Ramanujan s modular equations and Weber Ramanujan class invariants G n and g n

Transcription

1 Bull. Math. Sci. 202) 2: DOI 0.007/s Ramanujan s modular equations and Weber Ramanujan class invariants G n and g n Nipen Saikia Received: 20 August 20 / Accepted: 0 November 20 / Published online: 8 December 20 The Authors) 20. This article is published with open access at SpringerLink.com Abstract In this paper we use Ramanujan s modular equations and transformation formulas to find several new values of Weber Ramanujan class invariants g n.wealso give new approach to some known values of class invariants G n. Keywords Weber Ramanujan class-invariants Modular equations Transformation formulas Mathematics Subject Classification 2000) Primary 33D90; Secondary F20 Introduction The Dedekind eta-function ηz) and Ramanujan s function f q) are defined by f q) := q; q) = q /24 ηz) q = e 2πiz Imz) >0..) wherea; q) = n=0 aq n ). Now Weber Ramanujan class invariants G n and g n [4 p. 83.3)] are defined by G n = 2 /4 q /24 χq) and g n = 2 /4 q /24 χ q) q := e πn.2) where n is a positive rational number and Communicated by S.K. Jain. N. Saikia B) Department of Mathematics Rajiv Gandhi University Doimukh 792 Arunachal Pradesh India nipennak@yahoo.com

2 206 N. Saikia χq) = q; q 2 ) = f q) f q 2 )..3) In his notebooks [] and paper [0] Ramanujan recorded a total of 6 class invariants. The table at the end of Weber s book [2 p ] contains the values of 07 class invariants. There are many applications of Weber Ramanujan class invariants G n and g n. Weber primarily was motivated to calculate class invariants so that he could construct Hilbert class fields. On the other hand Ramanujan calculated class invariants to approximate π and probably for finding explicit values of Rogers Ramanujan continued fractions theta-functions etc. For further applications of class invariants see [5 9]. An account of Ramanujan s class invariants and applications can also be found in Berndt s book [4]. In 200 Yi [3] evaluated several new class invariants G n and g n by finding explicit values of the parameter r kn see [3 p. 2..)] or [4 p. 4.)]) defined as r kn := f q) k /4 q k )/24 f q k ) q = n/k e 2π.4) where n and k are positive real numbers. In particular she established the result [3 p. 8 Theorem 2.2.3] g n = r 2n/2..5) Adiga et al. [] and Baruah [2] also evaluated some new values of G n and g n. In this paper we find further new values of Ramanujan s class invariant g n and also give new approach to some known values of G n. We consider the parameter A n defined by A n = f q) 2 /4 q /24 f q 2 q := n/2 e 2π.6) ) where n is a positive rational number. Clearly the parameter A n is the particular case k = 2 of the Yi s parameter r kn defined above. In Sect. 2 we record some transformation formulas and modular equations. We also prove four eta-function identities which are also particular type of modular equations. In Sect. 3 we find several values of Weber Ramanujan class invariants g n. Finally in Sect. 4 we use some new values of g n evaluated in Sect. 3 to prove some known values on G n. Since modular equations are key in our evaluations we now define a modular equation. Let K K L and L denote the complete elliptic integrals of the first kind associated with the moduli k k l and l respectively. Suppose that the equality n K K = L L.7) holds for some positive integer n. Then a modular equation of degree n is a relation between the moduli k and l which is implied by.7). Ramanujan recorded his

3 Ramanujan s modular equations 207 modular equations in terms of α and β where α = k 2 and β = l 2. We say that β has degree n over α. Denoting z r = φ 2 q r ) where q = exp π K /K ) φq) = f q q) q < ; the multipliers m associated with α β is defined by m = z /z n. 2 Transformation formulas and modular equations The section is devoted to record some transformation formulas and modular equations which will be used in next section. The eta-function identities in Lemmas are found by Yi [3] byusinggarvan s Maple q-series package and then proved by verification. We give direct proofs of these identities. In our proofs it is not necessary to know the identities in advance and employ Ramanujan s modular equations and transformation formulas. Eta-function identity in Lemma 2.0 is new. Lemma 2. [3 p. 43 Entry 27 iii)]) If α and β are such that the modulus of each exponential argument is less than and αβ = π 2 then e α/2 4 α f e 2α ) = e β/2 4 β f e 2β ). 2.) Lemma 2.2 [3 p. 24 Entry 2ii) iii) & iv)]) f q) = z2 /6 α) /6 α /24 q / ) f q 2 ) = z2 /3 α α)) /2 q /2. 2.3) f q 4 ) = z2 2/3 α) /24 α /6 q /6. 2.4) Lemma 2.3 [3 p. 230 Entry 5ii)]) If β has degree 3 over α then αβ) /4 + α) β)) /4 =. 2.5) Lemma 2.4 [3 p. 280 Entry 3i) & x)]) If β has degree 5 over α then αβ) /2 + α) β)) /2 + 2 {6αβ α) β)} /6 =. 2.6) {α β)} /4 + {β α)} /4 = 4 /3 {αβ α) β)} / ) Lemma 2.5 [3 p. 34 Entry 9i)]) If β has degree 7 over α then Lemma 2.6 [4 p. 387 Entry 62]) Let αβ) /8 + α) β)) /8 =. 2.8) P = αβ α) β) ) Q = 64 αβ + α) β) αβ α) β)

4 208 N. Saikia and then if β has degree 9 over α R = 32 αβ α) β) P 6 R4P 3 + PQ) 3R 2 = ) Lemma 2.7 [4 p. 387 Entry 62]) Let P Q and R be as defined in Lemma 2.6 then if β has degree 3 over α PP 3 + 8R) RP 2 + Q) = ) Lemma 2.8 [4 p. 387 Entry 62]) Let P Q and R be as defined in Lemma 2.6 then if β has degree 7 over α P 3 R /3 0P 2 + Q) + 3R 2/3 P + 2R = 0. 2.) Lemma 2.9 [4 p. 386 Entry 58]) Let and P = αβ) /4 { α) β)} /4 Q = 6 αβ) /4 +{ α) β)} /4 {αβ α) β)} /4) then if β has degree 9 over α R = 6{αβ α) β)} /4 P 5 7P 2 R QR = ) Lemma 2.0 [4 p. 386 Entry 59]) Let P Q and R be as defined in Lemma 2.9 then if β has degree 27 over α P 9 RP 2 29P 4 + P 2 Q + Q 2 ) 7R 2 P 3 3R 2 PQ + R) = ) Lemma 2. [4 p. 385 Entry 53]) Let and P = + αβ) /8 +{ α) β)} /8 Q = 4 αβ) /8 +{ α) β)} /8 +{αβ α) β)} /8) R = 4{αβ α) β)} /8

5 Ramanujan s modular equations 209 then if β has degree 5 over α PP 2 Q) + R = ) Lemma 2.2 [3 p. 2 Theorem 3.2.2]) If P = f q 2 ) q /2 f q 4 ) f q) q /24 f q 2 ) and Q = ) 4 4 then PQ) 4 + = PQ Proof Transcribing using Lemma 2.2 we get ) Q 2. P P = 2 /6 α) /2 α /24 and Q = 2 /3 α) /24 α /2. 2.5) From 2.5) we find that and α = PQ) 8 2.6) ) Q 24 = 2 4 {α α)}. 2.7) P Applying 2.6) in2.7) and simplifying we get 6P 8 + P 6 Q 8 Q 6) 6P 8 + P 6 Q 8 + Q 6) = ) By examining the behavior near origin it can be shown that the second factor of the left hand side of 2.8) is non-zero in a neighborhood of the origin. Thus the first factor vanishes in that neighborhood. Hence by the identity theorem this factor vanishes identically i.e. 6P 8 + P 6 Q 8 Q 6 = ) Dividing 2.9) byp 2 Q 4 we complete the proof. Lemma 2.3 [3 p. 37 Theorem 3.5.2]) If P = f q 3 ) q 3/8 f q 2 ) f q) q /8 f q 4 ) and Q = then PQ + 4 PQ = ) Q 2 + P ) P 2. Q

6 20 N. Saikia Proof Transcribing P and Q by using Lemma 2.2 and simplifying we get α = where β has degree 3 over α. It follows that α = Using 2.20) and 2.2) in Lemma 2.3 wearriveat P 8 and β = 6 + Q ) P 8 Q P 8 and β = 6 + Q ) 4 + P 2 Q 2 ) 4 = 6 + P 8 )6 + Q 8 ). 2.22) Factorizing 2.22) usingmathematica we get P 4 4PQ P 3 Q 3 + Q 4 )P 4 + 4PQ + P 3 Q 3 + Q 4 ) = ) Since the second factor is non-zero in a neighborhood of the origin we deduce P 4 4PQ P 3 Q 3 + Q 4 = ) Dividing above equation by P 2 Q 2 we complete the proof. Lemma 2.4 [3 p. 38 Theorem 3.5.3]) If P = f q 5 ) q 5/8 f q 20 ) then ) P 3 + Q Proof By using Lemma 2.2 we find that α = where β has degree 5 over α. It follows that α = f q) q /8 f q 4 ) and Q = ) Q 3 ) 4 2 P = PQ) P PQ Q + Q ). P P 8 and β = 6 + Q ) P 8 Q P 8 and β = 6 + Q ) Now combining 2.6) and 2.7) we obtain { 2 αβ) /2 + { α) β)} /2} { = 2 {α β)} /4 + {β α)} /4} )

7 Ramanujan s modular equations 2 Employing 2.25) and 2.26) in2.27) and simplifying we obtain 6 + P 4 Q 4 ) P 8 )6 + Q 8 ) = {6 + P 8 )6 + Q 8 ) 8P 2 + Q 2 ) 4} 2 Factorizing 2.28) usingmathematica we obtain 2.28) P 2 + Q 2 ) 2 P 6 6PQ 5P 4 Q 2 5P 2 Q 4 P 5 Q 5 + Q 6 ) P 6 6PQ 5P 4 Q 2 5P 2 Q 4 + P 5 Q 5 + Q 6 ) = ) Now proceeding as in the proof of Lemma 2.3 it can be shown that the first and last factors of 2.29) are non-zero in a neighborhood of zero. Thus we have P 6 6PQ 5P 4 Q 2 5P 2 Q 4 P 5 Q 5 + Q 6 = ) Dividing above equation by P 3 Q 3 and rearranging the terms we complete the proof. Lemma 2.5 If P = then ) P 4 + Q f q) q /8 f q 4 ) and Q = f q 7 ) q 5/8 f q 28 ) ) Q 4 4 = PQ) 3 + P PQ PQ + PQ ) ) Proof Proceeding as in the proof of Lemma 2.3 we obtain α = ) ) PQ) 2 PQ P 8 β = Q 8 α = P 8 Q P 8 and β = 6 + Q 8. where β has degree 7 over α. Employing 2.3) in Lemma 2.5 and simplifying we deduce that Simplifying 2.32) we get 2.3) 2 + PQ) 8 = P 8 + 6)Q 8 + 6). 2.32) P 8 64PQ 2P 2 Q 2 2P 3 Q 3 70P 4 Q 4 28P 5 Q 5 7P 6 Q 6 P 7 Q 7 + Q 8 = ) Dividing above equation by P 4 Q 4 and rearranging the terms we complete the proof.

8 22 N. Saikia 3 Evaluation of class invariants g n Theorem 3. If A n as defined in.6) then i)a /n = /A n ii)a = and iii)a n = g 2n. Proof To prove i) we use the definition of A n and Lemma.4.Wesetn = in i) to prove ii). iii) follows directly from the definitions of A n and g n from.6) and.2) respectively. Theorem 3.2 One has α = where β has degree k over α. Proof For positive number kset + A n A 4n ) 8 and β = + A k 2 n A 4k 2 n) 8 P = f q) q /8 f q 4 ) and Q = f q k ) q k/8 f q 4k ). 3.) Transcribing using Lemma 2.2 we obtain 6 α = 6 + P 8 6 and β = 6 + Q ) Setting q := e 2π n/2 in 3.2) and using the definition of A n we complete the proof. Next theorem is due to Yi [3 p. 42 Theorem 4..]. Theorem 3.3 We have ) 4 A n A 4n ) 4 + A n A 4n ) 4 = A4n A n ) 2. Proof follows easily from Lemma 2.2 and the definition of A n. Theorem 3.4 We have ) An A 2 4n A9n A 36n + A 9n A 36n A n A 4n ) 2 = 2 { A n A 4n A 9n A 36n + A n A 4n A 9n A 36n ) }. Proof We set q := e 2π n/2 in Lemma 2.3 and use the definition of A n.

9 Ramanujan s modular equations 23 Theorem 3.5 We have A 6 = g 2 = 2 / ) /8 A /6 = g /3 = 2 /6 2 3) /8 A 3/2 = g 3 = 2 / ) /8 A 2/3 = g 4/3 = 2 /6 2 3) /8. Proof Setting n = /6 in Theorem 3.4 and simplifying using Theorem 3.i) we obtain ) 4 ) 4 A2/3 /A 6 + A2/3 /A 6 = ) Solving 3.3) fora 2/3 /A 6 ) we get A 2/3 /A 6 ) = 2 3) /4. 3.4) Now setting n = /6 in Theorem 3.3 and simplifying by using Theorem 3.i) we obtain { A2/3 ) 4 ) } 4 4 /A 6 + A2/3 /A 6 = ) 2 A 6 A 2/3 3.5) Using 3.3) in3.5) we deduce that A 6 A 2/3 = 2 /3. 3.6) From 3.4) and 3.6) we easily deduce the values of A 6 and A 2/3. The values of A /6 and A 3/2 immediately follow from Theorem 3.3i). Theorem 3.6 We have 4 {A n A 4n A 25n A 00n ) + A n A 4n A 25n A 00n ) } ) A25n A 3 ) 00n An A 3 4n A25n A 00n = A ) n A 4n. A n A 4n A 25n A 00n A n A 4n A 25n A 00n Proof We set q := e 2πn/2 in Lemma 2.4 and use the definition of A n. Theorem 3.7 We have A 0 = g 20 = 2 / ) / /4 5) A 5/2 = g 5 = 2 3/8 5 2) / /4 5) A /0 = g /5 = 2 /8 5 2) / /4 5) A 2/5 = g 4/5 = 2 3/ ) / /4 5)

10 24 N. Saikia Proof Setting n = /0 in Theorem 3.6 and simplifying using Theorem 3.i) we obtain { A 0 A 5/2 ) 6 + A 0 A 5/2 ) 6 5 A 0 A 5/2 ) 2 + A 0 A 5/2 ) 2} = ) From 3.7) we deduce that Solving 3.8) fora 0 A 5/2 we obtain A 0 A 5/2 ) 2 + A 0 A 5/2 ) 2 = ) A 0 A 5/2 =. 3.9) 2 Now setting n = /0 in Theorem 3.3 and simplifying using Theorem 3.i) we deduce that { 4 A 0 A 5/2 ) 4 + A 0 A 5/2 ) 4} = ) 2 A 0 /A 5/2. 3.0) Squaring 3.8) we deduce that A 0 A 5/2 ) 4 + A 0 A 5/2 ) 4 = ). 3.) Employing 3.) in3.0) and solving the resulting equation we obtain A0 /A 5/2 ) = 2 / ) /2. 3.2) Combining 3.9) and 3.2) we derive the values of A 0 and A 5/2. Then values of A /0 and A 2/5 follow from Theorem 3.i). Theorem 3.8 We have An A 4n ) 4 An A 4n + ) 4 A 49n A 96n A 49n A 96n = 8 {A n A 4n A 49n A 96n ) 3 + A n A 4n A 49n A 96n ) 3} +28 {A n A 4n A 49n A 96n ) 2 + A n A 4n A 49n A 96n ) 2} { +56 A n A 4n A 49n A 96n + A n A 4n A 49n A 96n ) } Proof We set q := e 2π n/2 in Lemma 2.5 and use the definition of A n.

11 Ramanujan s modular equations 25 Theorem 3.9 We have A 4 = g 28 = 2 / ) /6 A 7/2 = g 7 = 2 / ) /6 A /4 = g /7 = 2 / ) /6 A 2/7 = g 4/7 = 2 / ) /6. Proof Setting n = /4 in Theorem 3.8 and simplifying using Theorem 3.i) we deduce that Solving 3.3) we obtain A 4 A 7/2 ) 8 + A 4 A 7/2 ) 8 = ) A 4 A 7/2 = ) /8. 3.4) Now setting n = /4 in Theorem 3.3 and simplifying using Theorem 3.i) we arrive at { 4 A 4 A 7/2 ) 4 + A 4 A 7/2 ) 4} = ) 2 A 4 /A 7/2 3.5) Squaring 3.5) and then using 3.3) we deduce that ) A4 /A 7/2 = ) Combining 3.4) and 3.6) we derive the values of A 4 and A 7/2. Then the values of A /4 and A 2/7 follow from Theorem 3.i). Theorem 3.0 We have A 8 = g 36 = 2 / ) / ) /8 3 A 9/2 = g 9 = 2 / ) / ) /8 3 A /8 = g /9 = 2 /8 7 4 ) / ) /8 3 A 2/9 = g 4/9 = 2 /8 7 4 ) / ) /8 3. Proof Setting k = 9 in Theorem 3.2 we get α = + A n A 4n ) 8 and β = + A 8n A 324n ) ) Now setting n = /8 in 3.7) and simplifying using Theorem 3.i) we obtain α = A 8 A 9/2 ) 8 + A 8 A 9/2 ) 8 and β = + A 8 A 9/2 ) )

12 26 N. Saikia From 3.8) it is readily seen that α = A 8 A 9/2 ) 8 β α = β and β = α. 3.9) Employing 2.3) in Lemma 2.6 and simplifying we get where P = 2x Q = 64x2 x) and R = 32x ) Employing 3.20) in2.9) and factorizing we obtain x = A 8 A 9/2 ) 4 β. 3.2) + 8x 4x 2 ) 2 28x + 4x 2 ) = ) The first factor in 3.22) is not zero and do not give real value of x such the 0 < x < so we have 4x 2 28x + = ) Solving 3.23) and noting 0 < x < and A 8 A 9/2 ) 4 > is real we get Combining 3.8) 3.2) and 3.24) we deduce that A 8 A 9/2 ) 4 = x = 7 4 3)/ ) ) ) ) Now setting n = /8 in Theorem 3.3 and simplifying using Theorem 3.i) we arrive at { 4 A 8 A 9/2 ) 4 + A 8 A 9/2 ) 4} = ) 2 A 8 /A 9/ ) Using 3.25) in3.26) and solving the resulting equation we obtain ) A8 /A 9/2 = 2 / ) / ) From 3.25) and 3.27) we easily deduce the values of A 8 and A 9/2. Then the values of A /8 and A 2/9 follow from Theorem 3.i). Theorem 3. We have A 26 = g 52 = 2 /8 5 ) / /8 3 8) A 3/2 = g 3 = 2 /8 5 ) / /8 3 8)

13 Ramanujan s modular equations 27 A /26 = g /3 = 2 /8 5 ) / /8 3 8) A 2/3 = g 4/3 = 2 /8 5 ) / /8 3 8). Proof Setting k = 3 in Theorem 3.2 and then setting n = /26 and simplifying using Theorem 3.i) we obtain so that α = A 26 A 3/2 ) 8 + A 26 A 3/2 ) 8 and β = + A 26 A 3/2 ) ) α = A 26 A 3/2 ) 8 β α = β and β = α. 3.29) Employing 3.29) in Lemma 2.7 and simplifying we get + 2x) + 28x + 4x 2 ) x + 4x 2 ) = ) where x = A 26 A 3/2 ) 4 β. 3.3) Since first two factors are not zero so we have 4x x = ) Solving 3.32) we obtain x = 5 3 8)/ ) From 3.29) 3.3) and 3.33) and noting A 26 A 3/2 > we deduce that ) 4 A26 A 3/2 = ) ) ) Setting n = /26 in Theorem 3.3 simplifying using Theorem 3.i) employing 3.34) and then solving the resulting equation we arrive at ) A26 /A 3/2 = 2 /4 5 ) / ) From 3.34) and 3.35) we easily calculate the values of A 26 and A 3/2. Then the values of A /26 and A 2/3 follow from Theorem 3.i).

14 28 N. Saikia Theorem 3.2 We have A 34 = g 68 = 2 20 / ) / ) 2 7 A 7/2 = g 7 = 2 20 / ) / ) 2 7 A /34 = g /7 = 2 20 / ) / ) 2 7 A 2/7 = g 4/7 = 2 20 / ) / ) 2 7 /8 /8 /8 /8. Proof Setting k = 7 in Theorem 3.2 and then setting n = /34 and simplifying using Theorem 3.i) we obtain so that α = A 34 A 7/2 ) 8 + A 34 A 7/2 ) 8 and β = + A 34 A 7/2 ) ) α = A 34 A 7/2 ) 8 β α = β and β = α. 3.37) Employing 3.37) in Lemma 2.8 and solving the resulting equation we get x = ) 7 / ) where x = A 34 A 7/2 ) 4 β. 3.39)

15 Ramanujan s modular equations 29 From 3.36) 3.38) and 3.39) and noting A 34 A 7/2 > we arrive that ) 4 A34 A 7/2 = ) ) ) Setting n = /34 in Theorem 3.3 simplifying using Theorem 3.i) employing 3.40) and then solving the resulting equation we find that ) A34 /A 7/2 = 2 / ) / ) From 3.40) and 3.4) we easily calculate the values of A 34 and A 7/2. Then the values of A /34 and A 2/7 follow from Theorem 3.i). Theorem 3.3 We have ) /8 A 38 = g 76 = 2 5/24 3 /2 r / r 2 ) /8 A 9/2 = g 9 = 2 /24 3 /2 r / r 2 ) /8 A /38 = g /9 = 2 5/24 3 /2 r / r 2 ) /8 A 2/9 = g 4/9 = 2 /24 3 /2 r / r 2 where r = ) / ) 2/ ) / ) 2/3. Proof Setting k = 9 in Theorem 3.2 and then setting n = /38 and simplifying using Theorem 3.i) we arrive at so that α = A 38 A 7/2 ) 8 + A 34 A 9/2 ) 8 and β = + A 38 A 9/2 ) ) α = A 38 A 9/2 ) 8 β α = β and β = α. 3.43) Employing 3.43) in Lemma 2.9 and factorizing we get + 2y) 2 8y y 2 + 4y ) = )

16 220 N. Saikia where Since the first factor in 3.44) is non-zero so we have Solving 3.46) we obtain y = y = A 38 A 9/2 ) 2 β. 3.45) 8y y 2 + 4y = ) ) / ) /3) / ) From 3.42) 3.45) and 3.47) and noting A 38 A 9/2 > we deduce that ) 4 ) / A38 A 9/2 = r 2 2r 3.48) wherer = ) / ) 2/ ) / ) 2/3. Setting n = /38 in Theorem 3.3 simplifying using Theorem 3.i) employing 3.48) and then solving the resulting equation we find that ) A38 /A 9/2 = 2 2/3 3 /6 r / ) From 3.48) and 3.49) we easily find the values of A 38 and A 9/2. Then the values of A /38 and A 2/9 follow from Theorem 3.i). Theorem 3.4 We have ) /8 ) /6 A 54 = g 08 =2 / / / /3 2 4/3 7 ) /8 ) /2 A 27/2 = g 27 =2 / / / /3 2 4/3 7 ) /8 ) /6 A /54 = g /27 =2 / / / /3 2 4/3 7 ) /8 ) /2. A 2/27 = g 4/27 =2 / / / /3 2 4/3 7 Proof Setting k = 27 in Theorem 3.2 and then setting n = /38 and simplifying using Theorem 3.i) we arrive at so that α = A 54 A 27/2 ) 8 + A 54 A 27/2 ) 8 and β = + A 54 A 27/2 ) ) α = A 54 A 27/2 ) 8 β α = β and β = α. 3.5)

17 Ramanujan s modular equations 22 Employing 3.5) in Lemma 2.0 and factorizing Mathematica we get where + 6x 4x 2 + 8x 3 ) y 2y 2 + 8y 3 ) = ) Since the first factor in 3.52) is non-zero so we have Solving 3.54) we obtain y = A 54 A 27/2 ) 2 β. 3.53) + 30y 2y 2 + 8y 3 = ) y = + 2 /3 ) 2 / ) From 3.50) 3.53) and 3.55) and noting A 54 A 27/2 > we deduce that ) A54 A / /3 27/2 = 6 2 2/3 2 4/ ) 7 Setting n = /54 in Theorem 3.3 simplifying using Theorem 3.i) employing 3.56) and then solving the resulting equation we find that ) ) /2 A54 /A 27/2 = 2 / /3 2 4/ ) From 3.56) and 3.57) we easily find the values of A 54 and A 27/2. Then the values of A /54 and A 2/27 follow from Theorem 3.i). Theorem 3.5 We have A 30 = g 60 = 2 6 7/ ) / ) /6 5 A 5/2 = g 5 = 2 6 7/ ) / ) /2 5 A /30 = g /5 = 2 6 7/ ) / ) /6 5 A 2/5 = g 4/5 = 2 6 7/ ) / ) /2 5. Proof Setting k = 5 in Theorem 3.2 and then setting n = /30 and simplifying using Theorem 3.i) we arrive at α = A 30 A 5/2 ) 8 + A 30 A 5/2 ) 8 and β = + A 30 A 5/2 ) )

18 222 N. Saikia so that α = A 30 A 5/2 ) 8 β α = β and β = α. 3.59) Employing 3.59) in Lemma 2. we get where Solving 3.60) we obtain 4z 2 + 2z = ) z = A 30 A 5/2 )β /4. 3.6) z = + 5)/ ) From 3.58) 3.6) and 3.62) and noting A 30 A 5/2 > we deduce that ) 4 A30 A 5/2 = )/ ) ) Setting n = /30 in Theorem 3.3 simplifying using Theorem 3.i) employing 3.63) and then solving the resulting equation we find that ) A30 /A 5/2 = 2 7/2 7 3 ) / ) From 3.63) and 3.64) we easily deduce the values of A 30 and A 5/2. Then the values of A /30 and A 2/5 follow from Theorem 3.i). 4 Evaluation of class invariants G n In his paper [0] and also in page 294 of his second notebook [ Vol. II] Ramanujan recorded two simple formulas relating the class invariants g n and G n namely for n > 0 and g 4n = 2 /4 g n G n 4.) g n G n ) 8 G 8 n g8 n ) = ) Thus if we know g n and g 4n or only g n then the corresponding G n can be calculated by the above formulas. We now find some values of G n by using 4.) and the new values of g n and g 4n evaluated in above section.

19 Ramanujan s modular equations 223 Theorem 4. We have i) G 3 = 2 /2 ii) G 5 = 5 + 2) /2 iii) G 7 = 2 /4 iv) G 9 = ) /2 v) G 3 = ) /2 vi) G 7 = /2 7) vii) G 9 = 2 5/2 3 /6 r /2 where r is given in Theorem 3.3. viii) G 27 = 2 / /3 2 4/3 7) /2 ix) G 5 = 2 / ) /2. Proof For i) we use the values of g 2 and g 3 from Theorem 3.5 in 4.). To prove ii) we employ the values of g 20 and g 5 from Theorem 3.7 in 4.). To prove iii) employ the values of g 28 and g 7 from Theorem 3.9 in 4.). To prove iv) we use the values of g 36 and g 9 from Theorem 3.0 in 4.). Employing the values of g 52 and g 3 from Theorem 3. in 4.) we arrive at v). Employing the values of g 7 and g 68 from Theorem 3.2 in 4.) we prove vi). For vii) we use the values of g 9 and g 76 from Theorem 3.3 in 4.). To prove viii) we use the values of g 27 and g 08 from Theorem 3.4 in 4.). To prove ix) we employ the values of g 5 and g 60 from Theorem 3.5 in 4.). Open Access This article is distributed under the terms of the Creative Commons Attribution License which permits any use distribution and reproduction in any medium provided the original authors) and source are credited. References. Adiga C. Naika M.S.M. Shivashankara K.: On some P Q eta-function identities of Ramanujan. Indian J. Math. 443) ) 2. Baruah N. D.: On some of class invariants of Ramanujan. J. Indian Math. Soc. 68 4) ) 3. Berndt B.C.: Ramanujan s Notebooks Part III. Springer New York 99) 4. Berndt B.C.: Ramanujan s Notebooks Part V. Springer New York 998) 5. Berndt B.C. Chan H.H.: Some values for the Rogers Ramanujan continued fraction. Can. J. Math. 475) ) 6. Berndt B.C. Chan H.H. Kang S.-Y. Zhang L.-C.: A certain quotient of eta-funtions found in Ramanujan s lost notebook. Pac. J. Math. 2022) ) 7. Berndt B.C. Chan H.H. Zhang L.-C.: Ramanujan s class invariants with applications to the values of q-continued fractions and theta-functions. In: Ismail M. Masson D. Rahman M. eds.) Special Functions and q-series and Related Topics Fields Institute Communications Series vol. 4 pp Am. Math. Soc. Providence 997) 8. Berndt B.C. Chan H.H. Zhang L.-C.: Ramanujan s class invariants and cubic continued fraction. Acta Arith ) 9. Berndt B.C. Chan H.H. Zhang L.-C.: Ramanujan s remarkable product of theta-functions. Proc. Edinburg Math. Soc ) 0. Ramanujan S.: Modular equations and approximations to π. Quart.J.Math ). Ramanujan S.: Notebooks 2 volumes). Tata Institute of Fundamental Research Bombay 957) 2. Weber H.: Lehrburg der Algebra II. Chelsea New York 96) 3. Yi J.: Construction and application of modular equation. Ph. D thesis University of Illionis 200) 4. Yi J.: Theta-function identities and the explicit formulas for theta-function and their applications. J. Math. Anal. Appl )