SOME PROPERTIES OF GENERALIZED HYPERGEOMETRIC FUNCTION

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    Commun. Korean Math. Soc. 28 (2013), No. 2, pp. 303317http://dx.doi.org/10.4134/CKMS.2013.28.2.303

    SOME PROPERTIES OF GENERALIZED

    HYPERGEOMETRIC FUNCTION

    Snehal B. Rao, Amit D. Patel, Jyotindra C. Prajapati,

    and Ajay K. Shukla

    Abstract. In present paper, we obtain functions R t(c,,a,b) and R t(c,,a,b) by using generalized hypergeometric function. A recurrence re-lation, integral representation of the generalized hypergeometric function

    2R1(a, b; c; ; z) and some special cases have also been discussed.

    1. Introduction and preliminaries

    The special functions play very important role, particularly the hypergeo-metric function in solving numerous problems of mathematical physics, engi-neering and applied mathematics, is well-known ([4], [6], [9], [15]). This facthas inspired many mathematicians for investigations of several generalizationsof hypergeometric function ([5], [11], [16], [17], [18], [19], [20]).

    The Gauss hypergeometric function is defined [12] as,

    (1) 2F1(a, b; c; z) =

    k=0

    (a)k(b)k(c)kk!

    zk (|z| < 1, c = 0, 1, 2, . . .)

    and the generalized hypergeometric function, in a classical sense, has beendefined [3] by

    pFq

    a1, . . . , ap; z

    b1, . . . , bq

    = pFq [a1, . . . , ap; b1, . . . , bq; z](2)

    =

    k=0

    (a1)k . . . (ap)k(b1)k . . . (bq)k

    zk

    k!(p = q+ 1, |z| < 1)

    and no denominator parameter equals zero or negative integer.

    Received May 9, 2012; Revised December 26, 2012.2010 Mathematics Subject Classification. Primary 33C20, 33E20, 26A33.Key words and phrases. generalized hypergeometric function, recurrence relation, integral

    representation, fractional integral and differential operators.

    c2013 The Korean Mathematical Society

    303

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    304 S. B. RAO, A. D. PATEL, J. C. PRAJAPATI, AND A. K. SHUKLA

    E. Wright [21] has further extended the generalization of the hypergeometricseries in the following form:

    (3) pq(z) =n=0

    (1 + 1n) . . . (p + pn)

    (1 + 1n) . . . (q + qn)

    zn

    n!,

    where r and t are real positive numbers such that 1+q

    t=1 tp

    r=1 r > 0.When r and t are equal to 1, Equation (3) differs from the generalized

    hypergeometric function pFq by a constant multiplier only. This generalizedform of the hypergeometric function has been investigated by M. Dotsenko [2],V. Malovichko [8] and others. One of the interesting special case considered in[2] has the following form:

    2R,1 (z) = 2R1 (a, b; c; , ; z) =

    (c)

    (a) (b)

    n=0

    (a + n)

    b +

    n

    c +

    n

    zn

    n!;(4)

    ( (a) > 0, (b) > 0, (c) > 0) .

    Here , both either positive or negative simultaneously, |z| < 1.The function 2R

    ,1 (z) is not symmetric with respect to the parameters a

    and b. By letting

    = > 0 in Equation (4), Virchenko et al. [18] defined the

    generalized hypergeometric function in a different sense as:

    (5) 2R1 (z) = 2R1 (a, b; c; ; z) =

    (c)

    (b)

    k=0

    (a)k (b + k)

    (c + k) k!zk; > 0, |z| < 1.

    If = 1, then (5) reduces to a Gausss hypergeometric function 2F1(a, b; c; z).

    For > 0, on |z| = 1, the function 2R1 (a, b; c; ; z) is defined provided (c a b) > 0, as discussed in the proposition of Appendix A.Rao et al. [13], [14] studied various properties of generalized hypergeometric

    function in the light of fractional calculus.The Riemann-Liouville fractional integral of order is defined as [10]:For () > 0,

    (6) If (t) =1

    ()

    t0

    (t )1 f() d

    and the fractional differential operator of order defined as [10]:

    (7) Df(t) = Dn

    Inf(t)

    ,

    where () > 0, and n is the smallest integer with the property that n > .The Laplace transform of the function f(z) is defined as [1]:

    (8) L {f(z)} =

    0

    eszf(z) dz.

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    SOME PROPERTIES OF GENERALIZED HYPERGEOMETRIC FUNCTION 305

    2. Fractional op erators and the generalized hypergeometricfunction 2R1 (a, b; c; ; z)

    Consider the function f(t) = 1(b)

    k=0(a)

    k(b+k)(ct)k

    (k!)2= 2F1 (a, b; 1; ct),

    where a, b C ( (a) > 0, (b) > 0) and c is arbitrary constant such that|ct| < 1.

    On applying the fractional integral operator (6) of order on f(t), we give

    If(t) =1

    ()

    t0

    (t )

    1f()

    d

    =1

    ()

    t0

    (t )

    1 1

    (b)

    k=0

    (a)k (b + k) (c)k

    (k!)2

    d

    =t

    (+ 1) (+ 1)

    (b)

    k=0

    (a)k (b + k) (ct)k

    (+ 1 + k) k! ,

    one can easily write this in following form:

    (9)t

    (+ 1)2R1 (a, b; + 1;1; ct) =

    t

    (+ 1)2F1 (a, b; + 1; ct) .

    Here we denote (9) as Rt (c,,a,b), i.e.,

    Rt (c,,a,b) =t

    (+ 1)2R1 (a, b; + 1;1; ct)(10)

    =t

    (+ 1)

    2F1 (a, b; + 1; ct) .

    Now, applying the fractional differential operator (7) of order on f(t), weget

    Df(t) =

    d

    dt

    n In

    1

    (b)

    k=0

    (a)k (b + k) (ct)k

    (k!)2

    = Dn

    tn

    (b)

    k=0

    (a)k (b + k) (ct)k

    (1 + n + k) k!

    which yields

    Df(t) = t

    (1 )2R1 (a, b; 1 ; 1; ct)(11)

    =t

    (1 )2F1 (a, b; 1 ; ct) .

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    306 S. B. RAO, A. D. PATEL, J. C. PRAJAPATI, AND A. K. SHUKLA

    On denoting (11) as

    Rt (c, ,a,b) = t

    (1 )2R1 (a, b; 1 ; 1; ct)(12)

    =t

    (1 )2F1 (a, b; 1 ; ct) .

    3. Properties of the functions Rt (c , , a , b) and Rt (c, , a , b)

    Theorem 3.1. If a, b C ( (a) > 0, (b) > 0) and c is arbitrary constantsuch that |ct| < 1, then

    (13) IRt (c,,a,b) = Rt (c, + ,a,b) ,

    (14) DRt (c,,a,b) = Rt (c, ,a,b) .

    For n N and as any constant;The Laplace transform of Rt (c,, n, + n 1) is given as

    (15) L {Rt (c,, n, + n 1)} =1

    s+1 yn (c; , s) ,

    where yn (c; , s) is the generalized Bessel polynomial [7].

    Proof. From (6) and the left-hand side of (13), we get

    I

    Rt (c,,a,b) =

    1

    ()t

    0 (t )

    1

    R (c,,a,b) d

    =1

    ()

    t0

    (t )1

    (+ 1)2R1 (a, b; + 1;1; c)

    d

    =1

    ()

    t0

    (t )1

    (b)

    k=0

    (a)k (b + k) (c)k

    ( + 1 + k) k!

    d

    which gives

    I Rt (c,,a,b)

    =1

    () 1

    0

    (1 x)1

    t1 (xt)

    (b)

    k=0

    (a)k (b + k) (cxt)k

    ( + 1 + k) k! t dx=

    t+

    ( + + 1)2R1 (a, b; + + 1; 1; ct) = Rt (c, + ,a,b) .

    This is the proof of (13).

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    SOME PROPERTIES OF GENERALIZED HYPERGEOMETRIC FUNCTION 307

    From (7) and left hand side of (14);

    D Rt (c,,a,b) = Dn InRt (c,,a,b)

    = Dn [Rt (c, n + ,a,b)]

    = Dn

    tn+

    (n + + 1)2R1 (a, b; n + + 1;1; ct)

    =t

    ( + 1)2R1 (a, b; + 1;1; ct)

    = Rt (c, ,a,b)

    which is (14).On setting n N, replacing a by n and b by + n 1 in Rt (c,,a,b),

    where as any constant and then taking Laplace transform of (10), it yields

    L {Rt (c,, n, + n 1)}

    = L

    t

    (+ 1)2R1 (n, + n 1; + 1; 1; ct)

    =

    0

    est

    t

    (+ 1)2R1 (n, + n 1; + 1; 1; ct)

    dt

    =1

    s+1

    nk=0

    (n)k ( + n 1)kk!

    cs

    k

    =1

    s+12F0

    n, + n 1; ;

    c

    s

    =1

    s+1 yn (c; , s) .

    Thus, the Laplace transform of Rt (c,, n, + n 1) is

    L {Rt (c,, n, + n 1)} =1

    s+1 yn (c; , s) ,

    where yn (c; , s) is the generalized Bessel polynomial [7].This proves (15).

    Theorem 3.2. Let a, b C ( (a) > 0, (b) > 0) and c is arbitrary constantsuch that |ct| < 1, () < 1. Then

    IRt (c, ,a,b) = Rt (c, ,a,b) ,(16)

    DRt (c, ,a,b) = Rt (c, ,a,b) .(17)

    For n N and as any constant;

    The Laplace transform of Rt (c, , n, + n 1) is given as

    (18) L {Rt (c, , n, + n 1)} =1

    s1 yn (c; , s) ,

    where yn (c; , s) is the generalized Bessel polynomial [7].

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    308 S. B. RAO, A. D. PATEL, J. C. PRAJAPATI, AND A. K. SHUKLA

    Proof. From (6) and left hand side of (16), we get

    IRt (c, ,a,b) =1

    ()t

    0(t )

    1

    R (c, ,a,b) d

    =1

    ()

    t0

    (t )1

    (b)

    k=0

    (a)k (b + k) (c)k

    (1 + k) k!

    d

    which, upon substituting = xt, yields

    IRt (c, ,a,b)

    =1

    ()

    10

    (1 x)1

    t1

    (xt)

    (b)

    k=0

    (a)k (b + k) (cxt)k

    (1 + k) k!

    t dx

    =t

    ( + 1)2R1 (a, b; + 1; 1; ct) = Rt (c, ,a,b) .

    This is the proof of (16).From (7) and left hand side of (17), we get

    DRt (c, ,a,b)

    = Dn

    InRt (c, ,a,b)

    = Dn [Rt (c, n ,a,b)]

    = Dn

    tn

    (n + 1)2R1 (a, b; n + 1; 1; ct)

    =t

    ( + 1)

    ( + 1)

    (b)

    k=0

    (a)k (b + k)

    ( + 1 + k)

    (ct)k

    k!

    = t

    ( + 1)2R1 (a, b; + 1;1; ct) ,

    this can also be written as Rt (c, ,a,b) . This leads to (17).On setting n N, replacing a by n and b by + n 1 in Rt (c, ,a,b),

    where as any constant and then taking Laplace transform of (12), it yields

    L {Rt (c, , n, + n 1)}

    = L

    t

    (1 )2R1 (n, + n 1; 1 ; 1; ct)

    =

    0

    est

    t

    (1 )2R1 (n, + n 1; 1 ; 1; ct)

    dt

    = 1s+1

    nk=0

    (n)k ( + n 1)kk!

    cs

    k

    =1

    s12F0

    n, + n 1; ;

    c

    s

    =

    1

    s1 yn (c; , s) .

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    SOME PROPERTIES OF GENERALIZED HYPERGEOMETRIC FUNCTION 309

    Thus, the Laplace transform of Rt (c, , n, + n 1) is given as

    L {Rt (c, , n, + n 1)} =

    1

    s1 yn (c; , s) ,where yn (c; , s) is the generalized Bessel polynomial [7].

    This is the proof of (18).

    4. Recurrence relation for the generalized hypergeometric function

    2R1 (a, b; c; ; z)

    Theorem 4.1. If (k) > 0, (a) > 0, (b) > 0, (m) > 0, |z| < 1, then

    (m + 1) 2R1 (a, b; c + s + 1; k; z) 2R1 (a, b; m + 2; k; z)(19)

    =

    (k)2

    (m + 2)

    z22R1 (a, b; m + 3; k; z) +

    (k)

    (m + 2){(k) + 2 (m + 1)} z

    2R1 (a, b; m + 3; k; z) + (m) 2R1 (a, b; m + 3; k; z) ,

    where 2R1 (a, b; c; ; z) =ddz 2R1 (a, b; c; ; z) and 2R1 (a, b; c; ; z) =

    d2

    dz2 2R1 (a,b; c; ; z).

    Proof. On applying the fundamental relation of the Gamma function (z + 1)= z (z) to (5), we can write

    (20) 2R1 (a, b; m + 1; k; z) = (m + 1)

    (b)

    n=0

    (a)n (b + kn)

    (m + kn) (m + kn)

    zn

    n!

    and(21)

    2R

    1(a, b; m + 2; k; z) =

    (m + 2)

    (b)

    n=0

    (a)n (b + kn)

    (m + 1 + kn) (m + kn) (m + kn)

    zn

    n!.

    On writing equation (21) as:

    2R1 (a, b; m + 2; k; z)

    (22)

    = (m + 2)

    (b)

    n=0

    1

    (m + kn)

    1

    (m + 1 + kn)

    (a)n (b + kn)

    (m + kn)

    zn

    n!

    = (m + 1) 2R1 (a, b; m + 1; k; z) (m + 2)

    (b)

    n=0

    (a)n (b + kn)

    (m + 1 + kn) (m + kn)

    zn

    n!.

    For our convenience, we, denote the last summation in (22) by S as:

    S = (m + 2)

    (b)

    n=0

    (a)n (b + kn)

    (m + 1 + kn) (m + kn)

    zn

    n!(23)

    = (m + 1) 2R1 (a, b; m + 1; k; z) 2R1 (a, b; m + 2; k; z) .

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    On applying a simple identity 1u

    = 1u(u+1) +

    1u+1

    , (u = kn + m + 1) to (23), we

    obtain

    S = (m + 2)

    (b)

    n=0

    (m + kn) (a)n (b + kn)

    (m + 3 + kn)

    zn

    n!

    + (m + 2)

    (b)

    n=0

    (c + s + kn) (m + 1 + kn) (a)n (b + kn)

    (m + 3 + kn)

    zn

    n!

    and

    S =k

    (m + 2)

    (m + 3)

    (b)

    n=1

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!

    (24)

    +m

    (m + 2) (m + 3)

    (b)

    n=0

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    n! +

    k2

    (m + 2)

    (m + 3)

    (b)

    n=1

    n (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!

    +

    (m + 2)

    (m + 3)

    (b)

    n=1

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!

    +

    (m + 2)

    (m + 3)

    (b)

    n=0

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    n!

    ,

    where = k (2m + 1) and = m (m + 1).We now express each summation in the right hand side of (24) as follows:

    d2dz2

    z22R1 (a, b; m + 3; k; z)

    (25)

    = 2 2R1 (a, b; m + 3; k; z) + 4z 2R1 (a, b; m + 3; k; z)

    + z2 2R1 (a, b; m + 3; k; z)

    and

    d2

    dz2

    z22R1 (a, b; m + 3; k; z)

    (26)

    = (m + 3)

    (b)

    n=0

    (n + 2) (n + 1) (a)n (b + kn)

    (m + 3 + kn)

    zn

    n!

    =

    (m + 3)

    (b)

    n=1

    n (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!

    + 3 (m + 3)

    (b)

    n=1

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!+ 2 2R1 (a, b; m + 3; k; z) .

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    SOME PROPERTIES OF GENERALIZED HYPERGEOMETRIC FUNCTION 311

    (25) and (26) imply

    (m + 3) (b)

    n=1

    n (a)n (b + kn) (m + 3 + kn)

    zn

    (n 1)!(27)

    = z22R1 (a, b; m + 3; k; z) + 4z 2R1 (a, b; m + 3; k; z)

    3 (m + 3)

    (b)

    n=1

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!.

    Let

    d

    dz(z 2R1 (a, b; m + 3; k; z))(28)

    = 2R1 (a, b; m + 3; k; z) + z 2R1 (a, b; m + 3; k; z)

    andd

    dz(z 2R1 (a, b; m + 3; k; z))(29)

    = (m + 3)

    (b)

    n=0

    (n + 1) (a)n (b + kn)

    (m + 3 + kn)

    zn

    n!

    = (m + 3)

    (b)

    n=1

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!+ 2R1 (a, b; m + 3; k; z) .

    From (28) and (29), we get

    (30) (m + 3)

    (b)

    n=1

    (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!= z

    2R

    1(a, b; m + 3; k; z) .

    Combining (27) and (30), it yields

    (m + 3)

    (b)

    n=1

    n (a)n (b + kn)

    (m + 3 + kn)

    zn

    (n 1)!(31)

    = z22R1 (a, b; m + 3; k; z) + z2R1 (a, b; m + 3; k; z) .

    Now applying (30) and (31) to (24), we get

    S =

    k2

    (m + 2)

    z22R1 (a, b; m + 3; k; z)(32)

    +

    k2 + k + (m + 2)

    z2R1 (a, b; m + 3; k; z)

    +(m + )

    (m + 2)2R1 (a, b; m + 3; k; z) .

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    312 S. B. RAO, A. D. PATEL, J. C. PRAJAPATI, AND A. K. SHUKLA

    From (22), (23) and (32), we arrive at

    (m + 1) 2R1 (a, b; m + 1; k; z) 2R1 (a, b; m + 2; k; z)

    =

    k2

    (m + 2)

    z22R1 (a, b; m + 3; k; z) +

    k

    (m + 2){k + 2 (m + 1)}

    z2R1 (a, b; m + 3; k; z) + m2R1 (a, b; m + 3; k; z) .

    5. Some integral representations of the generalized hypergeometricfunction

    Theorem 5.1. For k > 0, (a) > 0, (b) > 0, (m) > 0, (m a b) > 0we get

    (33)

    10

    tm2R1

    a, b; m; k; tk

    dt =2R1 (a, b; m + 1; k; 1)

    (m) 2R1 (a, b; m + 2; k; 1)

    (m) (m + 1) .

    Proof. On putting z = 1 in (23), it yields

    (m)

    (b)

    n=0

    (a)n (b + kn)

    (m + 1 + kn) (m + kn) n!(34)

    =2R1 (a, b; m + 1; k; 1)

    m

    2R1 (a, b; m + 2; k; 1)

    (m + 1) m.

    Let

    z

    0

    tm2R1 a, b; m; k; tk dt = z

    0

    tm (m)

    (b)

    n=0

    (a)n (b + kn)

    (m + kn)

    tkn

    n! dt(35)=

    (m)

    (b)

    n=0

    (a)n (b + kn) zm+1+kn

    (m + 1 + kn) (m + kn) n!.

    Now, comparing (34) and (35) after setting z = 1 in (35), we have10

    tm2R1

    a, b; m; k; tk

    dt =2R1 (a, b; m + 1; k; 1)

    m

    2R1 (a, b; m + 2; k; 1)

    (m) (m + 1).

    This is the proof of Theorem 5.1.

    Theorem 5.2. If a,b,c, C such that (a) > 0, (b) > 0, (c) > 0, (c) > 0, k > 0 and |z| < 1, then

    2R1 (a, b; c; ; z)

    = kzc

    0

    exp

    tk

    zk

    tc1

    n=0

    (b + n) (c) (a)n tn

    (b) (c + n) c+n

    k

    dt.

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    SOME PROPERTIES OF GENERALIZED HYPERGEOMETRIC FUNCTION 313

    Proof. Consider,

    0exp

    t

    k

    zktc1

    n=0(b+n)(c)(a)

    ntn

    (b)(c+n)( c+nk )

    dt and su-

    bstituting tk

    zk = u, we get

    =

    0

    euzc1uc1

    k

    n=0

    (b + n) (c) znun

    k (a)n

    (b) (c + n) n!

    c + n

    k

    zk u 1kk du

    =zc

    k

    n=0

    (b + n) (c) zn (a)n (b) (c + n) n!

    c+n

    k

    c + nk

    which, on further simplification, gives

    0

    exptk

    z

    ktc1

    n=0

    (b + n) (c) (a)n tn

    (b) (c + n) c+n

    k dt

    =zc

    k2R1 (a, b; c; ; z) .

    This is the proof of Theorem 5.2.

    Theorem 5.3. If a,b,c, C such that (a) > 0, (b) > 0, (c) > 0, (c) > 0 and |z| < 1. Then

    2R1 (a, b; c; ; z) = (c)

    () (c )

    10

    1 t 1

    c1

    2R1 (a, b; ; ; tz) dt.

    Proof. Let 10

    1 t

    1

    c12R1 (a, b; ; ; tz) dt

    =

    10

    1 t 1

    c1 n=0

    (a)n (b + n) () (tz)n

    (b) ( + n) n!

    dt.

    On substituting t1 = u, we get

    10

    1 t 1

    c1

    2R1 (a, b; ; ; tz) dt

    =

    n=0

    (a)n (b + n) () (z)n

    (b) ( + n) n! (c , + n)

    = () (c )

    (c)2R1 (a, b; c; ; z) .

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    314 S. B. RAO, A. D. PATEL, J. C. PRAJAPATI, AND A. K. SHUKLA

    This can easily be written as

    2R1 (a, b; c; ; z) =

    (c)

    () (c )1

    0

    1 t

    1c1

    2R1 (a, b; ; ; tz) dt.

    This is the proof of Theorem 5.3.

    Theorem 5.4. If a,b,c, C such that (a) > 0, (b) > 0, (c) > 0, (c) > 0 and |z| < 1, then

    2R1 (a, b; c; ; z)

    = (c)

    () (c )

    10

    t1 (1 t)c1 2R1 (a, b; c ; ; z (1 t)

    ) dt.

    Proof.

    1

    0

    t1 (1 t)c12R1 (a, b; c ; ; z (1 t)

    ) dt

    =

    10

    t1 (1 t)c1

    (c )

    (b)

    n=0

    (a)n (b + n)

    (c + n)

    zn (1 t)n

    n!

    dt

    = (c )

    (b)

    n=0

    (a)n (b + n)

    (c + n)

    zn

    n!

    10

    t1 (1 t)c+n1

    dt

    = () (c )

    (c)

    (c)

    (b)

    n=0

    (a)n (b + n)

    (c + n)

    zn

    n!

    = () (c )

    (c)2R1 (a, b; c; ; z) .

    This gives

    2R1 (a, b; c; ; z)

    = (c)

    () (c )

    10

    t1 (1 t)c1

    2R1(a, b; c ; ; z (1 t)

    )dt,

    which proves Theorem 5.4.

    Appendix: A

    Proposition. For > 0, and (a) > 0, (b) > 0, (c) > 0; on |z| = 1, thefunction 2R1 (a, b; c; ; z) is defined provided (c a b) > 0.

    Proof. For > 0, and (a) > 0, (b) > 0, (c) > 0 the function 2R1(a, b; c; ;z) is

    (36) 2R1 (z) = 2R1 (a, b; c; ; z) =

    (c) (b)

    n=0

    (a)n (b + n) (c + n) n!

    zn.

    We have to prove its convergence condition for |z| = 1 by using the comparisontest.

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    SOME PROPERTIES OF GENERALIZED HYPERGEOMETRIC FUNCTION 315

    For |z| = 1,

    n=0 un =

    n=0

    (c)(b)

    (a)n

    (b+n)(c+n)n!

    zn

    = 1 +

    n=1

    (c)(b)

    (a)n

    (b+n)(c+n)n!

    and for (c a b) > 0, n=1 vn = n=1 1

    n

    1+ ; = 12

    (c a b) > 0,limn

    un

    vn

    = limn

    (c) (b) (a)n (b + n) (c + n) n!

    1

    n1+

    = limn

    n1+ (c) (b) (a)n (b + n) (c + n) n!

    = limn

    (a)n(n 1)!na (b + n) (b) (n 1)!nb (c) (n 1)!nc

    (c + n)

    (n 1)!n1+

    n!ncab

    = lim

    n 1

    (a)

    (b + n)

    (b) (n 1)!nb (c) (n 1)!nc

    (c + n) limn 1

    ncab = (Finite Number) lim

    n

    1ncab (Explanation is given below in Section I)

    = 0, because (c a b ) = 2 > 0.

    Therefore

    n=0 un= 1+

    n=1

    (c)(b) (a)n(b+n)(c+n)n! is convergent, since n=1 vnis convergent.

    Thus the series in (36) is absolutely convergent on |z| = 1 when (c a b)> 0.

    Hence, 2R1 (z) = 2R1 (a, b; c; ; z) =

    (c)(b)

    n=0(a)

    n(b+n)

    (c+n)n! zk is convergent

    for |z| = 1 when (c a b) > 0.

    Section I : For (a) > 0, n N, 1, 0 < | (a + n)| | (a + n)|. There-

    fore, 0 < 1(a+n) 1(a+n) .

    Thus

    0 < limn

    (n 1)!na (a + n) limn

    (n 1)!na (a + n) 0 < limn

    (n 1)!na (a + n) 1.

    (37)

    Also, for (a) > 0, n N, 0 < 1, 0 < | (a + n)| | (a + n)|.Which implies that

    0 0.

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    Snehal B. Rao

    Department of Applied Mathematics

    The M.S. University of Baroda

    Vadodara-390 001, India

    E-mail address: sbr [email protected]

    Amit D. Patel

    Department of Applied Mathematics and Humanities

    S.V. National Institute of Technology

    Surat-395 007, India

    E-mail address: [email protected]

    Jyotindra C. Prajapati

    Department of MathematicsCharotar Institute of Technology

    Changa, Anand-380 421, India

    E-mail address: [email protected]

    Ajay K. Shukla

    Department of Applied Mathematics and Humanities

    S.V. National Institute of Technology

    Surat-395 007, India

    E-mail address: [email protected]