Certain Transformations Involving q Series

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©IJRASET (UGC Approved Journal): All Rights are Reserved

Certain Transformations Involving q-Series

Rajesh Pandey1, Pravesh Chandra Srivastava2

1

Department of Applied Science, Institute of Engineering & Technology, Sitapur Road, Lucknow 226021 India.

2

Department of Applied Science, IMS Engineering College, Ghaziabad 201013, India.

Abstract : In this paper, certain transformations for basic hypergeometric series have been established by making use of a known identity.

Keywords: Hypergeometric functions, Summation ,Transformation, Polybasic, Converges

I. INTRODUCTION

In the following identity

n m n n n 1 r

m r r r 1

m 0 r 0 r 0 m 0 r 0 m 0

m m

 

     

       



(1.1)

If we take  _j zj, (1.1) takes the following form :

n n n 1 r

m r

m m

m 0 m 0 r 0 m 0

z (1 z) z m



   

     



 

(1.2)

II. DEFINITIONSANDNOTATIONS

For real or complex q( q 1), we get

j

j 0

( ; q) (1 q )

 



 



  (2.1)

and let ( ; q) _be defined by

( ; q) ( ; q)

( q ; q)



 

 

 

 (2.2)

For arbitrary parameters λ and μ, so that

n _{n 1}

1 , n 0

( ; q)

(1 )(1 q)...(1 q  ), n (1, 2, 3..)  



 _{ }

      

 

(2.3)

A truncated basic hypergeometric series is defined as

r

n i n N

1 2 r i 1

r s _s

1 2 s _N _{n 0}

j n n

j 1

[a ; q] z a , a ,...a ; q; z

b , b ....b

[b ; q] (q; q)



 

 _ _ 

 







(2.4)

Where q 1, z 1 and no zero appear in the denominator.

The truncated polybasic hypergeometric series of one variable is defined as

m

1 1 m 1

m

N n

1 2 r 1,1 m,1 m,r 1 m 1 2 r n j,1 r j n

r s

1 2 s 1,1 1s ms ms N n 0 1 2 s n j 1 j,1 j,s j n

a , a ,...a ;c ,...c ;...c ; q, q ,....q ; z [a , a ,...a ;q] z [c ,...a ; q ] b , b ....b : d ,..., d ;....;d ;...d : [q, b , b ,...b ; q] [d ,...d ;q ]

 

 

   

 

 





(2.5)

The series (2.5) converges for q q ,... q1 m 1, z 1.

The other notations appearing in this paper shall stand for their usual meaning. We shall use the following summations of truncated series in our analysis

n 2 1

n n

a, y; q [aq, yq;q] ayq [q, ayq;q]

 

 _ _ 

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n 4 3 n n n 1 a, q a , q a , e;q;

[aq, eq; q] e

aq aq _[q, _{;q] e} a , a ,

e e           _      (2.7) n 6 5 n n

a, q a , q a , b, c,d; q;q

[aq, bq, cq, dq;q] aq aq aq _{aq aq aq} a , a , , , _[q, _, _, _{; q]}

b c d _b _c _d

 _       _      (2.8)

Provided a= bcd.

k k k n

n

k k n n

n n

k 0

k

(1 ap q )[a; p] [c; q] c [ap; p] [cq; q] c ap [q; q] [ap / c; p] (1 a)[q; q] ; p

c         _ _  



(2.9)

k k k k k

k n

n

k n

k 0

k k n n

a aq

(1 ap q )(1 bp q )[a, b; p] c, ; q q [ap, bp; p] cq, ; q

bc bc

aq ap aq ap

(1 a)(1 b) q, ;q , bcp; p q, ;q , bcp; p

b c b c

        _ _ _ _                _ _{ } _ _ _{ } _        



(2.10)

2

k k k k k

k n k k 0 k k b ad

(1 adp q )(1 p q )[a, b; p] c, ; q q

d bc

b adq adp bcp

(1 ad)(1 ) dq, ; q , ; p

d b c d

      _ _           _ _{ } _    



=

2 2 n n 2 n n ad q ad

[ap, bp; p] cq, ; q (1 a)(1 b)(1 c) 1

bc

bc _{(b ad)(c ad)(d} _{bc)(1 d)}

b c ad adq adp bcp _{d(1 a)(1 b)(1 c)(bc ad )}

d(1 ad) 1 1 1 dq, ; q , ; p

d d bc b c d

           _  _               _ _         _        _  _  _{ }  _{ }  _ _ _{ } _               (2.11)

Which is m=0 case of [Gasper and Rahman 3; App. II (II.36)]

k k k k k k

k k k k

k k k k k k

n

k 0

d P Q b p P ad p Q (1 adp q P Q ) 1 1 1

cp q dq Q bcq P d b ad

(1 ad) 1 1 1 c d bc

         _   _{ }   _{ }  _              _  _{ }  _{ }  _      



˟

2 2 2 2 2k

k k k

k

k k

ad

[a, p ] [c, q ] [b, P ] ; Q q bc

qPQ qPQ ad pPQ pPQ ad pqQ pqQ bc pqP pqP

d ; ; ; ;

p p c q q b P P d Q Q

                  _ _{ }       = 2 ad (1 a)(1 b)(1 c) 1

bc b ad (1 ad)(c d) 1 1

d bc      _  _         _  _{ }  _    

˟

2

2 2 2 2 2 2 2 2

n n n

n

2

k k

n k

a d [ a p ; p ] [ c q ; q ] [ b P ; P ] Q ; Q

b c ₁ _{( b} _{a d ) ( c} _{a d ) ( d} _{b c ) (1} _{d )} b c p q P p q P

q P Q q P Q a d p P Q p P Q a d p q Q p q Q d (1 a ) (1 b ) (1 c ) ( b c a d ) ;

d ; ; ;

d Q Q

p p c q q b P P

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Which is m = 0 case of [Agarwal et. al.4 ; (18) p. (89)]

III.MAINRESULTS

If we take

m m

m _m

m

[a, y; q] q [q, ayq;q] q

  in (1.2) and make use of (2.6) we get :

n

2 1 2 1

n

n n 1

a, y;q; qz z [aq, yq; q] aq, yq;q; z (1 z)

ayq [q, ayq;q] ayq _

   

 _ _     _ _

   

,

z 1

 (3.1)

As n  (3.1) yields the following transformation

\ ₂ ₁ a, y;q; qz (1 z)₂ ₁ aq, yq; q; z , z 1 ayq ayq

   

 _ _   _ _ 

   

(3.2)

(i) Taking m m_m

m

a, q a , q a; e; q

[q, a , a , aq / e;q] e

  

 

 

 in (1.2) and making use of (2.7) we get

n

4 3 _n 2 1

n 1 n

n

a, q a , q a , e; z / e z [aq, eq; q] aq, eq, q; q; z / e (1 z)

e aq / e e [e,aq / e;q]

a , a aq / e 

            _ _        (3.3)

For z 1 and e 1 (1.1) yields the following transformation, when n  ,

4 3 2 1

a, q a , q a , e; z / e aq, eq, q; q; z / e (1 z)

e aq / e a , a aq / e

          _ _        (3.4)

(ii) Taking m m

m

a, q a , q a , b, c, a / bc;q

[q, a , a , aq / b, aq / c, bcq; q]

  

 

 

 in (1.2) and making use of (2.8) we get





n n

6 5 2 1

n 1 n

n

a, q a , q a , b, c, a / bc;q; zq [aq, bq, cq, aq / bc;q; z] z aq, bq, cq, aq / bc;q; z (1 z)

aq / b, aq / c, bcq q, aq / b, aq / c, bcq; q

q, a , a aq / b, aq / c, bcq 

    

       _ _

  

 

 

, z 1

(3.5) As n , (3.5) yield the following transformation

6 5 4 3

a, q a , q a , b, c, a / bc;q; zq aq, bq, cq, aq / bc; q; z (1 z)

aq / b, aq / c, bcq q, a , a aq / b, aq / c, bcq

          _ _        (3.6) (iii) Taking m

m m m

m

m m m

[a, pq; pq] [a; p] [c;q] c [a; pq] [a; q] [ap / c; p]



  in (1.2) and making use of (2.9) we get



 





 



n

n n

3 2 _n 2 1

n _n _n n 1

z ap; p cq; q

c;a; apq; q, p, pq, z / c cq, ap; p; z / c (1 z)

; ap / c;a; _c _{q; q} _{ap / c; p} ; ap / c       _ _    _ _       (3.7)

For z 1 and c 1 and n , (3.7) yields

3 2 2 1

c;a; apq;q, p, pq, z / c cq, ap; p; z / c (1 z)

;ap / c;a; ;ap / c

     _ _   _ _       (3.8) (iv) Taking m

m m m m

m

m m m m

[apq; pq] [bp / q; p / q] [a, b; p] [c, a / bc;q] q [a; pq] [b; p / q] [q, aq / b; q] [ap / c, bcp; p]

  in (1.2) and making use (2.10), we get





n n n 6 5 n n n

c;aq / bc; ap, bp; q, p, pq, z [ap, bp; p] [cq, aq / bc; q] z aq / b; ap / c, bcp; q, aq / b;q [ap / c, bcp; p]

 

 _ _ 

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+ ₄ ₃

n 1

cq, aq / bc; ap, bp;q, p; z (1 z)

ap / c;ap / c, bcp _

 

 _{ } _

 

, z 1 (3.9)

As n , (3.9) yields the following transformations

6 5 4 3

c;aq / bc; ap, bp; q, p, pq, z cq, aq / bc; ap, bp; q, p; z (1 z)

aq / b; ap / c, bcp; ap / c; ap / c, bcp

   

 _ _   _ _

   

(3.10)

(v) Taking

2

m

m m

m

m m

b p p ad

[adpq; pq] ; [a, b; p] c, ;q q

d q q bc

b p adp bcp

[a; pq] ; [q, aq / b;q; q] , ; p

d q c d

 

 

   

 

   

  _ _

 

In (1.2) and making use of (2.11)

2

7 6

n

c ; a d / b c, q ; a , b; a d pq ; b / d q ; q , p , p q , p / q ; zq d q , ad q / b; ad p / c, b c p / d ; ad ; b / d

 

  

 

 

=

n 2 2

n n

z (1 a)(1 b)(1 c)(1 ad / bc)[ap, bp; p] [cq, ad q; q] d(1 ad)(1 b / d)(1 c / d)(1 ad / bc)[dq, adq / b;q] [adp / c; bcp / d; p]

   

2

(1 a d )(c a d )(d b c )(1 d )

b c d (1 a d )(1 b / d )(1 c / d )(1 a d / b c )

   



   

+

2 2

5 4

n 1

(1 a)(1 b)(1 c)(1 ad / bc) cq;ad q / bc;q; ap, bp; q, p; z (1 z)

d(1 ad)(1 b / d)(1 c / d)(1 ad / bc) dq, adq / b, adp / c, bcp / d _

 

   

 _{  } _

    _ _ , z 1 (3.11)

As n ,(3.11) yields

2

7 6

c ;ad / bc, q; a, b;adpq; b / dq;q, p, pq, p / q; zq dq, adq / b; adp / c, bcp / d; ad; b / d

 

  

 

 

= 2 2

5 4

(1 a )(1 b)(1 c)(1 ad / bc) cq; ad q / bc; q; ap, bp; q, p; z (1 z )

d (1 ad )(1 b / d )(1 c / d )(1 ad / bc) dq, adq / b, adp / c, bcp / d

 

   

    

    _ _

2

(1 ad)(c ad)(d bc)(1 d) bcd (1 ad)(1 b / d)(1 c / d)(1 ad / bc)

   



    (3.12)

IV.CONCLUSION

In this paper, an attempt has been made to establish six certain transformation formulae for basic hypergeometric series by making use of the identity (1.2)

V. ACKNOWLEDGEMENT

My special thanks to my Ph.D supervisor Dr. Rajesh Pandey Department of Applied Science, Institute of Engineering & Technology (IET) Lucknow for his guidance and encouragement. It was not possible to complete this research paper without his necessary support. I am extremely grateful for his constructive support.

REFERENCES

[1] Srivastava, A.K. (1997): Proc. Indian Acad.Sci. 107-1-12.

[2] Agrawal , R.P. : Generalized hypergeometric series and its application to the theory of combinatorial analysis and partition theory (Unpublished monography). [3] Gasper, G. and Rahman, M. (1990) : Basic hypergeometric series, Cambridge University Press.

[4] Agarwal, R.P., Manocha,H.L. and Rao, K.Srinivasa (2001) : Selected topics in special functions, Allied Publishers Limited, New Delhi. [5] L.J. Slater: Generalized Hypergeometric Functions, Cambridge University Press,( 1966).

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