international journal of pure and applied mathematics ...−2p2q4)(25 −10p2 −10q2 +p4...

International Journal of Pure and Applied Mathematics

Volume 74 No. 4 2012, 413-435ISSN: 1311-8080 (printed version)url: http://www.ijpam.eu

PAijpam.eu

NEW MODULAR RELATIONS FOR

RAMANUJAN’S PARAMETER µ(q)

M.S. Mahadeva Naika1 §, B.N. Dharmendra2, S. Chandankumar3

1,2,3Department of MathematicsBangalore University

Central College CampusBangalore, 560 001, INDIA

Abstract: In his ‘lost’ notebook, S. Ramanujan introduced the parameterµ(q) := R(q)R(q4) related to the Rogers - Ramanujan continued fraction R(q).In this paper, we establish some new P − Q modular equations of degree 5.We establish some general formulas for the explicit evaluations of the ratiosof Ramanujan’s theta function ϕ. We obtain several new modular relationsconnecting µ(q) with µ(qn) for different positive integer n > 1, reciprocitytheorems and also compute several new explicit evaluations.

AMS Subject Classification: 33D20, 11F55, 11S23Key Words: continued fraction, modular equation, reciprocity theorem,theta function

1. Introduction

The Rogers-Ramanujan continued fraction is defined by

R(q) :=q1/5

1 +

q

1 +

q2

1 +

q3

1 +· · · , |q| < 1, (1)

was first studied by L.J. Rogers [19]. Later, this continued fraction was redis-

Received: August 18, 2011 c© 2012 Academic Publications, Ltd.url: www.acadpubl.eu

§Correspondence author

414 M.S.M. Naika, B.N. Dharmendra, S. Chandankumar

covered by S. Ramanujan and recorded may interesting results involving R(q).For more details on R(q) one can see [2], [3], [6] and [20].

In his ‘lost’ notebook Ramanujan [18], introduced the parameters µ(q) andκ(q) := R(q)R2(q2) which are related to Rogers-Ramanujan continued fraction.Ramanujan stated several interesting identities involving the parameters µ(q)and κ(q). These results were studied in detail by S. -Y. Kang [9]. A similarwork has been carried out by M. S. Mahadeva Naika [10], for Ramanujan cubiccontinued fraction. Recently, C. Gugg [8] established certain identities of Ra-manujan using the parameter κ(q). S. Cooper [7], also systematically studiedseveral results involving the parameter κ(q).

In Chapter 16, of his second notebook [17] Ramanujan defined his thetafunction as

f(a, b) : =∞∑

n=−∞an(n+1)/2bn(n−1)/2, |ab| < 1,

= (−a; ab)∞(−b; ab)∞(ab; ab)∞.

(2)

Three special cases of f(a, b) are as follows:

ϕ(q) := f(q, q) =

∞∑

n=−∞qn

2

=(−q;−q)∞(q;−q)∞

, (3)

ψ(q) := f(q, q3) =∞∑

n=0

qn(n+1)/2 =(q2; q2)∞(q; q2)∞

, (4)

f(−q) :=∞∑

n=−∞qn(3n−1)/2 = (q; q)∞, (5)

where

(a; q)∞ :=∞∏

n=0

(1− aqn), |q| < 1.

Now we define a modular equation in brief. The ordinary hypergeometricseries

2F1(a, b; c;x) is defined by

2F1(a, b; c;x) :=∞∑

n=0

(a)n(b)n(c)nn!

xn,

NEW MODULAR RELATIONS FOR... 415

where (a)0 = 1, (a)n = a(a+1)(a+2) · · · (a+ n− 1) for any positive integer n,and | x |< 1.Let

z := z(x) := 2F1

(

1

2,1

2; 1;x

)

(6)

and

q := q(x) := exp

(

−π 2F1(12 ,

12 ; 1; 1 − x)

2F1(12 ,

12 ; 1;x)

)

, (7)

where 0 < x < 1.Let r denote a fixed natural number and assume that the following relationholds:

r2F1

(

12 ,

12 ; 1; 1 − α

)

2F1

(

12 ,

12 ; 1;α

) =2F1

(

12 ,

12 ; 1; 1 − β

)

2F1

(

12 ,

12 ; 1;β

) . (8)

Then a modular equation of degree r in the classical theory is a relation between

α and β induced by (8). We often say that β is of degree r over α andm :=z(α)

z(β)is called the multiplier. We also use the notations z1 := z(α) and zr := z(β) toindicate that β has degree r over α.

In [21], J. Yi introduced two parameters hk,n and h′k,n as follows:

hk,n =ϕ(e−π

√n/k)

k1/4ϕ(e−π√nk)

, (9)

h′k,n =

ϕ(−e−π√

n/k)

k1/4ϕ(−e−π√nk)

(10)

and established several properties as well as explicit evaluations of hk,n and h′k,n

for different positive rational values of n and k by using the modular equationsof Ramanujan. Mahadeva Naika and S. Chandankumar [11], Mahadeva Naika,Chandankumar and M. Manjunatha [13], Mahadeva Naika, K. Sushan Bairyand Chandankumar [15] have established several new modular equations of de-gree 2. They also established general formulas for the explicit evaluations ofh2,n and several new explicit evaluations for Ramanujan -Gollnitz-Gordon con-tinued fraction, Ramanujan-Selberg continued fraction and a continued fractionof Eisenstein. In [16], Mahadeva Naika, Bairy and Manjunatha have establishedseveral new modular equations of degree 4 and established general formulas forthe explicit evaluations of h4,n. In [14], Mahadeva Naika, Bairy and Chandanku-mar have established several new modular equations of degree 9 and established


general formulas for the explicit evaluations of h9,n and they have establishedsome explicit evaluations of Ramanujan’s cubic continued fraction.

In Section 2, we collect the identities which are useful in proving our mainresults. In Section 3, we establish some new modular equations of degree 5. InSection 4, we establish some new modular relations connecting h5,n with h5,l2nfor l = 2, 4, 5, 7 and 11. In Section 5, we establish several new modular relationsconnecting µ(q) with µ(qn). In Section 6, we establish some new reciprocitytheorems for µ(q) and obtain some new explicit evaluations of µ(q) using thevalues of h5,n.

2. Preliminary Results

In this section, we collect some identities which are useful in establishing ourmain results.

Lemma 1. (see [18, p.56], [9]) We have

f3(−q)f3(−q5) =

ψ(q)

ψ(q5)

(

ψ2(q)− 5qψ2(q5)

ψ2(q)− qψ2(q5)

)

, (11)

f3(−q2)qf3(−q10) =

ϕ(q)

ϕ(q5)

(

5ϕ2(q5)− ϕ2(q)

ϕ2(q)− ϕ2(q5)

)

, (12)

f6(−q)qf6(−q5) =

ϕ4(−q)ϕ4(−q5)

(

5ϕ2(−q5)− ϕ2(−q)ϕ2(−q)− ϕ2(−q5)

)

. (13)

Lemma 2. (see [18, p.26], [9]) Let µ := µ(q), then

ϕ(q)

ϕ(q5)=

1 + µ

1− µ. (14)

Lemma 3. Let ω := ω(q) := −µ(−q), then

ϕ(−q)ϕ(−q5) =

1 + ω

1− ω. (15)

Proof. Change q to −q in the equation (14), we arrive at the equation(15).

Lemma 4. (see [18, Entry 1.6.2(i), p.50]) We have

16qf2(−q2)f2(−q10) = (ϕ2(q)− ϕ2(q5))(5ϕ2(q5)− ϕ2(q)). (16)


Lemma 5. (see [5, Ch. 16, Entry 24 (i), p.39]) We have

ψ2(q)

ψ2(−q) =ϕ(q)

ϕ(−q) . (17)

Lemma 6. (see [5, Ch. 16, Corollary (ii), p.74]) We have

ψ(q5)ψ(q11)− q5ψ(q)ψ(q55) = ψ(−q5)ψ(−q11) + q5ψ(q)ψ(q55). (18)

Lemma 7. (see [18, p.55]) If x =f(−q)

q1

6 f(−q5)and y =

f(−q2)q

1

3 f(−q10), then

xy +5

xy=

(

x

y

)3

+(y

x

)3. (19)

Lemma 8. (see [18, p.55]) If x =f(−q)

q1

6 f(−q5)and y =

f(−q4)q

2

3 f(−q20), then

(xy)3 +

(

5

xy

)3

=

(

x

y

)5

+(y

x

)5− 8

{

(

x

y

)3

+(y

x

)3}

+ 4

(

x

y+y

x

)

+4

(

x

y+y

x

) .(20)

Lemma 9. (see [18, p.55]) If x =f(−q)

q1

6 f(−q5)and y =

f(−q5)q

5

6 f(−q25), then

(xy)2 +

(

5

xy

)2

+ 5

(

xy +5

xy

)

+ 15 =(y

x

)3. (21)

Lemma 10. (see [5, Ch 20, Entry 18 (vi) and (vii), p.423]) If β, γ and δare of degrees 5, 7 and 35 respectively over α, then

(

αδ

βγ

)1/8

+

(

(1− α)(1 − δ)

(1− β)(1 − γ)

)1/8

−(

αδ(1 − α)(1 − δ)

βγ(1− β)(1 − γ)

)1/8

+ 2

(

αδ(1 − α)(1 − δ)

βγ(1− β)(1 − γ)

)1/12

=

√

m′

m,

(22)

(

βγ

αδ

)1/8

+

(

(1− β)(1 − γ)

(1− α)(1 − δ)

)1/8

−(

βγ(1− β)(1 − γ)

αδ(1 − α)(1 − δ)

)1/8

+ 2

(

βγ(1− β)(1 − γ)

αδ(1 − α)(1 − δ)

)1/12

= −√

m

m′ .

(23)


Lemma 11. (see [5, Ch. 17, Entry 10(i) and Entry 11(ii), pp. 122-123])We have

ϕ(q) =√z, (24)√

2q1/8ψ(−q) =√z{α(1− α)}1/8. (25)

Lemma 12. (see [5, Ch. 16, Entry 27 (i), (iv), p.43]) If αβ = π, then

√αϕ(e−α2

) =√

βϕ(e−β2

). (26)

e−α2/12√αf(−e−2α2

) = e−β2/12√

βf(−e−2β2

). (27)

Lemma 13. (see [1, Theorem 5.1]) If P =ψ(−q)

q1

2ψ(−q5)and Q =

ϕ(q)

ϕ(q5),

then

Q2 + P 2Q2 = 5 + P 2. (28)

Lemma 14. (see [3, Ch. 25, Entry 66, p.233]) If P =ϕ(q)

ϕ(q5)and Q =

ϕ(q3)

ϕ(q15), then

PQ+5

PQ= −

(

P

Q

)2

+

(

Q

P

)2

+ 3

(

P

Q+Q

P

)

. (29)

Lemma 15. (see [4, Theorem 2.16]) If P =ϕ(−q)ϕ(−q5) and Q =

ϕ(q)

ϕ(q5),

then(

P

Q

)

+

(

Q

P

)

=

(

PQ+5

PQ

)

− 4. (30)

3. P–Q Modular Equations

In this section, we establish some new P −Q modular equations of degree 5 forthe ratios of Ramanujan’s theta fuction.

Theorem 16. If P :=ψ(q)

q1

2ψ(q5)and Q :=

ϕ(q)

ϕ(q5), then

(

P

Q

)2

+

(

Q

P

)2

+ 4 = P 2 +5

P 2. (31)


Proof. Cubing both sides of the equation (19) and using the equations (11)and (12), we find that

(P 4 − 5P 2 + 4P 2Q2 − P 2Q4 +Q4)(P 4Q2 − P 4 − 4P 2Q2

+ 5Q2 −Q4)(P 2Q2 − P 2 + 5−Q2)(25 − 10P 2 − 10Q2

+ P 4 − 2P 4Q2 + P 4Q4 − 4P 2Q2 − 16P 3Q+ 16Q3P +Q4

− 2P 2Q4)(25 − 10P 2 − 10Q2 + P 4 − 2P 4Q2 + P 4Q4 +Q4

− 4P 2Q2 + 16P 3Q− 16Q3P − 2P 2Q4) = 0.

(32)

By examining the behavior of the above factors near q = 0, we can find aneighborhood about the origin, where the second factor is zero; whereas otherfactors are not zero in this neighborhood. By the Identity Theorem secondfactor vanishes identically. This completes the proof.

Theorem 17. If P :=ϕ(q)

ϕ(q5)and Q :=

ϕ(q2)

ϕ(q10), then

(

P

Q

)2

+

(

Q

P

)2

+ (PQ)2 +

(

5

PQ

)2

+ 16

(

P

Q− Q

P

)

= 2

(

P 2 +5

P 2

)

+ 2

(

Q2 +5

Q2

)

+ 4.

(33)

Proof. Cubing both sides of the equation (20), we deduce that

22500a6b24 + 1020a24b18 + a30b24 + 7420a12b24 + 22500a24b6

+ 1020a18b24 + 7420a24b12 + 1953125a12b6 + 127500a12b18a36

+ 1953125a6b12 + 127500a18b12 + 391b6a30 + 391a6b30 + a24b30

+ 937500a12b12 + 180a30b12 + 375000a18b6 + 16380a18b18 − b36

+ 60a24b24 + 24a18b30 + 24a30b18 + 180a12b30 + 375000a6b18 = 0.

(34)

where

a =f(−q)

q1

6 f(−q5)and b =

f(−q4)q

2

3 f(−q20).


Using the equations (12) and (13) in the above equation (34), we find that

(P 2Q4 −Q4 − 4P 2Q2 + 5P 2 − P 4)(25 + 16Q3P − 10Q2 − 10P 2

− 2P 2Q4 + P 4Q4 +Q4 − 4P 2Q2 − 16QP 3 + P 4 − 2P 4Q2)(25

− 16Q3P − 10Q2 − 10P 2 − 2P 2Q4 + P 4Q4 +Q4 − 4P 2Q2 + P 4

+ 16QP 3 − 2P 4Q2)(30P 4Q4 − 25P 2Q4 − P 2Q6 + P 2Q8 − 25P 4Q2

− 5P 4Q6 − P 6Q2 − 5P 6Q4 + P 6Q6 + 5P 6 + 5Q6 −Q8 + 25P 2Q2

− P 8 + P 8Q2)(250000P 2Q6 − 17976P 12Q6 − 78125Q2 + 368P 10Q14

− 112000P 2Q8 − 325000P 4Q2 + 600000P 4Q4 + P 16 + 270400P 4Q8

+ 230000P 6Q2 − 512000P 6Q4 + 527600P 6Q6 − 339840P 6Q8

− 5120P 2Q12 + 33200P 2Q10 + 109375Q4 − 65625Q6 + 21875Q8

− 4375Q10 + 525Q12 + 250000P 2Q2 + 14336P 4Q12 − 89880P 4Q10

− 35Q14 + 304P 2Q14 + 10816P 12Q8 − 67404P 8Q10 − 856P 4Q14

− 469000P 4Q6 − 17536P 6Q12 + 110672P 6Q10 + 277000P 8Q4 +Q16

− 337020P 8Q6 − 21P 16Q10 − 100500P 8Q2 + 27600P 10Q2 − 7P 16Q2

+ 110672P 10Q6 − 4280P 12Q2 + 14336P 12Q4 − 804P 8Q14

− 4096P 10Q12 + 21104P 10Q10 − 104P 12Q14 + 960P 12Q12

− 1024P 14Q4 − 896P 14Q8 + 400P 14Q10 − 128P 14Q12 − 35P 16Q6

+ 7P 16Q12 + 16P 14Q14 + 21P 16Q4 + 1328P 14Q6 − P 16Q14

+ 11080P 8Q12 + 1104P 6Q14 + 304P 14Q2 + 217648P 8Q8

+ 35P 16Q8 − 67968P 10Q8 − 3752P 12Q10 − 87680P 10Q4

− 400000P 2Q4) = 0.

(35)

By examining the behavior of the above factors near q = 0, we can find aneighborhood about the origin, where the third factor is zero; whereas otherfactors are not zero in this neighborhood. By the Identity Theorem third factorvanishes identically. This completes the proof.



ϕ(q5)and Q :=

ϕ(−q4)ϕ(−q20) , then

P 4

Q4+Q4

P 4+ 24

[

P 2

Q2+Q2

P 2

]

+ 8

[

P 2Q2 +52

P 2Q2

]

+ 120

+ 3

[

Q4 +52

Q4

]

= 20

[

P 2 +5

P 2

]

+ 32

[

Q2 +5

Q2

]

+

[

P 2Q4 +125

P 2Q4

]

+ 3

[

5P 2

Q4+Q4

P 2

]

.

(36)

Proof. Using the equations (13) and (20), we arrive at the equation (36).This completes the proof.

Theorem 19. If P :=ϕ(q)ϕ(q4)

ϕ(q5)ϕ(q20)and Q :=

ϕ(q)ϕ(q20)

ϕ(q5)ϕ(q4), then

Q4 +1

Q4− 112

[

Q3 +1

Q3

]

+ 1440

[

Q2 +1

Q2

]

− 3184

[

Q+1

Q

]

+ 7316 = 8

[

P +1

P

] [

22

[

Q2 +1

Q2

]

− 31

[

Q+1

Q

]

+ 170

]

− 2

[

P 2 +52

P 2

] [

3

[

Q2 +1

Q2

]

+ 24

[

Q+1

Q

]

+ 64

]

+ 4

[

P 3 +53

P 3

] [[

Q+1

Q

]

+ 4

]

.

(37)

Proof. Using the equations (36) and (30), we arrive at the equation (37).


ϕ(q5)and Q :=

ϕ(q5)

ϕ(q25), then

Q3

P 3− 5Q2

P 2− 15Q

P+ 5

(

PQ+5

PQ

)

+ 5

(

Q2 +5

P 2

)

= P 2Q2 +52

P 2Q2+ 15.

(38)

Proof. Using the equations (13) and (21), we arrive at the equation (38).


Theorem 21. If P :=φ(q)φ(q7)

φ(q5)φ(q35)and Q :=

φ(q)φ(q35)

φ(q5)φ(q7), then

Q4 − 1

Q4− 14

[(

Q3 +1

Q3

)

−(

Q2 − 1

Q2

)

+ 10

(

Q+1

Q

)]

+ P 3 +53

P 3= 7

{(

P 2 +52

P 2

)(

Q+1

Q

)

−(

P +5

P

)

×[

2

(

Q2 +1

Q2

)

+ 9

]}

.

(39)

Proof. Using the equations (22), (23), (24) and (25), we deduce that

1 + r − 2Ar + sr − s+ 2A2r = 0, (40)

where

r :=q3ψ(−q)ψ(−q35)ψ(−q5)ψ(−q7) , s :=

ϕ(q)ϕ(q35)

ϕ(q5)ϕ(q7)and A := (s/r)1/3.

On simplification of the equation (40), we find that

2A = 1 +m, where m = ±√

2s − r − 2sr − 2

r. (41)

Cubing both sides of the equation (41) and eliminating m, we deduce that

15cf2e− 6e2cf3 − e4cf3 + 8e4cf + 6e3df2 + e3df4 − 8edf4

− 14f3de2 − 6cf4e+ 3cf4e3 − 14cf2e3 + f5de2 + cf5e2

− 4f5d− 2cf5 − 4e5c+ 2e5d+ 5f3d+ 5cf3 + 5e3c− 5e3d

+ 15fde2 + e5cf2 − e5df2 − 6fde4 + 3f3de4 + 5edf2 − 5e2cf = 0,

(42)

where

c :=ψ(−q)

q1/2ψ(−q5) , d :=ψ(−q7)

q7/2ψ(−q35) , e :=ϕ(q)

ϕ(q5)and f :=

ϕ(q7)

ϕ(q35).

Collecting the terms containing c on one side of the above equation (42) andsquaring both sides and using the equation (28), we arrive at the equation(39).


Theorem 22. If P =φ(q)φ(q11)

φ(q5)φ(q55)and Q =

φ(q)φ(q55)

φ(q5)φ(q11), then

Q6 +1

Q6+ 33

[

Q5 +1

Q5

]

− 99

[

Q4 +1

Q4

]

+ 1529

[

Q3 +1

Q3

]

− 1683

[

Q2 +1

Q2

]

+ 8800

[

Q+1

Q

]

= 6534 +

[

P 5 +55

P 5

]

− 11

{[

P 4 +54

P 4

] [

Q+1

Q

]

−[

P 3 +53

P 3

] [

11 + 4

[

Q2 +1

Q2

]]

−[

P 2 +52

P 2

] [

18 − 56

[

Q+1

Q

]

+ 3

[

Q2 +1

Q2

]

− 8

[

Q3 +1

Q3

]]

−[

P +5

P

] [

−126

[

Q+1

Q

]

+ 160

[

Q2 +1

Q2

]

− 18

[

Q3 +1

Q3

]

+324 + 9

[

Q4 +1

Q4

]]

−[

P 3 +53

P 3

] [

11 + 4

[

Q2 +1

Q2

]]}

.

(43)

Proof. Replacing q by −q in the equation (16), we deduce that

− 16qf2(−q2)f2(−q10) = ϕ4(−q5)

×[

ϕ2(−q)ϕ2(−q5) − 1

] [

5− ϕ2(−q)ϕ2(−q5)

]

.(44)

Using the equations (44) and (16), we find that

ϕ4(q5)

ϕ4(−q5) =

[

ϕ2(−q)ϕ2(−q5) − 1

] [

ϕ2(−q)ϕ2(−q5) − 5

]

[

ϕ2(q)

ϕ2(q5)− 1

] [

5− ϕ2(q)

ϕ2(q5)

] . (45)

Replacing q by q11 in the above equation (45), we deduce that

ϕ4(q55)

ϕ4(−q55) =

[

ϕ2(−q11)ϕ2(−q55) − 1

] [

ϕ2(−q11)ϕ2(−q55) − 5

]

[

ϕ2(q11)

ϕ2(q55)− 1

] [

5− ϕ2(q11)

ϕ2(q55)

] . (46)

Employing the equation (17) along with the equations (45) and (46), we deduce


that

(

ψ(q5)ψ(q55)

ψ(−q5)ψ(−q55)

)8

=

(

ϕ2(−q)ϕ2(−q5) − 1

)(

ϕ2(−q)ϕ2(−q5) − 5

)

(

ϕ2(q)

ϕ2(q5)− 1

)(

5− ϕ2(q)

ϕ2(q5)

)

×

(

ϕ2(−q11)ϕ2(−q55) − 1

)(

ϕ2(−q11)ϕ2(−q55) − 5

)

(

ϕ2(q11)

ϕ2(q55)− 1

)(

5− ϕ2(q11)

ϕ2(q55)

) .

(47)

The equation (18) can be re arranged as,

[

ψ(q5)ψ(q55)

ψ(−q5)ψ(−q55)

]8

=

[

ψ(−q11)q

11

2 ψ(−q55)+

ψ(−q)q

1

2ψ(−q5)

]8

[

ψ(q11)

q11

2 ψ(q55)− ψ(q)

q1

2ψ(q5)

]8 . (48)

Using the equations (47) and (48), we deduce that[

ϕ2(−q)ϕ2(−q5) − 1

] [

ϕ2(−q)ϕ2(−q5) − 5

]

[

ϕ2(q)

ϕ2(q5)− 1

] [

5− ϕ2(q)

ϕ2(q5)

]

[

ϕ2(−q11)ϕ2(−q55) − 1

] [

ϕ2(−q11)ϕ2(−q55) − 5

]

[

ϕ2(q11)

ϕ2(q55)− 1

] [

5− ϕ2(q11)

ϕ2(q55)

]

=

[

ψ(−q11)q

11

2 ψ(−q55)+

ψ(−q)q

1

2ψ(−q5)

]8

[

ψ(q11)

q11

2 ψ(q55)− ψ(q)

q1

2ψ(q5)

]8 .

(49)

Using the equations (28), (30) and (31) in the above equation (49), we arriveat the equation (43). This completes the proof.

4. Explicit Evaluations for h5,n and h′

5,n

In this section, we establish several general formulas for the explicit evaluationsof h5,n by employing the P − Q modular equations of degree 5 established inSection 3.


Theorem 23. If X := h5,n and Y := h5,4n, then

(

X

Y

)2

+

(

Y

X

)2

+ 5

(

(XY )2 +1

(XY )2

)

+ 16

(

X

Y− Y

X

)

= 2√5

{(

X2 +1

X2

)

+

(

Y 2 +1

Y 2

)}

+ 4.

(50)

Proof. Employing the equation (33) with k = 5 in the equation (9), wearrive at the equation (50).

Corollary 24. We have

h5,2 = (√2− 1)

√√5 + 2, (51)

h5,1/2 = (√2 + 1)

√√5− 2, (52)

h5,4 =

√

17 + 7√5− 4

√

29 + 13√5

2

−

√

15 + 7√5− 4

√

29 + 13√5

2,

(53)

h5,1/4 =

√

17 + 7√5− 4

√

29 + 13√5

2

+

√

15 + 7√5− 4

√

29 + 13√5

2,

(54)

h5,8 =

√

37√5 + 58

√2 + 26

√10 + 82

−√

34√5 + 76 + 24

√10 + 54

√2,

(55)

h5,1/8 = (3− 2√2)(

√5− 2)

[√

37√5 + 58

√2 + 26

√10 + 82

+

√

34√5 + 76 + 24

√10 + 54

√2

]

.

(56)

Proofs of (51) and (52). Putting n = 1/2 in the equation (50) and using thefact that h5,2h5,1/2 = 1, we deduce that

(h45,2 − 12h25,2 − 6√5h25,2 + 9 + 4

√5)(h25,2 − 2 +

√5)2 = 0. (57)


Since the roots of the second factor are imaginary and h5,2 > 0, we deduce that

h45,2 − (12 + 6√5)h25,2 + 9 + 4

√5 = 0. (58)

On solving the above equation (58), we arrive at the equations (51) and (52).

Proofs of (53) and (54). Putting n = 1 in the equation (50) and using the factthat h5,1 = 1, we deduce that

x2 − (6 + 2√5)x− 2

√5− 2 = 0, where x := h5,4 −

1

h5,4. (59)

Since 0 < x < 1, we deduce that

h5,4 −1

h5,4= 3 +

√5− 2

√

4 + 2√5. (60)

On solving the above equation (60), we arrive at the equations (57) and (58).

Proofs of (55) and (56). Using the equation (51) in the equation (50), we ar-rive at the equations (55) and (56).

Theorem 25. If X = h5,n and Y = h′5,16n, then

X4

Y 4+Y 4

X4+ 24

(

X2

Y 2+Y 2

X2

)

+ 40

(

X2Y 2 +1

X2Y 2

)

+ 120

+ 15

(

Y 4 +1

Y 4

)

=√5

[

20

(

X2 +1

X2

)

+ 32

(

Y 2 +1

Y 2

)

+5

(

X2Y 4 +1

X2Y 4

)

+ 3

(

X2

Y 4+Y 4

X2

)]

.

(61)

Proof. Using the equation (36) and then employing the equations (9) and(10) with k = 5, we arrive at the equation (61).

Corollary 26.

h′5,2 =

√√5− 1, (62)

h′5,4 =

√√5 + 1−

√2√√

5 + 1√2

, (63)

h′5,8 =

√√2− 1, (64)


h′5,16 =

√

11 + 5√5

2−

√

(√

11 + 5√5− 4)

√

11 + 5√5

2− 1

1/2

, (65)

h′5,32 =

[√

34 + 11√10 + 15

√5 + 24

√2

−√

32 + 10√10 + 14

√5 + 22

√2

]1/2

.

(66)

Proof. Putting n = 1/8, 1/4, 1/2, 1, 2 in the equation (61) and using thevalues of h5,1/8, h5,1/4, h5,1/2, h5,1 and h5,2, we arrive at the equations (62), (63),(64), (65) and (66) respectively. This completes the proof.

Theorem 27. If X = h5,n and Y = h5,25n, then

Y 3

X3− 5Y 2

X2− 15Y

X+ 5

√5

(

XY +1

XY

)

+ 5√5

(

Y 2 +1

X2

)

= 5

(

X2Y 2 +1

X2Y 2

)

+ 15.

(67)

Proof. Using the equation (9) with k = 5 in the equation (38), we obtain(67).


h5,5 =

√

5− 2√5, (68)

h5,1/5 =

√

5 + 2√5√

5. (69)


(h25,5 − 5 + 2√5)(h25,5 −

√5)2 = 0. (70)

Since 0 < h5,5 < 1, Hence h25,5 − 5 + 2√5 = 0.


Theorem 29. If X = h5,nh5,121n and Y =h5,nh5,121n

, then

Y 6 +1

Y 6+ 33

[

Y 5 +1

Y 5

]

− 99

[

Y 4 +1

Y 4

]

+ 1529

[

Y 3 +1

Y 3

]

− 1683

[

Y 2 +1

Y 2

]

+ 8800

[

Y +1

Y

]

= 25√5

[

X5 +1

X5

]

− 11√5

{

5√5

[

X4 +1

X4

] [

Y +1

Y

]

− 5

[

X3 +1

X3

]

[11

+4

[

Y 2 +1

Y 2

]]

−√5

[

X2 +1

X2

] [

18− 56

[

Y +1

Y

]

+3

[

Y 2 +1

Y 2

]

− 8

[

Y 3 +1

Y 3

]]

−[

X +1

X

] [

324 − 126

[

Y +1

Y

]

+160

[

Y 2 +1

Y 2

]

− 18

[

Y 3 +1

Y 3

]

+ 9

[

Y 4 +1

Y 4

]]}

+ 6534.

(71)

Proof. Employing the equation (9) with k = 5 in the equation (43), weobtain (71).


h5,11 =

√

3 +√5−

√

6√5− 2

2, (72)

h5,1/11 =

√

3 +√5 +

√

6√5− 2

2. (73)


(h45,11 + (24 + 10√5)h25,11 + 1)(2h45,11 − (3 +

√5)h25,11 + 2)

(2h85,11 − (3 + 3√5)h65,11 − (6

√5− 30)h45,11 − (3 + 3

√5)h25,11 + 2)

(h45,11 + (6− 4√5)h25,11 + 1)2 = 0.

(74)

By observing the behaviour of the factors of the above equation (74), the second

factor vanishes for specific value of q = e−π√

11/5, where as the other factorsdoes not vanish. Hence

2h45,11 − (3 +√5)h25,11 + 2 = 0. (75)

On solving the above equation (75) and 0 < h5,11 < 1, we obtain the equation(72).


5. Modular Relations between µ(q) and µ(qn)

In this section, we establish several new modular relations connecting µ(q) withµ(qn) using the P −Q modular equations obtained in Section 3.

Theorem 31. If u := µ(q) and v := ω(q), then

(v − 1)u2 + (v2 − 4v + 1)u− v2 + v = 0. (76)

Proof. Using the equations (30), (14) we arrive at the equation (76).

Theorem 32. If u := µ(q) and v := µ(q2), then

(u4 − 3u3 + 5u2 − 3u)v3 − (u4 + 10u2 − 5u− 5u3 + 1)v2

− (3u3 − 5u2 + 3u− 1)v − u2 + u3 + (u− u2)v4 = 0.(77)



(3u2 − 3u3 + u4 + 3u)v3 − 3(u+ u3)v2

+ (1− 3u+ 3u3 + 3u2)v = uv4 + u3.(78)



u7 − u8 + (−u− u5 + 1 + 6u2 − 10u3 + 5u4)v7 + (7u5 − 31u2 + 6u

+ 45u3 − 27u4)v6 + (7u6 − 42u5 − 91u3 − u7 + 92u4 − 10u+ 45u2)v5

+ (u− 1)v8 + (92u5 − 27u6 − 27u2 + 92u3 − 142u4 + 5u+ 5u7)v4

+ (7u2 − 42u3 + 45u6 + 92u4 − u− 10u7 − 91u5)v3 + (45u5 − 31u6

+ 7u3 − 27u4 + 6u7)v2 + (u8 − u3 + 6u6 + 5u4 − 10u5 − u7)v = 0.

(79)


Theorem 35. If u := µ(q) and v := ω(q4), then

(u8 − 4u7 + 6u2 + 6u6 + 1− 4u− 14u4)(v7 + v) + (6u8 + 32u3

+ 6− 32u− 136u4 + 48u2 + 32u5 − 32u7 + 48u6)(v6 + v2) + (15

+ 80u3 + 15u8 + 80u5 − 434u4 + 202u6 − 108u − 108u7 + 202u2)

× (v5 + v3) + (16u + 16u7 − 2− 32u2 − 16u3 + 75u4 − 32u6

− 2u8 − 16u5)10v4 + u4 + u4v8 = 0.

(80)




(11v2 − 6v3 + v4 − 6v + 1)u5 + (−35v2 − 5v4 + 25v3 + 10v)u4

+ (5v + 20v2 − 25v3 + 10v4)u3 + (−5v4 − 20v3 − 10v + 25v2)u2

+ (35v3 + 5v − 25v2 − 10v4)u− v + 6v2 + 6v4 − v5 − 11v3 = 0.

(81)



v + u8v7 − v8u− u7 + 7{

(v − 3v2 + 4v3 − 4v5 + 2v6 − v7)u7

+ (3v7 − 2v + 14v2 − 26v3 + 29v5 − 14v6)u6 + (−5v4 − 55v5

+ 4v − 29v2 + 55v3 + 26v6 − 4v7)u5 + 5(v5 + v3)u4 + (4v7 − 4v

− 29v6 + 55v5 − 5v4 + 26v2 − 55v3)u3 + (14v6 − 14v2 + 29v3

+3v − 26v5 − 2v7)u2 + (v7 − v + 2v2 − 4v3 − 3v6 + 4v5)u}

= 0.

(82)



v12 − uv − u11v11 + 11{(14u8 − 4u+ 8u2 + u10 − 21u7 − 32u4

+ 21u5 + 8u3 + 9u6 − 5u9)v11 + (8u− 214u8 + 83u9 − 15u10

+ 247u7 + 6u2 − 211u5 − 39u6 + u11 − 128u3 + 268u4)v10

+ (8u+ 538u3 − 489u9 + 83u10 − 81u6 + 1292u8 − 1393u7

+ 1321u5 − 1166u4 − 5u11 − 128u2)v9 + (268u2 − 4284u5

− 214u10 − 3638u8 + 1292u9 − 1166u3 + 4176u7 + 3251u4

+ 14u11 − 32u+ 378u6)v8 + (4176u8 − 211u2 − 5452u7 + 21u

+ 5452u5 − 1393u9 − 21u11 − 4284u4 + 247u10 + 1321u3

+ 72u6)v7 + (9u− 81u9 − 81u3 − 39u2 + 72u7 + 72u5 − 39u10

+ 378u8 + 378u4 + 9u11 − 594u6)v6 + (21u11 + 247u2 − 21u

(83)


− 5452u5 + 5452u7 + 4176u4 + 1321u9 − 1393u3 − 211u10

− 4284u8 + 72u6)v5 + (14u − 3638u4 + 378u6 − 32u11 − 214u2

+ 3251u8 + 268u10 + 1292u3 + 4176u5 − 4284u7 − 1166u9)v4

+ (1292u4 − 128u10 − 5u− 489u3 − 1393u5 + 1321u7 + 8u11

− 1166u8 + 83u2 − 81u6 + 538u9)v3 + (8u11 − 15u2 − 211u7

+ 6u10 + 83u3 − 214u4 + 268u8 + u+ 247u5 − 39u6 − 128u9)v2

+ (8u10 + 21u7 − 32u8 − 5u3 − 21u5 + 9u6 + u2 − 4u11 + 8u9

+ 14u4)v}+ u12 = 0.


6. Reciprocity Theorems and Explicit Evaluations for µ(q)

In this section, we establish new reciprocity theorems for µ(q) and establishseveral new explicit evaluations of µ(q) by using the values of h5,n establishedin Section 4.

Theorem 39. If 5αβ = 1, then

µ(e−πα) + µ(e−πβ)

1 + µ(e−πα)µ(e−πβ)=

3−√5

2. (84)

Proof. Set q = e−πα in the equation (14), we deduce that

ϕ(e−πα)

ϕ(e−5πα)=

1 + µ(e−πα)

1− µ(e−πα). (85)

Replace α by β in the above equation (85), we find that

ϕ(e−πβ)

ϕ(e−5πβ)=

1 + µ(e−πβ)

1− µ(e−πβ). (86)

Using the equations (26), (85) and (86), we arrive at the equation (84).

Theorem 40. If 5αβ = 1, let x := φ(e−πα) and y := φ(e−πβ)where φ(q) = µ(−q), then

1 + x+ x4y − 4x2 − 4y2 + 6x3y − 14x3y2 − 14x2y − 64x2y2

− 14xy2 − 4x4y2 + 6x3y3 + x3y4 − 14x2y3 − 4x2y4 + 6xy3

+ 6xy + xy4 + x4y3 + x4y4 + x3 + x4 + y3 + y4 + y = 0.

(87)


Proof. Putting q = e−πα in (12) and change q to −q, we deduce that

f3(−e−2πα)

e−παf3(−e−10πα)=

(

1− x

1 + x

)(

x2 + 3x+ 1

x

)

. (88)

Replacing α by β in (88), we find that

f3(−e−2πβ)

e−πβf3(−e−10πβ)=

(

1− y

1 + y

)(

y2 + 3y + 1

y

)

. (89)

Using (27) in (88) and (89), we find that

(x2y2 + 3x2y + x2 + 3x+ 1 + 4xy + 3y + 3xy2 + y2)

(1 + x+ y + x4y − 4x2 − 4y2 + 6x3y − 14x3y2 − 14x2y

− 64x2y2 + 6xy − 14xy2 − 4x4y2 + 6x3y3 + x3y4 − 14x2y3

− 4x2y4 + 6xy3 + xy4 + x4y3 + x4y4 + x3 + x4 + y3 + y4) = 0.

(90)

As x2y2 + 3x2y + x2 + 3x + 1 + 4xy + 3y + 3xy2 + y2 6= 0, hence the secondfactor is zero. This completes the proof

Lemma 41. We have

µ(e−π√

n/5) =51/4h5,n − 1

51/4h5,n + 1, where n is any positive rational. (91)

Proof. Using the equations (9) with n = 5 and (14), we arrive the equation(91)

Lemma 42.

ω(e−π√

n/5) =51/4h′

5,n − 1

51/4h′5,n + 1

, where n is any positive rational. (92)

Proof. Using the equations (10) with n = 5 and (15), we arrive the equation(92)

We establish some explicit evaluations of µ(q) by using the values of h5,n inthe following Theorem.

Theorem 43. We have

µ(e−π/√5) =

51/4 − 1

51/4 + 1,


µ(e−π√

2/5) =51/4(

√2− 1)(

√

2 +√5)− 1

51/4(√2− 1)(

√

2 +√5) + 1

,

µ(e−π/√10) =

51/4(√2 + 1)(

√√5− 2)− 1

51/4(√2 + 1)(

√√5− 2) + 1

,

µ(e−π) =51/4(

√

5− 2√5)− 1

51/4(√

5− 2√5) + 1

,

µ(e−π/5) =

√

5 + 2√5− 51/4

(√

5 + 2√5) + 51/4

,

µ(e−π√

11/5) =51/4(3 +

√5−

√

6√5− 2)1/2 − 2

51/4(3 +√5−

√

6√5− 2)1/2 + 2

,

µ(e−π/√55) =

51/4(3 +√5 +

√

6√5− 2)1/2 − 2

51/4(3 +√5 +

√

6√5− 2)1/2 + 2

.

Remark 1. Using the equation (92) and values of h′5,n, one can evaluate

the explicit evaluations of ω(q).

Acknowledgments

Research work supported by University Grants Commission, India.

References

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