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Malaysian Journal of Fundamental & Applied Sciences

available online at http://mjfas.ibnusina.utm.my

Partial Sums For Class Of Analytic Functions Defined By Integral Operator

Nagat. M. Mustafa* and Maslina Darus

School of Mathematical Sciences, Faculty of Science and Technology KM Bang 43600 Selangor D. Ehsan, Malaysia Received 29 November 2011, Revised 10 June 2012, Accepted 15 August 2012, Available online 20 August 2012

ABSTRACT

In the present paper, we study the class of analytic functions involving generalized integral operator, which is defined by means of a general Hurwitz Lerch Zeta function denoted by

s bα,

f z ( )

with negative coefficients. The aim of the paper is to obtain the coefficient estimates and also partial sums of its sequence

αs b,

f z ( )

. | Univalent functions | uniformly starlike functions | Hadamard product | partial sums | fractional derivatives and fractional integrals |

® 2012 Ibnu Sina Institute. All rights reserved.

http://dx.doi.org/10.11113/mjfas.v8n4.146 1. INTRODUCTION

Let

A

denote the class of all analytic functions in the open unit disk

U {z ∈ =  : z 1} <

given by the normalized power series of the form

( )

2

(z)

n n

, ( ) . 1

n

f z

a z z U

=

=

+ ∑ ∈

1 . 1 Definition: (Owa and. Srivastava [3]) Let the function

f

be analytic in a simply connected domain of the z-plane containing the origin. The fractional derivative of

f

of order

α

is defined by

0

( )

1

( )

( ) d t , (0 1),

(1 )

z

z f t

D f z d

dz z t

α α

α

= α ≤ <

Γ − ∫ −

where the multiplicityof

(z — t)

αis removed by requiring

log(z — t)

to be real when

z — t 0 >

.

Using Definition 1.1 and its known extensions involving fractional derivatives and fractional integrals, Owa and Srivastava [3] introduced the operator

α

: A — A

as follows:

f (z) (2 )

Zα

D f z

zα

( ), ( 2,3,4,...)

α

α α

Ω =Γ − ≠

*Corresponding author at:

n n

2

(n 1) (2 )

z a z ,( ).

(n 1 )

n

α z U α

=

Γ + Γ −

= + ∈

Γ + −

1.2 Definition:

(

Srivastava and Choi [2

])

A general Hurwitz Lerch Zeta function

Φ ( , , ) z s b

defined

by

( , , ) =

=0

,

( )

n n s

z s b z

n b Φ

∑ +

where (

s ∈  ,b ∈ −  

0when (|z| < 1),

(b) 1

ℜ >

when (|z| = 1).

We can write this function as :

( , , ) ( z ( , , )) Φ

*

z s b = Φ b

s

z s bf (z) ,

then

*

=2

( , , ) = .

( 1)

s n n

n s

z s b z b a z

n b

Φ +

∑ + −

By using Definitions (1.1) and (1.2), the authors [1]

introduced the generalized integral operator

s bα,

:A A →

as the following:

For

0 ≤ < α 1

and

s ∈  ,b ∈ −  

0

,

( ) = ( ) (2 )

s *

( , , ), ( 2,3,4,...)

s b z

α

f z b α z D

α α

z s b α

ℑ Γ − Φ ≠

(2)

n n 2

(n 1) (2 ) b

z a z ,( ).

(n 1 ) (n b 1)

n

s

α z U α

=

 

Γ + Γ −

= +

Γ + −  + −  ∈

Note that :

0

0,b

(z) ff z ( ) .

Special cases of this operator includes:

0,b

f z ( ) f z ( )

α α

ℑ ≡ Ω

is Owa and Srivastava operator [3].

0

, 1

(z)

s,

(z),

s b+

fJ f

b

is Srivastav and A. Attiya

integral operator[5].

0

1,1

f (z) A( ) (z) , ≡ f

is Alexander integral operators [6].

0

1,1

(z) L( ) (z) ,

s+

ff

is Libera integral operators [7].

0

1,δ

(z) fL

δ

( )(z); f

δ > − 1,

is Bernardi integral operator [8].

0

,

(z) I (z);

sσ

f

σ

f

(σ > 0) is Jung- Kim-Srivastava integral operator [9].

0

,1

(z)

n

;

n

f S

(n ∈ N) is the Salagean derivative operator [13].

0

, 1

(z)

s b

f

+

(s ∈ Z) multiplier transformations studied by Flett [14].

Finally, for different choices of

s, b

and

α

, several operators investigated earlier by other authors Cho and Kim [15], and Lin and Owa [16] are obtained.

By using our integral operator we introduce the following class of

A

.

A function

f ∈ A

is said to be in the class denoted by

,

( , ),( 1 1), 0

SP

s bα

β γ − ≤ γ < β ≥

and satisfy

, ,

, ,

( ( )) ( ( ))

( ) ( ) ( ) ( ) 1 ,

s b s b

s b s b

z f z z f z

f z f z

α α

α

γ β

α

 ℑ ′  ℑ ′

 

ℜ ℑ − ≥ ℑ −

where

s ∈  ,b ∈ −  Z

0

.

By suitably specializing the values of

α β γ , ,

and

s

the class

SP

s bα,

( , ) β γ

reduces to the classes introduced and studied by various authors for example: For

00,b

SP ( , ) SP( , ) β γ ≡ β γ

was introduced Rφnning [10], [2].

2. RESULTS & DISCUSSION

2.1 Coefficient Estimates:

Before stating and proving our main results, we derive a sufficient condition giving the coefficient estimates for the function

f

to belong to the class

,

( , )

SP

s bα

β γ

.The result is contained in the following:

2.2 Theorem

A sufficient condition for a function

f

of the form (1) to be in

SP

s bα,

( , ) β γ

is that

[ ]

=2

( 1) (2 )

(1 ) ( ) | | 1 ,

( 1 ) 1

s n n

n b

n a

n n b

β γ β α γ

α

+ − + Γ + Γ −Γ + −   − +  ≤ −

where

s ∈  ,b ∈ −  

0and

( 1 − ≤ γ 1), < β ≥ 0

.

Proof: It suffices to show that

, ,

, ,

( ( )) ( ( ))

( s bs b) ( ) ( s bs b) ( ) 1

z f z z f z

f z f z

α α

α

γ β

α

 ℑ ′  ℑ ′

 

ℜ ℑ − − ℑ −

, ,

( ( ))

(1 ) 1

( ) ( )

s b s b

z f z f z

α

β

α

≤ + −

n 2

n 2

(n 1) (2 ) b

(1 ) ( 1) a

(n 1 ) (n b 1)

(n 1) (2 ) b ,

1 ( 1) a

(n 1 ) (n b 1)

s

n

s n

n n β α

α α α

=

=

 

Γ + Γ −

+ − Γ + −  + − 

≤ − − ΓΓ + Γ −+ −  + − 

[ ]

=2

( 1) (2 )

(1 ) ( ) | | 1 ,

( 1 ) 1

s n n

n b

n a

n n b

β γ β α γ

α

+ − + Γ + Γ −Γ + −   − +  ≤ −

the proof is complete.

2.3 partial sums

In this section we will examine the ratio of a function of the form (1) to its sequence of partial sums defined by

1

( ) =

f z z

and

f z

k

( ) = z + ∑

kn=2

a z

n nwhen the

coefficients of

f

are sufficiently small to satisfy the condition (2). We will determine sharp lower bounds

for

( ) , ( ) , ( ) ( ) .

( )

k

( ) ( )

k

( )

k k

f z f z

f z f z and

f z f z f z f z

     ′   ′ 

ℜ      ℜ     ℜ  ′   ℜ   ′  

2.4 Theorem

Let

f

be given by (1) satisfying (2), then

(3)

1 , ( 2) (2 ) b

( 2 ) b

(1 )( 1) ( )

( ) ( ) 1

s

k k

k k

k

f z f z

γ α

β γ β α

 

  −

 

  Γ + Γ −

 

Γ + − +

+ + − +

ℜ ≥ −

( z U ∈ ). (3) ( )

k

( ) f z f z

 

ℜ   ≥

 

( 2) (2 ) b

[ ( 2 ) b

( 2) (2 ) b

] ( 2 ) b

(1 )( 1) ( )]

[(1 ) [(1 )( 1) ( ) ,

s s

k k k

k k k

k k

αα αα

β γ β

γ β γ β

Γ + Γ −

Γ + − +

Γ + Γ −

Γ + − +

+ + − +

− + + + − +

(z U∈ ). (4)

The results are sharp for every

k

with the function given by

1 ,

( ) = (1 )

( 2) (2 ) b

(1 )( 1) ( )

( 2 ) b

k

s

f z z z

k k k k

γ

β γ β α

α

+

 

   

   

+ −

Γ + Γ −

+ + − +

Γ + − +

(z U k∈ , ≥0). (5) Proof:

We prove (3).Let

f

be given by (1) satisfying (2), by sitting

1 ( 2) (2 )

( 2 )

( )

1 ( )

[(1 )( 1) ( ) ]

(

s

k

k k

b

k b f z f z

k

αα

γ

β γ β

 

 

− Γ + Γ −

Γ + −

+

+ + − +

( 2) (2 )

( 2 )

1 (1 )( 1) ( )

s

)

k k b

k

αα k b

γ

β γ β

  

Γ + Γ −

Γ + − +

+ + − +

1 ( ) 1 w z( )

.

w z +

=

We find that ( ) w z =

[

n-1 n 1

nn-1 n n-1

2 1

( 2) (2 ) b

[ ( 2 ) b z

1 ,

( 2) (2 ) b a z( 2 ) b

2 2 a z 1

(1 )( 1) ( )]

(1 )( 1) ( )]

s

n k s

k

n n k

kk k

kk k

k k

αα γ

αα γ

β γ β

β γ β

= +

= = +

Γ + Γ −

Γ + − +

Γ + Γ −

Γ + − +

+ + −

+ + − +

+ + − +

∑ ∑

( ) w z =

1

1 [

2 n

( n

[ n

( 2) (2 ) b 1 )( 1) ( )] ( 2 ) b

( 2) (2 ) b (1 )( 1) ( )] ( 2 ) b

2 2 a 1

a 1

a

.

n k

n k

s

s k

n

k kk k

k k k k

β γ β αα

β γ β αα

γ γ

= +

= +

+

=

Γ + Γ −

+ + − + Γ + − +

Γ + Γ −

+ + − + Γ + − +

− −

∑ ∑

Now

| w(z) | 1 ≤

if and only if

1 n

( 2) (2 ) b

( 2 ) b

1

2((1 )( 1) ( )]

a

s

n k

k k k

k α α

γ

β γ β

= +

Γ + Γ −

Γ + − +

+ + − +

2 n

2 2

k

a ,

n =

≤ − ∑

which is equivalent to

1 n

k bb ( 2) (2 ) [(1 )( 1) ( )] ( 2 )

1 n k a

k s

k k

α

β γ β α

γ

= +

Γ + Γ − +

+ + − + Γ + −

+

2 n

k

a 1 . (6)

n =

∑ ≤

This will hold if we show that left-hand side of (6) is bounded above by

2

b n

n b 1

1 .

(n 1) (2 )

[(1 ) ( )] (n 1 ) a

n

n

s

γ

β γ β α α

=

+ −

Γ + Γ −

+ − +

Γ + −

This is equivalent to showing that

2 1 an

(n 1) (2 ) b

[(1 ) ( )] (n 1 ) n b 1 1

n

n s γ

γ

β γ β αα

=

− + Γ + Γ −

+ − + Γ + − + − − −

1

(n 1) (2 )(n 1 ) n b 1b ( 2) (2 )( 2 ) k bb

1

(1 )( 1 ) ( ) s s

n k

kk

n k αα αα

γ

β γ β

= +

Γ + Γ −Γ + − + − Γ + Γ −Γ + − +

+ − − − +

a

n

0 , (7)

To see that

( )

(1 ) 1

( ) = ,

( 2) (2 ) b

( 2 ) b

(1 )( 1) ( )

k

s

f z z z

k

k k

k

γ

α

β γ β

α

+

+ Γ + Γ −  

 

Γ + −  + 

+ + − +

( z U k ∈ , ≥ 0),

gives sharp result, we observe that for

i

z ren

π

that =

(4)

Re ( )

k( ) f z f z

 

 

 

 

 ≥

( 2) (2 ) b ,

( 2 ) b

[

1 (1 )( 1) ( )]

1

s zk

k k k

k

αα

γ

β γ β

Γ + Γ −

Γ + − +

+ + − +

+

1 1 ,

n b 1b (n 1) (2 )

[(1 )( 1) ( )] (n 1 )

k s

γ

β γ β αα

+ −

Γ + Γ −

+ + − +

Γ + − when z → 1.

To prove the second part of this theorem, we write bb 1

( 2) (2 )

( 2 ) ( )

( )

(1 ) [(1 )( 1) ( ) ]

(

s

k k

k k f z

f z

k

αα

γ

γ β γ β

+

Γ + Γ − Γ + −

− −

− + + + − +

[ ( 2) (2 ) b

( 2 ) b

( 2) (2 ) b

( 2 ) b

]

( ]

) 1 ) [(1 )( 1) ( )

(1 )( 1) ( )

s

s

k k k

k k k

k

k

αα

αα

γ β γ β

β γ β

Γ + Γ −

Γ + − +

Γ + Γ −

Γ + − +

− + + + − +

+ + − +

1 ( ) 1 w z( )

.

w z +

=

we find that

( ) w z =

1

n 1 2

. 1

( 2) (2 ) b

( ( 2 ) b n

( 2) (2 ) b

( ( 2 ) b n

2 2 a

1 ) [(1 )( 1) ( )]

1 1 ) (1 )( 1) ( )]

a

[ a

n k

n k

s

s k

n

kk k

kk k

k k

αα

αα

γ β γ β

γ

γ β γ β

γ

= +

= = +

Γ + Γ −

Γ + − +

Γ + Γ −

Γ + − +

+

− + + + − +

− + + + − +

∑ ∑

Now

| w(z) | 1 ≤

if and only if

n 1

( 2) (2 ) b

2 ( 2 ) b

1

(1 ) [(1 )( 1) ( )]

a

s

n k

kk k

k αα

γ

γ β γ β

= +

Γ + Γ −

Γ + − +

− + + + − +

n 2

2 2 k a ,

n =

≤ −

which is equivalent to

1 n

( 2) (2 ) b

( 2 ) b

1 a

(1 ) [(1 )( 1) ( )]

n k

k s

k k

k αα

γ

γ β γ β

= +

Γ + Γ −

Γ + − +

− +

− + + + − +

2 n

k

a 1 .

n =

∑ ≤

Making use of (2) to get (7). Finally, equality holds in (4) for the extremal function

f

given by (5).This completes the proof.

2.5 Theorem

Let

f

be given by (1) satisfying (2), then Re '( ) 1

' ( )k f z f z

 

 ≥ −

 

( 1)(1 ) , (8)

( 2) (2 ) b

[(1 )( 1) ( )] ( 2 ) b

s

k

k k k k

γ

β γ β αα

+ −

Γ + Γ −

+ + − + Γ + − +

' ( ) Re

k

'( )

f z f z

 

  ≥

 

1

( 2) (2 ) b

( 2 ) b

( 2) (2 ) b

([(1 )( 1) ( )] ( 2 ) b

.

( ) [(1 )( 1) ( )]

(9)

1 )( 1

n k

s s

k k k

k k k k

k

k αα

β γ β αα

β γ β

γ

= +

Γ + Γ −

Γ + − +

Γ + Γ −

+ + − + Γ + − +

+ + + − +

− +

The results are sharp for every

k

with the function

give 1

(1 ) , ( ) =

( 2) (2 ) b

(1 )( 1) ( ) ( 2 ) b

k

z s

f z z

k k k k

γ

β γ β αα

+

+ −

Γ + Γ −

+ + − +

Γ + − +

( z U k ∈ , ≥ 0).

Proof: To prove the result (8), define thefunction w(z) by

( 2) (2 ) b

( 2 ) b

(

1 )

[ ]

f ' ( ) ( '( )

(1 )( 1) ( )

s

k

k k k

z

f z

k

αα

γ

β γ β

Γ + Γ −

Γ + − +

− −

+ + − +

( ) )

( 2) (2 ) b

( 2 ) b

1- )( 1 [(1 )( 1) ( )]

1

s

,

k k k

k

k

αα

γ

β γ β

 

 

 

 

 Γ + Γ − 

 Γ + − + 

 

 

+

+ + − +

where

(5)

w z = ( )

1

1

n n-1

( .

n n-1 n n-1

2 (

] a z

1 )( 1)

( 2) (2 ) b [(1 )( 1) ( )] ( 2 ) b a z 2 2 a z

( 2) (2 ) b

[(1 )( 1) ( ) ( 2 ) b

1 )( 1)

n k

n k

s k

n n s

k

n k k k k

n

k kk k

k γ

β γ β αα

β γ β αα

γ

= +

= + +

=

− +

Γ + Γ −

+ + − + Γ + − +

+

Γ + Γ −

+ + − + Γ + − +

− +

Now

| ( ) | 1 w z ≤

if and only if

n

) 1

( 2) (2 ) b

[(1 )( 1) ( )] ( 2 ) b

(1 )( 1 n k a

k s

k k k

k n

β γ β αα

γ

= +

Γ + Γ −

+ + − + Γ + − +

− +

n 2

k a 1 .

n

n

=

+

Since the LHS of (8) is bounded above by

2 1 n

n b 1b

(n 1) (2 )

(1 ) ( )] (n 1 ) a .

s

n

n

γ

β γ β α α

=

Γ + Γ −

+ −

+ − +

Γ + −

The proof is complete.

To prove the second part of this theorem, we write ( 2) (2 ) b

( 2 ) b 1 (

(1 )( 1) [(1 )( 1) ( )]

f ' ( ) '( )

s

k

k k k

k k

f z z

αα

γ

γ β γ β

Γ + Γ − Γ + − +

− −

− + + + + − +

( 2) (2 ) b

( 2 ) b

( 2) (2 ) b

( 2 ) b

[(1 )( 1) ( )]

1 ( ) , 1 ( ) (1 )( 1) [(1 )( 1) ( )]

s

s

kk k

kk k

k w z

k k w z

αα

αα

β γ β

γ β γ β

Γ + Γ −

Γ + − +

Γ + Γ −

Γ + − +

+ + − +

= +

− + + + + − + −

we find that

( ) w z =

1

1

n-1

1 (

n-1 1

n )

2

( 2) (2 ) b n ( 2 ) b 1 )( 1)

( 2) (2 ) b( 2 ) b n n-1

2 2 a z (1 )( 1

a z

[(1 )( 1) ( )]

[(1 )( 1) ( )]

a z .

n k

n k

s

s k

n

k k k

k

kk k

n k

k

k

αα

γ

αα

γ

β γ β

β γ β

∞∞

= +

= +

+

+

=

Γ + Γ − Γ + − +

− +

Γ + Γ − Γ + − +

+ + − +

+ + − +

+ + − +

∑ ∑

Now

| ( ) | 1 w z ≤

if and only if

1 n

( 2) (2 ) b

( 2 ) b

2 1 [(1 )( 1) ( )]

(1 )( 1)

a

s

n k

k k k

k

k

n

αα

β γ β

γ

= +

Γ + Γ −

Γ + − +

+ + + − +

− +

n 2

2 2 k a ,

n

n

=

≤ −

which is equivalent to

( 2) (2 ) b

[ ( 2 ) b

(1 1 n

(1 )( 1) ( )]

)( 1)

1 a

k s

k k

n k

k

k n

αα

β γ β

γ

Γ + Γ − Γ + − + ∞

= +

 

+ + − +

 

 

 

 

 

− +

+ ∑

2 n

k a 1

n

n

=

+

From the condition (2), it is sufficient to show that

1 ( ) n

( 2) (2 ) b

( 2 ) b

1 )( 1 ((1 )( 1) [(1 )( 1) ( )]( 1))

a

s n k

kk k

k

k k n k α

α γ

γ β γ β

= +

Γ + Γ −

Γ + − +

− +

− + + + + − + − −

(1 )( 1) n

( 2) (2 ) b

( 2 ) b

2 a 0 .

[(1 )( 1) ( )]

s

k

k k

k k

n

k

γ

αα

β γ β

− +

Γ + Γ −

Γ + − +

+ = ≥

+ + − +

which gives (9). The bound in ( 9) is sharp for all k ∈ N with the extremal function (5). Special cases of Theorems 2.4, 2.5 by setting

α = 0, s 0 and = β = 0

can be found in [11].

2.5 Corollary

Let

f

be given by (1) satisfying

2

( ) 1 ,

n

n γ γ

=

− ≤ −

Then

Re ( ) ( )

( ) 1

k

f z k z U

f z k γ

 

≥ ∈

  + −

 

,

( ) 1 ( )

( ) 2 2

f z

k

k z U

f z k

γ γ

  + −

ℜ     ≥ + − ∈

,

(6)

'( ) ( )

' ( )

k

1

f z k z U

f z k

γ γ

 

ℜ   ≥ ∈

  + −

,

and

'( ) 1

( )

' ( )

k

( 1 ) ( 1)(1 )

f z k z U

f z k k

γ

γ γ

  + −

ℜ     ≥ + − + + − ∈

. The results are sharp for every k with the function given by

( ) 1 ( ) .

f z z 1 z U

k γ

γ

= + − ∈

+ −

3. CONCLUSION

Our results certainly generalized several results obtained earlier. Therefore, partial sums of functions in the unit disk defined by a general integral operator are obtained The operator defined can be extended and can solve many new results and properties.

ACKNOWLEDGEMENT

The work here is support by the Universiti Kebangsaan Malaysia and the Libyan government

REFERENCES

[1] N.M.Mustafa and M. Darus, ”Vasile Alecsandri” University of Bac˘au Faculty of Sciences Scientific Studies and Research Series Mathematics and Informatics . 21 (2011), No. 2, 45 - 56 [2] F. Rønning, Proc. Amer. Math., Soc,118 (1) (1993), 189-

196

[3] F S. Owa and H. M. Srivastava, Canadian Journal of Mathematics,39(5)(1987),1057- 1077.

[4] H. M. Srivastava and S. OwaMath Japonica., 29(3) (1984),383- 389.

[5] H. M. Srivastava and A. A. Attiya, Integral Transforms and Special Functions, 18(3)(2007),207- 216

[6] W. Alexander, Annals of Mathematics., 17(1915),12-22.

[7] R.J. Libera, Pro of the Amer Math Soc., 135(1969),429-44 [8] S.D. Bernardi, Trans of Amer Math Soc., 135(1969),429-449..

[9] I.B.Jung, Y.C.Kim, and H.M.Srivastava, Journal of MathAnalysis and Applications, 176(1993),138- 147.

[10] F. Rønning, Annal. Polon. Math., 60 (1995), 289-297.

[11] H. Silverman, J. Math. Anal. Appl., .209(1)(1997), 221-227 [12] H.M. Srivastava and J.Choi, Series Associated with the Zeta and

Related Functions, (Dordrecht, Boston and London: Kluwer Academic Publishers), (2001).

[13] G. S. Salagean, Lecture Notes in Math Springer-Verlag. , 1013(1983), 362-372.

[14] T. M.Flett, Math. Anal. Appl.38 (1972), 746-765.

[15] T N. E.Cho, T. H. Kim, Bull. Korean Math. Soc. 40(3) (2003), 399- [16] L. J.Lin ,S. Owa, Georgian Math. J. 5(4) (1998), 361-366. 410

References

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