# Some New Explicit Values of Parametersλn and μn of Quotients of Eta-Function

Authors: Nipen Saikia
Journal: ISRN Computational Mathematics
 (2013)

---

## Abstract

We find some new explicit values of the parametersλn and μn of quotients of eta-function by using Ramanujan's class invariants.


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## Body

## 1. Introduction

Forq=e2πiz and Im(z)>0, the Dedekind eta-function η(z) and Ramanujan’s function f(-q) are defined by
(1)f(-q):=(q;q)∞=:q-1/24η(z),
where (a;q)∞=∏n=0∞‍(1-aqn).In page 212 of his lost notebook [1], Ramanujan defined
(2)λn:=f6(q)22qf6(q3),q:=e-πn/3
and provided a list of eleven recorded and ten unrecorded values of λn for positive integers n. All 21 values of λn and many more were established by Berndt et al. [2] by using the modular j-invariant, modular equations, Kronecker’s limit formula, and the explicit Shimura reciprocity law. An account of this can also be found in Chapter 9 of [3]. Closely related to λn is the parameter μn introduced by Ramanathan [4] and is defined as
(3)μn:=133f6(-q)qf6(-q3);q=e-2πn/3.
Yi [5, 6] also found several values of parameters λn and μn by finding the explicit values of her parameters rk,n and rk,n′, defined by
(4)rk,n:=f(-q)k1/4q(k-1)/24f(-qk),q=e-2πn/k,rk,n′:=f(q)k1/4q(k-1)/24f(qk),q=e-πn/k,
where n and k are positive real numbers. In fact, r3,n′6=λn, and r3,n6=μn. Baruah and Saikia [7, 8] also evaluated several new values of λn and μn. A generalization of λn has been studied by Naika et al. [9]. Here we give a list of the values of n for which λn and μn were evaluated in [1, 2, 5–8] without referring them specifically to avoid repetitions. The values of n for which λn were evaluated in the literature are n = 1, 3, 5, 7, 9, 11, 13, 15, 5/3, 17, 25, 33, 41, 49, 57, 65, 81, 89, 73, 97, 121, 169, 193, 217, 241, 265, 289, 361, 11/3, and 19/3, and the values of values of n for which μn were evaluated in literature are n = 1, 2, 3, 4, 5, 6, 3/2, 7, 8, 9, 10, 5/2, 11, 13, 14, 7/2, 15, 5/3, 17, 19, 20, 5/4, 22, 11/2, 25, 26, 13/2, 44, 11/4, and 49. We also note from [2, page 281] and [7, page 43, Theorem 4.2] that
(5)λ1/n=1λn,μ1/n=1μn,
respectively.In this paper we find further new values ofλn and μn by using Ramanujan’s class invariants which are defined as
(6)Gn=2-1/4q-1/24χ(q),gn=2-1/4q-1/24χ(-q);q:=e-πn,
where χ(q)=(-q;q2)∞. In particular, we evaluate values of λn for n = 23, 31, 47, 59, 71, 2, 4, 6, 10, 5/2, 14, and 7/2 and new values of μn for n = 23, 31, 47, 59, and 71. It worth to mention here that for the first time in this paper explicit values of λn for some even values of n are evaluated. Previously, Berndt et al. [2] calculated explicit values of λ1, λ3, and μ1 by using Ramanujan’s class invariants. An account of Ramanujan’s class invariants can be found in Berndt’s book [10]. For further references on Ramanujan’s class invariants refer see [11–14].The explicit values of the functionsλn and μn can be applied to find explicit values of Ramanujan’s cubic continued fraction G(q) defined by
(7)G(q):=q1/31+q+q21+q2+q41+q3+q61+⋯,|q|<1.
We refer to [7] for details. To this end we define a general parameter Ik,n [14, page 2, equation (10)] for all positive real numbers n and k and connected with parameters λn, μn, and Ramanujan’s class invariants as
(8)Ik,n:=χ(q)q(-k+1)/24χ(qk);q=e-πn/k.
Saikia [14] studied several properties of the parameter Ik,n and evaluated many explicit values of Ik,n by using Ramanujan’s modular equations. From [14, page 4, Theorem 16(iii)], we also note here that Ik,n=In,k.In Section2, we record some preliminary results, and in Section 3 we evaluate new values of the parameters λn and μn.


## 2. Preliminaries

Lemma 1 (see [3, page 203, Theorem 9.3.1]).
One has(9)λnμn=(Gn/3G3n)6.Lemma 2 (see [14, page 4, Theorem 22(i)]).
One has(10)Ik,n=Gn/kGkn.Lemma 3.
One has(11)λnμn=(Gn/3G3n)6=I3,n6.Proof.
We setk=3 in Lemma 2 and apply Lemma 1 to complete the proof.Lemma 4.
One has(12)9{(λnμn)2/3+(λnμn)-2/3}-16=(Gn/3G3n)12+(G3nGn/3)12.This readily follows from [2, page 9, equation (3.4)] or [3, page 203, equation (9.3.4)].Lemma 5.
One has(13)(Gn/3G3n)12+(G3nGn/3)12=8{(Gn/3G3n)6+(Gn/3G3n)-6}-18.Proof.
From [11, page 123, equation (4.14)], we note that
(14)(GnG9n)6+(G9nGn)6=22{(GnG9n)3-(GnG9n)-3}.
Replacing n by n/3 in (14) and squaring, we arrive at desired result.Lemma 6.
One has(15)G31/3=(39+7312)1/6(31-332)1/4,(16)G93=(39+7312)1/6(31+332)1/4,(17)G117=(33+232)1/6(3+12)1/4,(18)G59/3=(33-232)1/6(3+12)1/4,(19)(G69G23/3)6=(5+232)(33+232)3/2,(20)(G141G47/3)6=(7+472)(43+47)3/2,(21)(G213G71/3)6=(53+712)3/2(59+7712),(22)I3,2=(G2/3G6)6=(-44+273-3458-2643)1/12,(23)I3,4=(G4/3G12)6=((-2+36--6+36)2)1/2.For (15), (16), (17) and (18) see [10, page 342]. For (19), (20), and (21) see [10, page 227, equation (5.8)], [10, page 229, equation  (5.16)], and [10, page 257, equation  (6.61)]. For (22) and (23) see [14, page 5, Corollary 24(i & ii))].Lemma 7.
One has(24)g10/3g30=(19+610)1/6,(25)g2/15g6/5=(19-610)1/6,(26)g14/3g42=(5+212)1/2,(27)g2/21g6/7=(5-212)1/2,(28)g6/3g18=(2+3)1/3.For (24) to (27) see [12]. Equation (28) follows from the values of g2 and g18 from [10, page 200].Lemma 8.
One has(29)(GnG9n)2=(gng9n)3+(gng9n)6+22gng9n.Proof.
From [15, page 124], we note that
(30)χ(q)=21/6{α(1-α)q}-1/24,χ(-q)=21/6(1-α)1/12(αq)-1/24.
Employing (30) in (6) and simplifying, we deduce that
(31)1-α=(gnGn)8.
From [15, page 230, Entry 5(ii)], we note that
(32)(αβ)1/4+((1-α)(1-β))1/4=1,
where β has degree 3 over α.
Combining (31) and (32) and simplifying, we deduce that
(33){(Gn8-gn8)(G9n8-g9n8)}1/4+(gng9n)2=(GnG9n)2.
From [10, page 187, Entry 2.2], we note that
(34)(gnGn)8(Gn8-gn8)=14.
Employing (34) in (33), and simplifying, we find that
(35)2(GnG9n)4(gng9n)2-2(GnG9n)2(gng9n)4-1=0.
Solving (35) for (GnG9n)2, we arrive at the desired result.Lemma 9.
One has(36)(G10/3G30)2=(3+10+3(5+2))(10-3)1/32,(37)(G2/15G6/5)2=(-3+10+3(5-2))(10+3)1/32,(38)(G14/3G42)2=(6+33+27+21)(7-3)4,(39)(G6/7G2/21)2=(6-21-33+27)(7+3)4,(40)(G6/3G18)2=(1+2+3+6)(3-2)1/32.Proof.
We setn = 10/3, 2/15, 14/3, 6/7, and 6/3 in Lemma 8 and employ the corresponding values of gng9n from Lemma 7 to complete the proof.


## 3. New Values ofμn and λn

The section is devoted to evaluate new values ofμn and λn.Theorem 10.
One has(41)λ23=2-3/8(24-53)1/4(25-369)3/8×(167+963+4347+20043)3/4,μ23=2-3/8(24+53)1/4(25+369)3/8×(167+963+4347+20043)3/4.Proof.
Employing (19) in Lemma 3 with n=23, we have
(42)μ23λ23=(G69G23/3)6=(5+232)(33+232)3/2.
From (42), we deduce that
(43)(G69G23/3)12+(G23/3G69)12=2990+17283.
Setting n=23 in Lemma 4, employing (43), and solving the resulting equation, we find that
(44)μ23λ23=(167+963+4347+20043)3/2.
Multiplying (42) and (44) and simplifying, we easily calculate the value of μ23, and dividing (44) by (42) and simplifying we evaluate the value of μ23.Remark 11.
The values ofλ1/23 and μ1/23 also follow from (5) and the values of λ23 and μ23, respectively.Theorem 12.
One has(45)λ47=(48-747)1/4(7563-19147)1/4×(6983+40323+12677315+3910483)3/4,μ47=(48+747)1/4(7563+19147)1/4×(6983+40323+12677315+3910483)3/4.Proof.
Employing (20) in Lemma 3 with n=47, we have
(46)μ47λ47=(G141G47/3)6=(7+472)(43+47)3/2.
From (46), we deduce that
(47)(G141G47/3)12+(G47/3G141)12=125854+725763.
Setting n=47 in Lemma 4, employing (47), and solving the resulting equation, we obtain
(48)μ47λ47=(6983+40323+12677315+3910483)3/2.
Multiplying (46) and (48) and simplifying, we easily calculate the value of μ47, and dividing (48) by (46) and simplifying we evaluate the value of μ47.Remark 13.
The values ofλ1/47 and μ1/47 also follow from (5) and the values of λ47 and μ47, respectively.Theorem 14.
One has(49)
λ
71
=
2
-
1
/
4
(
59
-
7
71
)
1
/
2
(
180
3
-
37
71
)
1
/
4
×
(
120551
+
69600
3
+
20
72662559
+
41951748
3
)
3
/
4
,
μ
71
=
2
-
1
/
4
(
59
+
7
71
)
1
/
2
(
180
3
+
37
71
)
1
/
4
×
(
120551
+
69600
3
+
20
72662559
+
41951748
3
)
3
/
4
.Proof.
Employing (21) in Lemma 3 with n=71, we have
(50)μ71λ71=(G213G71/3)6=(53+712)3/2(59+7712).
From (50), we deduce that
(51)(G213G71/3)12+(G71/3G213)12=2169902+12528003.
Setting n=23 in Lemma 4, employing (51), and solving the resulting equation, we obtain
(52)μ71λ71=(120551+696003+2072662559+419517483)3/2.
Combining (50) and (52) we easily evaluate the values of μ71 and μ71.Remark 15.
The values ofλ1/71 and μ1/71 also follow from (5) and the values of λ71 and μ71, respectively.Theorem 16.
One has(53)λ31=2-3/2(1351+7803)3/4(31-33)1/4,μ31=2-3/2(1351+7803)3/4(31+33)1/4.Proof.
Combining (15), (16) and Lemma 3, we deduce that
(54)μ31λ31=(G93G31/3)6=(31+33)38.
From (54), we find that
(55)(G93G31/3)12+(G31/3G93)12=24302.
Setting n=31 in Lemma 4, employing (55), and solving the resulting equation, we obtain
(56)λ31μ31=(1351+7803)3/2.
Combining (54) and (56) we easily evaluate the values of λ31 and μ31.Remark 17.
The values ofλ1/31 and μ1/31 can easily be calculated by employing the values of λ31 and μ31, respectively, in (5).Theorem 18.
One has(57)λ59=2-1/2(359-23)(62423+4692177)3/4,μ59=2-1/2(359+23)(62423+4692177)3/4.Proof.
Combining (17), (18) and Lemma 3, we deduce that
(58)μ59λ59=(G177G59/3)6=(359+23)22.
From (58), we find that
(59)(G177G59/3)12+(G59/3G177)12=1123598.
Setting n=59 in Lemma 4, employing (59), and solving the resulting equation, we obtain
(60)λ59μ59=(62423+4692177)3/2.
Combining (58) and (60) we easily calculate the values of λ59 and μ59.Remark 19.
The values ofλ1/59 and μ1/59 can easily be calculated by employing the values of λ59 and μ59, respectively, in (5).Next, we evaluate some values ofλn and μn for even values of n. The values of λn are new. The values of μn are classical, but the method of evaluations is new.Theorem 20.
One has(61)λ2=(-4+33+42-243)3/4×(-44+273-3458-2643)1/4,μ2=(-4+33+42-243)3/4×(-44+273+3458-2643)1/4.Proof.
Combining (22) and Lemma 3 with n=2, we note that
(62)λ2μ2=(G2/3G6)6=(-44+273-3458-2643)1/2.
From (62), we deduce that
(63)(G2/3G6)12+(G6G2/3)12=-88+543.
Setting n=2 in Lemma 4, employing (63), and solving the resulting equation, we obtain
(64)λ2μ2=(-4+33+42-243)3/2.
Combining (62) and (64) we easily evaluate the values of λ2 and μ2.Remark 21.
The values ofλ1/2 and μ1/2 can easily be calculated by employing the values of λ2 and μ2, respectively, in (5).Theorem 22.
One has(65)λ4=2-9/4(-76+336+12306-50166)3/4×(-2+36--6+36)3/2,μ4=2-9/4(-76+336+12306-50166)3/4×(-2+36+-6+36)3/2.Proof.
Combining (23) and Lemma 3 with n=2, we note that
(66)λ4μ4=(G4/3G12)6=(-2+36--6+362)3.
From (66), we deduce that
(67)(G4/3G12)12+(G12G4/3)12=-700+2976.
Setting n=2 in Lemma 4, employing (67), and solving the resulting equation, we obtain
(68)λ4μ4=(-76+336+12306-501662)3/2.
Combining (66) and (68) we easily evaluate the values of λ4 and μ4.Remark 23.
The values ofλ1/4 and μ1/4 can easily be calculated by employing the values of λ4 and μ4, respectively, in (5).Theorem 24.
One has(69)μ6=133(s+t)1/4(u+v)3/4,λ6=133(s+t)-1/4(u+v)3/4,
where
(70)s=-900+6502+5253-3646,t=5131074-926642-756563+535086,u=-892+6502+5253-3646,v=3262434-23062002-18830003+13318766.Proof.
Employing (40) in Lemma 5 with n=6, solving the resulting equation, and noting the result λn/μn=(Gn/3/G3n)6 from Lemma 3, we obtain
(71)(μ6λ6)2=s+t,
where
(72)s=-900+6502+5253-3646,t=5131074-926642-756563+535086.
Employing (71) in Lemma 4 and solving the resulting equation, we obtain
(73)(λ6μ6)2/3=u+v,
where
(74)u=-892+6502+5253-3646,v=3262434-23062002-18830003+13318766.
Combining (71) and (73) we calculate the values of μ6 and λ6.Remark 25.
The values ofλ1/6 and μ1/6 can easily be calculated by employing the values of λ6 and μ6, respectively, in (5).Theorem 26.
One has(75)μ10=(a+b)1/4(c+d)3/4,λ10=(c+d)3/4(a+b)-1/4,
where(76)
a
=
-
11960
+
4914
6
-
3780
10
+
3105
15
,
b
=
3
30
(
2131205
-
870044
6
+
673946
10
-
550264
15
)
,
c
=
-
1328
+
546
6
-
420
10
+
345
15
,
d
=
6
(
1183609
-
483196
6
+
374290
10
-
305600
15
)
.Proof.
Employing (36) in in Lemma 5 with n=10, solving the resulting equation, and noting the result λn/μn=(Gn/3/G3n)6 from Lemma 3, we obtain
(77)(μ10λ10)2=(G30G10/3)12=a+b,
where(78)a=-11960+49146-378010+310515,b=330(2131205-8700446+67394610-55026415).Employing (77) in Lemma 4 and solving the resulting equation, we obtain
(79)(λ10μ10)2/3=c+d,
where(80)c=-1328+5466-42010+34515,d=6(1183609-4831966+37429010-30560015).Combining (77) and (79) we calculate the values of μ10 and λ10.Remark 27.
The values ofλ1/10 and μ1/10 can easily be calculated by employing the values of λ10 and μ10, respectively, in (5).Theorem 28.
One has(81)μ5/2=(p+q)1/4(r+s)3/4,λ5/2=(p+q)-1/4(r+s)3/4,
where(82)
p
=
-
11960
+
4914
6
+
3780
10
-
3105
15
,
q
=
3
30
(
2131205
-
870044
6
-
673946
10
+
550264
15
)
,
r
=
-
1328
+
546
6
+
420
10
-
345
15
,
s
=
6
(
1183609
-
483196
6
-
374290
10
+
305600
15
)
.Proof.
Employing (37) in in Lemma 5 with n=5/2, solving the resulting equation, and noting the result λn/μn=(Gn/3/G3n)6 from Lemma 3, we obtain
(83)(μ5/2λ5/2)2=(G15/2G5/6)12=p+q,
where(84)p=-11960+49146+378010-310515,q=330(2131205-8700446-67394610+55026415).Employing (83) in Lemma 4 and solving the resulting equation, we obtain
(85)(λ5/2μ5/2)2/3=r+s,
where
(86)r=-1328+5466+42010-34515,s=6(1183609-4831966-37429010+30560015).
Combining (83) and (85) we calculate the values of μ5/2 and λ5/2.Remark 29.
The values ofλ2/5 and μ2/5 can easily be calculated by employing the values of λ5/2 and μ5/2, respectively, in (5).Theorem 30.
One has(87)μ14=(e+f)1/4(g+h)3/4,λ14=(e+f)-1/4(g+h)3/4,
where(88)
e
=
-
98000
+
56700
3
+
4125
7
-
2376
21
,
f
=
15
171108062
-
98789040
3
-
64672608
7
+
37338840
21
,
g
=
-
10888
+
6300
3
+
4125
7
-
2376
21
,
h
=
6
(
79213469
-
45733800
3
-
29939800
7
+
17285796
21
)
.Proof.
Employing (38) in in Lemma 5 with n=14, solving the resulting equation, and noting the result λn/μn=(Gn/3/G3n)6 from Lemma 3, we obtain
(89)(μ14λ14)2=(G42G14/3)12=e+f,
where(90)e=-98000+567003+41257-237621,f=15171108062-987890403-646726087+3733884021.Employing (89) in Lemma 4 and solving the resulting equation, we obtain
(91)(λ14μ14)2/3=g+h,
where(92)g=-10888+63003+41257-237621,h=6(79213469-457338003-299398007+1728579621).Combining (89) and (91) we calculate the values of μ14 and λ14.Remark 31.
The values ofλ1/14 and μ1/14 can easily be calculated by employing the values of λ14 and μ14, respectively, in (5).Theorem 32.
One has(93)μ7/2=(l+m)1/4(u+v)3/4,λ7/2=(l+m)-1/4(u+v)3/4,
where(94)
l
=
-
98000
-
56700
3
+
4125
7
+
2376
21
,
m
=
15
171108062
+
98789040
3
-
64672608
7
-
37338840
21
,
u
=
-
10888
-
6300
3
+
4125
7
+
2376
21
,
v
=
6
(
79213469
+
45733800
3
-
29939800
7
+
-
17285796
21
)
.Proof.
Employing (39) in Lemma 5 with n=7/2, solving the resulting equation, and noting the result λn/μn=(Gn/3/G3n)6 from Lemma 3, we obtain
(95)(μ7/2λ7/2)2=(G21/2G7/6)12=l+m,
where(96)l=-98000-567003+41257+237621,m=15171108062+987890403-646726087-3733884021.Employing (95) in Lemma 4 and solving the resulting equation, we obtain
(97)(λ7/2μ7/2)2/3=u+v,
where(98)u=-10888-63003+41257+237621,v=6(79213469+457338003-299398007+-1728579621).Combining (95) and (97) we calculate the values of μ7/2 and λ7/2.Remark 33.
The values ofλ2/7 and μ2/7 can easily be calculated by employing the values of λ7/2 and μ7/2, respectively, in (5).Remark 34.
(i) From (36) and (77) we can easily deduce the values of class invariants G10/3 and G30. (ii) Combining (37) and (83) we can calculate the values of class invariants G15/2 and G6/5. (iii) The values of class invariants G14/3 and G42 can be calculated from (38) and (89). (iv) Combining (39) and (95) we can obtain the values of class invariants G21/2 and G7/6. (v) Combining (40), (71), and Lemma 3 we can calculate the values of class invariants G2 and G18.



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*Source: 435261-2013-02-18.xml*