Abstract. The present paper concerns with the continued fraction representation for
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1 italian journal of pure and applied mathematics n ( A NOTE ON CONTINUED FRACTIONS AND 3 ψ 3 SERIES Maheshwar Pathak Pankaj Srivastava Department of Mathematics Motilal Nehru National Institute of Technology Allahabad India mpathak81@rediffmail.com pankajs23@rediffmail.com Abstract. The present paper concerns with the continued fraction representation for basic bilateral hypergeometric series. Several special cases are also discussed Mathematics Subject Classification: 33D15. Keywords: continued fractions, basic hypergeometric series, basic bilateral hypergeometric series. 1. Introduction Continued fraction has been centre of attraction for applied mathematicians as well as pure mathematicians of previous centuries. In previous centuries there are so many results, which are established in terms of continued fraction. It is also a tool which act as a bridge between pure and applied mathematicians. So, the attraction of continued fraction for today s mathematicians has also amplified. R.P. Agrawal [1] has given the continued fraction representation for the basic hypergeometric series 2 φ 1. R.Y. Denis [4] has also developed a continued fraction representation for the ratio of two basic bilateral hypergeometric series 2 ψ 2. There are also a number of researchers like R.P. Agarwal [2], G.E. Andrews and D. Bowman, [3], R.Y. Denis and S.N. Singh [5], P. Rai [7], S.N. Singh [8], etc., who have established a number of interesting results for hypergeometric function, basic hypergeometric function and basic bilateral hypergeometric function in terms of continued fraction.in this paper we are developing results for 3 ψ 3 ratio s in terms of continued fraction and we also deduce some interesting special cases with the help of these results. These results may be helpful for further research in this area to developed more results in terms of continued fraction. S.N. Singh [8] has derived an interesting transformation formula which transforms a basic bilateral hypergeometric function into basic hypergeometric function. In this paper, we shall use this transformation formula for the development of our main results. We shall use the following results in order to establish main results.
2 192 m. pathak, p. srivastava (1.1 (1.2 (1.3 r3ψ r3 a b, c, d, b 1 b m 1 b r,..., b mr z b, c, d, b 1 b,..., b r ( b, b az, az b, a ; (1 bc(1 bd(b = ( b, az, az, b 1 ; m1...(b r ; ( ( mr a ; b1 (1 c(1 d b ; br... m 1 b ; m r a, bc, bd, b 1 m 1,..., b r m r r3 φ r2 z [6; (11] bc, bd, b 1,..., b r a, b, c d, e a, b, c = 1 (1 a d, e A 0 ( 1 de abc de abc de abc B 0 1 A ( n (1 a 1 de abc A n = (de/abc(1 b n (1 c n, n = 0, 1, 2,... B n = a[1 (d n /a][1 (e n /a], n = 0, 1, 2,... = a 4 x, a 3, x a 1, x a 2 a 4 x 2, a 3, x 2, a 1 T (xp (x S(xQ(x x 2 a 2 x a 3 a 4 x a 3 a 4 R(x/S(xQ(x T (xp (x S(xQ(x R(x/S(xQ(x T (x 2 P (x 2 S(x 2 Q(x 2... B n [1; (2], 1 [2; (3.15],
3 a note on continued fractions and 3 ψ 3 series 193 T (x = (1 x 2 /a 1 (1 x 2 /a 2 (1 x 2 /a 3 /a 4, S(x = (x/a 1 2 (x/a 2 2 (x/a 3 a 4 2, P (x = 1 xs 1 {(S 3 S 4 (1 S 4 2 }(x 2 2 x 4 5 S 4 x 5 7 S4S 2 1 x 6 9 S4, 3 S M (x = (1 x 1M /a 1 (x 1M /a 2 2 (x/a 3 a 4 2 (x/ 2, Q(x = 1 x 2 4 S 4, R(x = x(1 x 2 2 S 4 (1 x 2 / (1 x 2 /a 1 a Definition and notations A continued fraction is a ratio of the type: a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 a 10 a 11..., a 1, a 2, a 3, a 4,... are real or complex numbers. A Basic hypergeometric series is denoted and defined as: (a Aφ B (b z = n=0 [(a; ] n z n, ( z < 1, < 1, [(b; ] n (; n (a represents seuence of A parameters, (b represents seuence of B parameters. (a; n = { (1 a(1 a(1 a 2...(1 a n 1, when n 0; 1, n=0. A Basic bilateral hypergeometric series is denoted and defined as: (a rψ r z [(a; ] n = z n, ( z < 1, < 1, [(b; ] (b n= n (a and (b represent seuences of r parameters. All the parameters and variable may be real or complex numbers. The other notations appearing in this paper carry their usual meaning. 3. Main results We shall establish the following results.
4 194 m. pathak, p. srivastava (3.1 and (3.2 a b, c, d b, c, d a b, c, d b, c, d = (a 1 (a b 1 A ( 0 (1 a 1 1 a 2 B A ( n (1 a A n = 1 a (1 3 bcn1 (1 bd n1, n = 0, 1, 2,... B n = a (1 (1 bcn bdn, n = 0, 1, 2,... a a az a 1, az 2 a 1 b, az 2 b, az a 1 b, az az 2 a 1, az 3 a 1 b, az 3 b, az 2 a 1 b, az 2 az 3 az 5 B n 1..., = b(az ba 1a 2 (az (1 az/a 1(1 az/a 2 (1 az 2 /a 1 b / (1 az/a 1 b(1 az/(1 az 2 /a 1 (1 az2 (1 az 2 /a 2 T (azp (az S(azQ(az R(az/S(azQ(az T (azp (az S(azQ(az R(az/S(azQ(az, T (az 2 P (az 2 S(az 2 Q(az 2...
5 a note on continued fractions and 3 ψ 3 series 195 T (az = (1 az 2 /a 1 (1 az 2 /a 2 (1 az 2 /a 1 /a 2, S(az = (az/a 1 2 (az/a 2 2 (az /a 2 z 2 4 2, P (az = 1 azs 1 {(S 3 S 4 (1 S 4 2 }(a 2 z 2 2 a 4 z 4 5 S 4 a 5 z 5 7 S4S 2 1 a 6 z 6 9 S4, 3 S M (az = (1 az 1M /a 1 (az 1M /a 2 2 (az /a 2 z (az/ 2, Q(az = 1 a 2 z 2 4 S 4, R(az = az(1 a 2 z 2 2 S 4 (1 az 2 / (1 az 2 a 2 /a 1 az Proof of main results Proof of (3.1. Taking r = 0 in (1.1, we get a/b, c, d z = (, b/az, az/b, /a; (/b, /az, az, b/a; /b, c, d (4.1 a, bc, bd (1 bc(1 bd (1 c(1 d z bc, bd Replacing a by a in (4.1, we get a/b, c, d z = (, b/az, az/b, 1/a; (/b, 1/az, az, b/a; /b, c, d (4.2 a, bc, bd (1 bc(1 bd (1 c(1 d z bc, bd Taking z = 1/a 2 in (4.1 and z = 1/ in (4.2 and then taking ratio of (4.1 and (4.2 and using result (1.2, we get a b, c, d b, c, d a b, c, d b, c, d = (a 1 (a b 1 1 a 2 1 A ( 0 (1 a 1 1 B 0 1 A ( n (1 a 1 1 B n 1.
6 196 m. pathak, p. srivastava Proof of (3.2. Now, replacing a by az/, bc by az/a 1, bd by az/a 2, c by az/a 1 b, d by az/, z by /az 3 in (4.1, we get az/a 1, az 2 /a 1 b, az 2 / /b, az/a 1 b, az/ /az 3 (4.3 = (, b3, 1/b 2, /az; (/b, 3, 1/ 2, b /az; (1 az/a 1(1 az/a 2 (1 az/a 1 b(1 az/ az/, az 2 /a 1, az 2 /a 2 3 φ 2 /az 3. az/a 1, az/a 2 Replacing a by az 2 /, bc by az 2 /a 1, bd by az 2 /a 2, c by az 2 /a 1 b, d by az 2 /, z by /az 5 in (4.1, we get az 2 /a 1, az 3 /a 1 b, az 3 / /b, az 2 /a 1 b, az 2 / /az 5 (4.4 = (, b4, 1/b 3, /az; (/b, 4, 1/ 3, b /az; (1 az2 /a 1 (1 az 2 /a 2 (1 az 2 /a 1 b(1 az 2 / az 2 /, az 3 /a 1, az 3 /a 2 3 φ 2 /az 5. az 2 /a 1, az 2 /a 2 Now, taking ratio of (4.3 and (4.4, then simplifying and using the result (1.3, we get az a 1, az 2 a 1 b, az 2 b, az a 1 b, az az 2 a 1, az 3 a 1 b, az 3 b, az 2 a 1 b, az 2 az 3 az 5
7 a note on continued fractions and 3 ψ 3 series 197 = b(az ba 1a 2 (az (1 az/a 1(1 az/a 2 (1 az/a 1 b(1 az/ (1 az2 /a 1 b(1 az 2 / (1 az 2 /a 1 (1 az 2 /a 2 T (azp (az S(azQ(az R(az/S(azQ(az T (azp (az S(azQ(az R(az/S(azQ(az T (az 2 P (az 2 S(az 2 Q(az Special cases Here, we shall deduce certain interesting cases of our main results. Putting c = in (3.1, we get a/b, 2, d 1/ /b, d a/b, 2, d (5.1 1/a 2 /b, d = (a 1 (a b 1 P ( 0 (1 a Q 0 1 P ( n (1 a P n = 1 a (1 3 bn2 (1 bd n1, n = 0, 1, 2,... Q n = a (1 (1 bn1 bdn, n = 0, 1, 2,... a a Putting d = 1/b in (5.1, we get a/b, 2 2φ 1 1/ a/b, 1/b 2 (5.2 1/a 2 2φ 1 1/b = (a 1 (a b 1 L ( 0 (1 a M 0 1 L ( n (1 a M n Q n 1 1
8 198 m. pathak, p. srivastava L n = 1 (1 bn2 (1 n1, n = 0, 1, 2,... ( M n = a (1 bn1 1 n, n = 0, 1, 2,... a a Replacing 1/b by in (5.2, we get (5.3 a 2, 2 2φ 1 1/ a, 2 2φ 1 1/a 2 = (a 1 (a 1 1 R ( 0 (1 a 1 1 S 0 1 R n S ( (1 a 1 1 n 1 Putting b = 1 in (3.2, we get R n = 1 (1 n1 2, n = 0, 1, 2,... ( 2 S n = a 1 n, n = 0, 1, 2,... a (5.4 2 az/, az 2 /a 1, az 2 /a 2 /az 3 az/a 1, az/a 2 az 3 /a 1, az 3 /a az /, 2 /az 5 az 2 /a 1, az 2 /a 2 = T (azp (az S(azQ(az R(az/S(azQ(az T (azp (az S(azQ(az R(az/S(azQ(az T (az 2 P (az 2 S(az 2 Q(az 2...
9 a note on continued fractions and 3 ψ 3 series 199 Replacing az by z in (5.4, we get (5.5 2 z/, z 2 /a 1, z 2 /a 2 /z 3 z/a 1, z/a 2 z 3 /a 1, z 3 /a z /, 2 /z 5 z 2 /a 1, z 2 /a 2 = T (zp (z S(zQ(z R(z/S(zQ(z T (zp (z S(zQ(z R(z/S(zQ(z T (z 2 P (z 2 S(z 2 Q(z 2... Similarly, some other interesting special cases could be deduced. References [1] Agarwal, R.P., Pade approximants continued fractions and Heine s - series, J. Math. Phys. Sci., 3 26 (1992, [2] Agarwal, R.P., Resonance of Ramanujan s Mathematics III, New Age International Pvt. Ltd. Publishers, New Delhi, [3] Andrews, G.E. and Bowman, D., A full extension of the Rogers-Ramanujan continued fraction, Proc. Amer. Math. Soc., ( (1995, [4] Denis, R.Y., On generalization of continued fraction of Gauss, Internat. J. Math. & Math. Sci., (4 13 (1990, [5] Denis, R.Y. and Singh, S.N., Certain result involving -series and continued fractions, Far East J. Math. Sci. (FJMS, (5 3 (2001, [6] Gasper, G. and Rahman, M., Basic Hypergeometric Series, Cambridge University Press, New York, 1991.
10 200 m. pathak, p. srivastava [7] Rai, P., Certain bilateral extensions of the Rogers-Ramanujan continued fraction, Ganita, (1 57 (2006, [8] Singh, S.N., Certain transformation formulae for basic and bibasic hypergeometric series, Proc. Nat. Acad. Sci. India (III, 65 (A (1995, Accepted:
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