .>1/(2m—1)27™-? (9). Hence Bm+i _ (2m + 2)(2m + 1)oom+2 8-4 . (Qar) Cam - (Qm + 2)(2m + 1){1 + 1/(2m + 1)2?m*} (Qr){1 + 1/(2m— 1)} 2m) — 1 Baie act Hence Byii/Bn>1, provided m> /(x° +4), that is, if m> 3°16. Now B,> B,;, hence B32 2,3, Fee Again, it follows from (9) that Loy, =1 when m=, and L(2m)! /(27)?™ is obviously infinite ; hence LB,, is infinite. Cor. By/(2m)! ultimately decreases in a geometrical proportion having for its common ratio 1/4x°. From whach tt follows that the XXX CONVERGENCE OF SERIES FOR TAN 6, &c, 341 series for tan 6, Ocot 6, and O cosec 0, given in § 4, have for their radu of convergence 0=47, and x respectively. § 14.] Turning now to the secant series, we observe that 4=( — )"~1(2n — 1)r/{(Qn — 1)*x? — 462} does not, if treated in the above way as it stands, give a double series satisfying Cauchy’s criterion, for, although when mod 6 <4z the horizontal series are absolutely convergent after we replace 6 by mod 6, yet the sum of the sums of the horizontal series, namely, 42(— )"-1(2n — 1)z/ {(2n — 1)" — 4(mod 6)"}, is only semizconvergent. We can, how- ever, pair the positive and negative terms together, and deal with the series in the form 12) GRP Pet (19 (4n—3)n°—40 (4n—- 1)’ -— 46 (4n — 3)(4n —1)9? + 46 (11) {(4n — 3)’x* — 40°} {(4n — 1)” — 467} Since (11) remains convergent when for 6 we substitute mod 4, it is clear that we may expand each term of (10) in ascending powers of 6, and rearrange the resulting double series according to powers of 6. In this way we get that is, Sad s - 1 1 22m Q2mn sec = 4 | 24 (4n — 3)2+1 (4n — 1)241 jf ] em+i ? m=0 = > 2G a ROS lected aad (12), m=0 where tom41 = 1/12"41— 1/324+14 1 /52mt1_ | |. (13), Comparing (12) with the series sec 0=1 + SE,,6?"/(2m)!, obtained in § 3, we see that 22m+2(2m)! rom m — qrzmt+1 ? Q\2m+t ey 1 1 a 2(2m)! (S) i 2m+1 wv 32m-+1 u Ramt+l °° ii (14), which may be transformed into aye 2. 2m+1 / 1 ] 1 En = 2(2m)! ies) / (1 ot Aiea (1 = =o) € ata =) Se in the same way as before. . (15).* * See again Euler, Jntrod. in Anal. Inf., § 284. 342 PROPERTIES OF EULER'S NUMBERS CHAP. Cor. 1. Euler’s numbers are all positive ; they continually increase in magnitude, and have infinity for their wpper limit. For we have letontne Lo (16). Hence | Em+i (2m + 2)(2m + 1)47om+s En ¥ greg i (2in + 2)(2m + ve eer ce? r But this last constantly increases with m, and is already greater than 1, when m=1. Hence E, = 10020083928 .. . =2°/29749°35 . . . EXPANSIONS OF THE LOGARITHMS OF THE CIRCULAR FUNCTIONS. § 16.] From the formule of §§ 6 and 7, we get, by taking logarithms, log sin 6 = log 6+ S log (1 — &/n'x°), x n= = log 6 = © og, 0?" /mx2™ Ly m=1 since the double series arising from the expansions of the logarithms is obviously convergent, provided mod 6 » 22m- 1B 2" /m(2m)! aby) m=1 * Inst. Cale. Diff., chap. vi. 344° STIRLING’S THEOREM CHAP. The corresponding formule for cos 6 are log cos 9= — (2? — 1) ,6?"/ma?™ (2); = em 192m 1) Bm 2) ee The like formule for log tan 6, log cot 6, log sinh u, log cosh w, &c., can be derived at once from the above. If a table of the values of o,,, or of B,, be not at hand, the first few may be obtained by expanding log (sin 6/6), that is, log (1 - 6/3!+ 6/5!-—...), and comparing with the series — Loon 9?"/mar?™ For example, we thus find at once that ener. STIRLING’S THEOREM. § 17.] Before leaving this part of the subject, we shall give an elementary proof of a theorem of great practical importance which was originally given by Stirling in his ee Differen- tialis (1730). When n is very great, n! be equality with »/(2nm) (n/e); or, more accurately, when n is a large number, we have nm! = »/(2xn) (n/e)” exp {1/12n + 6} (hile where — 1/24n* <9 <1/24n(n — 1). Since log {n/(n — 1)} = — log (1 — 1/n), we have Toei ae eal re : : +. Paps Sn—1 nm Oe 3m dnt on™ We can deprive this expansion of its second term by multi- plying by n—4. We thus get | (n — 3) ope ee Te i + : ee Bree. AA ieee 12m" 12m’ 2m(m + 1)n™ Hence, taking the exponential of both ‘sides, and writing suc- cessively ”, n—1, n—2,..., 2 in the resulting equation, we deduce ( “ Nee (ate + ; i nei) |) oe 12n? 12n8 ‘- mm — 1 pies ) 2m(m +1)n™ “7 Xxx STIRLING’S THEOREM 345 | (1 1 ri 1 j n—2 Pus w= 1 evan 1p rt m— 1 vA 2m(m+1)(n—-1)™ 7° z: m— I e ) 2m(m+1)(n-2)m °° 77? sya-a ata i (5) =exp ( iota aa - m—1 % ) eta atee es oe g\203 1 1 (7) =exp (lt syste % i | ) Imim+ 1am" °°)" By multiplying all these ek we ae aan | (- 1)+ 3 Sn + ape | : m—1 Ee where S’,, = 1/2™+1/3"+ 1/44. 2.4 1/nm. Now aa. I/{m+1)™—-1f(n+2yr—. . , (3), where Sm = 1/2" 4+ 1/8™+. + 1/nm +... adoo. By the inequality (6) of chap. xxv., § 25, we have 1/(m — 1)n™-1 > 1/(n + 1)™ + L/(m+2ymr. ., >1/(m—-1) (n+ 1)m-1, ene | = 1/(m —1) (n + 1 Eel apes 1/(m = 1)nm—1, Pert 1 fa 1 ee Toe 2+ ake ae is Ronny ee aS: & (me — 1)Sm 42 1 pie ae A \ hme 7 m(m+1) 73 m(m + 1)n-1 (4) ; Ss: I)Sin ahs shy A \ ee - m(m + 1) 2% mm +1) (m+ 1)"-2 oY 346 STIRLINGS THEOREM CHAP. Since S,,<1/(m-—1), the series D(m—1)S,,/m(m+ 1) con- verges to a finite limit which is independent both of m and of n. Again, ca 1 z m(m + 1)n"—1 f 1 1 = ——— + ——_54+-——4H+... Oy ee 2.3n 3.4n? 4.5n? (8) BAB aah gga 6n 12n° hat patie |b: 1 1 BS: Pits 7). - <6n | 12n(m—1) 7) Also, by (6), @ 1 2 m(m +1) (n+ 1)™-1 _g (2 ae! ) et 9 \m m+) (n iar ee = (n+ 1)s - ibis : L 2 m(n + 1)” Ut o(m +1) (m+ 1)™4?” eal 1 =(w+1)| = 5 =be(1-5) } ihe 1 1 ) = (0+ 1)| Eas n+l \ 1 : eA va | = +54 n-(n'+n) log (1 +-); _ pe AD he en : Us ae De 5 ant en Qn 3n ‘ aeilleeaae 1 eoEn 3m 3.40 4,on' 1 1 2. 8). sm ~ 6n - 12n? (8) Combining (2), (4), (5), (7), and (8), we have nts qatm— 1) Salo) le n! ~exp{u-1+ 22 m+ 1) 12n 24n(n—- 1) (m-1)Sy 1 1 10 Ce-™n"+4 exp (ss sat) (12), < Ce-"n"+2 exp (nt a a) (13); or, since the meet function is continuous, 1 = Ce-®nrtz exp (= + a) (14), where — 1/24n’<6<1/24n(n—1). Hence, putting = « on both sides of (14), we have Ln! = CLe-"n"+4 (15). Hence, putting The constant C may be calculated numerically by means of the equation (11). Its value is, in fact, ,/(27), as may be easily shown by using Wallis’s Theorem, § 6 (18). Thus we have, when n= 0, T 22"(n!)2(2n + 1) 24°(n!)4(20 + 1) eee nel) Hence, using (15), we gét 94n o—4ny4n+2(9nH 4 1 g7 OL eee CL eC ae La + 1/2n)?"}241 + 1/2}? Ce pers Therefore, since C is obviously positive, C= /(27) (16). Using this value of C in (14), we get finally n! = »/(Qrn) (n/e)” exp {1/120 + 6} * Chines where — 1/24n’ <6 <1/24n(n—1). * An elementary proof that La!=La/(2rn) (n/e)” was given by Glaisher (Quart. Jour: Math., 1878). In an addition by Cayley a demonstration of the approximation (17) is also given ; but inasmuch as it assumes that series 348 EXERCISES XXIII CHAP. Cor. By combining (11) and (16), we deduce that 2(m—1)8m_ 4 1] " Lae 1 18 ih a7 a Cemeaa log 2+ 4 log (18), where Sy = 1/2%+.1/3™+1/4™+... ado. EXERcISsES XXIII. (1.) Show that, when moda>z, «cota can be expanded in the form Ag+ 2(Bnv-"+C,2”); and determine the coefficients in the particular case where r<2%<2r. (2.) Show that the sum of the products 7 at a time of the squares of the reciprocals of all the integral numbers is 7?”/(2r+1)!; and find the like sum when the odd integers alone are considered. (3.) Sum to » terms | tan @+tan (0+2/n)+tan (9+27/n)+. . .; tan 264+ tan 2(0+7/n)+ tan 7(0+ 27/n)+. Sum the following :— (4.) 1/(1?+ a?) +1/(2?+ a?) +1/(8?+27) ... (5.) Lf? —1/(a? — a?) + 1/(ae? — 22?) — (6.) 1/e+1/(@-1)+1/(a+1)+1/(7-2 )+1/(0-+2) es (7.) 1/1 -e) Teen ee . .$1/(n?-e)4+. (8.) 1/1.2+1/2.4+1/38.64+1/4.8+. .. Show that (9.) (mw? -6)/6=1/17. 24+1/27.3+41/37.44+. ; (10.) 7/8—1/3=1/1.3.5-1/3.5.7+1/5.7.9-. .. (11.) If f,(m) be an integral function of x whose degree is 7, show that >f,(2)/(2n — 1)?" can be expressed in terms of Bernoulli’s numbers, provided r>2mu-2; and > — )r-lf,(n)/(2n —1)?"*1 in terms of: Euler’s numbers, pro-: vided r+ 2m-—-1. In particular, show that 1 14+2 14248) =F (1-3). oo 2b" as 4 64 12 (12.) Show that > 1/(nm + 0)? =cosec 76 ; ae GD) 31 /(nm + 6)4= cosec 40 — 3 cosec 70, n=0 being included among the values to be given to x. (Wolstenholme.) of the form of 1/2"+1/3"+. . . can be expanded in powers of 1/m, it cannot be said to be elementary. The proofs usually given by means of the Mac- laurin-sum-formula are unsatisfactory, for they depend on the use of a series which does not in general converge when continued to infinity, and which can only be used in conjunction with its residue. See Raabe, Crelle’s Jowr., xxv. ax EXERCISES XXIII 349 (13. ) SL _ra/2 sinh.ran/2+sin.rar/2 1 Da a = = ae eR Ra : a yp Mita 423 cosh.ran/2—cos.raxr/2 2a? (Math. Trip., 1888. ) (14.) Show that aoe 1 if Te 1 ; nai ((2n)* — (2m — 1)7}?°16(2m—1)? 2(2m—1)2’ ao it 172 2 = : n=11(2n — 1)? — (2m)?}?~ 64m? Also that the sum of the reciprocals of the squares of all possible differ- ences between the square of any even and the square of any odd number is 1/384. (15.) If p + tan-! tanh a = } =tan “(tanh v cot w) ; m= nT —U NT + U Sa, tnt \ at “tanh v t (Q2n—1)r—2y “*™ pee eee (tanh v tan w). (Schlomilch, Handb. d. Alg. Anal., cap. xi.) (oa) 2) ban ih (17.) If A(w) Saxqp{1- (z/na), u(a)=T1 {1 — (2x/2n-1.a)?}, express 1 1 \(x+ a/2) in terms of u(x), and also u(x + a/2) in terms of d(a). Hence evaluate L 1.3.5... (2m —1)/(2m+1)/2™m !. ses (Math. Trip., 1882.) (18.) Show that, if 7 be a positive integer, Lb (1-3) (-2)"-- (0 — re sen, T=0 , i A (19.) Show that mat —ptaapto et \=3 L=n e+ 12 v2 + 2? ee + 32 Act ao REVERSION OF SERIES—EXPANSION OF AN ALGEBRAIC FUNCTION. _§18.] The subject which we propose to discuss in this and the following paragraphs originated, like so many other branches of modern analysis, in the works of Newton, more especially in his tract De Analysi per Afquationes Numero Terminorum Infinitas. 350 STATEMENT OF EXPANSION PROBLEM CHAP. Let us consider the function 2(m, n)xy” = (1, O)a + (0, 1)y + (2, O)a’ + (1, 1)ay + (0, 2)y? +. . ., where the indices m and n are positive integers, and we us; the symbol (m, m) to denote the coefficient of 2”y”, so that (m,n) is a constant. We suppose the absolute term (0, 0) to be zero ; but the coefficients (1, 0) (0, 1) are to be different from zero. The rest of the coefficients may or may not be zero; but. if the number of terms be infinite, we suppose the double series to be absolutely convergent when modw=mody=1.* From this it follows that the coefficient (m, m) must become infinitely small when m and n become infinitely great ; so that a positive quantity \ can in all cases be assigned such that mod (m, n) +2 waatever values we assign to m and n. It also follows (see chap. xxvi., § 37) that =(m, n)a™y” is absolutely convergent for all values of « and y such that mod «$1 mod y+ 1. We propose to show that one value, and only one value, of y as a function of « can be found which has the following properties :— 1°. y is expansible in a convergent series of integral powers of x for all values of a lying within limits which are not infinitely narrow. 2°. y has the initial value 0 when x = 0. 3°. y makes the equation =(m, n)xmy” = 0 (1) an intelligible identity. Let us assume for a moment that a convergent series for ¥ of the kind demanded can be found. Its absolute term must vanish by condition 2°. Hence the series will be of the form y=d,a+ bya + bya? +... (2). In order that this value of y may make (1) an intelligible identity, it must be possible to find a value of z<1 such that (2) gives a value of y<1. The series (1), when transformed by means of (2), will then satisfy Cauchy’s criterion, and may be arranged according to powers of , All that is further necessary * The more general case, when the series is convergent so long as moda+a and mod y+, can easily be brought under the above by a simple transform- ation. Xxx GENERAL EXPANSION THEOREM 351 to satisfy condition 3° is simply that the coefficients of all the powers of « shall vanish. It will be convenient for what follows to assume that (0, 1)=—1 (which we may obviously do without loss of generality), and then put (1) into the form— Y= {(1,.0)a + (2, O)ar + (3) Oe +. 5 . } pet he (2 ul eS, ye Ae ee ky + {(0, 2) + (1, 2)x+ (2, 2)a%4 (3, 2)aPt+.. . } +{(0, m)+ (1, na + (2, ne +(3, ner... han rs ee Rye (3). Using (2), we get or baw ba = (1, O)a + (2, 0)a? + (3, 0)a? +... + (1, 1)a + (2, 1)a" + (3,1)a° +... 110, + bya + igen ae +4(0,2)+ (1,2)0 + (2, 2)n*+ (3, 2)0° +... 110, + Ban + Boa? es + {(0,n) + (Lnyz + (2,m)at + (3,m)o?-+...}10, + Dyer + Oya? +. Jan Hence, equating coefficients, we have b, = (i 0), b,=(2,0)+(1,1)b, + (0, 2)b,°, b, = (3,0)+(1,1)b, +(2,1)b, + (1, 2)d,° + 2(0, 2), 0, + (0, 3),"*, by = (n, 0) + (1, 1)dn-.+(2,1)bn-g ts. . + (0, 1)b,” Here it is important to. notice that each equation assigns one of the coefficients as an integral function of all the preceding coefficients. Hence, since the first equation gives one and only one value for 0,, all the coefficients are uniquely determined. There is therefore only one value of y, if any. In order to show that (5) really affords a solution, we have to show that for a value of « whose modulus is small enough, but not infinitely small, the conditions for the absolute convergency of (2) and (4) are satisfied when b,, b,, .. . have the values assigned by (5), | 352 GENERAL EXPANSION THEOREM CHAP. This, following a method invented by Cauchy, we may show by considering a particular case. The moduli of the coefficients of the series (3) have, as we have seen, a finite upper limit A. Suppose that in (3) all the coefficients are replaced by A, and that a has a positive real value <1.° Then we have © yaAlorera+.s = tA{er eae +. cle fy 4Afleorrere+... ty (6). This series is convergent so long as x<1 and mody<1. It can, in fact, be summed ; for, adding 4 + Ay to both sides, we get (1 +Ajy+A=A/(1-2) (1-9), that is, (1 + A)? —y + Aw/(1 — #) = 0. Hence, remembering that the value of y with which we are concerned vanishes when 2 = 0, we have y=[1— / {1-40 + A)a/(1 — 2)}]/2( + 1) (7). Now, provided 4A(1 + A)z/(1 -#) <1, that is, x<1/(2A+ 1), the right-hand side of (7) can be expanded in an absolutely con- vergent series of integral powers of «, the absolute term in which vanishes. Also, when #<1/(2X+ 1)’, the value of y given by (7) is positive and <1, therefore the absolute convergency of (6) is assured. It follows that the problem we are considera can be solved in the present particular case. If we denote the series for y in this case by , y=CO,0+ C0 +Ca°+... (8), then the equations for determining C,, C., C,, . =e eee found by putting (1, 0) = (2, 0)=(1, 1)=.. . =A in (5), namely, C, a ae OC, =A(1+C0,+C,), CG, =A(1+ C0, +0, + C, + 2C,C, beg! Cy = AL + Cy ; Cane aa 20H Gy : (Die XXX GENERAL EXPANSION THEOREM B93) from which it is seen that GG rs Cee ee iare alltreall and positive, Returning now to the system (5), and denoting moduli by attaching dashes, we have, since (1, 0)’, (2, 0)’, &., are all less than X, Oo =(1, 0) . Now the coefficients of (3) are symmetric functions of the roots of (2); therefore (3) could be exhibited as an equation whose coefficients are integral functions of A,, A,,.. ., A,, and there- fore integral functions of z.* Hence (2) would be reducible, which is supposed not to be the case. Tt must, however, be carefully noticed that irreducibility in general (that is, so long as x is not specialised) does not exclude reducibility or multiplicity of roots for particular values of 2 In fact, we can in general determine a number of particular values of # for which (2) and (3) may have a root in common, In other words, i¢ may happen that the n branches of y have points in common ; but it cannot happen that any two of the n branches wholly coincide. When, for =a, the n values Deals ee G arecall different, a (or its representative point in an Argand-diagram) is called an ordinary point of the function y, and b,, b,,. . ., by single values. Libre <= b.Gach = b, say, then a is called an r-ple point of the function, and b an r-ple value. For every value of x (zero point) which makes A,=0, one branch of y has a zero value ; for every value of « (double zero point) which makes A, = 0 and A, = 0, two branches have a zero value; and so on. These are called single, double, . . ., zero values. For every value of z (pole) which makes A,, = 0, one branch of y has an infinite value; for every value of a (double sole) which makes A, = 0 and A,-, = 0, two branches have an infinite * See chap. xviii., § 4. 358 EXPANSION AT AN ORDINARY POINT : CHAP. value; and so on. These may be called single, double, . . ., infinities of the function. The main object of what follows is to show that every branch of an algebraic function is (within certain limits), in the neighbourhood of every point, expansible in an ascending or descending power series of a particular kind ; and thus to show that every branch is, eacept at a pole, continuous for all finite values of a. § 21.] If, at the point «=a, the algebraic function y has a single value y=b, then y—b is, within certain limits, expansible in an absolutely convergent series of the form y—b=O(a—a)+C,(a—a)+C0,(a-a)+.... (4). Let a=a+& y=)+7, then the equation (1) becomes, after rearrangement, (0, 0) + (1, 0)E + (0, 1) + (2, 0) + ke. =0 (5). Since y=0 is a single root of (1) corresponding to =a, it follows that when €=0 (5) must give one and only one zero value for 7. Therefore we must have (0, 0)=0 and (0, 1) +0. It follows, from the general theorem of § 18, that within certain limits the following convergent expansion, n=O E+ O,+O,F+..., and no other of the kind will satisfy the equation (5); that is, y=b+O0,(e—-—a)+C(x-a)y+O(a-a)+... (6) will satisfy (1). The function y determined by (6) is continuous so long as mod (z—«) is less than the radius of convergency of the series involved ; and it has the value y= 0 when «=a. If we suppose all the values of y, say 0,, b,, . . ., bn, corre- sponding to «=a to be single, then we shall get in this way for each one of them a value of the function y of the form (6). Hence we infer that Cor. So long as no two of the branches of an algebrare function have a point in common, each branch is a continuous function of x ; and the increment of y at any point of a particular branch is expan- - +. . .), where a is any one of the five 5th roots of 1. All the ten branches are thus accounted for. It should be observed that, if we form an integral equation by selecting from any given one a series of terms which form an effective group, the new equation gives an algebraic function. Those branches of this function that have zero initial values coincide to a first approximation (that is, as far as the first term of the expansion) with certain of the branches of the algebraic function determined by the original equation which have initial zero values. Thus, reverting to the example just discussed, from the group ABC we‘have A€™ + BENn + CE7’ = 0. This gives, when we drop out the irrelevant factor €’, Cn’ + BE’) + AE = 0, which breaks up into two equations, nt p&=0, +98 =0; XXX ALGEBRAIC FUNCTION ALWAYS EXPANSIBLE 365 and thus determines two functions, each of which has a branch coincident to a first approximation with a branch of n (as deter- mined by (16)) which has zero initial value. In like manner, DE gives Cé&+Dy*°=0; and DE gives Dé a8 Ey’ ==): ; We thus get a number of binomial equations, each of which gives an approximation for a group of branches of the function » determined by (16). We shall return to this view of the matter in § 24. § 23.] Before leaving the general theory just established, we ought to point out that Mewton’s Parallelogram enables us to obtain, at every point (singular or non-singular), convergent expansions for every branch of an algebraic function in ascending or descending power-series, as the case may be. To establish this completely, we have merely to consider the remaining cases where x or y or both become infinite. Ist. Let us suppose that the value of the function y tends towards a finite limit ) when z tends towards o. Then, if we put »=4—06, «=&, we shall get an equation of the form 2(m, n)E™n” = 0 (his which gives 7 = 0 when £= 0. Let us suppose that Fig. 1, as originally constructed, is the Newton-diagram for (17), and let & be the highest power of E that occurs in (17) so that OO,=%. Now in (17) put é= 1/€,, and multiply the equation by é*; we then get the equation Dm, N)E*- My” — 0 (18), which is obviously equivalent to (17). But the Newton-diagram for (18) is obviously still Fig. 1, provided O,X, and O,Y, be taken, instead of OX and OY, as the positive parts of the axes. Hence, if we make a boundary convex towards O, in the same way as we did for O, we shall obtain a series of branches of 7 all of which are expansible in ascending powers of &’, that is, In descending powers of €, and all of which give »=0 when €=a. For each such branch we have 366 EXPANSION AT POLES, &C. CHAP. nab(c+de* + ee" +... .), that is, (y— Dd) =c+dla*+efee +... (1:9); where X, a, B,. . . are all positive, and ¢ is finite both ways. 2nd. Suppose that «=a is a pole of y, so that y= oo when a=a; and put »=y¥, €=2-a, so that we derive an equation X(m, née” = 0 (20), for which Fig. 1 is the Newton-diagram withiOX and OY as axes. Then, putting 7 =1/n’, we get an equation of the form X(m, n)E"n'!-™ = 0 (21), I being the highest exponent of 7 in (20). The Newton-diagram for (21) is then Fig. 1 with O,X, and O,Y, as axes; and we construct, as before, a boundary, EFG say, convex towards O,, every part of which gives a series of branches of 7’, that is, of 1/7, expansible in ascending powers of €& For every such branch we shall have n& =1f(e+ dé’ + ef +. . .), where A, a, B, . .. are all positive, and c is finite both ways. Hence also, by the binomial theorem combined with § 1, ne = ie+rd& +e is ae that is, | y(a — a) =1/e+ d'(w— a)” + (a - ay +S mae where A, a, f’,. . . are all positive, and c is finite both ways. 3rd. Suppose that y has an infinite value corresponding to x= (pole at infinity). Then, if we put «= €=1/€, y = =1/n', we shall get, by exactly the same kind of reasoning as before, a boundary GHI convex to O,, each part of which will give a group of expansions of the form abet de + eee eT. Whence, as before, for every such branch yl =1/(e + d/a® + eft? +. . .), =1fe+d'/x* + fat ive (23), where A, a, f’,. . . are all positive, and ¢ is finite both ways. | | Xxx ALGEBRAIC ZEROS AND INFINITIES 367 If we combine the results of the present with those of the foregoing paragraphs, we arrive at the following important general theorem regarding any algebraic function Y:— Ify=0 whenz=a (a+ x), then L y/(a— a) is finite both ways. If y=0 when x= w, then L y/x* is finite both ways. fy = whenz=a(a+ o), then L y/(a - a) is finite both ways. =a Ify= x when v= x, then Li y/x* is finite both ways. L=O A 1s in all cases a finite positive commensurable number which may be called the ORDER of the particular zero or infinite value of y. This theorem leads us naturally to speak of algebraical zero- or infinity- values of functions in general, meaning such as have the property just stated. Thus sinz=0 when «=0; but Lsin x/x=1 when x=0; therefore we say that sin # has an algebraic zero of the first order when x—0. Again, tanz=o when x=$7 ; but Ltan2/(x—-42)-1 is finite when z= 3m ; the infinity of tana is therefore algebraical of the first order. On the other hand, e*=co when a= ; but, this is not an algebraical infinity, since no finite value of \ can be found such that Le*/x* is finite when z=0, (See chap. xxv., § 15.) § 24.] Application of the method of successive approximation to the expansion of functions, This method, when applied in con- junction with Newton’s diagram, greatly increases the practical usefulness of the general theorems which have just been estab- lished. The method is, moreover, of great historical interest, because it appears from the scanty records left to us that it was in this form that the general theorems which we have been dis- cussing originated in the mind of Newton. Let us suppose that the terms of an equation (which may be an infinite series) have been plotted in Newton’s diagram, and that an effective group of terms has been found lying on a line A; and let »”- £€" (the coefficients are taken to be unity for simplicity of illustration) be a factor in the group thus selected, repeated, say, a times, so that the whole group is bi(§ 0) — &")*. Let A be moved parallel to itself, until it meets a term or group of terms ¢,(é 7); then again until it meets a group ¢,(& 7); and so on. | 368 SUCCESSIVE APPROXIMATION " OHARE : The complete equation may now be arranged thus— bl, )(n" i a “e pl§ ”) 4g bal, ”) eg, 5 ey or thus— (n™ — n\n 4 bl§, ”) PAE, Dee, - See (24), d(&n) lS) say, (7™ — Ee") + rt aet. , 2 =U, Now, by the properties of the diagram, when = &"™, p(n), $(& 7), . - . are in ascending or descending order as regards degree in €, and the same is true of 7, 7,,... Let us suppose that € and 7 are small, so that 7,, 7;, ... are in ascending order. As we have seen, 7”™= &", that 1s va gives a first » 1 ed bes approximation. To obtain a sini we may nenlee Tax Tae and substitute in 7, the value of 7 as determined by the first approximation. To get a third approximation, neglect 7,,.. ., substitute in 7, the value of 7 as given by the second approxima- tion, and in 7, the value of 7 as given by the first approxima- tion. We may proceed thus by successive steps to any degree of approximation ; the only points to be attended to are never to neglect any terms of higher degree than the highest retained, and not to waste labour in calculating at any stage the co- efficients of terms of higher degree than those already neglected. Example 1. Taking the equation (16), to find a third approximation to one of the branches of the group CD. Next in order to C and D a parallel to CD meets successively B and A. Hence, putting, for simplicity, D=+1, C=B=A=~—1, the equation (16) may be written Si ee a ka len =U Whence ne — 8-2 [qn -P% P+. . . =0 (25). The first approximation is 7=£* ; hence, neglecting £'°/y? in (25), we get for the second no — &4— 7/43 =0, Whence q=F5(1 + PO) = 1 + 325%) (26). If we use this second approximation in £7/n, and the first approximation in £°/n? now to be retained, (25) gives for the third approximation a — &4 — §7/E49( (1 + 4/9) — 07 £8/3 — () Xxx EXAMPLES 369 Whence, if all terms higher than the last retained be neglected, oP — £4 — 17/3 _ 2¢225 — 0), which gives = E43(1 + 93 + 2 10/3) Sa 4a +4 1 5/3 +. $E10/3) (27), which is the required third approximation. This might of course be obtained by successive applications of the method of transformation employed in the demonstration of § 22, or by the method of indeterminate coefficients, but the calculations would be laborious. It will be observed on comparing (27) with the theoretical result in § 22 that dy = dy==-d3=d4=dg=d7,=dg=dg=0; a fact which, in itself, shows the advan- tages of the present method. Example 2. To find a second approximation for the branches corresponding to ABC in equation (16), in the special case where A= +1, B= -2, C= +1, D=-1. The terms concerned in this approximation are (ABC) and (D). We therefore write E7(y — £3)? — £3 y5=0, or (n — §° —9°/E4=0. The first approximation is 7=£&*; hence the second is given by (n- £9 E2/¢4=0, that is, (yn — £8)? -EN=0. Whence n—-&téEN?2=—0, which gives the two second approximations corresponding to the group. These are two, because to a first approximation the branches are coincident. This, therefore, is a case where a second approximation is necessary to dis- tinguish the branches. Example 3. To find a second approximation, for large values both of € and y, to the branch corresponding to HI in equation (16). Referring to Fig. 1, we see that, if HI move parallel to itself towards O, the next point which it will meet is G. Hence, so far as the approximation in question is concerned, we may replace (16) by (HEM? 4 TEM 7) + Ge5nU4=0. . For simplicity, let us put H=1, I=G= —1, and write the above equation in the form. n> —£4-77/E4=0. Geanning ourselves to one of the five first approximations, say »=£*”, we get for the second approximation no — E4— £990, which gives n=£45(14+4E-P), Example 4. Given e=yty?/2!+y/3!+yi/4!+... to find y to a fourth approximation. We have y=u—y?/2!-y/3!—-y4/4t-... bo oe] VOU. il 370 HISTORICAL NOTE CHAP, Hence: Ist approx. y=. 2nd. Ya ae 3rd oT) y=x—-3(u-30°)? — gx, =x —4har+ Lo. 4th ,, y=u-3(e— 3a? + ba)? — 3 (a — $x°)8 — gaat, =a — 4a? + d03 — dat Historical Note.—As has already been remarked, the fundamental idea of the reversion of series, and of the expansion of the roots of algebraical or other equa- tions in power-series, originated with Newton. His famous ‘ Parallelogram”’ is first mentioned in the second letter to Oldenburg; but is more fully explained in the Geometria Analytica (see Horsley’s edition of Newton’s Works, t. i., p. 398). The method was well understood by Newton’s followers, Stirling and Taylor ; but seems to have been lost sight of in England after their time. It was much used (in a modified form of De Gua’s) by Cramer in his well-known Analyse des Lignes Courbes Algébriques (1750). Lagrange gave a complete analytical form to Newton’s method in his ‘‘ Mémoire sur l’Usage des Fractions Continues,” Nouv. Mém. d. 0 Ac. roy. d. Sciences d. Berlin (1776). (See Quvres de Lagrange, t. iv.) Notwithstanding its great utility, the method was everywhere all but forgotten in the early part of this century, as has been pointed out by De Morgan in an interesting account of it given in the Cambridge Philosophical Transactions, vol. ix. (1855). The idea of demonstrating, a priort, the possibility of expansions such as the reversion-formule of § 18 originated with Cauchy ; and to him, in effect, are due the methods employed in §§ 18 and 19. See his memoirs on the Integration of Partial Differential Equations, on the Calculus of Limits, and on the Nature and Properties of the Roots of an Equation which contains a Variable Parameter, Exercices d’ Analyse et de Physique Mathématique, t. i. (1840), p. 827; t. ii. (1841), pp. 41, 109. The form of the demonstrations given in §§ 18, 19 has been borrowed partly from Thomae, Hl. Theorie der Analytischen Functionen einer Complexen Verdnderlichen (Halle, 1880), p. 107 ; partly from Stolz, Allge- meine Arithmetik, I. Th. (Leipzig, 1885), p. 296. The Parallelogram of Newton was used for the theoretical purpose of establish- ing the expansibility of the branches of an algebraic function by Puiseaux in his Classical Memoir on the Algebraic Functions (Liowv. Math. Jour., 1850). Puiseaux and Briot and Bouquet (Théorie des Fonctions Elliptiques (1875), p. 19) use Cauchy’s Theorem regarding the number of the roots of an algebraic equation in a given contour ; and thus infer the continuity of the roots. The demonstra- tion given in § 21 depends upon the proof, a priori, of the possibility of an expansion in a power-series; and in this respect follows the original idea of Newton. The reader who desires to pursue the subject further may consult Durége, Elemente der Theorie der Functionen einer Complexen Verinderlichen Grésse, for a good introduction to this great branch of modern function-theory. The applications are very numerous, for example, to the finding of curvatures and curves of closest contact, and to curve-tracing generally. A number of beautiful examples will be found in that much-to-be-recommended text-book, Frost’s Curve Tracing. EXERCISES XXIV. Revert the following series and find, so far as you can, expressions for the coefficient of the general term in the Reverse Series :— i, 2 —1)(n-2 a yar tty MOaDang MOD Dg, Xxx EXERCISES XXIV awed! 5 7 (2.) y=u—40P + 42° -tal +. . 3 a eh (3.) Y=e— site ait (4.) y=uta?/2? + 23/3? + 8/424. . (5.) If y=sin a/sin (w+a), expand x in powers of y. # and y being determined as functions of each other by the following equations, find first and second approximations to those branches, real or imaginary, for which mod # or mod y, or both, become either infinitely small or infinitely great :— (6.) y?-2Qy=at— 2. (7.) @(y+x)-2a?a(y+x)+at=0, (F. 69%). (8.) (w—y)?—(x@—y)2° — Za4 - Sy*=0, (F. 82). (9.) a(y? — 2) (y — 2a) -y*=0, (F. 88). ot ) axly — x)? -y*=0, (I. 96). (11.) a(y-2x)?-a=0, (Pec li5). (12.) xy? — 2a?x? Eee Ge) (FE) 121). (13.) y(y —x)*(y + 2x) = 9c23, (rerial), (14.) {a(y-—a)-a?}*y=a7, (F. 140). (15.) a? — arty? + abyt — axy® = 0, (F.«143). (16.) a(x’ +y°) — aary + 2°yt=0, (F143). (17.) ay*+ ax?y? + baty+cx+dy?=0, where a, b, c, d are all positive y I , Clee) Bh (18.) If e, be any constant whatever when 7 is a prime number, and such that ¢,=épé,e,... when m is composite and has for its prime factors D, J, T, .*. -, then show that If a, b, c, . . . bea given succession of primes finite or infinite in number, s any integer of the form a%b®c”..., ¢ any integer of the forms a, abd, abe, . . . (where none of the prime factors are powers), and if F(x) =e, f(z"), then t= oat — ee He"), where wu is thé number of factors in t¢. (This remarkable theorem was given by Mobius, Credle’s Jour., ix. p. 105. For an elegant proof and many interesting consequences, see an article by J. W. L. Glaisher, Phil. Mag., ser. 5, xviii. p. 518 (1884). ) * F. 69 means that a discussion of the real branches of this function, with . the corresponding graph, will be found in Frost’s Curve Tracing, § 69. CHAPTER XXXI. Summation and Transformation of Series in General. THE METHOD OF FINITE DIFFERENCES. § 1.] We have already touched in various connections upon the summation of series. We propose in the present chapter to bring together a few general propositions of an elementary character which will still further help to guide the student in this somewhat intricate branch of algebra. It will be convenient, although for our immediate purposes it is not absolutely necessary, to introduce a few of the elementary conceptions of the Calculus of Finite Differences. We shall thus gain clearness and conciseness without any sacrifice of simplicity ; and the student will have the additional advantage of an intro- duction to such works as Boole’s Finite Differences, where he must look for any further information that he may require regarding the present subject. Let, as heretofore, uv, be the nth term of any series ; in other words, let w, be any one-valued function of the integral variable N; Un—1, Un—2) ++ +, U, the same functions of n—1, n—2,..., 1 respectively. Farther, let Az,, Atinnry 6 + 4 Au, denote Meni ns Ua Uae ee ep Use ‘also A(Au,,):° ACN 2s) pee oa which we may write, for shortness, | f — a an a —-e ne end CHAP, XXXi DIFFERENCE NOTATION O73 Aun; NL Vat Ss A*u,, denote Dene ee Unig cay fs sy Aly = Ay and so on. ‘Thus we have the successive series, Un, Us y eth oe enna a Ghat ig aN tea a aS ree i A) ee (2); MN UN tw ook te Ny on eee (Bs A*u,, EE ee BARON Thy 8 (4) ; where ah term in any series is apenned i siheewaatis the one immediately above it from the one immediately above and to the right of it. The series 2 (3), (4), . . . are spoken of as the series of Tst, 2nd, 3rd, . differences corresponding to the primary series (1). Example 1. If nT", the series in question are ee OE Tiles se he ee Oy De ils Oy A ere aS ye Hes ae efi ciip ion Pat ane where, as it happens, a second differences are all equal, and the third and all higher differences all vanish. Cor. If we take for the primary series Eee ee WEA ales AUR ie) then the series of Ist, 2nd, 3rd, . . . differences will be EME bet, fe LN tLe ied» TAttly is y: em ise tACEAD ES es WATE i vp: me Ne ACT Sy ti a AN 0 : In other words, we have, in general, A”A®u, = A’**u,. This is sometimes expressed by saying that the difference operator A obeys the associative law for multiplication. Although we shall only use it for stating formule in concise and easily-remembered forms, we may also introduce at this stage the operator EK, which has for its office to increase by unity the variable in any function to which it is prefixed. Thus old EXAMPLES CHAP. Ed(n)=¢(n+1); Eu, =tn4,; Eu, =u; and so on. In accordance with this definition we have E(Ew,), which we contract into E’u,, = Etn4,=Un+d.2; and, in general, Eu, = Unsm: We have also, as with A, E”E*%u, = E”*+*u,, for each of these is obviously equal t0 Upyrys. Example 2. E”n?=(n+7)3, Example 3. The mth difference of an integral function of » of the rth degree is an integral function of the (r—2)th degree if mr. Let p(n) =an" + bn’1+en"™ +... .; then A¢d,(n)=a(n +1)" +0(n+1)14+e(n+1)"-74 . — ant — bnr-1 — ent Fs, =raw" + thr(r-lat+(r—1)b}n72 +... ., = pr—i(2), say, where ¢,-3(7) is an integral function of n of the (7—1)th degree. Then, in like manner, we have A¢,_i(n)=¢,-2(2). But Ad,-1(n)=A?p,n ; hence A’$,(n) =$y-2(n). ° Similarly, A*%¢,(n)=¢,-s(n); and, in general, A”d¢,(n) = r-m(n). We see also that A"¢,(n) will reduce to a constant, namely, r!a ; and that all differences whose order exceeds 7 will be zero. ‘The product of a series of factors in arithmetical progression, such as aa+tb)...(a+(m-—1)b), plays a considerable part in the summation of series. Such a product was called by Kramp a Faculty, and he introduced for it the notation @”!>, calling a the base, m the exponent, and b the difference of the faculty. This notation we shall occasionally use in the slightly modified form a!™!°, which is clearer, especially when the exponent is compound. Since aa+b)...(a+(m-1)b)=b™(a/b) (a/b+1)... (a/b+m-1), any faculty can always be reduced to another whose difference is unity, that is, to another of the form ¢!™!1, which, omitting the 1, we may write ¢!”!. In this notation the ordinary factorial m! would be written 1!”!, The reader should carefully verify and note the following properties of the differences of Faculties and Factorials. In all. cases A operates as usual with respect to x. Example 4. A(a + bn) | m|b— ap fa + (n+ 1)}1m-1 |b. Example 5. Af1/(a+bn)!™| by = —mb/(a+bn)|mt1 10, Example 6. { gis _a-c (a—b)|"+110 aay ee: el vt1|b EL a, OO XXXI FUNDAMENTAL DIFFERENCE THEOREMS 375 Example 7. A cos (a+fn)= —2 sin $8 sin (a+48+ Bn); A sin (a+ Bn)= +2 sin $6 cos (a+38+4+ Bn). § 2.| Fundamental Theorems. The following pair of theorems* form the foundation of the methods of differences, both direct and inverse :— i A” Un = Un+m — POR) anata fe NOt week Spas rd ( 7 hia Te, ay nin nad Unit oa + Ay, To prove I. we observe that Avy, = Un+i— Un; 2 An =Unte- Unti — Untit Un, ies > Line + Un 3 hence 3 ae Un+te cis 2Un-+1 as Uns = Unts — 8Unte + 3Un41 — Un 5 and so on. Here the numerical values of the coefficients are obviously being formed according to the addition rule for the binomial coefficients (see chap. iv., § 14); and the signs obviously alter- nate. Hence the first theorem follows at once. To prove II. we observe that we have, by the definition of Atm, Um+i = Um + Am. Hence, since the difference of a sum of functions is obviously the sum of their differences, we have, in like manner, Uy+o= Um-+-1 + AUmti = Un + Athy + Ay + Atm) = Wm + Alm + AUm + A’. We therefore have in succession * The second of these was given by Newton, Principia, lib. iii., lemma v. (1687) ; and is sometimes spoken of as Newton’s Interpolation Formula. See his tract, Methodus Differentialis (1711); also Demoivre, Miscellanea Analytica, p. 152 (1730), and Stirling, Methodus Differentialis, &c., p. 97 (1730). 376 SUMMATION BY DIFFERENCES CHAP. Une Uy hr Ny, Um+2 = Um + Ati + Athy + ‘A°Um, Um + 2AUm + A°ubm, ; Ung = Um + 2A, + A’, + Atl + 20° + A°Umy Um + 3AUm + BAU + A°ttm 3 and so on. The second theorem is therefore established by exactly the same reasoning as the first, the only difference ‘being that the signs of the coefficients are now all positive. If we use the symbol EH, and separate the symbols of oper- ation from the subjects on which they operate, the above theorems may be written in the following easily-remembered symbolical forms :— Pal AMgg) = Hr —a1 yee i a) tha, = (lt Ai § 3.] The following theorem enables us to reduce the sum- mation of any series to an inverse problem in the calculus of finite differences. Tf vy be any function of n such that Avy = Uy, then 12 D Ay = Uy, ey 0; (1), N=Ss This is at once obvious, if we add the equations Un, = Avy, = Unti— Un) Un —1 = Dens; Un = Mays Ug =A, SVjag 05. The difficulty of the summation of any series thus consists entirely in finding a solution (any solution will do) of the finite difference equation Av, = Un, OF Un4,— Un =Un. This solution can be effected in finite terms in only a limited number of cases, some .of the more important of which are exemplified below. On the other hand, the above theorem enables us to con- \ | XXXI EXAMPLES 377 struct an infinite number of finitely summable series. All we have to do is to take any function of » whatever and find its first difference ; then this first difference is the nth term of a summable series. It was in this way that many of the ordinary summable series were first obtained by Leibnitz, James and John Bernoulli, Demoivre, and others. Example 1, 2 ee fa+(n+1)b}... (a+ (m+ m — 1)d}. Using rare s notation, we have here to solve the equation Avn= fa+nbd\|ml® (2). Now we easily find, by direct verification, or by putting m+1 for m and n—1 for nin § 1, Example 4, that Al {a+(n—1)b}!™411O/(m+1)o]= fat nd}! 12, Hence v= fa+(n—1)ot1™411 m+1)b is a value of v, such as we require. Therefore > fa-tnb}l™ ie fanny ae My Hence the well-known rule nN L{a+nb\ fat+(n+1)b}... fa+(n+m-—1)b} =C+ {a+nb} {at+(n+1)b}... {a+(n+m-—1)b} {a+ (n+m)bt /(m4+1)b (4), where C is independent of n, and may be found in practice by making the two sides of (4) agree for a particular value of n. Example 2. To sum any series whose nth term is an integral function of n, say f(2). By the method of chap. v., § 22 (2nd ed.), we can express /(n) in the form a+bn+en(n+1)+dn(n+1)(n+2)+... Hence N ; : . Tf(n) =C + an + $bn(n +1) + gen(n+1)(n+ 2)+4dn(n +1)(n+2)(n+8)+ ; where the constant C can be determined by giving x any particnlar value tn (5). Example 3. 1/{a+bn}!™1®, Proceeding exactly as in Example 1, and using § 1, Example 5, we deduce 1 _If{atbs}l™-11P-1/{a+d(n4+1)}l™—110 s fat+bnil™l> (m —1)b Hence a rule for this class of series like that given in Example 1. Example 4. To sum the series Sf(n)/{a+bn}!™!®, f(r) being an integral function of n. “ins EXAMPLES CHAP. Decompose f(n), as in Example 2, into a+B(a+bn)!11044(a+bn) 12194 s(a+0n)!81O4. (7). Then we have to evaluate v7 v1) aD1/{atbn} 119+ Bx1/ {a+d(n4+1)} 1-194... (8), which can at once be done by the rule of Example 3.* Example 5. N q|n| 0 a (a+b)|"10 (a+b) 18-118 : elnlo a-—c+b cl 2|d - cls-1]b \ (9). This can be deduced at once from § 1, Example 6, by writing a+) for b and m—1 for n. Example 6. To sum the series whose terms are the Figurate Numbers of the mth rank. The figurate numbers of the 1st, 2nd, 3rd, . . . ranks are the numbers in the Ist, 2nd, 3rd, . . . vertical columns of the table (II.) in chap. iv., § 25. Hence the (n+1)th figurate number of the mth rank is anu =ntm-1Cn=m(m+1)...(m+n-1)/n!. Hence we have to sum the series" n y ae Teen ...(M+n 1) 1 A ae! Now if in (9), Example 5, we put a=m, b=1, c=1, we get 2 oy || _(m+1)!"! m+1 21!" It 1 Hence m(m +1) mm+1):..(m+n-1) 1 aa ee: GC Ge a ae eee _(m+1)(m4+2)...(m+1+4+n-1) } = 1S a (10) 5 that is to say, the swm of the first n figurate numbers of the mth rank is the nth Jigurate number of the (m+1)th rank. This theorem is, however, merely the property of the function ,,H,, which we have already established in chap. xxiii, § 10, Cor. 4. The present demonstration of (10) is of course not restricted to the case where m is a positive integer. Many other well-known results are included in the formula of Example 5, some of which will be found among the exercises below. * The methods of Examples 1 to 4 are all to be found in Stirling’s Methodus Differentialis. He applies them in a very remarkable way to the approxi- mate evaluation of series which cannot be summed. (See Exercises AXVIE 17.) XXXI DIFFERENCE SUMMATION FORMULA By!) Example 7. To sum the series Sn=cosa+cos(a+B)+...+cos(a+(n—1)8) ; T,=sina+sin(a+8)+... +sin(a+(n-1)6). From § 1, Example 7, we have cos(a+fn)=A {sin (a - $84 n)/2 sin $8}. Hence Sn= (sin (a — 48+ Bn) - sin (a — $8)}/2 sin 48, _ sin $6n ‘ mg c08 {a-+48(0-1)}. Similarly, sing6n . =——— sin {a+4P(n-1)}. n= ag sin (a+ 36(n—1)} § 4.| Expression for the sum of n terms of a series in terms of the jist term and its successive differences. Let the series be u,+u.+...+U»,; and let us add to the beginning an arbitrary term u,. Then if we form the quantities S, Gis) y= Upk Uie | Sg Uy FUE Uy, ee Dye iy eg te ob SE se), we have ee a, 7 2, AMS, AM ly. Hence, putting n= 0, eee Nes ASS AMI 2 (Ly Now, by Newton’s formula (§ 2, II.), B= Petr lpAd a+ aed Sok ow + A'S, (2). If, therefore, we replace S,, AS,, A°S,, . . . by their values according to (1), we have 2 gee Ot, + nga, + jC, u, +... +A, .(3)% 0 or, if we subtract u, from both sides, iD Bea + CAA ier oe A ty, (4).* 1 The formula (4) is simply an algebraical identity which may be employed to transform any series whatsoever ; for example, in the case of the geometric series 22” it gives * This formula, which, as Demoivre (Miscell. An., p. 153) pointed out, is an immediate consequence of Newton’s rule, seems to have been first explicitly stated by Montmort, Journ. d. Savans (1711). It was probably independently found by James Bernoulli, for it is given in the Ars Conjectandi, p. 98 (1718). 380 MONTMORT’S THEOREM CHAP, ee ps, Se a ae TAN) ey eae te ae a) (oh a = Ne + 2! + x(% — 1)-1, which can be easily verified independently by transforming the right-hand side. The transformation (4) will, however, lead to the sum of the series, in the proper sense of the word sum, only when the mth differences of the terms become zero, m being a finite integer. The sum of the series will in that case be given by (4) as an integral function of of the mth degree. Since the nth term of the series is the first difference of its finite sum, we see conversely that any. series whose sum to nm terms is an integral function of n of the mth degree must have for its nth term an integral function of n of the m— 1th degree. We have thus reproduced fr om a more general point oy view the results of chap. xx., § 10. Example. Sum the series _ vi) =(2 +1) (m+ 2) (n+ 8). 1 If we tabulate the first few terms and the successive differences, we get I es 3, 4, 5 ty, |) D4y gO 120 £210... 1330, Ate |) (06, 00,8 905” =126. A°u, |} 24,2°30; 36; Leu CG: Attn 0. Hence, by (4), 117 =(n +1) (n+ 2) (n+3) 1 n(n — 1) n(n—1)(n-2) 5, , n(n—1) (n—-2) (n—-8) so 36 + 6 244 94 0, =4H(n*+10n? + 35n? + 500). =7.24+ § 5.] Montmorts Theorem regarding the summation of Suna. - elegant formula for the transformation of the power- series 2u,v" may be obtained as follows. Let us in the first i) . place consider S = 2v,7”, which we suppose to be convergent when 1 modz<1; and let us further suppose that mod # tan (6/2”) [an (15.) 2 tan! {(na-—nm+1)a"-"/(1+n(n-1)a?"-1)}, (16.) Stan-1{2/n?}. (17.) m!+(m-+I1) !/1!+(m+2)/2!4... (18.) 1!/m!4+2'/(m+1)!43!/(m4+2)!4... * The sums to 7 terms of arithmetical progressions whose first terms are all unity, and whose common differences are 0, 1, 2,. . .,(r—1),.. . respect- ively, are called the nth polygonal numbers of the Ist, 2nd, 3rd,..., rth,... order. The numbers of the first, second, third, fourth, . . . orders are spoken ~ of as linear, triangular, square, pentagonal, . . . numbers. 5. 16. Ly 1 384 EXERCISES XXV CHAP, (19.) 1—mCit+mC2—. . 2 (—)"mCn. (20.) Show that the figurate numbers of a given rank can be summed by the formula of § 3, Example 1. 1 1.2 ieee ely) bi rian 1) ee a ee = cir a(a@+1)... (a+r), (a+). Pt (atrt ly c c(e+1) (22.) a a(a+1) ot, devi). 2: (ctr) cerl) ...(e+7r+1) (24.) Z(a+n)!™—21/(c-4+m)1™, : 1.3 123.25 1.8.5.7 (oT. 84t 1 28.4.0 Le ee (l+r)(142r). (L+r)(1+2r)(1+8r) Ce, 1.2.3.4.5 1.2.3.4.5.6 27.) 2m m— e m(m — 1) +7 mm—1)(m-2)-.. . sites | 1.3 1.3.8 oy Show that [PE (od) B/E OD eh/dd Gd) (ga a= Pail (+4): (Glaisher.) (29.) Show that 1+2(1-a)+3(1-a)(1-2a)+. . .+n(1-a)(1-2a)...(1-(n—1)a) =a-1{1—(1-a)(1-2a)...(1-na)}. 1 1 2" 3 1 00) ge1e-1 @=1)@—2) * @-1)@-2) 3) (- ae n+ i); (a—1)(a—-2)... (a@- ie z+1/ . (81.) Ifa+b4+2=c+d, then geroet a aso" ee sq al (a1) (b+1)-cd itl ginl pit gis (32.) qt q(q-1).r(r-1) — +- en mmmmen>areien (p—qt1).(p+7—-1) (p-gt1) (p-g+2).(ptr—-1) (p+r— 2) | _(p-9).(pt7) | p.(p- qtr) (Educational Times Reprint, vol. xli., p. 98.) (33.) Transform the equation | log2=1-3+4-4+... by § 5, Cor. 2. ' (34.) Show, by means of § 2, I., that, if m be a positive integer, then aa—-1) , aa-1)(a—-2) L= mig +m m2 a5 =3) my? mC3 55-1) (6-2) 1 oe = (1:5) 55)" peace XXXI DEFINITION OF RECURRING SERIES 085 RECURRING SERIES. § 6.] We have already seen that any proper rational fraction such as (a + ba + ca*)/(1 + px + qa’ +72°)* can always be expanded in an ascending series of powers of x. In fact, if modz be less than the modulus of that root of 712° + qa” + px +1=0 which has the least modulus, we have (see chap. xxvii. §§ 6 and 7 ) a+ ba + cx” 1 + pa + ga? + rx? =U tUj Gt Ue +s. . tUya™+... (1). We propose now to study for a little the properties of the series (1). It we multiply both sides of the equation (Ls by - 1 + px + gx’ + 12x’, we have a + bx + ca? = (1 + pa t+ gx? + 12°) (uy + Ut + Ut +... Una +.. -) (2). Hence, equating coefficients of powers of 2, we must have Uy = a (3p; Uy + Puy = b (GPE Uy + PU, + JUy = (one Us + Py + Qu, + 7U, = 0 OAK Un, + Mn + Yn + 1Un—z = 0 (35-61). Any power-series which has the property indicated by the equation (3,+,)is called a Recurring Power-Series;+ and the equation (3n41) is spoken of as its Scale of Lelation, or, briefly, its Scale. The quantities p, g, r, which are independent of n, may be called the Constants of the Scale. According as the scale has 12; 3,25. ., 7, . . . constants, the recurring series is said to be of the 1st, 2nd, 3rd, .. ., rth, . . . order. When z= 1, so that we have simply the series Uget Utter 6 5. Unt 6. ce, with a relation such as (3n41) connecting its terms, we speak of “ For simplicity, we confine our exposition to the case where the de- nominator is of the 8rd degree; but all our statements can at once be generalised. + The theory of Recurring Series was originated and largely developed by Demoivre. VOL. II 2C 386 MANIFOLDNESS OF RECURRING SERIES CHAP. ¢ the series as a recurring series simply ;* so that every recurring series may be regarded as a particular case of a recurring power- series. It is obvious from our definition that all the coefficients of a recurring power-series of the rth order can be calculated when the values of the first r are given. Hence @ recurring ser ies of the rth order depends upon 2r constants ; namely, the r constants of its scale, and'r others. From this it follows that if the first 27 terms of a series be given, it can be continued as a recurring series of the 7th order in one way only ; as a recurring series of the (7 + 1)th order in a two-fold infinity of ways ; and so on. On the other hand, if the first 27 terms of the series be given, two conditions must be satisfied in order that it may be a recurring series of the (7 — 1)th order ; four in order that it may be a recurring series of the (r — 2)th order; and so on. Example. Show that a + Qa? + Bar? + dart + Bac’ + Gar + is a recurring series of the 2nd order. Let the scale be un + PUn-1 + (Un-2= 9- Then we must have 84+2n+q=0, 44+3p4+2q¢=0, 54+4p+3q=0, 6+5p+4q=0. The first two of these equations give p= — 2, g= +13 and these values are consistent with the remaining two equations. Hence the theorem. § 7.] The rational fraction (a + bx + ca”)/(1 + pa + qa" + ra), ot which the recurring power-series W)+,%+ Ut +... 18 the development when mod z is less than a certain value, is called the Generating Function of the series. We may think of the series and its generating function without regarding the fact that the one is the equivalent of the other under certain restric- tions. If we take this view, we must look at the denominator of the function as furnishing the scale, and consider the co- a aaa * We might of course regard a recurring power-series as a particular case of a recurring series in pel Thus, if we put U,=2,x”, we might regard the series in (1) as a recurring series whose scale is Un + px Unei + qa U,.25 + ra ios = 0. XXXI . GENERATING FUNCTION 387 efficients as determined by the equations (3,), (3.),. . ., (3n4,)-* No question then arises regarding the convergence of the series. Gwen the scale and the first r terms of a recurring power-series of the rth order, we can always find its generating function. Taking the case r = 3, we see, in fact, from the equations (3,), Poe oni,)) - - sof § 6, that (Mo + (Uy + PUly)% + (Ug + pt, + Quiy)a?}/{1 + pa + ga? +10") is the generating function of the series Ug t+ UE+ Ue +. . whose scale is WU, + pup_, + Gin cr oO: Cor. 1. Every recurring power-series may, of moda be small enough, be regarded as the expansion of a rational Fraction. Cor. 2. The general term of any recurring series can always be found when its scale is given and a sufficient number of its imitial terms. For we can find the generating function of the series itself or of a corresponding power-series ; decompose the generating function into partial fractions of the form A(%—a)-*; expand each of these in ascending powers of 2; and finally collect the coefficient of a” from the several expansions. ig Example. Find the general term of the recurring series whose scale is Un — 4Un—-1 + 5Un—2 — 2Un-3=0, and whose first three terms are 1 +0-5. Con- sider the corresponding power-series. Here p= —4, q=b, r= —2; so that 4=MW=1, b=m+pm=-4, c=UWZ + pu, +quo=0. The generating function is therefore 1-4 re 1-4 1 — 40+ 5x? — 2x? ~ (1 — x)%(1 - 22)’ oe 3 4 “T-# \(i—a)' (1 Bay Expanding, we have i = Dar? : mn § Son wn) [ode aboot + ze $+3{1+2(n+1)a”} 441+ 22%}, =1+2(3804+ 5 — Q+2)qn, The general term in question is therefore 8n +5 —2”+2, § 8.] If wu be any function of an integral variable n which satisfies an equation of the form Un, + My + QUin— + Tn, = 0, or, what comes to the same thing, Unts + PUnte + Unt, +7Un = 0 (1), * We might also regard the series as deduced from the generating function by the process of ascending continued division (see chap. v., § 20). 388 LINEAR DIFFERENCE-EQUATION CHAP. we see from the reasoning of last paragraph that uw, is uniquely determined by the equation (1), provided its three initial values Uy Uy, Uz are given; and we have found a process for actually determining U,. It is not difficult to see that we might assign any three values of w, whatever, say u,, Ug, U,, and the solution would still be determinate. We should, in fact, by the process § 7, determine w,, as a function of 7 linearly involving three arbitrary constants Wp, Ur) Urs say f(t, Uy U2, 2); and UZ, %, % would be uniquely determined by the three linear equations F(tos Urs Us 4) =Uny f(oy ti; Ue, B) = Ug, f(y, Uy, Uo, y) = Uy (2). An equation such as (1) is called a Linear Difference-Equation of the 3rd order with constant coefficients ; and we see generally that a linear difference-equation of the rth order with constant coefficients has a wnique solution when the values of the function pel are given for r different values of its integral argument. Example. Find a function wv, such that p13 — 4Uinte+ 5tinty — 2%, =0; and U=1, m=0, w= —- 5. We have simply to repeat the work of the example in § 7. § 9.] Lo sum a recurring series to n +1 terms, and (when con- vergent) to infinity. Taking the case of a power-series of the 3rd order, let Sn = Ut + Ugh +. . . + Ue”, then PUSy = PUgl + PU,e? + 66+ Py, 2"+ pupa”, oD = Mgt? +... + Qing + Qin tt) + Quine t2, io. = vee bh Ui gl" + Uy gt! + 1g 042 4 py, eet 8 Hence adding, and remembering that w, + Un =, + Gina +7U,-,=9 for all values of n which exceed 2, we have (1 + pa + qa? + 13°)S_ = Uy + (Uy + pUio)& + (Uy + pry + qtty)a0® + (Pin + QU —1 + 1 —2)8"F} + (Gti + TU — 1)a 2 + rU_art3 ie whence §, can in general be at once determined by dividing by 1+patqu + rx’. The only exceptional case is that where for the particular value of z in question, say 7 =a, it happens that 1+pa+ga'+ra> = 0. XX SUMMATION OF RECURRING SERIES 389 In this case the right hand of (1) must, of course, also ‘vanish, and 8, takes the indeterminate form 0/0. S, may in cases of this kind be found by evaluating the indeterminate form by means of the principles of chap. xxv. This, however, is often ‘much more troublesome than some more special process applicable to the particular case. If the series Du,7” be convergent, then Lu,z”=0 when n=; therefore the last three terms on the right of (1) will become infinitely small when n= a0. We therefore have for the sum to infinity in any case where the series is convergent Up + (Uy + PUp)& + (ty + PU, + Yl) ae” | Sa 1 + put qa’ + 72% i The particular cases | Uy + +Ugt. ..t+Unt... (3), Uy —U,+Ug—. . . +(—) yt... (4), are of course deducible from (1) and (2) by putting «= +1 and «= —1. Exceptional cases will arise if 1+p+q+7r=0, or if l—-p+q-r=0. It is needless to give an example of the above process, for Examples 1 and 2, chap. xx., § 14, are particular instances, Yn7x" and 1 + =(— )”~-12nx” being, in fact, recurring series whose Scales-are U,, — dU, 7+ Olin — Un-g= 0 and Uy + 2uq s+, = 0 respectively. EXERCISES XXVI. Sum the following recurring series to n+1 terms, and, where admissible, to infinity :— (1.) 24+54+134+354+97+... . (2.) 24104+12-24424+10+4124... 3.) 24+17¢+95e7+46lae+... = «| (4.) 5+12x%+380x?+ 7803 4+2100'+... (5.) 14+40+172?+ 7623 + 3538z4+. . (6.) 1+ 4xe+10a?+ 2203+ 4604+... (7.) If a series has for its rth term the sum of 7 terms of a recurring series, it will itself be.a recurring series with one more term in the scale of relation. Find the sum of the series whose 7th term is the sum of 7 terms of the recurring series 1+6+40+288+... 390 EXERCISES XXVI CHAP. (8.) IfTr, Troi, Tr4+2 be consecutive terms of the recurring series whose scale is Tr42=aTn41 —bTy, then CT257 = al, fe “ OTL peta 3) al nor ee ae Olea) = OF, (9.) Form and sum to x terms the series each term in which is half the difference of the two preceding terms. (10.) Show that every integral series (chap. xx., § 4) is a recurring series ; and find its scale. (11.) If pn =2¢n1+Un—2, and %=ar,, show that Un? — UntaUn—1 = ( — )"(a? — a — 1202. (12.) If the series %, tw», wg, . . .) Un,.. . be such that in every four consecutive terms the sum of the extremes exceeds the sum of the means by a constant quantity c, find the law of the series; and show that the sum of 2m terms is gm(m — 1) (42m — 5)e— m(m — 2)ey + mry-+m(m — 1)us. (13.) If Unyo=UnyitUn, %=1, w=1, sum the series i 2 Unt2 (14.) By French law an illegitimate child receives one-third of the portion of the inheritance that he would have received had he been legitimate. If there be Z legitimate and n illegitimate children, show that the portion of inheritance 1 due to a legitimate child is 1 n n(n —1) mn—A) ss. 2.1 7 81041) * 30+ 1\0+2)- — (— ani(T$ 1), ..(d+n) (Catalan, Nouv. Ann., ser. ii., t. 2.) WARINGS METHOD FOR SUMMING THE SERIES FORMED BY TAKING EVERY cTH TERM FROM ANY POWER-SERIES WHOSE SUM IS KNOWN. § 10.] This method depends on the theorem that the swm of the pth powers of the kth roots of unity is k if p be a multiple of k, but otherwise zero. This is easily seen to be true; for, if w be a primitive th root of 1, then the & roots are O50, Oy « . «. OFT LE heme then (w*)? = wi" = (whys = 1. If p be not a multiple of &, then ” we have (0)? + (w!)P +... + (ot!) P = 1+ (oP) + (oP) +. . + (oP) HD, = {1 =(w?)9/(1 — w?), = 0, for (w?)* = (w*)P = 1, and w? +1. XXXI WARING’S THEOREM 391 Let us suppose now that f(z) is the sum of m terms of the power-series Uy -+ Duna", m being finite, or, it may be, if the series is convergent, infinite. Consider the expression Une (°F fw") + (cw )e™ f(w'at) + (ie en f (w 2) +e..+ ea (w* 12) k (1); where m is 0 or any positive integer < k. The coefficient of «” in the equivalent series is ela ye +t xe (G eee ae ‘Gg See eee (ied eee i (2). Now, by the above theorem regarding the /th roots of unity, the quantity within the crooked brackets vanishes if /—m+r be not a multiple of /, and has the value & if kL-m+r be a multiple of &. Therefore we have ahh 1h Spl Oe ele St Tea le ed ere (3), where the series extends until the last power of a is just not higher than the nth, and, in particular, to infinity if f(z) be a sum to infinity.* If we put m=0, we get { flax) + flor) +flw'a) +... + f(w*-12)} [hk = Uy + UpUe + Uy t™ + Upc e+... (4). Example 1. L+oetert+. . .+2%=(1—a"t)/(1-2). Hence, if w be a primitive cube root of 1, we have T= wrtHyntl ee W2nt2yn+1 \ 1+ + 05+ aay {Py aah fe ~*\ 1-2 1- wx lee J’ where 3s is the greatest multiple of 3 which does not exceed 7. Example 2. To sum the series a oe gill ait Ta SHOR ad oo. We have ; 2 ian ana ad oo aitait °° : on rE EEE Ta * This method seems to be due to Waring. See Phil. Trans. R.S.L. (1784). o92 . MISCELLANEOUS METHODS CHAP. Hence, if w be a primitive 4th root of unity, say w=7, then, since here k=4, m=3, k-m=1, w*=-1, w®= —71, we get gs gt gl lt le ial Mlle edgar es pri Farah ol 4 : . ye gt pl that is, asin @— Be) Seta nak MISCELLANEOUS METHODS. § 11.] When the nth term of a series is a rational fraction, the finite summation may often be effected by merely breaking up this term into its constituent partial fractions; and even when summation cannot be effected, many useful transformations can be thus obtained. In dealing with infinite series by this method, close attention must be paid to the principles laid down in chap. xxvi., especially § 13; otherwise the tyro may easily fall into mistakes. As an instance of this method of working, ‘see chap. xxviil., § 14, Examples 1 and 2. Example. Show that { : Ss - ah I + (%-+1)?(a+2) © (w+2)?(a+3) ' (#+3)2(a4+4) °° ° J +{ : fe : ot : + } as (@+1) (+2)? © (+2) (+8)? “ (@4+3) (2442 7° °° (2#+1)2" Denote the sums of n terms of the two given series by S, and 'T,, respect- ively, and their nth terms by w,, and v, respectively. Then Un= —1f(@t+n)+1/(e+n)?+1/(e@+n4+1); m=1/(e+n)-1/(e@+n+1)?-1f(@+n+1). Whence we get at once . Sn +Tn=1/(e@+1)? -1/(v2+n4+1)2. Therefore | Soo + Ty. =1/(e+1) § 12.] Buler’s Identity. The following obvious identity * l—a, + ,(1 —a,) + a,0,(1 —a,) +270, bags. An(1 — Gn) is often useful in the summation of series. It contains, in fact, * Used in the slightly different form, (1+) (1+ a2) (1+a3)(1+a4)... =1+ a +49(1 +a) +a3(1 +41) (1+) +a4(1 +47) (1442) (1+a3)+. . ., by Euler, Nov. Comm. Petrop. (1760). XXXI EULER’S IDENTITY 393 as particular cases a good many of the results already obtained above. . If in (1) we put a, =~, ett pee es an anes ston oe aaa y Yt, Y + Ds Y + Pn and multiply on both sides by y/(y— x), we get ny a(x + p,) 2% + pi)... (+ Pn-1) + $+... 4+ y+ pr (y+ Dr) (Y + Ps) (y + pr) y+ Bs) »-» (Y + Pn) eth 2 ie +p) (@+ py)... (+ Pa) (2). y-% yu (Y+ py) (y+ Po) +.» (YF Pn) If the quantities involved be such that | (w+ pi) (@ + ps)» -- (@ + Pn) i n=a(¥ + pi) (Y + Po) «+» (Y + Pn) (3), then w wa + pr) y ~ ‘ ..ado=—*— (4). y+D, Y*D) YP) y-« If in (2) we put y= 0, we get 2 eet py) ig eee) Pas) Py Pips Pipe ++ Pn = (1+ 5) (1 +2) a (1+=) (5). Pr Ps Pn From (5) a variety of particular cases may be derived by putting n= 0, and giving special values to »,,-7,, ... Thus, for instance, if the infinite series 21/p, diverge to + 0, then (see chap. xxvi., § 24) we have a2 HERP) _ Pr Pi Ps In general, if the continued product H( + 2/Pn) converge to any eee aioe a=) | (6). definite limit, then the series 1 + a(x HO Eee Pnen) svat: Don converges to the same limit. Example. Find when the infinite series ~ g=14— x(x + p) aoa + pp) (a + 2p) iy: (7) yp (yt+p)(y+ 2p) (y+p) (y+ 2p) (y + 8p) converges, and the limit to which it converges. 394 . EXERCISES XXVII CHAP, If in (2) above we put y:=p, po=2p, &e., . . ., we have Sey ee L (w@+p)(@+2p)... (a+np) (8). YB Y- Lye w(Y+p) (yt2p)... (y+np) Now the limit in question may be written tees 1 1l+y/np J? but this diverges to ~ if (w-y)/p be positive, and converges to 0 if (w—y)/p be negative (chap. xxvi., § 24). Hence, if p denote in all cases a positive quantity, we see that a w(x +p) y 1 + — TE Ge ad o= ytp (y+p)(y+2p) y-a ify>a; and = Sd, +. ,/ado =i2e, + aeeeretege ree att y-p (y—p)(y-2p) y-%x if y(m?—17) m?(am? — 1°) (m? — 2?) l-t+— oF 12.92, 32 +...ado; 1 ae m ue oe ) m(m fa +3?) Choe (15.) Show that enac ae Pride a Piprts a+ pr (a +21) (dat P2) | (t+ Pr) (A2+P2) (gt Ds) Pipa. + + Pn-1%n Pips... p "(@+tn) (d+ Ps) (G_,+Pn) (G+ P1) (2 +P2) «++ (Gm + Pn)’ (16.) Show that 14— (12-22)? — (32-22)? cay tee ee (Glaisher, Math. Mess., 1873, p. 138.) tan?5 we= (17.) Show that 1 1 1 hope 7m nel) n+l) (n+2)' n(nt1)(n+2) (m+3)° aide and apply this result to the approximate calculation of 7° by means of the formula w?/6=1/127+1/27+1/37+.. . (Stirling, Methodus Differentialis, p. 28.) (18.) Show that 21/(m"-1)=1 and 21/(a"-1)=log2, where m and 2 have all possible positive integral values fetes from aiviors @ is any even positive integer, and each distinct fraction is counted only once. (Goldbach’s Theorem, see Liowv. Math. Jour., 1842.) (19.) If m have any positive integral value except unity, and 7 be any positive integer which is not a perfect power, show that 2(n- ae 1) =7/6; and, if d(n) denote the number of divisors of », that (a(n) — 1)/r* = 165 also that X(n —1)/r=Z1/(r-1)? (Jb. ) CHAPTER XXXII. Simple Continued Fractions. NATURE AND ORIGIN OF CONTINUED FRACTIONS. § 1.] By a continued fraction is meant a function of the form 2 % Qs + a, Sear (1) ; the primary interpretation of which is that 0, is the ante- cedent of a quotient whose consequent is all that lies under the line immediately beneath b,, and so on. There may be either a finite or an infinite number of links in the chain of operations; that is to say, we may have either a terminating or non-terminating continued fraction. hn Ceae In the most general case the component fractions 2, —, Uwe, b, . . . . —, + +. as they are sometimes called, may have either positive or Ws negative numerators and denominators, and succeed each other without recurrence according to any law whatever. If they do recur, we have what is called a recurring or periodic continued fraction. For shortness, the following abbreviative notation is often used instead of (1), UE fae a, + — rua!” 2 hg te ee (eee (2), the signs + being written below the lines to prevent confusion with EU tyhe / ae: | J a fy ? bp \e A j . 7] & : A »uAty’ KAM BAAD CHAP, XXXII SIMPLE CONTINUED FRACTIONS 397 b, b h ee eee Mt As WW Examples have already been given (see chap. ii., Exercises III., 15) of the reduction of terminating continued fractions ; and from these examples it is obvious that every terminating continued fraction whose constituents a,, Gs,» + +> Os, Os, ... Oe commensurable numbers reduces to a commensurable number. § 2.] In the present chapter we shall confine ourselves mainly to the most. interesting and the most important kind of continued fraction, that, namely, in which each of the numer- ators of the component fractions is +1, and each of the denominators a positive integer. When distinction is necessary, this kind of continued fraction, namely, eh cae e 4a gp Uy + Ms + Ay + may be called a simple continued fraction. Unless it is otherwise stated, we suppose the continued fraction to terminate. In this case, for a reason that will be understood by and by, the numbers @,, ds, 3, . . . are called the first, second, third, . . . partial quotients of the continued fraction. § 3.] Every number, commensurable or incommensurable, may be expressed uniquely as a simple continued fraction, which may or may not terminate. For, let X be the number in question, and a, the greatest integer which does not exceed X ; then we may write K=qty (1), where X,>1, but is not necessarily integral, or even commensur- able. Again, let a, be the greatest integer in X80 that G1 is then we have 1 where X,> 1, as before. % . by * bs 4 bs te . . The notation oe ea frequently used by Continental 2 a writers. 398 CONVERSION OF ANY NUMBER INTO §.C.F. CHAP, Again, let a; be the greatest integer in X,: then 1 DG mne X (3) ; and so on. This process will terminate if one of the quantities X, say X,-1, 18 an integer ; for we should then have PONE = An . (1). Now, using (2), we get from (1) I X =a, + , rae Ay 2 X, Thence, using (3), we get 1 X=a,+ ly + ——. d,+— 3 xe and so on. Finally, then, lites 1 X= 0, $2 Sa (a). Ay + Ug + On, It may happen that none of the quantities X comes out integral. In this case, the quotients a,, @,, .. . either recur, or go on continually without recurrence; and we then obtain in place of (a) a non-terminating continued fraction, which may be periodic or not according to circumstances. To prove that the development is unique, we have to show that, if 1 eel 1 1 a, + ee (tn Dorey es hoe «3 (8), Ay + Us + Ay! + Us + i / / then a, = dy’, dz = dq’, ds = ds, &e. Now, since a, and a,’ are positive integers, and + eT ? als + 1 PRA. hea | 1 —— ... are both positive, it follows that ..and — 1 ;, +++ are both proper fractions. Hence, by chap. ui, § 12, XXXII CONVERSION UNIQUE 399 we must have and eel 1 Ne Gz+ds+ a Again, from (6), we have Te, fl Sai es dy + 2 =O + ——... (€). Ay + dy + Gs + Ay + From («), by the same reasoning as before, we have (ly = Ay (¢), Lae 4 Latent 15 al ay ea and soe Rea GPE iy area aE Tk a (7). Ay + Ay + Us + Uy + Ay + As + Proceeding in this way, we can show that each partial quotient in the one continued fraction is equal to the partial quotient of the same order in the other.* This demonstration is clearly applicable even when the continued fraction does not terminate, provided we are sure that the fractions in (f), (0), (m), &c. have always a definite meaning, This point will be settled when we come to discuss the question of the convergency of an infinite continued fraction. Cor. If a,, Gs, . - -, Gn, 0,05, . . ., dy be all positive integers, T+. INE Yn+, any positive quantities rational or irrational each of which is greater than unity, and of , 1 ie gil 1 reel ete - = b, =f Se ume ons ’ he + An -- nti b, = Gn + Yn+1 a, + then must OU Oa U5 as, On Un, ONE O80 Ba = Hy §$ 4.] As an example of the general proposition of § 3, we may show that every commensurable number may be converted into a terminating continued fraction. Let the number in question be A/B, where A and B are integers prime to each other. Let a, be the quotient and C the remainder when A is divided by B; a, the quotient and D the * We suppose, as is clearly allowable, that, if the fraction terminates, the last quotient is>1. It should also be noticed that the first partial quotient may be zero, but that none of the others can be zero, as the process is arranged above. 400 CASE OF COMMENSURABLE NUMBER CHAP, remainder when B is divided by C; a, the quotient and E the remainder when C is divided by D; and so on, just as in the arithmetical process for finding the G.C.M. of A and B. Since A and B are prime to each other, the last divisor will be 1, the last quotient @,, say, and the last remainder 0. We then have A. C ae ee le B ih Ca BAG sec Ds C E ine Dit? thee ine Xe. ; Hence A 1 1 | Eyer et Fa ae er Or It should be noticed that, if A1, show that . _ 8a : pal) 1 2 edit lee * * (20.) Show that ele 3 20) ee es ee AM 2 ee gas (21.) Show that every rational algebraical function of x can be expanded y) and that in one way only, as a Ge sate continued fraction of the form 1 1 Qit+.~— ve ree Qs+ Qn g where Qi, Q2, . . -, Qn are rational integral functions of 2. Exemplify with (7?+a?+a+1)/(a!+3a3 + 2a?+2+1). eal (22. If aeyiaen ive 5 apy * * ban ae iaigeic. ae x show that “x-y=a-b, 404 COMPLETE QUOTIENTS AND CONVERGENTS CHAP, PROPERTIES OF THE CONVERGENTS TO A CONTINUED FRACTION. -§ 6.] Let us denote the complete eudletatin! pease by 2, so that , = + eke ‘Lye te ROY eile Ae he Us (1); and let shy rg 1 ye ig se 2); inet Og + Oy + Ws (2) 1 1 4 — a eee oO ° - Chae ibs (3) and so on. ‘ Then 2, 2, . . . are called the complete quotients corresponding to My, ds, » » . Or, Simply, the second, third, . . . complete quotients. The fraction itself, or 2,, may be called the first complete quo- tient. It will be observed that a,, a, a3, . .. are the integral PAL OL teed, , yes Let us consider, on the other hand, the fractions which we obtain by first retaining only the first partial quotient, second by retaining only the first and second, and so on; and let us denote the fractions thus obtained, when reduced (without simplifica- tion, as under) so that their numerators and denominators are integral numbers, by p,/q,, Pe/%2) -.. Then we have . ab al; = — aes (a), 1 Or Aydy + I a, +— =? etn oe me: (B), , As Jo ai 1 Lays + +s _ Ps - (y) ee es Gag tol oo} : i] 1 a, + : ee &e. ala @ (8), Ug + y+. Un dn and so on, where / Pi= thy n=l (0, Ds = yg + 1. Qo =e (2’), Pz = MNAghs + @, +43, Y3 = Ags + fi (y’), so on. XXXII RECURRENCE-FORMULA FOR CO VVERGENTS 405 The fractions p,/q,, p./q@2, . . . are called the first, second, . . . convergents to the continued fraction. — Cor. If the continued fraction ternvinates, the last convergent is, by its definition, the.continued fraction itself. § 7.] It will be seen, from the expressions for p,, p,, p, and Gy Gs Js in § 6 (a’), (B’), (y’), that we have Ds = G3 P2 + Py (1) ; Ys =%Jo + Hh (2). This suggests the following general formule for calculating the numerator and denominator of any convergent when the numerators and denominators of the two preceding convergents are known, namely, Pn = Pn-1 + Pn-» (3) ; Gn = 4m In-1 + Yn-2 (4). Let us suppose that this formula is true for the nth con- vergent. We observe, from the definitions (a), (8), . . ., (8) of § 6, that the n+ 1th convergent, p,+,/¢n4:, 1s derived from the mth if we replace a, by a,+1/@n4,. Hence, since Y__1, Qn-1 Pn-2) In-2 Ao not contain a,, and since, by hypothesis, Pn _UMPn-1 + Pn-2 See as 5) Gn Qn-1 Ft Qn-2 it follows that Pnti _ (an 42 Lie Das + Pn-2 ee ) Gn+i1 (Gn sie 1 /@n+1) An-1 rg In-2 or, after reduction, Pn+i He On+i(AnPn ~1 ona) + Pn-1 In+1 a Dn CAG “H i) a In -1 Ag Anti Pn t+ Pn-1 * dn+i In + Yn-1 by (3) and (4). Hence it is sufficient if we take Pati = %4i Pn + Pn-1 5 vs Qn+1 = Unt1 In + YIn-1- In other words, if the rule hold for the nth convergent, it holds for the n+1th. Now, by (1) and (2), it holds for the third ; hence, by what has just been proved, it holds for the fourth ; hence for the fifth; and so on. That is to say, the rule is general. 406 PRO} ERTIES OF CONVERGENTS CHAP. Cor. 1. Since a, isa positive integral number, it follows from (3) and (4) that the numerators of the successive convergents form an increasing series of integral numbers, and that the same is true of the denominators. Cor. 2. From (3) and (4) it follows that 1 LE (5) ; Pn-1 Uy —1 + Apn-2 + ay : and Bae ae: cr ae i (6). Qn -1 An—-1 + An—-2 + Qe For, dividing (3) by Pn-1, and writing successively »— 1, n— 2, +. 3 in place of n, we have Pal Dre = Ay + Delage ‘ . . 1 | i Psecerpaee Ps/Do = as +P, /De 3 Lal = eM From these equations, by successive substitution, we derive (5); y and (6) may be proved in like manner. . an 7 Example 1. The continued fraction which represents the ratio of the circumference of ; : . a I il 1 Lee | circl t ; *s cogtee — ee, | a circle to ne diameter 18 8 7s re oe aes , quired to calculate the successive convergents. It is re- bo i Qn» and since Qn+2/4n+2 = (n4+o9n41 a dn) | One = Yn4i + In| Ante < Inti + In (Qn4. being + 1), it follows that the upper and lower limits of the error committed by taking the nth convergent instead of the whole continued fraction may be taken to be 1 ihe and 1/¢n(qn + Yn-+1)- These, of course, are not so close as those given above, but they are simpler, and in many | cases they will be found sufficient. Cor. 2. In order to obtain a good approximation to a continued fraction, it is advisable to take that convergent whose corresponding partial quotient immediately precedes a very much larger partial quotient. For, if the next quotient be large, there is a sudden increase in Goan so that 1/¢ndn+1 18 @ very small fraction. The same thing appears from the consideration that, in taking pp/dn instead of the whole fraction, we take a, instead of 412 CONDITION THAT p,/q,; BE A CONVERGENT TO 2 CHAP. 1 1 Ay + ———.. . ., that is, we neglect the part ———. . . of the Anti + Ani + complete quotient. Now, if a+, be very large, this neglected part will of course be very small. Cor. 3. The odd convergents form an increasing series of rational fractions continually approaching to the value of the whole continued fraction ; and the even convergents form a decreasing series having the same property.* Cor. 4. Lf pn/dn-% <1/dn(Qn + n-1) Where dn. ts the denomi- nator of the penultimate convergent to pn/qn when converted into a simple continued fraction having an even number of quotients, then n/In is one of the convergents to the simple continued fraction which represents a,,; and the like holds of 2 — pp/dn<1/dn(Qn + Qn-1) where dn-, 8 the denominator of the penultimate convergent to Pn/dn when converted into a simple continued fraction having an odd number of quctients. Let 1, a, . . -, dm, be the n partial quotients of p,/¢, when converted into a simple continued fraction having an even num- ber_of quotients, and let p,_,/¢n-, be the penultimate convergent. Then Pn9n-1 — Pn-19n = 1. Let z,, be determined by the equation 1 ‘he sSyl dy + nf bn + Uti >» Ny = A, + Then we have t= (Gren + Pn- WACO rie + Qn- ae whence Un+1 = (2, Ona — Pn-1)|(Pn — 2 4n)s * The value of every simple continued fraction lies, of course, between 0 and o ; and we may, in fact, regard these as the first and second converg- | ents respectively to every continued fraction. If we write 0=%, and © =4,. and denote these rae and a so that we understand p_; to be 0, p» to be =I 0 1, g-1 to be 1, and q to be 0, then p_; and will be found to fall into the series 1, P2, ps3, &c., and g_; and q into the series 91, q2, g3, &c. It will be found, for example, that py=a pot+p-1, M=%9+9-1, Po9-1—-P-190=(- 1)? =1, and so on. xxx CONDITION THAT p,/7, BE A CONVERGENT TO z, 413 or, if we put = pp/qn — %, nts = {(PnGn-1 — Pn-1%n) [In In-1€}/ On = (1/¢n — Qn-1§)/4n6- Hence the necessary and sufficient condition that z,,,> 1 is that 1/dn met In-1 7 Qn; that is, E<1/Gn(Qn + Qn-1)s which is fulfilled by the condition in the first of our two theorems. } 9 Let now 0,, b,, . . ., dy, be the first m partial quotients in the simple continued fraction that represents 7. Then we have a, =b, + pile de ok z ae bs oa pag Dn re Yn+1 where Y%,4, > 1. Hence i, Let 1 1 1 , + Ene Say a ae ero y+ Un + Xpy4i bs os Dn + Ynt+i Therefore, by § 3, Cor., we must have a= b,, Uy = ba, Bovey ae gy = es Un+1 = Yn+i 1 ae Hence a, + ..+—, that is, Pn is the nth convergent to 2 Un In v. The second theorem is proved in precisely the same way, (©. ~ ath: Since gn-1 Pen-1/Gen-1- Hence, since both are positive, each of the two must approach a certain finite limit. Also the two limits must be the same; for by § 8, Cor. 2, Pon/qon — Pon-1/Yen-1 = 1/denJen-1, and by the recurrence formula for ¢, it follows that Gen aNd gon, Increase without limit with n; therefore pon/qon — Pon-1/Jon-, May be made as small as we please by sufficiently increasing 1. It appears, therefore, that every simple continued fraction has a definite finite value. Example. To obtain a good commensurable approximation to the ratio of the circum- XXXII EXERCISES XXIX 415 ference of a circle to the diameter. Referring to Example 1, § 7, we have the following approximations in defect :— 3 333 103993 1’ 106’ “33102 ’ and the following in excess :— 22 355 104348 7 abhi Sg 882152) 5 ois Two of these,* namely, 22/7 and 355/113, are distinguished beyond the others by preceding large partial quotients, namely, 15 and 292. The last of these is exceedingly accurate, for in this case 1/qndn41 =1/113 x 33102 = -0000002673, and an+42/QnGn42= 1/118 x 33215 = 0000002665. The error therefore lies between ‘000000266 and *000000267 ; that is to say, 355/113 is accurate to the 6th decimal place. In point of fact, we have m= 3°'14159265358 . . . 355/113 =3 14159292035 . Difference= ‘00000026677 .. . EXERCISES XXIX. fod (1.) Calculate the various convergents to aN and estimate the errors / 111 committed by taking the first, second, third, &c., instead of the fraction. Le aed (2.) Find a convergent to the infinite continued fraction Rowers which shall represent its value within a millionth. (3.) Find a commensurable approximation to ,/(17) which shall be accurate within 1/100000, and such that no nearer fraction can be found not having a greater denominator. (4.) The sidereal period of Venus is 224°7 days, that of the earth 365°25 days ; calculate the various cycles in which transits of Venus may be expected to occur. Calculate the number of degrees in each case by which Venus is displaced from the node, when the earth is there, at the end of the first cycle after a former central transit. (5.) Work out the same problem for Mercury, whose piderenl period is 87°97 days. (6.) According to make Northampton table of mortality, out of 3635 persons who reach the age of 40, 3559 reach the age of 41. Show that this is ex- pressed very accurately by saying that 47 out of 48 survive. Se ee A * The first of them, 22/7, was given by Archimedes (212 B.c.) The second, 359/113, was given by Adrian Metius (published by his son, 1640 A.D.): it is in great favour, not only on account of its accuracy, but because ‘it can be easily remembered as consisting of the first three odd numbers each repeated twice in a certain succession. 416 ‘EXERCISES XXIX CHAP, (7.) Find a good rational approximation to 4/(19) which shall differ from it by less than 1/100000 ; and compare this with the rational approximation obtained by expressing «/(19) as a decimal fraction correct to the 6th place. (8.) If a be any incommensurable quantity whatever, show that two integers, m and n, can always be found, so that 0 qq'( pb — qa) ; and b> (pb —qa)q’. Now p/g — a/b is positive, hence ph — qa 1s a positive integer. It follows, therefore, that ) >q’. Similarly it follows from (2) that b> g. Hence no fraction can lie between pl¢ and p'/q unless its denominator is greater than both g and q. In other words, if PY — p'g=1, no commensurable number can lie between p/g and. p'/q which is not more complex than either of them. § 13.] The nth convergent to a continued fraction is a nearer approximation to the value of the complete fraction than any fraction whose denominator is not greater than that of the convergent. For any fraction a/b which is nearer in value to the continued fraction than p,/q, must, a fortiori, be nearer than Pn-1/Qn-1. Hence, SINCE Ppn/qn and Pn_, /q.-1 include the value of the continued fraction between them, it follows that a/b must lie between these two fractions. Now we have, by § 8, either pagn_, — Dn aig ale OF Pn-19n — Pn9n-1=1. Hence, by § 12, b must be greater than Qn, Which proves our proposition. Example. Consider the continued fraction aie S414 95 F 84.15 64 143 779 Ld AMAT" 788 cee any one of these, say 64/17, the statement is, that no fraction whose denom- inator does not exceed 17 can be nearer in value to a, than 64/17. The successive convergents. are If we take § 14.] The result of last section is a step towards the solution of the general problem of § 12; but something more is required. Consider, for example, the successive convergents py-.2/¢n—25 Pn-1/In-15 Pn] Mm to 2, and let n be odd, say. Then Pn-2 Pr 2%, Pn-1 2 In ie Yn In = are in incréasing order of magnitude. We know, by last XXXII INTERMEDIATE CONVERGENTS 419 section, that no fraction whose denominator is less than Qn —1 Can lie in the interval Pn-2/In-2) Pn-1/Qn-1, and also that no fraction whose denominator is less than Qn can lie in the interval Pn/In; Pn-1/In-1; but we have no assurance that a fraction whose denominator is less than Qn May not lie in the interval Dice Onna, Pn/ Ins for PnQn-2 —Pn-29n = (In, Where Gm May Isto, ale This lacuna is filled by the following proposition :— 1°. The series of fractions : Pn-2 Pn-2+ Pn-1 Pn-2t 20-3 ’ p) b) Qn-2 Qn-2 t+ In-1 In-3 20 He 3 Pn-2 + Ay — 1pp-y eat On ne ( =") (1) Qn-2 + An — Lda’ Qn-2 t An In-1 Qn ; Jorm (according as n is odd or even) an increasing or a decreasing serves, 3 | 2°. Lach of them is at its lowest terms; and each consecutive pair, say P/Q, P’/Q’, satisfies the condition PQ’ - P’Q= +1 ; so that - no commensurable quantity less complex than the more complex of the two can be inserted between them. The first and last of these fractions (formerly called Con-. vergents merely) we now call, for the sake of distinction, Principal Convergents ; the others are called Intermediate Convergents to the continued fraction. To prove the above properties, let us con- sider any two consecutive fractions of the series (1), say P/Q, P’/Q’; then PP! Pn-st?Pn-1 Dns 7 + 1Dn-1 Q) (’ Gn-2 + 1n-1 In-s+f +1 On =; Gy ereia Oecral sore say abe: OF a) — 1), as _ (Ones Qn-2 — Pn-2 Ini) (Qn-2+ 19n—1) (Qn-2+ 7 +1Gn-1)' =n | (Ong ~ meres} Cae as r+1 ise ae sd ee = QQ’ if nm be odd, | Saat! if n be even. Q’ wep} (2). 420 COMPLETE SERIES OF CONVERGENTS CHAP. Hence PQ’ - P’'Q= — 1 if m be odd, 1 = +1 if beeven. J (2) and (3) are sufficient to establish 1° and 2°. 3°. Since P/Q — Pn-1/dn-1= + 1/Gn-i(Gn—2 + 7 9n-1), and Since 2, obviously lies between P/Q and py_,/gn_,, it follows that the intermediate convergent P/Q differs from the continued fraction by less than 1/qn-,Q, a fortiori by less than 1/qn-/. § 15.] Lf we take all the principal convergents of odd order with ther intermediates wherever the partial quotients differ from unity, and form the series (3); Sr ey ha My Be Pe ee ye hh qs In-2 In and likewise all the principal convergents of even order with their wutermediates, and form the series 1 Pe De ee Dian re Vi ee Oy ha ee B), 0 Qo U4 Un-3 dn-1 ( ) then (A) is @ series of commensurable quantities, increasing in com-— pleaity and mereasing im magnitude, which continually approach the continued fraction ; and (B) is a series of commensurable quantities, imereasing im complexity and decreasing in magnitude, which con- tinually approach the same; and it is impossible between any con- secutive pair of either series to msert a commensurable quantity which shall be less complex than the more complex of the two. If the continued fraction be non-terminating, each of the two series (A) and (B) is non-terminating. If the continued fraction terminates, one of the series will terminate, since the last member of one of them will be the last convergent to x; that is to say, ~, itself. The other series may, however, be prolonged as far as we please ; for, if Dna Ganoe Pnl qn be the last two convergents, the series of fractions Pn-1 Pn-1t Pn Pn-1 t+ 2Dn ) ’ ’ Gn-1 Qn-1 + Un Qn-1 + 29n forms either a continually increasing or a continually decreasing series, XXXII CLOSEST RATIONAL APPROXIMATION 431 in which no principal convergent occurs, but whose terms approach more and more nearly the value Py/dn, that is, a,.* § 16.] We are now in a position to solve the general problem of § 12.t Suppose, for example, that we are required to find the fraction, whose denominator does not exceed D, which shall approximate most closely by defect to the quantity 2, What we have to do is to convert x, into a simple continued fraction, form the series (A) of last section, and select that fraction from it whose denominator is either D, or, failing that, less than but nearest to D, say P/Q. For, if there were any fraction nearer to a, than P/Q, it would lie to the right of P/Q in the series; that is to say, would fall between P/Q and the next fraction P’/Q’ of the series, or between two fractions still more complex. Hence the denom- inator of the supposed fraction would be greater than Q’, and hence greater than D. Similarly, the fraction which most nearly approximates to a, by excess, and whose denominator does not eaceed D, is obtained by taking that fraction in series (B) of last section whose denominator most nearly equals without exceeding D. N.B.—If the denominator in the (A) series which most * This may also be seen from the fact that the continued fraction : nes Ze may also be written rH cae : bey that is to say dig+ Gy, a + An+ © we may consider the last quotient to be «, and the last convergent (Pn-1 = © Pn)/(Yn—1 + Yn) ° + The first general solution of this problem was given by Wallis (see his Algebra (1685), chap. x.); Huyghens also was led to discuss it when designing the toothed wheels of his Planetarium (see his Descriptio Automati Planetarii, 1682). One of the earlier appearances of continued fractions in mathematics was the value of 4/7 given by Lord Brouncker (about 1655). While discussing Brouncker’s Fraction in his Arithmetica Infinitorwm (1656), Wallis gives a good many of the elementary properties of the convergents to a general peeiinuad fraction, including the rule for their formation. Sanriderdore Euler, and Lambert all helped in developing the theory of the subject. See two interesting bibliographical papers by Giinther and Favaro, Bulletino di Bibliographia e di Storia delle Scienze Mathematiche e Fisiche, t. vii. In this chapter we have mainly followed Lagrange, who gave the first full exposition of it in his additions to the French edition of Euler’s Algebra (1795). We may here direct the attention of the reader to a series of comprehensive articles on continued fractions by Stern, Credle’s Jour., x., xi, XVill. 499 EXAMPLES CHAP, nearly equals without exceeding D be the denominator of an intermediate convergent, the denominator in the (B) series which most nearly equals without exceeding D will be the denominator of a principal convergent. Example 1. To find the fraction, whose denominator does not exceed 60, which at 779 . approximates most closely to 507" 779 Ie batoe Ea eS | We have 7 = Sal ys 3 207 tla B4 44 24 5 Or 38 1b las. The odd convergents are i ee ake awe Lie 4 04e ae the even. convergents 0 oT? 17? 907 The two series are Ue ree aaseey 11 ib 754 Tae 922 1701 2480 Ie te ap? a aT? ge ig Magee enene 1 4 19 84 49 64 207 350 493 636 779 py 0’ ik Te 9’ U 13’ a 55? 93’ 1317 169’ 207 “ Hence, of the fractions whose denominators do not exceed 60, 143/38 is the closest by defect and 207/55 the closest by excess to 779/207. Of these two it happens that 143/38 is the closer, although its denomin- ator is less than that of 207/55 ; for we have 143/38=3°76315 . . -» 207/55 =3°76363 . . ., and 779/207=3-76328 ... Fora rule enabling us in most cases to save calculation in deciding between the closeness of the (A) and (B) approximations, see Exercises XXX., 10. Example 2. Adopting La Caille’s determination of the length of the tropical year as 3654 5h 48’ 49”, so that it exceeds the civil year by 52 48’ 49”, we are required to find the various ways of rectifying the calendar by intercalating an integral number of days at equal intervals of an integral number of years. (Lagrange. ) ; ; 9294 ‘ The intercalation must be at the rate of pa per year; that is to say, at (A), the rate of 20929 days in 86400 years. If, therefore, we were to intercalate 20929 days at the end of every 864 centuries we should exactly represent La Caille’s determination. Such a method of rectifying the calendar is open to very obvious objections, and consequently we seek to obtain an approximate rectification by intercalating a smaller number of days at shorter intervals. If we turn 86400/20929 into a continued fraction and form the (A) and (B) series of convergents, we have (omitting the earlier terms) MRE a fg Se (A), 1 8 39 694 2043 3392 y¥ 21 25 29 62 95 198 989 4500 uaa 2.3" 47 8’ 6’ 7? 16? 93) Si? 40 aoe 772 = 983)—«1094 187), 256" pepe re] On hd OO Sea, XXXII EXAMPLES 423 Hence, if we take approximations which err by excess, we may with increas- ing accuracy intercalate 1 day every 4 years, 8 every 33, 39 every 161, and so on;* and be assured that each of these gives us the greatest accuracy obtainable by taking an integral number of days less than that indicated in the next of the series. The (B) series may be used in a similar manner.t Example 3. An eclipse of the sun will happen if at the time of new moon the earth be within about 13° of the line of nodes of the orbits of earth and moon. The period between two new moons is on the average 29°5306 days, and the mean synodic period of the earth and moon is 346°6196 days. It is required to calculate the simpler periods for the recurring of eclipses. Suppose that after any the same time from a new moon the moon and earth have made respectively the multiples z and y of a revolution, then x x 29°5306 = Tage licl ee opie y x 346°6196. Hence y/a=295306/3466196 = pray pa ewe ET RRS The successive convergents to this fraction are 1/11, 1/12, 3/35, 4/47, 19/223, 61/716. Suppose we take the convergent 4/47, the error incurred thereby will be <1/47 x 223 in excess, and we may write on the most unfavourable supposition y 4 1 a% 47° 47x 223° Hence, if e=47, y=4—1/223. But 360°/223=1°°61. Hence 47 lunations after total eclipse new moon will happen when the earth is less than 1°°61 from the line of nodes, 47 lunations after that again when the earth is less than 3°'2 from the line of nodes, and soon. Hence, since 47 lunations =1388 days, eclipses will recur after a total eclipse for a considerable number of periods of 1388 days. If we take the next: convergent we find for the period of recurrence 223 lunations, which amounts to 18 years and 10 or 11 days, according as five or four leap years occur in the interval. The displacement from the node in this case is certainly less than 360°/716, that is, less than half a degree, so that * The fraction 4/I corresponds to the Julian intercalation, introduced by Julius Cesar (45 B.c.). 33/8 gives the so-called Persian intercalation, said to be due to the mathematician Omar Alkhayami (1079 A.p.), The method in present use among most European nations is the Gregorian, which corrects the Julian intercalation, by omitting 3 days every 4 centuries. This corresponds to the fraction 400/97, which is not one in the above series ; in fact, 70 days every 289 years would be more accurate. The Gregorian method has, how- ever, the advantage of proceeding by multiples of a century. The Greeks and Russians still use the Julian intercalation, and in consequence there isa differ- ence of 12 days between their calendar and ours. See art. “ Calendar,” Encyclopedia Britannica, 9th ed. + See Lagrange’s additions to the French aiton of Euler’s Algebra (Paris, chai t, ile, Dy 312. 494. EXERCISES xxx CHAP, XXXII this is a far more certain cycle than the last ; in fact, it is the famous “‘ saros” of antiquity which was known to the Chaldean astronomers, Still more accurate results may of course be obtained by taking higher convergents, . ' EXERCISES XXX, , P Diol ela (1.) Find the first eight convergents to Leoer 5a Pn ih ee and find the fraction nearest to it whose denominator does not exceed 600. (2.) Work out the problem of Exercise XXIX., 4, using intermediate as well as principal convergents, (3.) Work out all the convergents to 2m whose denominators do not exceed 1000. (4.) Solve the same problem for the base of the Napierian system of logarithms e=2°71828183 ... , (5.) Two scales, such that 1873 parts of the one is equal to 1860 parts of the other, are superposed so that the zeros coincide: find where approximate coincidences occur and estimate the divergence in each case, (6.) Two pendulums are hung up, one in front of the other. The first beats seconds exactly ; the second loses 5 min. 37 sec. in 24 hours. They pass the vertical together at 12 o’clock noon. Find the times during the day at which the first passes the vertical, and the second does so approximately at the same time. (7.) Along the side AB and diagonal AC of a square field round posts are erected at equal intervals, the interval in the two cases being the same. A person looking from a distance in a direction perpendicular to AB sees in the perspective of the two rows of posts places where the posts seem very close together (‘‘ ghosts”), and places where the intervals are clear owing to approximate coincidences. Calculate the distances of the centres of the ghosts from A, and show that they grow broader and sparser as they recede from A. , (8) Show that between two given fractions pl¢ and p'/q', such that pq —p'q=1, an infinite number of fractions in order of magnitude can be inserted such that between any consecutive two of the series no fraction can be found less complex than either of them. (9.) In the series of fractions whose denominators are 1, 2,3,..., » there is at least one whose denominator is v, say, such that it differs from a given irrational quantity a by less than 1 /nv. (For a proof of this theorem, due to Dirichlet, not depending on the theory of continued fractions, see Serret, Alg. Aup., Ame éd:-t. 1., p. 27.) (10.) If the nearest rational approximation in excess or defect (see § 16) be an intermediate convergent P/Q, where Q=)gn-1+Qn-2, show that the approximation in defect or excess will be nearer unless Q > 39n + Qn—1/ 2244. (11.) If zero partial quotients be (contrary to the usual understanding) admitted, show that every continued fraction may be written in the form 1 ; shatas rap esata Where @, dy, @3, . . . are each either 0 or 1. Show the bearing of this on the theory of the so-called intermediate convergents, ~~ CHARTER XX XIIT On Recurring Continued Fractions. EVERY SIMPLE QUADRATIC SURD NUMBER IS EQUAL TO A RECURRING CONTINUED FRACTION. § 1.] We have already seen in two particular instances (chap. xxxil, § 5) that a simple surd number can be expressed as a recurring continued fraction. We proceed in the present chapter to discuss this matter more closely.* Let us consider the simple surd number (P, + “R)/Q,. We suppose that its value is positive ; and we arrange, as we always may, that P,, Q,, R shall be integers, and that VR shall have the positive sign as indicated. R will of course always be positive ; but P, and Q, may be either positive or negative. It is further supposed that R — P,’ is exactly divisible by Q,. This is allowable, for, if R — P,’ were, say, prime to Q,, then we might write (P, + VR)/Q = (P,Q, + VQ/’R)/Q. = (Py + VR)/Qi, where R’ — P,”{ = Q,'(R — P,’) = (R- P,’)Q,} is exactly divisible by Q,’. For example, to put : ( 2 - AY 4) into the standard form contemplated, we must write 1(p_ Ye ee acl i I SN 4 eee A Nee Otis tes Sam ee 89 so that in this case P)= —16, Qi= -32, R=96; R- Py?=96 — 256= — 160, which is exactly divisible by Q;= — 32. * The following theory is due in the main to Lagrange. For the details of its exposition we are considerably indebted to Serret, Alg. Sup., chap. ii. al 426 RECURRENCE-FORMULA FoR P,, AND Q, CHAP. § 2.] If we adopt the process and notation of chap. Xxxil., §§ 3 and 5, the calculation of: the partial and complete quotients of the continued fraction which represents (P, + “R)/Q, proceeds as follows :— Pow RB 1 l= Q = Ay, + — 2 1 2 Paw Fe , vs = aC) Ol =A, + ihe q ae P,+NR on l te Qn is Un+1 where it will be remembered that a,, a,, ... are the greatest integers which do not exceed 2,, %,.. . . Yespectively ; and ®, %, ... are each positive, and not less than unity. It should be noticed, however, that since we keep the radical /R unaltered in our arrangement of the complete quotients, it by no means follows that P,, Q,, P;, Q,, &., are integers, much less that they are positive integers. The connection between any two consecutive pairs, say P,, Qn and Pps, Qn4:, follows from the equation Pte 1 See fos = (Pee 15 VR)/Qn4s (2), or Ue ce Gna) To A 6 ees ty R} 7 es, ri OnQn ng Paar VR = 0 (3). It follows from (3), by chap. xi, § 8, that (EA —aAp Qi) Ens a Qn Qn+: += 0, Noe, ini OnQn ay Pia = 0 > begat = OnQn — lend (4), Ie ay! re Qn Qn =h (5). If we write n—1 for n in (5), we have Ree - Qn—1Qn =Rh (6). whence XXXII EXPRESSIONS FOR P,, AND Q,, 4.27 From (5), by means of (4) and (6), we have Qn Qn+1 =R- (ay Qn = epee oe ag hp One 1 Qn wat (ay Qn = tei) Wace (A ee yar yaa 3 — On, Able aa =i au On( Pn ae eal (7). The formule (4) and (7) give a convenient means of calculat- ing P., P;, Q;, Ps, Q,, &c., and hence the successive complete so that quotients 7, %, ... Q, is given by the equation Bs: + Q. Q, =R namely, Q): = git Rl Es Q: 2 “SS ne 1— A "Qh From this last equation it follows, since by hypothesis (R — P,’)/Q, is an integer, that Q, is an integer. Hence, since P,, Q, are integers, it follows, by (4) and (7), that P,, P,,. . ., Pn; Q;, - - -, Q, are also all integers. § 3.] We shall now investigate formule connecting P, and Q, with the numerators and denominators of the convergents to the continued fraction which represents (P, + / R)/Q,. We have (chap. xxxii., § 9) ley VR — Pn-1*n + Pn-2 Q, ix Qn-1%n + In-2 whe _ Pn-1Pn + Dn-2Qn+ Pn NR Oneal ay doa en 1 In-1 VR (A), Hence (P, + /R) Meta + On-Qn + On} /R) = Q,(Pn-1Pn +Pn-2Qn + Pn-1 VR) Ob), From (1) we derive | Onciin zy Gaon oe as eal se In-1 (2) 5 R-P/ Pn- dee + Pn- 2Qn =P > Pn whee Q, ~ n- 1 (3). 428 Paan Ry Qala) Ree te CHAP, From (2) and (3) we obtain, since Pn1Unes pene — iif 2a Ga, (i ie Wee 1 Gis 19n-2 +t Pn-2 In- 1) R-P/ af Sia In-19n-2 — Qi Pn-1Pn-2 (4); R- Ba 2 2 ( 1)? ?Qn = os 2Pn-19n-12 = Q Qn-1 + Q,Pn-1 (5). The formule (4) and (5) give us the required expressions, and furnish another proof that P,, P,,. . ., Pry.Q5-0e eee are all integral. § 4.] If in equation (2) of last paragraph we replace P, by its ain Q,(Pn-1%n + Pn-s) | (Qn-12n + Gna) — VR, derived from equa- tion (A), we have ( = LE ewe /D In- itn + Qn- a) z fr) Also, since %,=(Py+ awiOs we have Ase oe AS aa ho VR (2). From equations (1) and (2) we derive, by direct calculation, the following four :— Cn lbp hh ot nos a = Lin Yn-2 rT o| (Gr-itn—dn-a) (Ze x tnt) VR + [= 1)¥-1Q); } (3) ; CRS in i YVn-2 Qn ae 1 ¢: (Yn- Ln + Inca) rnin ae Gnea)2 /R < ( — 1)#=16@) 3} (4) ; / Ria Pees i 2a Ate es n=) VR -(- ya,} (5) ; (Cet, + In- neg 2 VR SS Qn ay 1 oe oe (qe In + ey {(y- Mga 7 Gna) Onan £9 Qn-2)2 /R-(-1)"1Q,} (6). The coefficients of WR and 2R in these four formule are positive, and increase without limit when n is increased without limit. Hence, since Q, is a fixed quantity, it follows that for XXXII CYCLE OF (P,+ “R)/Q, 4.29 some value of n, say n=v, and for all greater values, P,, Ge VJR-P,, 2VR-Q, will all be positive. Jn other words, on and after a certain value of n, n=v say, Py and Qn will be positwe ; and P,< VR, and Qn <2 JR. | Cor. 1. Since P, and Q, are integers, it follows that after n=v Py cannot have more than VR different values, and Qn cannot have more than 2VR different values; so that @ tha + J R)/ Q,, cannot have more than JR x 207 = 2R different values. In other words, after the vth complete quotient, the complete quotients must recur within 2h steps at most. Hence the continued fraction which represents (P, + “R)/Q, must recur in a cycle of 2K steps at most. Since ever after n =v P,, and Q, remain positive, it is clear that in the cycle of complete quotients there cannot occur any one m which PR, and Q, are not both positive. It should be noticed that it is merely the fact that P,, and Q, ultimately become positive that causes the recurrence. If we knew that on and after n =v P, remains positive, then it would follow, from § 2 (4), that Q, and all following remain positive ; and it would follow, from § 2 (5), that P,,, and all following are each v, P, and Q, are both positive, and P,, < VR, and Q,>1, it follows that, if n>v, dn<2VR. It follows, therefore, that none of the partial quotients in the cycle “.. can exceed the greatest integer im 2 VR. Cor. 3. By means of (3) and (4), we can show that ultimately Pa + Qn> VR (7). Cor. 4. From § 2 (5), we can also show that ultimately Pa + Qn-1> JR (8). 430 PURE RECURRING C.F. CHAP. Cor. 5. Since VR> Pin, it follows from Cor. 3 and Cor. 4 that ultimately Pn = Py < Qn: > Qn = (9). EVERY RECURRING CONTINUED FRACTION IS EQUAL TO A SIMPLE QUADRATIC SURD NUMBER. § 5.] We shall next prove the converse of the main _pro- position which has just been established, namely, we shall show that every recurring continued fraction, pure or mixed, is equal to a simple quadratic surd number. First, let us consider the pure recurring continued fraction Sete ce (1), * Og a Oy ae Let the two last convergents to ] a, + ease ny Pos Ag + Oy be p'/q and p/q. From (1) we have 1 a %=a,+——.., —, Og + Op + Uy _ pat p 2G whence qx + (q' —p)a,—p' =0 (2). The quadratic equation (2) has two real roots ; but one of them is negative and therefore not in question, hence the other must be the value of «, required. We have, therefore, Pa~ + M(p-q') + 4p 3) t= ( ); ‘ 29 [igs Way ne ey a which proves the proposition in the present case. XXXIII MIXED RECURRING C.F. 431 It should be noticed that, since a,+0, p/g>1; so that p>q>q. Hence p—q’ cannot vanish, and a pure recurring fraction can never represent a surd number of the form “N /M. Next, consider the general case of a mixed recurring con- tinued fraction. Let oe 1 @,=A,+——...- artes (4). A, + Ap + A, + Ag + Og + % % Also let 1 Be ark Sioa ge es (5). * Then, by (3), ir. L+W/N Y, — M . From (4) we have 1 Eira %=A+——... —, Uy + Ap t+ Yy whence, if P’/Q’ and P/Q be the two last convergents to wl 1 ere, Ay + Uy | owe Py, + P’ \ a Qy, = Oe i _PL+PM+PNVN (6) QL+QM+QVN Hence, rationalising the denominator, we deduce ee EEN 5 W Example 1. Ve Ve Evaluate Ce eo Tee pres laa The two last convergents to 1 +5 7 are 3/2 and 4/3 ; hence _ 4ar+ 3 4 ap 3X1 +2 ‘ We therefore have : 8212 — 2a, -3=0, the positive root of which is ees: 10 hemes ° 3 432 C.F. FoR ¥(C/D) CHAP. Example 2. Evaluate Uae Ty Te a4 Te ey fe * * The two last convergents to 845 are 3/1 and 13/4; and, by Example 1 above, We have, therefore, mS) 4+ (14N/ 10/3. _18(1+V10)/3 +3 4(1+/10)/3 +1’ 22 +13/10 = 7+4/10 _ 866-310 Tee be i X _ 122-10. i. 37 ON THE CONTINUED FRACTION WHICH REPRESENTS ¥(C/D). § 6.] The square root of every positive rational number, say /(C/D), where C and D are positive integers, and C/D is not the square of a commensurable number, can be put into the form N/M, where N=CD and M=D. Since N/M=C is an integer, we know from what precedes that / N/M can be developed, and that in one way only, as a continued fraction of the form 1 PAL: 1 a re . . . a Li a re i ae . . . Fe . . . th: 2 o fe il 2 s * * X, = A, + We have, in fact, merely to put P,=0, R= N, Q, = Minos previous formule. ‘< We suppose that N/M is greater than unity, so that a, + 0. If “N/M were less than unity, then we have only to consider M/ VN = VMPN/N, which is greater than unity. 1 ; ing must consist of one term at 2 fa The acyclic part a, + 7 XXXIII ACYCLIC PART OF /N/M 433 least, for we saw, in § 5, that a pure recurring continued fraction cannot represent a surd number of the form VN/M. Let us suppose that there are at least two terms in this part of the fraction ; and let P’/Q’, P/Q: be the two last convergents to 1 a, + aa ..—3; and p’/q', p/q the two last convergents to aii . Cy il 1 : a, + tees. : pele ree en, is A, + eri, Oy Oe Og Y, =O, + ’ Dy th Reig Re se we have . 1 Us il 1, = dy + -_—- = ly + Ap + Yy 1 rebel eon = tt, + —, Ag + Op Oy Og st Agr Y, Hence P Pp’ ’ , ps % ay = PY +p | (2y; Ly i , ! Qni+Q amt Eliminating y, from the equations (2), we have (Qq7’ — Q’q)a," — (Qp' — Q’p + Pq’ — P’g)a, + (Pp'- P’p)=0 (3). Now, if z,= “N/M, we must have Max,’ -N=0 (4). In order that the equations (3) and (4) may agree, we must have Qp — Qp + Pq’ - P'g=0 (5); and Pp’ — P'p N Q7-Qo = M (8). It is easy to show that equation (6) cannot be satisfied. We have, in fact, Pp’ — P’p - a P/P! — p/p’ (7) Q7- Qe V7 Q/Q - 9/4 But, by. chap. xxxii., § 7, ; | em 1 1 1 1 Bap 6 ail ° PM S/T ee van! = Uy — Og == where f is a proper fraction. VOL, II 2F 434. CYCLE OF QUOTIENTS FOR N/M OHAP, Similarly Qed 1 1 1 1 Q’ phat Nig eee eck ty Og og ee Selanne a? = Up — Os +f’, where /’ is a proper fraction. Now a, —as cannot be zero, for, if that were so, we should have a, = as, that is to say, the cycle of partial quotients would begin one place sooner, and would be ag, a,, a2, . . ., ag-;, and not Gy, Ug,+ + +) Gs, AS Was supposed. It follows then that a,—a, is a positive or negative integral number. Hence the signs of P/P’—p/p’ and Q/Q’-¢q/q are either both positive or both negative, and the sign of the quotient of the two is positive. Hence the left-hand side of (6) is positive, and the right-hand side negative. There cannot, therefore, be more than one partial quotient in the acyclic part of (1). Let us, then, write =a +— : Farad age Pe (8), @, + Ge + Qg + a, + * 4* - ome 1 1 1 | - G+ det. agt 1/(a, - a) Hence _ pia, - a) +p Of(@ — a)+@q’ which gives | qx,’ — (p' + ga — qx, — (p — ap’) = 0 (9): From (9) we obtain yeah yeast Te 'a oa ee ee Pee en Chee PRA eS PS) (10). 2¢q’ 2q In order that (10) may agree with z,= V N/M, we must have pt+da-q=0 (11); and gq N/M’ = (p ink ap’ ya’ (1 2). Cor. 1. By equation (11) we have P'/q +a=9/q. XXXIII CYCLE OF QUOTIENTS FOR N/M 435 Hence, by chap. xxxii., § 7, Cor. 2, ] 1 1 1 ae. = Ag+ ee O, + Og + Og—1 Og, om 2a + It follows, therefore, by chap. xxxii., § 3, that eee Om ge OS Og Gat ty Ao” dy = Obs In other words, the last partial quotient of the cyclical part of the continued fraction which represents N/M is double the unique partial quotient which forms the acyclical part; and the rest of the cycle is reciprocal, that is to say, the partial quotients equidistant from the two extremes are equal. In short, we may write /N Tale 1 —— =4, —...— M a, + dg+ Ag+ a,+ 2 * * pa Tet 1 1 pe Pte) Cor. 2. If we use the value of qa given by (11), we may throw (12) into the form g N/M’ = pa’ — p'(q- 7’); whence gq" N/M — p" = pq’ - pg, aus: (14), the upper sign being taken if p/q be an even convergent, the lower if it be an odd convergent. § 7.] All the results already established for (P,+ /R)/Q, apply to VN/M. For convenience, we modify the notation as follows :— a, =a, © 2 =(P,+ VR)/Q,=(0+ VN)/M; , =a, %=(P,+ VR)/Q,=(L,+ VN)/M,; % =o, «-(P,+ VR)/Q,=(1,+ VN)/M,; dg = Ag-1, m= (Ppt VB)/Q= (Ly. + VN)/M,_.; As4, = 2a, Ug+-2 = A) . . From § 2 (4), we then have Lin = On -_Mye,-— Ly -, Chie and, in particular, when n = 1, L, = aM (1’). 436 CYCLES OF DIVIDENDS AND DIVISORS CHAP. From § 2 (5), we have 1 ate M,-.M, =N (2) ; and, in particular, L, +MM,=N (2’). From § 3 (4) and (5), we have | (— 1)" Ly = (N/M) gn gn-1 - Mn pa-r (3) ; (- )"Mn= Mop,’ - (N/M) qn’ (4). These formule are often useful in particular applications. It will be a good exercise for the student to establish them directly. § 8.] Let us call L, L,, &c., the Rational Dividends and M, M,, M,, &c., the Divisors belonging to the development of V N/M. Then, from the results of § 4, we see that None of the rational dividends can exceed NN ; none of the partial quotients and none of the divisors can eaceed 2/N. : All the rational dividends, and all the divisors, are positive. It is, of course, obvious that the rational dividends and the divisors form cycles collateral with the cycle of the partial and total quotients ; namely, just as we have Og4+1 =, OAgto= Ag, so we have MBPs a Lis lias = Liss (1), and My4, ms M,, Mg1. a M,, (2). We can also show that the cycles of the rational dividends and of the divisors have a reciprocal property like the cycle of the partial quotients ; namely, we have L, =L, M, =M; L,.,=L, M,.,= M,; (3). Ls_. = L,, M;-.= M,; | For, by § 7 (2), Ls41 + M,4,M,=L,°+ MM ; but L,,, = L, and M,,,=M,, hence M,=M (4), XXXIII THE COLLATERAL CYCLES 437 Again, by § 7 (1), 1 =a,M,—-L, WY but L,,,=L,, a,= 2a, M,=M, hence we have YX ~ on Tea 1 = Now, by § 7 (1’), L, = aM, hence p= 20; — Le therefore eels (5). Again, by § 7 (2), L; + M,M,_, =,’ + MM, whence, bearing in mind what we have already proved, we have My-y= Ms (6). Once more, by § 7 (1), L; = as—,Mg-, mi Ls. iD L, = a,M, - L,. Now M,_, = M, and a,_, =a,, hence Lj, 0,=L, —1,-,. But L,=1L,, hence We abs Proceeding step by step, in this way, we establish all the equations (3). ) It appears, then, that we may write the cycles of the rational dividends and of the divisors thus— ibys L, Ibis eka Le is L, 3 M,, M,, VES vain SD Nie M., M,, M. Since M precedes M,, we may make the cycle of the divisors commence one step earlier, and we thus have for partial quotients, rational dividends, and divisors the following cycles :— Gy, Orgy gy ss 2, gy Ogy Oy 2h5 ay. L,, ie Li, idee Ta | li li. L, > L, : M, M,, M,, M,, sons Me Me 1M, That is to say, the cycle of the rational dividends is collateral with the cycle of the partial quotients, and is completely reciprocal ;— the 438 TESTS FOR MIDDLE OF CYCLE CHAP. cycle of the divisors begins one step earlier® (that is, from the very beginning), and is reciprocal after the first term. § 9.] The following theorem forms, in a certain sense, a converse to the propositions just established regarding the cycles of the continued fraction which represents VN /M. If Lm = Lin-ay Mm = M,, Gm = On; then Lin —1 = Linton Mn, i Mn+; Am —-1 = In+1 (Ve We have, by § 7 (2), Pe a Min Mmn-1 im ace + Maas M,, whence, remembering our data, we deduce Mmn-1 = Mn+: (2). Again, by § 7 (1), Lim 3.5 Lim 1 = Can —1 Mig 1, Ln+s ote Lins = Ont My +i, whence, since L,, = L,,, by data, Lin-1 os Ln+s “a (Gina = Cutie a (age ms ti, 4) Vie (3). Tf Lina > Ln+. we may write (3) (lie, oa Lape = Am-1 — An41 (4) > If Lim-; < Lyn4., we may write (lies ie Linea) Nunes = On+1 — Am-1 (5). But, by § 4 (9), the left-hand sides of (4) and (5) (if they differ from 0) are each < 1, while the right-hand sides are each positive integers (if they differ from 0). It follows, then, that each side of equation (3) must vanish, so that Lin-1 = Ln+s (6), Am-1 = On+1 (7), which completes the proof. : * The fact that the cycle of the divisors begins one step earlier than the cycles of the partial quotients and rational dividends is true for the general recurring continued fraction. Several other propositions proved for the special case now under consideration have a more general application. The circumstances are left for the reader himself to discover. XXXII TESTS FOR MIDDLE OF CYCLE 439 Cor. 1. Starting with the equations in the second line of (1) as data, we could in like manner prove that Lim-2 = Ln+25 Mm-2 = Mats AIm-2 = In+e 5 and so on, forwards and backwards. Cor. 2. If we put m=n, the conditions in (1) become Ly = Lins Mn=Mn, on=an3 in other words, the conditions reduce to Ly = Lint ; and the conclusion becomes Lin-y ae Dint-s5 Mn-1 aa Mansy An -1 = On+ie Hence, if two consecutive rational dividends be equal, they are the middle terms of the cycle of rational dividends, which must therefore be an even cycle; and the partial quotient and divisor corresponding to the first of the two rational dividends will be the middle terms of their respective cycles, which must therefore be odd cycles. Cor. 3. If we put m=n +1, the conditions in (1) reduce to Masi =M,, On+i = On 5 and the conclusion gives Ly, = Ln+s, M, = Mnsis On = On+1° Using this conclusion as data in (1), we have as conclusion | Lin-1 =Lniss Mn-1 = Mn+, An-1 = On+e 5 and so on. Hence, if two consecutive divisors (My, Mn+) be equal, and also the two corresponding partial quotients (an, an+4,) be equal, these two pairs are the middle terms of their respective cycles, which are both even; and the rational dividend (Liy+,) corresponding to the second member of either pair is the middle term of its cycle, which is odd. These theorems enable us to save about half the labour of calculating the constituents of the continued fraction which represents VN /M. In certain cases they are useful also in reducing surds of the more general form (L+ “N)/M to con- tinued fractions. Example 1. Express 8463/39 as a simple continued fraction ; and exhibit the cycles of the rational dividends and of the divisors. 440 EXAMPLES CHAP. We have < 8463) | —78+V/8463 __ ‘ 1 39 39 (78 +4/8463)/61 ’ 78+V8463_ | — 4444/8463 _ 1 ; lao 61 — (44-4-4/8463)/107 ’ 44+ V8463__ —63+/8463 __ 1 07a 107 (63 + -78463)/42? 63+ V8463_, | ~ 63+ 8463 _, 1 , 420 4 (68+ N18 4685/1079 SEEMED - Since we have now two successive rational dividends each equal to 63, we know that the cycle of partial quotients has culminated in 3. Hence the cycles of partial quotients, rational dividends, and divisors are— Partial quotients. . 2, de oats Lit ow eae Rational dividends . 78, 44, 68, 63, 44, 78; Divisors, +. . 5 2) 89,5 61;) 107, 4910; and we have A/ BAGS ee Leet eto Tse Tey oe OC D394 tear baad ok Me el he hae * * Example 2. If c denote the number of partial quotients in the cycle of the continued fraction which represents N/M, prove the following formule :— lfc=27, | | Po_ Ptyde t+ PeGe-1 (I.) ; ; Je 4e(9et1 + 9-1) ie=2¢+1, Pe _ Pte TPO (II.) : Lo ht — Wc Qe + Oe" if m be any positive integer, 2 Omen + N M? ine Pansg ine NM ene (IIL). 2me 2me qd me For brevity we shall prove III. alone. The reader will find that I. and II. may be proved in a similar manner. For a different kind of demonstra- tion, see chap. xxxiv., § 6. We have : 5 ae 1 1 1 1 Esme ee ee me CT, ee J2me a+ a+ 2a+ oa at 1 1 1 1 1 SS me ass hee et ee ay+ ajy+ 2a+ a+ pura) y ) xs (a + Del Yme) Pme + Pme-1 ~ (@+Pmel Ime) Yme + Yme-1" = (ADme +Pme-1) Yme + Dine? (a) Yme( Ame + Yme-1 +Pme) XXXII EXERCISES XXXI 44] Now the equations (2) and (8) of § 3 give us Ame Dad ar Yme-1 Onectt = Mitins Pme Prime + Pme-1 Qin == (N/M) Yme } (P). In the present case, Pmeti= Pep =Le=l=aM, Qme-H a Qet4 =M, = M. The equations (8) therefore give AY me + Yme-1 = Pme APme + Pme-1 = (N/M?) dime } (7). From (a) and (y) (III.) follows at once. The formule (I.), (II.), (III.) enable us, after a certain number of con- vergents to N/M have been calculated, to calculate high convergents with- out finding all the intermediate ones. Consider, for example, 8463 _ (i ae SES a 39 Te cere BE PTs Fa. ge * * Here c=6, ¢=3, and we have for the first four convergents 2/1, 5/2, 7/3, 26/11; hence P6_ Paqst+Psqe qe 93(qat qe) _26x384+7x2 92 eto tees) sO p) Also Piz _ pe’ + (N/M*)g6" ; qe 27696 _ 927+ (8463/397).39? 16927 , a 2x 92x 39 ee? 1 Ge Pr prs’ + (N/M?)q12? 24 2912912 : _ 169272 x 392+ 8463 x 71762 2 x 39% x 16927 x 7176 The rapidity and elegance of this method of forming rational approximations cannot fail to strike the reader. EXERCISES XXXI, Express the following surd numbers as simple continued fractions, and exhibit the cycles of the partial quotients, rational dividends, and divisors :— (1.) 4/(101). (2.) 3/(63). (3.) /(7#). 1 2+/(29) ; (4.) w/(61)" (i eeg ve (6.) T+a/%. (7.) Express the positive root of x?-2-4=0 as a continued fraction, and find the 6th convergent to it. (8.) Express both roots of 2a°-—6%-1=0 as continued fractions, and point out the relations between the various cycles in the two fractions. 449 EXERCISES XXXI CHAP, (9.) Show that b a+ - 7 * A/ (a? +b) =a+ b /(a? — 6) =a - 5 — es * (10.) Express 4/(a?+1) as a simple continued fraction, and find an expres- sion for the nth convergent. Evaluate the following recurring continued fractions, and find, where you can, closed expressions for their nth convergents; also obtain recurring formule for simplifying the calculation of high convergents :— it ee, Rikers oy aa * i (12.) rem * iby (Mal uP) arb * * Show, in this case, that Prnt2 — 2Pon + Pon—2= abpan. Loot 1 (14.) tee ee «6 6 24 oe 29, * * where the cycle consists of n units followed by 2. (15.) Show that ee a Suge ene * * * is independent of x. (16.) Show that * < x < (bok jhe Ba * * E sae | ( m 1 Oe a ab aa ) MS en ane x pape ; * show that 2atyt+z2)-(a+b+c) 1 1 1 2u —(a+b+c)—abe a ER RS Ps at (18.) Show that a 2 a (35>: +) =p * * XXXIII EXERCISES XXXI 443 (19.) If p be the numerator of any convergent to «/2, then 2p?+1 will also be the numerator of a convergent, the upper or lower sign being taken according as p/q is an odd or an even convergent ; also, if g, g’ be two con- secutive denominators, g?+4q’? will be a denominator. (20.) Evaluate 1 1 a Tees . gyre) os Be * " fu where the cycle consists of »—1 units followed by n. Lapel (21.) In the case ere a+ oy * Pon = G2n+1 — {(/2 ay 1) ses 23 (A/2 -y Leer} [2/2, Pen-1= 49 on — {(a/2 3 Le 7 (A/2 = ad [4A/2. (22.) Convert the positive root of axz*+ubz—b=0 into a simple con- tinued fraction; and show that p, and q» are the coefficients of 2” in (x + ba? — x4)/(1-ab+2.2%+a4) and (aw+ab+1.2?+24)/(1-ab+2.2?+24) re- spectively. Hence, or otherwise, show that tif a, 8 be the roots of 1—(ab+2)z+2?=0, then .» prove that pon = bYan—1 = ab— = : ; AC ea) ad Chad sae, PmH=In = a—B (23.) If the number of quotients in the cycle of JN _ 1 Reet be ¢ it ee apres a2+ oe. Giae Chis 7 Oe ai ; * * show that : a + Eis ee a (m cycles) = Ndme a+° °° ayt+ 2a+ m+ °° “ay+ a y M?0mne (24.)* If c¢ be the number of quotients in the cycle of ,/N/M, show that if c=2¢+1, 2 PP t—r-1 +0724» oh N geet gem aE! Poe techy 6 —215 and iif ¢=2, Pt—r—2 Pt-r-1 + Pttr—1 Petr _ N U—r—29t—r—-1 T+ Vt4+7-1Ut-t-r ~ M? 1 gle 1 1 1 ., and if the convergent * For solutions of Exercises 24 and 26-29 see Muir’s valuable little tract on The Expression of a Quadratic Surd as a Continued Fraction, Glasgow (Maclehose), 1874. t In connection with Exercises 25 and 30-32 see Serret’s Cours d’ Algébre Supérieure, 3m¢ éd., t. i, chaps. i. and ii. 444 EXERCISES XXXI CHAP, obtained by taking 1, 2, ..., ¢ periods, ending in each case with a1, be Littles es ly Zapand at 4;=Pi/Qi, . . ., Zi=P,/Q;, Py and Qy being integers prime to each other as usual, then Pi— QV Z= (Pia — Qi-a/Z) (Pa - Q/Z), = (Pi - Qin/Z)*; ; Li+”/Z_ (Zy+°/Z\i Zi- JL \by— JZ)” (26.) If N be an integer, and if a cyclical partial quotient occur in the development of »/N equal to the acyclic partial quotient a, that quotient will be the middle term of the reciprocal part of the cycle ; and no cyclical partial quotient can occur lying between a and 2a. (27.) When N is a prime integer, the cycle of partial quotients is even, and the middle term of the reciprocal part of the cycle is a or a—1 , according as a is odd or even. (28.) If N be an integer, and the cycle of \/N be odd, then A is the sum of the squares of two integers which are prime to each other. Exhibit 365, as the sum of two squares. (29.) The general expression for every integer whose square root has a cycle of c terms, the reciprocal part of which has the terms a, ao, . . ., Go, Ay, is (pm —(-1)%p'q')? + p'm - ( -1)q"2, where m is any positive integer, and p'/q’, p/q are the two last convergents to fee ss : Ce dg+ ay Find an expression for all the integers that have 1, 2, 1 for the reciprocal part of the cycle of their square root. (30.) If two positive irrational quantities, a and 2’, can be developed in continued fractions which are identical on and after a certain constituent, show that a =(ax+b)/(a'x+0'), where a, b, a’, b’, are integers such that ab!—a/b=+1 ; and that this condi- tion is sufficient. (31.) The equation of the 2nd degree with rational coefficients which is satisfied by a given recurring continued fraction has its roots of opposite signs if the fraction is purely recurring, and of the same sign if it is mixed and has more than one acyclic partial quotient. (32.) Investigate the relation between the cycles of the partial and com- plete quotients of the two continued fractions which represent thé numerical values of the two roots of an equation of the 2nd degree with rational co- efficients. Illustrate with 27x? - 97x”+77=0. XXXIII DIOPHANTINE PROBLEMS 445 APPLICATIONS TO THE SOLUTION OF DIOPHANTINE PROBLEMS. § 10.] When an equation or a system of equations is in- determinate, we may limit the solution by certain extraneous conditions, and then the indeterminateness may become less in degree or may cease, or it may even happen that there is no solution at all of the kind demanded. Thus, for example, we may require (I.) that the solution be in rational numbers; (II.) that it be in integral numbers ; or, still more particularly, (III.) that it be in positive integral num- bers. Problems of this kind are called Diophantine Problems, in honour of the Alexandrine mathematician Diophantos, who, so far as we know, was the first to systematically discuss such problems, and who showed extraordinary skill in solving them.* We shall confine ourselves here mainly to the third class of Diophantine problems, where positive integral solutions are required, and shall consider the first and second classes merely as stepping-stones toward the solution of the third. We shall also treat the subject merely in so far as it illustrates the use of continued fractions: its complete development belongs to the higher arithmetic, on which it is beyond the purpose of the present work to enter.T Equations of the 1st Degree in Two Variables, -§ 11.] Since we are ultimately concerned only with positive integral solutions, we need only consider equations of the form aa + by =c, where a, b, ¢ are positive integers. We shall suppose that any factor common to the three coefficients has been * See Heath’s Diophantos of Alexandria (Camb. 1885). tT The reader who wishes to pursue the study of the higher arithmetic should first read the late Henry Smith’s series of Reports on the Theory of - Numbers, published in the Annual Reports of the British Association (1859- 60-61-62); then Legendre, Théorie des Nombres; Dirichlet’s Vorlesungen diber Zahlentheorie, ed. by Dedekind ; and finally Gauss’s Disquisitiones Arith- metice. He will then be in a position to master the various special memoirs in which Jacobi, Hermite, Kummer, Henry Smith, and others have developed | this great branch of pure mathematics. 446 ax — by =c¢ CHAP, removed. We may obviously confine ourselves to the cases where a is prime to 0; for, if # and y be integers, any factor common to @ and 6 must be a factor inc. In other words, if a be not prime to 0, the equation ax + by =¢ has no integral solution. § 12.] Zo find all the integral solutions of ax —by=c ; and to separate the positive integral solutions. We can always find a particular integral solution of ax — by =c (1). For, since a is prime to ), if we convert a/b into a continued fraction, its last convergent will be a/b. Let the penultimate convergent be p/g, then, by chap. xxxil., § 8, ag—-pb= +1 (2). Therefore a( + cq) — 0( + ep) =¢ (3). Hence e=+c, y= top (4) is a particular integral solution of (1). Next, let (#, y) be any integral solution of (1) whatever. Then from (1) and (3) by subtraction we derive aia —( + cg)} — bly — ( = ep)} = 0. {e-(+cq)}/{y-(+ep)}=b/a (5). Since @ is prime to J, it follows from (5), by chap. iii, Exercises Ve ethat Therefore x—(40eq)=bt, y—-(+ cp) =at, where ¢ is zero or some integer positive or negative. Hence every integral solution of (1) is included in w= teqt+bht, y= tept+at (6), where the upper or lower sign must be taken according as the upper or lower sign is to be taken in (2). Finally, let us discuss the number of possible integral solu- tions, and separate those which are positive. 1°. If a/b>p/q, then the upper sign must be taken in (2), and we have a=cqt+bt, y=cp+at (6’). XXXIII ax + by =e 447 There are obviously an infinity of integral solutions. To get positive values for # and y we must (since cp/a cq/b, we must write a= —cq+bi, y= —cp+at (62). All our conclusions remain as before, except that for positive solutions we must have cp/api$ + a. We see, therefore, that aa —by=c has in all cases an infinite number of positive integral solutions. § 13.] To find all the integral solutions of au + by =c¢ is and to separate the positive integral solutions. We can always find an integral solution of (7); for, if p and q have the same meaning as in last paragraph, we have : (+ cg)a + (¥ ep)b =e (8), that is, x’ = + cq, y’ = + cp is a particular integral solution of (7). By exactly the same reasoning as before, we show that all the integral solutions of (7) are given by w= tceq—bi, y= =ept+at (Os so that there are in this case also an infinity of integral solutions. To get the positive integral solutions :— 1°, Let us suppose that a/b>~p/q, so that cp/a cq/d, then t= —¢q-bi, y=cpt+at (Gay: or for positive integral solutions we must have —cp/a}t — o/b. 448 EXAMPLES CHAP. In both these cases the number of positive integral solutions is limited. In fact, the number of such solutions cannot exceed 1 + mod (cg/b — cp/a); that is, since mod (aq — pb) = 1, the number of positive integral solutions of the equation ax +by=c cannot exceed 1 +¢/ab. Example 1. To find all the integral and all the positive integral solutions of 8%+13y=159. We have _ oo ek “bE — + ok + _ ae bole The penultimate convergent is 3/5; aud we have 8x5-13x3=1, 8(795) +13( — 477) =159. Hence a particular solution of the given equation is #’=795, y'/= —477; and the general solution is e=795-13t, y= -477+8t. For positive integral solutions we must have 795/13 +¢<+477/8, that is, 61475 4¢+4593. The only admissible values of ¢ are therefore 60 and 61; these give x=15, y=3, and 2=2, y=11, which are the only positive integral solutions. Example 2. Find all the positive integral solutions of 83%+2y+3z=8. We may write this equation in the form 3% + 2y=8 — 32, rom which it appears that those solutions alone are admissible for which e012 or 2, The general integral solution of the given equation is obviously e=8-32-2t, y= —-84324 38. In order to obtain positive values for « and y, we must give to ¢ integral values lying between +4-—$2 and +22-—z. The admissible values of ¢ are 3 and 4, when z=0; 2, when z=1; and 1, when z=2. Hence the only positive integral solutions are ; , ae TO Le Ue y=l, 4, 1,- 13 ee.) DOs a: In a similar way we may treat any single equation involving more than two variables. §14.] Any system of equations in which the number of variables exceeds the number of equations may be treated by methods which depend ultimately on what has been already done. XXXIII SYSTEM OF TWO EQUATIONS 449 Consider, for example, the system az+by+c=d (1), axt+by+cz=a' (2), where a, 0, c, d, a’, &c. denote any integers positive or negative. This system is equivalent to the following :— — (ca')a + (be')y = (de’) (3), ax+ by+cz=d (4), where (ca’) stands for ca’ — ca, &c. . Let 5 be the G.C.M. of the integers (ac’), (bc). Then, if 6 be not a factor in (dc’), (3) has no integral solution, and conse- quently the system (1) and (2) has no integral solution. If, however, 6 be a factor in (dc’), then (3) will have integral solutions the general form of which is_ . w= x + (be')t/s, y=y"' + (ca’)t/s (5), where (x, y") is any particular integral solution of (3), and @ is any integer whatever. If we use (5) in (4), we reduce (4) to cz — c(ab’)t/8 = d — ax” — by” (6), where c(ab’)/5 is obviously integral. In order that the system (1), (2) may be soluble in integers, (6) must have an integral solution. Let any particular solution of (6) be z=2,¢=2'. Then ) z—2 (ab’) tt 8 Hence, if « be the G.C.M. of (ad’) and 6, that is, the G.C.M. of (bc’), (ca’'), (ab’), then - g=2'+(ab'u/e t= + du/e (7), where w is any integer. . From (5) and (7) we now have e=a' +(be'\u/e, y=y' + (ca'jufg. z=2'+(ab'ju/e (8), where a’ =a" + (bc')t'/8, y= y" + (ca’)t'/6. i If in (8) we put u=0, we get =a’, y=y', z=2'; therefore (x', y', z) is a particular integral solution of the system (1), (2). A little consideration will show that we might replace (2’, 7’, 2’) by any particular integral solution whatever. Hence (8) gives all VOL. II 2G 450 FERMAT’S PROBLEM CHAP. the integral solutions of (1), (2), (a, y', 2) being any particular wmtegral solution, « the G.C.M. of (bc’), (ca’), (ab'), and u any integer whatever. | | The positive integral solutions can be found by properly limiting w. Example. 3a@+4y+272=384, 8x+5y+212=29. Here (bc’)= — 51, (ca’)=18, (ab')=3. Hence e=3; a particular integral solution is (1, 1, 1); and we have for the general integral solution e=1-17u, y=1+6u, z=1+4u. The only positive integral solution is z=1, y=1, z=1. Equations of the 2nd Degree in Two Variables. § 15.] It follows from § 7 (4) that, if pp/gn be the nth con- vergent and M, the (n+ 1)th rational divisor belonging to the development of /(C/D) as a simple periodic continued fraction, then | Dpn' — Can’ = (— "Mn (1). Hence the equation Da’ — Cy’ = +H, where C, D, H are positive integers, and C/D is not a perfect square, admits of an infinite number of integral solutions provided its right-hand side oceurs among the quantities (—)"M, belonging to the simple continued fraction which represents J(C/D); and the same is true of the equation Dx’ — Cy’ = — H. The most important case of this proposition arises when we - suppose D=1. We thus get the following result :— The equation a* — Cy? = + H, where C and H are positive integers, and © is not a perfect square, admits of an infinite number of integral solutions provided its right-hand side occurs among the quantities (—)"M,, belonging to the development of VC as a simple continued fraction. Cor. 1. The equation x — Cy’ =1, where C is positive and not cm perfect square, always admits of an infinite number of solutions.* “ By what seems to be a historical misnomer, this equation is commonly spoken of as the Pellian Equation. It was originally proposed by Fermat as a challenge to the English mathematicians. Solutions were obtained by xxxi1 LAGRANGE’S THEOREM REGARDING a°-Cy?=+H 451 For, if the number of quotients in the period of ./C be even, = 2s say, then (— )*M.,, will be +1 (since here M= +1). Therefore we have (ee a Oday. = +1, where ¢ is any positive integer; that is to say, we have the system of solutions LH =Pots, Y = Vets | (A), for the equation 2 — Cy’ =1. If the number of quotients in the period be odd, = 2s — 1 say, then (—)*-*M,,_, will be — 1, but (—)*-?M,,_., (—)®*-4Myy_,, . . will each be +1. “Hence we shall have the system of solutions UL =Pats—-oty Y = Vats-ct (B), for the equation 2° — Cy’ = 1. Cor. 2. The equation a — Cy’ = — 1 admits of an infinite number of integral solutions provided there be an odd number of quotients in the period of JC. § 16.] In dealing with the equation a — Oy = + 1 (1) we may always confine ourselves to what are called primitive solutions, that is, those for which a is prime to y. For, if # and y have a common factor 6, then 6° must be a factor in H, and we could reduce (1) to #” —Cy”= +H/6*. In this way, we could make the complete solution of (1) depend on the primitive solutions of as many equations like #” — Cy = + H/6° as H has square divisors. We shall therefore, in all that follows, suppose that @ is prime to y, from which it results that # and y are prime to H. With this understanding, we can prove the following im- portant theorem :— If H< JC, all the solutions of (1) are furnished by the con- vergents to /C according to the method of § 15. This amounts to proving that, if x = p, y=q be any primitive integral solution of (1), then p/q is a convergent to /C. equation is merely a part, was given by Lagrange in aseries of memoirs which form a landmark in the theory of numbers. See especially Qwvres, t. ii., Proll: 452 GENERAL SOLUTION OF #—Cy’=+1,0R +H cmap. Now we have, if the upper sign be taken, p - Cg =H. Hence pla- VC =H/q(p + VCq), 1. Hence pla- NO <1/2¢° (3). It follows, therefore, by chap. xxxii., § 9, Cor. 4, that p/q is one of the convergents to VC. If the lower sign be taken, we have ¢ - (1/C)p* = H/C, where H/C <,/(1/C). We can therefore prove, as before, that q/p is one of the convergents to ./(1/C), from which it follows that p/q is one of the convergents to /C. Cor. 1. All the solutions of e —Cy=1 (4) are furnished by the penultimate convergents in the successive or alternate periods of /C. Cor. 2. If the number of quotients in' the period of »/C be even, the equation Cy = -1 (5) has no integral solution. If the number of quotients in the period be odd, all the integral solutions are furnished by the penultimate convergents in the alternate periods of JC. § 17.] We have seen that all the integral solutions of the equation (4) are derivable from the convergents to /C; it is easy to give a general expression for all the solutions in terms of the first one, say (p, qg). If we put a+y /C=(p+g NC)” z—y/O=(p—q vor} a — Cy’ = (p' — Cg’) = 1. Hence (6) gives a solution of (4). (6), we have In like manner, if ~ be any integer, and (p,,q) the first solution of (5), a more general solution is given by o+yVC= elem) x—y NO=(p—q NO)-1 (7). a XXXIII EXAMPLES 453 Finally, if (p, g) be the first solution of (1), we may express all the solutions derivable therefrom* by means of the general solution (6) of the equation (4). For, if (7, s) be any solution whatever of (4), we have . Deed = =H, r—Cs =1; (p' — Cg") (7? — Cs") = +H, (pr + Cys)* — C(ps + gr)’ = +H. “= pr + Cqs y = ps + qr Therefore (8) is a solution of (1). The formule (6), (7), (8) may be established by means of the relations which connect the convergents of /C (see Exercises XXXI,, 25, and Serret, Alg. Sup., § 27 et seq.). This method of demonstration, although more tedious, is much more satisfactory, because, taken in conjunction with what we have established in '§ 16, it shows that (6), (7), and (8) contain all the solutions m question. Example 1. Find the integral solutions of a? —13y?=1. If we refer to chap. xxxii., § 5, we find the following table of values for 4/13 :— aT On Pn In Me 1 3 3 1 4 a, ang De 4 1 3 3 1 7 2 3 4 1 11 3 4 5 1 18 5 1 6 6 119 33 4 uv 1 137 38 3 8 1 256 Pie es 9 1 393 109 4 10 1 649 180 1 tis. 1. 6 4287 | 1189 4 Hence the smallest solution of «?-13y?=1 is z= 649, y=180. We have, in fact, 649? — 13.1802= 421201 — 421200=1. * It must not be forgotten that there may be more than one solution in the first period. For every such primary solution there will be a general group like (8). 454, a’ —Cy’?= +H, wHen H> VC CHAP. From (6) ae we see that the general solution is given by =${ (649 + 180/13)” + (649 — 180\/13)”"} y=4{ (649 + 180/13)" — (649 — 1804/18) yn} 1 /[/18, where 7 is any positive integer. In particular, taking n=2, we get the solution x= 6497+ 13.180?= 842401, y= 2.649.180 =233640. Example 2. Find the integral solutions of x? -18y?= -1. The primary solution is given by the 5th convergent to 4/13, as may be seen by the table given in last example. The general solution is, by (7), I {(18 + 54/18)?”-1 + (18 — 54/13)?” = Sa a ae 2n-1 Veet (18 + 5a/13)2"-1 — (18 — 54/18) when 1 is any positive integer. Example 3. Find all the integral solutions of 2? — 13y?=38. The primary solution is z=4, y=1, as may be seen from the table above. The general solution is therefore, by (8), e=4r£13s, y=4s=r, where (7, s) is any solution whatever of x? -13y?=1. In particular, taking r=649 and s=180, we get the two solutions, x= 256, y=71, and «=4936, y=1369. § 18.] Let us next consider the equation = Oy = (9), where C is positive and not a perfect square, and H is positive but > /C. We propose to show that the solution of (9) can always be made to depend on the solution of an equation of the same form in which H<./C; that is, upon the case already completely solved in §§ 15-17. Let (x, y) be any primitive solution of (9), so that x is prime to y. ‘Then we can always determine (z,, y,) so that ry, — yt, = +1 (10).* In fact, if p/g be the penultimate convergent to «/y when con- verted into a simple continued fraction, we have, by § 12, a, =e Ep, Yom ty eg (12). * There is no connection between the double signs here and in (9). XXXIII LAGRANGE’S CHAIN OF REDUCTIONS 4B5 If we multiply both sides of (9) by 2," — Cy,’, and rearrange the left-hand side, we get (xx, — Cyy,)’.— Clay, — ya)" = + H(a," — Cy’). This gives, by (10), (ax, — Cyy,)’ -C = + H(z,’ - Cy,’) (12). Now ww, — Cyy, = ta" — Cy") + (xp — Cyq) (13). But we may put ap — Cyg=SH=+K,, where K,+$H. Hence a2, — Cyy, = (t= S)H + (+K,) (14). Now ¢ and the double sign in (13) are both at our disposal ; and we may obviously so choose them that aa, — Cyy, = K, (15), where K, + 4H. (16). We therefore have, from (12), K’-C= + H(a,’ - Cy,’) Clie _ Now, by hypothesis, ./C VC, then we find all the values of K, < 2H, for which (K,’— C)/H, is integral, XXXIII EXAMPLE ABT = H, say, and, if H,<./C, solve the equation x,’ —- Cy, = +H, ; then pass back to z through the two transformations (20) and (22); and, finally, select the integral values of ~ and y thus resulting, if there be any. By proceeding in this way until each branch and twig, as it were, of the solution is traced to its end, we shall get all the possible integral solutions of (9), or else satisfy ourselves that there are none. | The straightforward application of these principles is illus- trated in the following example. Into the various devices for shortening the labour of calculation we cannot enter here. Example. Find the integral solutions of a2 — 15y2=61 (9’). Let (Ky? - 15)/6.=Hy (18"); where K,+ 30. Then K,?=15+ 61H). Since K,?+900, we have merely to select the perfect squares among the numbers 15, 76, 137, 198, 259, 320, 381, 442, 503, 564, 625, 686, 747, 808, 869. The only one is 625, corresponding to which we have K;=25 and H,=10. Since H;>/15, we must repeat the process, and put (Ko? — 15)/10 = He (18%); where K+ 5, and therefore Ky? + 25. Since Ks?=15+10Hb, the only values of Ky? to be examined here are 5, 15, 25. Of these the last only is suitable, corresponding to which we have Ks=5, He= 1; We have now arrived at the equation a? — 15 yo? = (21’), the first solution of which is easily seen to be (4, 1). Hence the general solution of (21’) is n= {(4-+ x/15)"-+ (4/15) n= 5g (A+ V5)" - (4 9/15) The general solution of (9’) is connected with this by the relations (24), a= (5a 15y2)/1, y= (5y27Far2)/1 (22’) ; = (2527715y)/10, y=(25y,;2)/10 (20’). Hence ' e@=14aop 45ye, Y = F3x2+14y2 (25); where x and yz are given by (24). The question regarding the integrality of «and y does not arise in this case. As a verification put a2,=4, y2=1, and we get the solutions (11, 2) and (101, 26) for (9’), which are correct. 458 REMAINING CASES OF BINOMIAL EQUATION CHAP. § 19.] There remain two cases of the binomial equation x — Cy’= +H which are not covered by the above analysis— aim Oy ese Hl (26), where C is a perfect square, say C = R’; and a +Cy= +H (27), The equation (26) may be written (a — Ry) (a + Ry) = +H. Hence we must have a—-Ry=u : ea (28), where w and v are any pair of complementary factors of +H. We have therefore simply to solve every stich pair as (28), and select the integral solutions. The number of such solutions is clearly limited, and there may be none. In the case of equation (27) also the number of solutions is obviously limited, since each of the two terms on the left is positive, and their sum cannot exceed H. The simplest method of solution is to give y all integral values + 4/ (H/C), and examine which of these, if any, render H — Cy’ a perfect square. § 20.] In conclusion, we shall briefly indicate how the solu- tion of the general equation of the 2nd degree, aa’ + Lhay + by? + 2ga + Wy +c=0 (29), can be made to depend on the solution of a binomial equation. By a slight modification of the analysis of chap. vii. § 13, the reader will easily verify that, provided a and b be not both zero, and ¢ be not zero, (29) may be thrown into one or other of the forms (Cy + F)’ - Caw + hy +9)? = —aA (30) ; or (Cx + GY — C(ha + by + fy = - dA (31), where A=abe + 2fgh —af* —bg’- ch’, C=h?-ab, F=gh-af, G=hf—bg; say into the form (30). If, then, we put yet eer 32 ax+hy+g=nJ (32), ae XXXIII GENERAL EQUATION OF 2ND DEGREE 459 (30) reduces to & = Cx’ = = 0 (33), which is a binomial form, and may be treated by the methods already explained. If h’>ab, then C is positive, and, provided C be not a perfect square, we fall upon cases (1) or (9). If C be a positive and a perfect square, we have case (26). It should be noticed that, if either a=0 or b=0, or both a=0 and b=0, we get the leading peculiarity of this case, which is that the left-hand side of the equation breaks up into rational factors (see Example 2 below). If h* < ab, then C is negative, and we have case (27). If h’ = ab, then C = 0, and the equation (29) may be written (ax + hy)’ + 2aga + 2afy + ac =0 (34), which can in general by an obvious transformation be made to depend upon the equation n = QE (35), which can easily be solved. Example 1. Find all the. positive integral solutions of 3x7 — 8xy + Ty? — 4a+2y=109. This equation may be written (8u — 4y — 2)? + 5(y-1)?=336, say £24 5y?= 336. Here we have merely to try all values of 7 from 0 to 8, and find which of them makes 336 — 57? a perfect square. We thus find f= 16) y= c4- fa y= Hence 38a—-4y-2=416, y-1=+4 (irs 8a—4y-2=44, y-1=+8 (2). It is at once obvious that in order to get positive values of y the upper sign must be taken in the second equation in each case. Hence y=5 or y=9. To get corresponding positive integral values of «, we must take the lower sign in the first of (1), and the upper sign in the first of (2). Hence the only positive integral solutions are Pa De Yao ang o—14, 4-9, 460 EXAMPLES CHAP, Example 2. Find the positive integral solutions of dxy + 2y? — 4a - 8y=12. This is a case where the terms of the 2nd degree break up into two rational factors. We may put the equation into the form (9x + 6y —1)(8y — 4)=112. Since 38y—4 is obviously less than 9%+6y—1 when both x and y are positive, 3y—4 must be equal to a minor factor of 112, that is,/6 jee aoe or 8; the second and the last of these alone give integral values for y, namely, y=2 and y=4. To get the corresponding values of 2, we have 92+ 6y—1 =56 and 9x+6y—1=14, that is to say, 9x=45 and 92=-9. Hence the only positive integral solution is z=5, y=2. Example 3. Find all the integral solutions of 9x? — 12xy + 4y? + 382+ Qy=12. Here the terms of the 2nd degree form a complete square, and we may write the equation thus— (8a — 2y)? + (8a — 2y)+4y=12, or 4(3x— 2y)? + 4(3a - 2y)+1+16y=49 ; that is, (6a — 4y+1)?=49 — 16y. Hence, if u=6x-4y+1 ' (1), so that w is certainly integral, we must have ' ; y =(49 — w)/16 (2). Now we may put w=16u+s, where s is a positive integer +8, It then appears that y will not be integral unless (49 — s*)/16 be integral. The only value of s for which this happens iss=1. Therefore t= LOuce | ; (3). Hence, by (1), (2), and (3), we must have e=2+ 4u(1 — 8u)/8, Y=3-2n-16u? (4), or w= 4u+(5—382u*)/3, y=34+2Qu—16p2 (5). It remains to determine yu so that x shall be integral. Taking (4), we see that y(1—8)/3 will be integral when and only when M=8v or w=3y-1. Using these forms for yu, we get x=2+4 4y — 96y?, y=8 — 6v -144p? (6); x= —10+68y-96r?2, y= -114 90» -—144y? (7). Taking (5), we find that (5—32u2)/3 is integral when and only when M=3v4+1 or w=8r-1, Using these forms, we get from (5) “x= —-5-—52v-96r2, y=-11-90v-144p2 (8); x= -—13+76v-96y2, y= —-15+102y—144y? (9). The formule (6), (7), (8), (9), wherein p may have any integral value, positive or negative, contain all the integral solutions of the given equation. XXXIII " EXERCISES XXXII | 461 EXERCISES XXXII. Find all the integral and also all the positive integral solutions of the following equations :— (1.) 5¢+7y=29. (2.) 162-17y=27. (3.) Lla+7y=1103. (4,) 1367” -1013y=16246, (5.) If £2, ys. be double Ly, ws., find x and y. (6.) Find the greatest integer which can be formed in nine different ways and no more, ie adding together a positive integral multiple of 5 and a positive integral multiple of 7. (7.) In how many ways can £2:15:6 be paid in settee and florins ? (8.) A has 200 shilling-coins, and B 200 frane-coins. In how many ways can A pay to B a debt of 4s, ? (9.) 4 apples cost the same as 5 plums, 3 pears the same as 7 apples, 8 apricots the same as 15 pears, and 5 apples cost twopence. How can I buy the same number of each fruit so as to spend an exact number of pence and spend the least possible sum ? (10.) A woman has more than 5 dozen and less than 6 dozen of eggs in her basket. If she counts them by fours there is one over, if by fives there are four over. How many eggs has she ? (11.) A woman counted her eggs by threes and found that there were two over ; and again by sixes and found there were three over. Show that she made a mistake. (12.) Find the least number which has 3 for remainder when divided by 8, and 5 for remainder when divided by 7. (13.) Find the least number which, when divided by 28, 19, 15 respect- ively, gives the remainders 15, 12, 10 respectively. (14.) In how many ways can £2 be paid in half-crowns, shillings, and sixpences ? (15.) A bookcase which will hold 250 volumes is to be filled with 3-volumed novels, 5-volumed poems, 12-volumed histories. In how many ways can this be done? If novels cost 10s. 6d. per volume, poems 7s. 6d., and histories 5s., show that the cheapest way of doing it will cost £129, 15s. Solve the following systems, and find the positive integral solutions :— (16.) %+2y+3z2=120. (17.) at+y+ze+u= 4, (18.) 22+ 5y+ 32=324, 5y + 624+ 9u=18. Rees e190) (19.) enn (20.) 17~+19y+21z=400. 17a —4y+3z=510. dx+-2y+42+ w=63, 2e+ 3y+224+ 4u=74. (22.) Show how to express the general integral solution of the system (21.) w+ y+ 2+ n=03, | 41%, +Ajg%et. . . + 0in%,=41, Ag] XH] + Ag9X%et+. . . +Aan%n= dy, G&n—-1,1 XY a An—1,2 0 eee An-1,n Ln =An—1 by means of determinants, when a particular solution is known. 462 EXERCISES XXXII CHAP, XXXIII Find the values of « which make the values of the following functions integral squares :— (23.) Qa?+2a, (24.) (a®-w)/5. (25.) a+1land #+20, simultaneously. (26.) 7x+6 and 4x%+8, PAE ARE: (27.) a? +x+8. Solve the following equations, giving in each case the least integral solution, and indicating how all the titer integral solutions may be found :— (28.) x?-44y?= -8. (29.) x? -44y?= +5, (30.) a?-449?= —7, (31.) pee k= +4, (32.) 22+ 3y?= 628, (33.) 22 —69y2= —11. (34.) v-47y?= +1. (35.) a2?—47y?= —1. (386.) 2? - 26y?= — 1105. (37.) x?-7y?=186. (38.) x -(a?+1)y?=1. (39.) 2? - (a?-1)y?=1. (40.) a?-(a?+a)y?=1. (41.) x?-(a?-a)y?=1. (42.) a? + Say — 2x + 3y = 853. (43.) xy —2a-3y=15. (44.) x?—y?+ 4a —- 5y=27. (45.) 3x? + 2ay + 5y?=390. © (46.) 2? + dary — 11y? + 2x — 86y —- 140=0. (47.) x? — ay — 72y?+ 2a — 440 — 659=0. (48.) a+ 2Qey -17y?+72y—75=0. (49.) 61a? + 28ay + 251y? + 264x + 526y + 260=0. (50.) Show that all the primitive solutions of Dz?-Cy?=+H are furnished by the convergents to ,/(C/D), provided H<,/(CD). Show also how to reduce the equation Dz? - Cy?= +H, when H>,/(CD). (51.) Find all the solutions of 47 —7y?= — 8, and of 4a? — 7y?=58. (52.) If D, E, F, H be integers, and H<,/(E _ DF) (real), show that all the solutions of Da? — 2Eay + Fy?= +H are furnished by the convergents to one of the roots of Dz? — 2Ez+ F=0. (See Serret, Alg. Sup., § 35.) (53.) If Un=pn-*qn, where is a periodic fraction having a cycle of ¢ quotients, and p, and gq» have their usual meanings, then Rete = (a = Bite)? Uy i} where Cpt] =O + ose r+ Bea Gaia Oita +H ’ * and ey, u (cee fey Oye + hie: Opte In particular, if z=/(C/D), then Dpnctr~ V(CD)gnctr= {aM — BLy - Bx/(CD)}"(Dp, — 4/(CD)qy)/M, Point out the bearing of this result on the solution of Da? —G,2=+H. ose ae COLT ale eX XX LV General Continued Fractions. FUNDAMENTAL FORMULA. § 1.] The theory of the general continued fraction Gao J My + Oz + Muereed G20, . ~;. 0a Dy «> Are any quantities whatever, is inferior in importance to the theory of the simple continued fraction, and it is also much less complete. There are, how- ever, a number of theorems regarding such fractions so closely analogous to those already established for simple continued fractions that we give them here, leaving the demonstrations, where they are like those of chap. xxxii., as exercises for the reader. ‘There are also some analytical theories closely allied to the general theory of continued fractions which will find an appropriate place in the present chapter. In dealing with the general continued fraction, where the numerators are not all positive units, and the denominators not necessarily positive, it must be borne in mind that the chain of operations indicated in the primary definition of the right- hand side of (A) may fail to have any definite meaning even when the number of the operations is finite. Thus in forming 1 Pali AEST, 1-I-1-'"* 1 +1/(1 -1) ;’ and in forming the fourth to 1 + 1/{1-1/(1 -1)}. It is obvious that we could not suppose the convergents of this fraction formed by the direct process of chap. xxxii., § 6 (a), (8), the third convergent of 1 + we are led to 464 C.FF. OF FIRST AND SECOND CLASS CHAP, (y). It must also be remembered that no piece of reasoning that involves the use of the value of a non-terminating continued fraction is legitimate till we have shown that the value in question is finite and definite. In cases where any difficulty regarding the meaning or convergency of the continued fraction taken in its primary sense arises, we regard the form on the right of (A) merely as representing the assemblage of convergents P,/q, P2/de - + +» Pn/In whose denominators are constructed by means of the recurrence-formule (2) and (3) below. That is to say, when the primary definition fails, we make the formule (2) and (3) the definition of the continued fraction. In what follows we shall be most concerned with two varieties of continued fraction, namely, Oy 05 a Ba Rely ah . (B), Felgen) and ay + PPS (C), where @,, Mo, Qa - - -» 0, 0, . .. are all real and positive. We shall speak of (B) al (C) as continued fractions of the first ee second class respectively. § 2.] If p,/d, po/q, &e. be the successive convergents to aaa ih Oi ee rite ts: (1); then Pn = Pn-1 + OnPaoe (2) 3 In = On Qn-1 + balance . (3), with the initial conditions p,=1, p,=a@,; ¢,=1, q=4. Cor. 1. In a continued fraction of the first class py, and dy are both positive ; and, provided dy+ 1, each of them continually increases with n.* In a continued fraction of the second class, subject to the restriction Ant1t+ On, Pn and dn are positive, and each aie them continually in- creases with n.* * It does not necessarily follow that Ly,=o and Lg,=0, for the suc- cessive increments here are not positive integral numbers, as in the case of ‘— simple continued fractions. a es oie XXXIV PROPERTIES OF CONVERGENTS 465 These conclusions follow very readily by induction from such formulz as Pn -Pn-1 = = (Gn ig Pn 2 Var be Dn (4). Cor. 2. Rate, ts by, res b, (5) n e . ° Pn-1 An-) + Ane. + a, Gn —@qQ, + Dn ee cea | b, (6). Qn- 1 7 An-1 + Un-2 <0 nee be § 3.] From (2) and (3) we deduce Pn4n- 1 n= 1Gn = (- Ni bab, . dk (1). Cor. 1. The convergents, as calculated bn y the recurrence-rule, are not necessarily at their lowest terms. Cor. 2. Pu _Bn-1 _¢ _ypbabo- + bn (2), Qn Yn-1 dnfn—1 Cor.3: b, sb s Png Pegs 6G), ; Yn VIVE) waa In- in Wor s4: PnUn2 ~ Pn-29n = ( ms De Oy, Cag wirean Oe (4) > Pn _ Pn-2 Fe =(- Ne 1Unbabs ete ae (5). In QWn-2 InYn-2 Oreo: (2: Tao) / (Pax - Pat) _ _nQn-2 Qn Yn-1 ein Giants a UOnaa (6). Ann =i a Ondn =f Cor. 6. In a continued fraction of the first class, the odd con- vergents form an increasing series, and the even convergents a decreas- img series; and every odd convergent is less than, and every even convergent greater than following convergents. In a continued fraction of the second class, subject to the restriction in t1+bp, all the convergents are positive, and form an increasing series, VOL. II 2H 466 CONTINUANT DEFINED CHAP. These conclusions follow at once from (2) ‘and (5), if we remember that, for a fraction of the second class, we have to Teplice 0, $Y. UntDy. — bau aie eee CONTINUANTS. § 4.] The functions pp, gn of a, dy . . ., n> 0s bon which constitute the numerators and denominators of the con- tinued fraction 2 ee tik Dn a, + Qg+ Ag+ °°” On belong to a common class of rational integral functions. * In fact, p, is determined by the set of equations P2 = Up, ae be Dog Ps = Ug + bsp, sty Pn =n Pn-1 + On Dn—s (1), together with the initial conditions p,=1, p,=a,; while ‘i eae determined by the system V3 = AsGe + bs, Vs = U3 + DsQo sty Yn = UMQn-1 + Un Gaes (2), together with the initial conditions ¢, = 1, q, = dp. It is obvious, therefore, that Gn 1s the same function Of Ay Mg . . «, On 3, Om 0a ys 24 Op GS:in 18 OF G0 eee On; Oy Dea We denote the function p, by * PR ars eae <= K ( 29 “3 " aa My Ug, + « +5 Uy (3) and speak of it as a continuant of the nth order whose denominators AL thy My + » + Gy, and whose numerators ared,, . . ., Dy». We have then ees b : a K ( 3) pena 2 ; i Hh zy Uz, » » + An (4) aaa * This was first pointed out by Euler in his memoir entitled « Specimen Algorithmi Singularis,” Nov. Comm. Petrop. (1764). Elegant demonstrations of Euler’s results were given by Mobius, Crelle’s Jour. (1830). The theory has been treated of late in connection with determinants by Sylvester and Muir. XXXIV FUNCTIONAL NATURE OF CONTINUANT 467 When the numerators of the continuant are all unity, it is. usual to omit them altogether, and write simply K(a, a, . . ., ai). A continuant of this kind is called a simple continuant. When it is not necessary to express the numerators and denominators it is convenient to abbreviate both K( bs, S 50a i and K(a, eg dere An) My, Uy - + +» into K(1, 7). In this notation we should have, if r p/és, each terminating so that the last partial quotient >1. Then each of these continued fractions has for its last convergent the value K(aj, de, . . +5 Gn)/K(d2, a3, . . +, &m), Where the two continuants are of course prime to each other, and q@>1, a@>1. From this it appears that there are as many ways, and no more, of representing p by a simple continuant (whose constituents are positive integers the first and the last of which are each greater than unity) as there are integers prime to p and <4». ; PMOWESIUCEN Kg, da, . « .¢ Gn)=K(Gn; «+ +» Gg; 1), and Qa>1, It is obvious that K(an, . . ., a2, a) must arise from one of the other fractions p/w. Hence, given any fraction p/u, it is possible to find another also belonging to the series which shall have the same partial quotients in the reverse order. Let » be a prime of the form 4X+1, then the greatest integer in $p is 2n, which is even. Since, therefore, the number of continuants which are equal to p must be even, and since K(p) is one of them, there must, among the remaining odd number, be one at least which gives rise to no new fraction when we reverse its constituents, that is to say, which is reciprocal. Now the reciprocal continuant in question cannot be of the form K(a, a, .. ., Qi-1, i, U1, . . », da, a), for it follows from (B) that such a continuant cannot represent a prime, unless 7=1, or else 7=2, and a=1, all of which are obviously excluded. We must therefore have an equation of the form T= (Ay, 2, « « +5 Wis My + « «5 May ay), K(m, Ag, + + % a) +K(a, U2, + © sy 1)", by (A), which proves the theorem in question. As an example, take 13=3x4+1. 13 13 Lats pe ee eels loePat W h — — — —_= -: —= -:; —-= —— —— —- ° a e ne 1 ee 2 eae a? 3 4433 4 Beale 5 SCTE Le 2 Guit;. So that 13=K(13)= K(6, 2)=K(4, 3)=K(3, 4)=K(2, 1, 1, 2) = K(2, 6); and, in particular, 13=K(2, 1, 1, 2)=K(2, 1)?+K(2)?=8?+ 27. § 7.] By considering the system of equations (1) of § 4, it is easy to see that, if we multiply a,, 0,, 6,4, by ¢,, the result is the same as if we multiplied the continuant K(1, n) (n>71) by c.. Hence we have’ K( Dee GaGa) © 050404, 2s 0a> > aris Oh, Colla, C335 Cy, + + 49 Cnbn ee ea Ostet onK( i Daa (15). Ay, Ces . . +> An * The following elegant proof of this well-known theorem of Fermat’s was given by the late Professor Henry Smith of Oxford (Credle’s Jowr., 1855). Le aaa 472 REDUCTION TO SIMPLE CONTINUANT CHAP, We may so determine ¢,, ¢,, . . ., Cy that all the numerators of the continuant become equal. In fact, if we put C09 = dN, CaCgD3 = Asm trae Cn—16n On = X, - we get C2 = X/bs, C3 = b,/bs, C4 = Nb,/b, bs, c; =), b,/, bs, Cg = Ab,b,/b,b, 05, « Hence OS SO aS is gy + 6 oy i) Xara pei. PLEA) Onn ane SS ha «K(, da,/b,, dsb,/bs, Nai,b,/b, bo, oe ‘) (16), where p is the number of even integers (excluding 0) which do not exceed n. Cor. Every continuant can be reduced to a sumple continuant, or to a continuant each of whose numerators is — 1. Thus, if we put A= +1 and A= - 1, we have K( DM haf ") Of Un Meee Nee =OUnbn-2. . .x KQ@, a/b, a,b4/b,, ,0,/0,0, ae On On—yOn-3.++/bnbn—2..-) (17), =p — 1, ad =( — JPO, Sue) eh & ( ) Un Ons x K a - Ay/do, As b,/ ds, a d4d;/b4 bo, om fe Mae —] ( a ) eatin eee o [bn Dase “ie ) eh) § 8.] The connection between a continuant and a continued Sraction Jollows readily from (11). For we have, provided K(2, n), K(3, n), K(4, n), . . . are all different from Zero, | val nee a b, : KQ2, 2)?" K(2, 0)/K(3, ny | K(2 7) ee | K(3, x)?” K(3, »)/K(4, n) . Hence ; K(1, n) _ b, Dg b, ER) Tena K(r, n)/K(r + 1, 2) on . XXXIV C.F. IN TERMS OF CONTINUANTS 473 If in this last equation we put =n, and remember that here K(n+1, )=K(_ ) =1, we get ee a0, WORDS Ag + Wz + 1 a result which was Wee from the considerations of § 4. § 9.] When the continuant equation K(1, n) =a,K(1, m — 1) +d,K(1, n — 2), or Pn =4nPn-1 + On Pn—sy which may be regarded as a finite difference equation of the second order, can be solved, we can at once derive from (20) an expression for . 4 Ueade tae) ed, Hoth ea When a and },, are constants, the problem is simply that of finding the general term of a recurring series, already solved in chap. xxxi., § 7. Example. To find an expression for the nth convergent to 5 rae | ys 1+ 14> °° 140°" Here we have to solve the equation pyp=py—1+pn—2, With the initial con- ditions p=1, m=1. The result is K(1, 2) =pn= {(1 + /5)et — (1 — 9/5)" 4} /2e+14/5. Pn _K(1, m)_ {+ /5)et — (1 = 9/5} ot 1/5 Qn K(2, m) {CL + /5)"—(1= 4/5)" 2/5” Bey oT = (eA oye ete (LN /5) P= (LA 5 ie From the expression for K(1, 7) (all the terms in which reduce in this case to +1) we see incidentally that the number of different terms in a continuant of the nth order is 1+ /5)P—(1—a/5)r41 1 ; : w Sa: = Su ntti + Sagas + BngiCs + me ves Hence § 10.] When two continued fractions F and F’ are so related that every convergent of F is equal to the convergent of F’ of the same order, they are said to be equivalent.* * We may also have an (m, n)-equivalence, that is, Remi Ven len | opae See Exercises XXXIII., 2, 17, &c. 474 REDUCTION TO SIMPLE G.F. CHAP. It follows at once from §§ 7 and 8 (and is, indeed, otherwise obvious, provided the continued fraction has a definite meaning according to its primary definition) that we may multiply a,, b,, and b,4, by any quantity m (+ 0) without disturbing the equivalence of the fraction. Hence we may reduce every continued fraction to an equivalent one which has all its numerators equal to + 1 or to —1. Thus we have de Can Me b Q, + = = ; : . . . = Gy Ug tee Oy gti 1 1 1 1 1 Mera S . . . = = fg As] by + Ozb,/ b, + yds) bby + dn) n—-1On =p aie nOn—s (a1), § 11.] If we treat the equations (1) as a linear system to determine K(1, 1), K(1, 2), . . ., K(1, 7), and use the determin- ant notation, we get CGE VF ape Ree Uni tke 0) 7 aere =~ De g.2 bo O50 e | Oe Onn Ot ge eee 0 0 ao 0 0-1 Os b, Ov 0 ’ O70 00 20.0; tee etree O30:0.0".0 er GeO O.=cleimone which gives an expression for a continuant as a determinant. The theory of continuants has been considered from this point of view by Sylvester and Muir;* and many of the theorems regarding them can thus be proved in a very simple and natural manner. EXERCISES XXXIII. (1.) Assuming that both the fractions SOP y= a 6b ¢ at b+eop 9” b+ c+ d+ °° are convergent, show that j xa+l+y)=at+y. tp i * See Muir’s Theory of Determinants, chap. iii. — show that a+ eon : . to n periods = _l. : =| pa : a ag a (+pp ¢ 1 where the quotient q’+p is repeated »—1 times, and the upper or the lower sign is to be taken according as p/q is an even or an odd convergent. i ‘ - : (3.) Evaluate a+ oe ~ . . . ton quotients, a being any real quantity "positive or negative. Show from your result that the continued fraction in ‘question always converges to the numerically ‘greatest root of x -axz-—1=0.* (4.) Deduce from the results of (2) and (3) that a recurring continued fraction whose numerators and denominators are real quantities in general converges to a finite limit ; and indicate the nature of the exceptional cases. (5.) Evaluate 2 - AS * 2... to. terms, 2- 2- 2- Leen os ek (6.) Show that the nth convergent to Sea CGE Hae sub- sequent component being = is (2"-—1)/(2"+1). 2 grtl — x (7.) Show that. co mae oe to n terms =a (8. ) om + ed . . . (v+1 components) =l+a+a(atl1)+...+a(atl1)...(a+n-1). (9.) If d(2) == = .. . n quotients, then $(m +n) = {o(m) + $(n) - apm) G(n)} /{1 + om) o(m)}. . (Clausen. ) (10.) Show that K(0, Mn, Ag, « « 5 (my) = K(dz, “3 Gn) 5 ME Ay ly Uribe 15 Oy « NK eee UU SCrr ey fas. 6 2) more ern. c,.0, 0, Ope Ss gas y= Kl SRA ig OP in K(... a, b,¢,0,0,¢f,...)=K(...4,5,46f,...-). (Muir, Determinants, p. 159.) (11.) Show that the number of terms in a continuant of the nth order is 1+(m-1)+ eas (21-3) Wes (n —5) a ee (Sylvester. ) Close if pa=K( ba, Bs,» s+ On i show that there exists a relation of 1, M2, M3, - + +5 An the form i Apr? + Bpn—1? + Cpn-2 + Dpn-3" = 9, where A, B, ©, D are integral functions of dn, bn, Gn—1, Oni. en ee erase Se" * This is a particular case of the theorem (due to Euler?) that the numerically greatest root of x*-px+q=0 is p— ~ is sah 476 EXERCISES XXXIII CHAP. (13.) Show that K( by, (b1 +a )bo, (b2 + e)bs, Ate 8b a; A; a2, Cascais =(b,+a) (b2 + de) (03+ ds) ote: oc and deduce the theorem of § 19. : (Muir, Zc.) Lit iome Taking (a, b,c, . ., k) to denote the continued fraction Pia, . o and [a, 6, ¢,.. ., J, or, when no confusion is likely, [a, &], to denote K( 7 iy i‘ eee ee 3k prove the following theorems * :— CCR Re Fae We ay é, y), then y=(e,..., ¢, b, a, x); EY ( 6, say C/E (Bye wo ey +(e, .. ., &)(@, ewe ene CESS @)(@, «os » O)=(e,.. =>) Ags naag d) ; ELE tea CA (0 ee ee a) =(¢, 21. ., a)(d, oy (15.) (@.. +) €) — (a,.. » A)=(e.. 9 a) (d, "9 (16.) [a, b, ¢, d, e]=1/(a, b, «, d, €) (b, c, d, e)(e, i e (17.) Prove the following equivalence theorem :-— CIT AS aga eps GE ari Met OSs 0", Seay ee ae aoa a). Sate Wigton a)... (a). (d, e) (0). g Pees ? [a’, e 4 [a, e] fat [@, e][a”, e’] [a’, e i [ae hae [a e”7| ~ fa, aaqil vale i [a, e e'|- WPS: e"\- [a’”, ge fae CT fe [a a” (18.) (a, f, a Lar. a” thes ve Ay ane .) aa iz a! Ped va } a afa' ay en ae a'f'al’ —a’ = A: Chae Mt agit a= Fe: aes reg et eg ee. Oo ey Ae bt mte+tm+ =" 1 1 th } a to ina oe 1 7 M Bees } feasts ae ae} (21...) (0,7. ay Get eager one Sir 9). Osos ere ad co ) (G5 in Re NO TS 2, cies ee eee ee ‘y Oy fC, ss BLOOD =(@,.. . ¢)=(@a eer (22.) Show that the successive constituents a, iy, -. «4 rm, » may be omitted from the continued fraction(. . . a, b, G; By Ys: fay An ee 2) without peter its value, provided 1B, Pee yet leeeioae ene es and y=+[@, . . ., \]; and construct examples. (20.) /2=14 +» ], (23.) If a=(a ee Pah 7, PAnt ihe other root of the quadratic equation to which this leads is x=(f, Cnet dt PE cst CAITR PEE ha. Ue) 1 : by + + ln + na . be one root of a quadratic ¥& “ The notation and the order of ideas used in (14) to (23), as well as some of the special results, are due to Mobius (Crelle’s Jour., 1830). XXXIV CONVERGENCE OF A C.F. ATT equation, the other is 1 Ae! 1 1 i 1 1 bt bm —Am— Amat Om—2+ G+ Omt "a+ | * (Stern, Crelle’s Jour., 1827.) * b+ (25.) If g>p, show that 12 —P pua—p) paq-p)? OU Gon iat ati A lacy cake _Pyg- py 7 275 YB Rea La oo cA eo 8 #y (9-p)q=P -P? CONVERGENCE OF INFINITE CONTINUED FRACTIONS. § 12.] By the value or limit of an infinite continued fraction is meant the limit, if any such exist, towards which the con- vergent p,p/gn approaches when is made infinitely great. It may happen that this limit is finite and definite ; the fraction is then said to be convergent. It may happen that L p,/q, fluctuates Nn=e0 between a certain number of finite values according to the in- tegral character of m ; the fraction is then said to oscillate. Finally, it may happen that L p,/q, tends constantly towards + 0; in N=” this case the fraction is said to be divergent. We have already seen that all simple continued fractions are convergent. eee L The fraction eS is an obvious example of oscillation, its Z ji 1 a) . value being 1, 0, or — © according as n=8m+1, 3m+4+2, or 8m+3. : a} Be aH aw asiae La) dneiraction | -——— —_— --_--__-__., ... diverges to—o , for —— $44/5- 141414 1+ 141+ . converges to—4+4,/5, as may be easily seen from the expression for its nth convergent given in § 9. The last example brings into view a fact which it is important to notice, namely, that the divergence of an infinite continued fraction is something quite different from the divergence of an infinite series. The divergence of the fraction is, in fact, an accidental phenomenon, and will in general disappear if we modify the fraction by omitting a constituent. It is therefore * (23) and (24) are generalisations of an older theorem of Galois’. See Gergonne Ann. d. Math., t. xix. 478 PARTIAL CRITERION FOR C.F. OF FIRST CLASS CHAP, not safe in general to argue that a continued fraction does not diverge because the continued fraction formed by taking all its constituents after a certain order converges. With the exception of simple continued fractions and recur-’ ring continued fractions (whether simple. or not), the only cases where rules of any generality have been found for testing con- vergency are continued fractions of the “first” and “second class.” To these we shall confine ourselves in what follows.* § 13.] A continued fraction of the first class cannot ‘be divergent ; and tt will be convergent or oscillating if any one of the residual STACHONS %, Xz,» . +» Lp, . . . converge or oscillate. The latter part of this proposition is at once obvious from the equation 4 A, + Az + Ln ; = Oy + Ba Pn Again, since (§ 3, Cor. 6) the odd convergents continually in- crease and the even convergents continually decrease, while any even convergent is greater than any following odd convergent, it follows that Lpn/dn =A and Lpyn_1/qon-; = B, where A and Bare two finite quantities, and ACB. If A=B, the fraction is con- vergent ; if A> B, it oscillates ; and no other case can arise. $14.) A continued fraction of the first class is convergent of the Serves LAy-1An/bn be divergent. We have, since all the quantities involved are positive, In = Un Qn-1 + Ostia 3 In-1 = On-19n-3 + On-19n-s; Qn-1 > An-1Yn-s 5 In-2 = An-29n-3 + by a0 as In-2> In-29n-3 5 U4 = UqJ3 + O49, Ya > 9s 5 Y3 = A39o + bs Ys > U34e 5 Jo = Ag. * Our knowledge of the convergence of continued fractions is chiefly due to Schlomilch, Handb. d. Algebraische Analysis (1845) ; Arndt, Disquisitiones Nonnullw de Fractionibus Continuis, Sundie (1845) ; Seidel, Untersuchungen tiber die Convergenz und Divergenz der Kettenbriiche (Habilitationsschrift Miinchen, 1846); also Abhandlungen d. Math. Classe d. K. Bayerischen Akad. d. Wiss., Bd. vii. (1855) ; and Stern, Credle’s Jour., xxxvii. (1848). .. lend XXXIV COMPLETE CRITERION FOR C.F. OF FIRST CLASS 479 . Hence dn > (dae 1% Dn) In-25 Qn-1 > (Gn tas 45 Ga) Ica In-2> CARES 13 by %) Un-4) Qs > (ds + 04) G2, ds = (Ally + b3)q,. Therefore ~— In Gn-1 > 1,920; Ly Ay As) (d, ate As (t,) sae (b,, + GnAiln)s and, since g, = 1, 9. = 4, | YnQn-1 as eo) ( st An-1 ‘n) Tae ee Cl h, 1+ i (a+ ee (1). “= Now, since 24 _14n/bp is divergent, II(1 + d,_,¢,/bp) diverges to + oo (chap. xxvi., § 23), therefore Lg,¢,-,/b.b3... bn = +0. Hence 1, (22 Pex) ae eee bon _ Yon Yen -1 Yon Yen -1 that is, the continued fraction is convergent. Cor. 1. The fraction in question is convergent if Lan —,4n/bn> 0. Cor. 2. Also if Lan/b,>0, and ay, be divergent. Wargo 180 27° Lid, £, 0g! Ono On+y > 1. The above criterion is simple in practice; but it is not complete, inasmuch as it is not proved that oscillation follows if 2An-1%n/bn be convergent. The theorem of next paragraph supplies this defect. §$ 15.] Lf a continued fraction of the first class be reduced to the ) form laa 1 d, ig d, 5 ds, a dl, 2% yore dy, 3 malta (4), so that As Os b, . Ws b, dy = a, se 55? d, = b, ’ rar ) An ln On-s ce dn = UU (9), then it is convergent if at least one of the series d,+d,+d,+... (6) djt+d,t+d;t+... (7) be divergent, oscillating if both these series be convergent. 480 COMPLETE CRITERION FOR C.F. OF FIRST CLASS CHAP. This proposition depends on the following inequalities be- tween the q’s and d’s of the fraction (4) :— 0 e+ 0, + tS +d, (9) ; ip tee | (10). These follow at once from Euler’s law for the formation of the terms in g,, which, in the present case, runs as follows :— Write down d,d,...d, and all the terms that can be formed therefrom by omitting any number of pairs of consecutive d’s. - We thus see that g, contains fewer terms than the product (1 +d,) (1 +d;)...(1+d,); and, since the terms are all positive, (8) follows. Again, in forming the terms of the Ist degree in qn, we can only have letters that stand in odd places in the succession d,d,d,...ds,; hence (9); and (10) is obvious from a similar consideration. To apply this to our present purpose, we observe that, since the numerators are all equal to 1, we have Pon _Pan-1___ 1 (11). Jon don tt | don Yon =i If we suppose d, + 0, neither ¢., nor g.,-, can vanish. Hence, if both Lg, and Lg.,_, be finite, the fraction will oscillate, and if one of them be infinite it will converge. Now, if both the series (6) and (7) converge, the series d,+d,+d,+...+d, will converge; and the product on the right of (8) will be finite when x=. In this case, therefore, both gq, and gon-, will be finite; and the fraction (4) will oscillate. | If the series d,+d,+d,+... diverge, then by (9) Len = 0 , and the fraction (4) will converge. By the same reasoning, if the series d;+d,+d,+... diverge, then the fraction 9 1, d, Tee 6 Ce te OO will converge ; and consequently the fraction (4) will converge. i oe XXXIV EXAMPLES 48] Remark.—We might deduce the criterion of last paragraph from the above. For we have dy ds — A, A,/be, d,ds 5 Ay Ats/ D5, a dy - an = Cae Bal Oa Now, if the series Yd, converge, the series formed by adding together the products of every possible pair of its terms must, by chap. xxx., § 2, converge: @ fortiori, the series Ldn-1dn, that is, Lan —14n/bn, must converge. Hence, if this last series diverge, Sd, cannot converge. 2d, must therefore diverge, since it cannot oscillate, all its terms being positive. Therefore either (6) or (7) must diverge, that is to say, the fraction (4) must converge. Example 1. Consider the fraction 12 22 32 Oe Fe ie _ 2(2n—1)(2n—-3)?.. . 37.1? ~~ (an)(Qn- 2)... 4.2? It may be shown, by the third criterion of chap. xxvi., § 6, Cor. 5, that the series Zdon4, is divergent. Or we may use Stirling’s Theorem. Thus, when 2 is very great, we have very nearly don} = 2(2n ! P/2(n De = 2[ {/(2w2n) (2nfe)?n} / { 2-n(2en) (nJe)" 51, =2/an. Here don+i The convergence of Zd2n41 is therefore comparable with that of 21/n, which is divergent. Hence the continued fraction in question converges. Example 2. a aes Gta+a+ oscillates or converges according as x>1 or +1. Example 3. ical 2+ 344+ Here Lain—14n/[bn= L(n—-1)n/(nt+l)=o, therefore the fraction is convergent. § 16.] There is no comprehensive criterion for the con- vergence of fractions of the second class; but the following theorem embraces a large number of important cases :— If an infinite continued fraction of the second class of the form ee pelo (1) lg — Us — Un — bo a VOL. Il 482 CRITERION FOR C.F. OF SECOND CLASS CHAP, be such that bin, = On F1 (2) for all values of n, it converges to a jinite limit F not greater than unity. Lf the sign > occur at least among the conditions (2), then F <1. If the sign = alone occur, then F=1—1 /S, where S=1+06,4+6,0,+0.b,b,+... + 05054. Opa are (A), so that F= or <1 according as the series in (A) is divergent or convergent. These results follow from the following characteristic pro- | perties of the restricted fraction (1) :— Pn ~ Pn-1 a bb, Oe) On (3); Pn Z 6, + bab, + bbgbt . . . + O,b,... Bp (4); Qn — Wn-1 =U ark (5); inci + b+ Dob; + 01 e+ BOs ee (6); In ~ Pn = Wnsi— Pana b,6;...0n, where the upper sign must be taken if it occur anywhere among the conditions to the right of the vertical line. 7 To prove (4), we have merely to put in (3) n-1, n—2Q, -. + 3 in place of n, adjoin the equation p,=0b,, and add all tae resulting equations. (5) and (6) are established in precisely the same way. It follows, of course, that Pn and q, both remain finite or both become infinite when n= oo, according as the series in (6) is convergent or divergent. =b,+1,; hence the theorem. It is Important to observe that the first > that occurs among the relations ad,20,+1,a,26,+1, . . . determines the first > that occurs among the relations (7): all the signs to the right of this one will be =, all those to the left >. The convergency theorems for the restricted fraction of the second class follow at once. In. the first place, as we have already seen in § 3, the convergents to (1) form an increasing series of positive quantities, so that there can be no oscillation. Also, since gn — n= 1, it follows that Pn/Qn = 1 a 1/¢n (8). Therefore, since ¢,> 1, it follows that F converges to a finite limit +> 1. If the sign > occur at least once among the relations (2), the sign < must be taken in (8); that is, F <1. If the sign = occur throughout, we have Lpn/dn = 1 — L1/g, =1—- 18, where S is the sum to infinity of the series (6). Hence, if (6) converge, F <1; if it diverge, F=1. If we dismiss from our minds the question of convergency, and therefore remove the restriction that 6,, b,, ..., 0b, be Posten Oust putd, — 0, + lydyty =n, tl, vo ., aged,’ + 1, a, =b, + 1, we get by the above reasoning Prl{r=1—-1/dn — . (8’); Vraele-0, (14 0, Os home Ogle s Oye (6’). 484 INCOMMENSURABLE C.FF. CHAP. Now (8’) gives us gn=1/(1—pn/qn). Hence the following remarkable transformation theorem :— Cor. 1. If b,,. . ., by be any quantities whatsoever, then 1+b,+b.6,+. ..+5,0,... 0, Rae b, b, bi, (9) 1.— 64+1- &+1-— by +1 ; from which, putting | u,=0,, w.=.b,0,, .. %, Une tpl we readily derive : L+U+U+.. .+Uy 1 U, Uy U, Us UUs L— 1+, — ty + Ug = Ug + Ug = Wg ge Un = 3Un =i Un = 2Un (1 0), Un—-2 +t Un-1— Un-; + Uy an important theorem of Euler’s to which we shall return presently. INCOMMENSURABILITY OF CERTAIN CONTINUED FRACTIONS. §17.] If a,, ay). » -, On, 0g, 05,. . «, Dy, be all positive integers, then I. The infinite continued fraction AS ‘2 3 nv converges to an incommensurable limit provided that after some finite value of n the condition dnb, be always satisfied. Il. The infinite continued fraction ee (2) converges to an incommensurable limit provided that after some finite value of n the condition a, >b,+ 1 be always satisfied, where the sogn > need not always occur but must occur infinitely often. * To prove II. let us first suppose that the condition An2b,+1 holds from the first. Then (2) converges, by § 16, EEE eee * These theorems are due to Legendre, Eléments de Goémétrie, note iv. XXXIV INCOMMENSURABLE C.FF. 485 to a positive value <1. Let us assume that it converges to a commensurable limit, say A,/A,, where A,, A, are positive integers, and A,> 3. Let now met Ps Os ols 4 an wm ; Since the sign > must occur among the conditions a,= 0, + 1, a,2b,+1,.. ., ps must be a positive quantity <1. Now, by our hypothesis, dy/ dy, — b/d = Ris)> therefore Pz = (dry — 0,A4)/Xg, aa d,/ Ass Say, where A, =4,A, — 0, A, 1s an integer, which must be positive and <,, since p, 1s positive and <1. Next, put en oth O,— As — Then, exactly as before, we can show that p,=A,/A,, where X, 1s a positive integer occurs infinitely often among the conditions Ay = by +1, this process can be repeated as often as we please. The hypothesis that the fraction (2) is commensurable therefore requires the existence of an infinite number of positive integers Meee Ae such that A,> Aj>A,=Ay> . . .; but this.is impossible, since A, is finite. Hence (2) is incommensurable. Next suppose the condition a,20),+1 to hold after =. Then, by what has been shown, b m+ Din sae Yy Se Cm+1 = Om+e 7; is incommensurable. Now we have jie be bs ee ty — 3 — io mn — y Sr \in ele 5. consequently r= (in Y)Pm 1 — °mPm -2 (Gin = Yn ine OmYm—s — Pm — YPm-1 (3) Ym — Y4m-1 ; 486 EULER'S TRANSFORMATION CHAP, where Pm/¢ms Pm-1/Gm-1 ave the ultimate and penultimate con- vergents of MAT ois ieee) ee gd 4 3) ha It results from (3) that Jd = Ong) ae EG, —Pm (4). Now Fam-1-Pm-, and Fgn—pm cannot both be zero, for that would involve the equality Pm Un = Pn 1! Un <1 inconsistent with the equation (2) of § 3. Hence, if F were commensurable, (4) would give a commensurable value for the incommensurable y. F must therefore be incommensurable. The proof of I. is exactly similar, for the condition dnc oe secures that each of the residual fractions of (1) shall be positive and less than unity. These two theorems do not by any means include all cases of incommensurability in convergent infinite continued fractions. heme vd: Pe ae ae converges to the incommensurable value 4/7, and yet violates the condition of Proposition I. Brouncker’s fraction, for example, 1 + ie | CONVERSION OF SERIES AND CONTINUED PRODUCTS INTO CONTINUED FRACTIONS. § 18.] Zo convert the series Uy + Ugh ss. Pilly taeaeas ito an “equivalent” continued fraction of the form q A continued fraction is said to be “equivalent” to a series when the xth convergent of the former is equal to the sum of n terms of the latter for all values of n. Since the convergents merely are given, we may leave the denominators 9,, %, . .., gm arbitrary (we take g,=1, as usual). XXXIV EULER'S TRANSFORMATION A87 For the fraction (1) we have Pnl In — Dn-1/In-1 —_ b, b, te Denar (2); M=%, Go=%:9,-90,, - - -y. In=4nIn-1 — OnQn-s (3); Diln = 51/4 (4). Since | DnlUn = Uy t+ Ugt . . . + Un (5), we get from (2) and (5) Un = b, b. as § Diner as Un—-1 = bb, ¢ehp aac ae Pea Uz = b, b./0, 923 Ue Uf 0h. From (6), by using successive pairs of the equations, we get b, =U, b, = JoU/Uty bs = dsUs/Qy Ug, + + 2 Ox = UnUn/In-2Un-1 (7). Combining (3) with (7), we also find =, Wes aU aE Us) [qi Ur, Co Q3( Uz 3 Us)/GoUo, re so Un = dha ay DTG mei vse (8). Hence a Ug ts. Uns hy Gott/'U, Js Us/ Qi Ue Gy — Galt, + Us) (Grey — Ja(Uy + Us) [Gog — °° InUnl Ino Un ; Onl Uae + Un Gn sting It will be observed that the g’s may be cleared out of the fraction. Thus, for example, we get rid of g, by multiplying the first and second numerators and the first denominator by 1/g,, and the second and third numerators and the second denominator by g,; and so on. We thus get for 8, the equivalent fraction oe Uy Uy Uy UW,/ Ug hae Un] Un 1 (10), 1 — (u, + tp)/t — (Uy + Ug)/ty — (thy + Un) [Un which may be thrown into the form Uy, Ug UUs Un —-oUn L— U+Ue— UgtUg—- ~ Un-, + Un Ss (11). 488 EXAMPLES—-BROUNCKER’S FRACTION CHAP. This formula is practically the same as the one obtained incidentally in § 16; it was first given, along with many applica- tions, by Euler in his memoir, “De Transformatione Serierum in Fractiones Continuas,” Opuscula Analytica, t. ii. (1785). It is important to remark that, since the continued fraction (10) or (11) is equivalent to the series, it must converge if the series converges, and that to the same limit. By giving to w,, %,. . ., Mm various values, and modifying the fraction by introducing multipliers as above, we can deduce a variety of results, among which the following are specially useful :— 2 n VU UG +. . . + Unt ke VX VU,0 Up stat (12); — ; » 6 » al 5) 2 x hd ae + a... +5 . ee + a V, Un x VX Vex Un TX (13) V— UL+%-— ULEV— ~~" " Wy _ 12+ Uy 7 ey athe ri +e ae Th Ti b, bb, ei ean: _ a, he b,a,2 bn Un & (14) Example 1. If -47 —-1, ole < nee 1(m — 1) 2(m — 2) 3(m— 8) Ns: 1- m+1—- m+1l- m4+1- and, if m>0, _ m 1(m—1) 2(m— 2) 8(m - 8) 1+ 3-m+ 5-m+ 7-m+ — Vest $ 19.] The analysis of last paragraph enables us to construct a continued fraction, say of the form (1), whose first n convergents shall be any given quantities f,, fo, . . .5 fn respectively. : All we have to do is to replace w,, %, . . ., U, in (10) or Dy fe fi; ~~ + In —Jn-1 respectively. The required fraction is, therefore, fi fa “fi fils — fo) (fs =Ji) (fi zis) ; te fa - ih ag be Veolia Cie = ns) (Fn a) cay I n = Ins Cor. Hence we can express any continued product, say A) ee as a continued fraction. We have merely to put f,=d,/e,, fa=d,d./e,¢, . . ., effect some obvious reductions, and we find _ a, & (d,—€,) dyl:(ds— 5) dses(da— €2) (ds &4) Ays(ds- €3) (Us— es) z Ci d, hia d, ds— 63 03— ds dy— €3€4— dy, ds = Cys — SOs GP or = cae) (do, ri ex) 3 2 LG) dy, dy — Cn— ren ( ) Pr § 20.] Instead of requiring that the continued fraction be equivalent to the series, or to the function /(n, 7), which it is to represent, we may require that the sum to infinity of the series (or f(x", z)) be reduced to a fraction of a given form, say ‘ee eae where f,, B,, . - -» By are all independent of 2. There is a process, originally given in Lambert’s Beytrdge * A similar formula, given by Stern, Cred/e’s Jowr., x., p. 267 (1833), may be obtained by a slight modification of the above process. 490 LAMBERT’S TRANSFORMATION CHAP. (th. i, p. 75), for reducing to the form (1) the quotient of two con- vergent serves, say F(1, x)/F(0, 2). We suppose that the absolute terms of F(1, z) and F(0, x) do not vanish, and, for simplicity, we take each of these terms to be 1. Then we can establish an equation of the form EC @) 7 F(0, a“) = 2, 2F(2, 7) (2), where F(2, 2) is a convergent series whose absolute term we suppose again not to vanish, and B, is the coefficient of z in F(1, x) — F(0, ), which also is supposed not to vanish.* In like manner we establish the series of equations F(2, x) i FYI, x) a B.xF(3, «) (25), F(3, x) — F(2, x) = B,xF(4, «) (25), F(n + 1, 2) — F(n, 2) = Bn4,2F(n + 2, 2) (ara) Let us, in the meantime, suppose that none of the functions F' becomes 0 for the value of x in question. We may then put G(n, x) = F(n + 1, x)/F(a, a) (3), where G(n, x) is a definite function of » and x which becomes neither 0 nor for the value of «in question. The equation (2,,1,) may now be written G(n, ) -— 1 = Bas, 2G(n + 1, a)G(n, x), that is, G(n, )=1/{1 — Bn4.eG(n + 1, 2)} (4), If in (4) we put successively n=0,n=1,. . . we derive the following :— es sls Pra Bnet B\. Ce) a ere TC ayeGaes (5) ; ae 1 a Bnti® : Bn+m% (6) Gin, a) 1- ~1-(1-1/G(n+m, 2)) ; “ The vanishing of one or more of these coefficients would lead to a more general form than (1), namely, Box”? Byar 1- l- General expressions have been found for By, Bi, . . . by Heilermann, Credle’s Jour. (1846), and by Muir, Proc. L.M.S. (1876). XXXIV LAMBERT’S TRANSFORMATION 49] In order that we may be able to assert the equality Be ee fade (7), it is necessary, and it is sufficient, that it be possible by making m sufficiently great to cause 1 —1/G(n, x) to differ from the mth convergent of the residual fraction Bnti% Bn+2% Pn+-m® ‘ kd ee (8) by as little as we please. -Let us denote the convergents of (8) by p,/q, peo/qe, - - Pm|Im Then, from (6), we see that {1 — 1/G(n, )} — Pin/m Pm —Pm-111 — 1/G(n +m, 2)} Pm Ym — Gm -141 is 1/G(n + mM, a) Ym x {1 26 1/G(n + mM, 2)}( Pin| Gn =Din—1|Im-1) (9) Gm/Im—-1— 11 = 1/G(n + m, x)} ~ {1 = 1/G(n +m, 2)}Bn4iBnte--- B+me™ Yml9m — Im—-1{1 — 1/G(n + m, 2) The necessary and sufficient condition for the subsistence of (7) is, therefore, that the right-hand side of (9), or of (10), shall vanish when m= co. Concerning these conditions it should be remarked that while either of them secures the convergence of the infinite continued fraction in (7), the convergence of the fraction is not necessarily by itself a sufficient condition for the subsistence of the equation (7). In what precedes we have supposed that none of the func- tions F(n, ~) vanish. This restriction may be partly removed. It is obvious that no two consecutive F'’s can vanish, for then (by the equations (2)) all the preceding F’s would vanish, and G(0, 2) would not be determinate. Suppose, however, that F(r + 1, a’) = 0, so that G(r, z’)=0; then (5) furnishes for G(0, a’) the closed continued fraction \ ee 1 Ba Bee G(0, “=~ — (10). \ 499 EXAMPLE CHAP, In order that this may be identical with the value given by (7), it is necessary and sufficient that G(r +1, 2’), as given by (6), should become o, that is, it is necessary and sufficient that the residual fraction By +o% Prist Lets] ee should converge to 1; but this. condition will in general be satisfied if the relation (4) subsist for all values of n, and the condition (9) be also satisfied when nr + 2. .ad a § 21.] As an example of the process of last paragraph, let 9 cz 4 Ve Bn, YTS Ga) SIGS AY G cee > a agelae Then i) — = {22 Se 7 ? Qf f vei a) F(n, a) Gan) (a ae , &) ( ue an G(n, a) = 12 -- GED Gass + Al, »} (4’), where G(n, x) = F(n + 1, a)/F(n, 2). Hence | (0, «) = 1 ayy + 1) x/(y a 1) yr 2) al(y +n—-1)(y +n) fe Lt ec (5') ; and 3 . en 1 _ _ ay +n) (y+n+1) G(n, 2) 1+ to hg aly +n+m—1)(y+n+m) (6’ 1-{1-1/G(n+m, z)} The series (11) will be convergent for all finite values of g, and for all positive integral values of n, including 0, provided y be not 0 or a negative integer. Hence we have obviously, for all finite values of 2, LG(n +m, z)=1 when m=. Let us suppose that a is positive. Then the residual con- tinued fraction | XXXIV EXAMPLE 493 aly +n) (y +n+ 1) a/(y +n +1) (y+ + 2) es eo ie aly +n+m—-1)(y+n+m) 1+ (ie is (by the criterion of § 14) evidently convergent. Hence the factor Pin/Gm—Pm-1/Gm-, in the expression (9) vanishes when i207; Also, since the q’s continually increase, L¢m/q¢m_,< 1. Therefore we may continue the fraction to infinity when « is positive. Next suppose x negative, = — y say ; we then have yy G(0,-g)=_ Hrs) GVO?) yy +n-1) (y +1) (5”) ; 1 — {1 - 1/G(n, - y)} ; and fee Ny ee) y +0 +11) CT) a y/(y +m +m —-1)(y +n +m) 6") 1 - {1-1/G(n, +m, - y)} The fraction (8) in this case is “equivalent” to l { y y yy \ ytnl(ytn+1l—- y+n+2- vy tnt Mm — J 8"), which is obviously convergent (by § 16), if y have any finite value whatever. Hence the factor Pm/¢m—m- /dm-, belonging to the equivalent fraction (8) must vanish. Again, by § 2 (6), Im-1 Sy yy +n +m—1)(y+n+m) yf(y +n +m—2)(y+n+m-1) Wy +m) Cy +04 I) . 1 ees || Tike A y ey y+ntm\|y+u+m—-1- ytn+m—-2-—- °° ytny (12). 494 C.FF. FOR TAN &@ AND TANH& CHAP. If only x be taken large enough, the fraction inside the brackets satisfies the condition of § 16 throughout : its value is therefore <1, however great m may be; and it follows from (12) that Lam/dm-1=1 when m= o. | Since LG(n +m, - y)=1 when m= , it follows that all the requisite conditions are fulfilled in the present case also. We have thus shown that 1 a) iE aly(y +1) al(y + 1)(y + 2) 28 BLO.) el adit 1+ ACA ENG) 50 .s adtegs (13), I whence, by an obvious reduction, FQ, 2) _ _ = e (14) HO, 2) y+ ytl+ yt24+ °° yene a result which holds for all finite real values of x, except such as render F(0, x) zero,* and for all values of y, except zero and negative integers. : If we put +2*/4 in place of z in the functions F(0, z) and F(1, x), and at the same time put y = 4, we get F(0, — 2°/4)= cosa, (1, —a*/4) =sin i: H(0, a'/4) =cosha, (1, 27/4) = sinh a/z. Cor. 1. Hence, from (14), we get at once gg? x t (= eee os Lie erp L433. 95= Pye ap (15); go . x tanh # = 2. 16). = 1+ 385 => 2n+1 + ( ) Cor. 2. The numerical constants x and x? are imcommensurable, For, if + were commensurable, z/4 would be commensurable, say =A/p. Hence we should have, by (15), * In a sense it will hold even then, for the fraction 1 { ts x } ey grey Parse as % which represents F(0, «)/F(1, «) will converge to 0. Of course, two consecu- tive functions F(n, x), F (x+1, ~) cannot vanish for the same value of CMe otherwise we should have F(co , «)=0, which is impossible, since F(o, ip) =A; * ———— XXXIV INCOMMENSURABILITY OF 7 AND e 495 Pe Ue Sie tall g tAi/ fel! ae eee te Rees] i age r — “ree 3 aie oes (Ona ab Ba vad Be (ea) pe Bp 5p (20 + 1) ) Now, since A and p are fixed finite integers, if we take n large enough we shall have (2n+1)u>.’+1. Hence, by § 17, the fraction in (17) converges to an incommensurable limit, which is impossible since 1 is commensurable. That z* is also incommensurable follows in like manner very readily from (15). By using (16) in a similar way we can easily show that Cor. 3. Any commensurable power of ¢ is incommensurable.* § 22.] The development of last paragraph is in reality a particular case of the following general theorem regarding the hypergeometric series, given by Gauss in his classical memoir on that subject (1812)+ :— If ss a fp ; ala + 1)6(8 = I)ey F(a, B, y, %)=1+ ieee 1.9 barly el vin ate and G(a, B, $6: x) = F(a, B+ 1, Ps + 1, x)/F(a, p, Y> x), then +. 1 Bx Bx Bon OT am 1- 1-°°' 1/Ga+n, B+n, y + 2n) (18), where . 9 == 2) g, B+ DG 1-0) ‘yy +1)’ ee Lie?) g, + Ny +1-B) g, B+) 42-0 (y+ 2)(y +3)” eet Ae B Pei +8-1-8) emcee) Pho) i (y 9 — 2) (y +9n—- yas ais (y + 2n—- 1)(y Ae 2n) * The results of this paragraph were first given by Lambert in a memoir which is very important in the history of continued fractions (Hist. d. 7. Ae. Roy. d, Berlin, 1761). The arrangement of the analysis is taken from Legendre (l.c.), the general idea of the discussion of the convergence of the fraction from Schlimilch. t+ Werke, Bad. iii., p. 134. 496 GAUSS’S 0.F. FOR HYPERGEOMETRIC SERIES CHAP. After what has been done, the proof of this theorem should present no difficulty. The discussion of the question of convergence is also com- paratively simple when a is positive ; but presents some difficulty in the case where z is negative. In fact, we are not aware that any complete elementary discussion of this latter point has been given. Cor. If im (18) we put B= 0, and write y — 1 in place of y, we get the transformation: get aa + I) 2 ala + I)(a + 2) is 710 maa OAC dual 916 daae)) _1 Be Bex SS where pe ae bes py y’ B, yy +1) (a + l)y ¥ 2(y + 1 - a) a G+ DG +3) Pe G+ Ny + 8)’ steed (a+n—I1)(y+n-2) ny +n— 1 —a) “(y+ 2n—8)(y+an—2) ° on = oye Gauss’s Theorem is a very general one; for the hypergeometric series includes nearly all the ordinary elementary series. Thus, for example, we have, as the reader may easily verify, (l+2)"=K(-—™m, B, B, -2); log (1 + 2) =2F(1, 1, 2, -2); sinha=a7 L L E(k, hk, $a j4heye snl =P sin =e lee ele Bee seg) k=o0 kK =o0 sin-*¢ = «F(, td, 3, 2’); =2/(1 =a) FQ, 1, 8 22); tan-4¢=aF(4, 1, 3, — 2’). XXXIV EXERCISES XXXIV 497 EXERCISES MXXLY, Examine the convergence of the following :— 1 1 1 12 92 32 Be TPE PE P+ CaM Ee se ra = 14917, 2%.27:8? De Leama eee ie te Caos ey ie Lei ee a * », m2 (m+n)? (m4+2n)P _ nk leks dl 6. UC On SS eT 1+ 1+ ( ee m+ n+ Ape 8: (eae sl wey: (7.) & 305 Sepa (COE Tat orerh rte res (ony ae, LES 2h ob doyess a 25 Tee lpi Plea 1 me bf (11.) Show that the fraction of the second class, a -= - g— a3 — verges to a positive limit if, for all values of , g/d, bo+ a3] b2 bs GPG 0 + Ant4/bn On+4 opie (Stern, Gétt. Nach., 1845.) (12:) Show that Pa a = . . +, Where @,>0, converges if dn41+ An+1. tg (18.) Show that the series of fractions (~n—Pn-1)/(Yr-Qm—1) forms a descending series of convergents to the infinite continued fraction of the second class, provided a@>b,+1, and the sign > occurs at least once among these conditions. (14.) Show that z e eo pol ele o-l ee where «> 0, is equal to x or 1 according as w Be agiions Sees Se Oh) Tae ne 2 2 2 (05. \ aaah eae (26) 122 eee 1+14+14+1+ 2+384+ 4454 (27.) Show that tan 2 and tanhz are incommensurable if 2 be commen- surable. Establish the following transformations :— (28.) «= _! oe © © ew & HE 12 2 2 2, 2, The. 2+ 34+ 44+ 54+ 64 ai (30.) tana= 1%? 2x? Bia 1+ 34+ 54+ 74° °° tahoe 1747 D242 3272 \.3 5 ah eee tanz (n?—1?)tan 2x (n?— 2?)tan Pat ict ee ROY a) Oe ial act ipsa (31.) tan nx Te ae Bie . (Euler, Mem. Acad. Pet., 1813.) (32. ) sith Cia Npay 008 @ ae ae dice sin nx 2cos%— 2cosx— where there are » partial quotients. (33.) If p(a, B, y, x) =14°-¢-», ,@- nig -n@P-ng-1 (q-1)(q"-1) — (q- De ~1)(q” -1)(q”** -1) g(a, B+1, yell Bix Box p(a, B, Y a) 1 ied then - ale Le ¥ 4 7 a z + Te Pee, > ee XXXIV EXERCISES XXXIV 499 where PP -yg*=1) trai Bar Ere 1)(q y+2r 244 J pee | Ag O51) fist Md 2r-+- (eae aa eon t Ls com Crelle’s Jour., XXxil.) (34.) Show that a= {a-1+7 os a } Zz 2(a—1)+ 2(a—1)+ 2(@-1)+ °° 1 Be Sd | 1 ee rie x {at + Sa+l)+ e+) + Xarl)+ Wallis (see Muir, Phil. Mag., 1877). CHAPTER: XXXV. General Properties of Integral Numbers. NUMBERS WHICH ARE CONGRUENT WITH RESPECT TO A GIVEN MODULUS. § 1.] Lf m be any positive integer whatever, which we call the modulus, two integers, M and N, which leave the same remainder when divided by m are said to be congruent with respect to the modulus m.* In other words, if M = pm+7,-and N = qm +7, M and N are said to be congruent with respect to the modulus m. Gauss, who made the notion of congruence the fundamental idea in his famous Disquisitiones Arithmetice, uses for this relation between M and N the symbolism M=N(mod m) ; or simply M=N, if there is no doubt about the modulus, and no danger of con- fusion with the use of = to denote algebraical identity. Cor. 1. Lf two numbers M and N be congruent with respect to modulus m, then they differ by a multiple of m ; so that we have, say, M=N +pm. Cor. 2. If either M or N have any factor in common with mM, then the other must also have that factor ; and if either be prime tom, | the other must be prime to m also. In the present chapter we shall use only the most elementary consequences of the theory of congruent numbers. * To save repetition, let it be understood, when nothing else is indicated, that throughout this chapter every letter stands for a positive or negative integer. CHAP. XXXV PERIODICITY OF INTEGERS 501 Our object here is simply to give the reader a conspectus of the more elementary methods of demonstration which are employed in establishing properties of integral numbers ; and to illustrate these methods by proving some of the elementary theorems which he is likely to meet with in an ordinary course of mathematical study. Further developments must be sought for in special treatises on the theory of numbers. § 2.] If we select any “modulus” m, then it follows, from chap. ii., § 11, that all integral numbers can be arranged into successive groups of m, such that each of the integers in one of these groups is congruent with one and with one only of the set Dee ee oy Ait 2). (7 — 1) (A), or, if we choose, of the set OR ae ist oo, 1 (B), where there are m integers. Another way of expressing the above is to say that, if we take any m consecutive mtegers whatever, and divide them by m, their remainders taken im order will be a cyclical permutation of the integers (A). Example. If we take m=5, the set (A) is 0, 1, 2, 8,4. Nowif we take the 5 consecutive integers 63, 64, 65, 66, 67 and divide them by 5, the remainders are 3, 4, 0, 1, 2, which is a cyclical permutation of 0, 1, 2, 3, 4. § 3.] A large number of curious properties of integral numbers can be directly deduced from the simple principle of classification just explained. Example 1. Every integer which is a perfect cube is of the form 7p, or 7p+1. Bearing in mind that every integer N has one or other of the forms (i, TLL, Ree eS, ~ also that (7mxr)? = (7m) £3(7m)r + 3(7m)r+£r°, = (72m 121 mr + 8mr2)7 £73, = Myce, we see that in the four possible cases we have Ne (imme me NE Sa forte) Pty ibe Ne=(7m2)%, —M7+8=(M+1)7+1; Ne=(/3) =(M24)7=— 1. 502 EXAMPLES CHAP. _In every case, therefore, the cube has one or other of the forms 7p or (p21. Example 2. Prove that 3°”+1+4 2+ is divisible by 7 (Wolstenholme). We have Bent 4 Ont — (7 — 4)antl 4 ont2, Now (see above, Example 1, or below, § 4) (7 —4)t1—= M7 — 42nt1, Hence Santi 4 Qet8— M7 — 42ntl 4 ont2, = M7 — 2"+°(28" — 1), But 23”—1 is divisible by 23-1 (see chap. v., § 17), that is, by 7. Hence geta(2" — 1) == NT, Finally, therefore, g2ntl + Qnt+2—(M—N)7, which proves the theorem. Example 3, The product of 3 successive integers is always divisible by 1 yess Let the product in question be m(m +1) (m+). Then, since ™ must have one or other of the three forms, 3m, 3m-+1, 83m-—1, we have the following cases to consider :— 37(3m + 1) (8m + 2) (1); (3m +1) (8m +2) (3m +8) (2); (8m —1)3m(3m-+1) (3). In (1) the proposition is at once evident; for 3m is divisible by 3, and (3m +1) (8m+2) by 2. The same is true in (2). In case (3) we have to show that (3m —1)m(3m+1) is divisible by 2. Now this must be so; because, if m is even, m is divisible by 2; and if m be odd, both 3m—-1 and 8m+1 are even; that is, both 3m—1 and 3m+1 are divisible by 2. In all cases, therefore, the theorem holds. Example 4. To show that the product of p successive integers is always divisible by 1.2.8... », Let us suppose that it has been shown, Ist, that the product of any p-1 successive integers whatever is divisible by 1.2.3. . . p—1; 2nd, that the product of p successive integers beginning with any integer up to x is divisible Dy sl aot ite poy P real Consider the product of p successive integers beginning with +1. We have (w+1)(w@+2)...(a+p-1) (a+p) =p(%+1) (+2)... (@+p—1)+a(a+1) (#+2)... (+p 1ho" Se Now, by our first supposition, (@+1)(w+2)... (a+ p-—1) is divisible by 1.2... p-—1; and, by our second, x(a+1)(a+2)... (x+p-—1) is divisible Len gee DR as ori oi Hence each member on the right of (1) is divisible by 1.2.3 seen It follows, therefore, that, if our two suppositions be right, then the pro- duct of p successive integers beginning with «+1 is divisible byl. 273 ee en But we have shown in Example 3 that the product of 3 consecutive integers is always divisible by 1.2.3; and it is self-evident that the product of 4 con- XXXV PYTHAGOREAN PROBLEM 503 secutive integers beginning with 1 is divisible by 1.2.3.4. It follows, there- fore, that the product of 4 consecutive integers beginning with 2 is divisible by 1.2.3.4. Using Example 3 again, and the result just established, we prove that 4 consecutive integers beginning with 8 is divisible by 1.2.3.4; and thus we finally establish that the product of any 4 consecutive integers whatever is divisible by 1.2.3.4. Proceeding in exactly the same way, we next show that our theorem holds when p=5; andsoon. Hence it holds generally. This demonstration is a good example of ‘‘ mathematical induction.” Example 5. Ifa, 0, ¢ be three integers such that a?+6?=c?, then they are represented in the most general way possible by the forms a=(m?—n?), b=2\mn, c=r(m?2+7n?), First of all, it is obvious, on account of the relation a?+6?=c?, that, if any two of the numbers have a common factor A, then that factor must occur in the other also ; so that we may write a=)a’, bD=Nb', c=dc’, where a’, 0’, c’ are prime to each other, and we have | a+ b7=c¢" (1). No two of the three, a’, b’, c’, therefore, can be even ; also both a’ and 6’ cannot be odd, for then a+ 6” would be of the form 4n+2, which is an im- possible form for the number c”. It appears, then, that one of the two, a’, b’, say b’ (=28), must be even, and that a’ and c’ must be odd. Hence (c’ +a’)/2 and (c’ —a’)/2 must be integers ; and these integers must be prime to each other ; for, if they had a common factor, it must divide their sum which is ¢’ and their difference which is a’ ; but c’ and a’ have by hypothesis no common factor. Now we have from (1) +a’ ef —a'\) 4, Greer) 2) Therefore, since (c’+a’)/2 is prime to (c’—a’)/2, each of these must be a perfect square ; so that we must have whence am (3), Fan? (4), =mn (5), where m is prime to 2. From (3) and (4), we have, by subtraction and addition, a’=m?—n?, c=m'+n?; and, from (5), b'=28=2mn. Returning, therefore, to our original case, we must have generally a=Nm2—n?), b=2Qdmn, c=m?+n?). This is the complete analytical solution of the famous Pythagorean problem—to find a right-angled triangle whose sides shall be commensurable. 504 PROPERTY OF AN INTEGRAL FUNCTION CHAP § 4.] The following theorem may be deduced very readily from the principles of § 2. Let f(z) stand for Pot PrU + po% + - +Pn&", where ~, Pi, .~. +, Mm are positive or negative integers, and z any positive integer ; then, if x be congruent with r with respect to the modulus m, I(x) will be congruent with f(r) with respect to modulus m. By the binomial expansion, we have m+ ry" = (qm)™ + nC,(qm)?— r+... + Cn .(gm)r™-1 4 9m q q q = (grm™-14 ,O,q* lm 2r 4+. + nCn-1gr™ Dm + 7%, =M,m+r"; where M,, is some integer, since all the numbers nis nee nUn-, are, by § 3, Example 4, or by their law of formation (see chap. iv., § 14) necessarily integers. Similarly (gm + 7)"-1 = M,_ m+ 70-1 Hence, if «= gm +7, I(®) = Pot PT + Dr +. . + Dy + (p,M,+p,M,+...+ PrMy)m, = f(r) + Mm. Hence f(z) is congruent with /(r) with respect to modulus m. Cor. 1. Since all integers are congruent (with respect to modulus m) with one or other of the series Oe eee caer Ls it follows that to test the divisibility of f(x) by m for all integral values of x, we need only test the divisibility by m of f(0), f(1), I(2), Premera) (it <1), Example 1. Let /(7) =a(u+ 1)(2e+1); and let it be required to find when J(x) is divisible by 6. We have 7/(0)=0, /(1) =6, J(2)=30, 7(3)=84, 7(4)=180, J(5)=3830. Each of these is divisible by 6; and every integer is congruent (mod 6) with one of the six numbers 0, 1, 2, 3, 4,5; hence x(x +1) (2e+1) is always divisible by 6. Cor. 2. f{q f(r) + r} as always divisible by f(r); for Sighr) +r} - = Mf(r) + f(r) = (M+ 1)f0). Hence an infinite number of values of « can always be found which will make f(x) a composite number. ow XXXV DIFFERENCE TEST OF DIVISIBILITY 5O5D This result is sometimes stated by saying that no integral function of x can furmsh prime numbers only. Example 2. Show that a*-1 is divisible by 5 if x be prime to 5, but not otherwise. With modulus 5 all integral values of are congruent with 0, £1, +2. i fiej—2—1, 7(0)=—1, f(=1)=0, f(2)=15. —Now 0 and 15 are each divisible by 5; but —1 is not divisible by 5. Hence a1 is divisible by 5 when z is prime to 5, but not otherwise. Example 3. To show that 2?+a+17 is not divisible by any number less than 17, and that it is divisible by 17 when and only when is of the form 17m or 17m —1. Here J(0)=17, A(+1)=19, f(+2)=28, f(+3)=29, f(+4)=37, f(+5)=47, f(+6)=59, A+7)=73, fl+8)=89, fl-1)=17, A-2)=19, f(-3)=23, f(-4)=29, f(-5)=37, f(-6)=47, f(-7)=59, f(-8)=73. These numbers are all primes, hence no number less than 17 will divide x“*+a+17, whatever the value of 2 may be; and 17 will do so only when x=ml17/ or «=mi17 - 1. § 5.| Method of Differences.—'There is another method for testing the divisibility of integral functions, which may be given here, although it belongs, strictly speaking, to an order of ideas somewhat different from that which we are now following. Let f,(z) denote an integral function of the nth degree. Trl@ +1) —fnl(e) =p t+ pletrl)+.. .+pn-(e +1)" + pyle + 1)” Pee Del oa On — Oye (1). Now on the right-hand side the highest power of a, namely a", disappears ; and the whole becomes an integral function of the n—1th degree, f,-,(z), say. Thus, if m be the divisor, we have IAG ur 1) — Fn) = Jn-i{@) (2). Mm 1% It may happen that the question of divisibility can be at once settled for the simpler function f,,,(z). Suppose, for example, that it turns out that f,-,(z) is always divisible by m, whatever « may be; then /,(a+1)—//,(x) is always divisible by m, whatever « may be. Suppose, further, that 7,,(0) is divisible by m; then, since f,(1) —/,,(0), as we have just seen, is divisible by m, it follows that f,(1) is divisible by m. Similarly, it may be shown that /,(2) is divisible by m; and so on. 506 : EXERCISES XXXV CHAP. - If the divisibility or non-divisibility of Jn-\(@) be not at once evident, we,may proceed with Jn-(%) as we did before with Snr{#), and make the question depend on a function of still lower degree ; and so on. Example 1. /5(x)=2> — a is always divisible by 5. Jo(% + 1) — fol) = (e+ 1) —(@+1)-a +2, = 5a*+ 10x? + 10x? + 5a, = M5. Now ‘Ts 10, therefore S52) —75(1) =Mo5, and JFs(2) =M5. Similarly, 75(8) —f5(2) =M)5, therefore Js(3) =(Mo+My)5 ; and-so on. Thus we prove that /;(1), f5(2), Jx(3), &c., are all divisible by 5; in other words, that ~°—« is always divisible by 5. EXERCISES XXXYV. (1.) The sum of two odd squares cannot be a square. (2.) Every prime greater than 3 is of the form 6n+1. (3.) ( 4.) Every integer of the form 42-1 which is not prime has an odd number of factors of the form 4n—-1. (5.) Every prime greater than 5 has the form 30m+n, where n=1, 7, 11, 13, 17, 19, 23, or 29. (6.) The square of every prime greater than 3 is of the form 24m +1 ; and the. square of every integer which is not divisible by 2 or 3 is of the same form. (7.) If two odd primes differ by a power of 2, their sum is a multiple of 3. (8.) The difference of the squares of any two odd primes is divisible by 24. (9.) None of the forms (3m +2)n?+3, 4ann—m-— 1, 4mn—m—n can repre- sent a square integer. (Goldbach and Euler.) (10.) The nth power of an odd number greater than unity can be presented as the difference of two square numbers in » different ways. (11.) If N differ from the two successive squares between which it lies by wand y respectively, prove that N-ay isa square, ‘ (12.) The cube of every rational number is the difference of the squares of two rational numbers. (18.) Any uneven cube, n3, is the sum of » consecutive uneven numbers, of which n? is the middle one. (14.) There can always be found n consecutive integers, each of which is not a prime, however great n may be, tip 53 a ia XXXV EXERCISES XXXV 507 (15.) In the scale of 7 every square integer must have 0, 1, 2, or 4 for its unit digit. (16.) The scale in which 34 denotes a square integer has a radix of the form n(3n +4) or (7+ 2) (82+ 2). (17.) There cannot in any scale be found three different digits ‘such that the three integers formed by placing each digit differently in each integer shall be in Arithmetical Progression, unless the radix of the scale be of the form 3p+1. If this condition be satisfied, there are 2(—1) such sets of digits ; and the common difference of the A.P. is the same in all cases. (18.) If a>2, x*— 4a + 5x7 - 2x is divisible by 12. (19.) 2/5 +-a:4/2 + 03/3 — 7/30, and «9/6 + #°/2+ 5a4/12~ 27/12 are both in- tegral for all integral values of x. (20.) If x, y, 2 be three consecutive integers, (Zx)?— 82x? is divisible by 108. (21.) «—a is divisible by 6. (22.) Find the form of x in order that z*+1 may be divisible by 17. (23.) Examine how far the forms z?+#+41, 2z?+29 represent prime numbers. (24.) Find the least value of x for which 2*-1 is divisible by 47. (25.) Find the least value of x for which 2% —1 is divisible by 23. (26.) Find the values of « and y for which 7*—y is divisible by 22. (27.) Show that the remainder of 22°*”+1 with respect to 2°°+1 is 2. (28.) 37% ~ 2°4 is divisible by 5, if x~y=2. (29.) Show that 27+141 is always divisible by 3. (30.) 4941+ 994141] is divisible by 7. (31.) at”+22"+41 never represents a prime unless x=0 or e=1. (32.) If P be prime and =a?+06?, show that P” can be resolved into the sum of two squares in $n ways or $(n+1) ways, according as 7 is even or odd, and give one of these resolutions. (33.) If 2+y?=22, a, y, z being integers, then wyz=0 (mod 60); and if x be prime and >3, y=0(mod12). Show also that one of the three numbers = 0 (mod 5). (34.) The solution in integers of x?+y?=2z" can be deduced from that of a? +y?=2. Hence, or otherwise, find the two lowest solutions in integers of the first of these equations. (35.) If the equation z?+y?=z* had an integral solution, show that one of the three x, y, z must be of the form 7m, and one of the form 3m. (36.) The area of a right-angled triangle with commensurable sides cannot be a square number. (37.) The sum of two integral fourth powers cannot be an integral square. (38.) Show that (3+ 4/5)* + (3 — 4/5)* is divisible by 2*. (39.) If 2 be any odd integer, not divisible by 3, prove that the integral part of 4% -(2+/2)* is a multiple of 112. (40.) If m be odd, show that 1+,Cuyt+tnCg+nCigt+.. . is divisible by 2(n — 8)/2 5O8 LIMIT AND SCHEME FOR DIVISORS oF N CHAP, ON THE DIVISORS OF A GIVEN INTEGER. § 6.] We have already seen (chap. iii, § 7 ) that every com- posite integer N can be represented in the form alc’. . .» where a, b,¢,... are primes. If N be a perfect square, all the indices must be even, and we have N =a2”J2"¢2"... ; so that VN=a"leo’... In this case N is divisible by VN. If N be not a perfect square, then one at least of the indices must be odd; and we have, say, Neg pee | =o ic?) a tee = so that N is divisible by a*b%c”..., which is obviously less than VN. Hence Every composite number has a factor which is not greater than its square root. This proposition is useful as a guide in finding the least factors of large numbers. This has been done, once for all, in a systematic, but more or less tentative, manner, and the results published for the first nine million integers in the Factor Tables of Burckhard, Dase, and the British Association. * § 7.] The divisors of any given number N =a*d°e’... are all of the form a*l°c”..., where a’, B', y',. .. may have any values from 0 up to a, from 0 up to f, from 0 up to y,.. . respectively. Hence, if we include 1 and N itself among the divisors, the divisors of N =a"'c’... are the various terms obtained by distributing the product } (l+asiott. 2. 4a") x(1+b4+0+... 48%) «(LHC Fightin aimee) io Soi oh (1). “ For an interesting account of the construction and use of these tables, see J. W. L. Glaisher’s Report,. Rep. Brit. Assoc. (1877). XXXV SUM AND NUMBER OF FACTORS 509 Cor 1. Since a+l im 1 l+a+a + +0" = ae : eel ARS ie. Sar mts Pere cu: a Be and so on, It follows that the sum of the divisors of N = a*b'c’. .. is (a*** — 1)(**—1)... (a—1)(b-1)... If in (1) we puta=1, b=1,c=1, . . ., each divisor, that is, each term of the distributed product, becomes unity ; and the sum of the whole is simply the number of the different divisors. Hence, since there are «+1 terms in the first bracket, 8 +1 in the second, and so on, it follows that | Cor. 2. The number of the divisors of N=arl8er... is (a+1)(6+1)(y+1)... Cor. 3. The number of ways in which* N =a*becev... can be resolved into two factors is ${1+(a+1)(B+1)(y+1)...}, or a(a+1)(8+1)(y+1)..., according as N is or is not a square number. For every factor has a complementary factor, that is to say, every factorisation corresponds to two divisors; unless N be a square number, and then one factor, namely VN, has itself for complementary factor, and therefore the factorisation N= VN x WN corresponds to only one divisor. Cor. 4. The number of ways in which N =a%bPer... can be resolved into two factors that are prime to each other is 2”-1, n being the number of prime factors a, b°, cv, . . For, in this kind of resolution, no single prime factor, a” for example, can be split between the two factors. The number of different divisors is therefore the same as if a, B, y,.. . * This result is given by Wallis in his Discourse of Combinations, Alterna- tions, and Aliquot Parts (1685), chap. iii., $12. In the same work are given most of the results of §§ 6 and 7 above. 510 EXAMPLES CHAP. were each equal to unity. Hence the number of ways is 3(1+1)(1+1)(1 +1)... (a factors) = 4. 9” = gn-1, Example 1. Find the different divisors of 360, their sum, and their number. We have 360 = 23325, The divisors are therefore the terms in the distributed product (1+2+ 27+ 23) (14+3+482) (145); that is to say, 1, 2; 4, 8, 3, 6, 12, 24,:9, 18, 36, 72, 5, 10, 20, 40, 15, 30, 60, 120, 45, 90, 180, 360. Their sum is (2*— 1) (83-1) (5?-1)/(2-1) (3-1) (5- LJ = pio: Their number is (1 +8) (1+2)(1+1)=24. Example 2. Find the least number which has 30 divisors. Let the number be N=a%d°c”, There cannot be more than three prime factors ; for 30=2x3x5, which has at most three factors, must = (a+1)(8+1)(y+1). There might of course be only two, and then we must have 30— (a+1)(8+1); or there might be only one, and then 30=a+1. . In the first case a=1, B=2, y=4. Taking the three least primes, 2, 3, 5, and putting the larger indices to the smaller primes, we have a2". Ded 20 In the second case we should get 24.3, 25.34, or 29.32 In the last case, 279. It will be found that the least of all these is 24.37.53; so that 720 is the required number. Example 3. Show that, if 2"-1 be a prime number, then 2”-1(2”—1) is equal to the sum of its divisors (itself excluded).* Since 2”—1is supposed to be prime, the prime factors of the given number are 2”-1 and 2”—1, Hence the sum of its divisors, excluding itself, is, by Cor. 1 above, (2"-1){(2"-1)-1} @=1){(2"=1)-Hj — 2n-K(2n — 1) =(2"—1){(2"—1) 41} — 22-190 1), = (2"— 1) {Qn— gna) =2-12"—1) (2-1), = anon = 1) : as was to be shown. ON THE NUMBER OF INTEGERS LESS THAN A GIVEN INTEGER AND PRIME TO IT. i § 8.] If we consider all the integers less than a given one, N, a certain number of these have factors in common with N, and the rest have none. The number of the latter is usually denoted “ In the language of the ancients such a number was called a Perfect Number. 6, 28, 496, 8128 are perfect numbers. XXXV EULER'S THEOREMS REGARDING ¢(N) 511 by ¢(N). Thus (N) is taken to denote the number of integers (including 1) which are less than N and prime to N. We have the following important theorem, first given by Euler :— Tf N= 0,70, .... a, then HN) =N(1-2) (1 -=) (1-=) a (-=) (1). The proof of this theorem which we shall give is that which follows most naturally from the principles of § 7. Proof.—Let us find the number of all the integers, not greater than N, which have some factor in common with N. That factor must be a product of powers of one or more of the PLIMeS C,, Uy, O3, - . -> Ory Now all the multiples of a, which do not exceed N are li, 2d,, o0,, --., (N/a,)a,, N/a, in number (3); all the multiples of a, which do not exceed N are ifeetd. od, .. ¥, (Nija,)a,, N/a, in number (4); and so on. All the multiples of a,a, which do not exceed N are lame eds.) 00,0,, .. 4 (N/a,0,)0,0,, N/a,a,in number (5); and so on. Similarly, for a,a,a, we have Wg naif, 30,glta, . . 2, (N/d,d,0s)0, OMe, N/a,a,a4, in number (6). Let us now consider the number N N N = +> $500 C+ N N N a, Qs A, as A, Us N N N +—— + —— + 4 N IRONSIDE we bee (7). 512 EULER'S THEOREMS REGARDING ¢(N) CHAP. The number of terms in the first line is nl, The number in the second line is ,0,, since every possible group of 2 out of the m letters a,a,...d, occurs among the denominators, The number in the third line is ,C, for a similar reason. And so on. Now consider every multiple of the 7 letters Oy Ugg a Op which does not exceed N; in other words, every number, not exceeding N, that has in common with it a factor of the form M10."2 a, H'r, This multiple will be enumerated in the first line, once as a multiple of a,, once as a multiple of a,, and so on; that is, once for every letter in it, that is, ,C, times. In the second line the same multiple will be enumerated once as a multiple of a,a,, once as a multiple of a,a,, and so on; that is, once for every group of two that can be formed out afi the r letters a,a,... d,; that is, ,C, times. And so on. Hence, paying attention to the signs, the multiple in question will in the whole expression (7) be enumerated pVirra Cg ck Une aly nek Oe #,0,= 1 = (eae times ; that is, just once. This proof holds, of course, whatever the 7 lect in the group may be, and whether there be Le. a or any number up to 7 in the group. It follows, therefore, that (7) enumerates, without repetition or omission, every integer which has a factor in common with N. But, from formula (1), chap. iv., § 10, we see that (7) is simply N-N(1-2) (17) 1. (i=) (8). To obtain the number of integers less than N which are prime to N, we have merely to subtract (8) from N. We thus obtain . wN)=N(1 -=) (1 -=) er ( -=), which establishes Euler’s formula. Example. N=100=2°.5?; $(100)=22.52(1— —$)(1—-+2)=40. ae § 9.] Jf M= PQ, where P and Q are prime to each other, then $(M) = $(P) $(Q) (1). ' =a SS XXXV DEQ )= b(P)6(Q)d(R) as 513 | For, since P and Q are prime to each other, we must have bade it ae ec ie Q=b,PtbP2 where none of the,prime factors are common ; and therefore ES ged GPL OeEe ee ca milere'a,, a,,.. . ., 0,0, ... are all primes. But, by § 8, we then have $(M) =m(1-7) (0-7)... (1-3) @-) =M(1 : 1 et moth Ashe tny Spee (=). ). Pa. 1-~) (1-2)... Cy Us Si b, b, = $(P) $(Q). oraiy PORS ua; be prume to each other, then PPQRS . . .) = H(P) (Q) P(R) (8)... (2). For, since P is prime to Q, R, 8, . . ., it follows that P is prime to the product QRS... Hence, by the above proposition, A(PQRS. ..) = 6(P) d(QRS. . .). Repeating the same reasoning, we have PQRS ...) = o(Q)G(RS...) ; and so on. _ Hence, finally, H(PQRS .. .) = g(P) 6(Q) G(R) $(S) . Remark.—There is no difficulty in establishing the theorem $(PQ) = o(P) 4(Q) @ priori. This may be done, for example, by means of § 13 below (see Gross’ Algebra, § 230), The theorem of § 8 above can then be deduced from $(PQR...) = ¢(P)4(Q) 4(R)... The course followed above, though not so neat, is, we think, more instructive for the learner. Example. 567% 8, (56) = 24, p(7)=6, $(8)=4 ; $(56) = (7) x (8) VOL. II rae 514 GAUSS’S THEOREM REGARDING THE DIVISORS OF N cuap. § 10.] Ff d,, d,, d,, . . ., &., denote all the divisors of the integer N, then * (dy) + o(d.)+ o(d,)+...=N... (ci (Gauss, Disg. Arith., § 39.) For the divisors, d,, dy, ds, . . ., are the terms in the dis- tribution of the product (l+aq+4'+...+a,%)(l+atag+...+4,%)... If we take any one of these terms, say d, = a,%'a,® .. ., then, by § 9, Cor., d(d,) = bay" a, di fi = (a, h(a,” ) SINC d,, G, . . . are primes. It follows that the left-hand side of (1) is the same as tl + bm) + (a) +... + $(4,%)} x {1+ d(a,) + f(a.) +... + p(a,%) } ; : (2). 1 But d(a,") = a?(1 = =| =a," —a,7—". 1 Hence 1 + f(a) + p(a) +... + d(a,™) =l+a,-l+a,/—-4,+.. /+¢%=g ee = a7 3 and so on. It appears, therefore, that (2) is equal to a,%a,%.. ., that is, equal to N. Example. N=315=8?.5.7, The divisors are 1, 8, 5, 7, 9, 15, 21, 35, 45, 63, 105, 815, and we have P(1)+ $(3)+ (5) +. . . +(315) . =14+2+44+6464+8+412424424+364484144=315. OO ae * Here and in what follows 1 is included among the divisors, and, for con- venience, $(1) is taken to stand for 1. Strictly Seal #(1) has no meaning at all. XXXV PRIME DIVISORS OF m! , 515 PROPERTIES OF m! § 11.] The following theorem enables us to prove some im- portant properties of m ! :— The highest power of the prime p which divides m! exactly is 1(™) +1(3) +1(3) ae he P P p where 1("), 1(*), - . . denote the integral parts of m/p, mp, . . .; and the series is continued until the greatest power of -p as reached which does not eaceed m. _ To prove this, we remark that the numbers in the series VS3 ee) e778 which are divisible by p are evidently ep moe. , 1D, where kp is the greatest multiple of pm. In other words, k=I(m[p). Hence I(m/p) is the number of the factors in m! which are divisible by p. If to this we add the number of those that are divisible by p, namely I(m/p’), and again the number of those that are divisible by p*, namely I(m/p*), and so on, the sum will be the power in which p occurs in m! Hence, since p is a prime, the highest power of p that will divide m! exactly is 1(7) +1(4) +1(S) +... P P P It is convenient for practical purposes to remark that m m 1(%) =1/1(4) /pl. p { Cia aka m/p"-\= 4+ k/pt-1 (k (p —1)/p + (p"-} 2 1)/p", adam ow yy sen 5 a4 te m 4 — il (“) . . : pe But, since i/p =j + I/p, j=1(5) =111(55) /o, by (2). We may therefore proceed as follows :—Divide m byp; take the integral quotient and divide again by p; and soon ; until the integral quotient becomes zero; then add all the integral quotients, and the result is the highest power of p which will divide m! exactly. Hence, by (3), Example 1. To find the highest power of 7 which divides 1000! exactly. In dividing successively by 7 the integral quotients are 142, 20, 2; the sum of these is 164. Hence 71 is the power of 7 required. Example 2. To decompose 25! into its prime factors. Write down all the primes less than 25 ; write under each the successive quotients ; and then add. We thus obtain 2 PUL e 7 te t1! | 18 ey ae See 12 8 5 3 2 1 1 1 1 : 6 2 1 3 1 22° | 10 6 3 9 1 1 1 1 Hence 25 !=2”.3",58,78,112,13.17.19.23, Example 3, Express 39!/25! in its simplest form as a product of prime factors. Result, 2!°: 35067577 11.137.17 219 329) Siea7e Example 4, Find the highest power of 5 that will divide 27.28.29 . . . 100 exactly. Result, 518, Example 5. If m be expressed in the scale of p, in the form M=potpiptpop +. . .+pnp”, . ‘ the highest power of » that will divide m! exactly is the ‘M—Po-Pr-P2r~. +» ~ Pay p-l Example 6. If mao +P pore... (z terms), where ad,{+ then a aire Z) 6 h | (4 +I(= +1(7 ae 2). d d dl (2) * This theorem might, of course, be inferred from the fact that m!/f!g!h!.. . represents the number of permutations of m things / of which are alike, g alike, h alike, &c. + If m be the number of the letters f, g, h, . . ., the utmost value of f' +o +h'+... isn(d—-1). Hence the utmost difference between the two sides of (2) is I{n(d@-1)/d}. 518 EXERCISES XXXVI CHAP. It appears, therefore, that, even if m= ftgth+ a Ry 1(7)«1(4) +1(8) + a (3). A fortiori is this so if m>ft+g+h+... If now we give d the successive values p, y*, . . ., and com- bine by addition the inequalities thus obtained from (3), athe a truth of (1) is at once established. Cor. 1. Jf f+gt+h+...+m, and none of the numbers LG h, . . . ts equal to m, the integer m!/flgih!.. . is divisible by m if m be a prime. Cor. 2. The product of r successive integers is exactly divisible by r!. The proofs of these, so far as they require proof, we leave to the reader. Cor, 2 has already been established by a totally different kind of reasoning in § 3, Example 6. EXERCISES XXXVI. (1.) What is the least multiplier that will convert 945 into a complete square ? | (2.) Find the number of the divisors of 2160, and their sum. (3.) Find the integral solutions of vy =1002+10y+1 (a); xy = 12" (8) ; y® =108x (7). (4.) No number of the form a+ 4 except 5 is prime. (5.) No number of the form 2+?+1 except 5 is prime. (6.) To find a number of the form 2”.3.a (a being prime) which shall be equal to half the sum of its divisors (itself excluded). (7.) To find a number N of the form 2abe...(a, b, ¢ being unequal primes) such that N is one-third the sum of its divisors. (8.) Show how to obtain two ‘‘amicable” numbers of the forms 2"pq, 2", where p, g, 7 are primes. (Two numbers are amicable when each is the sum of the divisors of the other, the number itself not being reckoned as a divisor. ) (9.) To find a cube the sum of whose divisors shall be a square. (One of Fermat’s challenges to Wallis and the English mathematicians. Var. Op. Math., pp. 188, 190.) (10.) If N be any integer, 2 the number of its divisors, and P the product of them all, then Nx= P2., -S XXXV EXERCISES XXXVI 519 (11.) The sum and the sum of the squares of all the numbers less than N and prime to it are $N(@-1)(b-1)(c-1)... and §N%(1-1/a) (1-1/2) . +4N(1-a)(1-6)... respectively. (Wolstenholme. ) (12.) Ifp, g, 7, . . . be prime to each other, and d(N) denote the sum of the divisors of N, show that Upgr .. .)=a(p) aq) ar) . (13.) If N=abc, where a, b, ¢ are prime to en other, then the product of all the numbers less than N and prime to N is (abe — 1)!IL {(@— 1) !/(be - 1) !a-De-D, (Gonv. and Caius Coll., 1882.) (14.) The number of integers less than (7?+1)" which are divisible by r but not by 7? is (r—1){(7?+1)"- 1} /7?. (15.) Prove that Ae (3)7 ato); area. en oe pt sar al 00 = ne (16.) In a given set of N consecutive integers beginning with A, find the number of terms not divisible by any one of a given set of relatively prime integers. (Cayley.) (17.) If m—1 be prime to n+1, show that ,,C,, is divisible by n+1. (18.) (@+1)(a@+2)...2ax0(b+1)... 2b/(a+56)! is an integer. (19.) The product of any 7 consecutive terms of the series x-1, 2?-1, 2—1,... is exactly divisible by the product of the first r terms. (20.) If » be prime, the highest power of » which divides n! is the ereatest integer in {n—S(n)/(p—1)”, where S(n) is the sum of the digits of m2 when expressed in the scale of p. If S(m) have the above meaning, prove that S(m—-) + S(m)-—S(n) for any radix. Hence show that (n+1)(n+2)... (+m) is divisible by m!. (Camb. Math. Jour. (1839), vol. i, p. 226.) (21.) If (nm) denote the sum of the uneven, and F(n) the sum of the even, divisors of m, and 1, 3, 6, 10,.. . be the ‘‘ triangular numbers,” then Sin) +fin-1)+f(r-3)+f(n-6)+... =F(n)+ Fn -1)+F(n-3)+F(n-6)4+. . ., it being understood that f(n-n)=0, F(n—-1)=n. ON THE RESIDUES OF A SERIES OF INTEGERS IN ARITHMETICAL PROGRESSION. § 13.] The least positive remainders of the series of numbers hk, k+a, k+2a, ..., k+(m—-l1)a with respect to m, where m is prime to a, are a permutation of the numbers of the series QMely eds on calcein el). 520 PROPERTIES OF AN INTEGRAL A.P. CHAP, All the remainders must be different ; for, if any two different numbers of the series had the same remainders, then we should have ‘ k+ra=pm+ p, and k+sa='m + p, ; whence (7 — s)a = (4 —- p')m, and (r= s)a/m = p— pl. Now this is impossible, since a is prime to m, and r and ¢ are each hE AM, where k= 0 or a multiple of m, and a prime to m as before, are all prime tom , and their remainders with respect to m are a permutation of eet deat kit ee For, as we have seen already, all the m remainders are unlike, and every remainder must be prime to m; for, if we had k+7,4= pm + p, where p is not prime to m, then ra = pm + p—k would have a factor in common with m, which is impossible, since 7; and a are both prime to m. Hence the remainders must be the numbers 7, 7,, . in some order or other. § 14.] Lf m be not prime to a, but have with it the G.C.M. g, so that a = ga’, m=gm’, the remainders of the series ky k+a,. k+2a, ..., k+(m—l1)a with respect to m will recur im a shorter cycle of m’. Consider any two terms of the series out of the first m’, say k+vra,k+sa. These two must have different remainders, otherwise (r —s)a would be exactly divisible by m; that is, (7 —s)ga'/gm’ would be an integer; that is, (r—s)a’/m’ would be an integer ; which is impossible, since a’ is prime to m’ and r —s < m7’. Again, consider any term beyond the m’th, say the (m’ + 7)th, then, since sty Tr {k + (m' + 7r)a} — {k + ra} = ma, = 9m, =m, it follows that the (m’'+7r)th term has the same remainder with respect to m as the 7th. In other words, the first m’ remainders:are all different, and after that they recur periodically, the increment being ga’, where a” is the remainder of a’ with respect to m’, subject to diminution by m as in last article. Example. If k=11, a=25, m=15, we have the series 11, 36, 61, 86, 111, 136, 161, 186, 211, 236, 261, . . . e' 3 522 FERMAT’S THEOREM CHAP, and here y=5; a’=5; m'=3; a”’=2; k'=11; ga’=10. Hence-the re- mainders are 11,*6,)1,11,°6, 1,11 6,1 oe Cor. If the G.O.M,, 9, of a and m divide k exactly, and, in particular, if k = 0, the remainders of the series hk, k+a, k+ 2a, are the numbers Og, ly, 29, 39, 5 wagees (mm! — l)g continually repeated in a certain order. For, in this case, since k = gx, we have (k + ra)/m = (x + 1a’) |m’, hence the remainders are those of the series Ky EK Ose eketed, angeee with respect to m! which is prime to a’, each multiplied by g. Hence the result follows by § 13. Example. Let k=10, a=25, m=15; then the series of numbers is 10, 35, 60, 85, 110, 135, 1605185," eae We have g=5; a’=5; m/=3; k=2; and the remainders are 10,/.3)-0),10,75. 00, 11005 eee : that is to say, 2x5, 1x5, 0x65, § 15.] From § 13 we can at once deduce FERMAT’S THEOREM, * which is one of the corner-stones of the theory of numbers. If m be a prime number, and a be prime to m, a™-1_1 is divisible by m. ; If a be prime to m, then we have la=p,.m + p,, 20 = [gM + pay (m Bas 1) a= Pm-1M + pm-1; where the numbers Pir Poy «+ + Pm-1 are the numbers 1, 2, . . ., (m—-1) written in a certain order. * Great historical interest attaches to this theorem. It was, with several other striking results in the theory of numbers, published without demonstra- tion among Fermat’s notes to an edition of Bachet de Meziriac’s Diophantus (1670). For many years no demonstration was found. Finally, Euler (Com- ment. Acad. Petrop., viii., 1741, and Comment. Nov. Acad. Petrop., vii., 1761) gave two proofs. Another, due to Lagrange (Nowy. Mém. de?’ Ac. de Berlin, 1771), is reproduced in § 18. The proof given above is akin to Euler’s second and to that given by Gauss, Disqg. Arith., § 49. Cr bo Co xxxv EULER'S GENERALISATION OF FERMAT’S THEOREM Hence Pe (ne — 1) a1 = (n,m + p,) Gum + p,). - = (m—1 + pm.-1); =Mm+ PiPo+++ Pm-=19 =Mm+1.2...(m-1). We therefore have 1.2... (m—1) (a”-1-1)=Mm. Now, m being a prime number, all the factors of 1.2... (m-— 1) are prime to it. Hence m must divide a™~-1- 1. It is very easy, by the method of differences, explained in § 5, to establish the following theorem :— Tf m be a prime, a” — a is exactly divisible by m.* Since a” —a=a(a™-1—1), if a be prime to m, this is simply Fermat’s Theorem in another form. § 16.] By using Cor. 4 of § 13 we arrive at the following generalisation of Fermat’s Theorem, due to Euler :— If m be any integer, and a be prime to m, then a%™ — 1 is exactly divisible by m. Here ¢(m) denotes, as usual, the number of integers which are less than m and prime to it. For, if 7,, 72) . - -,%nm be the integers less than m and prime to it, we have, by the corollary in question, 7,4 = pM + pj, 1 = pg + Pay Tn = fin ™ + Pas where the numbers p,, p:, . + -, pn are simply 7, %, - - Tn written in a certain order. We have therefore, just as in last paragraph, fits... Ta — 1) — Mim, whence, since 1,7, . - ., 7, are all prime to m, it follows that a” — 1, that is, a — 1, is divisible by m. § 17.] The famous theorem of Wilson can also be established by means of the principles of § 13. * For another proof of this theorem see § 18 below. 524 WILSON’S THEOREM—GAUSS’S PROOF CHAP, Any two integers whose product has the remainder + 1 with respect to a given modulus m may be called, after Euler, Allied Numbers, Consider all the integers, I, 2,3, «4 -i(m— 1); less than any prime number m (the number of them is of course even). We shall prove that, if we except the first and last, they can be exhaustively arranged in allied pairs. 7 For, take any one of them, say 1, then, since 7 is prime to m, the remainders of ) JOR SSD as delat ie Jah 1 pet are the numbers Rava pA ab in some order. Hence, some one of the series, say 77’, must have the remainder 1; then rr’ will be allies. The same number r cannot have two different allies, since all the remainders are different. Nor can the two, r and r’, be equal, unless 7=1 or =m-—1 : for, if we have | r= pM + 1% then 7°-1= pin; that is, (r +1) (r— 1) must be divisible by m. But, since m is prime, this involves that either r+ 1 or r—1 be divisible by m, and, since 7 cannot be greater than m, this involves in the one case that 7 =m — 1, in the other that r= 1. Excluding, then, 1 and m-—1 , we can arrange the series 2, 3, ..., (m—2) in allied pairs. Now every product of two allies is of the form pm+t13; hence the product 2.3... (m—2) is of the form (mm + 1) (um +1)..., which reduces to the form Mm + 1. Hence | 2.3...(m—2)=Mm+1; and, multiplying by m—1, we get 1.2.3... (m—2)(m— 1) = Mm(m—1)+m-1. Whence 1.2.3...(m—1)+1=Nm. ~ weal XXXV THEOREM OF LAGRANGE 525 That is, ee m be a prime, (m—1)!+1 as divisible by m, which is WILSON’S ‘THEOREM. * It should be observed that, af m be not a prime, (m—1)!+1 is not divisible by m. For, if m be not a prime, its factors occur among the numbers 2, 3, . . .. (m—1), each of which divides (m-—1)!, and, there- fore, none of which divide (m-— 1)! +1. § 18.] The following THEOREM oF LAGRANGE embraces both Fermat’s Theorem and Wilson’s Theorem as particular cases :— If (@+1)(@+2)...(@+p-1) Sh STN esta mel en a here a and p be prime, then A,, A,, . . ., Ap-s are all divisible by p. We have ope Awe? 44... + AD_.@ + A,_,} Se) ee) Ps A (et 1)P2 +... . + Ay.g(e + 1) + Ap-,}. Hence eee iE PAP oe pA, «t+ pAy-, = {(v+ 1)? —7P} + Af(a+1)Pt—aP th + A f(at+1)PA-aP} +. Therefore pA, =U, ay aero iy 5 ey Oe Ay ty a0 AG, pA, = 70, ie Me es ae ane Hence, since “Os Aut aC, . .. are not divisible by p if p be prime, we get, by successive steps, the proof that A,, Ag, A,, ... are all divisible by p. * This theorem was first published by Waring in his Meditationes Alge- braice (1770). He there attributes it to Sir John Wilson, but gives no proof. The first demonstration was given by Lagrange (Nouv. Mém. de lAc. de Berlin, 1771) ; this is reproduced in § 18. A second proof was given by Euler in his Opuscula Analytica (1783), vol. i., p. 329, depending on the theory of the residues of powers. The proof above is that given by Gauss (Disg. Arith., $$ 7 7, 78), who generalises the theorem as follows:—‘‘ The product of all the numbers less than m and prime to it is congruent with —1, if m=p* or =2p", where p is any prime but 2, or, again, if m=4; but is congruent with +1 in every other case.” This extension depends on the theory of quadratic residues. 526 EXERCISES XXXVII CHAP, Cor. 1. Put =1, and we get 23.0 = io (Ap Ape -+Apy_,) + Ap. Therefore A,_, + 1, that is, (p-1)!+1 is divisible by p. Cor. 2. Multiplying by x and transposing, we get eP—g=a(r+1)...(¢+p—1) —(1+A,_,)¢ -(A,?-14 A 7p-2 Apa But z(z+1)...(@+p-1), being the product of p consecutive integers, must be divisible by p. Also, if p be prime, 1 + AC is divisible by p. . Therefore, «? —a is divisible by p if p be prime. From which Fermat’s Theorem follows at once if x be prime to p. EXERCISES XXXVII. (1.) a? —a is divisible by 2730. (2.) Ifa be prime greater than 13, #—1 is divisible by 21840. (3.) If the nth power of every number end with the same digit as the number itself, then n= 4p +1. Give a rule for determining by inspection the cube root of every perfect cube less than a million, (4.) If the radix, 7, of the scale of notation be prime, show that the rth power of every integer has the same final digit as the integer itself, and that’ the (r-1)th power of every integer has for its final digit 1. (5.) If be prime, and x prime to n, then either a-D/2_ 1 or a(-)/ 247 is divisible by n. (6.) Ifn be prime, and x prime to n, then either 2("- D2 _ 1 or a™("- D244 is divisible by n2, (7.) If m and-n be primes, then m”—* +-1"™-1=1 (mod. mn). (Si) Lig. 0 ae he primes, and N=afy..., then =(N/a)*-1=1(mod Cae a ae (9.) If be an odd prime, show that (a+1)"-(a"+1)=0 (mod. 27). Hence show that, if » be an odd prime and p an integer, then any integer expressed in the scale of 2n will end in the same digit as its (pn—p+1)th power. Deduce Fermat’s Theorem. (Math. Trip., 1879.) (10.) If x be prime and >a, then gn—2tgn-34 | ttl =0 (mod. 7). (11.) Ifm be an odd prime, then 1+2(m+1)4+22(m+1)?4. . 4 2"-°(n + 1)"-2=0 (mod. 2). (12.) If be odd, "+24... 4 (7 —1)"=0 (mod. n). XXXV NOTATION FOR NUMBER OF PARTITIONS oo (13.) If be prime, and p 1, then 12m A 92m a +(25)" == () 0 (mod. p). (17.) If neither @ nor d be divisible by a prime of the form 4-1, then ain — hin will not be exactly divisible by a prime of that form. Hence show that a#”~* + 64-? is not divisible by any integer (prime or not) of the form 4n—1. Also that a?+ 6? is not divisible by any integer of the form 4n—1 which does not divide both a and b. Also, that any divisor of the sum of two integral squares, which does not divide each of them, is of the form 4n+1. (Euler. ) (18.) Show, by means of (17), that no square integer can have the form 4mn—m-—n™, where m, ”, a are positive integers. (Euler.) PARTITION OF NUMBERS. Euler’s Theory of the Enumeration of Partitions. § 19.] By the partition of a given integer is meant the division of the integer into a number of others of which it is the sum. Thus 1+2+2+3+3, 1+34+7, are partitions of 11. There are two main classes of partitions, namely, (I.) those in which the parts may be equal or unequal; (II.) those in which the parts are all unequal. When the word “ Partition” is used without qualification, class (I.) is understood. We shall use a quadripartite symbol to denote the number of partitions of a given species. Thus P(| |) and Pu(| |) are used to denote partitions of the classes (I.) and (II.) respectively. In the first blank inside the bracket is inserted the number to be partitioned ; in the second, an indication of the number of the parts ; in the third, an indication of the magnitude or nature of 528 EXPANSIONS AND PARTITIONS CHAP, the parts. It is always implied, unless the contrary is stated, that the least part admissible is 1; so that +m means any integer of the series 1, 2, . . ., m. An asterisk is used to mean any integer of the series 1, 2, . . ., ©, or that no restriction is to be put on the number of the ete other than what arises from the nature of the partition otherwise. Thus P(x|p|q) means the number of partitions of into p parts the greatest of which is ¢; P(n|p|q) the number of partitions of m into p parts no one of which exceeds q; P(n| * |) the number of partitions of m into any number of parts no one of which is to exceed g; Pu(n|pp|*) the number of partitions of m into p or any less number of unequal » parts unrestricted in magnitude ; Pu(n|p|odd) the number of partitions of m into.p unequal parts each of which is an odd integer ; P(n|*|1, 2, 2°, 2°, .) the number of partitions of m into an number of parts, Aa part being a number in the series 1,2, 2", 2°)°. ...;7 and 80’ on. The titer of partitions has risen into great importance of late in connection with the researches of Sylvester and his followers on the theory of invariants. It is also closely con- nected with the theory of series, as will be seen from Euler’s enumeration of certain species of partitions, which we shall now briefly explain. § 20.] If we develop the product (1 + zz) (1 + 20) ca eee it is obvious that we get the term 2?2” in as many different ways as we can produce n by adding together p of the integers 1, 2,.. ., q each to be taken only once. Hence we have the following equation :— (1 +20) (1 + 20°)... (1 + 222) = 1+ 2Pu(n|p|pg)z?a" (1). Again, if to oe pasa on the left of (1) we adjoin the factor l+zt+27+2+.... ado (that is, 1/(1-2)), we shall evidently get 2?z” as often as we can produce 1 by adding together any number not exceeding p of the integers 1, 2,.. ., ¢. Therefore (1 + 2a) (1 +207)... (1 + 222)/(1 — 2) = 1+ Pu(n|+p|+q)z?a” (2). it et el ——— a ee XXXV EXPANSIONS AND PARTITIONS 529. In like manner, we have (+2) (1+2°)...(1 +29) =1+4 2Pu(n| * |p9)o (3) ; (1 + 2x) (1+227)...adao =1 + =Pu(n |p| * )zPa™ (4); (+2) (1+27)...adao =1 + 2Pu(n|*|* ja” — (5), Also, as will be easily seen, we have | 1/(1 — 2x) (1 — 297)... (1-20) =14 =P(n|p|bq)zPa” (6); 1/(1 —2)(1—2z)... (1 — 20%) =14 =P(n |p |p g)zP2” (7) ; Vila) (l= a")... (1 —22) = 14+ ZP(n | * |g)a” (8) ; I/(1 — 2x) (1 — 22°)... ad 0 =1+ =P(n|p|* )ePa” (9); 1/(1—z) (1—za) (1—20")... ado =14 =P(n |p| * )zPa” (10); 11-2) (1-2)... adeo =14EP(n| #|*)a” ab i- and so on. . . By means of these equations, coupled with the theorems given in chap. xxx. § 2, and Exercises XXI., a considerable number of theorems regarding the enumeration of partitions can be deduced at once. § 21:] To find a recwrrence-formula for enumerating the parti- tions of n into any number of parts none of which exceeds qs and thus to calculate a table for P(n | «|+4q). By (8), we have If =) (1 -—2°)... (1 -at) =14 =P(n| * |g)a”. Hence, multiplying on both sides by 1 -a4, and replacing 1/(1 —2)(1—a°)... (1 —29-) by its equivalent, we derive 1+ =P(n| * |g -1)a” = 1+ 2{P(e| «| 9) — P(r —9|* |p g}2” (12), where we understand P(0,| * |-q) to be 1. Hence, ifn+tq, ~ P(n| * |) =P(n| * |[$g-1)+P(m—g|*|$g) (13); and, if n ¢) CHAP. 1 ; 6p 7¢8 9 10 11 12.18 14 15 16717 Reese i) es) et eet Sy Ss oC ee ee roo OC . et | — 1/1 1 11 1 1 1 1) i 415 56 6 6 4% 7 °8 8 9 ORO 8 10|12 14 16 19 21 24 27 30 38 87 40 44 11 15 18|28 27 34 39 47 54 64 72 84 94 108 13 18 23 30/37 47 57 70 84 101 119 141 164 192 14 20 26 35 44/58 71 90 110 136 163 199 235 282 15 21 28 88 49 65|82 105 131 164 201 248 300 364 22 29 40 52 70 891116 146 186 230 288 352 434 30 41 54 73 94 123|157 201 252 318 393 488 42 55 75 97 128-164|212 267 340 423 530 56 76 99 131 169 219|278 355 445 560 F bo | bo eo po | Ww oP we) SIO OVC HH] OH I eH Oo oO E a ad Take a rectangle of squared paper BAC ; and enter the values of » at the heads of the vertical columns, and the values of 4 at the ends of the horizontal lines. We remark, first of all, that it follows from (14) that all the values in the part of any vertical column below the diagonal AF are the same. We therefore leave all the corresponding spaces blank, the last entry in the column being understood to be repeated indefinitely. Next, write the values of P(1|*|+1),P(2|*|#1),.- 39m that is, 1, 1, . . ., in the row headed 1. To fill the other rows, construct a piece of paper of the form abed. Its use will be understood from the following rule, which is simply a translation of (13) :— To fill the blank immediately after the end of any step, add to the entry above that blank the number which is found at the left-hand end of the step. Thus, to get the number 23, which stands at the end of the step lying on the fourth horizontal line, we add to 14 the number 9, which lies to the immediate right of ab in the same line as the blank. Again, in the ninth line 157 = 146 + 11; and sO on. By sliding abcd backwards and forwards, so that bc always lies on AD, we can fill in the table rapidly with little chance of error. We shall speak of the table thus constructed as Euler’s Ot i oe et el oe ees Re XXXV ENUMERATIONS REDUCIBLE TO EULER’S TABLE 531 Table. It will be found in a considerably extended form in his Introductio, Lib. I., chap. xvi. A variety of problems in the enumeration of partitions can be solved by means of Euler’s Table, as we shall now show. § 22.) Zo find by means of Euler's Table the number of partitions of n into p parts of unrestricted magnitude. Let us first consider P(n|p|*). By (9) above, we have 1+2P(n|p|* )a"zP =1/(1 — 2&)(1 — 20°)... ad oo, = 1+ 2aPeP/(1 — 2) (1 —2%)... (1 — 2), by Exercises XX]. (18). Hence ! =P(n |p| * ja” = Sar /(1 — x) (1 — 2")... (1 — 22), =3P(n|* |p), by (8) Pn |p|*)=P(m-p| * |p) (15). Therefore, Again, 1+ 2Pu(n|p|* )a”2? = (1 + 20) (1 + 20)... ad o, 3 = 1+ 2atmle Ve /(] — 2) (1 — x)... (1-2), by chap. xxx., § 2, Example 2. Hence =Pu(n| p | * ju” = al(P+0)/(1 — 2) a) aed @ Usage a = =P(n | # |p p)art2P(+1), by (8). Therefore Pu(n|p|*)=P(n—4n(p+1)|*|bp) (16). Example 1. P(20|5| *)=P(15| «|+5)=84, Example 2. Pu(20|5|*)=P(5| x [+5)=7. § 23.] If we take any partition of into p parts in which the largest part is g, and remove that part, we shall leave a parti- tion of n—g into p— 1 parts no one of which exceeds g. Hence we have the identity Pin |p | a) =P(n—g | p—1|+q) (17) ; and, if we make p infinite, as a particular case, we have Pu | * | g)=P(w—q | *|49) (18). It will be observed that (18) makes the solution of a certain class of problems depend on Euler’s Table. 532 THEOREMS OF CONJUGACY CHAP. By comparing (15) and (18), we have the theorem P(w| *|q)=Pm|q|*), which, however, is only a particular case of a theorem regarding conjugacy, to be proved presently. § 24.] Theorems regarding conjugacy. (L.) P(n|$p| $4) =P(n|$a| PP) (19). (IL) P(n-p|q-1|+p)=Pm-g|p-1|Fq (20). (IIL.) P(n|p |g) =P(@|¢@|P) (21). To prove (I.) we observe that, by (7), we have 1+3P(n|$p|$ Qe?ar=1/-z)( —zx)... (1 — 22%), its sont — g9t1)(1—a9+?),..(1—a9*?) (1 —2) (1 = 2) ee Hence : fe =P(n |p | qa" = C TT a d. : “a 2 (1—2)(1—2) .<: (1 (1—a)(1—2’)...(1—#9)(1 —2)(1—2")...(1—a?) Since the function last written is symmetrical as regards p and g, it must also be the equivalent of =P(n|-q| + p)2”. ‘Hence Theorem (I.). Theorem (II.) follows from (6) in the same way. Since, by (17), we have P(n|p|q)=P(n-g|p-1| +49), P(n|q|p)=P(-pla-1| FP); therefore, by (II.), 4 P(n|p|q)=P(r|a|P); i which establishes Theorem CURES) ‘ The following particular cases are obtained by making p or g infinite :— P(n|>p|*)=Pm|*|pp) - (22); P(n |p| *) = P| * |p) (23). XXXV FURTHER REDUCTIONS TO EULER’S TABLE 533 § 25.] The following theorems enable us to solve a number of additional problems by means of Euler’s Table :-— ‘(P| 9)=Pin—p|* |p) -2P(n—p,—p] s | +p) + 2P(1 = po —p| * |p) — 2P(n — ps — p| * |p) (24). Here the summations are with Tespect to w,, w.,...; and , 18 any one of the numbers %WqQ+1,...,¢9+p-1, p, the sum of any two of them, fs the sum of any three, and so on, The series of sums is to be continued so long as h— fy-9t0. If P(n|p|+q) come out 0 or negative, this indicates that the partition in question is impossible. Pn |p| +9) =P(n| *|$p) - BP(n—v, | «| +p) + 2P(n — v, | * |} p) — 2P(nm — vs | * | bp) (25), Heronisv., . ...t have the same meanings with regard to Cee +p as formerly p,, po. . . with regard to Orgiat.. oe — |. P(n | * | * ) =P(n-1| x [> 1)+P(n-2| x |p 2)+...+PO|*|pn) (26). The demonstrations will present no difficulty after what has already been given above. CONSTRUCTIVE THEORY OF PARTITIONS. § 26.] Instead of making the theory of partitions depend on series, we might contemplate the various partitions directly, and develop their properties from their inherent character. Sylvester has recently considered the subject from this point of view, and has given what he calls a Constructive Lheory of Partitions, which throws a new light on many parts of the subject, and greatly simplifies some of the fundamental demonstrations.* Into this * Amer. Jour. Math. (1882). 534 GRAPH OF A PARTITION - CHAP. theory we cannot within our present limits enter ; but we desire, before leaving the subject, to call the attention of our readers to the graphic method of dealing with partitions, which is one of the chief weapons of the new theory. By the graph of a partition is meant a series of rows of asterisks, each row containing as many asterisks as there are units in a corresponding part of the partition. Thus % & is the graph of the partition 3+ 95+ 3 of the number 11. For many purposes it is convenient to arrange the graph so that the parts come in order of magnitude, and all the initial asterisks are in one column. Thus the above may be written— The graph is then said to be regular. The direct contemplation of the graph at once gives us intuitive demonstrations of some of the a a) AY my AY ¥ 1 7 7 “- foregoing theorems. For example, if we turn the columns of the graph last written into rows, we have where there are as many asterisks as before. The new graph, therefore, represents a new partition of 11, which may be said to be conjugate to the former partition. Thus to every partition of n into p parts the greatest of which is q, there is a conjugate partition into q parts the greatest of which is p. Hence P(n|p| a) =P@|a|p) a we ar = ee eS an old result. Again, to every partition of n into p parts no one of which eaceeds q, there will be a conjugate partition into q or fewer parts the greatest of which is p. Hence P(n|p | 4) =P(n| |p) (27), a new result ; and so on.* we ee * According to Sylvester (J.c.), this way of proving the theorems of con- jugacy originated with Ferrers. a een —. = a Sk — OY a ae eee XXXV EXTENSION AND CONTRACTION OF GRAPHS 535 § 27.] The following proof, given by Franklin, of Euler’s famous theorem that (1 -a)(1—a')(1—2°).. .ad @ = 3 (-) PHP) (9g) # p=0 is an excellent illustration of the peculiar power of the graphic method. The coefficient of 2” in the expansion in question is obviously Pu(n | even | * )— Pu(n| odd | « ) (29). Let us arrange the graphs of the partitions (into unequal parts) regularly in descending order. Then the right-hand edge of the graph will form a series of terraces all having slopes of the same angle (this slope may, however, consist of a single asterisk), thus— A B * * + * * & * * * Va ie 4 * 6 * *¥ *& *¥ KX & r * ¢ * oo to ko hae eo ee © kX * We can transform the graph A by removing the top row and placing it along the slope of the last terrace, thus— id At We then have a regular graph A ae representing a partition into unequal parts. AS: _ This process may be called contraction. ES eo We cannot transform B in this way ; * * % % x ey » Dut we may extend B by removing the slope of its last terrace, and placing it above the top row, thus— . We then have a regular graph B’ repre- ee He senting a partition into unequal parts. Ae tiare Every graph can be transformed by con- ieee traction or by extension, except when the top % % % « « . Tow meets the slope of the last terrace; and in % % & % x% x this case also, provided it does not happen that the number of asterisks in the top row is equal * Euler originally discovered this theorem by induction from particular cases, and was for long unable to prove it. For other demonstrations, see Legendre, Théorie des Nombres, t. il., § 15, and Sylvester (J.c.). 536 FRANKLIN'S PROOF OF EULER’S EXPANSION CHAP. to the number in the last slope or exceeds it only by one, as, for example, in * * %* #& & & *¥ *& & x %* & * ¥ % + & x %¥ ¥ € & Contraction or extension in the first of these would produce an irregular graph ; contraction in the second would produce an irregular graph; and extension would produce a graph which corresponds to a partition having two parts equal. ‘These two cases may be spoken of as wnconjugate ; they can only arise when the p parts of the partition are p, D+), 12,9 yep ener eee and the number n=pt+(p+l)t+...+(2p-1)=3(3p' —p); or when the p parts are p+1, p+2, p+3, .-.-, 2p, and n=(p+1)+(p+2)+.. . + 2p=3(3p" + p). Since contraction or extension always converts a partition having an even or an odd number of parts into one having an odd or an even number of parts respectively, we see that, unless » be a number of the form 4(3p" + p), Pu(n| even | * ) = Pu(n| odd | + ). ) When n has one or other of the forms 4(3p’ + p), there will be one unconjugate partition which will be even or odd according as p is even or odd; all the others will occur in pairs which are conjugate in Franklin’s transformation. Hence Pu(}(3p" + p)| even | * ) - Pu(Z(3p" + p)|odd| * )=(— 1)? (30). Euler’s Theorem follows at once. Exercises XXXVIII. (1.) Show how to evaluate Pu(n|+p | ) by means of Euler’s Table. Evaluate (2.) P(13| 5| +8). (3.) P(13 |+5 |+ 3). (4.) P(10| «| « ). (5.) P(20|9|+8). XXXV EXERCISES XXXVIII 537 Establish the following :— (6.) Pu(n|*|*)=P(n- 29(7+1)|*|+4q), where 39(7+1) just} n. (7.) Pu(n|p|* )=P(n-4p(p—1) |p| «). | (8.) P(n |p| *)=Pu(n+4p(p-1) |p| x ). (9.) Pu(n |p |+9)=P(n-4p(p-1) |p |+q-p+1). (10.) Is the theorem P(n-p|g=1|*)=P(n-¢ |p-1|«) universally true ? (11.) Show how to form a table for the values of P(n| x (2ueeeres aay (See Proc. Edinb. Math. Soc., 1883-4.) | (12.) Show how to form a table for the number of partitions of » into an indefinite number of odd parts. Establish the following :— (13.) P(n|*|1, 2,22 93 . | , ae (14.) Pu(n|p|1, 3. . ., 2q-1)=P(n-p*+p|p|1, 3, .. ., 29-1). (15.) P(m |p |2; 4, . . +» 29)=P(n-pfp|1, 3, .. ., 2q-1). (16.) P(m|* | odd)=Pa(n |*]*). Pin leplQ4, . . 2, 2q)=P(n|+q|2,4,..., 2p). (18.) P(w+p|p|1, 3, sey 2G4+1)=P(n+¢/¢@]1, 3, ..., 2p +1). (19.) Pu(n+p?|p|1, 3, ..., 2¢+1)=Pu(n+q?|q|1,3,..., 2p+1). (20.) P(m+2p|p|2,4,..., 2q¢+2)=P(n+2¢]¢|2,4,..., 2p + 2). (21.) Show that P(n |p| « )=P(n-1|p-1|*)+P(n-p|p|*); and hence - construct a table for P(n | p [*). (See Whitworth, Choice and Chance, chap. ili.). CHAPTER XXXVI. Probability, or the Theory of Averages. § 1.] An elementary account of the Theory of Probability, or, as we should prefer to call it, the Theory of Averages, has usually found a place in English text-books on algebra. This | custom is justified by several considerations. The theory in question affords an excellent illustration of the application of the theory of permutations and combinations which is the funda- mental part of the algebra of discrete quantity ; it forms in its elementary parts an excellent logical exercise in the accurate use of terms and in the nice discrimination of shades of meaning ; and, above all, it enters, as we shall see, into the regulation of some of the most important practical concerns of modern life. The student is probably aware that there are certain occur- rences, or classes of events, of such a nature that, although we ~ cannot with the smallest degree of certainty assert a particular proposition regarding any one of them taken singly, yet we can assert the same proposition regarding a large number N of them with a degree of certainty which increases (with or without limit, as the case may be) as the number N increases. For example, if we take any particular man of 20 years of age, nothing could be more uncertain than the statement that he will live to be 25; but, if we consider 1000 such men, we may assert with considerable confidence that 96 per cent of them will live to be 25; and, if we take a million, we might with much greater con- fidence assign the proportion with even closer accuracy. In so doing, however, it would be necessary to state the limits both of habitat and epoch within which the men are to be taken ; and, even with a million cases, we must not expect to be able to assign CHAP. XXXVI DEFINITION OF PROBABILITY 539 the proportion of those who survive for 5 years with absolute accuracy, but be prepared, when we take one million with another, to find occasional small fluctuations about the indicated percentage. We may, for illustration, indicate the limits just spoken of by saying that “man of 20” is to mean a healthy man or woman living in England in the 18th century. The “event,” as it is technically called, here in question is the living for 5 years more of a man of 20; the alternative to this event is not living for 5 years more. The whole, made up of an event and its alternative or alternatives, we call its universe. The alternative or alternatives to an event taken collectively we often call the Complementary Event. 'The living or not living of all the men of 20 in England during the 18th century we may, following Mr. Venn,* call the series of the event. It will be observed that on every occasion embraced by the series the event we are consider- ing is in question ; and we express the above result of observa- tion by saying that the probability that a man of 20 living under the assigned conditions reached the age of 25 is -96. We are thus led to the following abstract definition of the Probability or Chance of an Event :— If on taking any very large number N out of a series of cases im which an event A is in question, A. happens on PN occasions, the probability of the event A is said to be p. In the framing of this definition we have, as is often done in mathematical theories, substituted an ideal for the actual state of matters usually observed in nature. In practice the number p, Which for the purposes of calculation we suppose a definite quantity, would fluctuate to an extent depending on the nature of the series of cases considered and on the number N of specimen cases selected. Moreover, the mathematical definition contains no indication of the extent or character of the series of cases. * Logic of Chance. + We might take more explicit notice of this point by wording the definition thus :—‘‘ If, on the average, in N out of a series of cases, &c.” But, from the point of view of the ideal or mathematical theory, nothing would — thus be gained. 540 REMARKS ON THE DEFINITION CHAP. How far the possible fluctuations of p, the extent of the series, and the magnitude of N will affect the bearing of any con- clusion on practice must be judged by the light of circumstances. It is obvious, for instance, that it would be unwise to apply to the 14th century the probability of the duration of human life deduced from statistics taken in the 18th. This leads us also to remark that the application of the theory of probability is not merely historical, as the definition might suggest. Into most of the important practical applications there enters an element of induction.* Thus we do in fact apply in the 19th century a table of mortality statistics deduced from observations in the 18th century. The warranty for this extension of the series of cases by induction must be sought in experience, and cannot in most cases be obtained a priori. There are, however, some cases where the circumstances are so simple that the probability of the event can be deduced, without elaborate collecting and sifting of observations, merely from our definition of the circumstances under which the event is to take place. The best examples of such cases are games of hazard played with cards, dice, &c. If, for example, we assert regarding the tossing of a halfpenny that out of a large number of trials heads will come up nearly as often as tails—in other words, that the probability of heads is 5, what we mean thereby is that all the causes which tend to bring up heads are to neutralise the causes that tend to bring up tails. In every series of cases in question, the assumption, well or ill justified, is made that this counterbalancing of causes takes place. That this is really the right point of view will be best brought home to us if we reflect that undoubtedly a machine could be con- structed which would infallibly toss a halfpenny so as always to land it head-up on a thickly sanded floor, provided the coin were always placed the same way into the machine; also, that the coin might have two heads or two tails ; and so on. In cases where the statement of probability rests on grounds so simple as this, the difficulty regarding the extension of the * In the proper, logical sense of the word. XXXVI COROLLARIES ON THE DEFINITION 541 series by induction is less prominent. The ideal theory in such cases approximates more closely than usual to the actual circum- stances. It is for this reason that the illustrations of the elementary rules of probability are usually drawn from games of hazard. The reader must not on that account suppose that the main importance of the theory lies in its application to such cases; nor must he forget that its other applications, however important, are subject to restrictions and limitations which are not apparent in such physically simple cases as the theory of cards and dice. Before closing this discussion of the definition of probability as a mathematical quantity, it will be well to warn the learner that probability is not an attribute of any particular event happening on any particular occasion, It can only be predicated of an event happening or conceived to happen on a very large number of “occasions,” or, in popular language, of an event ‘on the average” or in the “long run.” Unless an event can happen, or be conceived to happen, a great many times, there is no sense in speaking of its probability, or at least no sense that appears to us to be admissible in the following theory. The idea conveyed by the definition here adopted would be better expressed by substituting the word frequency for the word probability; but, after the above caution, we shall adhere to the accepted term. § 2.] The following corollaries and extensions may be added - to the definition. Cor. 1. If the probability of an event be p, then out of N cases in which it is in question it will happen pN times, N being any very large number. This is merely a transposition of the words of the definition. As an example, let it be required to find the number out of 5000 men of 20 years of age who will on the average live to be 25. The probability of a man of 20 living to be 25 may be taken to be ‘96; hence the number required is "96 x 5000= 4800. Cor. 2. If the probability of an event be p, the probability of its failing is 1 — p. For out of a large number N of cases the event will happen on PN occasions; hence it will fail to happen on N-pN 542 COROLLARIES ON THE DEFINITION CHAP. =(1-yp)N occasions. Hence, by the definition, the probability of the failing of the event is 1 — p. Cor. 3. If the universe of an event be made up of n alter natives, or, in other words, if an event must happen and that mm one out of n. ways, and if the respective probabilities of its happening in these ways De DI Ma; Les ny HON Dg LO Dye ly For on every one of N occasions the event will happen; and it will happen in the first way on p,N occasions, in the second on w,N occasions, and soon. Hence N=p,N+p,N+...+p,N; that18, 1 = ye, aD, Cor. 4. If an event is certain to happen, its pr opabaan Y is Teayiat is certain not to happen, its probability is 0. For in the former case. the event happens on 1.N cases out of N cases; in the latter on 0.N cases out of N. The probability of every event is thus a positive number lying between 0 and 1. Cor. 5. If an event must happenin one out of n ways all equally probable, or if one out of n events must happen and all are equally probable, then the probability of each way of happening m the first case, or of each event happening in the second, is 1/n. This follows at once from Cor. 3 by making p,=p,=. . .= Pp. As a particular case, it follows that, if an event be equally likely to happen or to fail, its probability is 4. Definition.—The ratio of the probability of the happening of an event to the probability of its failing to happen is called the odds in Favour of the event, and the reciprocal of this ratio as called the odds against it. Thus, if the probability of an event be p, the odds in favour are p:1—p;,the odds against 1-p:p. Also, if the odds in favour be m:n, the probability of the event is m/(m+n). If the probability of the event be 4, that is, if it be equally likely to happen or to fail, the odds in favour are 1:1, and are said to be even. Cor. 6. If the universe of an event can be analysed into m+n cases cach of which in the long run will oceur equally often,* and if * This is usually expressed by saying that all the cases are equally likely. XXXVI DIRECT CALCULATION OF PROBABILITIES 543 in m of these cases the event will happen and in the remaining n fail to happen, the probability of the event is m/(m + n). After what has been said this will be obvious. DIRECT CALCULATION OF PROBABILITIES. § 3.] The following examples of the calculation of proba- bilities require no special knowledge beyond the definition of probability and the principles of chap. xxiii. Example 1. There are 5 men ina company of 20 soldiers who have made up their minds to desert to the enemy whenever they are put on outpost duty. If 3 men be taken from the company and sent on outpost duty, what is the probability that all of them desert? The 3 men may be chosen from among the 20 in 0, ways, all of which are equally likely. Three deserters may be chosen from among the 5 in 503 ways, all equally likely. The probability of the event in question is therefore Clas 93 [eg =U. Example 2. If n people seat themselves at a round table, what is the chance that two named individuals be neighbours ? There are (see chap. Xxlil., § 4) (n—1)! different ways, all equally likely, in which the people may seat themselves, Among these we may have A and B or B and A together along with the (7-2)! different arrangements of the rest ; that is, we have 2(n— 2)! cases favourable to the event and all equally likely. The required chance is therefore 2(n — 2)!/(n—1)!= Aimee ie When n=3, this gives chance =1, as it ought to do. The odds against the event are in general n—3 to 2 3 the odds will therefore be even when the - number of people ig 5. Example 3. If a be a prime integer, and n=a", and if any integer I+ 7 be taken at random, find the chance that I contains a as a factor s times and no more. The integer I must be of the form ha*, where \ is any integer less than a”~* and prime to a’-s, Now, by chap. Xxxv., § 8, the number of integers less than a7-* and prime to it is a-*(1—1/a). Also the number of integers +n is a", Hence the required chance is a’-*(1 — 1/a)/at =a-4(1 - 1/a)=1/as | ~lfarh, Example 4. Find the probability that two men A and B of m and n years of age respectively both survive for Pp years, The mortality tables (see § 15 below) give us the numbers out of 100,000 individuals of 10 years of age who complete their mth, nth, m+pth, n+ pth years. Let these numbers be by Uns lmtp y n+p» The probabilities that A and B live to be m+p and n+ P years of age respectively are L-tol ems Untp|en respectively. Consider now two large groups of men numbering M and N respeetively. We suppose A to be always selected from the first and B always 544 DIRECT CALCULATION OF PROBABILITIES CHAP. from the second. In this way we could select altogether MN pairs of men who may be alive or dead after p years have elapsed, The number out of the M men living after p years is MJn4p/lm, by § 2, Cor. 1. Similarly the number living out of the N men is Nlpi,//n. Out of these we could form MNimtplitp/tmln pairs. This last number will be the number of pairs of survivors out of the MN pairs with which we started. Hence the probability required is Um4plntp/lmln=(lmtp/lm)(ln+p/ em) ; In other words, it is the product of the probabilities that the two men singly each survive for p years. The student should study this example carefully, as it furnishes a direct proof of a result which would usually be deduced from the law for the multiplication of probabilities. See below, § 6. Example 5. A number of balls is to be drawn from an urn, 1, 2,..., ” being all equally likely. What is the probability that the number drawn be even ? We can draw 1, 2, .. ., m respectively in »C), nCo, . . ., nCn ways respectively. Hence we may consider the universe of the event as consisting of nCitnOot. » »tnCn=(14+1)"-1=2"-1 equally likely cases. The number of these in which the drawing is even is ,Co+nC4+...=${(1+1)"+(1-1)"- 2} =4(2”—2)=2"-!_-1, The number of ways in which an odd drawing can be made is n»0y+nC3t+.. .=${(1+1)"-(1-1)"} =42"=2"-1. Hence the chance that the drawing be even is (2”-1-1)/(2"-—1), that it be odd 2”-1/(2"—1), The sum of these is unity, as it ought to be ; since, if the drawing is not odd, it must be even. In general, an odd drawing is more’likely than an even drawing, the odds in its favour being 2"-1:2”-!-1; but the odds become more nearly even as 7 increases. Example 6. A white rook and two black pawns are placed at random on a chess-board in any of the positions which they might occupy in an actual game. Find the ratio of the chance that the rook can take one or both of the pawns to the chance that either or both of the pawns can take the rook. Let us look at the board from the side of white ; and calculate in the first place the whole number of possible arrangements of the pieces. No black pawn can lie on any of the front squares ; hence we may have the rook on any of these 8 and the two pawns on any two of the remaining 56; in all, 8 x 25gC2=8 x 56x 55 arrangements. Again, we may have the rook on any one of the 56 squares and the two pawns on any two of the remaining 55 squares ; in all, 56x 55x54 arrangements. The universe may therefore be supposed to contain 62 x 56 x 55 equally likely cases. Instead of calculating the chance that the rook can take either or both of the pawns, it is simpler, as often happens, to calculate the chance of the complementary event, namely, that the rook can take neither of the pawns. If the rook lie on one of the front row of squares, neither of the pawns can lie on the corresponding column, that is, the pawns may occupy any two out of 49 squares; this gives 8 x 49x48 arrangements. If the rook lies in any one of the remaining 56 squares, neither of the pawns must lie in the row or column belonging to that square ; hence there are for the two pawns 42 x 41 positions, We thus have 56x 42x41 arrangements. Altogether we have a XXXVI DIRECT CALCULATION OF PROBABILITIES 545 8x 49 x 48 +56 x 42 x 41=56 x 49 x 42 arrangements in which the rook can take neither pawn. Hence the chance that the rook can take neither pawn is 56 x 49 x 42/62 x 56 x 55=1029/1705. The chance that the rook can take one or both of the pawns is therefore 1 — 1029/1705 = 676/1705. Consider now the attack on the rook. If he is on a side square, he can only be attacked by either of the two pawns from one square. For the side Squares we have therefore only 24 x 54 arrangements in which the rook can be taken. There remain 36 squares on each of which the rook can be taken from two squares, that is, in 6 ways. For the 86 squares we therefore have. ea Pn, 36 x 2+36 x 4 x 53 arrangements in which the rook can: be taken by one or by both the pawns. Altogether there are 9000 arrangements in which the rook may be taken. Hence the chance that he be in danger is 9000/62 x 56 x 55 = 225/4774. The ratio of the two chances is 9464 : 1125. § 4.] A considerable number of interesting examples can be solved by the method of chap. xxiii, § 15. Let there be r bags, the first of which contains a,, b,,%, .. ., %, counters, marked with the numbers a,, (,, Yi» + ++ k3 the second, a,, b,, ¢, ... les Eaeodada 0. 9,0". «, ks; and-so on.» If a counter be drawn from each bag, what is the chance that the sum of the numbers drawn is n? By chap. xxiii, § 15, the number of ways in which the sum of the drawings can amount to n is the coefficient, A, say, of x” in the distribution of the product (2,2 + bot +... + heyackt) X (44% + byaP2+ |. + Hyak) X (aya + byaPr +... + kyaokr), Again, the whole number of drawings possible is the sum of all the coefficients ; that is to say, (a, +0,+...+h) x (da tb,+. . . +h) X (dp + bp +. . . +h) =D, say. Hence the required chance is Ar LD. Example 1. A throw has been made with three dice. The sum is known to be 12; required the probability that the throw was 4,4, 4. The number of ways in which 12 can be thrown with three dice is the co- efficient of x}? in i (al + 027 + 223 + act + a5 + 916)3, VOL. II 2N y ‘ 546 DIRECT CALCULATION OF PROBABILITIES CHAP. that is to say, of a in (L+otart+ae + at+ 2)8, Now the coefficients in (l+a+.. .+2°)? up to the term in 2° are (see chap. iv., §15)1+2+3+4+4+54+6+4+544+4+38+42. Hence the coefficient of 2° in the cube of the multinomial is 5+6+5+4+3+42=25.* The required probability is therefore 1/25. Example 2. One die has 3 faces marked 1, 2 marked 2, and 1 marked 8 ; another has 1 face marked 1, 2 marked 2, and 3 marked 3. What is the most probable throw with the two dice, and what the chance of that throw ? The numbers of ways in which the sums 2, 3, 4, 5, 6 can be made are the coefficients of x?, a, a4, xv, x6 in the expansion of (8a + 2u? +23) (w+ 2a? + 823), Now this product is equal to 3x? + 8a? + 14x4 + 82> + 3x8, The sum that will occur oftenest in the long run is therefore 4. The whole number of different ways in which the different throws may turn out is (83+2+1)(1+2+3)=36. Hence the probability of the sum 4 is 14/36 1/18. | Example 3. An urn contains m counters marked with the numbers 1, 2, ...,m. A eounter is drawn and replaced 7 times ; what is the chance that the sum of the numbers drawn is n ?+ The whole number of possible different drawings is m”. The number of those which give the sum » is the coefficient of a” in (ete?+...+a™)", that is to say, of a in (l+a+...+2a™—1)r, Now 1+au+...+a"-l=(1-a™)/(1-a). We have therefore to find the coefficient OfaRe =" In. (1- ids aa - Lt = 1 — Oy em + 050? — Cs a3™ + ae .} ret) go Tr Flee } x {14tet 12 OT rea oe ae 0? A ee The coefficient in question is n art) 3. (nad) ee ey (n—7)! (n-r—m)!1! r(irt+1)...(~-2m—-1)r(r-1) ~ "(= 7—2m)21 + The required probability is A,—,/m”. Example 4. If m odd and m even integers (7 6 b y a Ol XXXVI ADDITION RULE 547 integers into the spaces so that there shall always be one at least in every one of the m—1spaces. A little consideration will show that the number of ways, irrespective of order, is the coefficient of 2” in (lteta?+. .. adwo)%u+a24+, . ; ad co )m—1 . that is, of a—"tin (l+ado2+.. PL +atar+ . . .)m-i, that is, of a-m+1 jn (1 —a)-(t+)), This coefficient is (m+1)(m+2)...(n+1) (n+1)! (n-—m+1)! mm \(nm—m+1)! If we remember that every distribution of the n integers among the m-+1 Spaces can be permutated in x! ways, we now see that the number of ways in which the m+v7 integers can be arranged as required is mim !(n+1)!/m !(n-m+1) !=n1(n+1)!/(w-m+1)!. The whole number of ways in which the m+n integers can be arranged is (m-+7)!, hence the probability required is 2!(n+1)!/(n—m+ 1)!(m+n)!. ADDITION AND MULTIPLICATION OF PROBABILITIES. § 5.] In many cases we have to consider the probabilities of a set of events which are of such a nature that the happening of any one of them upon any occasion excludes the happening of any other upon that particular occasion. A set of events so related are said to be mutually exclusive. The set of events con- sidered may be merely different ways of happening of the same event, provided these ways of happening are mutually exclusive. In such cases the following rule, which we may call the Addition Rule, applies :-— Lf the probabilities of n mutually exclusive events be Dit Dak Pn; the chance that one out of these n events happens on any particular occasion on which all of them are in question is Diels tie sa. Dy, To prove this rule, consider any large number N of occasions where all the events are in question. Out of these N occasions the n events will happen on p,N, p,N, . . ., PrN occasions respect- ively. There is no cross classification here, since no more than one of the events can happen on any one occasion. Out of N occasions, therefore, one or other of the n events wil] happen on Voge). .+9,N=(p,4+0,4... + Ym) N occasions. Hence the probability that one out of the.» events happens on any one occasion is p,+P.+. ..+ Dn 548 MULTIPLICATION RULE CHAP. It should be observed that the reasoning would lose all force if the events were not mutually exclusive, for then it might be that on the p,N occasions on which the first event happens one or more of the others happen. We shall give the proper formula in this case presently. As an illustration of the application of this rule, let us suppose that a throw is made with two ordinary dice, and calculate the probability that the throw does not exceed 8. There are 7 ways in which the event in question may happen, namely, the throw may be 2, 3, 4, 5, 6, 7, or 8; and these ways are of course mutually exclusive. Now (see § 4, Example 1) the probabilities / of these 7 throws are 1/36, 2/36, 3/36, 4/36, 5/36, 6/36, 5/36 respectively. Hence the probability that a throw with two dice does not exceed 8 is (14243444546 +5)/36 = 26/36 =13/18. § 6.] When a set of events is such that the happening of | any one of them in no way affects the happening of any other, we say that the events are mutually independent. For such a set of events we have the following Multiplication Rule :— If the respective probabilities of n independent events be p,, po, 45 Pny the probability that they all happen on any occasion in which all of them are in question 18 Pp, p.+.+ Pn In proof of this rule we may reason as follows :—Out of any large number N of cases where all the events are in question, the first event will happen on p,N occasions. Out of these p, N occasions the second event will also happen on p,(p,N) =p, p.N occasions; so that out of N there are p,p,N occasions on which both the first and second events happen. Continuing in this way, we show that out of N occasions there are Pip2-++PnN occasions on which all the n events happen. The probability that all the m events happen on any occasion 1s there- fore P,P.--+ Pn . It should be noticed that the above reasoning would stand if the events were not independent, provided p, denote the probability that event 2 happen after event 1 has happened, p, the probability that 3 happen after 1 and 2 have happened, and sO on. It must be observed, however, that the probability calculated is then that the events happen in the order 1, 2, 3,..., n. Hence the following conclusion :— XXXVL EXAMPLES OF ADDITION AND MULTIPLICATION 549 Cor. If the events 1, 2,.. ., m be interdependent and p, denote the probability of 1, Pz the probability that 2 happen after 1 has happened, p, the probability that 3 happen after 1 and 2 hawe happened, and so on, then the probability that the events LeR2 yt ape te happen in the order indicated is Oper eve. | As an illustration of the multiplication rule, let us suppose that a die is thrown twice, and calculate the probability that the result is such that the first throw does not exceed 3 and the second does not exceed 5. The probability that the first throw does not exceed 3 is, by the addition rule, 3/6 ; the probability that the second does not exceed 5 is 5/6. The result of the first throw in no way affects the result of the second ; hence the probability that the result of the two throws is as indicated is, by the multiplication rule, (3/6) x (5/6) =5/12. As an example of the effect of a slight alteration in the wording of the question, consider the following :—A die has been thrown twice : what is the probability that one of the throws does not exceed 3 and the other does not exceed 5 ? Since the particular throws are now not specified, the event in question happens—lIst, if the first throw does not exceed 3 and the second does not exceed 5 ; 2nd, if the first throw is 4 or 5 and the second does not exceed 3, These cases are mutually exclusive, and the respective probabilities are 5/12 and 1/6. Hence, by the addition rule, the probability of the event in question 18. 7/12. § 7.] The following examples will illustrate the application of the addition and multiplication of probabilities, Example 1. One urn, A, contains m balls, pm being white, (1—p)m black; another, B, contains » balls, gn white, (1-q)n black. A person selects one of the two urns at random, and draws a ball ; calculate the chance that it be White; and compare with the chance of drawing a white ball when all the m+n balls are in one urn. There are two ways, mutually exclusive, in which a white ball may be drawn, namely, from A or from B. The chance that the drawer selects the urn A is 1/2, and if he selects that urn the chance of a white ball is p. Hence the chance that a white ball is drawn from A is (§ 6, Cor.) dp. Similarly the chance that a white ball is drawn from B is ¢. The whole chance of drawing a white ball is there- fore (p +q)/2. If all the balls be in one urn, the chance is (pm + qn)/(m+n). Now (pm +qn)/(m+n)> = <(p+q)/2, according as 2(pm + qn) > = <(p+q) (m+n), according as (m—n)(p-q)>= <0. Hence the chance of drawing a white ball will be unaltered by mixing if either the numbers of balls in A and B be equal, or the proportion of white balls in each be the same. +. chi w-—-) FO awe 550 * EXAMPLES OF MULTIPLICATION AND ADDITION CHAP. If the number of balls be unequal, and the proportions of white be un- equal, then the mixing: of the balls will increase the chance of drawing a white if the urn which contains most balls have also the larger proportion of white ; and will diminish the chance of drawing a white if the urn which contains most balls have the smaller proportion of white. De Morgan* has used a particular case of this example to point out the danger of a fallacious use of the addition rule. Let us suppose the two urns to be as follows: A (3 wh., 4 bl.); B (4 wh., 3 bl). We might then with some plausibility reason thus :—The drawer must select either A or B. If he select A, the chance of white is 3/7; if he select B, the chance of white is 4/7. Hence, by the addition rule, the whole chance of white is 3/7+4/7=1. In other words, white is certain to be drawn, which is absurd. The mistake* consists in not taking account of the fact that the drawer has a choice of urns and that the chance of his selecting A must therefore be multiplied into his chance of drawing white after he has selected A. The chance should there- fore be 3/14+ 4/14=1/2. The necessity for introducing the factor 1/2 will be best seen by reasoning directly from the fundamental definition. Let us suppose the drawer to make the experiment any large number N of times. In the long run the one urn will be selected as often as the other. Hence out of N times A will be selected N/2 times. Out of these N/2 times white will be drawn from A (3/7) (N/2) =N(3/14) times. Similarly, we see that white will be drawn from B N(4/14) times. Hence, on the whole, out of N trials white will be drawn (3/14+4/14)N times. The chance is therefore 3/14+4/14. Example 2. Four cards are drawn from an ordinary pack of 52 ; what is the chance that they be all of different suits ? We may treat this as an example of § 6, Cor. The chance that the first card drawn be of one of the 4 suits is, of course, 1. The chance, after one suit is thus represented, that the next card drawn be of a different suit is, since there are now only 3 suits allowable and only 51 cards to choose from, 3.13/51. After two cards of different suits are drawn, the chance that the next is of a different suit is 2.13/50. Finally, the chance that the last card is of a different suit from the first three is 13/49. By the principle just mentioned the whole chance is therefore 3.13.2.13.13/51.50.49 = 137/17. 25.49 =1/10 roughly. Example 8. How many times must a man be allowed to toss a penny in order that the odds may he 100 to 1 that he gets at least one head ? Let x be the number of tosses. The complementary event to “ one head at least” is ‘‘ all tails.” Since the chance of a tail each time is 1/2, and the result of each toss is independent of the result of every other, the chance of ‘all tails” in x tosses is (1/2)*. The chance of one head at least is therefore 1—(1/2)". By the conditions of the question, we must therefore have 1 —(1/2)*=100/101 ; * Art. ‘Theory of Probability,” Zney. Metrop. Republished Eney. Pure Math. (1847), p. 399. XXXVI EXAMPLES OF MULTIPLICATION AND ADDITION 551 hence 2%=31 01 “=log 101/ log 2, =2-0043/°3010, 26°66 ioe It appears, therefore, that in 6 tosses the odds are less than 100 to 1, and in 7 tosses more, Example 4. A man tosses 10 pennies, removes all that fall head up ; tosses the remainder, and again removes all that fall head up ; and so on. How many times ought he to be allowed to repeat this operation in order that there may be an even chance that before he is done all the pennies have been removed ? Let x be the number of times, then it is clearly necessary and sufficient for his success that each of the 10 pennies shall have turned up head at least once. The chance that each penny come up head at least once in z trials is 1—(1/2)". Hence the chance that each of the 10 has turned up heads at least once is {1 —(1/2)"}}°, By the conditions of the problem we must therefore have . {1 - (1/2)*}10=1/2 ; (1/2)"=1 — (1/2)¥10 = -06697 ; «= — log :06697/ log 2, =3°9 very nearly. Hence he must have 4 trials to secure an even chance. Example 5. A man is to gain a shilling on the following conditions. He draws twice (replacing each time) out of an urn containing one white and one black ball. If he draws white twice he wins. If he fails a black ball is added, he tries twice again, and wins if he draws white twice. If he fails another black ballis added ; and go on, ad infinitum. What is his chance of gaining | the shilling? (Laurent, Calcul des Probabilités (1873), p. 69.) The chances of drawing white in the various trials are 1 [240 Leer oe 1/n?,... The chances of failing in the various trials are 1-1 /2?, Delis. 1, 1—1/n?,.7 . Hence the chance of failing in all the trials is (1-1/2?) (1 — 1/37)... (1-1/n?)... ad wo, Now 1 1 1 c(t 2)(1-a)-- 0-3) ae eae dd Qed bp Un = 3)(n=1)} {(m = 2)n} {(m~-1)(n+1)} ° 5 ? aes Tene ene n(n +1) > N=n 2n2 nN 1 1 1 — aoe By 3(1 +5)=5: Sy ny 2 The chance of failing to gain the shilling is therefore 1/2. Hence the chance of gaining the shilling is 1/2. We might have calculated the chance of gaining the shilling directly, by et 552 EXAMPLES OF MULTIPLICATION AND ADDITION CHAP. observing that it is the sum of the chances of the following events: 1°, gaining in the first trial; 2°, failing in 1st and gaining in 2nd; 3°, failing in 1st and 2nd and gaining in the 38rd; and soon. In this way the chance presents itself as the following infinite series :— 1 iby Les 1 7 at(1- 5 Un opaehe recast) eee acest ae The sum of this series to infinity must therefore be 1/2. That this is so may be easily verified. The present is one example among many in which the theory of probability suggests interesting algebraical identities. Example 6. A and B cast alternately with a pair of ordinary dice. A wins if he throws 6 before B throws 7, and B if he throws 7 before A throws 6. If A begin, show that his chance of winning: B’s=30: 381. hick De Ratiociniis in Ludo Alex, 1657.) Let p and q be the chances of throwing and of failing to throw 6 at a single cast with two dice ; 7 and s the corresponding chantves {Ort A may win in the Piva ae ways: 1°, A succeed at 1st throw; 2°, A fail at Ist, B fail at 2nd, A succeed at 3rd; and soon. His chance is therefore represented by the following infinite series :— ptasp+qsqgspt.. .=p{1+(qs)+(qs?+.. .t, =p/(1- 4s). B may win in the following ways:—1°, A fail at 1st, B succeed at 2nd ; 2°, A fail at Ist, B fail at 2nd, A fail at 8rd, B succeed at 4th; and so on. His chance is therefore qr +qsqr + qsqsqr+. . . =gr{1+(qs)+(gsP+.. .}, =qr/(1 - 9s). Hence A’s chance: B’s=p: qr. Now (see § 4, Example 1) p=5/36, g=31/36, r=6/36; hence A’s chance : B’s=5/36 : 6.31/36?, =30 2:31. For Huyghens’ own solution see Todhunter, Hist: Prob., p. 24. Example 7. A coin is tossed m+n times (m>n). ° Prove that the chance of at least m consecutive heads appearing is (n+2)/2™+1, The event in question happens if there appear—lIst, exactly m; 2nd, exactly m+1;...3 (~+1)th, exactly m+n consecutive heads. Now a run of exactly m consecutive heads may commence with the Ist, 2nd, 8rd, n—1th, nth, n+1th throw. Since m>n, there cannot be more than one run of m or more consecutive heads, so that the complication due to repetition of runs does not occur in the present problem. The chances of the first and last of these cases are each 1/2”+1, the chances of the others 1/2™+2, Hence the chance of a run of exactly m consecutive heads is 2/amr2 +(m—1)/2"+? = (n+ 3)/2™+, In like manner, we see that the chance of a run of m+1 consecutive heads is (1 +2)/2™+8 ; and so on, up to m+n-2. Also the chances of a run of exactly m+n—-1 and of exactly m+ consecutive heads are 1/2”+"-1 and - 1/2”+ respectively. XXXVI PROBABILITY OF COMPOUND EVENTS 553 Hence the chance p of a run of at least m heads is given by _n+38 n+2 5 4 ] PP eas Qm+2 Qm+3 Orla y Qm+n “of 9m+n+1 Qm-+n E The summation of the series on the left-hand side is effected (see chap. xx., § 13) by multiplying by (1-1/2)?=1/4. We thus find _nt+38 n+2 n+1 4p Daas amet om+3 E oma t gM YS 9m-+n+1 2(n+38) 2(n+2) mn 2.4 vi 9m+3 os om+4 a. ps ee Qm+n+1 , Qm+n+2 n+3 6 5 4 ie x Oyama TS 2 F gQm+n+1 S, gm+n+2 i 9m+n+3 Ni: Qm+n+2 ? n+3 n+4 3 Y's 1 ap a Qm-+2 + Qm-+3 me Qm-+n+2 ar Qm+n+2 a Qm-+n+2? +2 pu: 9m-+3 , Hence p=(n+2)/2"4, GENERAL THEOREMS REGARDING THE PROBABILITY OF COMPOUND EVENTS. § 8.] The probability that an event, whose probability is p, happen on exactly r out of n occasions in which it is in question is »Cpp'g”-", where ¢= 1 —~p is the probability that the event Sail. The probability that the event happen on 7 specified occasions and fail on the remaining n—r is by the multiplication rule Ppqpqq ... where there are r p’s and n-+r q's, that is, p"g™-". Now the occasions are not specified ; in other words, the happen- ing, and failing, may occur in any order. ‘There are as many ways of arranging the r happenings and n—,r failings as there are permutations of n things r of which are alike and n —r alike, that is to say, n!/ ri(n-r)l= aba There are therefore ,,C, mutually exclusive ways in which the event with which we are concerned may happen ; and the probability of each of these is pg". Hence, by the addition rule, the probability in question iA OS a aaa It will be observed that the probabilities that the event happen exactly n,n-—1,..., 2, 1, 0 times respectively, are the Ist, 2nd, 3rd,. . ., (w+ 1)th terms of the expansion of (p+q)”. Since, if we make 1 trials, the event must happen either 0, 554 PROBABILITY OF COMPOUND EVENTS CHAP. or 1, or 2,. . ., or m times, the sum of all these probabilities ought to be unity. This is so; for, since p+ q=1, (p+q)"=1. It will be seen without further demonstration that the pro- position just established is merely a particular case of the following general theorem :— If there be m events A, B, C,. . . one but not more of which must happen on every occasion, and tf their probabilities be », q, 7, . respectively, the probability that on n occasions A happen exactly a times, B exactly B times, C exactly y times, . . . is ni p’g't”...falBly!.. whereatBPtryt+....=N It should be observed that the expression just written is the general term in the expansion of the multinomial Go aor e erat ys" | Example 1. The faces of a cubical die are marked 1, 2,2, 4, 4, 6; required the probability that in 8 par g Ne 1, 2, 4 turn up ores 3, 2,3 Himes respectively. By the general theorem just stated the probability is atin) G8 aryl approximately. Example 2. Out of m occasions in which an event of probability p is in question, on what number of occasions is it most likely to happen ? We have here to determine 7 so that ,,C, p”q"-" may be a maximum. Now nln a" /,.C,ag tg Hh S(n=9 + 1 pre. Hence the probability will increase as 7 increases, so long as (n-r+1)p>rq, that is, (n+1)p>r(p+q), that is r<(n+1)p. If (1+1)p be an integer, =s say, then the event will be equally likely to happen on s—1 or ons occasions, and more pst to happen s—1 or s times than any other number of times. If (1 +1)p be not an integer, and s be the greatest integer in (n+1)p, then the event is most likely to happen on s occasions.* * When is very large, (n+1)p differs inappreciably from np. Hence out of a very large number 7 of occasions an event is most likely to happen on pn occasions. This, of course, is simply the fundamental principle of § 2, Cor. 1, arrived at by a circuitous route starting from itself in the first instance. XXXVI PASCAL’S PROBLEM 555 As a numerical instance, suppose an ordinary die is thrown 20 times, what is the number of aces most likely to appear ? Here m=20; p=1/6; (n+1)p=88. The most likely number of aces is therefore 3. § 9.] The probability that an event happen on at least r out of n occasions where it is in question ts WP + Cp pr rigr 714. + Cn p™-lg +p . . (1). For an event happens at least + times if it happen either exactly 7; or exactly r+1;...; or exactly nm times. Hence the probability that it happens at least 7 times is the sum of the probabilities that it happens exactly 7, exactly r+1,.. ., exactly times ; and this, by § 8, gives the expression (1), Another expression for the probability just found may be deduced as follows:—Suppose we watch the sequence of the happenings and failings in a series of different cases. After we have observed the event to have happened just 7 times, we may withdraw our attention and proceed to consider another case ; and so on. Looking at the matter in this way, we see that the r happenings may be just made up on the rth, or on the r + 1th, . ., or on the uth occasion. If the r happenings have beeh made up in just s occasions, then the event must have happened on the sth occasion and on any *—1 of the preceding s—1 occasions. The probability of this contingency is px AO iy [ATs ces = Peep Hence the probability that the event happen at least 7 times in nm trials is eee 710,09 +... ty Cy rp g?-? SO pa Cog 63.) tet Onarga ty (2): As the two expressions (1) and (2) are outwardly very different, it may be well to show that they are really identical. Todo this, we have to prove that 1 + Cig t+r41Cog?+ ni ei 8 Shine Cn Oe SH + 1+.C (2) +ca(£) +. s#Cn( 2)" “ { *\p p Tom Selly et anar oy Sie Hee a a \ ee ee (1 7) { 1 ele = -) taal 4 a _) toes tok nme 5 a -) \ 556 GENERAL FORMULA FOR COMPOUND EVENT CHAP. The expression last written is, up to the (x —7)th power of g, identical with (lag) {1 +9/(l— 9) }"9= (nae ay re Now, as may be readily verified, (L—g)-7=14+,Ci¢gtenCeg?+...; “bin-1 Uno The required identity is therefore established. Example. A and B play a game which must be either lost or won; the probability that A gains'any game is p, that B gains it 1-p=q; what is the chance that A gains m games before B gains 2? (Pascal’s Problem.)* The issue in question must be decided in m+n-1 games at the utmost. The chance required is in fact the chance that A gains m games at least out of m+-n—1, that is, by (1) above, F Sdschassame apt Pe | pment Metin On ee (1°: We might adopt the second way of looking at the question given above, and thus arrive at the expression psd timCig + mpeg? +. . « Himtn—2Cn-19" 4} (2’). for the required chance. “§ 10.] The results just arrived at may be considerably general- ised. Let us consider » independent events A,, As, «aeons whose respective probabilities are Dir Pas + + By Dns In the first place, in contrast to §§ 8,9, let us calculate the chance that one at least of the n events happen. The complementary event is that none of the n events happen. The probability of this is (1 — p,) (1 —p,)...(1-pn). Hence the probability that one at least happen is Boe Ur) Clima tos Mota = Dp, — 2p; Pati DsDs— » . - (—)"-1 9) D5 wee Next let us find the probability that one and no more of the n events happen. ‘The probability that any particular event, say A,, and none of the others happen is p,(1—p,)(1—>p,)...(1—pp,). Hence the required probability is =p(1 — pe) (1 — ps) ee (1 — Pn) aaa =p, us 2 2D De 3 gC, 2D, DoDs eS aie a® ( - yr) Gass ee -Pn (2). * Famous in the history of mathematics. It was first solved for the - particular case y=q by Pascal (1654). The more general result (1') above was given by John Bernoulli (1710). The other formula (2’) seems to be due to Montmort (1714). See Todhunter, Hist. Prob., p. 98. XXXVI GENERALISATION OF PASCAL’S PROBLEM Hos For the products two and two arise from — =Pi( po + Ps + + py), and each pair will come in once for every letter in it. Again, the products three and three arise from Sp, (Dds Pep, oe ye hence each triad will come in once for every pair of Teeter that can be selected from it; and so on. By precisely similar reasoning, we can show that the pyr obability y that r and no more of the n events Ree as 2D; Pe - : Se Mec pat) (Drive) (1 — pn) ey id Uf el Pein OD) 1 a en + dia ieee Uete e Ny r+s C29; 0, Was Pr+s ( oe Oe »++ Dn (3). a We can now calculate the probability that r at least out of the n events happen. To do so we have merely to sum all the values of (3) obtained by giving r the values 7,7 +1,7r+2,..., n iyie Se In this summation the coefficient of Sp, p,... D+ is ( ee Pires Dy, rtsCs—1 “5 r+sUg—s ae he ye rsU sr ( a ve Now the expression within the praceaee is the coe of a in (1 + #)'t* x (1+2)-}, that is to say, in (1+2)'t*-1 This coefficient is »,,_,C,. Hence the coefficient of By Da ts a0, es ( a eee s The probability that r at least out of the m events happen is therefore | Fa Oey Oia OND (ay ama (Pea a ra P Ds s+ + Pyte ( } 7-8 — a =P ++ Pr+s (- Jn ne) Oi rPiPo-++PpPn (4). Since the happening of the same event on n different occasions may be regarded as the happening of n different events whose 558 THIRD SOLUTION OF PASCAL’S PROBLEM CHAP. probabilities are all ae the formule (3) and (4) above ought, when 7, =9,= =P, each = p, to reduce to ,C,p"9"-" and the expression (1) ¢ or (2) of § 9 respectively. If the reader observe pat: when 9,=9,= . = peg, LP, s- --Pr=a0, p'; &e,, he will have no aifionleys in iano that (3) is actually identical with ,C,p"q"-" in the particular case in question. The particular result derived from (4) is more interesting. We find for the probability, that an event of probability p will happen 7 times at least out of m occasions, the expression Ar" — OyqQres Pte oe oO) eps Or pee ( de \0-? 3, Ce (5), Here we have yet another expression equivalent to (1) and (2) of § 9. It is not very difficult to transform either of the two expressions of § 9 into the one now found; the details may be left to the reader. Example. The probabilities of three independent events are p, g, 7; re- quired the probability of happening— Ist. Of one of the events but not more ; 2nd. Of two but not more ; ard. Of one at least ; Ath. Of two at least ; 5th. Of one at most ; 6th. Of two at most. The results are as follows :— Ist. ptqg+r—2(patpr+qr) + 3pqr ; 2nd. pat+prt+qr—dspqr ; srd. pt+q+r—(pat+pr+aqr)+paqr ; 4th. pat+pr+qr—2pqr ; 5th. 1 -(pat+prt+qr)+2pqr ; 6th. 1—~pgqr. The first four are particular cases of preceding formule; 5 is comple- mentary to 4; and 6 is complementary to ‘‘ of all three.” § 11.] The Recurrence or Finite Difference Method for solving problems in the theory of probability possesses great historical and practical interest, on account of the use that has been made of it in the solution of some of the most difficult questions in the subject. The spirit of the method may be explained thus. XXXVI RECURRENCE METHOD 559 Suppose, for simplicity, that the required probability is a function of one variable a; and let us denote it by uz. Reasoning from the data of the problem, we deduce a relation connecting the values of uw, for a number of successive values of a; say the relation LC) AU TAK) (A). We then discuss the analytical problem of finding a function Uz, Which will satisfy the equation (A). It is not by any means necessary to solve the equation (A) completely. Since we know that our problem is definite, all that we require is a form for u, which will satisfy (A) and at the same time agree with the conditions of the problem in certain particular cases. The following examples will sufficiently illus- trate the method from an elementary point of view. Example 1. A and B play a game in which the probabilities that A and B - win are a and @ respectively, and the probability that the game be drawn is y. To start with, A has m and B has ” counters. Each time the game is won the winner takes a counter from the loser. If A and B agree to play until one of them loses all his counters, find their respective chances of winning in the end.* Let wv, and v, denote the chances that A and B win in the end when each has x counters. If we put m+n=p, the respective chances at any stage of the game are wz and Up_,. Consider A’s,chance when he has x+1 counters. The next round he may, Ist, win; 2nd, lose; 8rd, draw the game. The chances of his ultimately winning on these hypotheses are atiz42; Biz ; Yat respectively. Hence, by the addition rule, Ug] = AUy49 + Bu, + YUz41. If we notice that a+8++y=1 (for the game must be either won, lost, or drawn), we deduce from the equation just written ; AUz+9 — (a+ B)e41+ Buz=0 (1). It is obvious that w,=AX*, where A and X are constants, will be a solution of (1), provided ah? —(a+8)X\+B=0 (2), that is, provided \=1 or X=8/a. Hence wy=A and Uz = B(B/a)* are both solutions of (1); and it is further obvious that U,=A + B(B/a)* is a solution of (1). We have now the means of solving our problem, for it is clear from (1) that, if we knew two particular values of w,, say wo and w, then all other * First proposed by Huyghens in a particular case; and solved by James Bernoulli. See Todhunter, Hist. Prod., p. 61. 560 PROBLEM REGARDING DURATION OF PLAY CHAP. values could be calculated by the recurrence formula (1) itself. The solution U,=A +B(8/a)*, containing two undetermined constants A and B, is therefore sufficiently general for our purpose.* We may in fact determine A and B most simply by remarking that when A has none of the counters his chance is 0, and when he has all the counters his chance is 1. We thus have A+B=0, A+B(8/a)?=1, whence A=a?/(a?—B?), B= -a?/(aP—B?), We therefore have Uy, = aP—*(a% — B*)/(aP — BP) ; 0_ = B?-*(a* — B*)/(a? — 8”). The chances at the beginning of the game are given by Um = an(a rt p™)/(aP a; B?), tin = Bal" — B")|(a” — B). Cor. 1. Jf a=B, then (see chap. xxv., § 12) Um=M/P, Uy,n=n/p. The odds on A in this particular case are m ton. It might be supposed that when the skill of the players is unequal this could be compensated by a disparity of counters. There is, however, a limit, as the following proposition will show :— and, in like manner, Cor. 2. The utmost disparity of counters cannot reduce the odds in , this can - ene z imereaninn N 3 we since L (6/a)"=0, it cannot become N=0 less than (a —)/6:1, that is, a—6:f. Hence we see that, if A be twice as skilful as B(a=28), we cannot by any disparity of counters (so long as we give him any at all) make the odds in his favour less than even. Example 2. A pack of diferent cards is laid face downwards. A person names a card ; and that card and all above it are removed and shown to him. He then names another ; and so on, until none are left. Required the chance that during the operation he names the top card once at least.+ Let uw, be the chance of succeeding when there are cards ; so that w%,»—} is the chance of succeeding when there are n—1; andsoon. At the first trial the player may name the Ist, 2nd, 38rd, . . ., or the mth card, the chance of each of these events being 1/n. Now his chances of ultimately succeeding in the n cases just mentioned are 1, tn-2, Ung, . » «5 %, 0 respectively. Hence Un=1/0+Un—2/0 + Un—slV+ . . . +Ue/n+Uy/n We have therefore NUn=1L+Uytugt. . . +Upn-g se * This piece of reasoning may be replaced by the considerations of chaps, X5x1L, 9c. t+ Reprint of Problems from the Ed. Times, vol. xlii., p. 69. XxxvI EVALUATION OF PROBABILITIES INVOLVING FACTORIALS 56] From (1) we deduce (2 —1)tlpiy=1+mtuet... +p (2). From (1) and (2) NUn — (0 — 1)tn-17=Un_o, that is, 2(Un — Un—1) = — (Un—1 — Un—2) (3). Hence (n = 1) (Un—1 ay Un—2) he (Un—g = Un—3)s (1 5 2) (Un—2 in Un—3) ary (Un—s = Und); 3(W3 — U2) = — (uy — Uy). Hence, multiplying together the last n—2 equations, we deduce 30! (Un — Un—1)=(- 1)”(us — Uy). Since w#=1, w2=4, this gives Un a Un-1 = ( 6 Leen ! (4), Hence, again, ¥ Un—1 — Un—2=(—1)"-*/(n -1)1, U2 -W=(- 1)1/2 r oe Wat From the last x equations we derive, by addition, Um=1—T/2!+1/a1—. .. +(-1)Y/n! (5). Introducing the sub-factorial notation of chap. xxiii., § 18, we may write the result obtained in (5) in the form w,=1- 1; /n!. From Whitworth’s Table * we see that the chance when »=8 is °632119. When n= the chance is 1 — 1/e= 632121; so that the chance docs not diminish greatly after the number of cards reaches 8. EVALUATION OF PROBABILITIES WHERE FACTORIALS OF LARGE NUMBERS ARE INVOLVED. § 12.] In many cases, as has been seen, the calculation of probabilities depends on the evaluation of factorial functions. When the numbers involved are large, this evaluation, if pursued directly, would lead to calculations of enormous length,+ and the greater part of this labour would be utterly wasted, since all that is required is usually the first few significant figures of the probability. The difficulty which thus arises is evaded by the use of Stirling’s Theorem regarding the approximate value of a! * Choice and Chance, chap. iv. tT In some cases the process of chap. xxxv., § 11, Examples 2 and 3 is useful. VOL. II 20 562 EXERCISES OKT CHAP, when z is large. In its modern form this theorem may be stated thus— i ji != ./(2Qriev)a%e-* : Bea |) x J (2rt)a%e (1 tpg aoe ) (see chap. xxx., § 17). From this it appears that, if « be a large number, #! may be replaced by ./(27a)a%e-*, the error thereby committed being of the order 1/12zth of the value of z!.: As an example of the use of Stirling’s Theorem, let us consider the follow- ing problem :—A pack of 4n cards consists of 4 suits, each consisting of n cards. The pack is shuffled and dealt out to four players; required the chance that the whole of a particular suit falls to one particular player. The chance in question is easily found to be given by . p=(8n)!n1/(4n)!. Hence, by Stirling’s Theorem, we have es A/ (2m 8n) (82)? e-F"/(Qarn) nr e-” en A/ (2m 4n) (4n)e~- i the error being comparable with 1/11nth of py. Hence, approximately, p=rn/(8rn/2) (27/256)". Example. Let 4n=52, n=18, then p=/(8x 3'1416 x 13/2) (27/256), This can be readily evaluated by means of a table of logarithms. We find p=156/104. The event in question is therefore not one that would occur often in the experience of one individual. EXERCISES XX XIX. (1.) A starts at half-past one to walk up Princes Street; what is the probability that he meet B, who may have started to walk down any time between one and two o’clock? Given that it takes A 12 minutes to walk up, and B 10 minutes to walk down. (2.) A bag contains 3 white, 4 red, and 5 black balls. Three balls are drawn ; required the probability—Ist, that all three colours ; 2nd, that only two colours ; 3rd, that only one colour, may be represented. (3.) A bag contains m white and 7 black balls. One is drawn and then a second ; what is the chance of drawing at least one white—Ist, when the first ball is replaced ; 2nd, when it is not replaced ? (4.) If m persons meet by chance, what is the probability that they all have the same birthday, supposing every fourth year to be a leap year ? (5.) If a queen and a knight be placed at random on a chess-board, what is the chance that one of the two may be able to take the other ? XXXVI EXERCISES XXXIX 563 (6.) Three dice are thrown ; show that the cast is most likely to be 10 or 11, the probability of each being }. (7.) There are three bags, the first of which contains 1, 2, 1 counters, marked 1, 2, 3 respectively ; the second 1, 4, 6, 4, 1, marked 1, 2, 3, 4, 5 re- spectively ; the third 1, 6, 15, 20, marked 1, 2, 3, 4 respectively. A counter is drawn from each bag ; what is the probability of drawing 6 exactly, and of drawing some number not exceeding 6 ? (8.) Six men are bracketed in an examination, the extreme difference of their marks being 6. Find the chance that their marks are all different, (9.) From 2n tickets marked Osi eur (20 1). 0 are drawh ; find the probability that the sum of the numbers is 2n. (10.) A pack of 4 suits of 13 cards each is dealt to 4 players. Find the chance—Ist, that a particular player has no card of a named suit ; 2nd, that there is one suit of which he has no card. Show that the odds against the dealer having all the 13 trumps is 158, 753,389,899 to 1. (11.) If I set down any r-permutation of n letters, what is the chance that two assigned letters be adjacent ? (12.) There are 3 tickets in a bag, marked 1, 2,3. A ticket is drawn and replaced four times in succession ; show that it is 41 to 40 that the sum of the numbers drawn is even. (13.) What is the most likely throw with n dice, when n>6 2 (14.) Out of a pack of n cards a card is drawn and replaced. The opera- tion is repeated until a card has been drawn twice. On an average how many drawings will there be 2 (15.) Ten different numbers, each $100, are selected at random and multi- plied together ; find the chance that the product is divisible by 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively. (16.) A undertakes to throw at least one six in a single throw with six dice ; B in the same way to throw at least two sixes with twelve dice ; and C to throw at least three sixes with eighteen dice. Which has the best chance of succeeding ? (Solved by Newton ; see Pepys’ Diary and Correspondence, ed. by Mynors Bright, vol. vi., p. 179.) (17.) A pitcher is to be taken to the well every day for 4 years. If the odds be 1000 : 1 against its being broken on any particular day, show that the chance of its ultimately surviving is rather less than ., (18.) Five men toss a coin in order till one wins by tossing head; calculate their respective chances of winning. ; (19.) A and B, of equal skill, agree to play till one is 5 games ahead. Calculate their respective chances of winning at any stage, supposing that the game cannot be drawn. (Pascal and Fermat. ) (20.) What are the odds against throwing 7 twice at least in 3 throws with 2 dice ? (21.) Show that the chance of throwing doublets with 2 dice, 1 of which is loaded and the other true, is the same as if both were true. 564 EXERCISES XXXIX CHAP. (22.) A and B throw for a stake’; A’s die is marked 10, 13, 16, 20, 21, 25, and B’s 5, 10, 15, 20, 25, 30. The highest throw is to win and equal throws to go for nothing ; show that A’s chance of winning is 17/33. (23.) A pack of 2n cards, 7 red, n black, is divided at random into 2 equal parts and a card is drawn from each ; find the chance that the 2 drawn are of the same colour, and compare with the chance of drawing 2 of the same colour from the undivided pack. (24.) 4m cards, numbered in 4 sets of m, are distributed into m stacks of | 4 each, face up; find the chance that in no stack is a higher one of any set above one with a lower number in the same set. (25.) Out of m men in aring 3 are selected at random ; show that the chance that no 2 of them are neighbours is (m — 4)\(m — 5)/(m —1)(m - 2). (26.) If m things be given to a men and 6 women, prove that the chance that the number received by the group of men is odd is {E(0-+a)"—4(b—a)"} [(b+a)”. (Math. Trip., 1881.) (27.) A and B each take 12 counters and play with 3 dice on this condi- tion, that if 11 is thrown A gives a counter to B, and if 14 is thrown B gives a counter to A; and he wins the game who first obtains all the counters. Show that A’s chance is to B's as 244,140,625 : 282,429,536, 481. (Huyghens. See Todh., Hist. Prob., p. 25.) (28.) A and B play with 2 dice; if 7 is thrown A wins, if 10 B wins, if any other number the game is drawn. Show that A’s chance of winning is to B’sas 13:11. (Huyghens. See Todh., Hist. Prob., p. 23.) (29.) In a game of mingled chance and skill, which cannot be drawn, the odds are 8 to 1 that any game is decided by skill and not by luck. If A beats B 3 games out of 2, show that the odds are 3 to 1 that he is the better player. If B beats C 2 games out of 3, show that the chance of A’s winning 3 games running from C is 103/352. (30.) There are m posts in a straight line at equal distances of a yard apart. A man starts from any one and walks to any other; prove that the average distance which he will travel after doing this at random a great many times is 4(m+1) yards. (31.) The chance of throwing f named faces in » casts with a p+1-faced die is -1 {weap Lm F@Z Dp -ayp—. be taye (Demoivre, Doctrine of Chances.) (32.) If m cards be thrown into a bag‘and drawn out’ successively, the chance that one card at least is drawn in the order that its number indicates is 1=1/21f1/8)=. 6 a ee (This is known as the 7'reize Problem. It was originally solved by Mont- mort and Bernoulli.) XXXVI VALUE OF AN EXPECTATION 565 (33.) A and B play a game in which their respective chances of winning area and 8. They start with a given number of counters p divided between them ; each gives up one to the other when he loses ; and they play till one is ruined. Show that inequality of counters can be made to compensate for inequality of skill, provided a/8 is less than the positive root of the equation xP —2eP-14+1=0. If p be large, show that, to a second approximation, this ie p= 9 p-1 22p-1" root is 2— MATHEMATICAL MEASURE OF THE VALUE OF AN EXPECTATION. § 13.] If a man were asked what he would pay for the privi- lege of tossing a halfpenny once and no more, with the under-_ standing that he is to receive £50 if the coin turn up head, and nothing if it turn up tail, he might give various estimates, accord- ing as his nature were more or less sanguine, of what is some- times called the value of his expectation of the sum of £50. It is obvious, however, that in the case where only one trial is to be allowed the expectation has in reality no definite value whatever—the player may get £50 or he may get nothing ; and no more can be said. . If, however, the player be allowed to repeat the game a large number of times on condition of paying the same sum each time for his privilege, then it will be seen that £25 is an equitable payment to request from the player; for it is assumed that the game is to be so conducted that, in the long run, the coin will turn up heads and tails equally often ; that is to say, that in a very large number of games the player will win about as often as he loses. With the above understanding, we may speak of £25 as the value of the player's expectation of £50 ; and it will be observed that the value of the expectation is the sum expected multiplied by the probability of getting it. This idea of the value of an expectation may be more fully illustrated by the case of a lottery. Let us suppose that there are prizes of the value of £a, £), £c, . . ., the respective prob- abilities of obtaining which by means of a single ticket are P 4% 7%... It the lottery were held a large number N of times, the holder of a single ticket would get £a on pN ocea- 566 ADDITION OF EXPECTATIONS CHAP. sions, £5 on gN occasions, £¢ on 7N occasions, . . . Hence the holder of a single ticket in each of the N lotteries would get £(pNa+qNb+7Nce+.. .). . If, therefore, he is to pay the same ~ price £¢ for his ticket each time, we ought to have, for equity, | Ni= pNa+qNb+rNe+.. ., that is, . f=pa+qo+re+. Hence the price of his ticket is made up of parts correspond- ing to the various prizes, namely, pa, qb, rc, . . . . These parts are called the values of the expectations of the respective prizes ; and we have the rule that the value of the expectation of a sum of money is that sum multiplied by the chance of getting tt. The student must, however, remember the understanding upon which this definition has been based. It would have no meaning if the lottery were to be held once for all. Example. A player throws a six-faced die, and is to receive 20s, if he throws ace the first throw ; half that sum if he throws ace the second throw ; quarter that sum if he throws ace the third throw ; and so on. Required the value of his expectation. The player may get 20, 20/2, 20/22, 20/2%, . . . shillings. His chances of getting these sums are 1/6, 5/62, 52/6%, 53/64, . . . Hence the respective values of the corresponding parts of his expectation are 20/6, 20.5/6?.2, 20.5?/6%.2, 20.57/64.2%, . . . shillings. The whole value of his expectation is therefore fr 8 a(S) a(E) a... atm pa (red B lmeicce a UN Ree seaee = Fe 12 that is, 5s. 84d. ) =2 shillings, § 14.] It is important to notice that the rule which directs us to add the component parts of an expectation applies whether the separate contingencies be mutually exclusive or not. Thus, if Dry Po» Ps, ++ be the whole probabilities of obtaining the separate sums dy, Mz, 3, . . -, then the value of the expectation is P,Q, + Polly + Pz +. . ., even if the expectant may get more than one of the sums in question. Observe, however, that p, must be the whole probability of getting a,, that is, the probability of getting the sum «@, irrespective of getting or failing to get the other sums. If the expectant may get any number of the sums @,, @,, XXXVI ADDITION OF EXPECTATIONS - 567 - +, My, we might calculate his expectation by dividing it into the following mutually exclusive contingencies :—a,, Castine ir cba ieee eG, COC, 5 0, FO, + ee... es HOt. +O, Hence the value of his expectation is 2a, p,(1 =.) (1 ar) we G (1 = Dn) + 2(a, + de), p(1 — ps)... (1 — pn) + D(a; + a, + a5) p,p.p;(1 —p,)... (1 — pn) TO, + Gy. . s+ On) D,Da0e ass Dy By the general principle above enunciated the value in question is also Ya,p,. The comparison of the values gives a curious algebraic identity, which the student may verify either in general or in particular cases. Example. A man may get one or other or both of the sums a and 8. The chance of getting @ is p, and of getting b is g. Required the value of his expectation. He may get w alone, or d alone, or a+b; and the respective chances are P(1— q), q(1-p), pg. Hence the value of his expectation is ap(1-¢)+ ba(1-p) +(a+b)pq, which reduces to ap+bg, as it ought to do by the general principle. NV. £.—If the man were to get one or other, but not both of the sums a and 6, and his respective chances were p and q, the value of his expectation would still be ap+bq; but p and q would no longer have the same meanings as in last case. LIFE CONTINGENCIES. § 15.] The best example of the mathematical theory of the value of expectations is to be found in the valuation of benefits which are contingent upon the duration or termination of one or more human lives. The data required for such calculations are mainly of two kinds—1st, knowledge, or forecast as accurate as may be, of the interest likely to be yielded by investment of capital on good and easily convertible security ; 2nd, statistics regarding the average duration of human life, usually embodied in what are called Mortality Tables. The table printed below illustrates the arrangement of mortality statistics most commonly used in the calculation of life contingencies :— 568 MORTALITY TABLE CHAP. The H™ Table of the Institute of Actuaries. Number | Decre- Number | Decre- Number | Decre- ae Living. | ment. nee Living. | ment. Age Living. | ment. x Bs A& x bee ae x Bs de 10 |100,000 490 || 40 | 82,284 848 || 70 | 38,124 | 2371 11 | 99,510 | 397 |] 41 | 81,436 | 854 || 71 | 35,753} 2433 12 | 99,113 | 329 || 42 | 80,582 | 865 || 72 | 33,320| 2497 13 | 98,784 | 288 || 43 | 79,717 | 887 || 73 | 30,823 | 2554 14 | 98,496 | 272 || 44 | 78,830 | 911 || 74 | 28,269 | 2578» 15 | 98,224 | 282 || 45 | 77,919 | 950 || 75 | 25,691 | 2597 16 | 97,942 | 318 || 46 | 76,969 | 996 || 76 | 23,164 | 2464 17 | 97,624 | 379 || 47 | 75,973 | 1041 || 77 | 20,700] 2374 18 | 97,245 | 466 || 48 | 74,932 | 1082 || 78 | 18,396 | 2258 19 | 96,779 | 556 |] 49 | 73,850 | 1124 || 79 | 16,068 | 2138 20 | 96,223 | 609 || 50 | 72,726 | 1160 | 80 13,930 | 2015 21 | 95,614 | 643 || 51 | 71,566 | 1193 || 81 | 11,915 | 1883 22.| 94,971 | 650 || 52 | 70,373 | 1235 || 82 | 10,082 | 1719 23 | 94,321 | 638 |, 53 | 69,138 | 1286 || 83 | 8,313 | 1545 24 | 93,683 | 622 || 54 | 67,852 | 1339 || 84 | 6,768 | 1346 25 | 93,061 | 617 || 55 | 66,513 | 1399 || 85 | 5,422] 1138 26 | 92,444 | 618 || 56 | 65,114 | 1462 || 86 | 4,284| 941 27 | 91,826 | 634 || 57 | 63,652 | 1527 || 87 | 3,343] 773 28 | 91,192 | 654 || 58 | 62,125 | 1592 || 88 | 2,570] 615 29 | 90,538 |—678-]| 59 | 60,533 | 1667 || 89 | 1,955] 495 30 | 89,865 | 694 || 60 | 58,866 | 1747 || 90 | 1,460] 408 31 | 89,171 | 706 |] 61 | 57,119 | 1830 || 91 | 1,052] 3929 32 | 88,465 | 717 || 62 | 55,289 | 1915 || 92 723| 254 33 | 87,748 | 727 || 63 | 53,374 | 2001 || 93 469] 195 34 | 87,021 | 740 || 64 | 51,373 | 2076 || 94 974 | 139 35 | 86,281 | 757 || 65 | 49,297 | 2141 || 95 135| 86 36 | 85,524 | 779 || 66 | 47,156 | 2196 || 96 49] 40 37 | 84,745 | 802 || 67 | 44,960 | 2243 || 97 9 9 38 | 83,943 | 821 || 68 | 42,717 | 2274 || 98 0 39 | 83,122 £5, 69 | 40,443 | 2319 In the first column are entered the ages 10, 11, 12, . . Opposite 10 is entered an arbitrary number 100,000 of children that reach their tenth birthday ; opposite 11 the number of these that reach their eleventh birthday ; opposite 12 the number that reach their twelfth birthday ; and soon. We shall denote these numbers by /,,, 1,, 2, . .. In a third column are entered the differences, or “decrements,” of the numbers in the second column; these we shall denote by dy, d,,,d.,... It is obvious that d, gives the number out of the 100,000 that die between their zth and «+ 1th birthdays. It is impossible here to discuss the methods employed in constructing a table of mortality, or — a ‘ —- XXXVI USES OF MORTALITY TABLE 569 to indicate the limits of its use; we merely remark that in applying it in any calculation the assumption made is that the — lives dealt with will fall according to the law indicated by the numbers in the table. This law, which we may call the Law of Mortality, is of course only imperfectly indicated by the table itself ; for although we are told that d, die between the ages of « and a+1, we are not told how these deaths are distributed throughout the intervening year. For rough purposes it is sufficient to assume that the distribution of deaths throughout each year is uniform; although the variation of the decrements from one part of the table to another shows that uniform decrease * is by no means the general law of mortality. § 16.] By means of a Mortality Table a great many interesting problems regarding the duration of life may be solved which do not involve the consideration of money. The following are examples. Example 1. By the probable duration n of the life of a man of m years of age is meant the number of years which he has an even chance of adding to his life. To find this number. By hypothesis we have bimtn|lm=1/2. Hence bmtn=lm{2. bn {2 will in general lie between two numbers in the table, say 7, and/,.;. Hence m+n must lie between p and p+1. We can get a closer approximation by the rule of proportional parts (see chap. xxi., § 13). Example 2. To find the “mean duration” or “ expectancy of life” for a man of m years of age. By this is meant the average N (arithmetical mean) of the number of additional years of life enjoyed by all men of m years of age. Let us take as specimen lives the 7 men of the table who pass their mth birthday ; suppose them all living at a particular epoch ; and trace their lives till they all die. In the first year Jn —Jm4i die. If we suppose these deaths to be equally distributed through the year, as many of the Z»— Imi Will live any assigned amount over half a year as will live by the same amount under half a year. Hence the Uy, —Jm41 lives that have failed will contribute 3(lm — (m4) years to the united life of the J,, specimen lives. Again, each of the lmii who live through the year will contribute one year to the united life. Hence the whole contribution to the united life during the first year 18 $(Um — Uma) +lmsa =3(Im+lm4i). Similarly, the contribution during the second year is amtatlms+e); and soon. Hence the united life is 3(Lin + lmtp) + ¥(Emta + lm) + 6 s = blatlnttlinget..~. ale * Demoivre’s hypothesis. 570 EXAMPLES CHAP. the series continuing so long as the numbers in the table have any significant value. If we now divide the united life by the number of original lives, we find for the mean duration N=$+(lmar ate bmte + ros Stas (2). Owing to our assumption regarding the uniform distribution of deaths over the intervals between the tabular epochs, this expression is of course merely an approximation. Example 8. A and B, whose ages are a and 0 respectively, are both living at a particular epoch ; find the chance that A survive B. The compound event whose chance’ is required may be divided into mutually exclusive contingencies as follows :— 1st. B may die in the first year, and A survive ; ; 2nd. fr second cs ; and so on. The 1st contingency may be again divided into two :— (a) A and B may both die within the year, B dying first ; (8) B may die within the year, and A live beyond the year. The chance that A and B both die within the first year is (lg —Jq+1) (7; —ls41)/lals. Since the deaths are equally distributed through the year, if A and B both die during the’ year, one is as likely to survive as the other ; hence the chance of A surviving B on the present hypothesis is 3. The chance of the contingency (a) is therefore (lg —la+1) (lo — lo41)/2laly. The chance of (8) is obviously le4i(ls — lo41)/lu lo. Hence the whole chance of the 1st contingency, being the sum of the chances of (a) and (8), is (q+ la+41)(2, — lo41)/2la le. In like manner, we can show that the chance of the and contingency is (Zata + late) (loa — 1 ay) lal. Hence the whole chance that A survive B is given by Sao= {(Ca+ lati) (Zo — lo41) + (lata + late) (loti — lege) +... }/2lat (1). The reader will have no difficulty in seeing that (1) may be written in the following form, which is more convenient for arithmetical computation :— T=0 . Ses=st{ 2 ete\lotra — Us4n41) — lalssi} /2lals (2), poe where stands for the greatest age in the table for which a significant value of 7, is given. If we denote by Sp the chance that B survive A, we have, of course, So,a sai ie Site If a=), it will be found that (2) gives S,.,=1/2 ; as it ought to do. § 17.] Let us now consider the following money problem in life contingencies :— What should an Insurance Office ask for under-. taking to pay an annuity of £1 to a man of m years of age, the first ~~ XXXVI ANNUITY PROBLEMS——AVERAGE ACCOUNTING 571 payment to be made n+ 1 years hence,* the second n +2 years hence ; and so on, for t years, if the annuitant live so long. We suppose that the office makes no charges for the use of the shareholders’ capital, for management, and for “margin” to cover the uncertainty of the data of even the best tables of mortality. Allowances on this head are not matters of pure calculation, and differ in different offices, as is well known. We suppose also that the rate of interest on the invested funds of the office is £i per £1,s0 that the present value, v, of £1 due one year hence is £1/(1 +7). The solution of the problem is then a mere matter of average accounting. Let nj¢@m Aenote the present value of the annuity ; and let us suppose that the office sells an annuity of the kind in question + to every one of /,, men of m years of age supposed to be all living at the present date. The office receives at once ,, jt4mlm pounds. On the other hand, it will be called upon to pay Sole tas Bb wence ch tye RE) Pi es a n+ 1, | Og Dee a ietiere n+t > years hence respectively. Reducing all these sums to present value, and balancing outgoings and incomings on account of the lm lives, we have, by chap. xxii., § 3, y: 2 n| tinea a i bmn anaes lmtnts fame eke lm tnt Hence | n-+2 n|tCm = Uieor; bmn-+1 nea mints oe dae bien ee) ene r=t =" 2 lmtntrl [ban (1). r=1 The same result might be arrived at by using the theory of expectation. * This is what is meant by saying that the annuity begins to run n years hence. | Tt The annuity need not necessarily be sold to the person (‘‘ nominee ”) on whose life it is to depend. ‘The life of the nominee merely concerns the definition of the “‘ status” of the annuity, that is, the conditions under which it is to last. 572 PROBLEMS SOLVABLE BY ANNUITY TABLE CHAP, The annuity whose value we have just calculated would be technically described as a deferred temporary annuity. If the annuity be an immediate temporary annuity, that is, if it commence to run at once, and continue for ¢ years provided the nominee live so long, we must put n=0. Then, using the actuarial notation, we have Y=t [tm = = Inte lm (2). T= If the annuity be complete, that is, if it is to run during the whole life of the nominee, the summation must be continued as long as the terms of the series have any significant value; this we may indicate by putting = 0. Then, according as the annuity is or is not deferred, we have T=0 n|Cm = vy” > bnt-nrY™ Moe (3). r=1 ee . bm = p> mer bin (4). r=1 § 18.] The function a,,, which gives the valle an im- mediate complete annuity on a life of m years, a lameéntal importance in the calculation of contingenéies which depend on a single life. Its values have been deduced from yz of mortality, and tabulated. By means of such 4 we can readily solve a variety of problems. Thus, for example, »/Qn, 1t4ny njt@m can all be found from the annuity tables; for we have nim = 0" lman Om+n| bm *(5) ; [tn = Gm — v! ln-+t in 1 an (6) 3 n|tlm = (Ulan Onm+n — paid PGR ig Onn+-n-+tt)/ bm (7) ; as the reader may easily verify by means of formulz (1) to (4). These results may also be readily established a priori by - means of the theory of expectation. § 19.] Let us next find ay, the present value of an immedi complete annuity of £1 on the joint lives of two nominees of k m years of age respectively. The understanding here is that the annuity is to be paid XXXVI SEVERAL NOMINEES—-METHOD OF EXPECTATIONS 573 Jong as both nominees are living and to cease when either of them dies. ; The present values of the expectations of the 1st, 2nd, ord, . instalments are 4 2 3 n+ beable lie v Lyte bretol Ui te v lie+s brutal Ux ihe &e., Hence we have : Ok ,m = = (psy b m+it¥ “Tytgl m+e2 7 Sia my T=0 : vee vy Tear b m+ [Ug d m (1). i] as ; . | Just as in § 18," we obviously have . n|%k,m = O Oban, men pre veal: bn3 |t%&,m = km — Opes lit lnetl Ue bmn 3 nitlk,m = Ci eae lean bmn a Taped 2 Sea ee lea ntt ee xe), Li ban 5 . and it will now be obvious that all these formule can be easily extended to the case of an annuity on the joint lives of any number of nominees. | ; Tables k,m have been calculated ; and, by combining | | them with s for a», a large number of problems can be solved. l. To find the present value of an immediate annuity on the last surviyol wo liveSwn and n, usually denoted by lana t] bilities that the nominees are living 7 years after e probability that one at least is living r years here- a) aOm,n— rN Prt Ur — Pr r)s = 20" pp + 2" Vp — ZU" Dr Gr 5 =Am + An- Am, ne This is also obvious from the consideration that, if we paid an annuity on each of the lives, we should pay £1 too much for every year that both lives were in existence. Example 2. Find the present value azmm» of an annuity to be paid so long as any one of three nominees shall be alive, the respective ages being AM, 7. e If ps, Ys, 7s be the chances that the respective nominees be alive after s ears, then Ok, m, n= = {1 —(1—ps)(1—-gs) (1-75)}, = ZU(Ps+Ist1s— YUs1s—1sPs— PsUs + DsIs"'s)y = An + Am + Gn - Om ,n— On,% Ak,m+ Ok, m,n« e numerical solution of this problem would require a table of annuities ree joint lives, or some other means of calculating a; m,n. a ee 574 LIFE INSURANCE PREMIUM CHAP. § 20.] A contract of life insurance is of the following nature:—A man A agrees to make certain payments to an insurance office, on condition that the office pay at some stated time after his death a certain sum to his heirs. As regards A, he enters into the contract knowing that he may pay less or more than the value of what his heirs ultimately receive accord- ing as he lives less or more than the average of human life; his advantage is that he makes the provision for his heirs a certainty, so far as his life is concerned, instead of a contingency. As regards the office, it is their business to see that the charge made for A’s insurance is such that they shall not ultimately lose if they enter into a large number of contracts of the kind made with A; but, on the contrary, earn a certain percentage to cover expenses of management, interest on shareholders’ capital, &c. The usual form of problem is as follows :-— What annual premium P», must a man of m years of age pay (in advance) during all the years of his life, on condition that the office shall pay the sum of £1 to his heirs at the end of the year in which he dies ? 7 Pm 1s to be the “net premium,” that is, we suppose no allowance made for profit, &c., to the office. Suppose that the office insures /,, lives of m years, and let us trace the incomings and outgoings on account of these lives alone. The office receives in premiums £Plm, LPindmyi, . . . ab the beginning of the Ist, 2nd, . . . years respectively. It pays out on lives failed, £(lm — bmi), (lima: —lnss); ~ « » at the end seman temie 2nd, ... years respectively. Hence, to balance the account, we must havé, when all these sums are reduced to present value, Panklon + bing. + Lents Teen , i? oF (bin bint1)Y ae (lint il Lm+2)0 a (lms 7 lena) ore (1), the summation to be continued as long as the table gives signi- ficant values of /,. Since din = lm —Im4i, we deduce from (1) Pre Arn v + dm 40 + din 420" Tee ees bt binaaY + bral ti «.. — XXXVI RECURRENCE METHOD FOR ANNUITIES 575 Dividing by 1,,, we deduce from (1) ace ie (Are, a Linke? te lite? + es -) {Ln} =U+ Ulm. ,v + Lima eee) ibe, = bintiY + binget” +...) {lin Hence Pin(1 + dn) =¥ + Clim — Om, Po =0— Om] (1 + Gm) (3). The last equation shows that the premium for a given life can be deduced from the present value of an immediate com- plete annuity on the same life. In other words, life insurance premiums can be calculated by means of a table of life annuities, § 21.] It is not necessary to enter further here into the details of actuarial calculations; but the mathematical student will find it useful to take a glance at two methods which are in use for calulating annuities and life insurances. They are good specimens of methods for dealing with a mass of statistical information. Recurrence Method for Calculating Life Annuities. The reader will have no difficulty in showing, by means of the formule of § 17, that ; Um = v(1 is Coun baesiion (1). From this it follows that we can calculate the present value of an annuity on a life of m years from the present value on a life of m+1 years. We might therefore begin at the bottom of the table of mortality, calculate backwards step by step, and thus gradually construct a life annuity table, without using the com- plicated formula (4) of § 17 for each step. A similar process could be employed to calculate a table for two joint lives differing by a given amount. Columnar or Commutation Method. Let us construct a table as follows :— In the 1st column tabulate J, ; ea! r es Pee vord ¢ vl, =D,, say ; Wane bh a‘ mt manC susay, 576 COMMUTATION METHOD CHAP, Next form the 5th column by adding the numbers in the 3rd column from the bottom upwards. In other words, tabulate in the 5th column the values of No= Deer De yD, eee In like manner, in the 6th column tabulate M=— C40, Cy tee All this can be done systematically, the main part of the labour being the multiplications in calculating D, and C,. From a table of this kind we can calculate annuities and life premiums with great ease. Referring to the formule above, the reader will see that we have bm = Nontlan (2) ; n|ln = ING en (3) > [tam = NG - Ne en (4) ; n|tlm = (Ngee a Ninel (5) ; Ms m — 4%4m/4+Nm-1 (6) : § 22.] In the foregoing chapter the object has been to illustrate as many as possible of the elementary mathematical methods that have been used in the Calculus of Probabilities ; and at the same time to indicate practical applications of the theory. All matter of debatable character or of doubtful utility has been excluded. Under this head fall, in our opinion, the theory of @ priori or inverse probability, and the applications to the theory of evidence. The very meaning of some of the pro- positions usually stated in parts of these theories seems to us to be doubtful. Notwithstanding the weighty support of La Place, Poisson, De Morgan, and others, we think that many of the criticisms of Mr. Venn on this part of the doctrine of chances are unanswerable. The mildest judgment we could pronounce would be the following words of De Morgan himself, who seems, after all, to have “doubted”:—‘“ My own impression, derived from this [a point in the theory of errors] and many other cir- cumstances connected with the analysis of probabilities, is, that mathematical results have outrun their interpretation.” * * “An Essay on Probabilities and on their Application to Life Contin- gencies and Insurance Offices” (De Morgan), Cabinet Cyclopedia, App., DP Exvh, XXXVI GENERAL REMARKS—REFERENCES 517 The reader who wishes for further information should consult the elementary works of De Morgan (just quoted) and of Whitworth (Choice and Chance); also the following, of a more advanced character :— Laurent, Traité du Calcul des Pyo- babilités, (Paris, 187 3); Meyer, Vorleswngen iiber Wi ahrscheinlich- keitsrechnung (Leipzig, 1879); Articles, “ Annuities,” ‘‘ Insurance,” “ Probabilities,” Encyclopedia Britannica, 9th edition. The classical works on the subject are Montmort’s Essai @ Analyse sur les Jeux de Hazards, 17 08,1714; James Bernoulli’s Ars Conjectandi, 1713; Demoivre’s Doctrine of Chances, 1718, 1738, 1756; Laplace’s Théorie Analytique des Probabilités, 1812, 1820; and Todhunter’s History of the Lheory of Probability, 1865. The work last mentioned is a mine of information on all parts of the subject ; a perusal of the preface alone will give the reader a better idea of the historical development of the subject than any note that could be inserted here. Suffice it to say that few branches of mathematics have engaged the attention of so many distinguished cultivators, and few have been so fruitful of novel analytical processes, as the theory of probability. EXERCISES XL. (1.) A bag contains 4 shillings and 4 sovereigns, Three coins are drawn ; find the value of the expectation. (2.) A bag contains 3 sovereigns and 9 shillings. A man has the option, Ist, of drawing 2 coins at once, or, 2nd, of drawing first one coin and after- wards another, provided the first be a shilling. Which had he better do ? (3.) One bag contains 10 sovereigns, another 10 shillings. One is taken out of each and placed in the other. This is done twice ; find the probable value of the contents ofeach bag thereafter. (4.) A player throws x coins and takes all that turn up head ; all that do not turn up head he throws up again, and takes all the heads as before : and soon ry times. Find the value of his expectation ; and the chance that all will have turned up head in 7 throws at most. (St. John’s Coll., Camb., 1870.) (5.) Two men throw for a guinea, equal throws to divide the stake. A uses an ordinary die, but B, when his’ turn comes, uses a die marked 2, 3, 4, 5, 6, 6; show that B thereby increases the value of his expectation by 5/18ths. (6.) The Jew des Noyauax was played with 8 discs, black on one side and VOL. II 2P 578 EXERCISES XL CHAP. white on the other. A stake S was named. The discs were tossed up by the player ; if the number of blacks turned up was odd the player won §, if all were blacks or all whites he won 28, otherwise he lost S to*his opponent. Show that the expectations of the player and opponent are 1318/256 and 125S/256 respectively. (Montmort. See Todh., Hist. Prob., p. 95.) (7.) A promises to give B a shilling if he throws 6 at the first throw with 2 dice, 2 shillings if he throws 6 at the second throw, and so on, until a 6 is thrown. Calculate the value of B’s expectation. (8.) A man is allowed one throw with 2 ordinary dice and is to gain a number of shillings equal to the greater of the two numbers thrown ; what ought he to pay for each throw? Generalise the result by supposing that each die has n faces. (9.) A bag contains a certain number of balls, some of which are white. I am to get a shilling for every ball so long as I continue to draw white only (the balls drawn not being replaced), But an additional ball not white having been introduced, I claim as a compensation to be allowed to replace every white ball I draw. Show that this is fair. (10.) A person throws up a coin 2 times ; for every sequence of m (m > 7) heads or m tails he is to receive 2”—1 shillings; prove that the value of his expectation is n(z+3)/4 shillings. (11.) A manufacturer has n sewing machines, each requiring one worker, and each yielding every day it works g times the worker’s wages as net profit. The machines are never all in working order at once ; and it is equally likely that 1, 2, 3, . . ., or any number of them, are out of repair. The worker’s wages must be paid whether there is a machine for him or not. Prove that the most profitable number of workers to engage permanently is the integer next to nq/(q+1)-4%. (Math. Trip., 1875.) (12.) A blackleg bets £5 to £4, £7 to £6, £9 to £5 against horses whose chances of winning are 8, 3, $ respectively. Calculate the most and the least that he can win, and the value of his expectation. (13.) The odds against » horses which start for a race area:1; a+1:1; ..,@+n-1:1. Show that it is possible for a bookmaker, by properly laying bets of different amounts, to make certain to win if n>(a@+1)(e+1), and impossible if 7= 1/24.” °{19.)-loga.’ (20:) 1-9 (21) Tata (23.) 1. (24) 0. (25.) o if e=1+0,0 if =1-0. (26.) & (27.) 0 if 2 be negative, if m be positive 0 or o accordingasa<>1. (28.) 1° (29.) 1. (80.) 0 or © according as m>a. (7.) Conv. ifw<4; div. ifa<¢4. (8.) Conv. (9.). Div., (w<1). (10.) Conv. (11.) Div. , (12.) Cony. if dd; div. ifa+t1. (13.) Div. (14.) Div. (15.) Abs. conv. (16.) Div. (=e. 2.4... Or. (8.) : (87 — 5)273/12. 24.3 8..5:27. (2) 1.35 eee 12... 47, 1S oa 48 - 11, 2... (8r—4)al®—)/r1. (6, ils. Ds 3r — oe Ord. G2 0. 9) OTs) aan Ck) ee (Oe assed (nr-n—1)/r!. (8 my pes Ee are 2) \/(r/2)! if 7 be even ; 0 if7 be odd. (9.) (—)"n(m+1). Ric aon 1)/{F(r—n)}!. (10.) 14+ 3(a/a)+2(a/ayP +22(x/a)%.° (11.) The first. (12.) The third. (13.) The fourth and fifth. (14.) Theeighth. (15.) If n=1, the 2nd and 8rd ; if n=2, the 2nd ; if n+3, the lst. (19.):If m=0, S=a; ifm=1,8S=6; ifa>1, S=0; ifm@=iea the series is divergent. (22.) 1- ee (23.) Ifm +1, S=m(m—1)2"- ; if m=0, S=0. RESULTS OF EXERCISES - 583 X. (1.) Zl/a(c-a)(a—b). (2) 0. (3.) Z1/a7--2/(¢ — a)(a — b). (4.) 2r+1+1/2", (6.) 7, if rbe even ; r— lifr=4¢+1; r+1,if r=4t-1. (6.) HQ” — m1 mH 1 09 + Ce. n Ho ed mea te (15.) &( +1)(w+2)(n +8). (19.) 1- 1.3... (Qu-1)/2"m!. (20.) 7.10... (8n+1)/3.6... (8n—3), XI. (2.) 275/128, (3.) 869699/256. (4.) 48; 0. (5.) 11989305/2048, (6.) (=) (r—1)+(r+5)/27?}, 10.) 1°0001005084 ; 1:0004000805. (41.) 2m. (12.) 1+2a(1-rn)/(1-r). (18.) 14(—)ejon, XII. (1.) *367879. (2.) 04165. (5.) (l-a)e% (6.) 8(e-1). (7.) “e+1. (8.) I/e. (9.) 15e ° XIII. (4.) 917. (5.) 2log {(w@-1)/(a+ 1)} +log {(a+2)/(a - 2)}. (6.) log (12e). (7.) (1+1/x)log(1+z)-1. — (8.) 4(x%— a") log {(1+a)/(1—a)} +4, (9.) When x=1 the sum is 18—24 Og 2, (10: S$!) (12.05 {22"—2/(38n — 2) + 2"—1/(3n — 1) — 2237/3} , XXV, (1.) 4n(n+ 1)+4(r — 2)n(m +1)(n—-1). (2.) tn(w+1)(n+4)(m+5), (3. ) 3/4—1/2n-1/2(n+1). (4.) 1/15- 1/5(5n+8). (5.) 1/12- 1/4(2n +1) (2+ 8). (6.) 1/18 --1/3(7+1)(m+2)(n+ 3). (7.) a/2+b/4 -af/(n+2) — b/2(n+ 1)(n+2), (8.) 1/8 —(4n+8)/8(2n+1)(2n+3). 9.) 7/36 — (82+ 7)/(n +1) (n+2)(n +8), (10.) 11/180 — (6 +11)/12(2n+1) (2n+8) (2n+5). (11.) 3/4+n-(2n+8)/ 2(n +1) (+2). (12.) m=(n+1)(n+3)(w+5)/n(w+1)... (n+6); apply § 3, Example 4, (13.) sin @ sec(n+1)0 sec 0. (14. ) cot (8/2”) /2” — cot 0. (15.) tan~*na™, = (16.) tan-11 + tan-11/2 - tan-*1/n—tan-11/(n+1). — (17.) (m+n)!/(m+1)(n—-1)!. (18.) {1/(m—-1)!-(m4+1)!/(m+n-1) !} /(m — 2). (19.) (- "min. (21.) {n~1-(n+1) !/m'""} (m—-2). (22.) (a!ntrifelnt — a'™1"\(a-c+r+1). (23. ) (aintalielntrtl! — afelrl)i(e—e—r+1), (24. ) {(a- 1)!m=1Jelm—1! —(a+n)'™UW(e+n+ 1)!™=11,/ (a, —1)(a-c-1). (25.) Deduce from (24). (26.) Deduce from (24). (27.) 2m-{1-( — )”2(m -1) (m-2)...(m-n)/1.8...(2n- 1)} /(2m—1). XXVI. (1.) Qntl + 4(3ntl — 3), (2.) ${1+(-1)*} +6 - 3{ieti4(— ett} — Pe —(-a)n}. — (B.) 11 {1 -(4e)"1} / {1 —-42} 9 11 - (3a)"t?} / {1-32} ; (2+32%)/(1—7a2+122%), w<}3. 4.) 3{1—(2x)e+1} /{1 — 20} +2{1—(8a)"+}} | {1 - 8x}; (5-182)/(1- 5x +622), v<, (5.) ${1 — (Ba)r41} /{(1- 3x)+ %{1-(5x)"+1} {1-52}; (1 — 4x)/(1 — 8%+1527), w2. (8.) Converges. (9.) Oscillates, (10.) Oscillates. (15.) Each of the fractions converges tol. (23.) ¢ (24.) 1/(1—e). (25.) log,2. (26.) (3 —e)/(e-2). XXXIX., (1.) 11/80. (2.) 8/11, 29/44, 3/44. (3.) m(m + 2n)/(m+n)?, m(m+2n—-1)/ (m+n)(m+n-1). (4) (365.4"+1)/(1461)% (5.) 4/9. (7.) 55/672, 299/2688. (8.) 1/42. (9.) (~—1)/n(2n-1). (10.) (39 !)2/26!152!, 4(39 !)?/26152!. (11) 2(r—1)/n(n-1). (18.) 7n/2, or, if this be not integral, the two integers on n+ = 1 either side of it. (14.) 2 r(r—1)n(n—-1)... (n—r+2)/n™ (18.) 16/81, 2 1 8/31, 4/31, 2/31, 1/31. (19.) The chances in A’s favour are 6/10, 7/10, 8/10, 9/10, when he is 1, 2, 8, 4 up respectively. (20.) 25 to 2. (23.) (1 —1/n)/2, (1—1/n)/(2-1/n). XL. (1.) £1:11:6. (2.) His expectations are 11s. 6d. and 10s. 44d. respect- lvely. (8.) £8:5:94, £2:4:2%. (4) n(1—1/2"), (1 —1/2")*. (7) 7s. 24d. ; (n-+1)(4n—1)/6n. (12.) £6, £1, £4:2:24. ' nd , * = , : a ¢ ‘ ra ¢ c : , , “ . + F . : is > s Ps = ° \ . r ! . ‘ 4 fi “ » , . ’ £ .- j ~ <# . J i (ge eet * | a6, ; ay “aye 2 : ‘ ey j = i «<) : + ¥ t : eo F ae INDEX OF PROPER NAMES, PARTS I. AND IL The Roman numeral refers to the parts, the Arabic to the page. ABEL, ii. 119, 122, 127, 130, 263 Adams, ii. 219, 227 Alkhayami, ii. 423 Allardice, i. 441 Archimedes, ii. 415 Argand, i. 222, 254 Arndt, ii. 478 BABBAGE, li. 156 Bernoulli, James, ii, 204, 208, 252, 377, 379, 559, 577 Bernoulli, Johre i. 251, 274, 342, 377, 556 Bertrand, ii. 112, 118, 159 Bezout, i. 358 Blissard, i, 84 Bombelli, i. 201 Bonnet, ii. 118, 159 Boole, ii. 209, 372 Bourguet, ii. 159, 229 Briggs, i. 529 Briot and Bouquet, ii. 370 Brouncker, ii. 327, 421, 451, 488 Burckhardt, li. 508 Biirgi, i. 530 CARDANO, i, 253 Catalan, ii. 159, 196, 227, 229, 390 Cauchy, i. rie 254; ii. 42, 47, 83, 97, 98, 110, E08, L2750147, 164, 202, 204, 215, 263, 316, 320, 370 Cayley, ii. 33, 287, 289, 301, 347 Clausen, ii. 399, 475 Clerk- Maxwell, ii. 301 Cossali, i. 191 Cotes, i. 247 Cramer, ii. 370 DasE, ii. 508 De Gua, ii. 370 De Morgan, 1, 254, 346; ii, 112, 118, 360, 370, 394, 550, 576 Demoivre, i. 239, 247 ; ii, 274, 282, 375, 377, 379, 381, 385, 546, 564, DG69, 577 Desboves, ii. 63 Descartes, i, 201 Diophantos, ii. 445 Dirichlet, ii. 95, 126, 133, 424, 445 Du Bois Reymond, ii. 119, 130, 181, 160 Durége, ii. 370 ELY, ii. 209, 320 Kuclid, i. 47, 272 Kuler, i. 254; li, 81, 97, 164, 228, 256, 274, 309, "815, 317, 318, 319, 320, 321, 824, 334, 339, 341, 342, 382, 392, 421, 466, 468, 484, 487, 498, 511, 522, 523, 525, 527, 528, 535 FAavaro, li, 421 ‘| Fermat, ii. 450, 471, 518, 522, 563 ‘Ferrers, ii, 534 Fibonacci, i, 202 Fort, ii. 77 Fourier, ii. 122 Franklin, ii. 88, 535 Frost, ii. 99, 370, 371 GALOIS, ii. 477 Gauss, i, 46, 254 ; ii. 81, 160, 309, 321, 445, 495, 514, 522, 525 Glaisher, DeLee, 530 ; lin 216, 333, 347, 371, 384, 395, 508 Goldbach, ii. 395 Grassmann, i, 254 Gray, ii. 219 Greenhill, ii. 289 Gregory, ii. 97 Gregory, James, ii. 309, 327 Grillet, ii. 59 Gronau, ii. 289 Gross, ii. 513 Gudermann, ii. 287, 289 Giinther, ii. 288, 421 HAMILTON, i. 254 Hankel, i. 5, 254, Harriot, i. 201 Heilermann, ii. 490 Heine, ii. 95, 499 Heis, ii. 289 Herigone, i. 201 Hermite, li. 445 Hero, i. 83 Hindenburg, ii. 467 Horner, i. 346 Houel, ‘i, 288 588 ei Hutton, i. 201 Huyghens, ii. 421, 552, 559, 564 JACOBI, ii. 445 Jensen, ii. 160 Jordan, i. 76 ; ii. 32 Kony, ii. 119 Kramp, ii. 4, 374 Kronecker, ii. 213 Kummer, ii. 119, 160, 445 LAGRANGE, i. 57, 451 ; ii. 370, 421, 423, 425, 450, 522, 525 Laisant, ii. 289, 312, 334 Lambert, i. 176 ; ii. 288, 821, 421, 489, 495 Laplace, ii. 50, 577 Laurent, ii. 160, 551, 577 Legendre, ii. 445, 484, 495, 535 Leibnitz, ii. 309, 377 Lionnet, ii. 225, 228 Lock, ii. 247 Longchamps, ii. 97 MACDONALD, i. 530 Machin, ii. 309 Malmsten, ii. 80, 118 Mascheroni, ii. 81 Mayer, F. C., ii. 288 Mercator, ii. 288 Mertens, ii. 127 Metius, ii. 415 Meyer, ii. 577 Mobius, ii. 871, 466, 476 Montmort, ii. 379, 381, 556, 564, 577, 578 Muir, i. 358; ii. 311, 443, 466, 469, 474, 475, 476, 490 Napier, i. 171, 201, 254, 529 Netto, ii. 32 Newton, i. 201, 436, 472, 474, 479 ; ii. 256, 306, 327, 349, 862, 370, 375, 563 Nicolai, ii, 81 OUGHTRED, i. 201, 256 PAOcIo.i, i, ‘202 Pascal, i. 67; ii. 556, 563 Paucker, ii. 119 Peacock, i. 254 Pfaff, ii. 311 INDEX Pringsheim, ii. 161 Puiseanx, ii. 370 Purkiss, ii. 61 Pythagoras, ii. 503 RAABE, ii. 118, 348 Recorde, i. 216 Reynaud and Duhamel, ii. 49 Riemann, i. 254 ; ii. 241, 301, Rudolf, i. 200 SALMON, i. 440 Sang, i. 530 Scheubel, i. 201 Schlémilch, ii, 45, 51, 80, 98, 160, 186, 335, 349, 478, 495 Seidel, ii. 131, 478 Serret, i. 76; ii. 32, 416, 425, 448, 453 Shanks, ii. 310 Sharp, ii. 309 Smith, Henry, ii. 445, 471 Sprague, i. 531; ii. 88 Stainville, ii. 311 Stern, ii. 421, 469, 477, 478, 489, 497 Stevin, i: 171, 201 Stifel, i. 81, 200 Stirling, ii. 344, 375, 378, 395, 561 Stolz, ii. 157, 161, 370 Sutton, i. 531 Sylvester, i. 48, 176 ; ii. 466, 475, 528, 533 Talt, ii. 229 Tartaglia, i. 191 Tchebichef, ii. 159 Thome, ii. 160, 370 Todhunter, ii. 247, 252, 546, 552, 556, 577 VAN CEULEN, ii. 809 Vandermonde, ii. 9 Venn, ii. 539 Viete, i. 201; ii. 252 Vlacq, i. 530 WALLACE, ii. 288, 290, 291 Wallis, ii. 827, 421, 450, 499, 509 Waring, ii. 391, 525, 527 Weierstrass, i. 230; ii. 138, 144, 161 Whitworth, ii. 25, 38, 587, 561, 577 Wilson, ii. 523 Wolstenholme, i. 448 ; ii. 17, 83, 348, 519 Wronski, ii. 188 THE END, Printed by R. & R. CLarK, Edinburgh. " i . a eh lee {Ih ~~ > 5 oS N eo f=) we j= fo) N = — fo) IT l ll