a parameter for ramanujan s function q its explicit values and applications

We prove some general theorems for the explicit evaluations of the parameterI k,nand find many explicit values.. Some values ofI k,nare then used to find some new and known values of Ram

Trang 1

ISRN Computational Mathematics

Volume 2012, Article ID 169050, 14 pages

doi:10.5402/2012/169050

Research Article

Its Explicit Values and Applications

Nipen Saikia

Department of Mathematics, Rajiv Gandhi University, Rono Hills, Doimukh 791112, India

Correspondence should be addressed to Nipen Saikia,nipennak@yahoo.com

Received 3 May 2012; Accepted 28 June 2012

Academic Editors: L Hajdu, L S Heath, and H J Ruskin

Copyright © 2012 Nipen Saikia This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

We define a new parameterI k,ninvolving quotient of Ramanujan’s functionχ(q) for positive real numbers k and n and study its

several properties We prove some general theorems for the explicit evaluations of the parameterI k,nand find many explicit values Some values ofI k,nare then used to find some new and known values of Ramanujan’s class invariantG n

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

f (a, b) =

∞

n =−∞

a n(n+1)/2 b n(n −1) , | ab | < 1. (1)

After Ramanujan, for| q | < 1, we define

f

− q

:= f

− q, − q2

=

∞

n =−∞

(−1)n q n(3n −1) =q; q

∞, (2)

where (a; q) ∞:=∞ n =0(1− aq n) Ifq = e2πizwith Im(z) > 0,

then f ( − q) = q −1/24 η(z), where η(z) denotes the classical

Dedekind eta function

Ramanujan’s functionχ(q) is defined by

χ

q

:= f

q

f

− q2 =− q; q2

The function χ(q) is intimately connected to Ramanujan’s

class invariantsG nandg n, which are defined by

G n =2−1/4 q −1/24 χ

q

, g n =2−1/4 q −1/24 χ

− q

whereq : = e − π √

nandn is a positive rational number Since

from [2, page 124, Entry 12(v) & (vi)]

χ

q

=21/6

α(1 − α) q

−1/24

,

χ

− q

=21/6(1− α)1/12

α q

−1/24

,

(5)

it follows from (4) that

G n = {4α(1 − α) } −1/24,

g n =2−1/12(1− α)1/12 α −1/24

(6)

Also, ifβ has degree r over α, then

G r2n = 4β

1− β−1/24

, g r2n =2−1/12

1− β1/12

β −1/24

(7)

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 approximateπ and

probably for finding explicit values of Rogers-Ramanujan continued fractions, theta-functions, and so on An account

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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 invariantsG nfor odd values

ofn and g nfor even values ofn For the first time, Yi [13]

calculated some values ofg nfor odd values ofn by finding

explicit values of the parameterr k,n(see [13, page 11, (2.1.1)]

or [14, page 4, (1.11)]) defined by

r k,n:= f

− q

k1/4 q(k −1)/24 f

− q k , q = e −2π √

n/k (8)

In particular, she established the result [13, page 18, Theorem

2.2.3]

g n = r2,n/2 (9) However, the values ofG nfor even values ofn have not been

calculated The main objective of this paper is to evaluate

some new values ofG n for even values ofn We also prove

some known values ofG n For evaluation of class invariant

G n in this paper, we introduce the parameterI k,n, which is

defined as

I k,n:= χ

q

q(− k+1)/24 χ

q k , q = e − π √ n/k, (10) wherek and n are positive real numbers.

InSection 3, we study some properties ofI k,n and also

establish its relations with Ramanujan’s class invariantG n In

Section 4, by employing Ramanujan’s modular equations, we

present some general theorems for the explicit evaluations of

I k,n and find several explicit values of I k,n InSection 5, we

establish some general theorems connecting the parameter

I k,nand the class invariantG n We also evaluate some explicit

values of the productG nk G n/k by employing some values of

I k,nevaluated inSection 4 Finally, inSection 6, we calculate

new and known values of class invariantG nby combining the

explicit values ofI k,n and the productG nk G n/k evaluated in

Sections4and5, respectively.Section 2is devoted to record

some preliminary results

Since Ramanujan’s modular equations are key in our

evaluations ofI k,nandG n, we complete this introduction by

defining Ramanujan’s modular equation from Berndt’s book

[2] The complete elliptic integral of the first kindK(k) is

defined by

K(k) : =

π/2

0

dφ

1− k2sin2φ = π

2

∞

n =0

(1/2)2n

(n!)2 k

2n

= π

22F1

1

2,

1

2; 1;k2

,

(11)

where 0 < k < 1,2F1 denotes the ordinary or Gaussian

hypergeometric function, and

(a) n = a(a + 1)(a + 2) · · ·(a + n −1). (12)

The numberk 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

n K

K = L

holds for some positive integern Then, a modular equation

of degreen is a relation between the moduli k and l, which is

implied by (13)

If we set

q =exp

− π K K

, q =exp

− π L L

we see that (13) is equivalent to the relationq n = q Thus,

a modular equation can be viewed as an identity involving theta-functions at the arguments q and q n Ramanujan recorded his modular equations in terms ofα and β, where

α = k2andβ = l2 We say thatβ has degree n over α The

multiplierm connecting α and β is defined by

m = K

Ramanujan also established many “mixed” modular equa-tions in which four distinct moduli appear, which we define from Berndt’s book [2, page 325]

Let K, K ,L1, L 1,L2, L 2, L3, and L 3 denote complete elliptic integrals of the first kind corresponding, in pairs, to the moduli√

α, β, √ γ, and √ δ and their complementary

moduli, respectively Letn1,n2, andn3be positive integers such thatn3= n1n2 Suppose that the equalities

n1K

K = L 1

L1

, n2K

K = L 2

L2

, n3K

K = L 3

L3

(16) 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 thatβ, γ, and δ are of degrees

n1,n2, and n3, respectively, over α or α, β, γ, and δ have

degrees 1,n1,n2, andn3, respectively Denotingz r = φ2(q r), where

q =exp

− πK

K

q

= f

q, q

, q< 1 (17)

the multipliers m and m associated with α, β, and γ, δ,

respectively, are defined bym = z1/z n1andm = z n2/z n3

2 Preliminary Results

Lemma 1 (see [2, page 43, Entry 27(v)]) If α and β are such that the modulus of each exponential argument is less than 1 and αβ = π2, then

e α/24 χ(e − α)= e β/24 χ

e − β

Trang 3

Lemma 2 (see [15, page 241, Lemma 2.3]) Let X: =

q1/12(χ(q)/χ(q3)) and Y : = q1/6(χ(q2)/χ(q6)); then

X

Y

12

+

Y

X

12

+

X Y

6

+

Y X

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.

(19)

Lemma 3 (see [15, page 241, Lemma 2.8]) Let X : =

X

Y

3

+

Y

X

3

(XY )5+ (XY ) −5+ 8

(XY )3+ (XY ) −3

+19

XY +(XY ) −1

+

X

Y

6

+

Y

X

6

=(XY )6+ (XY ) −6

+ 13

(XY )4+ (XY ) −4

+ 52

(XY )2+ (XY ) −2

+ 82.

(20)

Lemma 4 (see [15, page 252, Lemma 2.13]) Let X : =

X

Y

12

+

Y

X

12

+16

X

Y

10

+

Y X

10

(XY )2+(XY ) −2

+ 8

X

Y

8

+

Y X

8

7

(XY )4+ (XY ) −4

−19

+

X

Y

6

+

Y X

6

×(XY )10+ (XY ) −10

+ 32

(XY )6+ (XY) −6

−81

(XY )2+ (XY ) −2

+

X

Y

4

+

Y X

4

16

(XY )8+ (XY ) −8

−288

(XY )4+ (XY ) −4

+ 352

+

X

Y

2

+

Y

X

2

296

(XY )2+ (XY ) −2

−256

(XY )6+ (XY ) −6

−8

(XY )10+ (XY ) −10

+ 1746

=(XY )12+ (XY ) −12

+ 145

(XY )8+ (XY ) −8

+ 496

(XY )4+ (XY ) −4

.

(21)

Lemma 5 (see [2, page 231, Entry 5(xii)]) Let P = {16αβ(1 − α)(1 − β) }1/8 and Q = (β(1 − β)/α(1 − α))1/4 ; then

Q + Q −1+ 2√

2

P − P −1

where β has degree 3 over α.

Lemma 6 (see [2, page 282, Entry 13(xiv)]) Let P = {16αβ(1 − α)(1 − β) }1/12 and Q = (β(1 − β)/α(1 − α))1/8 ; then

Q + Q −1+ 2

P − P −1

Lemma 7 (see [2, page 315, Entry 13(xiv)]) Let P = {16αβ(1 − α)(1 − β) }1/8 and Q = (β(1 − β)/α(1 − α))1/6 ; then

Q + Q −1+ 7=2√

2

P + P −1

Lemma 8 (see [5, page 378, Entry 41]) Let P =

21/6 { αβ(1 − α)(1 − β) }1/24 and Q =(β(1 − β)/α(1 − α))1/24 ; then

Q7+Q −7+ 13

Q5+Q −5

+ 52

Q3+Q −3

+ 78

Q + Q −1

−8

P −6+P6

=0,

(25)

For Lemmas9to15, we set

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

.

(26)

Lemma 9 (see [5, page 381, Entry 50]) If α, β, γ, and δ have degrees 1, 5, 7, and 35, respectively, then

R4+R −4−Q6+Q −6

+5

Q4+Q −4

−10

Q2+Q −2

+15=0.

(27)

Lemma 10 (see [5, page 381, Entry 51]) If α, β, γ, and δ have degrees 1, 13, 3, and 39, respectively, then

Q4+Q −4−3

Q2+Q −2

−T2+T −2

+ 3=0. (28)

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Lemma 11 (see [5, page 381, Entry 52]) If α, β, γ, and δ have

degrees 1, 13, 5, and 65, respectively, then

Q6+Q −6−5

Q + Q −12

T + T −12

−T4+T −4

=0.

(29)

Lemma 12 (see [16, page 277, Lemma 3.1]) If α, β, γ, and δ

have degrees 1, 3, 7, and 21, respectively, then

R2+R −2− Q4− Q −4+ 3=0. (30)

Lemma 13 (see [15, page 243, Theorem 2.5]) If α, β, γ, and

δ have degrees 1, 2, 3, and 6, respectively, then

R4+R −4+P2−2P −2=0. (31)

Lemma 14 (see [15, page 248, Theorem 2.10]) If α, β, γ and

R6+R −6+ 5

R2+R −2

=4P −4− P4. (32)

Lemma 15 (see [15, page 252, Theorem 2.12]) If α, β, γ and

2√

2

T3+T −3

=4P −1−2P + 4P3− P5. (33)

3 Properties of Ik,n

In this section, we study some properties of I k,n We also

establish a relation connecting I k,n and Ramanujan’s class

invariantsG n.

Theorem 16 For all positive real numbers k and n, one has

(i)I k,1 =1, (ii)I

k,1

n

= 1

I k,n, (iii)I k,n = I n,k (34) Proof Using the definition of I k,n andLemma 1, we easily

arrive at (i) Replacingn by 1/n in I k,nand usingLemma 1,

we find thatI k,n I k,1/n =1, which completes the proof of (ii)

To prove (iii), we use Lemma 1in the definition of I k,n to

arrive at (I k,n /I n,k) =1

Remark 17 By using the definitions of χ(q) and I k,n, it can

be seen thatI k,n has positive real value less than 1 and that

the values ofI k,ndecrease asn increases when k > 1 Thus, by

Theorem 16(i),I k,n < 1 for all n > 1 if k > 1.

Theorem 18 For all positive real numbers k, m, and n, one has

I k,n/m = I mk,n I −1

Proof Using the definition of I k,n, we obtain

I mk,n

I nk,m = χ

e − π √ n/mk

e π( √ m/nk − √ n/mk)/24 χ

e − π √ m/nk. (36) UsingLemma 1in 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

I k2 ,n = I nk,n I k,n/k (37)

Proof Setting k = n inTheorem 18and simplifying using

Theorem 16(ii), we obtain

I k2 ,m = I mk,k I k,m/k (38)

Replacingm by n, we complete the proof.

Theorem 20 Let k, a, b, c, and d be positive real numbers such

that ab=cd Then

I a,b I kc,kd = I ka,kb I c,d (39)

Proof From the definition of I k,n, we deduce that, for

positive real numbersk, a, b, c, and d,

I ka,kb I −1

e − π √

ab

e π(k √

ab − √ ab)/24 χ

e − kπ √

ab

I kc,kd I c,d −1= χ

e − π √

cd

e π(k √

cd − √ cd)/24 χ

e − kπ √

cd.

(40)

Now the result follows readily from (40), and the hypothesis thatab = cd.

Corollary 21 For any positive real numbers n and p, we have

I np,np = I np2 ,n I p,p (41)

Proof The result follows immediately from Theorem 20

witha = p2,b =1,c = d = p, and k = n.

Now, we give some relations connecting the parameter

I k,nand Ramanujan’s class invariantsG n.

Theorem 22 Let k and n be any positive real numbers Then

(i)I k,n = G n/k G − nk1, (ii)G1/n = G n (42)

Proof Proof of (i) follows easily from the definitions of I k,n

andG nfrom (10) and (4), respectively To prove (ii), we set

k =1 in part (i) and useTheorem 16(i) and (iii)

4 General Theorems and Explicit Evaluations of Ik,n

In this section, we prove some general theorems for the explicit evaluations ofI k,nand find its explicit values

Trang 5

Theorem 23 One has

I3,n

I3,4n

12

+

I3,n

I3,4n

−12

+

⎡

⎣

I3,n

I3,4n

6

+

I3,n

I3,4n

−6⎤

⎦

×

I3,n I3,4n

10

+

I3,n I3,4n

−10

+16

I3,n I3,4n

6

+(I3,n I3,4n) −6

+71

I3,n I3,4n

2

+

I3,n I3,4n

−2

=I3,n I3,4n

12

+

I3,n I3,4n

−12

+25

I3,n I3,4n

8

+

I 3,n I3,4n

−8

+ 200

I3,n I3,4n

4

+

I3,n I3,4n

−4

+ 550.

(43)

Proof The proof follows easily from the definition of I k,nand

Lemma 2

Corollary 24 One has

(i)I3,2=

−44 + 27√

3−3 458−264√

3

1/12

,

(ii)I3,4=

⎛

⎝

−2 + 3√

6−−6 + 3√

6

2

⎞

⎠

1/2

,

(iii)I3,1/2 =

−44 + 27√

3 + 3 458−264√

3

1/12

,

(iv)I3,1/4 =

⎛

⎝

−6 + 3√

6 +

−2 + 3√

6

2

⎞

⎠

1/2

.

(44)

Proof Setting n = 1/2 in Theorem 23 and using

I3,224+I3,2−24+ 176

I3,212+I3,2−12

−1002=0. (45) Equivalently,

where

B = I12 3,2+I −12

Solving (46) and using the fact inRemark 17, we obtain

B =54√

Employing (48) in (47), solving the resulting equation for

I3,2, and noting thatI3,2< 1, we arrive at

I3,2=

−44 + 27√

3−3 458−264√

3

1/12

. (49) This completes the proof of (i)

Again setting n = 1 in Theorem 23 and using

Theorem 16(i), we obtain

I6

3,4+I −6

3,4

I103,4+I −10 3,4

+16

I3,46 +I3,4−6

+ 71

I3,42 +I3,4−2

=25

I8 +I −8

+ 200

I4 +I −4

+ 550.

(50)

Equivalently,

D2+ 12

D4+ 4D2−50

where

D = I3,42 +I3,4−2. (52) Since the first factor of (51) is nonzero, solving the second factor, we deduce that

D =−2 + 3√

61/2

Employing (53) in (52), solving the resulting equation, and using the fact thatI3,4< 1, we obtain

I3,4=

⎛

⎝

−2 + 3√

6−−6 + 3√

6

2

⎞

⎠

1/2

. (54)

This completes the proof of (ii)

Now (iii) and (iv) follow from (i) and (ii), respectively, andTheorem 16(ii)

I5,n

I5,4n

6

+

I5,n

I5,4n

−6

+

⎡

⎣

I5,n

I5,4n

3

+

I5,n

I5,4n

−3⎤

⎦

×

I5,n I5,4n

5

+

I5,n I5,4n

−5

+ 8

I5,n I5,4n

3

+

I5,n I5,4n

−3

+19

I5,n I5,4n

+

I5,n I5,4n

"

=I5,n I5,4n

6

+

I5,n I5,4n

−6

+ 13

I5,n I5,4n

4

+

I5,n I5,4n

−4

+ 52

I5,n I5,4n

2

+

I5,n I5,4n

−2

+ 82.

(55)

Proof The proof follows fromLemma 3and the definition

ofI k,n.

(i)I5,2=

−14 + 5√

10− 445−140√

10

1/6

,

(ii)I5,4=

11 + 5√

51/4

− −4 +

11 + 5√

5

(iii)I5,1/2 =

−14 + 5√

10 + 445−140√

10

1/6

,

(iv)I5,1/4 =

11 + 5√

51/4

+ −4 +

11 + 5√

5

(56)

Proof Setting n = 1/2 in Theorem 25 and using

C2+ 56C −216=0, (57)

Trang 6

C = I6 5,2+I −6

Solving (57) and noting the fact inRemark 17, we obtain

C = −28 + 10√

Employing (59) in (58), solving the resulting equation, and

noting thatI5,2< 1, we obtain

I5,2=

−14 + 5√

10− 445−140√

10

1/6

. (60)

This completes the proof of (i)

Again, setting n = 1 in Theorem 25 and using

Theorem 16(i), we obtain

B8−22B4−4=0, (61)

where

B = I5,4+I4,5−1. (62)

Solving (61), we obtain

B =11 + 5√

51/4

Using (63) in (62), solving the resulting equation, and noting

thatI5,4< 1, we arrive at

I5,4=

11 + 5√

51/4

− −4 +

11 + 5√

5

This completes the proof of (ii)

Now (iii) and (iv) follow from (i) and (ii), respectively,

andTheorem 16(ii)

I7,n

I4,7n

12

+

I7,n

I4,7n

−12

+ 16

⎡

⎣

I7,n

I4,7n

10

+

I7,n

I4,7n

−10⎤

⎦

I7,n I7,4n

2

+

I7,n I7,4n

−2

+ 8

⎡

⎣

I7,n

I4,7n

8

+

I7,n

I4,7n

−8⎤

⎦

×7

I7,n I7,4n

4

+

I7,n I7,4n

−4

−19

+

⎡

⎣

I7,n

I

6

+

I7,n

I

−6⎤

⎦

×

I7,n I7,4n

10

+

I7,n I7,4n

−10

+ 32

I7,n I7,4n

6

+

I7,n I7,4n

−6

−81

I7,n I7,4n

2

+

I7,n I7,4n

−2

+

⎡

⎣

I7,n

I4,7n

4

+

I7,n

I4,7n

−4⎤

⎦

×16

I7,n I7,4n

8

+

I7,n I7,4n

−8

−288

I7,n I7,4n

4

+

I7,n I7,4n

−4

+ 352

+

⎡

⎣

I7,n

I4,7n

2

+

I7,n

I4,7n

−2⎤

⎦

×296

I7,n I7,4n

2

+

I7,n I7,4n

−2

−256

I7,n I7,4n

6

+

I7,n I7,4n

−6

−8

I7,n I7,4n

10

+

I7,n I7,4n

−10

+ 1746

=

I7,n I7,4n

12

+

I7,n I7,4n

−12

+ 145

I7,n I7,4n

8

+

I7,n I7,4n

−8

+ 496

I7,n I7,4n

4

+

I7,n I7,4n

−4

.

(65)

(i)I7,2=2−1/4

×

⎛

⎝ −8−5√

2 + 163 + 116√

2

−

#

−4+

−8−5√

2+ 163 + 116√

2

2⎞

⎠

1/4

,

(ii)I7,4= 3

−6 + 5√

2−−58 + 45√

2

√

(iii)I7,1/2 =2−1/4,

×

⎛

⎝ −8−5√

2 + 163 + 116√

2

+

#

−4+

−8−5√

2+ 163+116√

2

2⎞

⎠

1/4

(iv)I7,1/4 = 3

−6 + 5√

2 +

−58 + 45√

2

√

(66)

Trang 7

Proof Setting n = 1/2 and simplifying using Theorem 16

(ii), we obtain

I24

7,2+I −24

7,2

+ 32

I20 7,2+I −20 7,2

−40

I16 7,2+I −16 7,2

−96

I12

7,2+I −12

7,2

−192

I8 7,2+I −8 7,2

+64

I4 7,2+I −4 7,2

+462=0.

(67) Equivalently,

A2−4

A4+ 32A3−42A2−128A −191

=0, (68) where

A = I7,24 +I7,2−4. (69)

By using the fact inRemark 17, it is seen that the first factor

of (68) is nonzero, and so from the second factor, we deduce

that

A = −8−5√

2 + 163 + 116√

Combining (69) and (70) and noting thatI7,2< 1, we obtain

I7,2=2−1/4

⎛

⎝ −8−5√

2 + 163 + 116√

2

−

#

−4 +

−8−5√

2+ 163 + 116√

2

2⎞

⎠

1/4

.

(71) This completes the proof of (i)

To prove (ii), setting n = 1 and simplifying using

Theorem 16(i), we arrive at

E2

E2−4

E4+ 108E2−1134

=0, (72) where

E = I2 7,4+I −2

Using the fact inRemark 17it is seen that the first two

factors of (72) are nonzero, and so solving the third factor,

we obtain

E =3 −6 + 5√

Combining (73) and (74) and noting thatI7,4 < 1, we

deduce that

I7,4= 3

−6 + 5√

2−−58 + 45√

2

√

So the proof of (ii) is complete

Now (iii) and (iv) follow from (i) and (ii), respectively,

andTheorem 16(ii)

I7,n I7,25n

2

+

I7,n I7,25n

−2

−

⎧

⎨

⎩

I7,25n

I7,n

3

+

I7,25n

I7,n

−3⎫⎬

⎭

+ 5

⎧

⎨

⎩

I7,25n

I7,n

2

+

I7,25n

I7,n

−2⎫

⎬

⎭

−10

I7,25n

I7,n

+

I7,25n

I7,n

+ 15=0.

(76)

Proof Using (5) inLemma 9, we find that

Q =

* +qχq5

χ

q7

χ

q

χ

q35 , R =

* +q3/2 χ

q

χ

q5

χ

q7

χ

q35 . (77)

Settingq = e − π √

n/7and using the definition ofI k,n in (77),

we get

Q =

I7,25n

I7,n

1/2

, R =I7,n I7,25n

1/2

. (78)

Employing (78) in (27), we complete the proof

(i)I7,5=

h − √ −36 +h2

√

(ii)I7,1/5 =

h + √

−36 +h2

√

(iii)I7,25=

d − √ −144 +d2

(iv)I7,1/25 =

d + √

−144 +d2

(79)

where h = 5 + (62−6√

105)1/3+ (62 + 6√

105)1/3 and d =

12 + 22/3(135−15√

21)1/3+ 22/3(135 + 15√

21)1/3 Proof Setting n =1/5 inTheorem 29and simplifying using

I6 7,5+I −6 7,5−5

I4 7,5+I −4 7,5

+ 10

I2 7,5+I −2 7,5

−17=0. (80) Equivalently,

H3−5H2+ 7H −7=0, (81) where

H = I2 7,5+I −2

H =

5 +

62−6√

1051/3

+

62 + 6√

1051/3

3

(83)

Trang 8

Combining (82) and (83) and noting thatI7,5< 1, we deduce

that

I7,5=

h − √ −36 +h2

√

where

h =5 +

62−6√

1051/3

+

62 + 6√

1051/3

. (85)

Again settingn =1 and simplifying usingTheorem 16(i),

we arrive at

U3−6U2+ 7U −3=0, (86)

where

U = I7,25+I −1

Solving (86) and usingRemark 17, we get

U =

6 +

15

9− √21

/21/3

+

15

9 +√

21

/21/3

(88) Combining (87) and (88) and noting thatI7,25< 1, we obtain

I7,25=

d − √ −144 +d2

(89) where

d =12 + 22/3

135−15√

211/3

+ 22/3

135 + 15√

211/3

.

(90) This completes the proof of (ii) Now (ii) and (iv) easily

fol-low from (i) and (ii), respectively, andTheorem 16(ii)

I13,n

I13,9n

2

+

I13,n

I13,9n

−2

−3

⎧

⎨

⎩

I13,n

I13,9n

+

I13,n

I13,9n

−1⎫⎬

⎭

−

I13,n I13,9n

+

I13,n I13,9n

−1

+ 3=0.

(91)

Proof Proceeding as in the proof ofTheorem 29, using (5)

inLemma 10, settingq : = e − π √

n/13, and using the definition

ofI k,n, we arrive at

Q =

I13,9n

I13,n

1/2

, T =I13,n I13,9n

1/2

. (92)

(i)I13,3=

1 +√

13−−2 + 2√

13

(ii)I13,9=

2 +√

3−3 + 4√

3

(iii)I13,1/3 =

1 +√

13 +

−2 + 2√

13

(iv)I13,1/9 =

2 +√

3 +

3 + 4√

3

(93)

Proof Setting n =1/3 inTheorem 31and simplifying using

V2−3V −1=0, (94) where

V = I2 13,3+I −2

V = 3 +

√

13

Combining (95) and (96) and noting thatI13,3< 1, we obtain

I13,3=

1 +√

13−−2 + 2√

13

So we complete the proof of (i)

Again settingn =1 and usingTheorem 16(i), we obtain

J2−4J + 1 =0, (98) where

J = I13,9+I −1

J =2 +√

Combing (99) and (100) and noting thatI13,9< 1, we deduce

that

I13,9=

2 +√

3−3 + 4√

3

So the proofs of (ii) is complete Now the proof of (iii) and (iv) follow from (i) and (ii), respectively, andTheorem 16(ii)

I13,25n

I13,n

3

I13,25n

I13,n

−3

−5

⎧

⎨

⎩

I13,25n

I13,n

+

I13,25n

I13,n

−1

+ 2

⎫

⎬

⎭

×

I13, 25n I13,n

+

I13,25n I13,n

−1

+ 2

−

I13,25n I13,n

2

+

I13,25n I13,n

−2

=0.

(102)

Trang 9

Proof Using (5) in Lemma 11, setting q : = e − π √

n/13, and using the definition ofI k,n, we arrive at

Q =

I13,25n

I13,n

1/2

, T =I13,n I13,25n

1/2

(i)I13,5= 3 +

√

65−58 + 6√

65

(ii)I13,25=

c − √ −36 +c2

(iii)I13,1/5 = 3 +

√

65 +

58 + 6√

65

(iv)I13,1/25 =

c + √

−36 +c2

(104)

where c =6 + (1080−15√

39)1/3+ (1080 + 15√

39)1/3 Proof Setting n = 1/5 and simplifying using

Theorem 16(ii), we arrive at

where

L = I2 13,5+I −2

L =3 +

√

65

noting thatI13,5< 1, we obtain

I13,5=

3 +√

65−58 + 6√

65

A3−6A2−23A −18=0, (109) where

A = I13,25+I −1

A =

6 +

1080−15√

391/3

+

1080 + 15√

391/3

Employing (111) and (110), solving the resulting equation, and noting thatI13,25< 1, we obtain

I13,25=

c − √ −36 +c2

wherec =6 + (1080−15√

39)1/3+ (1080 + 15√

39)1/3 This completes the proof of (ii) Now the proofs of (iii) and (iv) follow from (i) and (ii), respectively, and

Theorem 16(ii)

I7,n I7,9n

+

I7,n I7,9n

−1

−

⎧

⎨

⎩

I7,9n

I7,n

2

+

I7,9n

I7,n

−2⎫⎬

⎭+ 3=0.

(113)

n/7, and using the definition ofI k,n, we arrive at

Q =

I7,9n

I7,n

1/2

, R =I7,n I7,9n

1/2

(i)I7,3=

5− √21 2

1/4

,

(ii)I7,9=

1 +√

21−6 + 2√

21

(iii)I7,1/3 =

5 +√

21 2

1/4

,

(iv)I7,1/9 =

1 +√

21 +

6 + 2√

21

(115)

Proof Setting n = 1/3 and simplifying using

Theorem 16(ii), we arrive at

I4 7,3+I −4

I7,3=

5− √21 2

1/4

where

D = I7,9+I −1

Trang 10

D =

1 +√

21 2

noting thatI7,9< 1, we deduce that

I7,9=

1 +√

21−6 + 2√

21

This completes the proof of (ii) Now the proofs of (iii) and

(iv) follow from (i) and (ii), respectively, andTheorem 16(ii)

5 General Theorems and Explicit Evaluations of

GnkGn/k

In this section we evaluate some explicit values of the

productG nk G n/kby establishing some general theorems and

employing the values ofI k,nobtained inSection 4 We recall

fromTheorem 22(ii) thatG1/n = G nfor ready references in

this section

(i)

I3,n I3,4n

2

+

I3,n I3,4n

−2

+ (G n/3 G4n/3 G9n/3 G36n/3) −1

−2(G n/3 G4n/3 G9n/3 G36n/3) =0,

(ii)I6

3,n+I −6

3,n+ 2√

2

(G n/3 G3n) −3−(G n/3 G3n)3

=0.

(122)

Proof To prove (i), using (5) in Lemma 13, setting q : =

e − π √

n/3, and employing the definitions of I k,n and G n, we

obtain

R =

* +q1/4 χ

q

χ

q2

χ

q3

χ

q6 =I3,n I3,4n

1/2

,

P =(G n/3 G4n/3 G9n/3 G36n/3) −1/2

(123)

Employing (123) in (31), we complete the proof (ii) follows

similarly from Lemma 5 and the definition of I k,n andG n

withq : = e − π √

n/3

(i)G6G3/2 =

#

1 +√

3

(ii)G12G4/3 =2−13/6 6 + 3√

6 + −2 + 3√

6

, (iii)G39G13/3 =21/6

3 +√

131/3

.

(124)

Proof Setting n = 1/2 in Theorem 37(i) and simplifying

usingTheorem 16(ii) and the resultG1/n = G n, we obtain

2(G G /2)2−(G G /2) −2−2=0. (125)

Solving (125) and noting thatG6G3/2 > 1, we complete the

proof of (i)

To prove (ii), setting n = 1 in Theorem 37(i); using

Theorem 16(i), and noting thatG1/n = G n, we obtain

I2 3,4+I −2 3,4+

G2G4/3 G12

−1

−2

G2G4/3 G12

=0. (126) Employing (53) in (126), solving the resulting equation, and noting thatG2G4/3 G12> 1, we obtain

G2G4/3 G12=

6 + 3√

6 +

−2 + 3√

6

Using the value G3 = 21/12 from [5, p 189] in (127), we complete the proof of (ii)

To prove (iii), setting n = 13 in Theorem 37(ii), we obtain

I6 3,13+I −6 3,13+ 2√

2

(G3/13 G39)−2−(G3/13 G39)2

=0 (128)

Cubing (96) and then employing in (128) and solving the resulting equation, we complete the proof

(i)

I5,n I5,4n

3

+

I5,n I5,4n

−3

+ 5

I5,n I5,4n

+

I5,n I5,4n

−1

=4(G n/5 G4n/5 G25n/5 G100n/5)2

−(G n/5 G4n/5 G25n/5 G100n/5) −2,

(ii)I5,3n+I5,− n3+ 2

(G n/5 G5n) −2−(G n/5 G5n)2

=0.

(129)

n/5, and employing the definitions ofI k,nandG n, we obtain

R =

* +q1/2 χ

q

χ

q2

χ

q5

χ

q10 =I5,n I5,4n

1/2

,

P =(G n/5 G4n/5 G25n/5 G100n/5)−1/2

(130)

Employing (130) in (32), we complete the proof of (i) Similarly, (ii) follows fromLemma 6and the definition ofI k,n

andG nwithq : = e − π √

n/5

(i)G10G5/2 =

3 +√

10 2

1/4

,

(ii)G20G5/4 =1

2

-−3

11+5√

51/4

+

11 + 5√

53/4

+M, (131)

where M denotes 16 +

11 + 5√

5(−3 +

11 + 5√

5)2 Proof Setting n = 1/2 in Theorem 39(i) and simplifying usingTheorem 16(ii) and the resultG1/n = G n, we obtain

4(G G /2)4−(G G /2) −4−12=0. (132)

q1/6(χ (q) /χ (q5)) and Y : = q1/3(χ (q2)/χ (q10));... Letn1,n2, andn3be positive integers such thatn3= n1n2 Suppose that the...

respectively, are defined bym = z1/z n1andm = z n2/z n3

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