+ + + + = q qq q qR

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Abstract: In his first and 'lost' notebook, Ramanujan recorded several values of the ... which was given by Ramanujan on page 366 of his 'lost' notebook [9]. But.
SOME NEW EXPLICIT EVALUATIONS OF RAMANUJAN’S CUBIC CONTINUED FRACTION Chandrashekhar Adiga, K. R. Vasuki & M. S. Mahadeva Naika*.

Abstract: In his first and ‘lost’ notebook, Ramanujan recorded several values of the Rogers – Ramanujan continued fraction and cubic continued fraction. In this paper we establish several new evaluations of Ramanujan’s cubic continued fraction using some modular equations.

1.INTRODUCTION

Ramanujan’s cubic continued fraction G(q) is defined by

q1 / 3 q + q 2 q 2 + q 4 , G (q ) := 1 + 1 + 1 + ...

q < 1.

H. H. Chan [4] has proved several elegant theorems for G(q), many of which are analogues of well-known properties satisfied by the Rogers – Ramanujan Continued Fraction:

R (q ) :=

q1 / 5 q q 2 q 3 ... , 1 +1+ 1 + 1 +

q < 1.

It is generally quite difficult to evaluate G (−e −π

n

) and G (e −π

n

) for specific values of

n. Recently B. C. Berndt et al. [4] have proved general formulas for G (−e −π G ( e −π by

n

n

) and

) in terms of Ramanujan – Weber class invariants Gn and gn which are defined

Gn := 2 −1 / 4 q −1 / 24 (−q; q 2 ) ∞ and g n := 2 −1 / 4 q −1 / 24 (q; q 2 ) ∞

, q = e −π

n

,

1

where n is a positive rational number. ∞

Here

(a; q) ∞ = ∏ (1 − aq n )

,

q < 1.

n =0

Using these general formulas, Berndt et al. [4] have evaluated G (−e −π 37 and G (e −π

n

n

) for n = 1,5,13,

) for n = 2, 10, 22, 58. Other values of G(q) can also be computed from

the following reciprocity theorems proved by Chan [5], [6, Section 6.3] :

[1 − 2G (−e −πα )] [1 − 2G (−e −πβ )] = 3

(1.2)

and

[1 + G (−e −

2πα

)] [1 + G (−e −

2πβ

)] = 3 / 2

(1.3)

where α, β > 0 and αβ = 1. K. G. Ramanathan [7] has proved that G (e −π

10

)=

9+3 6 − 7+3 6 (1 + 5 )

6+ 5

which was given by Ramanujan on page 366 of his ‘lost’ notebook [9].

But

Ramanathan’s proof depends upon Kronecker’s limit formula. The main purpose of this paper is to determine some new values of G (q) and

H(q) := -G(-q). For our evaluations we use some modular equations and transformation formulas stated by Ramanujan in his notebooks [8, pp.199, 241, 324].

2. MAIN THEOREMS Theorem 2.1. We have

H (e −5π / 3 ) =

1 1/ 3 1/ 3 (t1 + t 2 ) , 12

(2.1)

where

t1 = 432(5 + 2 5 ) and

t 2 = −864(2 + 5 ) .

2

Proof. To prove this theorem we shall employ the following transformation formula

[1, Entry 27 (iv) of Chapter 16, p.39]: If αβ = π 2

, α , β > 0 then

e −α / 24

4

α f (e −α ) = e − β / 24

4

β f (e − β ) .

(2.2)

Here f ( − q ) = ( q; q ) ∞ ,

Let P =

f (q) 1 / 12 q f (q 3 )

q < 1.

f (q 5 ) . q 5 /12 f (q15 )

and Q =

(2.3)

3

9 ⎛Q⎞ ⎛ P⎞ = ⎜ ⎟ − ⎜⎜ ⎟⎟ + 5 . Then ( PQ) + 2 ( PQ) ⎝ P⎠ ⎝Q⎠ 3

2

(2.4)

This appears as equation 62.2 of Chapter 25 [3, p.221]. Let q = e −π /

3

. Then putting α = π / 3 , β = 3π in (2.2), we obtain

P = 31 / 4 .

(2.5)

Using (2.5) in (2.4) and then setting Q = i31/ 4 T , we obtain the equation 3

3

⎛ ⎡− i + i 5 ⎤⎞ ⎛ ⎡ ⎤⎞ ⎟ ⎜T − ⎢ − i − i 5 ⎥ ⎟ = 0 . T 6 + 3iT 5 + 5iT 3 + 3iT + 1 = ⎜ T − ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ 2 2 ⎦⎠ ⎝ ⎦⎠ ⎣ ⎣ ⎝

Thus,

⎡− i ± i T = ⎢ 2 ⎣

5⎤ ⎥ . ⎦

⎡− i + i 5 ⎤ 1/ 4 ⎥ , we would find that Q = i3 T < 0. But clearly Q > 0, and 2 ⎣ ⎦

If we took T = ⎢

⎡1 + 5 ⎤ ⎥. 2 ⎣ ⎦

so we deduce that Q = 31 / 4 ⎢

(2.6)

Replacing q by –q in Entry 1 (iv) of Chapter 20 of Ramanujan’s second notebook [8, p.241], [2, p.345], we obtain

3

⎡ f 12 (q ) ⎤ ⎢27 − qf 12 (q 3 ) ⎥ ⎣ ⎦ Setting q = e −5π /

3

1/ 3

=

−1 + 4 H 2 (q) . H (q)

(2.7)

in (2.7) and using (2.6), we find that

12 ⎡ ⎛1+ 5 ⎞ ⎤ 3 ⎟ ⎥ 4t − ⎢27 − 27⎜⎜ ⎟ ⎥ 2 ⎢ ⎝ ⎠ ⎦ ⎣

1/ 3

t − 1 = 0,

where t = H (e −5π / 3 ). Solving this cubic equation, we find that

(

)

(

)

(

1 2 1/ 3 1 1/ 3 1/ 3 1 t1 + t 2 , ω t11 / 3 + ω 2 t 21 / 3 and ω t1 + ω t 21 / 3 12 12 12

)

are the roots. Since, the above cubic equation has at most one positive real root and t = H (e −5π / 3 ) > 0, we conclude that t =

(

)

1 1/ 3 1/ 3 t1 + t 2 . 12

Theorem 2.2. We have

G ( e − 2π ) =

[

]

1 3 − 1+ 3 − 2 3 . 2 4

(2.8)

Proof. To prove this theorem we use the following transformation formula [1, Entry

27(iii) of Chapter 16, p.39] : If αβ = π 2

e −α /12

, α , β > 0 , then 4

α f (−e −2α ) = e − β /12

4

β f ( − e −2 β ) .

(2.9)

Setting α = π / 3, β = 3π in (2.9), we find that

f (−e −2π / 3 ) = 3. e −2π / 9 f (−e −6π )

(2.10)

From Entry 1 (iv) of Chapter 20 of Ramanujan’s second notebook [8, p.241], [2, p.345], we have

f 3 ( − q1 / 3 ) 1 3 + 1/ 3 3 = + 4G 2 ( q ) . 3 q f (−q ) G (q ) This can be rewritten as

4

(2.11)

4Z 3 − (3 + 3 3 ) Z + 1 = 0 , where Z = G (e −2π ) . Setting Z = −

3 1 x + , we obtain the equation 2 2

x 3 − 3x 2 − 3x + 1 = 0 . The standard method of solving such an equation [10, pp.27-28] then gives us x = −1 or x =

( 3 + 1) ± 2 3 . 2

If we took x = -1, we would find that Z > 1. If we took x =

( 3 + 1) + 2 3 , we would 2

find that Z < 0. But clearly 0 < Z < 1, and so we deduce that x=

( 3 + 1) − 2 3 . 2

Hence, it follows that

Z=

1 3⎡ − 1+ 3 − 2 3 ⎤ . ⎥⎦ 2 4 ⎢⎣

Theorem 2.3. We have

(a) H (e −π / 3 ) =

1 , 3 4

(2.12)

(b) H (e −π / 5 3 ) =

1 1/ 3 1/ 3 (t1 + t 2 ) , 12

(2.13)

where

[

t 1 = 432 5 − 2 and

(

5

]

)

t 2 = 864 5 − 2 , (c) H (e −7π / 3 ) =

1 1/ 3 1/ 3 (t1 + t 2 ) , 12

(2.14)

where

5

(

t1 = 864 14 + 3 21 and

)

(

)

t 2 = −1296 9 + 2 21 , (d) H (e −π / 7 3 ) = where

1 1/ 3 1/ 3 (t1 + t 2 ) , 12

t1 = 864 14 − 3 21

(

)

(

)

and

(2.15)

t 2 = 1296 2 21 − 9 , (e) G (e −2π / 3 ) =

1 . 3 2

(2.16)

Proof. As the proof of this theorem is identical with the proof of Theorems 2.1 and 2.2,

we just give only references of the required transformation formula, modular equation and set of values of α, β and q. To prove (2.12) we employ the transformation formula (2.2) and the equation (2.7) with α = π / 3 , β = 3π and q = e −π /

3

.

To prove (2.13) we employ the transformation formula (2.2) and the equation (2.7) with α = π / 3 , β = 3π and q = e −π / 5 3 . To prove (2.14) we employ the transformation formula (2.2) and the modular equation [3, Entry 69 of Chapter 25] with α = π / 7 , β = 7π and q = e −π /

7

, and then

we use (2.7) with q = e −7π / 3 . To prove (2.15) we employ the transformation formula (2.2) and the modular equation [3, Entry 69 of Chapter 25] with α = π / 3 , β = 3π and q = e −π / 7 3 . Then we use (2.7) with q = e −π / 7 3 . To

prove

(2.16)

we

employ

the

transformation

α = π / 3, β = 3π and the equation (2.7) with q = e −2π / 3 .

6

formula

(2.9)

with

Acknowledgment

Authors are thankful to the referee for his valuable comments. This work has been supported by the Council for Scientific and Industrial Research, Govt. Of India, New Delhi for which we are thankful to the authorities concerned. REFERENCES

1. C. Adiga, B. C. Berndt, S. Bhargava and G. N. Watson, Chapter 16 of Ramanujan’s second notebook: Theta – functions and q – series, Mem. Am. Math. Soc., No. 315, 53(1985), American Mathematical Society, Providence, 1985. 2. B. C. Berndt, Ramanujan’s Notebooks, Part III, Springer – Verlag, New York, 1991. 3. B. C. Berndt, Ramanujan’s Notebooks, Part IV, Springer – Verlag, New York, 1994. 4. B. C. Berndt, H. H. Chan and L.-C. Zhang, Ramanujan’s class invariants and cubic continued fraction, Acta Arith. 73 (1995), 67-85. 5. H. H. Chan, On Ramanujan’s cubic continued fraction, Acta Arith. 73 (1995), 343 – 355. 6. H. H. Chan, Contributions to Ramanujan’s Continued Fractions, Class Invariants, Partition identities and Modular Equations, Doctoral Thesis, University of Illinois at Urbana Champaign, 1995. 7. K. G. Ramanathan, On Ramanujan’s continued fractions, Acta Arith. 43 (1984), 209 – 226. 8. S. Ramanujan, Notebooks (2 Volumes), Tata Institute of Fundamental Research, Bombay, 1957. 9. S. Ramanujan, The ‘Lost’ Notebook and Other Unpublished Papers, Narosa, New Delhi, 1988. 10. J. V. Uspensky, Theory of Equations, McGraw-Hill, New York, 1948.

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