ON A TWO VARIABLE RECIPROCITY THEOREM OF
RAMANUJAN
D. D. SOMASHEKARA
Department of Studies in Mathematics
University of Mysore,
Manasagangotri, Mysore-570006,
INDIA.
Email address : [email protected]
and
K. NARASIMHA MURTHY
Department of Mathematics
Government First Grade College,
Kuvempunagar, Mysore-570023,
INDIA.
Email address : [email protected]
Abstract
In this paper, we give a simple proof of the two variable reciprocity
theorem of Ramanujan by using Abel’s lemma on summation by parts.
Keywords : Ramanujan’ s reciprocity theorem, Abel’s lemma on summation by parts, difference operators.
MSC 2000 : 05A30, 33D15.
1
Introduction
In his lost notebook[13] S. Ramanujan has recorded the following beautiful
reciprocity theorem.
µ
¶
1 1 (aq/b)∞ (bq/a)∞ (q)∞
ρ(a, b) − ρ(b, a) =
−
,
(1.1)
b a
(−aq)∞ (−bq)∞
where
µ
¶ ∞
1 X (−1)n q n(n+1)/2 an b−n
ρ(a, b) = 1 +
b
(−aq)n
n=0
and as usual
(a)∞ := (a; q)∞ =
∞
Y
(1 − aq n ) ,
n=0
1
2
(a)n := (a; q)n =
(a; q)∞
,
(aq n ; q)∞
| q |< 1.
The first proof of (1.1) was given by G. E. Andrews[3]. Since then several proofs and applications of (1.1) were independently given by D. D.
Somashekara and S. N. Fathima [14], T. Kim, Somashekara and Fathima
[12], S. Bhargava, Somashekara and Fathima[7], P. S. Guruprasad and N.
Pradeep[10], C. Adiga and N. Anitha[2], B. C. Berndt, S. H. Chan, B. P.
Yeap and A. J. Yee[6], and S.-Y. Kang[11]. For more details one refer the
book by Andrews and Berndt[4]. Kang has also obtained a four variable generalization of (1.1) which is equivalent to the identity of Andrews[Theorem
6, 3], and was further generalized by Z. Zhang[15].
The main objective of this note is to give a simple and elegant proof of
(1.1). Using the Heine’s transformation[9],
∞
X
(α)n (β)n
n=0
(q)n (γ)n
zn =
∞
(γ/β)∞ (βz)∞ X (αβz/γ)n (β)n
(γ/β)n ,
(γ)∞ (z)∞
(q)n (βz)n
(1.2)
n=0
the Jacobi’ s triple product identity[9]
∞
X
(−1)n q n(n+1)/2 z n = (qz; q)∞ (1/z; q)∞ (q; q)∞ ,
(1.3)
n=−∞
and a reformulated Abel’s lemma[1] on summation by parts given by W.
Chu[8]. In his paper, Chu defines the backward and forward difference operator ∇ and 4 respectively for an arbitrary sequence {τk } as
∇τk = τk − τk−1 and 4τk = τk − τk+1 .
Then the Abel’s lemma was reformulated as
∞
X
Ak ∇Bk =
k=−∞
∞
X
Bk 4Ak ,
k=−∞
provided that the series on both sides are convergent and Ak Bk−1 → 0
k → ±∞.
2
as
A Proof of (1.1).
Setting α = −1/a, β = q, z = −bq and let γ → 0 in (1.2), we obtain
ρ(b, a) = 1 +
∞
X
(−1)n q n(n+1)/2 bn a−n−1
n=0
(−bq)n+1
¶
∞ µ
1X
1
=1+
−
(−bq)n .
a
a n
n=0
(2.1)
3
Next
µ
¶ ∞ µ
¶
1 X
1
ρ(a, b) = 1 + −
−
(−bq)−n ,
a
a −n
(2.2)
n=1
on using
(a)−n =
(−1)n q n(n+1)/2
.
an (q/a)n
On using (2.1) and (2.2) on left side of (1.1) and after some simplifications we obtain
¶
∞ µ
X
(a/b)∞ (bq/a)∞ (q)∞
1
.
(−bq)n =
−
a n
(−aq)∞ (−bq)∞
n=−∞
(2.3)
Now let
µ
¶
1
An = −
and Bn = (−bq)n .
a n
Then
∇An =
(−1/aq)n q n
and 4Bn = (−bq)n (1 + bq) .
1 + aq
Hence
¶
∞ µ
∞
X
X
1
1
−
(−bq)n =
An 4Bn ,
a n
(1 + bq) n=−∞
n=−∞
∞
X
1
Bn ∇An ,
=
(1 + bq) n=−∞
¶
∞ µ
X
1
1
=
−
(−bq)n q n .
(1 + aq)(1 + bq) n=−∞
aq n
Iterating ‘m’ times and letting m → ∞ we obtain
¶
∞
∞ µ
X
X
1
1
−
(−bq)n =
(−1)n q n(n+1)/2 (b/a)n . (2.4)
a
(−aq)
(−bq)
∞
∞
n
n=−∞
n=−∞
Using (1.3) with z = b/a, on the right side of (2.4) we obtain (2.3) and hence
(1.1).
References
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m
1x
+
m(m−1) 2
x
1.2
+ ....., J.
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4
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and
Other
Unpublished
pa-
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