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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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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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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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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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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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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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(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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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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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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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 msub@yahoo.com

Amit D. Patel

Department of Applied Mathematics and Humanities

S.V. National Institute of Technology

Surat-395 007, India

E-mail address: ptlamit83@gmail.com

Jyotindra C. Prajapati

Department of MathematicsCharotar Institute of Technology

Changa, Anand-380 421, India

E-mail address: jyotindra18@rediffmail.com

Ajay K. Shukla

Department of Applied Mathematics and Humanities

S.V. National Institute of Technology

Surat-395 007, India

E-mail address: ajayshukla2@rediffmail.com