NOTES ON RAMANUJAN'S SINGULAR MODULI. Bruce C. Berndt and Heng Huat Chan. 1. Introduction

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1 NOTES ON RAMANUJAN'S SINGULAR MODULI Bruce C. Berndt Heng Huat Chan 1. Introduction Singular moduli arise in the calculation of class invariants, so we rst dene the class invariants of Ramanujan Weber. Set 1Y (a q) 1 = (1? aq n ) jqj < 1 n=0 If (q) = (?q q ) 1 : (1.1) q = ex(? n) (1.) where n is a ositive integer, the class invariants G n g n are dened by G n :=?1=4 q?1=4 (q) g n :=?1=4 q?1=4 (?q): (1.3) In the notation of Weber [8], G n =:?1=4 f(?n) g n =:?1=4 f1(?n): As usual, in the theory of ellitic functions, let k := k(q) 0 < k < 1 denote n the modulus. The singular modulus k n is dened by k n := k(e? ) where n is a natural number. Following Ramanujan, set n = k : n The algebraic natures of G n g n n are described in the following theorem of Chan S.{S. Huang [4]. Theorem 1. (a) If n 1 (mod 4) then G n n are units. (b) If n 3 (mod 8) then?1=1 G n n are units. (c) If n 7 (mod 8) then?1=4 G n 4 n are units. (d) If n (mod 4) then g n n are units. Lastly, we dene a modular equation in the sense of Ramanujan. Let F1( x) be the ordinary or Gaussian hyergeometric function, which, for brevity, we denote by F1(x): Suose that, for some ositive integer n for = k = ` where k ` are moduli, n F1(1? ) F1() = F1(1? ) : (1.4) F1() Tyeset by AMS-TEX 1

2 BRUCE C. BERNDT AND HENG HUAT CHAN Then a modular equation of degree n is a relation between which is induced by (1.4). We often say that has degree n over : If, as customary, in the theory of ellitic functions, q := ex(? F1(1? )=F1()) =: F () (1.5) then [1,. 14] (q) = 1=6 f(1?)=qg?1=4 (?q) = 1=6 (1?) 1=1 (=q)?1=4 : It follows from (1.1), (1.3), (1.5) that G n = f4 n (1? n )g?1=4 g n = 4 n (1? n )??1=4 : (1.6) Thus, if G n or g n is known, then n can be easily determined from (1.6) by solving a quadratic equation. However, usually, the exression for n that one obtains in this manner is not very attractive does not reect the algebraic nature of n described in Theorem 1. Thus, further algorithms which more readily yield elegant reresentations of n are highly desirable. In his rst notebook [7], Ramanujan oers without roofs over 30 values for n briey indicates some formulas which are useful in calculating n when n is even. The authors L.{C. Zhang [3] roved these formulas of Ramanujan, develoed some new formulas for calculating n when n is odd, consequently established all of Ramanujan's values for n : Ramanujan obscurely describes two further methods for calculating n in his rst notebook [7]. In the rst, Ramanujan indicates that n may be calculated by solving a certain tye of modular equation of degree n: For several rime values of n the desired tye of modular equation exists many of these modular equations can be found in Ramanujan's notebooks are roved in [1]. This very novel method is described in Section, it is the only known method that does not require a riori the value of gn: Thus, the method is a new, valuable tool in the comutation of n: In the second, Ramanujan discloses a method for determining 3n arising from the denition of a modular equation of degree n: In Section 3 we give a rigorous formulation of this formula rove it by using a device the authors Zhang introduced in [] to calculate certain class invariants. Lastly, in Section 4, we follow our methods in Section 3 derive a similar formula for 5n: Although we utilize results of Ramanujan from his notebooks in our roof, this formula was not given by Ramanujan.. The determination of n from modular equations In the middle of age 9 in his rst notebook, Ramanujan claims that, \Changing to 4B=(1 + B) to 1? B we get an equation in 4B(1? B)=(1 + B) the value of B is for e? n :" We now state rove a rigorous formulation of this assertion. Theorem. Let have degree n over suose are related by a modular equation of the form F (() r f(1? )(1? )g r ) = 0 (.1)

3 NOTES ON RAMANUJAN'S SINGULAR MODULI 3 for some olynomial F some ositive rational number r: If we relace by 1? x by 4x=(1 + x) then (.1) becomes an equation of the form G(x) := g (f4x(1? x)=(1 + x)g r ) = 0 for some olynomial g: Furthermore, x = n Proof. Under the designated substitutions, F (() r f(1? )(1? )g r ) =F =F =0: Hence, the rst art of Theorem follows. Next, setting = 4x=(1 + x) we nd that F1(1? ) F1() F1 = F1 is a root of this equation. (1? x r 4x ) (1 + x) x 1? r 4x(1? x) 1 4x(1? x) 1 + x 4 r 1 + x 1? 4x (1 + x) = 4x (1 + x) 1 r 4x (1 + x) r! F1(1? x ) F1(x (.) ) by a fundamental transformation for F (x) [1,. 93], which actually arises from a secial case of Pfa's transformation. With relaced by 1? x we nd from (1.4) (.) that Therefore, n F1(x ) F1(1? x ) = 1 F1(1? x ) F1(x ) F1(1? x ) F1(x : ) = n: Now recalling the denition of a singular modulus (1.5), we deduce that x = n the roof is comlete. In each examle below we use the following denesting theorem in our calculations. If a? db = c then q r r a + c a? c a b d = : (.3) Examle.1. Let n = 3: Then () 1=4 + f(1? )(1? )g 1=4 = 1 (.4) which is originally due to Legendre was rediscovered by Ramanujan [1,. 30, 3]. With Ramanujan's substitutions, (.4) takes the form u + 1 u = 1 (.5)

4 4 BRUCE C. BERNDT AND HENG HUAT CHAN where u = 1=4 4x 1? x : (.6) 1 + x Solving (.5), we nd that u = 3? 1: Then solving (.6), we nd that x = 3? 3? + 6: Using two dierent modes of calculation, we nd that 6 = x = (? 3) ( 3? ) = 3 + 6? 3? : It should be remarked that there are two roots of (.6), namely, 6 =3: For if we set = 4x=(1 + x) = 1? x at the beginning of our roof of Theorem, we nd that x still satises (.1). However, in this case, F1(1? x ) F1(x ) = r n : That we have selected the correct root can be veried by a numerical check. The former reresentation for 6 two other reresentations as well, can be found in Ramanujan's rst notebook, they were roved by the authors Zhang in [3, Theorem.1]. Examle.. Let n = 5: Then we have Ramanujan's modular equation of degree 5 [1,. 80], () 1= + f(1? )(1? )g 1= + f16(1? )(1? )g 1=6 = 1: (.7) Using Ramanujan's substituions, we nd that (.7) can be ut in the shae where u u + u = 1 4 u + 3u = 1 (.8) u = 4x 1 1=? x : (.9) 1 + x Solving (.8), we deduce that u =?6 + 10: Next, solving (.9), we nd that x = 3 10? 9? : Lastly, by two distinct routes for calculation, 10 = x = ( 10? 3) (3? ) = ? 9? : The former reresentation was given by Ramanujan in his rst notebook roved by the authors Zhang in [3, Theorem.1]. Examle.3. Let n = 11: The modular equation () 1=4 + f(1? )(1? )g 1=4 + f16(1? )(1? )g 1=1 = 1 (.10)

5 NOTES ON RAMANUJAN'S SINGULAR MODULI 5 is indeendently due to Schroter Ramanujan [1,. 363{364]. With Ramanujan's substitutions, (.10) can be ut in the form where u + 6u? = 0 (.11) u = 1=4 4x 1? x : (.1) 1 + x Solving (.11), we nd that u =?3 + 11: Then solving (.1), we nd that x = 30 11? 99? : Hence, = x = (10? 3 11) (3 11? 7 ) = ? 99? : In his rst notebook Ramanujan oered the former value, which was roved in [3, Theorem.1]. 3. A formula for 3n Let q be given by (1.), suose that has degree n over : Thus, (1.4) holds. Now suose also that has degree 3 over 1? =: 0 : Then, by (1.4), Hence,, from (3.1) (3.), 3 F1() F1(1? ) = F1(1? ) F1() = n F1(1? ) : (3.1) F1() F1() F1(1? ) = r n 3 (3.) F1(1? ) F1() r n = 3 3 = 3n: (3.3) Next, from [1,. 37], since has degree 3 over 0 we have the arametrizations = = 3 + (3.4) where 0 < < 1: It follows from (3.4) that (1? 0 ) = (1? )3 (1 + ) (1 + ) = 3 1? (1 + )( + ) (3.5) (1? ) 0 = 3 (1? )(1 + )3 + = 1? (1 + ) (1 + )( + ) : (3.6) 1 +

6 6 BRUCE C. BERNDT AND HENG HUAT CHAN Next, set observe that It follows from (3.5){(3.8) that Now, set (1? t) = t = 1? 1 + (3.7) (1 + )( + ) : (3.8) 1 + = (1? 0 ) = 16t 3 (1? t) (3.9) (1? )(1? ) = (1? ) 0 = 16t(1? t) 3 : (3.10) Observe from (3.9), (3.10), (1.6), (3.), (3.3) that k = k = 4t(1? t): (3.11) G 6 n=3 G6 3n?1 : (3.1) We determine (3n) as a function of k: From (3.11), we nd that, with no loss of generality in choosing the minus sign in the rst equality below, t = 1? 1? k Thus, by (3.9) (3.10), we nd that 1? t = 1 + 1? k : (1? 0 ) = (1? 1? k) 3 (1 + 1? k) = k(1? 1? k) (3.13) (1? ) 0 = (1? 1? k)(1 + 1? k) 3 = k(1 + 1? k) : (3.14) Subtracting (3.13) from (3.14), we nd that 0 = 4k 1? k + : (3.15) Substituting (3.15) into (3.14), we deduce that + (4k 1? k? 1) + k(1? 1? k) = 0: Thus, q 3n = = 1? 4k 1? k (4k 1? k? 1)? 4k(1? 1? k) = 1? 4k 1? k (1? k) 1? 4k r 1? 4k 1? = 1? k (1? k)(1? 4k) : (3.16)

7 NOTES ON RAMANUJAN'S SINGULAR MODULI 7 This last formulation was that given by Ramanujan. We now resolve the sign ambiguity in (3.16). From (3.13) (3.14), it is clear that both 0 satisfy the equation y + y(4k 1? k? 1) + k(1? 1? k) = 0: (3.17) Since 0 = 1? it follows from (3.) (3.3) that the solutions of (3.17) are 3n 3=n: Thus, it suces to show that for n > 1: If then clearly 3=n > 3n (3.18) '(q) = 1X k=?1 q k ' (e? 3=n ) > ' (e? 3n ): From Entry 6 in Chater 17 of Ramanujan's second notebook [1,. 101], it follows that F1(3=n) > F1(3n): Since F1(x) is increasing on (0 1) (3.18) follows. Thus, we have roved the following theorem. Theorem 3. Let q be given by (1.), suose that has degree n over let have the arametrization (3.4), dene t by (3.7). Then, if k is dened by (3.11), 3n has the reresentations given in (3.16), where the minus sign must be chosen. Ramanujan's formulation of Theorem 3 at the bottom of age 310 in his rst notebook is a bit dierent. He rst gives (3.9) (3.10), but with the left sides switched. He then states (3.7), followed by the equality F 3 + = e? 3n 1 + which is a consequence of (3.3) (3.4), where F is dened in (1.5). He concludes by dening k in (3.11) by claiming that a more comlicated, somewhat ambiguous version of the right side of (3.16) equals e? 3n i.e., he forgot to write \F " in front of the right side of (3.16). (In recording secic values of F ( n ) Ramanujan frequently omitted arentheses about the arguments.) Examle 3.1. Let n = 1: Then from Weber's tables [8,. 71], G3 = 1=1 = G1=3 since G n = G1=n [5], [6,. 3]. Hence, by (3.1), k = 1=4 by (3.16), 3 =? 3 : 4 See our aer [3, Theorem 3.] for another roof further references.

8 8 BRUCE C. BERNDT AND HENG HUAT CHAN Examle 3.. Let n = 5: Then, from Weber's tables [8,. 71], G15 =?1=1 ( 5+ 1) 1=3 : It can also be veried that G5=3 =?1=1 ( 5? 1) 1=3 : Hence, it is easily seen from (3.1) that k = 1=16: Therefore, from (3.16), which is simler than the formula 15 = 16? 7 3? = ? 1! 4 (? 3) (4? 15) which was given by Ramanujan in his second notebook. For its roof further references, see [3, Theorem 3.]. 4. A formula for 5n We roceed as in Section 3 the calculations are easier, however. Let q be given by (1.), suose that has degree n over degree 5 over 1? =: 0 : Then, by the same argument as in Section 3, but with 3 relaced by 5, we nd that F1() F1(1? ) = r n 5 F1(1? ) F1() = 5n: (4.1) Now, from (4.1), (1.6), Entry 14(ii) in Chater 19 of Ramanujan's second notebook [1,. 88], if 0 < < (5 5? 11)= is dened by then Therefore, 1 + = m := F1() F1() (4.) k := (G n=5g5n)?4 = f4(1? )4(1? )g 1=6 =? 1 + : (4.3) Solving (4.3) for we nd that 1? k = : (4.4) = 1? k? k? 3k + 1: (4.5) Also, from [1,. 89, Entry 14(iii)], 1? 5n = (1 +? 1 + 1= ) : (4.6) 1 + Hence, using (4.4) (4.5) in (4.6), we nd that, after a modicum of calculation, 5n = 1? 1? k k? k + Thus, we have established the following theorem. 1? k k? 3k + 1 : (4.7)

9 NOTES ON RAMANUJAN'S SINGULAR MODULI 9 Theorem 4. Let q be given by (1.), let have degree n over dene by (4.), dene k by (4.3). Then 5n has the reresentation (4.7). Examle 4.1. Let n = 1: From Weber's tables [8,. 71], G5 =! 1= = G1=5 since G n = G1=n: Thus, from (4.3), k = (3? 5)=: Noting that k? 3k + 1 = 0 we nd from (4.7) that 5 = 1! 3= 5? 1? which can be comared with the value 5 = 1 5? 1! 3 s ? s 5? 1 given by Ramanujan in his rst notebook roved by the authors Zhang in [3, Theorem 3.]. Examle 4.. Let n = 5: From Weber's tables [8,. 7], 1 A 4 G5 = so, since G1 = 1 from (4.3), k = (7? 3 5)=: Thus, q k? 3k + 1 = 14? 6 5 = 3? 5 by an alication of (.3). After a moderate amount of calculation the use of (.3), we nd that (4.7) yields This can be comared with the value 5 = 1 (161? 7 5) 5 = =4 (4 5? 9): ? s imrecisely given in Ramanujan's rst notebook established in [3, Theorem 3.]. The authors are very grateful to the referee for very carefully reading our manuscrit, making valuable suggestions, uncovering some errors. 1 A 8

10 10 BRUCE C. BERNDT AND HENG HUAT CHAN References 1. B. C. Berndt, Ramanujan's Notebooks, Part III, Sringer{Verlag, New York, B. C. Berndt, H. H. Chan, L.{C. Zhang, Ramanujan's class invariants, Kronecker's limit formula, modular equations, Trans. Amer. Math. Soc. 349 (1997), 15{ B. C. Berndt, H. H. Chan, L.{C. Zhang, Ramanujan's singular moduli, The Ramanujan Journal 1 (1997), 53{ H. H. Chan S.{S. Huang, On the Ramanujan{Gordon{Gollnitz continued fraction, The Ramanujan Journal 1 (1997), 75{ S. Ramanujan, Modular equations aroximations to, Quart. J. Math. (Oxford) 45 (1914), 350{ S. Ramanujan, Collected Paers, Chelsea, New York, S. Ramanujan, Notebooks ( volumes), Tata Institute of Fundamental Research, Bombay, H. Weber, Lehrbuch der Algebra, dritter B, Chelsea, New York, Deartment of Mathematics, University of Illinois, 1409 West Green Street, Urbana, IL 61801, USA School of Mathematics, Institute for Advanced Study, Princeton, NJ 08540, USA

Bruce C. Berndt, Heng Huat Chan, and Liang Cheng Zhang. 1. Introduction

RADICALS AND UNITS IN RAMANUJAN S WORK Bruce C. Berndt, Heng Huat Chan, and Liang Cheng Zhang In memory of S. Chowla. Introduction In problems he submitted to the Journal of the Indian Mathematical Society