THE ELEMENTARY PROPERTIES OF THE ELLIPTIC FUNCTIONS WITH EXAMPLES BY ALFRED CARDEW DIXON, M.A. LATE FELLOW OF TRINITY COLLEGE, CAMBRIDGE; PROFESSOR OF MATHEMATICS AT QUEEN'S COLLEGE, GALWAY ' o nt b o it MACMILLAN AND CO. AND NEW YORK 1894 All rights reservecl PREFACE. THE object of this work is to supply the wants of those students who, for reasons connected with examinations or otherwise, wish to have a knowledge of "the elements of Elliptic Functions, not including the Theory of Transformations and the Theta Functions." It is right that I should acknowledge my obligations to the treatise of Professor Cayley and to the lectures of Dr. Glaisher, as well as to the authorities referred to from time to time. I am also greatly indebted to my brother, Mr. A. L. Dixon, Fellow of Merton College, Oxford, for his kind help in reading all the proofs and working through the examples, as also for his valuable suggestions. A. C. DIXON. DUBLIN, October, 1894. CONTENTS. CHAPTER I. PAGE INTRODUCTION. DEFINITION OF ELLIPTIC FUNCTIONS, - 1 CHAPTER II. FIRST DEDUCTIONS FROM THE DEFINITIONS. THE PERIODS. THE RELATED MODULI, -- 8 CHAPTER III. ADDITION OF ARGUMENTS, - - - - 25 CHAPTER IV. MULTIPLICATION AND DIVISION OF THE ARGUMENT, - 38 CHAPTER V. INTEGRATION, - - - - - 46 CHAPTER VI. ADDITION OF ARGUMENTS FOR THE FUNCTIONS E, I, - 53 CHAPTER VII. WEIERSTRASS' NOTATION, - 63 viii CONTENTS. CHAPTER VIII. PAGE DEGENERATION OF THE ELLIPTIC FUNCTIONS - - - 69 CHAPTER IX. DIFFERENTIATION WITH RESPECT TO THE MODULUS, - - 73 CHAPTER X. APPLICATIONS, - -- 82 APPENDIX A. THE GRAPHICAL REPRESENTATION OF ELLIPTIC FUNCTIONS, 129 APPENDIX B. HISTORY OF THE NOTATION OF THE SUBJECT, - - - 136 ELLIPTIC FUNCTIONS. CHAPTER I. INTRODUCTION. DEFINITION OF ELLIPTIC FUNCTIONS. ~ 1. In the earlier branches of mathematics functions are defined in various ways. Some are the results of the fundamental operations of algebra. x+l, 2x, x2 are such functions of x. Others are introduced by the inversion of those operations; such are x-1, l/x, ^/x; and others by conventional extensions of them, as x, ex. It is not easy to draw the line of distinction between the two last-named classes. Sometimes, again, geometrical constructions are used in the definition, as in the case of the trigonometrical functions. ~ 2. The elliptic functions cannot readily be defined in any of the foregoing ways; their fundamental property is that their differential coefficients can be expressed in a certain form, and as this is a somewhat new way of defining a function, we shall take one or D. E. F. A 2 ELLIPTIC FUNCTIONS. two examples to show that it is as effective as any of those above mentioned. ~ 3. Let us define the exponential function by the equation d exp u = exp u. This equation tells us what addition is to be made to the value of exp u when a small change is made in that of u, and would therefore enable us gradually to find the value of the function for every value of the argument u, provided we knew one particular value to start with. Suppose then that when 'z has the value 0, exp u has the value 1, that is, exp 0= 1. This equation combined with the former supplies a definition of the function exp u.~ 4. From the foregoing definition we can deduce the properties of the function exp u. First of all we can find an expression for exp(( + v). Let u +v =, and suppose w to be kept constant while u and v vary. cd d Then -exp v = - C exp v= - exp v. d Thus exp'u.,-lexpv +expv. — expu=O0, or tid(exp % exp v) = O. Hence exp uexpv is a constant as long as w is a constant, and has the same value whatever we may put for u and v so long as u+v=w. % Compare the construction of trigonometrical tables, as explained in works on Trigonometry. The sine, tangent, etc., of every angle are found by adding the proper increments to those of an angle slightly less. INTRODUCTION. 3 Put then v = 0, =w z, and we have exp u exp v = exp w exp 0 = exp(u + v), since exp = 1. ~ 5. We can also deduce the expansion of exp u in powers of it. anid Fnor -- exp u = exp u, which =1, when = 0. Thus Maclaurin's Theorem gives qt2 t r exp u = + qZ + +... +- +..., the convergency of which may be established in the usual way. ~ 6. As another example, define the sine and cosine by the equation cd sint............... - sn { 'M'=cos,................(.....1) where cos + in =,......................(2) and sin =0, cos0=1. ~ 7. Differentiating (2), we have cos ui- cos 7a + sin u cos u = 0, whence - cos u = -sin,...................(3) as cos ' is not zero in general. 4 ELLIPTIC FUNCTIONS. ~ 8. To find sin(u + v) and cos(u + v) put u + v = zw, a constant, as before. Consider a symmetrical function of u and v, such as sin u + sin v. d(sin u + sin v) = cos u - cos '. Wu/ In the same way C-d(cos u + cos v) = - sin zt + sin v. But cos2u + sin2u = cos2v + sin2v, so that (cos u - cos v)(cos u + cos v) = (- sin in u+sinv)in (siu + sin v).......(4) Hence (cos u + cos v)d -(sin u + sin v) = (sin u + sin v)-'- (cos t + cos '), ~,sin /U + sinz v sin(zu + v) so that sin = a const........ (5) cos U + cos v cos(u + v)+ l. ( putting w for +, and 0 for v. Then from (4) and (5) - sin u + sin v ------— =a const. also, Cos u - cos v sin( t+ v) 1 -cos(Iz + v)' And we find by solving s sin2n- sin2%,6 sin(zt + v)=. sim u cos v - sin v cos u =sin u cos v + sin v cos u by help of (2). Here again the functions may be expanded by Maclaurin's Theorem. INTRODUCTION. 5 ~ 9. The equations of definition are satisfied also if we change the signs of mt and of sin t. Thus sin( - u) = - sin Zt, cos(-a) = Cos K. The equations (1) and (2) are also satisfied if cos u is put for sin u and -singt' for coszta. The initial values however are now different and a constant must be added to a. Call this constant M. Then sin(u + a) = cos u, cos(u + a) = - sin 'a, if za is such that sin as = 1, cos = 0. Hence sin( u + 2m) - cos(' + a) = - sin tz, cos(u + 2u) = - sin(u + w) = -cos tu, sin(z + ~4) = - sin(u + 25) = sin u, cos(u' + 4a) = - cos(u + 2) = cos u. Hence the functions are unchanged when the argument u is increased by 4~, that is to say, they are periodic. ~ 10. Again, writing t for ^/-1, — _(cos u + t sin u) = L(cos u + i sin 'a), orl d (coss 'i + L sin u) = cos (a + l sin u, and cos O+t sin = 1, so that cos u + t sin = exp tu. This equation includes De Moivre's Theorem, and shows that exp u is also periodic, the period being 4it. These examples may be enough to show that functions which we know already can be defined in the way that was mentioned in ~ 2. 6 ELLIPTIC FUNCTIONS. ~ 11. Now the three elliptic functions sn u, cn U, dn zt * are defined by the equations du s n z=cnn dun, cn2U + sn2n = 1, dn2 + /c2sn2, = 1, snO=0, cnO=dn0=1. From these it follows at once that d d- cn u= -sn dnu, d- n u = - 2sn u cn e. clu The quantity k is a constant, called the qnodulus; ut is called the cargu ent. ~ 12. For different values of the modulus k (or, perhaps, rather of 1c2, as the first power of k does not appear in the definition) there will be different values of the elliptic functions of any particular argument, in fact, sn n, cn, dn z are really functions of two independent variables, and when it is desirable to call this fact to mind we shall write them sn(u, k), cn(u, k), dn(u, ic). We shall also use the following convenient and suggestive notation, invented by Dr. Glaisher:cn n/dn u = cd u, sn u/cn s = se u, dn u/cn u = dc u, 1/sn u = ns u, 1/cnz = nc z, etc. It is usual to write c' for (1 - k2)2, and k' is called the complementary modlits. * Read s, n, u-C, n1, u-d, n, u. + Here and elsewhere sn2l, etc., stand for (sn) u2, etc., as in Trigonometry. INTRODUCTION. 7 The reader will not fail to notice the analogy between the two functions sn u and sin u, as also that between cos t and either cn u or dn u. (Compare ~~ 74-75 below.) EXAMPLES ON CHAPTER I. 1. Find the value of tan( + v) in terms of tan u and tan v from the equations - tan u = 1 + tan2u, tan O = 0. du 2. Prove also that tan u is a periodic function of u, the period being twice that value of u for which tan u is infinite. 3. Find the value of sech(zf+v), given that sech u= - sech u tanh Z, du where sech2t + tanh2 = 1, and that sech = 1, tanh = 0. 4. Find the differential coefficients with respect to u of ns u, nc u, ndcl, sd t, sd, cs u, cd u, ds it, dcc. Ans. -cs u ds m, sci de, kc2sd it cd, nnc t dc u, nd u ecdu, -nsz ds /, - c'2sdc ndcu, -csunsu, /'2sc u nc u. 5. Differentiate with respect to u (1) sn u/(1 + en ). Ans. dn u/(l +cn u). (2) sn u/(1 + dn u). Azs. cn zu/(1 +dn a). (3) en u/(l + sn u). Ans. - dn u/(l + sn ). (4) dn u/(l + k sn t). Ats. -7c cnu/(l +lcsnz.). (5) arcsin sn u. Ans. dn 't. (6) sn t/(dn i - en u). Ans. 1/(cn u - dn u). CHAPTER II. FIRST DEDUCTIONS FROM THE DEFINITIONS. THE PERIODS. THE RELATED MODULI. ~ 13. It follows from the foregoing definitions that if a function S or S(v) of a variable v satisfies the equation dvr..... where C and D are other functions of v connected with S by the equations 2 + 2 =,.......................(2) 2 + 2S2=1;.............. (3) then S= sn(v a, X),....................(4) = cn(v+, ),....................(5) D = dn(v + a, X),.....................(6) where a is such a constant that sn(a, )= S(O ),.....................(7) cn(a, X)= C()....................(8) dn(a, X)=D( ),....................(9) these last equations being clearly consistent. NEGATIVE ARGUMENTS. 9 ~ 14. Now, in the first place, the foregoing conditions hold if we put S=-snu, C=cn, D)=dn u, X=k, v= ---, ca=O; and thus sn( - u) = - sn, u cn(-u)= c nu, -.....................(10) dn( -u)= dnm, J or cn and dn are even functions, and sn is an odd function. ~ 15. We have also d _ sc u = (cn2 dn lb + sn dn m)/cn2U dn iu/cn2u = dc t nc u, and in the same way d - nc u = sc u de u, d dc = /c'2sc u nc u, du -cs u = - ds ns u, du - ns u = - cs u, dcls u, du du -d su = cd undu, dc -cl 'd, = - '2sd u nd ut, d -- nd u = l2sd u cd u..................(11) By integrating these equations we shall deduce several important theorems. 10 ELLIPTIC FUNCTIONS. ~ 16. Take for instance du dcdc= -k'2sd u ndc. We have cn2% + sn2a = 1, dn2%- + kc2sn2 = 1; and dividing by dn2tb, cd2' + sd2% = ndcl', 1+ k2sd2% = nd2U. Hence k'2sdl2% +cd2' =1, by elimination of nd2%, and 7c'2nd2% + k2cd2=- 1, by elimination of sd2%. In the equations (1)... (6) of this chapter we may therefore put S= cd t, (= - c'sd,, D= c'nd u, = k, v = u. The value of a is such that sna=1, cnac=0, dn a ='. Let us write K for this value of a; then we have sn(u + K) = cd a, } cn(zt +K) = -k'sdc,.................(12) dn(u + K) = k'nd u,. J ~ 17. From these it further follows that sn(u + 2K) = cd(u + K) = - c'sd uz - 'nd ' = -sn u, cn( + 2K) = - c'sd(u f K) = - c'cd u- k'nd z' = -cn u, dn(t + 2K) = k'- dn(u+ K) = dn '. Also sn(, + 3K)= -sn( + K)= -cd u, cn(u + 3K) = c'sd u, dn(u + 3K) = k'nd z, sn( +4K) = -sn(a + 2K)= sn to, cn(' + 4K) = cn u, dn(u + 4K) = dn u. THE PERIODS. 11 Again, sn(K - u) = cd( - u) = ed m, cn(K - u) = 'sd u, dn(K - u) = k'nd u. Thus the function dn u is unaltered when its argument is increased by 2K; sn u and cn u are unaltered when the argument is increased by 4K, that is to say the functions are periodic. ~ 18. Take now the equation d = - c c-yans'a= -csu ds u, du where - Cs2f+ ns221, =1, - ds2U + ns2u = lc2. Here we may write S= -nsu,, C==dst, D=ecsu, X=kc, v==, but sna, cn a, dna are all infinite. We have, however, CSa=t, ds c= tC. Let this value of a be called L for the time being. 1 " Then sn('t + L)= ns U, cn(u + L) =,ds t, dn(u + L) =csu, sn(u + 2L) = sn u, cn(, + 2L)= -cn,...............(13) dn( + 2 L)= -dn tc, cn(u + 3L)= - ds zc, dn(u + 3L) = - cs u, cn(u + 4L)= cn u, dn(u + 4L) = dn u. 12 ELLIPTIC FUNCTIONS. ~ 19. Also 1 1 sn( + K+L)= ns(u + K)=7 de, k cn(zc+Z^+~)=le ds(u+ K) = t/,wu dn(u+K+L)= i cs(u+K)= -ck'sc,...(4) sn(u + 2K+ 2L)= - sn, cn(u + 2K + 2L) = cn, dn( + 2K+ 2L) = - dn u. ~ 20. Hence sn t has a period 2L as well as 4K, cn u has a period 2K+ 2L as well as 4K, dn z has a period 4L as well as 2K. We may also notice that sn(K+L)=1 o(K++ ) L= 1, c n(+ L) = 0. THE COMPLEMENTARY MODULUS. ~ 21. Now consider the first equation of the system (11). d -SC = dc n ne n, where lnc2U - sc2 = 1, dc2n - 1kc'SC2 = 1. Hence we may put S=tscu, n, D=dcnc, D=d =t, Xvt, ', in the equations (4), (5), (6); and as S(o)=0, C(0)=D(O)=1, we have c = 0. THE COMPLEMENTARY MODULUS. 13 Thus sn(tu, V') = L sc(u, k), Z cn(u,')= nc(, k),...............(15) dn(tu, V) = dc(u, k). J These equations are of great importance. They embody what is called Jacobi's ImaLyinary c Transfobrmation, and enable us to express elliptic functions of purely imaginary arguments by means of those of real arguments with a different modulus. ~ 22. In the equations (15) put L for bu. Then sn(QL, k')= sc(L, k) = 1, cn(tL, ') = O, dn(LL, i')= k. Thus tL stands to kc' in the same relation as K to ic, and we are naturally led to write tL=K', L=-tK'. Thus if mi and n are any two whole numbers sn(ntb+ 2nmK+- 2ntK') = ( - 1)sn u, cn(n+2mK+ 2nK)=(- 1) cn,....(16) dn('t + 2nmK + 2nLK') = ( - 1)ndn u. We have then the following scheme for the values of sn, en, dn, of u+nmK+nbK', m and n being integers: m= -0, m= 1, mn = 2, m = 3. sn,(, cd, - sn u, - cdt. n =O cn, - k'sd, - cn u, c'sd u. dn z, lc'nd iu, dn it, k'nd zt. (ns u)/7c, (de u)l/k, -(ns u)/lc, - (de z)/lc. i=l1 - t(ds-)/l, -t 't(ncu)/k, i(ds-)/lc, ikl'(ncn)l/k. - cs U, Lklsc U, -- I cs i, dc'se u, 14 ELLIPTIC FUNCTIONS. m =0, M -1, I-2, -m3. sn ut, cd, -n, -n u - cd. - 2 - cn, c'sd u, cn z, - k'scl. - dn, - k'nd c, -dn u, - c'nd u. (ns u)/k, (de u)/kl - s), -n)/, -(dc U)/k. -a 3 t(ds u)/1, tl'(ncu)/l,,- (dsc)//,k, - d'(ncu)/lc. c CS Z', - tc'SC ', t CS U, -- SC %. the modulus in the congruences being 4. ~ 23. These equations show that a knowledge of the values of sn u, cn t, dn u does not enable us to fix the value of u, and that accordingly the value of K is not perfectly defined since we have only assigned the conditions snK=l, cnK=0, dnK=7'. Writing x for sn u we have en t( = (1 - x2), dn ua = (1 - k22), dcx =(I - 2)2 72X2)2 Hence u= (1 - 2)2(1 - 7_22)-2 d, 0 the lower limit being 0 because u and x vanish together. Thus K= J'(1 2)-2(1 - k22-)- %2'. This is a function of 7 only. The variable C will be supposed in the integration to pass continuously from 0 to 1 through all intermediate real values and those only, and the initial value of the subject of integration will be supposed to be unity and positive. There is now no ambiguity in the value of K so long RELATED MODULI. 15 as 7c2 is less than 1. Also with the same provision K is a purely real positive quantity as every element in the integration is so. Further, k' is to be the positive value of (1-72)2, for dnu does not change sign within the limits of integration and k' = dn K. ~ 24. Again, so long as kc'2 is less than 1, K' is also a purely real positive quantity. Thus for values of the modulus between 0 and 1 the periods 4K and 4tK' are the one real, the other purely imaginary. We shall now show how to reduce elliptic functions in which the square of the modulus is real, but not a positive proper fraction, to others in which the modulus lies between 0 and 1. ~ 25. We have d- sn u = en dn u, cnU + sn2u = 1, dn2U + k2sn2a = 1, and we may put C= dn u,.) = n u, provided we have S= k sn u, X = 1/, v=k u. Furthermore ct = 0. Thus sn(ku, 1/k) = k sn(m, 7c), cn(ku, Il/k)= dn(, k),............(17) dn(lct, 1/c) = cn(tz, k,). J The equations (17) enable us to reduce the case of a modulus numerically greater than unity to that of one less than unity. 16 ELLIPTIC FUNCTIONS. ~ 26. From the equations (15) and (17) we deduce sn(tlk'u, 1/k7') =c'sn(tz, kc')= ck'sc(u, 7c), cn(o'nu, 1/7')= dn(tu, 7c')= dc(u, 7c), (.....(18) dn(t7'u, 1 /J')-= cn(Ltu, 7')= nc(u, 7c), J and also, since kc'/lk is the modulus complementary to l/7c, sn(dcko, I'lc/k) = t sc(7oa, 1/Jk) = Ik sdcl(, 1c),' cn((7u, tc'/7l)= nc(7cu, 1/7J)= nd(c, 7c),.....(19) dn(tl7, tL'/7c) = dc(7c, 1/7c)= cd(u,7c),J and from (19) by help of (15) sn(7c', di/kl')= - i7csd(iu, c') = kc'sd(uc, 7c), cn(7c', ll/7l') = nd(, lo') = cd(u,, ),..(20) dn(J7'u, o/7c') = cd(u(, 70') = nd(u, J7),J ~ 27. The quantities corresponding to K, K', the quarter-periods, are given in the following table for the group of six related moduli:First Second Modulus. Quarter-period. Quarter-period. ck, K, K', Ic', K', iK, 1lk, 7o(K-t'), 7cK', 1/7o 'K'-, tc'' K, to'/Jo, 7o', 7c(K'+ cK), c//c', kc'K 7'(+ tK'), the distinction being that sn 1 and dn=the complementary modulus for the first quarter-period, and that for the second sn, cn, dn are infinite and proportional to i, 1 and the modulus. ~ 28. We can prove that if the modulus is a real proper fraction the elliptic functions of a real argument are real. REALITY. 17 For as sn u increases from 0 to 1, while cn u decreases from 1 to 0, and dn u from 1 to k', the argument ', increases continuously from 0 to K, so that for any value of 'u between 0 and K, sn a, cn z have real values between 0 and 1, dn 'a has a real value between Ic' and 1. Also we see from ~~ 14, 17 that sn(2K -~) = sn, cn(2K- ) )= - cn u, dn(2K- u) = dn u, so that when u lies between K and 2K. sn t is real and between 0 and 1, cn,,,,, 0 and -1, dn,,, 1 and '. Again, sn(- a) = -sn u, cn( - u) = en u, cn( -t ) = dcn ', so that sn u, cn u, dn u are also real for values of u between 0 and - 2K. Also sn(u + 4K) = sn 'u, etc., so that, as any real quantity can be made up by adding a positive or negative multiple of 4K to a quantity between + 2K, sn qu, en u, dn 'a are all real if u is real. They are also real if u is a complex quantity whose imaginary part is a multiple of 2tK', for sn(u + 2K') = sn zt, cn(z + 2tK')= - cn B, dn( + 2K') = - dn u. ~ 29. Further, when the imaginary part of u is lK', or an odd multiple of it, sn u is real, en u and dn u are purely imaginary, D. E.F. B 18 ELLIPTIC FUNCTIONS. for sn(u + K') = 1/lc sn u, cn(u + tK')=- L dn u/lk sn u, dn(u + tK') = -L Cen /sn u. Again, since sn(Lu, I) = Lsc(u, k'), cn(Lu, k)= nc(u, k'), dn(tu, 7c)= dc(u, c'), it follows that for a purely imaginary argument or a complex argument whose real part is a multiple of 2K sn is purely imaginary, en and dn are real. Also, for a complex argument whose real part is an odd multiple of K sn and dn are real, en is purely imaginary, for sn(K+mu, k)= cd(ou, k)= nd(u, k'), cn(K+ t, k)= -k'sd(tu, k)== - c'sd(u, k'), dn(K+ L, kc)= k'nd(t,, k)= k'cd(m, k'). ~ 30. It is to be noticed that one of the periods at least is always imaginary or complex, and it may be proved that their ratio cannot be purely real. For let co and &2 be two periods of a function qp(u) so that 0(u) = O(u -+ i1) = p(u + (2) = O('t + amwol + n W2), m and n being any integers. Also let wJ/w2 be real. Two cases arise. If w1 and w2 have a common measure c let C)i =P-), w)2 = qW, p and q being two integers prime to each other. Then integral values of in and n can be found such that mp + nq = 1, so that <P( + o) = 0(u), and the two periods ol, w2 reduce to one, o. PERIODICITY IN GENERAL. 19 ~ 31. But if, on the other hand, wo and o2 are incommensurable we can prove that mnwo + n('2 may be made smaller than any assignable finite quantity. For let Xo2 be the nearest multiple of o2 to (0; then W1 - XWo2(= o3, say) is less than 2(02 -Let /c0 be the nearest multiple of o0 to w2; then 2 - 0W3, or (04, is less than 0o3, and so on. Then 1 (o0+21 is less than 2(02, which can be made smaller than any assignable finite quantity by taking r great enough. Also each of the quantities (04, 4,..., is of the form rzno1 + 02, so that the statement is proved. In this case then if p(u + noWI q-?o2)= 0(f)), the value of the function is repeated at indefinitely short intervals, and the function must be either a constant or have an infinite number of values for each value of its argument. ~ 32. It may be proved that the same kind of consequences will follow if a function is supposed to have three periods whose ratios are complex. We shall represent the argument of the function on Argand's diagram, in which the point P whose coordinates are (x, y) referred to rectangular axes OX, O Y, represents the complex quantity x + iy. The statement that a straight line AB is a period will be understood to mean that if from any point P a line is drawn parallel to AB and equal to any multiple of it the value of the function is the same at the two ends of the line. Now let OA, OB be two periods. Join AB. Through 0, A, B draw lines parallel to AB, BO, OA respectively. Through their intersections draw other lines in the 20 ELLIPTIC FUNCTIONS. same directions and continue the process till the whole plane is covered with a network of triangles, each equal in all respects to the triangle OAB. Then any line joining two vertices of triangles of the system is a period, since each side of any triangle is one. The triangles can be combined in pairs into parallelograms, all exactly alike, and similarly situated, and the values of the function at points similarly situated in different parallelograms will be the same. Such a parallelogram is called the 'parallelogram of' the periods.' Suppose, however, that there is a third period OC; then C must fall within or on the boundary of one triangle of the network. If it fall at an angular point then OC is not a new period, but is only a combination of OA and OB. If it fall on a side of a triangle, say DE, then DC and CE must be periods, and their ratio is real, since they are in the same direction; thus this case reduces to the one already discussed. If C fall within a triangle, say DEF, then CD, CE, CF are all periods. Let G be the point similarly situated within the triangle OAB, then OG, AG, BG are all periods being respectively equal to CD, CE, CF in some order. Any of the triangles OBG, BAG, AOG may now be taken as the foundation of another network covering the whole plane, and since there is still a third period, we can again find a point within the fundamental triangle with which to carry on the same process. We can prove that ultimately either the point will fall on the boundary of one of the triangles, which case has been discussed above, or a period can be found shorter than any assigned finite straight line. We shall form each triangle from the one before it as follows. Let Oab be a triangle of the series, and g the point found within it. Let Oa f Ob. Then we take Obg as the next triangle of the series. IMPOSSIBILITY OF THREE PERIODS. 21 Let e be any finite length, then we shall prove that a period can be found shorter than e. Suppose that none such can be found among the sides of such triangles as ABG,..., abg,..., which have not 0 for a vertex. The angle Oab is always acute, and can never be greater than ~ r- where / is some finite acute angle. For if there is no such limit, and Oab can be made to approach 7r/2 without limit, then since Obca Oab, aOb can be diminished without limit, and therefore ab can be made less than e. If Oh is drawn perpendicular to ab and g falls within the triangle Ohb then Og < Ob. 0 Fig. 1. If not, we have Oa- Og = ag sin l(Oga - Oag) cos -atOg>e sin ~/3, for ag Se, Oga >Oh Oa, Oag < -r-3. Thus Og is less than Oa by a finite quantity, and if Og > Ob it will be reduced by a finite quantity at the next step and so on, until after a finite number of steps we have a triangle in which Ob is the greater side. We can then replace Ob by a line which is less by at least e sin /3, and carry on the process, reducing this line again in the same way. Let, be the greatest integer in Ob e sin /3. Then after /u stages at most the shorter side Ob of the triangle Oab will be replaced by a line less than 22 ELLIPTIC FUNCTIONS. esin -/3, and therefore less than e. Each of these t. stages will consist of a finite number of steps by which the originally greater side of the triangle is gradually diminished till it becomes the less, followed by another step in which that which was the less originally is itself diminished. It is proved then that if there are three periods w1, w2, tC3, either they are not independent but satisfy an identity of the form lwt-+mw2+q —o= 0 with integral coefficients, or else a period can be found whose mocldus is smaller than any assignable finite quantity, so that the function has an infinite number of values for any single value of its argument. It might of course be a constant. EXAMPLES ON CHAPTER II. 1. Prove that each of the twelve functions sn, cn u, ns u,..., can be expressed as a multiple of the sn of an integral linear function of u with one of the six related moduli, in two ways, e.g. dn(u, kc)= k'sn(K'- K — Iu, V'). = sn(k'-K' - tc'K- tk'u, 1/i'). 2. What are the periods of the functions sc u, dc u, sn u cnn sn u ds u,snu cd u, sn2cd, 2n, n 1+cnu' 1+sn U IIl+sn2m 3. Putting S for sn u sn(u + K), verify that dS 1 d I = 2(dn2q-d dn2(u +K)},.............(1) {dn u +dcn(u + )}2+42= (1 + )2.......(2) {dn - dn(t +K)}2 + 4S2 = (l - ')2..(3) Deduce that (1 + ') = sn{ u( + '), l + ' EXAMPLES II. 23 and find the values of cn u(l+ 1), + and dn t(l + ), +} 4. Putting S for sn u de u, prove that (d)2 =1 -+ 2(7C'2 _ 72)2 + S4. 5. Verify that cn{(1 +lc)2, 1 +l 7)8 sn (I+ 1)~ 17-~~ I+ I + 1~2' d 1 + k, K = {i ks' dn (l+/)l, l+7-l+s2'=^ where s, c, d are sn(,', k), cn(u, k), dn(u, k), respectively. 6. If k = /2-1, prove that sn u(- 2) -.. ( - 2) scu~ ndu z, cn (- 2) = nc n 2) +2 c sc s sd u, dn ( - 2) = nc z nd u- k sc u sd L. Hence prove that for this value of Ic, K'/K= V/2. 7. If c = sin 75~, verify that sn u(- 3)t =( sc 4(4,/3 - 6- sn2t)/(4- 2/3 - sn2u), n u( - 3)' = (2 -^/3)(2 - /3sn2U)/cnu(4 - 2./3 - sn2/,), dn u( - 3)2 = (2 - /3)dc u(2 - sn2kf)/(4 - 2/3 - sn2m). Prove also that for this value of k, K/K' =/3. 24 ELLIPTIC FUNCTIONS. 8. Find the expansions of snu, cnn, dntz in ascending powers of u as far as 5. Ans. sn u= u- i-(1 + c2)U3 + _(1 + 141c2+k4)5..., en u = 1 - -u2 + -1(l + 47c2)U4..., dn u = 1 - 2%2 + 1-(4c2 + 4)4.... 9. Trace the changes in sign and magnitude of sn, en, dn for real and purely imaginary arguments for all real or purely imaginary values of D. CHAPTER III. ADDITION OF ARGUMENTS. ~ 33. We shall now show how to express the sn, cn, and dn of the sum of two arguments in terms of the elliptic functions of those arguments themselves. Let ul and u2 be the two arguments and let us write 8s, cl, d1 for sn ul, cn un, dn u, and s2, c2, d2 for sn u2, en u2, dn u2. This notation will often be found convenient. Suppose u1 and u2 to vary in such a way that their sum is constant, say a. Then 1 + u2- = a, d- -1. Consider now some symmetric functions of tt1 and Z2, as sn nl + sn u2, sn sncn u2 + sn u2cn t1, etc. We have d (81 + 82) = ldl c22, di (s12 + s2cl) = cldc2 - sdls2 + sls26d2 - c2d2c = (dC - d)(cIc - s182). d-(d( + d2) - 2(sCl- S2C2) -= -7c2sc(c22 + s2) + kS2C2(C12 + 812) = - (C1C2 - 812)(SC2 - S2Cl). 26 ELLIPTIC FUNCTIONS. Now - k2(s1222 - s22 2) - kc2(s2 - s22) = d- d2, and thus we have (d1 + d2)d (sdc2 + s2C1)=(s1C2 + 8,C1)cl (d1 + c2)From this it follows at once that s-2 + 2cl = a const. di + c2 so long as u + %2= a. The value of this constant may be found by putting sn a z1=O and u =a. It is 1 2 1 + dn a' Thus sn(u + %) s1c2 +sc1 1 + dn(1 + z2) d + d2 ~ 34. Again, s1- C2 - c _ dl+ d2 1i - (I2 1C2 + 82elC1 = a constant also sn a cn a - Thus sn(1l + t2-) S1= - 2S' dn(u +u2)-1 d1 —d2 Inverting these two relations and subtracting, we have 2 _ c-d2+ d - d2 sn (%1 + '2)- S1C2 + 2c1 S1i2 - 82C1 _ 2(sc2d2 - S9C11dl) 1222- 22C 2 812 82 2l so that 2 so that sn(ual + U2) = 1 2 81CAd - 82cidl, ADDITION OF ARGUMENTS. 27 By inverting and adding, we have ds(m+q) =Slc2d1 - s2cAd2 ds(36 + tt.2)- 812 _ 2 1 2 and dn(ul+ u) = Sqe2dl- s2Cd S1c2d2- 82Cd11 ~ 35. In the same way we could prove the following relations s1672 + s2 _ sn(u. + u2) c1 + c2 cn(u1+ ) +1' sd2 -sd1 _ sn(1 + 2t2) C1 -c2 cn( + U2)-1' cA12 + C2dc? _ cn(ul + a2) + dn(u, + 2) S1 + 82 sn(ul + u2) 1d9-c2d _ cn(Ul + 'u2)- dn((l + u2) S1- S2 sn(ul + '%,) which we shall leave to the reader to verify. ~ 36. Any one of them is enough to give the value of cn(~l+u2). Adding the last two we have cs(1 + II2) = S1Cld- s21d =?- S- ' C and hence cn(u1 + %2) = - s2c2dl S1C2C -- S2cl1 1 by help of the value given in ~ 34 for sn(Ul + -2). ~ 37. The formulae just found can be expressed in other ways. We know that sn( ')= n( + t) ns, cn( + K')= - dsu, dn(t + tK') = - cs u. 28 ELLIPTIC FUNCTIONS. Put then +l tK' for u, in the above formulae. We have sn(u, + tK' + U2) = -(2 -822) (cd 2 + d) sn(u, + 2) = k sn + + ) k+ ) sn(iu, $+ K'+?t2) S1c2d2 + s8cld1 1-k2S12s22 cs(j+ K'+ 2)= (- + 2 1) (2 2 -8/) IC2__2 81 __ dn(%u + Z2) =t cs(u1 + t + n%2) 1 - 1c2sl2s2 ds(i+ + LJ'f%~2)=(-s22 2 2 =cs/ /(128 2-82/ cn(us + u,2) = Ci.- Sl S2dd2 1 - k9's[2s9 These three forms, in which the denominator is 1-k72s2s22, are those generally quoted. It may be verified by multiplication that they are the same as the former set. Thus, in the case of dn(u, ~u2), (d1d2 - k72sls2c12)(s12d2- S2Cdl) = s1c2(d2 22 + 7'2822c12) - s82Cd2(d12 + k2812C22) = (s1c2d1 - s2cld2)(1 - k12S~222), for d2 + 7c2S22cl2 = 1 - c2s12s2 = ds + 7c2s82c22. The other verifications are left to the reader. ~ 38. By putting l + K for ul we may form another set from each of the two we have. The ADDITION OF ARGUMENTS. 29 four sets of formulae are embodied in the following scheme: Numerator of sn(u1 + uz): slc2d scd, 12 22, d + s2c2cd, sc22- 2, sc1dd +2. s d ld Numerator of cn(u1 + u2): C1C2-s182dld2, scd2-s, cd-s.cd l-s-2 k s12ss22, cIc2cld2- k' s1s2. Numerator of dn(u1 + ut2): dld2-k2Slssc lc,2, SlC2d1-scld2, cl2c2cld2 + k'2s1s, 1- kSs2-k2 22 + k2s1'2s2. Denominator of each: 1-k2s12s822 s, cslc2 + -sscsdd, Cdd. + s12d2c dc2d +.kcc ~ 39. The above formulae give the sn, en, dn of U1 - u2 by simply changing the sign of s8. Thus sn(-tb.- 2) 1 2 1c etc. 1- C2812822 2 By combining different formulae we easily find the following, writing A for 1 - /c2s22: A sn(u' + t,2) sn(Z1 - -2)= S12- s22, A cn(u1 + U2) cn(ul - U2) = 1 - s2 - s2 + 1c2s2, A dn(u,1 + u)dln(uzL - u) = 1 - 7c2s12 -_ 12s22 + 1c2s12s22 A Sn(ul + q62) cn(l-1 - a'2) = s81Cd2 + s2c2d1, A sn(,l + n2L)dn( - -u2) = S1c2Cd + s2c1d2, A cn(ul + u2)dn(q, - u2) = Clc2cld- k7c'2ss A{1~ sn(u,+u2)}{l ~ sn(u1-u2)} =(c2~s1c2)2, A{ 1 ~+ sn(1 a+ u2)} {1 ~ sn(1 - u2)} = (d + ksc2)2, A{dn(ul + '2) ~ cn(l + u2)} {dn(u —u,) cn(u,- U2)} = (c1d + C~,)2, A{dn(u + u2) + k cn(u + u2)} { dn(ul- z2) + ~ cn(ul- tb)} = (,d2~ + kccy2)2, A{1 +cn(ul+ 2)}{1 + cn(u1 —t2)} = (C + )2, Ai{1 + dn - )}{ +dn( -2)} =(dl~ )2, A {1 ~ cn(u1 + m2) } {1 T cn(tl - u) } = (sld, s2dl)2 30 ELLIPTIC FUNCTIONS. A{ 1 ~ dn(Q1 + 2) } { 1 T cln(nu - 2)= k2(S1C S2C)2, A {7 +~ dn(l + ~ 2)} {I + cln(mL - n2)} = (' + d, d)2, A { '~ d+ in(ul + u2)} {(c' T dn(l - U2) }= - c2(C2 ki's1s2)2, A dcln(u, + u2) ~ k'sn(ul + q2)} {dn(u- u2) ~+ 1C'sn(- U2)} = (cdl +~ 7 s)2, etc., etc. The verification of the above results will give the reader useful practice in the algebraical handling of the elliptic functions. ~ 40. Since u = v + a is the integral of the equation du= dv, a being the constant of integration, the different addition-formulae may be considered as forms of the integral of the same differential equation. Also if we write x for sn u, y for sn v, the differential equation becomes (I - 2)(l - kC22) f d = ( - y2i%(I _ y l y2)- l, which therefore has an integral that is algebraical in x and y, although neither side can be integrated by means of algebraical functions. This fact was known for a long time before elliptic functions were invented. Euler succeeded in integrating the equation X- dx + Y- -'dy = 0, where X is a quartic function of x and Y is the same function of y. Let X = cX4 + bx3 + cx2 + ex +f, Y= ay4 +by3 + cy2 + e +f. Then the integration is as follows:Write X', Y' for dX/dx and dY/dy. We have X-Y X - = a(x3 +2y xy2+ 3) + b(x2 xy+y2) + c(xfy) + e. x-y X'+ Y'= 4a(cv3+y3)+3b(cv2+y2)+2c(cc- y)+ 2e. EULER'S EQUATION. 31 Thus X Y- I-(X' +Y') = - C(x + y)(x-y)2 y- - y2)2 X-y = (x- y)2{t(x +y)+ b}. d IdI I 1 Also dX=X-X',x f =1 -.1y dcx dy 2 -xHe dy d( +cy) d(x- -y) d(X2- Y) Hence X _- -Y X- y- X(X' + (X+ y) (X - Y)dcl(x - y)-(x - y)cd(XZ - Y1) X- Y- -(X'+ Y')( — y) = 1 dF _ y1 (a-y){ct(x+y)+-b} V -y Therefore {aC( + y)+ b}d(+) = dX x~_ y x;_ y; and ( 2 ) =ac(x+y)2+b(x+ y)+ g, g being the constant of integration. This is the integral sought. Further information, with references, will be found in Forsyth's Diferenticd Equcations, pp. 237-247. ~ 41. Suppose in the addition-formulae that u1 is real, and u2 purely imaginary. Then s1, cl, de, c2, d2 are all real, and s2 is purely imaginary. Thus the imaginary part of sn(QL + U2) is s2clidl 1 -lc2128s2' This cannot vanish unless 2=0 or co, or c= 0 or dc =0. But d, cannot vanish as ui1 is real, and if cl =0 we have - = an odd multiple of K. 32 ELLIPTIC FUNCTIONS. Also since u2 is purely imaginary, if 82=0 or cc we have u2 = a multiple of tK'. If then a complex argument have a real sn, its real part must be an odd multiple of K, or its imaginary part a multiple of iK'. In the same way if the sn be purely imaginary, s8=0 or co, or c2=0 or d2=0. These are all impossible but the first, so that the real part must be a multiple of 2K. ~ 42. From this it follows that sn has no other period than 4K and 2tK'. For if A were such a period it must be complex, say Al+tA2. Then sn(zf+ Al+ LA2) is real or imaginary according as u is real or imaginary. If u is real we have A2 =a multiple of K', for u +A1 is not generally an odd multiple of K. If mt is imaginary we have A = a multiple of 2K. Hence there can be no periods other than those already found. The same holds for en and dn. ~ 43. Suppose now that there are two arguments U2 and un for which sn, en, and dn are all the same. Then it follows from the addition-formulae that sn(ul + %2)= sn(l + U%), etc., whatever ul may be. Hence u2- U3 is a period for sn, cn, dn, and must be a quantity of the form 4rnK+4ncK'. Thus all arguments having the same sn as u are included in the formula ( -1)mu + 2tmK+ 2tK'; UNIFORMITY. 33 all having the same en in the formula ~ + 4nmK + 2(K + K'); and all having the same dn in the formula ~ u + +2mK+ 4tK'. ~ 44. An important property of the elliptic functions, which has been assumed once or twice in the foregoing pages (as in ~ 41) is that they are tuniform, that is to say that each of them has one single definite value for each value of its argument. Many examples might be given of functions for which this is not the case; x2 is one. The property may be proved as follows:Suppose sn m= x, and let us examine the behaviour of u and x when x is in the neighbourhood of a value a. Put x = a + C, and let a be the value of u when x= a. Then = {-(a+ ) {-a+ )2}. The right hand side of this equation can be expanded in a series of powers of, which will always converge absolutely so long as I I (the modulus of 5) does not exceed the least of the quantities I 1 Ij-a,, +a ) -ca, k c Jr (See Chrystal, Algebra, ch. xxvii., ~ 11). By integrating every term on the right we get another absolutely convergent series since the term in er is multiplied by /(r +1), a constant (complex) multiple of a quantity that decreases as r increases. Hence the value of u is given as the sum of an absolutely convergent series. Therefore (see Chrystal, ch. xxx., 18) $ can be expanded in a convergent series of powers of u-a D. E. F. C 34 ELLIPTIC FUNCTIONS. within limits which are not infinitely narrow, and within those limits e is defined as a continuous uniform function of t (Chrystal, ch. xxvi., ~ 18, 19). This applies to every finite value of a but + 1, + 1/ic. If a has any of these values we may put x = a + e2, and deduce the same conclusion. Lastly, in order to consider very great values of x we put x= 1/$, and find that 1/x is in that region a continuous uniform function of u. Hence in all the plane there is no point where any branching-off of two or more values of x takes place, and therefore x is a uniform function of u. The uniformity of en u and dn m can be proved in the same way. EXAMPLES ON CHAPTER III. 1. Verify from the formulae of this chapter that -d sn(q, + b2) = en(u1 + U2)dn(ul + i2), cn2(Ul + U2) + sn2(l + + 2) = 1, dn2(u i + 2) + k2sn2(Ql + m2) = 1. 2. Find the sn, cn, dn of l,+u a2+fU3 in terms of those of 4l, Ia2, u3, and show that the results are symmetrical. 3. If 4l + + u,3 = O, show that dld23 - kc2ccc3 = lc'2, d + ]2c0ls2s3 = dcd3, dlc2c3 — c1d2C13 = 1c'2s2s3, s2cl + s3dl + s812d3 = 0, C1 ~ d1s2S3 = C2C3. EXAMPLES III. 35 4. If u1+u2+,t3+u4'= O, show that dld2d3d4 - kclc2c3c4 + k2k2ss2s3s =4 A', I Ce c11 1 =0, 82 02 12 1 S3 C3 4(3 1 84 c4 d4 1 (8s2 - s82c)(c13 - d4) + (3c4 - 84c3)(d1 - d2) = 0, (0Cld2- C2dl)(83 - 4) + (C3d4 - C4d3)(81 - 82) = 0, (81d2 - 82cl)( 3 - C4) + (S3c14 - 84d3)(Cl - 2) = 0 (These relations may be put in many more forms by such substitutions as u + K, %2, ut-i, u4 for u1, %2, m3, m64.) 5. If + u2 + 3=0, then 81 cdl s1 =0O. 83 cd 8 2 C22 2 S3 C3d3 83 6. If S(u) be written for sn a dc u and S'(u) for its differential coefficient then ( + ) S1S2' + S2S1'_ S2 - S22 S(ul ~ u2) 1- 222 - 81 -- Sa' 7. Verify the formulae of ~ 39. 8. Prove the following:cn(u - a)- cn(U + a) sn u sn a a= dn(u-a) + dn( + a) dn(u - a) -+ dn(t + a) dn(b - a) - dCn(u + a) k-2 cn(z( -a)+ cn(' + a)' cd(v; - a) + cd(tC + a) c n d(- - a) + nd(u + a) c'2 nd(u - a) - nd(u + C) c2 cd(u-C - -cd(zL + a)' 36 ELLIPTIC FUNCTIONS. dn, dn a=dc(6 - a) + dc(u + a) nc(u - a) + nc(,t + a) = ' nc(u' - a)- nc(z + a) 'dC(q(b - a) - dc(lt + a)' sd %u en a = sn( u + a) + sn(n/ - a) dcn(u + a) + cn(b - a) I cln( - a)c- dn(Nc + a) IC2 sn(/l + )- sn(.t-ca)' se n cm a sn(a + a) + sn(m - a) cn(,tb + a) + In(I - a) cn(zc -a)- cn(, + a) sn(mz + a) - sn(t - a)' sn l d a sd( tt + a) + sd(u. - a) ncl(eZ + a) + nld(u - a) 1 nd(u + a) - nd('( - a) - /k2 sd( + a)- sd(- a)' s sc(mt + a) + sc(zt - a) sn (u nu + a) + nc(u - a) _ nc(zC + a) - nc(u - a) - sc(t + a) - sc(tz - a)' dnu nd ca ds('t + Ca) + ds(c - a) dn u nd c =ns(u + a) + ns(u - a) ns(u + a) - ns(zt - a) ds(u + a) - ds(u - a)' ds(a, - a) + ds('la + a) sn uc ns a = cs(t - Ca)- cs( + a) cs(uC - a) + cs(u + a) cs(u - a) - ds( + a)' EXAMPLES III. 37 scunda sd( + a) + scld( - a) cd(u + a) + cd('u - a) 1 cd(m + a)- cd( - a) c'2 sd(Zt + a) - sd(L - a)' cn u ds a = dc(u + a) + dc(.u - a) cn u ds Ca = - sc(t a- ) - sc(u - a) _ 72. sc(U + a) + sc(Zt - a) dcc(u + a) - dlc(t - a)' en n ns(u - ) - ns(M + a) en X nc li = -- cs( - a) - cs( t + a) cs('a - a) + cs('. + a) iYs(Zt - aC) + ns(mt + a)' CHAPTER IV. MULTIPLICATION AND DIVISION OF THE ARGUMENT. ~ 45. By putting u = u in the addition-formulae we easily find the values of sn 2u, cn2u, dn2u in terms of sn u, cn u, dn u. Writing S, C, D, s, c, d for these quantities respectively, we have S = 2scd/(l - k2s4), C= (C2 - S2d2)/(l - 284) = (1 - 22 + k284)/(l- 712s4), D = ( - 2s2c2)/(1 - 2s4() = (1 - 2s2 + k284)/(l 24) ~ 46. Moreover, these equations can be solved for s, c, c if S, C, D are supposed known. We have D- C = 21'2s2/(l - s4), D - k2C= k'2(l + 7C284)/(1 _.28), D - 2C '2 = 2k'2/(1 - 24), D - 7c20- 7'2 = 2k'2k2s4/(1 - /c2s). Ds k2C_- k'2 D-C - 12(D - C) -D- k2 C+ '2 1-C 1= + ' by subtraction, 1-D =/2(1 + C)' by subtracting again. HALVING OF THE ARGUMENT. 39 Hence we find the following formulae for >u:l - Cn 2U 1 1-dn u2 sn ~u = d +- cnnu 2 \l+dnu/ k l+cn " (1 -en u)t(1- dn n)~ k sn um 1 1 d lnu+cnu2 ( 1-dnu C 2= 1 +dnu ) k \dn u -cnu (1 -dn u)2(dni u+cn u)! lc sn u lnu+cnu2 1-n 2 dn u cn -- 2 2 l+enu ) Vdnu-cnn (dn m +cn u,)(1 - en u)2 sn u ~ 47. In particular sn K = (1 + k')2, cn r/K=,c'(1 + k,)en 'K = k'-, dn 1K=k'1, sn2- j, sn I~i K'= i - being purely imaginary and of the same sign as its argument; cn ~K'= k-'/( + ' )I being a positive quantity; dn tK' = (1 + )2, being also positive. These three may also be deduced from the others by using the complementary modulus. 40 ELLIPTIC FUNCTIONS. Also sn (K + LK') + - (1 2 lo2 en /1 ((K+ + ki') 2- ( 1 - (') } = l -l (-CI dln (K + l )= 2 t{(1 + )2 + L((1-:)2 }. These three are most conveniently found from the former six by the addition-formulae. MULTIPLICATION OF THE ARGUMENT BY ANY INTEGER. ~ 48. By repeated use of the addition-formulae we can find the elliptic functions of 3t, 4t,..., in terms of those of u. We may prove the following facts about the formulae for sn nz, cn nuc, dn nu:Firstly, when n is odd, sn nu = sn u x a rational fractional function of sn2tu, cn nut = cn u x a rational fractional function of sn2,, dn nu = dn u x a rational fractional function of sn2u. dn nib=dntt x a rational fractional function of sn2U. In each case the denominator is the same function, and is of the degree n2 -1 in sn u; the numerators are different, but are of the same degree, mn2_ -. Secondly, when n is even, sn nt = sn Z en u dn u x a rational fractional function of sn2U, cn nu = a rational fractional function of sn2jt, dn nu = a rational fractional function of sn2m. MULTIPLICATION OF THE AEGUMENT. 41 In each case the denominator is the same, and its degree is n2 in sn u; this is also the degree of the numerators of cn n*, and dn ut; the numerator of sn -t sn, en u dn u is of the degree 2 - 4. Clearly we may say a rational function of cn2% or dn2 instead of sn2u without altering the meaning or the degree to be assigned. ~ 49. These statements are evidently true when = 1 or 2. Suppose them to be true for the values mt and m +1 of n; one of these values will be even, and the other odd. Write Sp, Cp, Dp, Np for the three numerators and denominators of snpu, cn mu, dn]pLt respectively, and s, c, d for sn u, en u, dn u. Then S2m -= 2SCmnDgnNAM = scd x a rational integral even function of s of degree 4.m2 -4, --- 7nn DI C2n = - C2N,2-S, =a rational integral even function of s of degree 41m2, D21n = DM nn-k SXG = a rational integral even function of s of degree 4m2, = - 4 _ Ji'^2 4 7V2m = N-k2 S2 =a rational integral even function of s of degree 4nz2. Also 2m+ +1,n - + - +i1N+'~ +Nl, ~ )i ) = a rational integral odd function of s of degree 2m2 + 2( + + 1)2-1, that is, (2m +1)2; C2n+1 = Csimlar function of c;+SnD nSm D = a similar function of c; 42 ELLIPTIC FUNCTIONS. D2m+l - = DnNn^Dm + IN,,, + 1 - 2SmGmSm + C, +1 = a similar function of d; N2 = N N2 NN,_ 72S2 + 1 2 =a rational integral even function of s of degree (2m +1)2-1. Hence, if the theorems hold for the values m, m + 1, they hold also for 2m and 2m+ 1. Now they hold for 1 and 2, and therefore for 2 and 3, 4 and 5, and universally. ~ 50. Also these expressions will be in their lowest terms. Consider for instance Cm, a rational integral function of c of degree m2. This must vanish whenever cn mue= 0, that is, whenever mu= K+ 2pK+2qK', p and q being any integers. Hence the roots of CG =0 as an equation for c are the values of cnK 2pK+ 2qK' This expression has ~'t m2 different values found by making p=0, 1... n-1, and q=O, 1... + (m -1) or +-m, in turn. Thus the degree of the numerator of en mu cannot possibly be lower than m2 and the expression we have found for cn nu is in its lowest terms. Also as CM~+ S N = N, D2 + S2=S2 = 2 and Con, N, have no common factor,,, and D,, can have no factor in common with either. ~ 51. We may notice that when Nm is expressed in terms of s, the coefficient of s2 in it vanishes. DIVISION OF THE ARGUMENT. 43 For N -9M =N-V2S4, N2=n = N N2 +-22 8X2X1 Now s is a factor in Sn and Sm+i, so that if the term in s2 is wanting in Nn and Nm+i it will be wanting in N2rn and N2m+1. Now N,= 1, N2 = 1 -k2s4, from which by induction the theorem follows. By changing u into u+tK' we find that the coefficient of s7m2-2 vanishes in Sm when m is odd and in N,, when n is even. DIVISION OF THE ARGUMENT BY ANY INTEGER. ~ 52. If we know the value of sn u, the multiplication-formula gives us an equation to find sn u/n. When n is odd, sn - is the root of an equation of the degree n2, n whose coefficients are rational in sn u. When n is even, sn2- is the root of a similar equation. We may show that the solution of these equations depends only on that of equations of the nth degree. ~ 53. Take the case when n is odd. Since snu =sn(u+L4pK+2qIK'), it follows that snl (u+4pK+2qcK') is also a root, and as this expression has n2 values it includes all the roots. Call it X(p, q). Then clearly any symmetrical function of X(p, 0), X(p, 1),..., X(p, n-1) will be unchanged by adding any multiple of 2LK' to u. Such a function then will have only n values, given by putting p = 0,1,..., n- 1 44 ELLIPTIC FUNCTIONS. in turn. It will therefore be a root of an equation of the nth degree only. Thus X(p, q) is the root of an equation of the nth degree whose coefficients are also given by equations of the 'th degree, rational in sn u. The same form of argument holds in the case when n is even, and also in the case when cn u or dn u is the function given and we have to find the sn, en, or dn of u/n. EXAMPLES ON CHAPTER IV. 1. Find the values of the sn, cn, and dn of (mntK+- nK') for all integral values of m and t. 2. Prove that sn 1K is a root of the equation 1- 2x+ 2k2X3 - c2x4 = O. What are the other roots, and which is the real one? Ans. sn( K~+ tK'), sn(3K+2tK'). The last is real. 3. With the notation of this chapter, show that -2mn+1 C~02m+i, expressed in terms of c, has 1 ~ c for a factor, the other factor being a perfect square. 4. Show that N2,n-C2,,, has 1-c2 for a factor, and that the other factor of it is a perfect square, as is also N2m + 2,C.a 5. Prove that when expressed in terms of d, N2,m,+1 D2,n+i has 1 ~ d for a factor, the other factor being a perfect square, that N2,, - D2,m has 1 - 2 for a factor, and that the other factor, as also N21, + D2s,,, is a perfect square. 6. Show that V2m,, ~+ S2,r can be expressed as a perfect square, as can also the quotient of nV2,,+i ~ S2zm+1 by 1 + (-1)'As. 7. Prove similar facts with regard to NA,,, ~kST,, 1CTmN + Dn, Dn ~ Cm, Dm +~ kC,, EXAMPLES IV. 45 8. Prove that (CNm- Cn)2 (N^ + 1 -,n + 1ni)(N,, -, - ) (dNm - D)2 (,+ - D, + - )(~Tm - 1 - Dm - ) are independent of the argument u. 9. If,u, v are any two nth roots of unity, show that the nth power of V-1 16-1 1 E EAPlqsn -(u+ 4pK + 2qt K') p=0 q=0 is a rational function of sn u and cn uq dn u. Hence show that the value of sn u/n may be found by the extraction of nth roots, if sn 2K/n and sn 2cK'/n are supposed known. 10. Use the last example to find expressions for sn nu, sn 1z. 11. When n is odd, prove that n sn nu=_ 1 sn (U+ 4gK 2tK') v=O /.=O n-1 n- ( 4iAK+ 2vJK') and that,sn2(n - 4Ksn2' / +~K' v =0 /U=O 12. When n is even, prove that ns2nun -= (ns2( ' — + ') V=0 /==0 CHAPTER V. INTEGRATION. ~ 54. We must now examine how far it is possible to integrate, with respect to u, any rational algebraic function of sn i, cn u, dn a, or, as we shall write them, s, c, d. In the first place, suppose the function to be ~(ScA) q(s,c,d)' 0 and r being rational integral algebraic functions. We may make the denominator rational in s by multiplying it and the numerator by (s, - c, d)VI(s, c, - d)V(s, - c, - cd), and by means of the relations c2= -2, d2 1- lc_28; by means of the same relations we may reduce the numerator to the form Xi(8) + CX2(8) + dX3(S) + cdx4(S), the denominator being X(s) and X, XI, X2, X3, X4, all rational integral algebraic functions. ~ 55 Now JCdXX4(8)d J>=X4('<)dsc ~ "5. Now J- (,Sa x (,4 ', INTEGRATION. 47 which can be integrated by the ordinary rules for rational fractions; dC. X3(s)(SU Jx)(1 x ( S) J xdO= / _ - I d s, and this can be reduced to the integral of a rational function by the substitution 2z 1 +z2' which gives (1 - 2) - -2 1 +z2' Also JSdu )= X( - 2s32) d8, J X(S) x(-) which can be reduced by putting 2z ks = 1 +z2' The problem is thus reduced to the integration of X(S)/X(S). ~ 56. The first step will naturally be the expression of X1(S)/X(S) as a series of partial fractions. When this has been done the expressions to be integrated will fall under one of the two forms sm, (s — a) -", a being any constant, real or imaginary. We will consider these in turn. Let Ir*dum = v,. Now d s - 3Cd) = (m - 3)8m - 4c2d2 - 8 - 2d2 - C2sm- 2C2 = (m - 1)c2s (m - 2)(1 + 72)sm- 2 + (r- 3)-4, 48 ELLIPTIC FUNCTIONS. and therefore, integrating, we have C + sm - 3Ced = (in - 1)k2V,, - (m - 2)(1 + 2)vm- 2 + (m -:3)v, _ 4 where C is a constant. Thus when m > 3, v? can be expressed by means of vm,-2 and v,-4; and in the case when = 3, t'3 can be expressed by means of v,. Thus when im is odd the integration of:,u depends only on that of vl, and when in is even on that of v2 and vo. ~57. Now vl-= sn u du = 2Jsn 2x dx, putting 2x= A, _ 4 sn x en x dn xdx J 1- 2sn4X =2 d z putting = -sn2x, 1 l+kz = - log 1 -7X, I -kIz 1 1 +t sn~2 1 - 1 l 1k sn2I ' Thus the integral of an odd power of sn u can always be expressed by means of the functions sn, en, dn, log. ~58. Again, = v ds=c u, V2= snu du. It is not possible to express v2 by means of known functions, and a new symbol has to be introduced. THE FUNCTION E. 49 The letter E is generally used, and the definition of its meaning is Eu = cln2 cu, 0 so that v = ( - Eu)/1k2. The value of Eu when u-=K is generally denoted by E simply, so that rK E= dn2u du. 0 The Greek letter Z was used by Jacobi for a slightly different function, defined as follows:Zu = u- uE/K. Thus ZK= 0. One advantage in the use of this notation is that there is not the same risk of confusing the product Eu with the function Eu. ~ 59. We now turn to (s - a) - ncd, which we shall call w,. Put s- a=t. du' / =(- m + 1 )(s - a)- 'c2d 2- ( - a )- n+ls(c2 + /22) =t-(- m + 1)- ( -n + 1)(1 +k2)(t +a)2 +(-m + 1)12(t + a)4 - t(t + a){ 1 + i2 - 2c2(t + a)2] = ( - ( )n -1 2)(1 - a2 C22)t -. + (2m - 3){ + c2 - 2c22a2}at -n+l +(m- 2){1 +12c- 6kI2a2}t-mn+2 -(2mn -5). 212 a. t-'t+3-(m-3). c2. t-'n+4. Integrating, we find that W,, can be expressed by means of known functions, and w,, -_, wi -- 2, Wi, -3, D. E.F. D 50 ELLIPTIC FUNCTIONS. W,,,_.4, provided always that (ma-1)(1- a2)(l1-k/a2) does not vanish. If a2=1 or 1/k2, then wu,,i_ can be expressed in terms of 't',,,-2, Wm,,-3 Wm-_4 for 2m-3 does not vanish. Hence for these special values of a the integral can be reduced to wQ, wzu, zv 2, that is to vo, vl, v2, and no new function need be introduced. But in general the reduction can only be carried on as far as wu, since when mn = 1 the coefficient of zt,,n in the formula of reduction vanishes. We must introduce a new function to express zl, and w2, Iz3... can be expressed by means of this and known functions. ~ 60. Now though ((s a)-1d and (s+a), l cannot be found in terms of known functions, their sum can. For by the addition-theorem 2 sn u cn ac dn c6 sn(u + a) + sn(u - a) = 1- - 2-sena sn Now each of the terms on the left can be integrated since we have found Jsnttdtn. Hence if a be so chosen that k sn a= l/a, we have an expression for 22S 2d 1 or (s-a) ~ + (s + a) ctlu. 2a2 J The new function that is introduced is therefore only needed to express (s-a) - 'dl~-(s + a) ldu), and the one actually chosen is 7u /2sn Ca cna dn c sn2. I1 - k2s n2'U sn2a 0 THE FUNCTION H.1 51 This is denoted by HI(u, a), and uc is called the argument, a the parameter. It has been shown then that any rational function of snu, cnu, dnr can be integrated by help of the new functions E and II. The properties of these will be considered in the next chapter. EXAMPLES ON CHAPTER V. 1. Prove that ic2j sn2n' du = ns2U du - c'/. K K 2 2 2. Prove that k'2 1 dL = llcdn + - '2 Eu. -sn 1 -sn u F cld du du dCu ld 3. Fnld Sl+ sn(, J '+dnn' l+cn 'L 1-dnu.1,l c en u dn u 1 -' 1/n sn u cn,1L A ns. EU + 2 — 2 --- 'A,. ic'2 1 + k sn n,' c/'E' k1c2- '(kl' + dn i)' sn u dn u, 1, -K I sn u cn 1+cn /2+ - ' 2 —dn u 4. Show that J* sn a cna dna dut = II(, a) - log sn(a +-u) Sn2nt- - s2 I(,a) l sna - S ) 0 5. Prove that ns du = log sn u - log cn u - log dn 2U., b 2 ~V 1 IV~ ~ill 2 W 2 cs du = log sn -z- +log en u - log dn |nU, ds u du = log sn Iu - log en ~ u + log dn lu. 52 ELLI PTIC FUNCTIONS. 6. Verify the formulae sn s n ac dn Ca dn2l d = I(, )+ log cn(u - a) cn2 -sn2a dn2 u cn(u + a)' o {u / sn en a cdn oda cnu2 ctu cl dn(u - a) lG1d2 n — -cn7 a cn29 9 = II(u, a) + 1 log c-n a). J dn 2 - 2 sna cn 2 ', dn(tb + oa) 0 7. Prove that II(/,c, ka, 1/k)= H(z, a, k), I(c, t, ) I(, a), + cn(k- ) =I(, a) + -l og ca(~+ - thn(e touls n at the modulus on the right being k throughout. CHAPTER VI. ADDITION OF ARGUMENTS FOR THE FUNCTIONS E, I. ~ 61. Expressions can be found for E(u +u 2) and II(z,1+u2, a) in terms of functions of mu and U2. As in the former case, suppose l + u2 = b, a constant. Take the function E1- + E~2. d l(FL + EU2) C=l2- d22 = — "(s12 -s2) - - k2sn(i, + t )(Ascc,d2 - s8cl11) = k2sn b. d(8180) Thus Ea1 + Em792 - 7c2s1ssn b is constant, and putting 1 = b, U2 = 0, we find its value to be Eb. Hence Eu1 + EUm - E(m1 + in2) = k2sn Z1lSn u2sn(1, + 2^), It follows that Zu1 + Zu2 - Z(U1 + i2)= k2sn i1sn mt2sn(q1- + U2)-. ~ 62. Putting u2 = K we have E(u + K) - EFt = E- k72sn zt sn(u + K) = 7E- k2sn i, cd v. 54 ELLIPTIC FUNCTIONS. E(, + 2K) - E(u + K) = E - k2sn(,t + K)sn(it + 2K) =E+-k2sn u cd u. E(u + 2K) -Eu = 2E. Hence E(v, + 2imK) - Eu = 2nE. Z(u + 2mKt)= Zr. ~ 63. Let us apply Jacobi's Imaginary Transformation (~ 21) to Eu. We have E(u,, 1k) Jcldn2(LU, /l0)clu = dc2(t, 7)dv. O O dc sn u dn z sn2, cdln2u Now _ = Cdn2 - d 2sn2t, + - dcl cn b/ en22U = dcl2i- 1k2sn2u. Hence t(czt, ') =n i d + t -(Eu, cnl i the modulus when not expressed being k; no constant is added for both sides vanish with u. Thus as cn K=O, IE(K, k') and therefore also E(,K', k) are infinite. Let us find the value of E(K +K', k). E(K+ u) = Eu + E - 72sn u sn(u + K) = Eu + E - k2sn n cd i. Thus tE(KI+ )+ E(t, ') = t(u + E) + t sn a(dc ' - l2cd u) =(u+E)+ clc'2sn cn u dn ' Put now ltK' for ut, and write E' for E(K', k'). Then 1E(K+tK)-E'=tE-K', E(K~+ t')=E+ (Ki'-E'). THE FUNCTION E. 55 ~ 64. Since E(K + ) = E + E- l-2sn u sn(u +K) we have E(K + mK) = E(mK) + E= E(nK- K)+ 2E.... Thus E(mK)= mE if m is any whole number. Also E(tt + 2nK) - Eu, = E(2mK) = 2mE. In the same way Ein(K+ IK') = mnE(K+ tK') = E + tm(K'- E'). E(t + 2rnmK + 2 1( K') - Eu = 2miE + 2mn (K' - E'). Thus E(tu + 2rK + 2nt K') = Eu + 2mE + 2n (K'- E'). This equation shows that the effect on the function Eu of adding any multiple of 2K or 2 K' to its argument is to add the same multiple of 2E or 2t(K' —') to the function. ~ 65. The quantities K, K', E, E' are connected by an important equation which we shall now prove. Clearly 0 K rK / K+IK' K'. E= ( IdnKd2v dudv, o K Thus EK K+tK' K. E(K+ K')-(K+ tK')= (dn2u - dn2v)dudcv. 0 K The right-hand side may be transformed by putting snusnv=x, dnutdnv=y. 56 ELLIPTIC FUNCTIONS. We have,(x, y)_ cnudn usnv, -k2sn ecn z dnv a(u, v) cn v dn v sn Z_, - c2sn v cn v dn d = -c 2cn z, cn v(sn2v cln2% - sn2t dn2v) = cn m en v(dn2v- dn2z). The subject of integration is then cn u cn v Now 7c2Cn2% cn2 - jy2 = k2722 -_ 1,2, so that the transformed integral is W Zk' dy dx (y2 + k2]1'22 _ k'2)2 As to the limits, sn v takes all real values from 0 to 1, and sn zL all real values from 1 to 1/k. Thus, if x has an assigned value > 1, sn u and sn v are nearest when sn = x, snv =, and furthest apart when sn u= l/k, snv kx. The value of y will therefore range from k'(1-k2x2)2 to 0. For y2 = 1 + k4z2 2c 2k2 - k2(sn sn v)2, which is least when sn ut and snv are furthest apart, and greatest when they are nearest. Also, if x has an assigned value < 1, snu and sn v are nearest when snz= 1, snv =, and furthest apart when sn u = 1/k, snv = kx. VALUE OF AN INTEGRAL. 57 The value of y will therefore range from k'(l- 2x2)2 to 0 still. The integral is therefore I'(l- 2, k) clc a d1 -x (y2 +J 27,2x2 _,2)' that is, -2 Any doubt there may be as to the sign of this result is removed by the consideration that in the original double integral cn v > k'> dn u, so that the subject of integration is always negative, while du is positive and dv has the sign + t. Hence K. E(K +K')-(K+ K')E= - 7r. Substituting the value that was found above for E(K +K'), we have EK'+ E'K-KK'=. ~ 66. The following result will be useful afterwards: JKEu du = 1 (KE- log '). 0 We may prove it thus J Eud c= E(K- i)du= {Eu+E(K-ui)}du 0 0 J K{E+ c2snsn Ksn(K-u)}cd o — KE,1 [aKs2sn u cn uc, =IKE+ K-g2 n Tdu 0 = KE- -1 log dn K = (KE- log k'). 58 ELLIPTIC FUNCTIONS. ADDITION OF ARGUMENTS FOR THE FUNCTION IH. ~ 67. Again if u1 t- + 2= b, du, {nH(n, (a)+ II(a, a)} k2sn a cn c dn as 2 1c2sn a cn a dn cas22 1 - k2s2sna 1 -- 2s22sn2ca k72n a cn a dn a(sl2 - s22) = (1 - k212Sn'2()(1 - k2s22sn2 ) Now we have seen that 812-s22 = -sn b- (818) What we have to do is therefore to express s12 + s,2 in terms of s192 and b. Now (1-1 72,i2s22)2cn b dn b = (cC2 - ss2cdid2c)(dd2 - lc2 s12c,,) = clcidld2( 1 + kl2si2 2) - s812( kl2c122 + dc12d22), (1 -7 2s 2s22)2sn2b = 2s1s2clc2dd2 + s12c222d2 + s22c12C12. So that (1 - 12s12s22)2{(1 + l22s2s22)sn2b - 2sls2cn b cn b} = ( 1 +2s12S22)(812 22dc22 +- 22cl2 12) + 28s2s2(k2c12c22+ d12( 122) which reduces to (1 - 12si2s22)2(sl2 +,22). Hence sl2 + s2 = (1 + 1k2s12822)sn2b - 2ls2cn b dn b, and d{ i('aI, a)+ I(n2, (a)} k2sn a cn a dn a sn b -1 -sn2ca {(1 +J12s2,s22)sn2b - 2scs2n b dn b } +kc4s22sin4a, cl1 2 THE FUNCTION II. 59, The denominator = (1 - k2sn2 sn2b) + 2k2s12. sn2a cnb dn b + 1k4sl2s22sn2a(sn2c - sn2b) = (1 - k1sn2a sn2b) { 1 + ks2sssn a sn(a + b)} { 1 + c2sls2sn a sn(a - b)}. The numerator = (1 - k2sn2a sn2b)k2sn a {sn(a + b) - sn(a - b)}. Hence — {nII(, ca)+ I(m2, a)} 1 7k2sn a sn(a + b) d 2 1 + kc28lssn a sn(a + b) dl 2 1 /c2sn a sn(a - b) c( 2 1 + 2Cls82sn a sn(a - 6) clU 1 _1 dl 1 + kc2slssn a sn(a + b) 2 di1 g 1 + kc2sl2sn a sn(a - b)' Integrating then, we have II(m1 +4 2, a) - I(u1, ) -11(2, a) = 1 log k2sn ulsn %tsn a sn(l + u2 + a) 2 " 1 - k2sn tsn uz2sn t Sn1( s(1 + U2 - ~)' ~ 68. There is another interesting property of the function I which we shall now prove. It connects II(u, a) with II(a, u), the same function with argument and parameter interchanged. We have 2d 9 2l2sn a en a dn a sn2t duII(,_ ) = 1- ck2sn2( snn2z = I2sn asn u{sn(u + a)+ sn(u - a)}. 60 ELLIPTIC FUNCTIONS. Thus a2 iH(, a) -= c2sn / {sn(ut + a)cn a dn a + sn a cn(t, + a)dn(u + a)} + lc2sn u { sn(u - a)cn a dn a - sn a cn( - a)dn(u - a)}. But by the addition-theorem sn2(mt + a) - sn2a sn(u + a)cn a dn a + sn a cn(u + a)dn(i( + a) _n2(,u - a) - sn2a% sn(u - a)cn a dn a - sn a cn(u - a)cln(~c - a)' for u = (u+a)-a= (u- a)+a. Hence 2 — I-(u, a) = a2sn2(m + c) + 2sn2(z - a) - 2ka2sn2 uoaa = 2 dna - dn2( - a) - dn2(u + a). In the same way 2 Dte (a, ') = 2 din2 - dn2(a - u) - dn2(a + u), DuDa so that -- {II(, a)-HI(a, u)} = dn2a -dn2U, auDa ~-{n(u, a)-II(a, tt)} = / dn2a-Eu, for 11(0, a) = n(a, 0)= 0. Finally then I(u, a) - II(a, z) = u. Ea - a. Eu. This may also be written zZa - aZet. EXAMPLES ON CHAPTER VI. 1. Prove that E(u + K) -Eu ~= E + - log dn c. EXAMPLES. VI. 61 2. Prove that E(t + K+IK') -Eu= E(K+ tK') + d log cn U. 3. Prove that E(uL + cK')-Eu=E(K+ K') —E+ I- log sn u. 4. Prove that kcE(kl, 1/k)= E(u, k)- '2 a. 5. Prove that kE'(ctlo, tic'/c)u - tE(u, ) c)+lc 2sn(u, c)cd(u, Jo). 6. Find the values of E~K, E-LK', E( IK+K'). Ans. (E~+ 1-/'), 1(K'-E,'+ l + l), I(E+ iK'- E'+ k+t '). 7. Show that I(K, a)= KZa, II(K+ tK', a) = (K+ K')Za + 7ra/2K. 8. Prove the formula 2II(m, a) = 2uEta - Ev dv. U -- 9. Verify that 2I(u, ~K) = u(1 - ') + log dn(u + K) - log Jo'. 10. Prove that the limit when a is indefinitely diminished of II(tb, a)- a is u -Ec. 11. Show that Eiut-nEu is equal to a rational fractional function of sn, multiplied by cn it dn u. By partial fractions or otherwise show that nEnu - n2eEu = d-m log V9,, where Nr denotes the common denominator in the expressions for sn nic, cn nit, dn nit. 62 ELLIPTIC FUNCTIONS. 12. In the same way prove the formula (n being odd) nE?2v- ^11 n- 2)uJ?+ 2vJt\ nEnu-E E ^+ — n ---.=o 0 v=0 = - n( - 1)(E+K'E'). 13. Prove the formula for addition of parameters in the function II, namely, II(u, a + b)- II(u, a)- II(u, b) _ lg 1 + /c2sn a sn b sn z sn(u + c + b) - 2 ~1 + k2sn a sn b sn U sn(u - a - b) - ]c2tC sn a sn b sn(a + b). 14. Find the value of II(C, a) and prove that 'aa II(u, u) = u~'u -UI- Ev iv. o 15. Prove, by putting t + v = 2, c- v = 2t, and integrating, that I(u, ct) + II(v, a) - I(u + v, a) log { 1 - ksn2(r - a)snsn } { 1-sn2 + a)sn2 {2 g1 - lc2sn2(r + a)snSt } {1 - k2sn2(r - a)sn 21} CHAPTER VII. WEIERSTRASS' NOTATION. ~ 69. For some purposes it is convenient to use the notation of Weierstrass, which we shall now explain shortly. We write Spu for a2ns2am+/3, where a is any constant and 3 is a constant which we shall determine. Differentiating, we have 'yl = - 2a3ns au cs au ds au. Also cs2aU = ns2a1 - 1, ds2aU = ns2au - c2. Thus (p%)2 = 4(pt - /)(ptb - - a2)(pmC - /3- a2/c2). Now choose / so that the coefficient of p2it on the right may vanish. Then _ - 3 a2(l + 1c2), and (p')2 = 4p3% - 2P a-g3, where 2= - 43(/3 + a2) - 4/(/ + ~ a22) - 4(/ + a2)( + a2,-2) g3 = 4( + a2)(3 + a2k2). The equation (p t)2= 4p'3 - gy2P - 64 ELLIPTIC FUNCTIONS. with the particular equation Li'm=o(2pl%) = 1 constitutes the definition of Weierstrass' function Apu. ~ 70. Conversely, if pu =x, V, = (4x3 - g - g3)-dx. x The periods of the function pu are 2K/a, 2tK'/a. They are denoted by 2w, 2w' respectively, and their sum by 2w'. We then have po) =/3+ a2 =el, say, o,, =/3 + a2lc2 = 2, say, gd =/3 = e3, say; and e, e2, e3 are the roots of the equation 4x - g2x -g3= 0 in descending order of magnitude. Thus PW = I'd' = 'W' == 0. ~ 71. We may write p(n,, g2, g3) for PgL when we wish to specify the quantities g2, g3. Thus if we put uxa for a in the original definitions pu is changed into U2p/U, and g2, g3 are changed into b412 and /6g3. Hence P(u, g2, g3)= 2p(fUl, 1-4g2,/ -693). In particular (t1u, ~2, g3)= - (n, 2, -y3). Also by a second differentiation we have 2p'up"' = 12ptp 22 - g2p', p"' = 6p2 - Ig. WEIEESTRASS' NOTATION. 65 ~ 72. The addition-formula for pu is easily found from the formula 8 2 2 sn(vl + 2) = s c 1 sICAd2 - 82(1(11 For ns2(v1 + 2)- (s1c2 l- 2l)2 S1S2 S12 =(2 _ Cl1)2 (l _ C2C22 CVI + 2) (-9 - -"(- e c )l) \sf2s2 s J 2 S 2/ (c2d2 _ 01 2 1 2 3 1 2 (C (7 ) (1\ 2 1) 2(17 2 i 2 \k 83 813/ \ / 82 24 2 1 9 2 1 This, translated into Weierstrass' notation, as explained in ~ 69, gives, if we take v,= au, v2= av, and remember that 1 + 1c2= - 3/3/a2, P(U, +)+ Pu+ v= (Pf -p) the formula sought. Again, a a- _ p' p'2 _ -'uS 'V 6p2- g2 - 1p- ) - -(- p) v + 6z~ - 222 2\= U-'S - 2 (_u-p+)2 pu-pv Now p/2 - p'2v = 4(p3m - P3/) - g2(u- pv), -= p(p ^ -p ) 2 " / - / ' so that 3 pu pr _l Pq-2(2g, p( -p ) 2 au / \ppr2 p-pr and p(1 + p) = — p' D2 tF pu - pv D. E. F. E 66 ELLIPTIC FUNCTIONS. ~ 73. Instead of the function E or Z, Weierstrass uses ~t, defined by the equation 0 Differentiating, we find u = - U. The term is put outside the sign of integration because pSu is infinite at the lower limit, but Vu -2 is finite. The value of ~(uf+v) is found as follows:5'(u + v) - ' = - P( + v) + Pg 1 D S't-I'v 2?u pu - v Hence (uM + v) - t - = P -%-V V 2 pu - v' where C is a quantity independent of u. Also u, - -= 0, when = = 0; and for the same value U 2 1 of u, 'u+-= =0, and gu- - is finite. Thus -1 y - + is zero when u = 0 and 2 pu - pv 't c= v. Hence 1( + v)-i —rv = 2 pu - v The definition of ~ shows that since p is an even function, ~ is an odd function. Thus (-,)= - r; and if +v+v=z= 0, EXAMPLES VII. 67 1 ^d'Lb- Sd 9.) we have + v + v I — -P 2 p',tn - p'v 1 'v - P'W 1 5'W - p 2 v- v-w 2 pw - pu = - (p + + w)2. The theory of these functions will be found developed in Halphen's Traite des Fonctions Elliptiques et de leurs Applications (Gauthier-Villars). EXAMPLES ON CHAPTER VII. 1. Prove that {S(m+w ) —} {V —~} = ( - /)( — P ). 2. If + v + 'w =0, show that quantities a and b may be found such that p'q = aCpt + b, p'v = —agv + b, p'w=apw + b. 3. In the last question prove that a= - 2(,+ + +v + w). 4. If the equation 4x3-g2x-g= 0 has only one real root, prove that one corresponding value of 7c is a complex quantity whose modulus is unity, and that in this case k2sn nc-2 is real if tU is real. 5. Show that 4p2u = — pi + p(u, + o) + Pd( + w') + p(n + w"). 6. Prove the formulae (1) {pTU+p(mo+w)} {p(h+~W' ) +p(n+w )} = - 4pop2u - 4pw'pw". (2) p-!=pu+ -e(p (-e)2 - 2 (2) 4 (+ - e(u2)i+(PS- - e3)+ (o 5 - e3)2 (Sd - e1) - + (2P5 - e1)2 (pT- e-) 68 ELLIPTIC FUNCTIONS. 7. Writing /y, ', " for e, i', ow", prove the formulae (1),7+,'= ", (2) (u 2w + 2o + 2m'c') = ~, + 2mq + 2mra'', if m and m' are integers; (3) t]'-~ = l T1, (4) +v dvAc=(-w'),+ Avdv+Ilog- e- - M~~~~u ~ ~C,~2eO' = (u + ) + I log(p W- el) - log(el- e2)(e- e3), (5) 2 2u = ~ + (uZ + ) + ( + )+(u - co). 8. Show that -c dx - =fv2' v _ - 2v,m. J x-Pv x- PU 0 0 9. If a and b have the same meaning as in Ex. 2, show that d log - + -+ ' t J. d a x'x + apx +b Sx- px t Sx - +p-pw CHAPTER VIII. DEGENERATION OF THE ELLIPTIC FUNCTIONS. ~ 74. For certain values of the modulus the elliptic functions degenerate into trigonometrical or exponential functions. Thus let k = 0, then dn L = 1 always, and d s-ll = en u. where cn22 + sn2% = 1, and sn0=0, n 0 = 1. Therefore sn u is sin u and cn u is cos u (~ 6), Eu =, K = E =r, Zu = 0. ~ 75. The six related moduli in this case are equal in pairs, the three values being 0, 1, cc. If k = 1, then dn u = cn u, and we have sn u 2t& sn2 = cn = n, n = 0. du Put sn u,= tanh 0 and we have sech2- = 1 - tanh20 = sech20. du Thus 0= au, as they vanish together. 70 ELLIPTIC FUNCTIONS. Hence sn(u, 1) = tanh u, cn(u, 1) = dn(uz, 1) = sech u, E(u, 1)== dn2(,, l)du = sn(m, 1) = tanh u. 0 K is the least positive value of u for which sech tu =0, that is K= oo, E= sn = 1. Z(u, 1)=tanh.* ~ 76. For the case when k7= oe we have / 7, 1 /. 1I I. 7 sn(u, k)= c sn( l, -= 7- sin eu, cn(, c) =dn (b/, 7)=1, (U I ) /0 dn(u, c) = en ui, ) cos cu. These formulae show the behaviour of snu, cn u, dn u when Au is a quantity comparable with 1/k. The table of periods for the related moduli (~ 27) shows that in this case both the periods are infinite, their ratio being -1. ~ 77. When c =0, the real quarter-period is finite, its value being — Tr; the imaginary period is infinite. When Ic=l, the imaginary quarter-period is finite and equal to -are; the real period is infinite. It may be shown that in this case the limit of K log c' is finite, and in fact -1. %The notation sgut, cgut for sn(iu, 1), cn(u, 1) is sometimes used, in honour of Gudermann. As however the functions have names already, being the hyperbolic tangent and secant, we have not used the others. The function arcsin tanh un is generally called the Gudermannian of u and written gd Iu. (See Chrystal's Algebra, chap. xxix., ~31, note.) DEGENERATION. 71 For we proved that ~(EK-log c')= Et du. 0 Thus ~(EK-2K-log ')= (Ett-1.)dm. 0 Also E(U, 1)- 1 = tanh u — 1 = -2e- 2/(1+e-2u), so that J{E(m, 1)-I}du=log(1+e-2U)) -log 2, between the limits 0 and oo. Hence Limk=(K/log c') = Lmlog 2- log log 7'= -1, as E = 1 in the limit. EXAMPLES ON CHAPTER VIII. 1. When 7c vanishes, prove that sin(a-u) HI(, a + ~K') = cot a + log sin(- ) 2 sin(c + m)' 2. Show that II(m, a, 1) = I log cosh(u - a)sech(z + a) + n tanh a. 3. Prove that the degeneration of dpa takes place when g23= 27Y32 4. Show that gd(t gd t) = l. 5. By the substitution b cot 0 - a tan 0 = (a + b)cot 4, prove that 7r 7r 2T 1 T2 (a2sin'2 + b2os20)-dO = (a2sin20 + bl2cos2O0)-1, 0 0 where 2ca=a+b, bl=cabt, and a, b, a1, bl are all positive. 72 ELLIPTIC FUNCTIONS. 6. If in the last question c2, b2 are formed from a, b1 as these from a, b, and if this process is carried on, show that in the limit, when n is increased indefinitely, 7r n= bn7 = X= (ct2cos20 + b2sin20).ct0. 0 (This quantity is Gauss' Arithmetico-Geometric Mean between c and b.) CHAPTER IX. DIFFERENTIATION WITH RESPECT TO THE MODULUS. ~ 78. The elliptic functions depend on two variables, the argument and the modulus. We must now show how to differentiate them with respect to the modulus. Wirite s, c, d for sn t, cnu, dnu, and let a, y, s denote -sn u, — cn nz, -dn z. ak 'a ak Since then- == cd dzn we have - ==, + c8. Since c2 + s2 = 2 + 2s2 = 1, we have cy + s=0, dd + ks2 + 28sa- =0. Eliminating y and 8, dn edd + so-(d2 + ) + c2 = O. Now dldcd = - s(d2 + k2), so that d l(-+) 2 0. du cd d+ 2 74 ELLIPTIC FUNCTIONS. dc se 7C2s2c2 Again -- = 2 - 2 + -- = c2 - 282d2. du d - — d2 c /T i cSC iC2 7c'2 - i2 Th d\cd tc-k'2d = -kJ'2= 7kc'2 c kCsc fU Eu and + k cd c'2d ic /tc'2' each side vanishing when u =0. Hence a sn u ic / Eu -- = sn u cn2a + -cn dn - cn cu dn u, ic'2 ic ic/c'2 cn nt k 2 EU -c- = - -,2sn2t cen u - -sn c dn ~ + 7 —sn u dn u, Ddn'm( ic c c = — c2sn2 ndn t-ic ksn tcnn+ 7 Em sn u cnn. ~ 79. From the last we may further find tcEU, as follows:D2Eu 3 27c 9 2 27l -, = - dn2tI = - -s2s i - 2kc. sCd + /,2T. scd. Now -d scd = Cd2 - s22 - 7c2s2c2 du -I. s2 = 2u. scd + s2. a2XE j fi i0cs2 2 cscd) Hence - Eu - ids2 8 ~ c2 Th3c= +k W+c) =c-in i- 2. S — 2C2 ~ is82 - 7(ic'282 -- C2c2) = -- DIFFERENTIATION OF THE PERIODS. 75 Integrating, -Eu =- cn2mEmE - lcu sn2% + 77 sn u en u dn v, a IC'2 Ic'2 since again both sides vanish with u. ~ 80. These equations enable us also to find dK d etc. idk' dcc' We have cn(K, k) = 0, and therefore dKK K E - sn K dn. d7_- sn K dn K+7c-o2 sn K dn K=0 d/dE cc Ilc'2K by differentiating. Thus d = E-1/2K Again, when u = K, -E, = - 7K. lc7 Thus dE - kKc+dn2K d d -lc dki E-7'2K E-K = -kK+ - k k l dK' E' - c2K' dE E' - K' Also dAso' Ick' 2 ' d ~ ic' 1 -;dK' k2K'-E' dE' k(K'-E') so that - = ckl'2 ' c - 7c'2 ~ 81. Again E-K 7,2 E- '2K =kK. 76 ELLIPTIC FUNCTIONS. Putting k2 = c, '2 = c' in this we have it in the form d(,dK\ dc\- dc) 4= K which is unchanged if c and c' are interchanged. It must therefore also hold when K is put for K, as can easily be verified. The most general solution of the equation d {(/c- _ 3dy} cy is accordingly y = AK+ BK', where A and B are any constants. ~ 82. In the same way df dE\ dE cdK E-K E- k'2K cE _7a__ _ __ - _c_ 2 __ d~(l )~d/ ~d l) k AC 7c 2= - c 2 This equation is not satisfied by E' also, but we saw (~ 63) that E(K + K')= E + (K'- E), so that K'- E' is suggested as a second solution. d El Now d-(') E Thus d7(K' ')} cE' E'). Hence the most general solution of the equation d (d\dz) (1- k2)A(l- +kz= o is z=E+ D(K'-E'), C and D being any constants. EXPANSION OF THE PERIODS. 77 ~ 83. The differential equations just found for K and E may be solved in series, and thus the expansions of K, K', E,' in powers of k may be found. Take d (i(k - k3C} = Icy, d /o = Co and put y = E 1 + 2r -=0 for the exponents of Ik in successive terms must clearly differ by 2. Then 7 - 7 j )d- - ky = s2o10S -1 + { (8+ 2r)2,_-(s + 2r- 2)(s + 2r),_ 1-i- 1}/ +2'-1. The coefficients are therefore given successively by the relation /r=-( + 2r ) 2 ) -1, and the values of s by the equation s2= 0. This equation has equal roots, so that we find the second solution by differentiating the first, namely S+ C (s +1)2( +3)2... (8 +2r-1)21-+2 1 (8 + 2)2(8 + 4)2... (s + 2)2 with respect to s before putting in the value of s. Hence, if y~=l+ =2+... + "2 2.4... 2r + { 1.3...(2r-1)) 2 and 2 = ilog k + 2 -...2r-1).1 2+ 4 -2... 2r x- I I+ + * 2* -r)7b2 the complete primitive is y=Ay1+By2. 78 ELLIPTIC FUNCTIONS. ~ 84. We may therefore choose A and B so that this expression shall be the value of K or K'. Now we have seen that when k1=0, K=7r. But 1 = 1, 2 = oo for this value of k. Thus K-= -ry-l7r 1 +,-4C2 4+...2. Suppose that K' = A y- + By2. ~ 85. In the same way, from the equation for E we may find series for E and K'-E', or we may use the formulae E = lc'2K + kll'2dK/dclc, K'- E' = k'2K' + ckk'IdcK'/dk. Putting =(1 - 72) 1Y + Ycl Z=(1- /2)(y2 + 75 C2), we find E= 7r1, K'- E' = AzB + Bz2, where 1 - / 12_ 3 12. ' ".(2r- 3)(2r-1) = -127, 1C4 - _ - - (2-...(2__ r)-k2 1-l 22 22.42.2.42...(21) zs = Zlog ]c + 1 r+,l 2 1 [ (2 ) 2rc 1 2'.4 (.. 2 -Hence E=A (y- )+ B(2 -,); and as when k= 0, E'=l, y1- z1=(, and y2-z2 =-l, we have B= -1. EXAMPLES IX. 79 ~ 86. A, as well as B, may be found as follows:We found (~ 66) that the limit of (EK- 2K- log k'), when k= 1, was -log 2. Thus in the limit, when kc= 0, {A (y,1- z1) + B(Y2 -z) }(Ay + By2) - 2(Ay +By2) - log k + 2 log 2 = 0. The coefficient of log on the left is -B2- 2B-1. This must vanish, so that, as we found before, B= -. The absolute term is - AB- 2A + 2 log 2. This must vanish, so that A = 2 log 2. Hence K'= 2ylog 2- y, E'= 2(y,- z)log 2- (Y2- Z). It is noticeable that the series y1, z are hypergeometric. Thus, in the notation of hypergeometric series, K —7F(1,t,72) E F(-2'2l, k2) EXAMPLES ON CHAPTER IX. 1. Prove that K increases with k so long as the latter is a positive proper fraction, while E decreases as lc increases. 2. Show that 9 (U 7 1 2 \u r u2 Sn2u Ev0 hec find / 2/i 2k' ' and hence find alII(u, c). 80 ELLIPTIC FUNCTIONS. 3. Prove that if N,, is the common denominator of sn nqu, cn nz, dn nt, and is equal to unity when = 0, then 2aNI a2N'* 2nJ\(ET _ k'2) 2n2 ki" ' + D23 + 2n " (E '2) + n2(n2 - 1)Nj2sn2 = 0. 4. Writing x for sn u, transform this differential equation into the following, in which x and k are the independent variables:2n2k'2 n+ +(1 - x2)(1- 2) 22?)k - / x 2?aN +x {(22- l)c)2(1 -x2)- 1+k/2}+n2(2-1 )NJ1cx2= -. (For Examples 3 and 4 use the result of Ex. 11, Chap. VI.) 5. Show that (g2 — 27g32)- 'a = (g2 2- 3 - 9g3u) - 9 22u + -2 2 '3 + -y (g2 - 2732) aPU = p'u(3g2Nu - g2 3) + 6g92p2 - 9gPt - g2 6. Prove also that (g23- 2732)7 'a = - u(g22 - 2g3) + 4g3I ' t + 9 422m - 3g23,ff and that (23 -27g3'2)e 'a = - u(3g92u- - ff9) - 2j'u - 23a + 1g2z2 EXAMPLES IX. 81 7. Show, by differentiating the equation p(u + 2w)=U, or otherwise, that 2 -24 2 (g23-27g2) =- 22 (g23 - 27g32) = g 3i-2' 8. Prove also that (9 3 g22 4 2 8Wg2g3, (g23-27g 2)-) = -g22-_ ~og2g2, 2 1/2 (23 - 2793 2)- =all -gt 2. (QY-27,32)=_g-21193 + 14 g 2 Dy3 9. Verify by differentiating that EK' + E'K-KK' and yow'- i-'w are constants. 10. Interpret the following differential equation, satisfied by lpu:22 g2 + a3 3g 3 t -i + u'f dK dE 11. Verify the values of d,7 and -k when one of the related moduli k', 1/k, /lk', ik/k', c'/lk is substituted for lc. 12. Deduce the expansions of K and E in powers of k by means of the equations 7r r K= (1 - k2sin20)d, E-= (1 - 2sin20)2dt 0 0 13. From the equations of Ex. 12 find the values dK dE of and. dck dic D. E. F. F CHAPTER X. APPLICATIONS. ~ 87. The usefulness of the Elliptic Functions consists chiefly in this, that by means of them two surds of the form (a + 23x + yx2)2 can be rationalized at once. One such surd could be made rational by an algebraical substitution: thus (1 - 2) becomes (-y2)/(l y2) if 2y/(l +y2) is put for x, and (1 +x 2) becomes (1l+y2)/( -y2) if 2y/(1-y2) is put for x; but generally speaking no rational algebraical or trigonometrical substitution will rationalize two such surds. ~ 88. Let the two surds be s2 and 5' where = a + 2bx + cx2, = a + 2^x + yx2, We shall suppose the coefficients in s and -c to be real. Also let S = A + 2Bx + Cx where A, B, C are found from the equations Ac - 2Bb + Ca=O, Ay-2B 3+Ca =, so that in fact S= I -x x2. c b a y f3 a APPLICATIONS. 83 Let C, r be the two roots of the equation S= 0. Then it is known that s and o- can both be expressed as sums of multiples of squares of x -, x -, and in fact it is easily verified, since a+b(+}C)+ c$ =0, a + 3(+ )+y = 0, that s()-=) = ( c+ b)(x-,)2-( Cl + b)(x- )2, and (-( = ) (y+ )(x- )2 (7y + 3)(x- _)2. Also by tracing the rectangular hyperbolas ab+3($+)+ cn==0, each of which has the line C= 7 for an axis, it is at once seen that the values of C and C which they furnish are real except when the line =r is the transverse axis in each, and.each hyperbola has one vertex lying between those of the other. This is the case in which = 0 and -= 0 have both real roots, arranged so that one root of each falls between those of the other. We see also that in the identity S(- I) = (CC+ b)(x - _)2 - (Cj + b)(x - )2, the product of the coefficients of the squares on the right is -{c2, + bc($+, )+ b2}, that is ac-b2. Hence s is expressed as the sum of two squares if s=0 has imaginary roots, as their difference if s=0 has real roots. The same holds for a-. If then $ and r are real we may by the real rational substitution y = (x - j)/(x - e), express s2 and o-2 in terms of y and two surds ( +1 ~ K2y2)2, ( + 1 + 22). ~ 89. Such a surd as (-1 - K2y2)2 will be imaginary for all real values of y. The other cases we shall take in turn. 84 ELLIPTIC FUNCTIONS. I. To rationalize (1-K2y2)2, (1-t2y22). (Take K>M.) Put Ky =sn(m, -), then (1 - K2y) =en, (1 - 2y2)= dn u. II. (1 - K2y2), (1 +,2y2). Put KY = cn +, (2 K2)-, o 1( then (1 - K22)2 = sn, (1 + M2y2) = (U)2 + K2)2dln,. III. (1 + K2y2), (1 + 2y2)2. (Take K > /.) Put K = sc( u,,2 i) then (1 + K2y2) = nc U, (1 + 2/22)2 = de u. IV. (K2y2-1)7, (1 - 2y2)2. Here K must > /, or both surds cannot be real. Put y =dn, (K-,) K then (1 - 1U2y2) = (K2 - 2)2cn, 12 (K y2 1 2 _ ) cn,. V. (K2y2-1 ), (I + )2 y2. Put Ky=ncf I, -t,}, then (K2y2 - 1)' = sc, (1 + L2y2)2 = (K2 + s2) de, K RATIONALIZATION OF SURDS. 85 VI. ( -2y2- 1), (2y2 1)2. (Take K >.) Put =ns(,, ), K then (Ky2 - 1)2 = ds z, (j2y2- 1)2 = cs U. In each case the value of x is given in terms of u by substituting for y in =( -)I/(y - 1). It hardly need be said that if e were infinite, we should put y =x - j, and then we could go on as before. ~ 90. If t=, the process fails. But in that case s and -c have a common factor x-. Let s=(x- )(cx +d), 0-=(x -)(yx + ). ex + Cxd dy2 - (d Put. ---d=y2, - yx + C-yY Thus s(c - 7y2)2 = (8y2- d - eC + 7Vy2)(c8 - d)y2 -(c - y2)2 ( (y2 - c - + ~y2)(6 - dy), so that s2 and c-' can be expressed by means of a single surd of the form (A+By2)t. This surd can again be rationalized by putting B 2im -A 1 1-2' Hence if ~=, the surds can be rationalized by an algebraical substitution. ~ 91. The above does not apply to the case when s = c(x - )(x - e), cr = y( - S)(x - C), d, S, e, e being real quantities in order of magnitude. 86 ELLIPTIC FUNCTIONS. In this case put x-d,y2 - ax-6 = y ~I. Then s= c(8-d)y2{(- e)y2- ((- - e)} (y2-1)2, =y7(8-d) {(8-e)2-( d _ )}(2 2 1)2. Thus s2 and o2 are expressed by means of two surds only, and those of the form (Ay2+ B)~, which we have already shown how to rationalize. ~ 92. It is easy to verify, and important to notice, dX. that in each case - is a constant multiple of sc2. ~ 93. An expression of the form ax4 + ~x3 + yx2 + 6X + e ( = X, say) can always be expressed as the product of two real quadratic factors by the solution of a cubic equation. Hence any expression which is rational in x and X2 can be rationalized by a substitution such as we have just discussed. The exceptional case of ~ 91 need not arise. It will not be possible unless the roots of X = 0 are all real. In that case there will be three ways of resolving X into real quadratic factors, and only one of the three will lead to the exceptional case. If a=0, X becomes a cubic instead of a quartic; but by a linear substitution for x of the form X Ky+X Kty + X the expression is made rational in y and Y~ where Y= X(uy + v)4, so that Y is a quartic in y having /Uy + v for one of GEOMETRICAL APPLICATIONS. 87 its linear factors. Thus there is no real distinction between the cases of the cubic and the quartic. ~ 94. It must not be supposed that the rationalizing of these surds can only be accomplished by the particular substitutions which we have used. The number of substitutions that might be used is unlimited. We have tried to choose the simplest. The comparison of the different substitutions that would rationalize the same surd or pair of surds belongs to the theory of Transformations, which is beyond our limits. APPLICATION IN THE INTEGRAL CALCULUS. ~ 95. When an expression has to be integrated which contains two surds, each the square root of a quadratic, or one surd which is the square root of a quartic, linear functions being counted as quadratic and cubic functions as quartic, then it follows from what we have proved that the integral can be expressed by means of the functions sn, cn, dn, 1, II. For the subject of integration can be made a rational function of sn zC, cn t, dn u by a properly chosen substitution, and such a function can be integrated as explained in Chapter IV. GEOMETRICAL APPLICATIONS. ~ 96. The elliptic functions have an important use in the theory of curves, plane and twisted. This depends on the following theorem:The coordinates of any point on a curve whose deficiency is 1 can be expressed rationally by means of elliptic functions of a single parameter. (Compare Salmon, Higher Plane. Curves, ~~ 44, 366.) Suppose the equation to the curve to be U= 0, and 88 ELLIPTIC FUNCTIONS. that it has multiple points of orders, k1, k2..., its degree being m. Then the deficiency is (m — 1)(n- ) 2)- 2 9i(/- I), and we have -7kc(lk - 1) = I-I(m - 3). Take a system of curves of the degree qn-2, each having a point of order - 1, where U = 0 has one of order I, and passing also through nm-2 other fixed points on the curve. The number of arbitrary coefficients in the equation to such a curve is (m + 1)(m - 2), and the number of conditions assigned is 1c(k- 1) +m- 2, that is 1(m + l)(mi - 2)- 1. Hence there will be one arbitrary coefficient left, and as all the equations to be satisfied by the coefficients were linear the equation to any curve of the system is S + XT= 0, X being the arbitrary coefficient and S, 2' determinate functions of the coordinates of the degree m -2, such that S =0, T=0 are two curves of the system. Of the m(nm-2) intersections of the curves U=0, S+XT=0, c(7c-1)+ m-2, that is m2-2_m-2, are fixed. Thus only two depend on X. Call these P and Q. Let A be one of the m -2 fixed intersections of S +XT= 0 with U= 0. Replace A by any other point Al taken at random on the curve. Then we have another system of curves S+XlTl=O, whose intersections with U=0 are all fixed but two. Choose X\ so that P may be one of these and let Q1 be the other. Q1 will not be the same as Q. For a curve of the degree m -2, satisfying all the conditions above prescribed for S + X T= 0 except that of passing through A, and also passing through both P and Q, will be altogether fixed, and all its intersections with U=O have been already specified but one. This one is A, and therefore it cannot be A1. Hence Q1 and Q are different. CURVES OF DEFICIENCY ONE. 89 The three equations U= O, S+XT=O0, + X T = 0 will therefore enable us to express the two coordinates of P rationally in terms of X, X,, and also to eliminate those coordinates and find the relation between X and X,. When X is given, there are two possible values for X,, found by substituting in -Sl/T1 the coordinates of P and Q respectively. In the same way when NX is given there are two possible values for N. The equation connecting them must then be of the second degree in each, and may be written X,2(AX2 + B\ + C) + Xl(DX2 + EN + F) + GX2 + HX + 1=0. This equation may be solved for X,, the only irrational element being the square root of a quartic in X. Hence this is the only irrational element in the expression of the coordinates of P in terms of X, and it may be removed by a substitution for X in terms of elliptic functions. Thus the theorem is proved. ~ 97. If the curve is not plane, but twisted, we may suppose S + XT =, S + X 1T1 = to represent not curves but cones, of a degree lower by 2 than that of the curve. Take U =0 to be a cone with any vertex standing upon the curve and S+ T=0 a cone with the same vertex, and having as a (7c-l)ple edge any multiple edge of order 7 on U =0 and also having m - 2 fixed edges in common with U = 0. S1 + Xl'l = ()0 may then be a cone drawn in the same way with another vertex and we may ensure that Q1 is not the same as Q as follows: Let the positions of P and Q when X = 0 be F and G. Through F and another point H draw a cone with the vertex that is proposed for 81 + XlTl = 0 and satisfying those of the conditions that S1 + X L= 0 must satisfy which are not at our disposal. Take the other mn-2 90 ELLIPTIC FUNCTIONS. simple intersections of this cone with the curve as defning the fixed edges of the system 81+XlXl'=0. Then as G is not the same as H, Q1 cannot in general be the same as Q. The rest of the argument goes on as before, the two equations to the curve taking the place of the single equation U= 0. The deficiency of a twisted curve is thus understood to mean that of its projection from an arbitrary point upon an arbitrary plane. In general the double points of the projection will not all be the projections of double points of the curve, but some at least will be the intersections with the plane of chords of the curve drawn from the vertex of projection. ~ 98. The simplest examples of curves of the kind in question are non-singular plane cubics, and among twisted curves the quartics which are the intersections of pairs of conicoids, and in particular sphero-conics. If X is the parameter of ~ 96, and u the elliptic argument, then it follows from ~ 9 2 that the coordinates are expressed rationally in terms of X and d-, which we may call X', and X'2 is a rational quartic in X. To each value of X there correspond two values of u and two points on the curve the two corresponding values of X' being equal with opposite signs. ~ 99. It may be proved that if a variable curve of any assigned degree meet the curve in points whose arguments are tz1 u%2..., U,, then 1+ L2 +... + i, * = a constant. For let 1= (), 2 = 0 be any two curves of the degree assigned. Then we can prove that for the intersections of the given curve with 1 + L 2= 0, lit is independent of Au. ABEL'S THEOREM. 91 In fpi and ~2 substitute the values of the coordinates in terms of u, and let f, f2 be the results of substitution. Then u is given by the equation fi+ f2= 0, f + Ai4, and da f (Cl, dif2) Now f, and f2 are rational functions of X and X', so that f2 +(/i+A /2) is also a rational function of them, say (X, X') X(X, X'). Its denominator may be rationalized by writing it AV(X, X')(X, -X') X(X, X')X(X, -X'). Thus since X'2 is rational in X we may write f A + B' A + W'2 C A, B, C being rational functions of X. Let X,, X..., X,, be the roots of the equation C(= 0, corresponding to the values j1, 2a,., I,*. Then A/C and B/C may be resolved into partial fractions, there being an absolute term in the first case because A and C are of the same degree. Hence we have an identity of the form J1+ + 19r +-X- +42 r: qr Now of the two points for which X = X,., only one is generally to be taken, suppose that for which X'= X,'. The left-hand side is therefore finite at the point for which X = X,. and X' = - X/'. Making this substitution after multiplication by X - Xr we find Pr - qX = 0. 1u p ~+ A.(X'+ X,.') Thus f-A +- AX-XA, fA+ tk X X-), 92 ELLIPTIC FUNCTIONS. If, however, the point (X,., - X/) is one of the intersections we must have X = X., X' = - /.' corresponding to Us, another of the series u1, un2,..., un,. Then the equation (C=0 has only one root corresponding to the two arguments, and there is only one fraction (pr + qX')/(X - X,.) for both. But in this case the equation p. - qI'.I = 0 does not hold, and we write Pr +.XN _ qXr, +?r N' + NX/ q.'+S +~r. X' + X,' X-X,. 2X,' X-X,. 2s ' X-X ' so that the final form is the same. The identity f + /f Po+ 0 EP X-,.' being thus proved to exist, we may find the value of q, in the usual way, by multiplying by X-X,. and putting n =,ur. Thus q,. 2X,. Lnn,, f( ) = value of J2X' (fi+ \ Mt1 ) when ur. is put for t, = -rX/CIn/dfL. That is,,r= -cl d,. /du. Now give in such a value that X becomes infinite. Then X' is infinite of a higher order; but as f, and f2 are of the same degree, f2 -(fA + f2) is finite. Thus Eq= 0, and Ed',j-/di/ = 0, so that ti,,. is independent of /u. AN EXAMPLE OF INTEGRATION. 93 Giving / the two values 0 and oo, we find that EZr is the same for the two curves pl = 0 and 2 =0. But these were taken to be any curves of the assigned degree. Hence the theorem is proved. It will clearly hold also if the given curve is not plane and 1=(), 2= 0 are any surfaces of the same degree. ~ 100. The facts proved in ~~ 96-8 may be applied to integration. If y is a function of x, and the relation connecting them is the equation to a curve of deficiency 1, then any rational function of x and y may be expressed rationally by means of the functions sn, en, dn of a single variable, and may be integrated with respect to x or? by means of these functions together with E and II. ~ 101. Take, for instance, (1- x3)-(cx. Put y=(1I-x3), so that x3 + y3 =. This is a cubic without singularity, so that the deficiency is 1. Put + y= z Then 3 -3yz = 1, z2 1 " 3 3z 4 z2 (-Y)^-=3z 3' The radical is therefore (4z -4)2. The real quadratic factors of z4 - 4z are z(z-2t) and 2 + 2 z+ 2 % Here z takes the place of the X of ~ 96, and the curves S=O, T=0 are respectively the straight line x + y = 0 and the line at infinity, the point of intersection of these two being clearly a point on the curve. 94 ELLIPTIC FUNCTIONS. The roots of the equation 2z-28, -23z =0 2z +2, 2z +2} are 2-(-1 +3) Hence we put t = (2z- /3+ 1)/(2z + /3+1), that is, z=-{t(3+1)+(,/3-1)} 2(t-1). Then (t - 1)2(z- 2s)= 2'/3(2 + /3){t2 -(2 - 3), (t- 1)2(z2 + 2z + 2') = 2. 3(t2 + 1). We therefore take t= (2- _/3)cn(, / 3- 1), using the substitution II of ~ 89, since is (4z - Z4), not (Z4-4z)i. Then the radical = 2- (3 - )(1 + cn u) - (2 -/3)cnu} =2'(l +cn m) -{(V/3 + )-(V/3-l)cn }), (t - 1 )2(z- 2) = -_ 2-3(2 - J/3)sn2tt, (t- 1)2(z2 + 2z + 2) = 2. 3(2 - ^3)cln2. Thus (t- l)42(x-y )2= 2;33(2- /3)2sn2 dn2u, - y = 2t34(2 - ~/3)sn u dn u 2- (/3- 1)(1 + cn)l- (2- )cn u}. f 1 = 2 3Tsn u dn m6 + (1 +cn ){(^/3 + )- (^/3-)cn,}. Also x + = 2(1 +cn t) ({(/3 +l - )-(3-l)n }. AN EXAMPLE OF INTEGRATION. 95 From these equations x and y can be found at once. Now if v be written for (1- x3) -dx, we have, since x3+y3 = 1, dx dy d(x + y) =v -- = _ 2 _ *' But du(x + y)=-23sn dn {(/3 + 1) + (/3-)} + {(/3+ 1)- (/3-1)cn }2 = -32-(x -y)(x + ), so that v = 2-3 3. u + const., that is to say, d x =2 -433n -,1(3 + 1) xt + (I-3) 3 2+ __~=^ 2-33cn-^1( ~) -'-e +const. (l 3) (3- 1)+ (1- 3)3} + 2W the modulus being (/3 - 1)/2/2. ~ 102. It should be noticed that when a is a constant, the equation connecting sn u and sn(u + a) is of the same doubly quadratic form as the one found between X, X, in ~ 96. For the two values of sn(zu+a) when sn u is given are sn(vJ+a) and sn(2K-it+ a). Their sum is 2 sn u cn a dn a- (1 - c2sn2u sn2L), and their product is (sn2 - sn2a_) - (1 - k2n2u s n2a). Hence sn2(i + a) {1 - ksn2a sn2%} - 2sn(u + a)sn u cn a dn a sn - sna = 0, that is, k2sn2a sn2u sn2( ( + a)- sn2(u + a) - sn2U + 2 sn( + a)sn cn a dn a + sn2a = 0, 96 ELLIPTIC FUNCTIONS. The same holds for any other of the elliptic functions sn, en, dn, sc, etc. This suggests another way of integrating Euler's equation (~ 40) which was given by Cauchy. Let p(x, y)=0 be an equation of the second degree both in x and y, and let (X, ( ) = Xoy + 2X1y + X2 Yo + 2 YLx + Y2. Then g= 2(Yo + Y), = 2(Xoy + X). ay But since +(x, y)= 0 we have (YoT+ YT)2= YL2 Y Y= Y _say, and (Xy + X1)2 X2- XX02 = X, say. Hence (p(x, y)=O is an integral of the equation X-cdx+ Y-c2dy=O, and X and Y are quartics in x and y respectively. Also if in (x, y) the coefficients of x2y and xy2 are equal, as also those of x2 and y2, and those of x and y, then O(x, y) will be symmetrical in x and y, and X will be the same function of x that Y is of y. Also the number of coefficients in p is still one more than the number in X or Y so that if the coefficients of X and Y are known, 0 = 0 will contain one and only one arbitrary constant, and will be the complete primitive. ~ 103. If in a doubly quadratic equation connecting x and y we transform x or y or both by substitutions of the form x = (e+f)/(g+ h), the transformed equation is still of the same form in the new variables, though with different coefficients. Now there are three arbitrary constants in such a transformation, and they may be so chosen as DOUBLY QUADRATIC EQUATIONS. 97 to make the transformed equation symmetrical, since symmetry is ensured if six coefficients are equal in pairs, namely those of x2y, x2, x to those of xy2, y2, y respectively.? When the expression has been made symmetrical, x and y can be rationalized by a substitution for either in terms of elliptic functions, the two substitutions being of the same form and having the same modulus but different arguments. It follows however from the differential form of the equation that if u and v are the two arguments, clu = ~ dv, mz ~ v = a constant. Hence transformations x= g+' y = - + — can be ge+h' u cq+d found such that $ and I are the same function (sn, cn, dn, sc, etc.), with the same modulus, of arguments differing by a constant. "-With the notation of ~ 102, it may be proved that the anharmonic ratio of the roots of X = 0 is always the same as that of the roots of Y = 0. For, by putting xy = z, p(x, y) may be made a quadratic function of x, y and z, so that the two equations xy - z = 0, 0 = 0 represent a twisted quartic curve. The cone standing on this curve whose vertex is any point of it will be a cubic cone and the anharmonic ratio of the four tangent planes to it drawn through any one of its edges is a constant. (Salmon, Higher Plane Curves, ~ 167.) Thus if A, B, C, D are any four points on the curve the four tangent planes through AB have the same anharmonic ratio as those through BC, and these have the same as those through CD. Now let AB, CD be the lines at infinity in the planes x = 0, y = 0 respectively, these being chords of the curve xy = z, 0. The equations X = 0, Y = 0 represent the two systems of tangent planes and the theorem follows. Another proof is given by Salmon (Higher Plane Curves, ~ 270). It follows that by a linear transformation of x the roots of X = 0 can be made the same as those of Y = 0. This is the transformation wanted, for it may be verified that 0 is symmetrical if the coefficients in X are proportional to those in Y. In carrying out this verification it is advisable to suppose X and Y reduced to their canonical form, in which the second and fourth terms are wanting. (See Salmon, Higher Algebra, ~ 203.) D. E.F G 98 ELLIPTIC FUNCTIONS. ~ 104. This applies to any case in which two parameters are connected by an algebraical relation, such that to each value of either there correspond two values of the other. There are two or three important cases of this which we shall now discuss. In the first place, let P, Q be two points on a conic, such that the line joining them touches another fixed conic. If P is given there are two possible positions of Q, one on each of the tangents from P to the other conic. The relation between P and Q is reciprocal, and the coordinates of each may be expressed rationally in terms of a single parameter. Hence the parameters of the two points are connected by a doubly quadratic equation of the form we have been considering. The same may be proved if the tangents at P and Q are to meet on another fixed conic, or if P and Q are to be conjugate points with respect to another fixed conic. It is in fact known that these three conditions are only the same stated in different ways. ~ 105. Jacobi has given a full discussion of the case when the two conics are circles, into which they can always be projected. Take any four points A, a, 3, B (Fig. 2), in order on a straight line, and on AB, acc as diameters describe circles. Let the centres be Q, 0, the radii R, r, and let 02 = 8. Let P, Q be two points on the outer circle, such that PQ touches the inner circle at T. Let P'T'Q' be a consecutive position of PTQ, meeting it in U. Also write O=BAP, rp=BAQ, O+dO=BAP', -+dp=BAQ'. Then B2P = 20, P2P' = 2d, PP'=2RdO, QQ'= 2Rda. JACOBI'S CONSTRUCTION. 99 The angle PUP= Q UQ', and the angle P'PQ = QQ'P'. PP' QQ' Thus P UQ dO de and in the limit pT _Q PT- TQ' But PT2= OP2- OT2 = R2+ 2 +2RScos 2 - 12, TQ2 R2 + 82 + 2R cos 2 - r2. Fig. 2. If then we write 7C2 = 4R6/{Q(R + )2 _ r2}, sin 0 = sn(u, k), sin b = sn(v, k7), we have cos 0 = n u, cos 0 = cn v, PT= {(R+ )2- 2}~dnm, Q= {(R + )2 r2}dn v. 100 ELLIPTIC FUNCTIONS. Also cos 0 dO = cn u dn u du, d0 =dn udu. Thus du = dv, v - u = a, a constant. ~106. If now we put = tan, = tan <, the coordinates of P and Q can be expressed rationally in terms of e and v respectively, and we can find the algebraical relation between C and Y that follows from the equation v - u = a. Take Q2B as axis of x, and a perpendicular to it from Q as axis of y. Then the equation to PQ is x cos(0 + (p) + y sin(O + b) = R cos(0 - P). The perpendicular drawn to it from 0 is r. Hence R cos(O - ) + ~ cos(O + )) = r, that is, R+ ( + (R- 6)ti = r sec 0 see, (R + 6)2 + 2( ti2 _- 2)e2 + (R _ 6)2e212 = r2( + $2)( +,2). Putting r/(R + 8) = cn(a, I), the value of cos p when 0 is 0, we find (R - 8)/(R + 8) = dn(a, k). Thus 1 + 2e dn a + C22dln2a = (1 + $2)(1 + r2)cn2a. Solving the quadratic for i, we find _ - dn +a sn a cn a(l + 2)(1 + l/22T) 92.'2sn2ca -cn2a As was to be expected, this is rationalized by the substitution s= sc(u, k), and becomes sn u cn uc dn a h sn a cn a dn u cn2 cn2a - c'sn2a sn2 u sn u cn u dn a + sn a cn a dn u so that sc(u +a)= c n2a - '2sn2a sn2 - cn2u cn2a(- /g' sn/a sn2% JACOBI'S CONSTRUCTION. 101 the lower sign being taken in order that the two sides may agree when u=(0. This is justifiable because a was found from its en and dn, and therefore the sign of sn a is as yet undetermined. The equation just found is one of the additionformulae. Others may be written down at once from the figure. For instance, PT+ TQ=2R sin(~- 0), that is, (R + ~)sn a {dn tt + dn v} = (R + 8)(1 + dn a)(sn v cn u- sn u en v), sn(U + a)cn u- sn cn(Z + n ca) sn a dn(u +c)+dcn) l+dn ct ~ 107. When the outer circle and AB, the axis of symmetry of the figure, are kept fixed, the quantities a and ic depend on the position and size of the inner circle. It is of some importance to know under what circumstances the modulus Ik will be constant. Now 1c2=4R/(= 4 R+ )2 - r}. But if s is the distance from Q2 of the radical axis of the two circles 82 - R2 (S _ )2 _,2, and 2s =R2 + 62 - r2 so that s=2 R/172- R. Hence if the inner circle vary so as always to have the same radical axis with the outer, the elliptic functions will have the same modulus. The quantity a is then the argument belonging to the other end of a chord of the outer circle drawn from B to touch the inner circle: ~ 108. An interesting case is that in which the inner circle has its radius zero, so that all the tangents to it 102 ELLIPTIC FUNCTIONS. pass through the inner limiting point of the coaxial system. In that case en a=0, so that a is an odd multiple of K, if real. Let L be the limiting point. Then if PL produced meet the outer circle again in P1, the argument u + K belongs to the point P,. Thus u + 2K belongs to P. It should, however, be noticed that when the argument u is increased by 2K in this way, 0 is increased by -r only, so that snu and en ut have signs opposite to those they had before. The signs of BP and AP are in fact changed, because the positive direction of measurement has been changed in each case by a rotation through two right angles. We have then sn It = BP/BA, cn u = AP/BA, dn u = LP/LB; and, travelling along the are PAP,, sn(u + K) = BP,/BA, cn(u + K)= - AP1/BA, dn(u + K) = LP1/LB. Now BPI = BA sinBPL = BA sin PBL x BL/PL = PA. BL/PL. Thus sn(u + K) = cd u. Also AP=.PB. AL/PL. Now AL/BL=dn K = k'. Thus cn(mu + K) = - k'scl t; and since PL. LP = BL. LA, dn(u + K) = k'nd '. ~ 109. The coaxial system of circles have a common self-polar triangle of which L is one angular point, the other two being L' the other limiting point and PONCELET'S POLYGONS. 103 the point at infinity in a direction perpendicular to AB, which we may call M. The figure shows that if L'P and MP meet the circle again in P2 and P3, the arguments belonging to P2 and P3 are K-tu and -it respectively, for P P3 passes through L. But since sc(2tK'- a) s)= st, every point on the circle has two distinct (that is, not congruent) arguments belonging to it, and the second arguments belonging to P2, P3 are respectively congruent to 2cK'+K+u and 2tK'+u (mod. 2K, 4tK'). It is now clear that if the inner circle in Jacobi's construction is replaced by a circle of the same coaxial sytem, but containing the other limiting point, then the quantity a is not purely real but has its imaginary part equal to an odd multiple of 2LK'. If on the other hand a is purely imaginary, its en and dn are real, so that the inner circle is to be replaced by a real circle of the system, but one which contains the original outer circle. ~ 110. By help of the foregoing we can answer the following question: Can a polygon of an assigned number of sides be inscribed in one given conic and circumscribed to another? Project the two conics into circles as before. Let lb be the argument of one angular point, tzc+a that of the next, then uc+ 2a will be that of the third, and so on, and if the polygon has n sides and is closed the argument. u+ na must belong to the first angular point. Hence z,+nta=t- or 2tK'- ' (lmod. 2K, 4tK'). Suppose first that -tb + na 2tK' - t, then ' +- a - 2K' - ' - (n1 - 1)(, zl-+ 2a - 2tK'- -- (n - 2)a, etc., 104 ELLIPTIC FUNCTIONS. so that the second angular point coincides with the nth, the third with the (z-l)th, and so on. Thus there is no proper polygon in this case. If on the other hand we take uz + a = u we find a= 0 (mod. 2K/n, 4t4i'/n). This condition does not assign any of the angular points, but only shows that unless the two conies are related in a particular way the problem has no solution. If the conics are so related, that is, if a has one of the values included in the formula (2rK+4stK')/n, then the value of u does not matter, and any point on the circumscribing conic may be taken as an angular point of the polygon. ARCS OF CENTRAL CONICS. ~ 111. It is most likely known to the reader that the length of any elliptic arc can be expressed in terms of the coordinates of its ends by means of the elliptic functions sn, en, dn, E, and that it is from this fact that the name " elliptic" arises. The ellipse x2/a2 +y2/b2= 1 is the locus of the point (a sn u, b en u) for different values of the argument 'a. If S is the length of the arc measured from one end of the minor axis (0, b) then S vanishes with u and (dS/ld)2 = (a2n2U + b2sn2U)dn2t, = a2(1 - e2sn2)dn2U. So far we have not assigned the value of k. If we take e for its value we have dS/du = a dn2m and S = aE(u, e), if x = a sn(u, e), y = b cn(z, e). ARCS OF CENTRAL CONICS. 105 This expression holds equally well for the hyperbola, but it is not so useful, as the modulus of the elliptic functions is then greater than 1 and the point from which the arcs are measured is imaginary, b being imaginary. ~ 112. In the hyperbola X2/a2 - y2/b2= we may however put y = b cs(K - U)= bk'sc A, x = a ns(K- u) = a de n. so that u vanishes for the point ((a, 0). If S is the length of the arc measured from this point we have (dS/du)2 = (aC2714SC2% + b21d2clc2U)nc2, = b21'2nc4U, if a2'/'2= b21c2, that is k=l /e. Thus dS/du = bkc'nc2, if 7k= 1/e, and S = ae{sc u dn u + '12v - _EU}. ~ 113. The equation Eum + Ev - E(u + v) = lc2sn u sn v sn(u + v) may be expected to furnish a geometrical theorem concerning arcs of a central conic. We must first find what geometrical condition is expressed by such an equation as u - v = t, connecting the arguments u and v of two points on the ellipse. It will be more convenient to put u=a+3, v=a-/3. The tangents at u, v are then sn(a 3)+ cn(a ~/3)=l, a ~~~b 106 ELLIPTIC FUNCTIONS. and at their intersection we have sn a cn dn 3 cn a cn = I- 2sn2a sn2i, a b sn/3 en a dn a = snsn a dn a dn/3, a b whence x= asn adc 3, y = b cn a nc. Eliminating a, we have x/aC2dC2.+ yl2/b2nc2/ =. Eliminating /3, we have, since e is the modulus, 2/ac2e2sn2a - y2/2e2cn2a = 1. Each of these conics is confocal with the original one. Thus if + ~ v is constant, the intersection of tangents at the points 'u, v traces a confocal conic. ~ 114. At a point on the tangent at u whose distance from the point of contact is z we have x-a sn u y-b cn u z acn u -b snu a dn ' so that x =asnu + zcd = asn + zsn( + K), y=b cnu+zcn(u+K). It is hence easily found that the lengths of the two tangents at (a +~ ) measured to their intersection are a sc /3 dn a dn(a /3). Call these t, t2. Then tl + t = 2a sc /3 dn2a dn //'(1 - kC2s12a sn12), t - t2= - 2ae2sn2/3 sn a cn a dn a/(l - 7isn2a sn2/). Now by the addition-formula for the function E E(a +/)- Ea-E = - c2sn a sn / sn(a +/3), E(a - /3)- Ea + E/3 = 2sn a sn 3 sn(a - 3), GRAVES' THEOREMS. 107 and by addition and subtraction E(a +/3) + E(a-/03)- 2E =- 212sn2/ sn a en a dn a/(l - k2sn2a sn2n/) =(t - t2)/a, E(a + /3)- E(a-/3)- 2E/ = - 2kc2sn2a sn /3 en 3 dn /3/( - 7c2sn2a sn2/3) = (t, + t)/a -2 sco 3 dn /. If then a +/3, a-/3 are the arguments of the two points P and Q the tangents at which meet in T, and if B is the point from which the arcs are being measured, we have, when T traces a confocal ellipse, so that /3 is a real constant, arc BP - arc BQ - TP - TQ = a constant, or TP + TQ- arc PQ = a constant; and when T traces a confocal hyperbola, so that a is a real constant, arc BP + arc BQ - TP + TQ = a constant = twice arc BR, if R is the point of intersection of the hyperbola and ellipse between P and Q. Thus TP- arc RP = TQ- arc RQ. ~ 115. This applies also to the hyperbola, but since in that case b is a pure imaginary the relation TP + TQ - ar PQ = a constant holds when T moves along a confocal hyperbola, and TP- arc RP = 'TQ- arc RQ when T moves along a confocal ellipse. For geometrical proofs of these theorems, which are due to Dr. Graves, see Salmon's Conic Sections Chap. XIX. 108 ELLIPTIC FUNCTIONS. It is noticeable that the system of confocal conies is the reciprocal of a system of coaxial circles with respect to one of the limiting points, so that this case is closely connected with that of ~~ 107-110. A CASE IN SPHERICAL GEOMETRY. ~ 116. Another case of a doubly quadratic relation between two parameters is afforded when an arc of a great circle moves on a sphere so as always to have its two ends on two fixed great circles, its length being constant. Let PQ, P'Q' be two consecutive positions of the movable arc, OPP', OQ'Q the two fixed arcs (Fig. 3). P <Q=PQ Fig. 3. Let OP =, OP'=O+dO, OQ =, OQ'= +do, POQ=A, PQ =a. Then the integral equation connecting 0 and b is cos 0 cos (~ + cos A sin 0 sin 0 = cos a. To form the differential equation, since PQ =P'Q', we have PP' cos OPQ= Q'Q cos OQP in the limit, that is, (l - sin2A cosec2a sin2p)~dc + (1 - sin2A cosec2u sin20)~dP = 0. THE AMPLITUDE. 109 We may then put sin =sn u, cos O=cn u, cos OQP=dn u, sin =snv, cos p=cn v, cosOPQ = dn v, the modulus being sin A cosec a, and we have du + dv =0, u + v = constant = w, say. Then zu is the value of v given by supposing 'a and therefore 0 to vanish, so that snw= sina, cn = cos a, dnw=-cosA, and we have cn wv = cn e cn v - dn w sn u sn v, that is, cn( + v) = cn u cn v - sn u sn v dn(Zt+v). This is one of the addition-formulae. We have also cos 0 = cos a cos 0 + sin a sin 0 cos OQP, or cn u = cn(u + v)cn v + sn(zt + v)sn v dn u, and cn v = cn(u + v)cn u + sn(u + ')sn u dn v. These three equations may be solved for sn('u+v), cn( + v), dn(u +v). If the modulus is to be real and less than unity and w real, we must have A obtuse and a+ A greater than two right angles. We may then write sin 0 = sn., cos 0 = cn u, sin = sn(w - u), cos 0 = cn(w - u), w being a constant. ~ 117. In this case we have dO/d = dn u or d/dO =(1 - c2sin20)-2. The function 0 of u, which satisfies this condition and vanishes with u, was called by Jacobi the ampli ELLIPTIC FUNCTIONS. tude of u, it being the upper limit on the right-hand side of the equation U= ( - k2sin20)~i2c0. 0 It was also customary to write A0 for (L - 2in20)2. Thus sn,, cn u, dn z were conceived as the sine, cosine and A of the amplitude of u,, and in Jacobi's notation were written sin am uA, cos am uA, A am 'A, the amplitude 0 being denoted by am 'l. The shorter notation, sn, en, dn, was suggested by Gudermann. The function am u is of no importance in the theory of elliptic functions, but it sometimes presents itself in the applications of the theory. In the case considered we may, for instance, write 0 = am n, =am(z - ). APPLICATIONS IN DYNAMICS. THE PENDULUM. ~ 118. There are certain problems in dynamics whose solution can be expressed by means of elliptic functions. The simplest is perhaps that of the motion of a pendulum. The equation of motion is 10= -y sin 0, where 0 is the inclination to the vertical of the plane through the axis of suspension and the centre of inertia and 1 is the length of the simple equivalent pendulum. A first integral is found by multiplying by 0, it is i102 =-g(K + COS ) = g(l +K-2 sin2~0), K being a constant. To integrate this put 0=2 am {', 2(1 +K)-}, THE PENDULUM. 111 so that sin 0 = sn cos 0= cn u, COS 1 0-cnm 1 + I - 2 sin2lO = (1 + K)dn2U. Then 2 = (1 +K)g/2l, and -= {(1 +)g/21}+ const. ~ 119. Let A, B be the highest and lowest points of the circle described by the centre of inertia of the pendulum, P its position at any time, h its distance from the fixed horizontal axis, and let (1 +K)g/2 = n2 Then BP = 2h sn nt, AP = 2h cnt, if the time is measured from the moment when P is at B. If PY is the perpendicular drawn from P to a horizontal plane at a distance Kh above the axis, that is, at the level of zero velocity, we have P Y = (1 + K)h dn2 t. Let BA, produced if necessary, meet this plane in C. Then let a circle be described having CY as its radical axis with the circle APB. The tangent from P to such a circle varies as P Y2, that is, as dn nt. Hence the figure is the same as that in Jacobi's construction (~ 105 above). ~ 120. The application of the addition-formula will then give us the following theorem:The envelope of the line which joins the position of the centre of inertia at any time to its position at a fixed interval afterwards is a circle of the coaxial system which has for radical axis the line of zero velocity, and includes the circle described by the centre of inertia. 112 ELLIPTIC FUNCTIONS. When the pendulum is performing complete revolutions K 3 1, and the elliptic functions have a modulus ~ 1. Thus if the fixed interval is half the whole time of revolution, the straight line joining the two positions will always pass through a fixed point, namely, the inner limiting point of the system of circles, whose depth below the radical axis is h(K2 - l). Further, the envelope of the line joining two variable positions of the centre of inertia, which are separated by equal intervals of time from any fixed position (one before, one after) is a circle of the same coaxial system; and if the revolutions are complete, and the fixed position is at a depth h(2(K - ) below the line of no velocity, the line always passes through the outer limiting point. The velocity of the centre of inertia varies as the tangent drawn from it to any fixed circle of the coaxial system, or in the case of complete revolutions as the distance from either limiting point. ~ 121. In the case when the pendulum oscillates, 1- K is positive, so that the modulus of the elliptic functions is greater than unity. The expressions may be transformed by the usual formulae; putting g = lmr2, we have B P= 2(1 + K)lh sn rt, AP = 2h dn nmt, the modulus being now 2-2(1 + )2. The velocity varies as en mt. The general theorems derived above from the addition-formula still hold, the system of coaxial circles having now real intersections, namely, the extreme points reached in the oscillation. The limiting points are however imaginary, and the line joining MOTION UNDER NO FORCES. 113 positions separated by an interval of half the period is always horizontal, as is also that which joins two that are separated by equal intervals from the lowest. The coaxial circle, which is the envelope in this case, consists of the radical axis and the line at infinity, and the tangents to it pass through their intersection. MOTION OF A RIGID BODY UNDER NO FORCES. ~ 122. Another interesting case is that of a rigid body in motion under the action of no forces. The centre of inertia will then move uniformly in a straight line or be at rest, and the motion of the body about its centre of inertia will be unaffected by the motion of the centre of inertia, which we will therefore suppose to be fixed. Let W, W2, W3 be the angular velocities of the body at any time t about its three principal axes of inertia, and let A, B, C be the three corresponding moments of inertia, and suppose that they are in descending order of magnitude. The equations of motion are then A l = (B - 0)Wo,03 Bo2( = ((7 - A))0l, OC(i3 = (A - B)OwIW. The form of these suggests a substitution = a n qt, )2 = - sn qt, 3=7 dn qt, since the sign of C-A is negative and opposite to those of B- C, A -B. Making the substitution we have Aqa =(B -C)/^y, - Bq3=(C - A)ya, Cqyk72 =(A - B)a3. D. E. F. H 114 ELLIPTIC FUNCTIONS. HenceA a2 B/32 Cy21 ay2 f Henlce B-L-A-= A-B = say. The equations are therefore satisfied if 7A-BC (73 q(t- t~ ) ABC where = -(-CABCA}A and the arbitrary constants of integration are j, the modulus k, and t0. The following two important equations are easily found either from the equations of motion or the integrals:AcW2 + B22 + C32 =j2(A _ ( '2B- k20)/k2 = T, say, A 2+2 + B2w22 + C232 =j2(k2AB + k'2A C-BC)/712=G2, say. ~ 123. Suppose now that (1, m, n) are the directioncosines of a straight line fixed in space. We then find I = lwl. - los, = lw - mw, and w), w2, 23, are now known functions of t. If these equations can be integrated the problem is completely solved. The equations give 11+m +n O= 0, and therefore 12 + ri2 + 2 = constant. The value of this constant is known to be 1. MOTION UNDER NO FORCES. 115 Also A w0l + B0o2A + Cw03, = (C - B)0203 + m(A - C)w30w + n(B -- A )w0e2 = -A l1 - B)n2 - (Cna3. Hence A lw, + Bmn2 + CCwO = K, a constant. This equation expresses that the line (1, n, 'n) makes a constant angle with that whose direction-cosines are (Aw1/G, Bo2/G, C03/G) and shows therefore that this latter is fixed in space. It is easily found that the equations are actually satisfied if l=AwI/G, m=Bo2/G, n = colG. ~ 124. We may now simplify the problem by supposing the line (1, m, n) to be perpendicular to this known fixed line, that is by putting K = 0. Let (X, ~, v) be the direction-cosines of another line perpendicular both to (1, m, n) and to (Aw1,/G, BO2/G, C(,3/G), so that GX = Cmw3 - Bnw2, etc. Then since (X, A, v) is also fixed in space we have X = JW0 3 - w0)2, and l - = (( - Xm)03 -(v - Xn)&w2 =- (Bo22 + C832)/G =-(T-Aw,2)/G. Also 12 + 2 + A2w12/G2= 1. dt Hence d' arctan l/\ = G(T- A }2j/G2-A2012). Thus I= X tan v, if G)dI Wl) t. if ~=SG1'-v = 2 A2-122 This integral can be expressed in terms of the function H, for the subject of integration is a known function of t. 116 ELLIPTIC FUNCTIONS. Then 1, Mr, n are given by the equations A low, + Bmw2 + Cnw3 -O GI cot v - Cmw + Bnwt2 = 0, 12 + m2 +,;2 = 1, _ _ m BoJ22 2 022 CGw3cot v - ABw1W2 - BGw2cot v- A Uwo13 1 G cosec v(B2022+ C2c 2)3' To find X,,/, v we need only change v into v + in these expressions. Referred to the three fixed axes, the directioncosines of the principal axis of greatest moment are (Ao1/G, I, X), those of the mean axis (Bw2/G, m, /u), and those of the third principal axis (Cw,/G, n, v). Hence the orientation of the body is completely determined at any time. The actual value of v is found to be o + G(t- to)/C+ HlH{q(t- t), a} if sn a= L{A(B-C)/C(A-B)},2 the values of en a, dn c being both positive, as well as that of - sna. vo is the value of v when t=t0, and it varies according as different straight lines in the " Invariable Plane" are considered. (t is a purely imaginary constant depending on the nature of the rigid body. k may be any real quantity. If it is numerically greater than unity the formulae may be reduced by the usual transformation to others in which the modulus is less than unity. The values of arctan n//u and arctan n/v might have been found in terms of II functions instead of that of ATTRACTION OF AN ELLIPSOID. 117 arctan 1/X; the formulae thus found must however reduce to those we have by means of the formula for addition of parameters in the function II. A further discussion of the motion, with references, may be found in Routh's Advancecl Rigid Dynamics (Chap. IV.). ATTRACTION OF AN ELLIPSOID. ~ 125. The potential of a solid homogeneous ellipsoid at any point may also be conveniently expressed in terms of elliptic functions. The expressions x2 = c2a12a2/(a2 - b2)(Ct2 - C2), y2 = b2 b2 b,2/(b2 _ c2)(b2 - 2), z = C2 C'2 C/(C2 - a2)(C - b), for the coordinates of any point in terms of the semiaxes of the three conicoids of a confocal system that pass through it, suggest that we make x, y, z constant multiples of S, C, D respectively where S = sn uq sn u 2 sn U3 = 8 8 s 3, say, C = cn u1 en u2c n {MS = c1c2 C3, D = dn u n dn z dn 3 = dld2d3. Since k2k'2s12s2 - k12c2c22 + c12d22 = k2, we have k2k'2S2/8s.2- c2C2/c,2 + D2/d,2 = k'2, where r = 1, 2 or 3. This equation is the relation that connects S, C, D when u,. is a constant. If then we put x=l. c272'S, y=l.72C, z==l.ID, 1 being any constant, the locus of (x, y, z) when rl, is a constant will be a conicoid whose semi-axes are the square roots of j272%C12852, -_ 2c21'12r2, - 12,'2Cr2. 118 ELLIPTIC FUNCTIONS. The differences of these quantities are constants, so that the different conicoids are all confocal. ~ 126. For an ellipsoid the imaginary part of U,. must be an odd multiple of tK'. It will be more convenient to have u,. real in this case; we therefore put ur+ tK' for u,. throughout, and we have x = lk'/kS, y =. ID/kS, z = -. C/S, the squares of the semi-axes being now 12%c'2/S.2, /21C'2d2/ST?2, 12i2Ck2/8r2. When u,. is constant and real, we now have an ellipsoid, when its real part is an odd multiple of K a hyperboloid of one sheet, and when its imaginary part is an odd multiple of tK' a hyperboloid of two sheets. In other cases the surface u = constant is imaginary. Since then one surface of each kind passes through any point, we may suppose Uh, t(U2-K), u,-tK' to be all real. The semi-axes of the focal ellipse are found, by putting u.= K, to be Vc' and lk'2; and, as I and V' are arbitrary, these may be made equal to any lengths whatever, so that any system of confocals whatever may be represented in this way. ~ 127. We must now transform the equation V2 V- 0, that is, a2-F 2V 12V aX2 3 D2 2 0. 3x- + ~y- + Z.- = o. Now, in the first place, if V is expressed in terms of S, C, D, V?V a V V 2 - c- = -S s j - CD ScdCc — ksld2C- SC 3S 1 3 2- a I 2 3) ATTRACTION OF AN ELLIPSOID. 119 a2V —+ 2 — 2Q 2 a1 2 2C- 2 22DV cS 4 2 2dl 2t 2 - z^'q 1 lCl82q 283s3-c8 3 - (12 + 7lc 2) _ _CS2C3(dl2 - Sdc282) _ D 2dd2d3(c12 _ s2) with symmetrical expressions for ~2V/~u22, ~2V/u^32. Thus (-S%)v)2 v/ + (S-S~)2 V/vt. + (S-s'2 )V/ - -S 2 2~3 2 )1 2 2 2)( 1- ~2 2 x [-32 /cS2 + c- V1s2S3(,T 2 _ 7, 2C4/2] all the other terms disappearing. If the symmetrical expressions for, z = / D, we have3 Thus (822 S2)(2 V/12 + (S2-2)2 /22 + (12-2)2 / 2 - (- (s )(83 - S2)(82 - ' W822) + [(2 -aS2 + V/-2 + l2 - k243 2 V/DD2], If then we put X = 172kZ'S, y = 172 z = ltD, we have (22 - S3)(S32- s12)(8-2 - s2)121V2V +2V = (s22 - s32)a2 V/iaU12 2+ ( 8312 - S) 2)D2 V/Dt3 2. If now we change nv. into u, + t K, this becomes 2(s2 832)(832 - 832)(S12 - 8 2)1c12IV2 -F 8 28282 = 812(822 - 832)a2 V/Din12 ~ 822(832 - s12)D2 V/Di22 2 + 832(812 - s22)D2 Vin t,32. The equation V2 V= 0 is therefore to be replaced by 812(s22 - S32)D2 VF/Di12 + 82(8 - s12)2 V/322 + s2(S1 - S22)a2 VT/a2 = 0. ~ 128. Now it is known that the equipotential surfaces of a thin homogeneous homoeoid (shell bounded by two similar, similarly situated and concentric ellipsoids) are the confocal ellipsoids that lie outside it, that is, the surfaces represented by l = constant 120 ELLIPTIC FUNCTIONS. if our confocal system is that to which the surface of the shell belongs. If V is the value of the potential it is a function of zb1 only, satisfying the equation just written, which now becomes 32 '7V/3t,2 = 0. Hence V= Qu1 z+R, Q and R being constants. Now V vanishes at infinity and at very distant points is in a ratio of equality to M/r where M is the mass of the shell and r the distance of the point from the centre. Also at infinity u =O0, and for small values of tz1 the surfaces may be regarded as spheres of radius 1k'/sr. Hence when ua1 is small we have Ms/llk'= Qz1 + R, that is, R =, Q=M/lk' The potential of the homoeoidal shell is therefore MUlk'. ~ 129. If now we have a homogeneous solid ellipsoid whose semi-axes in descending order of magnitude are a, b, c and whose density is p, it may be divided up into thin homoeoidal shells, to each of which the foregoing will apply. To get the different shells we need only suppose I to vary in the above expression from 0 to sn zv, its value for the outside surface, V1 being the constant value of vI for the outside surface referred to its own system of confocals. The sum of the volumes of all the shells up to any value of I is 4-Tr37'3cn vldn v1/sn3vl, so that we substitute for M the expression 47rpl2dl. /'3cn vldn vl/sn3v, ATTRACTION OF AN ELLIPSOID. 121 which is the differential of this with respect to I multiplied by p. The potential of the solid ellipsoid at an external point is therefore a Sll V1 4irp7c'2cn vldn vl [k' l7 dl, sn I 0 and 'u1 is given as a function of 1 by the equation x2sn21u, + y2sd2tq + z2Sc2ub = 12k'2, (x, y, z) being the coordinates of the external point. We find at once k'21 dl = (x2sccld + y2sIc1/d13 + zy2Sdl /c13)d ' 1. Thus if now we write u1 for the value of qt1 at (x, y, z) in the system of confocals to which the outside surface belongs we have for the potential 4rpcn vldn vl gu{2 + yS2nd4^t + z2nc4snucnadn iudu sn3v J o 27rp cn vldn vlr ( y 2 sd 21 ~ z2sc2U9) - (sn2Sl - +- y2sd2m + z2sc2)du. 0 Also by definition of ul, x sn2u, + y2sd2ul + z2sc2a1 = ct2sn2v1, and |sn2 du = 2 — 2Eu, o X,1 snucnu 1 dU'a' =- 2 dn ul -/2 +c2c'2 1' o U1 scd 1 snudnzu, 1 SC2^ du- - 1 1cnv -721. j ioc2 cn, //2 1 0 122 ELLIPTIC FUNCTIONS. Hence the potential 27rp en vldn v ln a r 2sn2V_ (X22 - y2)/C2 sn 1 a(y2sn2 - 2 12)/2} + '2cn uldn u y2cn2 - z2dn2m) + E (22 - y2 + c2z2)] Here C2 (a2 - b2)/(a2 _ C2), c'2 = (b2 - C2)/(a2 - C2), cn v1 = b/a, en v, = c/a, sn 1 - (2 - C2)~/CV, and u1 is the least real positive argument that satisfies the equation x2sn2u + y2sc2 + z2sc2 = L62 _ c2. If the point (x, y, z) lies on the outer surface, we have -, =vl ~ 130. If the point (x, y, z) lies inside the ellipsoid, the above formula ceases to hold. We may however describe through (x, y, z) a similar, similarly situated and concentric surface, and use the above expression for the volume contained. If Xa, Xb, Xc are the semi-axes of this one, its potential is 2'rp en vdn v. [- {Xa2sn 2V - (x2 - y2)/k2} sn v1 + s n Vv1 (Yn2v, - zl2dn 2V,) + '2cn vldn v 1 + ( ~2_ 1-X2 2)]Z -0/e2 y 2 EXAMPLES X. 123 We have then to deal with the outer shell. This may be divided into thin homoeoids as before. The potential of each is the same at all points inside it, and equal to 47rpl dl. k'2v1cn vldn vl/sn3v. This is to be integrated with respect to 1 between the limits Xa sn v/k' and a sn vlk', and added to the potential of the inner part. The integral is 27rpa2(1 - X2). vlcn vldn vl/sn v, and the potential of the whole ellipsoid at an internal point (x, y, z) is found to be 27rp cnvdn 1i-, vIasn (2 - y2)/ 2} +... sv{ sn%-(z- y)/l }n sn3~;1 + v1c2 (y2cn21 - z2dn2v) +Ev2/ 2 y2 ( + 2Z2)] The expression is the same as for an external point, but that the constant v, takes the place of the variable u1. EXAMPLES ON CHAPTER X. 1. Prove that (1 - 2x2cos 2a + x4)2 can be rationalized by putting x +-= 2 ns(2n, cos a), and that then x — = -2 cs(2zt, cos a), X 124 ELLIPTIC FUNCTIONS. Qx2- 2 cos 2a +-) = 2 ds(2v,, cos a), u = (1 - 2X2cos 2a + xy)-cdx. 0 2. Discuss the spherical figure of ~ 116 in the case when sin A > sin a and show that in that case we may put sin OPQ = sn(c, sin a cosec A), sin OQP = sn(w- u, sin a cosec A), where r - A = am w. sin a 3. If cos 0 = cos dn u, tan= si- sc U, where cos a = l' cos 3, prove that the point whose polar coordinates are (II, 0, 0), R being a constant, traces a sphero-conic whose semi-axes are a, /3 and that the area of a central sector of this sphero-conic is dn u clm R2sin a sin~3 I + cos t dn u 4. Prove that the chord joining the points u ~ a on this sphero-conic touches the sphero-conic whose equation is cot2 cn2a = cot2/3 dn2a cos2p + cot2a sin29, and that this has the same cyclic arcs as the former one. 5. Show that the sector bounded by the semidiameters to the points t +~ a differs from 2R2arctan dn u(sn2a + cn2a cos2/0) + dn a cos / sn a en a sin a sin /3 by a quantity independent of u. Prove also that the area of the spherical triangle EXAMPLES X. 125 formed by these two semi-diameters and the chord joining the points u ~ a is 2R2arcta^ n sin a sin 3 sn a cn a dn u 2Rarctan 1 -sna sin2a + dn u dn c cos 3' and that the area of the segment cut off by this chord is independent of v. 6. In the same sphero-conic (cot2O = cotsa sin2~ + cot2o/ cos2p) prove that by the substitution tan tan an a cot /3 sin a cosec f3 cs(zt, ic), where c' = sin 3 cosec a, the expression for the arc is reduced to R tan a tan / sin /3Jtan2/3 sn, + tan2a cnd ' tanl1 S,12,U + tan'a cn"n' 7. Prove that at the intersection of tangents to this sphero-conic at the points u ~ ac (as in Ex. 6, not Ex. 3) cot _ cot a sin; _ cot 3 cos cna dn u dna cn u c'sn u and that as u varies this point traces the confocal sphero-conic cot2O nc2a = cot2a sin2p nd2a + cot2/ cos2q5. 8. The length of the tangent at u+ac in the last example is aretan 2~ tan a tan / sn a sin /3 R arctan tan a cn a cn(u + a)dn a + tan2/ sn U sn(u + c) Find the differential coefficient of this expression with respect to u in the form R tan a tan /t sin 3F-( —+-n +1a R tan tand sin tan2a cn2(m + a,) + tan2/ sn2(u + a) tan2acn2 tan2~3 tan2a cn2m + tan2/3 sn2j' 126 ELLIPTIC FUNCTIONS. and prove that the sum of the two tangents exceeds the intercepted arc by a quantity independent of u. (Compare Salmon, Geometry of Three Dimensions, ~252.) 9. Verify that when sn2ou tan2/ + cn2u tan2a = 0, then sn u = / cos, cn u = + sin 3 cot a, dn u = + sin3, and the above expression for the length of the tangent becomes R arctan + t. 10. Prove that the following equations give the motion of a heavy particle constrained to move on a fixed smooth spherical surface:cos 0 = cos a sn2wt + cos 3 cn2wt, rt tdu j4 sin^2 a sn2U + sin2L/3 cn2 I0 o du h cos a snn+cosf te pacnrte o where 6 is the angular distance of the particle from the lowest point, a, / are the greatest and least values taken by 0 during the motion, p is the angle made by the vertical plane through the centre and the particle at time t with its initial position, t being measured from a time when 0 =, 1 is the radius of the sphere, and 1/2 = (cos2 o - cos2a)/(l + oo2f +2 cos a cos /3), 1 g(cos 3- cos ) n2 4 sin2a sin2//( 1 + cos22/ + 2 cos a cos /3). EXAMPLES X. 127 11. Reduce the above value of p to the form n{wt coseC2h3+ -Isc a cosec la cosec I3lHl(wt, a) +jsc b see la see I3II(wt, b)}, where dn a = sin I a/sin,3, dn b = cos 1 a/cos -S3. 2 2 What is the general character of the motion? 12. On a curve of deficiency 1 and degree n, the sum of the arguments of its intersections with a curve of degree in is a-. Show that if n > 3 the fact of the sum of the arguments of mn points on the curve being - does not ensure that the points lie on an qzio, but that if = 3 this condition is enough. 13. If the curve of intersection of two conicoids is projected from any point of itself on any plane, the projections will all be projections of the same plane cubic. [The anharmonic ratio of the four tangents drawn to any of the cubics from a point on itself is the same for all. It may be expressed as a function of the elliptic modulus.] 14. Verify that the expressions found (~~129, 130) for the potential of an ellipsoid satisfy Laplace's and Poisson's equations, and find the components of the attraction at any point. 15. In Jacobi's coaxial circle figure (Fig. 2, ~105), prove that when a= tK', 0 is at B, and when c=K+tK', at A. In general when 0 lies between L and L', so that the variable circle is imaginary, the real part of a is an odd multiple of K. 16. The arguments of the circular points at infinity are + K', and of the other common points of the coaxial system K ~+ tK. 17. If 1, qn, n are in descending order of magnitude show that the two ends of a chord of the circle x2 + y2 - which touches the ellipse x2/1,2 + y2/n2 = 1 128 ELLIPTIC FUNCTIONS. have for their coordinates 7c sn( +~ a), m dn(u + a), where 12 (m2- n2) i n k =n2(2 n2)) cn a =, dn a= —, M -n) In IM and m is a variable parameter. 18. If x+ty=sn(u+tv), the points on the curves t = const., v = const. at which the tangents are parallel to the axes of coordinates, lie either on one of those axes or on a rectangular hyperbola whose axes they are. (See Appendix A.) 19. If x+- iy = sn2(u + v) or cn2(u+ftv) or dn2(u+tv) or p(u++v), the curves t = const., v = const. are confocal Cartesian ovals, and for one value of each the oval becomes a circle. Distinguish between the outer and inner ovals. (Greenhill.) 20. Examine the curves qL = const., v = const. when x + y = sn(u + tv) dc( + iv). [The distances of the point (x, y) from the points (+~, ~ ') are found to satisfy two linear relations. Hence the curves are bicircular quartics having these points for foci. In the particular cases when u= + -K, or v = + 1K' they become arcs of the circle X2 + y2.] APPENDIX A. THE GRAPHICAL REPRESENTATION OF ELLIPTIC FUNCTIONS. ~ 131. The nature of the elliptic functions unfits them for representation by a linear graph as in the case of functions of a real variable. We may however get some idea of their variations by means of Argand's Diagram. Let x + y = sn(a + tv), x, y, u, v being real, and let us examine the curves u = constant, v = constant; we need not consider values of u outside the limits ~ 2K or of v outside ~ K'. Call the point (x, y)P and the points (1, 0), (-1, 0), (l/k, 0), (- 1/k, 0), A, B, C, D respectively. Then AP2 = {1 - sn(u + v)} {1 - sn(ut- v)} = (en tv - dn v sn u)2/(l - k2sn2, sn2tv), BP2 = (en tv + dn tv sn bb)2/(1 - k2nsn2t sn2tv), kc2CP2= {1 - k sn(u + v)} { - ksn (- v)} = (dn cv - k en Lv sn u)2/(1 - ksn2v sn2uv), k2DP2 = (cln cv + k, en tv sn u)2/(1 - k2sn2u sn2v). BP-AP BP+AP DP-CP k(DP+C,P) Hence -- - = dn tv snu cen cv en cv sn % dn tv D. E.F. I 130 ELLIPTIC FUNCTIONS. Thus the locus when v is a constant is given by BP - AP = (DP-CP)dc v, or the equivalent BP + AP = c(D)P + CP)cd cv. The locus when u is a constant is given by BP- AP = c(DP + C(P)sn u, or BP +AP= (DP- CP)nsu. The curves in each case are bicircular quartics having A, B, C, D for foci. They are symmetrical about both axes. The curves v= const. are found to be a series of ovals enclosing the points (+ 1, 0) but not the points (+, 0). The ends of the axes of these ovals are the points (~cdtv,0) and (0, tsniv). When v is indefinitely diminished the oval shrinks up into the straight line between A and B. As v increases in magnitude irrespective of sign the oval swells out. The points on the axis of x are points of undulation when 2 cd2V = 1 + 1/k2, and for greater values the oval swells out above and below the axis of x, and is narrowest at the axis. In the limit when v= + K' it becomes the part of the axis of x beyond (+ 1/k, 0), together with the line at infinity. The curves u = const. consist each of a pair of ovals, one enclosing the points (1, 0)(1/c, 0) the other the points (-1, 0)(- 1/k, 0). Each of these cuts each of the curves v = const. orthogonally. Of the two ovals, the one on the positive side of the axis of y belongs to the values u and 2K - (u being positive) and the other to the values - t and - 2K+,. GRAPHICAL REPRESENTATION. 131 When u= ~ K the corresponding oval shrinks into the straight line between (~1, 0) and (~ l/k, 0), the upper or lower sign being taken throughout. When u=0 the oval swells out until it becomes the axis of y with the line at infinity. The curve v= ~K' is the circle whose centre is the origin and radius c-. ~ 132. Since dn(, + Iv, k) = k7'sn(v - t u+ K'- tK, k'), the figures for the function dn will be of the same general nature as those for sn. The foci (~1, 0)(~ 1/l, 0) are replaced by ( ~ ', 0)( ~ 1, 0) respectively, and the single central ovals are now the curves L= const., the pairs of ovals belonging to the system v=const. The curve au= K is a circle of radius c'2. In the case of the function en the figures are different. Putting x + ty = cn(u + tv), we have x = en u en (v/(l - k72 sn2o sn2tv), y = sn u sn t dn dn cn /(1 - 12 sn2o sn2tv). The curves t= const., v= const. are still bicircular quartics but the four real foci are not collinear. They are the points (+1, 0)(0, ~+k'/k), each of these pairs being collinear with the antipoints of the other.* Each of the curves consists of a single oval. The curves u=const. enclose the foci (0, ~k'/l) and not (~ 1, 0). The curve u =0 consists of the parts of the axis of x beyond the points ( + 1, 0), the curve u= + K * This may be compared with ~ 131 by means of the formula cn(t, k) = sn(k'K- k'u, tk/k'), which follows from equations (20) of ~ 26. 132 ELLIPTIC FUNCTIONS. of the line between the points (0, ~ +7c'). As u decreases numerically from ~ K to 0, or increases from +K to ~2K, the oval swells out. It has points of undulation on the axis of x when 2 cn2 = 1 _- '2//c2 if k2 > 1'2. When cn2t is greater than the value thus given the oval is shaped rather like a dumb-bell, and the two ends of it expand to infinity as u diminishes to 0 or increases numerically to ~ 2K. Since k cn(u + t, k)= - tl'cn(v - u + K' - i K, c'), the general form of the curves t= const., v = const. is the same if one set is turned through a right angle. There will be points of undulation on one of the curves = const. if c'2> kc2, that is if there are not on any of the curves = const. ~ 133. These bicircular quartics are shown in figures 4ct, 5a, 6a, for sn, cn, dn respectively. They have been drawn to scale with some care for the value ^/2 - 1 of kc, and for values of u and v which are successive multiples of IK and K' respectively. In each case the curves it = const. are drawn thick, and the curves v =const. thin. The figures 4b, 5b, 6b show on the same scale the corresponding variations in the argument, corresponding lines in the two figures being numbered alike. Only one period-parallelogram has been drawn for each function. In each case the centre is at the origin. The figures 4b, 5b, 6b are reproduced on a smaller scale as 4c, 5c, 6c the parallelograms being divided into the regions that correspond respectively to the four quadrants in 4a, 5a, 6a. In figure 6a the curves v=0, t = + IK', v = +~K', v = + 2K' are too small to be shown. GRAPHICAL REPRESENTATION. 133 (a) 1 6 3 -5 (c) -4 5 - 3214 -I. -2 2 3 4 1 3 S 1 5 1 9 (b) Fig. 4. 134 ELLIPTIC FUNCTIONS. (a) (c) 1 Fig. 5. GRAPHICAL REPRESENTATION. (a) 135 (b) (c) 2 3 5 1 4 41 S 3 2 5 1 59 Fig. 6. M~ APPENDIX B. HISTORY OF THE NOTATION OF THE SUBJECT. ~ 134. The notation used by Legendre was as follows:* E(k, 0) = (1 - k2sin20)2d0, 0 F(ic, ') F,(k), E(lq -) =E2 ) I(kn,,0)= - s(l-l 2sin dj/( +~ si2), 0 AO = (1 - 12sin20). Jacobi and Abel proposed to take F(kc, 0) as the independent variable. Putting u for this, Jacobi called 0 the amplitude of u, or shortly am au. Then sin 0, cos 0, A0 were the sine, cosine, and A of the amplitude of m, or as he wrote them, sin am u, cos am u, A am rn. * The expressions F(k), (, ), E(k,),, 0, were called the First, Second, and Third Elliptic Integrals respectively. HISTORY. 137 He used the symbol coam u for am(K- u), and also tan am nb, sin coam u, etc. He changed the meaning of the symbols E, II to those we have given (Chap. V.), and also brought in the function Z. It was proposed by Gudermann to write sn, en, dn for sin am, cos am, A am, and the notation sc, cd, etc., was introduced by Dr. Glaisher. Sometimes tn is written for sc, ctn for cs. The function gd (see ~ 75, note) is the amplitude, the modulus being unity. For the notation of Weierstrass see Chap. VII. In the further development of the subject other symbols are wanted. Jacobi used the Greek capitals 0 and H; the functions u, Hu may be defined as follows:t = exp( Zv dv), Hit =,/k. Oe. sn u. The arbitrary constant in the value of 0 is not determined until a later stage. Some of the properties of the function Oue have been suggested in the examples to Chapter VI. MISCELLANEOUS EXAMPLES (FROM EXAMINATION PAPERS). 1. Prove that scl2x - scdly sd(x + y)sd(x - y) = 2 ksd sd2 y 1 + /k"2'sd2X sd&y sd x cn x - sd y en y cn(x +) = sd x cn y - sd y cn 2. Show that snasn cn a cn /3 - cn(a + 3) dn a dn - dn(a + ) dn(a + /3) k2cn(a + / 3. Two sets of orthogonal curves (Cartesian ovals) being defined by the equation x + ty = sn2 {-(u + tv), k}, show that the polar coordinates of any point (t, v) are given by cos 0= - cn(u, ) + dn(n, k)dn(v, k') dn(u, k) - cn(u, k)dn(v, k') sin 0= s'sn(u, k)sn(v, k') dn(I, k)- cn(m, k)dn(v, k')' r = I - cn(u, k)cn(v, k') dn(v, 7') + dn(u, k)cn(v, k)' MISCELLANEOUS EXAMPLES. 1.39 4. Prove that the functions (cs u cd u cn u - k'2sc Z sd u sn u)2 and (ds udc u dn u + k2k'sc sd i sn u)2 have periods K and iK'. 5. If U d ---—,\,l 5. If | (J + X)(b + X)(c + X) where a, b, c are positive quantities in descending order of magnitude, then e S112u = a cn2u - c, the modulus being {(a -- b)/(a - c)}. 6. Show that k72sn I (u1 + Uta + uT3 + "t4)s11 ~(u + ' - 's - t4) x sn -( 1 ' - it + It3 - *?4)sn 2 - 13 - 2 - 3 + 't4) dld2r,'d4 - k2 ccc2c3c4 + kk'12sls2s3s4 - k'2 cdld2d3cd4 - - cClC2C34 k k' s1s28s4 + k'2 7. Show that the form assumed by a uniform chain of given length whose ends are at two fixed points is represented by the equation k'2y 2kb sn - ' b when its moment of inertia about the axis of x has a stationary value. 8. Prove that cn(B - C)sn(C - A) + dn(B - C)sn(A - B) + sn(B - C)cn(C- A)dn(A - B) = 0. 9. Verify that {1 - k2sn2(c + d)sn2(a - b)}{ - k1n2s2(a + b)sn2(c - d)} { 1 - 12sn2(ca + b)sn2(C - b)} { 1- k/2n2(c + d)sn2(c - d)} is a symmetric function of a, b, c, d. 140 ELLIPTIC FUNCTIONS. 10. Prove that - / 1 j e\ cd-t(l(+~k)s 1+ks2 - &k-sn(u + ~ ') =cc - L(1 kS = 1 + k-s2 2t / l+/1 ks2 cd + (1 + k)s d - ksc c - isd c + sd d + tksc' where s, c, d denote sn u, cn uz, dn 't respectively. 11. Prove that - k sn2(I + tK) = - tkS C - t C + kS D + &S _C- kD - tk'2S D - kC D - C - kD + kD'2S' where S=sn 2u, C=cn 2u, D =d cn 2i(. 12. If xx,, denote sc(ux - uQ)cs(zx + 1,) then %41X42X43`12X23X31 ~+ 41X23 + X42X31 + X43X12 = 0. 13. If k2 = - (where )2 + o+ 1 = 0) then 1 - sn(o - o2)l 1 - sn ( 1 - sn \2 1 + sn(W - o02)t 1 + snzll, + tosnu ' 14. If Qu = pt + p(u + W) - Po, then (Q'u)2 = 4Q3u + 4(g9 - 15.p2o)Qt - 14g92p - 22g3. 15. Evaluate (pu - pv)2du, and express (pu, - pv)-2du in terms of J(gpt - )-'dit. 16. Find Jnd t d. Prove that 3idn4a dcu = 2(1 + c'2)E+ ' 2Esu + 2s n u dn i - k2, k,2 sn u du = E(n + K+ K') + dc i6, Jl1 + sn u 2k sn t, Ct = lo -- +k 0 MISCELLANEOUS EXAMPLES. 141 17. Show that Kl flog sn u du = - -rK' - ~Klog k, o Jlog dn u d = - iKlog k'. (In the first put am u = 0 and expand in powers of k.) 18. Prove the formulae log(1 - k2n)d= - 4rK' + Klog, jKlog(l + dn u)du- = rK' + 1Klog k. o 19. Prove that hI(u, a) + I(v, a) - I(it + v, a) = log { 1 - kc2sn2(u + a)sn2(v + a) } { 1 - k2sn2a sn2( + v - a) 4 {1 - ksn2(u - a) sn(v - ) - sn sn2(t + v + a)} 20. Expand sn" udu in ascending powers of k2; and o thence, or otherwise, prove that r1 fK, f'Ksnn"+2 C clkJ snze6a=A'| uduk d -du. dkA dlUd 0 o (Compare Ex. 12, Chap. IX.) 21. In Weierstrass' notation, if Jis the absolute invariant as given by the equations,T- I J 1 27g2 g23 a 142 ELLIPTIC FUNCTIONS. then the periods satisfy the differential equation fJ( 1 - J) (yA12) + i(4 - 7J) dJ ) 1 2 dJ2 W DJ 4 4 22. Verify that the expression of Ex. 19 agrees wit' that of Ex. 15, Chap. VI., and with that of ~ 67. 23. Find expressions for the arcs of the curves, k'2y = 2kb sn - + c y=bk'nc 2b b GLASGOW: PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSE.