# A Parameter for Ramanujan's Functionχ(q): Its Explicit Values and Applications

Authors: Nipen Saikia
Journal: ISRN Computational Mathematics
 (2012)

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## Abstract

We define a new parameterIk,n involving quotient of Ramanujan's function χ(q) for positive real numbers k and n and study its several properties. We prove some general theorems for the explicit evaluations of the parameter Ik,n and find many explicit values. Some values of Ik,n are then used to find some new and known values of Ramanujan's class invariant Gn.


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## Body

## 1. Introduction

In Chapter 16 of his second notebook [1], Ramanujan develops the theory of theta-function. Ramanujan's general theta-function is defined by
(1)f(a,b)=∑n=-∞∞‍an(n+1)/2bn(n-1)/2,|ab|<1.
After Ramanujan, for |q|<1, we define
(2)f(-q):=f(-q,-q2)=∑n=-∞∞‍(-1)nqn(3n-1)/2=(q;q)∞,
where (a;q)∞:=∏n=0∞‍(1-aqn). If q=e2πiz with Im(z)>0, then f(-q)=q-1/24η(z), where η(z) denotes the classical Dedekind eta function.Ramanujan's functionχ(q)  is defined by
(3)χ(q):=f(q)f(-q2)=(-q;q2)∞.
The function χ(q) is intimately connected to Ramanujan's class invariants Gn and gn, which are defined by
(4)Gn=2-1/4q-1/24χ(q),gn=2-1/4q-1/24χ(-q),
where q:=e-πn and n is a positive rational number. Since from [2, page 124, Entry 12(v) & (vi)]
(5)χ(q)=21/6{α(1-α)q}-1/24,χ(-q)=21/6(1-α)1/12(αq)-1/24,
it follows from (4) that
(6)Gn={4α(1-α)}-1/24,gn=2-1/12(1-α)1/12α-1/24.
Also, if β has degree r over α, then
(7)Gr2n={4β(1-β)}-1/24,gr2n=2-1/12(1-β)1/12β-1/24.
In his notebooks [1] and paper [3], Ramanujan recorded a total of 116 class invariants or monic polynomials satisfied by them. The table at the end of Weber's book [4, page 721–726] also contains the values of 107 class invariants. 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, and so on. An account of Ramanujan's class invariants and applications can be found in Berndt's book [5]. For further references, see [6–12].Ramanujan and Weber independently and many others in the literature calculated class invariantsGn for odd values of n and gn for even values of n. For the first time, Yi [13] calculated some values of gn for odd values of n by finding explicit values of the parameter rk,n (see [13, page 11, (2.1.1)] or [14, page 4, (1.11)]) defined by
(8)rk,n:=f(-q)k1/4q(k-1)/24f(-qk),q=e-2πn/k.
In particular, she established the result [13, page 18, Theorem 2.2.3]
(9)gn=r2,n/2.
However, the values of Gn for even values of n have not been calculated. The main objective of this paper is to evaluate some new values of Gn for even values of n. We also prove some known values of Gn. For evaluation of class invariant Gn in this paper, we introduce the parameter Ik,n, which is defined as
(10)Ik,n:=χ(q)q(-k+1)/24χ(qk),q=e-πn/k,
where k and n are positive real numbers.In Section3, we study some properties of Ik,n and also establish its relations with Ramanujan's class invariant Gn. In Section 4, by employing Ramanujan's modular equations, we present some general theorems for the explicit evaluations of Ik,n and find several explicit values of Ik,n. In Section 5, we establish some general theorems connecting the parameter Ik,n and the class invariant Gn. We also evaluate some explicit values of the product GnkGn/k by employing some values of Ik,n evaluated in Section 4. Finally, in Section 6, we calculate new and known values of class invariant Gn by combining the explicit values of Ik,n and the product GnkGn/k evaluated in Sections 4 and 5, respectively. Section 2 is devoted to record some preliminary results.Since Ramanujan's modular equations are key in our evaluations ofIk,n and Gn, we complete this introduction by defining Ramanujan's modular equation from Berndt's book [2]. The complete elliptic integral of the first kind K(k) is defined by
(11)K(k):=∫0π/2‍dϕ1-k2sin2ϕ=π2∑n=0∞‍(1/2)n2(n!)2k2n=π2F12(12,12;1;k2),
where 0<k<1, 2F1 denotes the ordinary or Gaussian hypergeometric function, and
(12)(a)n=a(a+1)(a+2)⋯(a+n-1).
The number k is called the modulus of K, and k′:=1-k2 is called the complementary modulus. 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
(13)nK′K=L′L
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 (13).If we set(14)q=exp⁡(-πK′K),q′=exp⁡(-πL′L),
we see that (13) is equivalent to the relation qn=q′. Thus, a modular equation can be viewed as an identity involving theta-functions at the arguments q and qn. Ramanujan recorded his modular equations in terms of α and β, where α=k2 and β=l2. We say that β has degree n over α. The multiplier m connecting α and β is defined by
(15)m=KL.
Ramanujan also established many “mixed” modular equations in which four distinct moduli appear, which we define from Berndt's book [2, page 325].LetK, K′, L1, L1′, L2, L2′, L3, and L3′ denote complete elliptic integrals of the first kind corresponding, in pairs, to the moduli α, β, γ, and δ and their complementary moduli, respectively. Let n1, n2, and n3 be positive integers such that n3=n1n2. Suppose that the equalities
(16)n1K′K=L1′L1,n2K′K=L2′L2,n3K′K=L3′L3
hold. Then, a “mixed” modular equation is a relation between the moduli α, β, γ, and δ that is induced by (16). In such an instance, we say that β, γ, and δ are of degrees n1, n2, and n3, respectively, over α or α, β, γ, and δ have degrees 1, n1, n2, and n3, respectively. Denoting zr=ϕ2(qr), where
(17)q=exp⁡(-πK′K),ϕ(q)=f(q,q),|q|<1
the multipliers m and m′ associated with α, β, and γ, δ, respectively, are defined by m=z1/zn1 and m′=zn2/zn3.


## 2. Preliminary Results

Lemma 1 (see [2, page 43, Entry 27(v)]).
Ifα and β are such that the modulus of each exponential argument is less than 1 and αβ=π2, then
(18)eα/24χ(e-α)=eβ/24χ(e-β).Lemma 2 (see [15, page 241, Lemma 2.3]).
LetX:=q1/12(χ(q)/χ(q3)) and Y:=q1/6(χ(q2)/χ(q6)); then
(19)(XY)12+(YX)12+[(XY)6+(YX)6]×[(XY)10+(XY)-10+16{(XY)6+(XY)-6}××+71{(XY)2+(XY)-2}]=(XY)12+(XY)-12+25{(XY)8+(XY)-8}++200{(XY)4+(XY)-4}+550.Lemma 3 (see [15, page 241, Lemma 2.8]).
LetX:=q1/6(χ(q)/χ(q5)) and Y:=q1/3(χ(q2)/χ(q10)); then
(20)[(XY)3+(YX)3][(XY)5+(XY)-5+8((XY)3+(XY)-3)[(XY)3+(YX)3]x+19(XY+(XY)-1)]+(XY)6+(YX)6=((XY)6+(XY)-6)+13(((XY)4+(XY)-4))==+52(((XY)2+(XY)-2))+82.Lemma 4 (see [15, page 252, Lemma 2.13]).
LetX:=q1/4(χ(q)/χ(q7)) and Y:=q1/2(χ(q2)/χ(q14)); then
(21)(XY)12+(YX)12+16[(XY)10+(YX)10][(XY)2+(XY)-2]+8[(XY)8+(YX)8][7{(XY)4+(XY)-4}-19]+[(XY)6+(YX)6]×[(XY)10+(XY)-10+[(XY)6+(YX)6]××+32{(XY)6+(XY)-6}+[(XY)6+(YX)6]×cc-81{(XY)2+(XY)-2}]+[(XY)4+(YX)4][16{(XY)8+(XY)-8}+[(XY)4+(YX)4]xxxxx-288{(XY)4+(XY)-4}+352]+[(XY)2+(YX)2][296{(XY)2+(XY)-2}+[(XY)2+(YX)2]cc-256{(XY)6+(XY)-6}+[(XY)2+(YX)2]cc-8{(XY)10+(XY)-10}]+1746={(XY)12+(XY)-12}+145[{(XY)8+(XY)-8}]=+496[{(XY)4+(XY)-4}].Lemma 5 (see [2, page 231, Entry 5(xii)]).
LetP={16αβ(1-α)(1-β)}1/8 and Q=(β(1-β)/α(1-α))1/4; then
(22)Q+Q-1+22(P-P-1)=0,
where β has degree 3 over α.Lemma 6 (see [2, page 282, Entry 13(xiv)]).
LetP={16αβ(1-α)(1-β)}1/12 and Q=(β(1-β)/α(1-α))1/8; then
(23)Q+Q-1+2(P-P-1)=0,
where β has degree 5 over α.Lemma 7 (see [2, page 315, Entry 13(xiv)]).
LetP={16αβ(1-α)(1-β)}1/8 and Q=(β(1-β)/α(1-α))1/6; then
(24)Q+Q-1+7=22(P+P-1)=0,
where β has degree 7 over α.Lemma 8 (see [5, page 378, Entry 41]).
LetP=21/6{αβ(1-α)(1-β)}1/24 and Q=(β(1-β)/α(1-α))1/24; then
(25)Q7+Q-7+13(Q5+Q-5)+52(Q3+Q-3)+78(Q+Q-1)-8(P-6+P6)=0,
where β has degree 13 over α.For Lemmas9 to 15, we set
(26)P:=(256αβγδ(1-α)(1-β)(1-γ)(1-δ))1/48,Q:=(αδ(1-α)(1-δ)βγ(1-β)(1-γ))1/48,R:=(γδ(1-γ)(1-δ)αβ(1-α)(1-β))1/48,T:=(βδ(1-β)(1-δ)αγ(1-α)(1-γ))1/48.Lemma 9 (see [5, page 381, Entry 50]).
Ifα, β, γ, and δ have degrees 1, 5, 7, and 35, respectively, then
(27)R4+R-4-(Q6+Q-6)+5(Q4+Q-4)-10(Q2+Q-2)+15=0.Lemma 10 (see [5, page 381, Entry 51]).
Ifα,β,γ, and δ have degrees 1, 13, 3, and 39, respectively, then
(28)Q4+Q-4-3(Q2+Q-2)-(T2+T-2)+3=0.Lemma 11 (see [5, page 381, Entry 52]).
Ifα,β,γ, and δ have degrees 1, 13, 5, and 65, respectively, then
(29)Q6+Q-6-5(Q+Q-1)2(T+T-1)2-(T4+T-4)=0.Lemma 12 (see [16, page 277, Lemma 3.1]).
Ifα,β,γ, and δ have degrees 1, 3, 7, and 21, respectively, then
(30)R2+R-2-Q4-Q-4+3=0.Lemma 13 (see [15, page 243, Theorem 2.5]).
Ifα, β, γ, and δ have degrees 1, 2, 3, and 6, respectively, then
(31)R4+R-4+P2-2P-2=0.Lemma 14 (see [15, page 248, Theorem 2.10]).
Ifα, β, γ and δ have degrees 1, 2, 5, and 10, respectively, then
(32)R6+R-6+5(R2+R-2)=4P-4-P4.Lemma 15 (see [15, page 252, Theorem 2.12]).
Ifα, β, γ and δ have degrees 1, 2, 7, and 14, respectively, then
(33)22(T3+T-3)=4P-1-2P+4P3-P5.


## 3. Properties ofIk,n

In this section, we study some properties ofIk,n. We also establish a relation connecting Ik,n and Ramanujan's class invariants Gn.Theorem 16.
For all positive real numbersk and n, one has
(34)(i)Ik,1=1,(ii)Ik,1n=1Ik,n,(iii)Ik,n=In,k.Proof .
Using the definition ofIk,n and Lemma 1, we easily arrive at (i). Replacing n by 1/n in Ik,n and using Lemma 1, we find that Ik,nIk,1/n=1, which completes the proof of (ii). To prove (iii), we use Lemma 1 in the definition of Ik,n to arrive at (Ik,n/In,k)=1.Remark 17.
By using the definitions ofχ(q) and Ik,n, it can be seen that Ik,n has positive real value less than 1 and that the values of Ik,n decrease as n increases when k>1. Thus, by Theorem 16(i), Ik,n<1 for all n>1 if k>1.Theorem 18.
For all positive real numbers k, m, and n, one has(35)Ik,n/m=Imk,nInk,m-1.Proof.
Using the definition ofIk,n, we obtain
(36)Imk,nInk,m=χ(e-πn/mk)eπ(m/nk-n/mk)/24χ(e-πm/nk).
Using Lemma 1 in the denominator of the right-hand side of (36) and simplifying, we complete the proof.Corollary 19.
For all positive real numbers k and n, one has(37)Ik2,n=Ink,nIk,n/k.Proof.
Settingk=n in Theorem 18 and simplifying using Theorem 16(ii), we obtain
(38)Ik2,m=Imk,kIk,m/k.
Replacing m by n, we complete the proof.Theorem 20.
Let k, a, b, c, and d be positive real numbers such that ab=cd. Then(39)Ia,bIkc,kd=Ika,kbIc,d.Proof.
From the definition ofIk,n, we deduce that, for positive real numbers k,a,b,c, and d,
(40)Ika,kbIa,b-1=χ(e-πab)eπ(kab-ab)/24χ(e-kπab)Ikc,kdIc,d-1=χ(e-πcd)eπ(kcd-cd)/24χ(e-kπcd).
Now the result follows readily from (40), and the hypothesis that ab=cd.Corollary 21.
For any positive real numbersn and p, we have
(41)Inp,np=Inp2,nIp,p.Proof.
The result follows immediately from Theorem20 with a=p2,b=1,c=d=p, and k=n.Now, we give some relations connecting the parameterIk,n and Ramanujan's class invariants Gn.Theorem 22.
Letk and n be any positive real numbers. Then
(42)(i)Ik,n=Gn/kGnk-1,(ii)G1/n=Gn.Proof.
Proof of (i) follows easily from the definitions ofIk,n and Gn from (10) and (4), respectively. To prove (ii), we set k=1 in part (i) and use Theorem 16(i) and (iii).


## 4. General Theorems and Explicit Evaluations ofIk,n

In this section, we prove some general theorems for the explicit evaluations ofIk,n and find its explicit values.Theorem 23.
One has(43)(I3,nI3,4n)12+(I3,nI3,4n)-12+[(I3,nI3,4n)6+(I3,nI3,4n)-6]×[(I3,nI3,4n)10+(I3,nI3,4n)-10+16((I3,nI3,4n)6+(I3,nI3,4n)-6)××+71((I3,nI3,4n)2+(I3,nI3,4n)-2)]=(I3,nI3,4n)12+(I3,nI3,4n)-12+25((I3,nI3,4n)8+(I3,nI3,4n)-8)=+200((I3,nI3,4n)4+(I3,nI3,4n)-4)+550.Proof.
The proof follows easily from the definition ofIk,n and Lemma 2.Corollary 24.
One has(44)(i)I3,2=(-44+273-3458-2643)1/12,(ii)I3,4=((-2+36--6+36)2)1/2,(iii)I3,1/2=(-44+273+3458-2643)1/12,(iv)I3,1/4=((-6+36+-2+36)2)1/2.Proof.
Settingn=1/2 in Theorem 23 and using Theorem 16(ii), we obtain
(45)I3,224+I3,2-24+176(I3,212+I3,2-12)-1002=0.
Equivalently,
(46)B2+176B-1004=0,
where
(47)B=I3,212+I3,2-12.
Solving (46) and using the fact in Remark 17, we obtain
(48)B=545-88.
Employing (48) in (47), solving the resulting equation for I3,2, and noting that I3,2<1, we arrive at
(49)I3,2=(-44+273-3458-2643)1/12.
This completes the proof of (i).
Again settingn=1 in Theorem 23 and using Theorem 16(i), we obtain
(50)(I3,46+I3,4-6)[(I3,410+I3,4-10)(I3,46+I3,4-6)xx+16(I3,46+I3,4-6)+71(I3,42+I3,4-2)]=25(I3,48+I3,4-8)+200(I3,44+I3,4-4)+550.
Equivalently,
(51)(D2+1)2(D4+4D2-50)=0,
where
(52)D=I3,42+I3,4-2.
Since the first factor of (51) is nonzero, solving the second factor, we deduce that
(53)D=(-2+36)1/2.
Employing (53) in (52), solving the resulting equation, and using the fact that I3,4<1, we obtain
(54)I3,4=((-2+36--6+36)2)1/2.
This completes the proof of (ii).
Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem16(ii).Theorem 25.
One has(55)(I5,nI5,4n)6+(I5,nI5,4n)-6+[(I5,nI5,4n)3+(I5,nI5,4n)-3]×[(I5,nI5,4n)5+(I5,nI5,4n)-5+8((I5,nI5,4n)3+(I5,nI5,4n)-3)××+19((I5,nI5,4n)+(I5,nI5,4n))]=(I5,nI5,4n)6+(I5,nI5,4n)-6+13((I5,nI5,4n)4+(I5,nI5,4n)-4)=+52((I5,nI5,4n)2+(I5,nI5,4n)-2)+82.Proof.
The proof follows from Lemma3 and the definition of Ik,n.Corollary 26.
One has(56)(i)I5,2=(-14+510-445-14010)1/6,(ii)I5,4=((11+55)1/4--4+11+55)2,(iii)I5,1/2=(-14+510+445-14010)1/6,(iv)I5,1/4=((11+55)1/4+-4+11+55)2.Proof.
Settingn=1/2 in Theorem 25 and using Theorem 16(ii), we obtain
(57)C2+56C-216=0,
where
(58)C=I5,26+I5,2-6.
Solving (57) and noting the fact in Remark 17, we obtain
(59)C=-28+1010.
Employing (59) in (58), solving the resulting equation, and noting that I5,2<1, we obtain
(60)I5,2=(-14+510-445-14010)1/6.
This completes the proof of (i).
Again, settingn=1 in Theorem 25 and using Theorem 16(i), we obtain
(61)B8-22B4-4=0,
where
(62)B=I5,4+I4,5-1.
Solving (61), we obtain
(63)B=(11+55)1/4.
Using (63) in (62), solving the resulting equation, and noting that I5,4<1, we arrive at
(64)I5,4=((11+55)1/4--4+11+55)2.
This completes the proof of (ii).
Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem16(ii).Theorem 27.
One has(65)(I7,nI4,7n)12+(I7,nI4,7n)-12+16[(I7,nI4,7n)10+(I7,nI4,7n)-10][(I7,nI7,4n)2+(I7,nI7,4n)-2]+8[(I7,nI4,7n)8+(I7,nI4,7n)-8]×[7{(I7,nI7,4n)4+(I7,nI7,4n)-4}-19]+[(I7,nI4,7n)6+(I7,nI4,7n)-6]×[(I7,nI7,4n)10+(I7,nI7,4n)-10xxx+32{(I7,nI7,4n)6+(I7,nI7,4n)-6}xxx-81{(I7,nI7,4n)2+(I7,nI7,4n)-2}]+[(I7,nI4,7n)4+(I7,nI4,7n)-4]×[16{(I7,nI7,4n)8+(I7,nI7,4n)-8}ccc-288{(I7,nI7,4n)4+(I7,nI7,4n)-4}+352]+[(I7,nI4,7n)2+(I7,nI4,7n)-2]×[296{(I7,nI7,4n)2+(I7,nI7,4n)-2}×+-256{(I7,nI7,4n)6+(I7,nI7,4n)-6}×+-8{(I7,nI7,4n)10+(I7,nI7,4n)-10}]+1746={(I7,nI7,4n)12+(I7,nI7,4n)-12}=+145{(I7,nI7,4n)8+(I7,nI7,4n)-8}=+496{(I7,nI7,4n)4+(I7,nI7,4n)-4}.Corollary 28.
One has(66)(i)I7,2=2-1/4(i)I7,2=×(-4+(-8-52+163+1162)2-8-52+163+1162(i)I7,2=222--4+(-8-52+163+1162)2)1/4,(ii)I7,4=3-6+52--58+4522,(iii)I7,1/2=2-1/4,(iii)I7,1/2=×(-4+(-8-52+163+1162)2-8-52+163+1162(iii)I7,1/2=222+-4+(-8-52+163+1162)2)1/4(iv)I7,1/4=3-6+52+-58+4522.Proof.
Settingn=1/2 and simplifying using Theorem 16(ii), we obtain
(67)(I7,224+I7,2-24)+32(I7,220+I7,2-20)-40(I7,216+I7,2-16)-96(I7,212+I7,2-12)-192(I7,28+I7,2-8)+64(I7,24+I7,2-4)+462=0.
Equivalently,
(68)(A2-4)(A4+32A3-42A2-128A-191)=0,
where
(69)A=I7,24+I7,2-4.
By using the fact in Remark 17, it is seen that the first factor of (68) is nonzero, and so from the second factor, we deduce that
(70)A=-8-52+163+1162.
Combining (69) and (70) and noting that I7,2<1, we obtain
(71)I7,2=2-1/4(-4+(-8-52+163+1162)2-8-52+163+1162I7,2=2-1/4vvv--4+(-8-52+163+1162)2)1/4.
This completes the proof of (i).To prove (ii), settingn=1 and simplifying using Theorem 16(i), we arrive at
(72)E2(E2-4)(E4+108E2-1134)=0,where(73)E=I7,42+I7,4-2.Using the fact in Remark17 it is seen that the first two factors of (72) are nonzero, and so solving the third factor, we obtain
(74)E=3-6+52.Combining (73) and (74) and noting that I7,4<1, we deduce that
(75)I7,4=3-6+52--58+4522.
So the proof of (ii) is complete.Now (iii) and (iv) follow from (i) and (ii), respectively, and Theorem16(ii).Theorem 29.
One has(76)(I7,nI7,25n)2+(I7,nI7,25n)-2-{(I7,25nI7,n)3+(I7,25nI7,n)-3}+5{(I7,25nI7,n)2+(I7,25nI7,n)-2}-10{(I7,25nI7,n)+(I7,25nI7,n)}+15=0.Proof.
Using (5) in Lemma 9, we find that
(77)Q=qχ(q5)χ(q7)χ(q)χ(q35),R=q3/2χ(q)χ(q5)χ(q7)χ(q35).
Setting q=e-πn/7 and using the definition of Ik,n in (77), we get
(78)Q=(I7,25nI7,n)1/2,R=(I7,nI7,25n)1/2.
Employing (78) in (27), we complete the proof.Corollary 30.
One has(79)(i)I7,5=h--36+h26,(ii)I7,1/5=h+-36+h26,(iii)I7,25=(d--144+d2)12,(iv)I7,1/25=(d+-144+d2)12,
where h=5+(62-6105)1/3+(62+6105)1/3 and d=12+22/3(135-1521)1/3+22/3(135+1521)1/3.Proof.
Settingn=1/5 in Theorem 29 and simplifying using Theorem 16(ii), we obtain
(80)I7,56+I7,5-6-5(I7,54+I7,5-4)+10(I7,52+I7,5-2)-17=0.
Equivalently,
(81)H3-5H2+7H-7=0,
where
(82)H=I7,52+I7,5-2.
Solving (81) and noting the fact in Remark 17, we obtain
(83)H=(5+(62-6105)1/3+(62+6105)1/3)3
Combining (82) and (83) and noting that I7,5<1, we deduce that
(84)I7,5=h--36+h26,
where
(85)h=5+(62-6105)1/3+(62+6105)1/3.
This completes the proof of (i).
Again settingn=1 and simplifying using Theorem 16(i), we arrive at
(86)U3-6U2+7U-3=0,
where
(87)U=I7,25+I7,25-1.
Solving (86) and using Remark 17, we get
(88)U=(6+(15(9-21)/2)1/3+(15(9+21)/2)1/3)3.
Combining (87) and (88) and noting that I7,25<1, we obtain
(89)I7,25=(d--144+d2)12,
where
(90)d=12+22/3(135-1521)1/3+22/3(135+1521)1/3.
This completes the proof of (ii). Now (ii) and (iv) easily follow from (i) and (ii), respectively, and Theorem 16(ii).Theorem 31.
One has(91)(I13,nI13,9n)2+(I13,nI13,9n)-2-3{(I13,nI13,9n)+(I13,nI13,9n)-1}-{(I13,nI13,9n)+(I13,nI13,9n)-1}+3=0.Proof.
Proceeding as in the proof of Theorem29, using (5) in Lemma 10, setting q:=e-πn/13, and using the definition of Ik,n, we arrive at
(92)Q=(I13,9nI13,n)1/2,T=(I13,nI13,9n)1/2.
Employing (92) in (28), we complete the proof.Corollary 32.
One has(93)(i)I13,3=(1+13--2+213)4,(ii)I13,9=(2+3-3+43)2,(iii)I13,1/3=(1+13+-2+213)4,(iv)I13,1/9=(2+3+3+43)2.Proof.
Settingn=1/3 in Theorem 31 and simplifying using Theorem 16(ii), we obtain
(94)V2-3V-1=0,
where
(95)V=I13,32+I13,3-2.
Solving (94) and using Remark 17, we get
(96)V=3+132.
Combining (95) and (96) and noting that I13,3<1, we obtain
(97)I13,3=(1+13--2+213)4.
So we complete the proof of (i).
Again settingn=1 and using Theorem 16(i), we obtain
(98)J2-4J+1=0,
where
(99)J=I13,9+I13,9-1.
Solving (98) and using Remark 17, we get
(100)J=2+3.
Combing (99) and (100) and noting that I13,9<1, we deduce that
(101)I13,9=(2+3-3+43)2.
So the proofs of (ii) is complete. Now the proof of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).Theorem 33.
One has(102)(I13,25nI13,n)3(I13,25nI13,n)-3-5{(I13,25nI13,n)+(I13,25nI13,n)-1+2}×{(I13,25nI13,n)+(I13,25nI13,n)-1+2}-{(I13,25nI13,n)2+(I13,25nI13,n)-2}=0.Proof.
Using (5) in Lemma 11, setting q:=e-πn/13, and using the definition of Ik,n, we arrive at
(103)Q=(I13,25nI13,n)1/2,T=(I13,nI13,25n)1/2.
Employing (103) in (29), we complete the proof.Corollary 34.
One has(104)(i)I13,5=(3+65-58+665)2,(ii)I13,25=(c--36+c2)6,(iii)I13,1/5=(3+65+58+665)2,(iv)I13,1/25=(c+-36+c2)6,
where c=6+(1080-1539)1/3+(1080+1539)1/3.Proof.
Settingn=1/5 and simplifying using Theorem 16(ii), we arrive at
(105)L3-23L-42=0,
where
(106)L=I13,52+I13,5-2.
Solving (105) and using the fact in Remark 17, we obtain
(107)L=3+652.
Employing (107) in (106), solving the resulting equation, and noting that I13,5<1, we obtain
(108)I13,5=(3+65-58+665)2.
This completes the proof of (i).
To prove (ii), settingn=1 and simplifying using Theorem 16(i), we arrive at
(109)A3-6A2-23A-18=0,
where
(110)A=I13,25+I13,25-1.
Solving (109) and using the fact in Remark 17, we obtain
(111)A=(6+(1080-1539)1/3+(1080+1539)1/3)3.
Employing (111) and (110), solving the resulting equation, and noting that I13,25<1, we obtain
(112)I13,25=(c--36+c2)6,
where c=6+(1080-1539)1/3+(1080+1539)1/3.
This completes the proof of (ii). Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem16(ii).Theorem 35.
One has(113)(I7,nI7,9n)+(I7,nI7,9n)-1-{(I7,9nI7,n)2+(I7,9nI7,n)-2}+3=0.Proof.
Using (5) in Lemma 12, setting q:=e-πn/7, and using the definition of Ik,n, we arrive at
(114)Q=(I7,9nI7,n)1/2,R=(I7,nI7,9n)1/2.
Employing (114) in (30), we complete the proof.Corollary 36.
One has(115)(i)I7,3=(5-212)1/4,(ii)I7,9=(1+21-6+221)4,(iii)I7,1/3=(5+212)1/4,(iv)I7,1/9=(1+21+6+221)4.Proof.
Settingn=1/3 and simplifying using Theorem 16(ii), we arrive at
(116)I7,34+I7,3-4-5=0.
Solving (116) and noting the fact in Remark 17, we obtain
(117)I7,3=(5-212)1/4.
This completes the proof of (i).
To prove (ii), settingn=1 and simplifying using Theorem 16(i), we arrive at
(118)D2-D-5=0,
where
(119)D=I7,9+I7,9-1.
Solving (118) and using the fact in Remark 17, we obtain
(120)D=(1+212).
Employing (120) in (119), solving the resulting equation, and noting that I7,9<1, we deduce that
(121)I7,9=(1+21-6+221)4.
This completes the proof of (ii). Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and Theorem 16(ii).


## 5. General Theorems and Explicit Evaluations ofGnkGn/k

In this section we evaluate some explicit values of the productGnkGn/k by establishing some general theorems and employing the values of Ik,n obtained in Section 4. We recall from Theorem 22(ii) that G1/n=Gn for ready references in this section.Theorem 37.
One has(122)(i)(I3,nI3,4n)2+(I3,nI3,4n)-2+(Gn/3G4n/3G9n/3G36n/3)-1(i)-2(Gn/3G4n/3G9n/3G36n/3)=0,(ii)I3,n6+I3,n-6+22{(Gn/3G3n)-3-(Gn/3G3n)3}=0.Proof.
To prove (i), using (5) in Lemma 13, setting q:=e-πn/3, and employing the definitions of Ik,n and Gn, we obtain
(123)R=q1/4χ(q)χ(q2)χ(q3)χ(q6)=(I3,nI3,4n)1/2,P=(Gn/3G4n/3G9n/3G36n/3)-1/2.
Employing (123) in (31), we complete the proof. (ii) follows similarly from Lemma 5 and the definition of Ik,n and Gn with q:=e-πn/3.Corollary 38.
One has(124)(i)G6G3/2=1+32,(ii)G12G4/3=2-13/6(6+36+-2+36),(iii)G39G13/3=21/6(3+13)1/3.Proof.
Settingn=1/2 in Theorem 37(i) and simplifying using Theorem 16(ii) and the result G1/n=Gn, we obtain
(125)2(G6G3/2)2-(G6G3/2)-2-2=0.
Solving (125) and noting that G6G3/2>1, we complete the proof of (i).
To prove (ii), settingn=1 in Theorem 37(i); using Theorem 16(i), and noting that G1/n=Gn, we obtain
(126)I3,42+I3,4-2+(G32G4/3G12)-1-2(G32G4/3G12)=0.
Employing (53) in (126), solving the resulting equation, and noting that G32G4/3G12>1, we obtain
(127)G32G4/3G12=(6+36+-2+36)4.
Using the value G3=21/12 from [5, p. 189] in (127), we complete the proof of (ii).
To prove (iii), settingn=13 in Theorem 37(ii), we obtain
(128)I3,136+I3,13-6+22{(G3/13G39)-2-(G3/13G39)2}=0.
Cubing (96) and then employing in (128) and solving the resulting equation, we complete the proof.Theorem 39.
One has(129)(i)(I5,nI5,4n)3+(I5,nI5,4n)-3+5{(I5,nI5,4n)+(I5,nI5,4n)-1}(i)=4(Gn/5G4n/5G25n/5G100n/5)2(i)=-(Gn/5G4n/5G25n/5G100n/5)-2,(ii)I5,n3+I5,n-3+2{(Gn/5G5n)-2-(Gn/5G5n)2}=0.Proof.
Using (5) in Lemma 14, setting q:=e-πn/5, and employing the definitions of Ik,n and Gn, we obtain
(130)R=q1/2χ(q)χ(q2)χ(q5)χ(q10)=(I5,nI5,4n)1/2,P=(Gn/5G4n/5G25n/5G100n/5)-1/2.
Employing (130) in (32), we complete the proof of (i). Similarly, (ii) follows from Lemma 6 and the definition of Ik,n and Gn with q:=e-πn/5.Corollary 40.
One has(131)(i)G10G5/2=(3+102)1/4,(ii)G20G5/4=12-3(11+55)1/4+(11+55)3/4+M,
where 𝔐 denotes 16+11+55(-3+11+55)2.Proof.
Settingn=1/2 in Theorem 39(i) and simplifying using Theorem 16(ii) and the result G1/n=Gn, we obtain
(132)4(G10G5/2)4-(G10G5/2)-4-12=0.
Solving (132) and noting that G10G5/2>1, we complete the proof of (i).
For proof of (ii), settingn=1 in Theorem 39(ii) and simplifying using Theorem 18(i) and the result G1/n=Gn, we obtain
(133)I5,43+I5,4-3+2{(G4/5G20)-2-(G4/5G20)2}=0.
Employing the value of I5,4+I5,4-1 from (63) in (133), solving the resulting equation, and noting that G4/5G20>1, we get
(134)G4/5G20=12-3(11+55)1/4+(11+55)3/4+M.
So the proof is complete.Theorem 41.
One has(135)22{(I2,nI2,49n)3/2+(I2,nI2,49n)-3/2}=4(Gn/2G4n/2G47n/2G196n/2)1/2=-2(Gn/2G4n/2G47n/2G196n/2)-1/2=+4(Gn/2G4n/2G47n/2G196n/2)-3/2=-(Gn/2G4n/2G47n/2G196n/2)-5/2.Proof.
Using (5) in Lemma 15, setting q:=e-πn/2, and employing the definitions of Ik,n and Gn, we obtain,
(136)R=q1/3χ(q)χ(q7)χ(q2)χ(q14)=(I2,nI2,49n)1/2,P=(Gn/2G4n/2G47n/2G196n/2)-1/2.
Employing (136) in (33), we complete the proof.Corollary 42.
One has(137)G14G7/2=(2+4+82)4.Proof.
Settingn=1/7 and simplifying using Theorem 16(ii) and the result G1/n=Gn, we get
(138)4(G14G7/2)6-42(G14G7/2)5-2(G14G7/2)4+4(G14G7/2)2-1=0.
Solving (138) and noting that G14G7/2>1, we complete the proof.Theorem 43.
One has(139)I7,n4+I7,n-4+7=22{(G7/nG7n)-3+(G7/nG7n)3}.Proof.
Using (5) in Lemma 7, setting q:=e-πn/7, and employing definitions of Ik,n and Gn, we arrive at
(140)P=(Gn/7G7n)-3,Q=(q1/4χ(q)χ(q7))4=I7,n4.
Using (140) in (24), we complete the proof.Corollary 44.
One has(141)(i)G7/4G28=(2+2)2,(ii)G7/3G21=2-1/6(3+7)1/3,(iii)G7/9G63=21/6(5+21)1/3.Proof.
Settingn=4 in Theorem 43, we get
(142)I7,44+I7,4-4+7=22{(G7/4G28)3+(G7/4G28)-3}.
Squaring (74) and simplifying, we obtain
(143)I7,44+I7,4-4=-56+452.
Employing (143) in (142), solving the resulting equation, and noting that G7/4G28>1, we complete the proof of (i).
To prove (ii), we setn=3 in Theorem 43, and employing (116), we obatin
(144)(G21G7/3)3+(G21G7/3)-3-32=0.
Solving (144) and noting that G21G7/3>1, we complete the proof.
To prove (iii), settingn=9 in Theorem 43, we get
(145)(I7,94+I7,9-4)+7=22{(G7/9G63)3+(G7/9G63)-3}.
Squaring (120) twice and simplifying, we obtain
(146)I7,94+I7,9-4=(31+721)2.
Employing (146) in (145), solving the resulting equation, and noting that G7/9G63>1, we complete the proof.Theorem 45.
One has(147)(I13,n)7+(I13,n)-7+13{(I13,n)5+(I13,n)-5}+52{(I13,n)3+(I13,n)-3}+78{(I13,n)+(I13,n)-1}-8{(Gn/13G13n)6-(GnG13n)-6}=0.Proof.
Using (5) in Lemma 8, setting q:=e-πn/13, and employing definitions of Ik,n and Gn, we arrive at
(148)P=(Gn/713G13n)-1,Q=(q1/2χ(q)χ(q13))4=I13,n.
Using (148) in (25), we complete the proof.Corollary 46.
One has(149)G117G13/9=(450+2603+405299+234003)1/6.Proof.
Settingn=9 in Theorem 45, we get
(150)(I13,9)7+(I13,9)-7+13{(I13,9)5+(I13,9)-5}+52{(I13,9)3+(I13,9)-3}+78{(I13,9)+(I13,9)-1}-8{(G9/13G117)6-(G9/13G117)-6}=0.
Employing (100) in (150) and simplifying, we obtain
(151)(G9/13G117)12-20(45+263)(G9/13G117)6+1=0.
Solving the resulting equation (151) and noting that G9/13G117>1, we complete the proof.


## 6. New Values of Class InvariantGn

In this section we find some new values of Ramanujan's class invariantGn by using explicit values of Ik,n and GnkGn/k evaluated in Sections 3 and 4, respectively. For ready references in this section, we recall from Theorem 22 that Ik,n=Gn/k/Gnk. We also recall from Theorem 22(ii) that G1/n=Gn.Theorem 47.
One has(152)(i)G3/2=2-1/4(1+3)1/4(i)G3/2=×(-44+273-3458-2643)1/24,(ii)G6=2-1/4(1+3)1/4(ii)G6=×(-44+273+3458-2643)1/24.Proof.
From Corollary24(i), we have
(153)I3,2=(G3/2G6)=(-44+273-3458-2643)1/12.
Also, from Corollary 38(i), we have
(154)G6G3/2=(1+32)1/2.
Multiplying (153) and (154) and simplifying, we complete the proof of (i). Dividing (154) by (153) and simplifying, we complete the proof of (ii).The proofs of the Theorems48–56 are identical to the proof of Theorem 47. So we give the references of the required results only.Theorem 48.
One has(155)(i)G4/3=2-4/3(-2+36--6+36)1/4(i)G4/3=×(6+36+-2+36)1/2,(ii)G12=2-4/3(6+36+-2+36)1/2(ii)G12=×(-2+36+-6+36)1/4.Proof.
We use Corollary24(ii) and Corollary 38(ii) and proceed as in Theorem 47.Theorem 49.
One has(156)(i)G5/2=2-1/8(3+10)1/8(i)G5/2=×(-14+510-445-14010)1/12,(ii)G10=2-1/8(3+10)1/8(ii)G10=×(-14+510+445-14010)1/12.Proof.
We employ Corollary26(i) and Corollary 40(i) and proceed as in Theorem 47.Theorem 50.
One has(157)(i)G5/4=12((11+55)1/4--4+11+55)1/2(i)G5/4=×(16+11+55(-3+11+55)2-3(11+55)1/4+(11+55)3/4(i)G5/4=××+16+11+55(-3+11+55)2)1/4(ii)G20=((11+55)1/4--4+11+55)-1/2(ii)G20=×(16+11+55(-3+11+55)2-3(11+55)1/4+(11+55)3/4(ii)G20=××+16+11+55(-3+11+55)2)1/4.Proof.
We employ Corollary26(ii) and Corollary 40(ii) and proceed as in Theorem 47.Theorem 51.
One has(158)(i)G7/2=2-9/8(2+4+82)1/2(i)G7/2=×(-4+(-8-52+163+1162)2-8-52+163+1162(i)G7/2=××--4+(-8-52+163+1162)2)1/8,(ii)G14=2-9/8(2+4+82)1/2(ii)G14=×(-4+(-8-52+163+1162)2-8-52+163+1162(ii)G14××=+-4+(-8-52+163+1162)2)1/8.Proof.
We employ Corollary28(i) and Corollary 42(i) and proceed as in Theorem 47.Theorem 52.
One has(159)(i)G7/4=2-3/4(2+2)1/2(3-6+52--58+452)1/4,(ii)G28=2-3/4(2+2)1/2(3-6+52+-58+452)1/4.Proof.
We employ Corollary28(ii) and Corollary 44(i) and proceed as in Theorem 47.Theorem 53.
One has(160)(i)G7/3=2-5/24(3+7)1/6(5-21)1/8,(ii)G21=2-5/24(3+7)1/6(5+21)1/8.
The values G7/3 and G21 can also be found in [5, 6].Proof.
We employ Corollary36(i) and Corollary 44(ii) and proceed as in Theorem 47.Theorem 54.
One has(161)(i)G9/7=2-11/12(5+21)1/6(1+21-6+221)1/2,(ii)G63=2-11/12(5+21)1/6(1+21+6+221)1/2.
The value G63 can also be found in [5, page 192].Proof.
We employ Corollary36(ii) and Corollary 44(ii) and proceed as in Theorem 47.Theorem 55.
One has(162)(i)G13/3=2-11/12(3+13)1/6(1+13--2+213)1/2,(ii)G39=2-11/12(3+13)1/6(1+13+-2+213)1/2.
The values G13/3 and G39 can also be found in [5, 6].Proof.
We use Corollary32(i) and Corollary 38(iii) and proceed as in Theorem 47.Theorem 56.
One has(163)(i)G13/9=2-1/2(450+2603+405299+234003)1/12(i)G13/9=×(2+3-3+43)1/2,(ii)G117=2-1/2(450+2603+405299+234003)1/12(ii)G117=×(2+3+3+43)1/2.
The values G39 and G117 can also be found in [5, page 193].Proof.
We employ Corollary32(ii) and Corollary 46 and proceed as in Theorem 47.



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