A New Numerical Procedure to compute the residues of a Complex ...

Raf. J. of Comp. & Math’s. , Vol. 5, No.1, 2008

A New Numerical Procedure to compute the residues of a Complex functions Abbas Y.Al-Bayati College of Computer sciences and Mathematics University of Mosul

Sasan A. Al-Shwani College of Science University of Kirkuk

Received on:27/8/2006

Accepted on:16/11/2006 ‫ﺍﻟﻤﻠﺨﺹ‬

‫ ﻜﺫﺍﻟﻙ‬. ‫ﻓﻲ ﻫﺫﺍ ﺍﻟﺒﺤﺙ ﺴﻭﻑ ﻨﺘﻁﺭﻕ ﺇﻟﻰ ﺤﺴﺎﺏ ﺍﻟﺭﻭﺍﺴﺏ ﻭﺍﻷﻗﻁﺎﺏ ﻓﻲ ﺍﻟﺩﻭﺍل ﺍﻟﻌﻘﺩﻴﺔ‬

‫ﺴﻭﻑ ﻨﺤﺎﻭل ﺍﻟﺘﻘﺼﻲ ﻓﻲ ﺃﺴﺘﺤﺩﺍﺙ ﻭﺴﻴﻠﺔ ﻋﺩﺩﻴﺔ ﺠﺩﻴﺩﺓ ﻨﻅﺭﻴﺎ ﻭﻋﻤﻠﻴﺎ ﻟﺤﺴﺎﺏ ﺍﻟﺭﻭﺍﺴـﺏ ﻟﻠـﺩﻭﺍل‬ ‫ ﻭﻴﺤﺘﺎﺝ ﻫﺫﺍ ﺍﻟﺒﺤـﺙ ﺇﻟـﻰ ﻤﻌﺭﻓـﺔ ﺤـﺴﺎﺒﺎﺕ‬.‫ﺍﻟﻌﻘﺩﻴﺔ ﺍﻟﻘﺎﺒﻠﺔ ﻟﻠﺘﺤﻠﻴل ﻟﻸﻗﻁﺎﺏ ﺫﺍﺕ ﺍﻟﺭﺘﺏ ﺍﻟﻌﺎﻟﻴﻪ‬ . ‫ﺍﻟﺘﻜﺎﻤﻼﺕ ﺍﻟﻌﻘﺩﻴﺔ ﺍﻟﻤﻌﺘﻠﺔ‬ ABSTRACT In this paper, we are going to deal with computations of Residues and Poles for the complex functions . We are also going to investigate a new numerical procedure theoretically and its implementation numerically to compute the residue of complex analytic functions with high order poles. The paper needs the knowledge of computing the complex improper integrations. 1-Introduction : A final bit of theory in this paper is devoted to Cauchy's residue theorem and its applications . Recall that Cauchy's theorem stated that ( ∫ f ( z)dz = 0 ) so long as f(z) was analytic every where inside C.What if f(z) is not analytic within C? The answer is provided by Cauchy's integral formula . Here we have proposed a new procedure to compute the complex improper integrations by calculating the residues of the function . To compute the residues of a complex function of the form ( f ( z ) = p ( z ) ) such that the function f(z) has a pole at ( z = zo ) , we are going q ( z)

to use the (short – cut - method) to estimate the residues[9]. If R is the real field and C is the complex field then consider the following definitions and theorems:

29

Abbas Y.AL-Bayati and Sasan A. Al-Shwani

1-1 poles :[3] If we find a positive integer (m) such that [ lim z→ zo ( z − zo ) m f ( z) ≠ 0 ] Then ( z = zo ) is called a pole of order (m) . If (m=1) simple pole.

( zo ) is called

a

1-2 Analytic functions :[9] If the derivative f / ( z) exists at all points (z) of a region R ,then f (z) is said to be analytic in R. 1-3 Taylor's theorem :[3] Let f(z) be analytic inside and on a simple closed curve C . Let (a) and (a+h) be two points inside C , then f ( a + h ) = f (a ) + h

/

f

f ( z) = f (a ) + ( z − a )

(a ) + h

2

2!

/

f

(a ) +

f

//

( a ) + .....

(z − a) 2!

or writing as

2

f

//

(a ) + .....

Where (z=a+h) The above series is called Taylor expansion for f(z). 1-4 Cauchy integral formula :[2] If f(z) is analytic inside and on a simple closed curve C and (a) is any point inside C then f (a ) =

1 2πi

∫ (z − a ) dz f ( z)

c

Where C is traversed in the positive sense also the nth derivative of (z=a) is given by

f

( n)

(a ) =

1 2πi

∫ c

f ( z)

(z− a )

n +1

;

dz

n=1,2 ,3,…..

1-5 Lemma (1) :[9] Let ( Zo ) be a pole of order (m) of a function f(z) then the residue of the function at a pole given by the form : Re s( f , Z o ) =

[

]

1 d m−1 lim z→ zo m−1 ( z − zo ) m f ( z) . (m − 1)! dz

The formula namely (short – cut – method ) .

30

A New Numerical Procedure to …

2- A New method to compute the residues of a complex function: If f(z) has a pole ( z = zo ) of order (m) then 1 2πi

∫

f ( z)dz =

c

[

1 d m−1 lim z→ zo ( z − zo ) m f ( z) m−1 (m − 1)! dz

]

[ f ( z) =

where

p ( z) ] q( z)

We know that : Re s( f , zo ) =

[

d m−1 1 lim z→ zo m−1 ( z − zo ) m f ( z) (m − 1)! dz

]

We find the value of the above integration by calculating the residues by this procedure and it can be found in [5].

[

]

2.1) If ( z = zo ) is a simple pole i.e q( z) = 0, q / ( zo ) ≠ 0 then the residue is given by the following form :

Re s ( f , z o ) =

2.2)

p ( zo )

( z = zo )

If

[q( z ) = 0, q ( z ) = 0, q /

o

….. (1)

q / ( zo )

//

o

is

]

( zo ) ≠ 0

a

pole

of

then the residue

order

(m=2)

is given by the

following form : Re s ( f , z o ) = 2 ×

p / ( zo )

2 p( zo ).q ( 3) ( zo ) − × q (2) ( zo ) 3 (q ( 2 ) ( z o )) 2

….(2)

See [5] for the details of these procedures. 3- In this paper we are going to follow a procedure for a general function of a pole of order (m): 3.1) Let us start with order (m=3): If ( z = zo ) is a pole of order (m=3) i.e q( zo ) = 0, q / ( zo ) = 0, q // ( zo ) = 0, q /// ( zo ) ≠ 0 then the residue is given by the following form :

[

]

Re s( f , zo ) = 3 × [

p (2) ( zo ) q ( 3) ( zo )

−

(5 ) 1 p( zo ).q ( zo ) × 10 (q ( 3) ( z o )) 2

1 q ( 4) ( z o ) p (1) ( z o ) 1 (q ( 4) ( zo )) 2 p( z o ) − × + × ] 2 8 (q ( 3) ( zo )) 2 (q ( 3) ( z o )) 3

31

i.e

….(3)

Abbas Y.AL-Bayati and Sasan A. Al-Shwani

Proof:

If ( z = zo ) is a pole of order (m=3) then by (short-cut-method) we

get : Re s( f , zo ) =

1 d2  p( z)  lim z→ zo 2  ( z − zo ) 3  2! q ( z)  dz 

…………(4)

We expand the analytic function q(z) into Taylar series valid in disk / z − zo / < r   p( z)( z − zo ) 3 1 d2  = lim z→ zo 2 2 ( z − z o ) 2 (2 ) dz  (1) q ( zo ) + .........  q ( zo ) + ( z − z o ) q ( z o ) + 2! 

= 3 lim z→ zo

     

d 2  p( z)    dz 2 U ( z) 

…….(5)

………….(6)

Where U ( z) = q (3) ( zo ) +

( z − zo ) (4) ( z − zo ) 2 (5) (z − zo ) 3 (6) q ( zo ) + q ( zo ) + q ( zo ) + .... 6 × 5× 4 4 5× 4

…..(7)

Then from (6) we get : = 3 lim z→ zo

= 3 lim z→ zo [ −

d  U ( z) p (1) ( z) − p( z) U (1) ( z)  dz  ( U ( z) ) 2

  

…………(8)

U ( z) p ( 2) ( z) − p (1) ( z) U (1) ( z) (U ( z)) 2

p( z) U ( 2 ) ( z) + U (1) ( z) p (1) ( z) (U ( z)) 2

− 2×

(U (1) ( z)) 2 p( z) (U ( z)) 3

32

……(9) ]

A New Numerical Procedure to … lim z→ zo p( z) = p( zo )

    lim z→ zo p (1) ( z) = p (1) ( zo )     ( z − zo ) ( 4 ) ( z − zo ) 2 ( 5) ( 3) U ( z) = q ( zo ) + q ( zo ) + q ( zo ) + ...  4 5× 4   lim z→ zo U ( z) = q ( 3) ( zo )    1 ( 4) 2  (1) (5 ) ( z − zo )q ( zo ) + ..... U ( z) = q ( zo ) +  4 5× 4  1 (4)  (1) lim z→ zo U ( z) = q ( zo )  4    2 3 × 2 ( z − zo )q ( 6 ) ( zo ) + ....... U ( 2) ( z) = q ( 5 ) ( zo ) +  5× 4 6×5× 4   1 ( 5) ( 2) lim z→ zo U ( z) = q ( zo )  10 

…….(10)

Then Substituting the above values in (10) in equation (9) we get the desired proof . 3.2) If ( z = zo ) is a pole of order (m=4) i.e

[q( z ) = 0, q (z ) = 0, q /

o

//

o

]

( zo ) = 0, q /// ( zo ) = 0, q //// ( zo ) ≠ 0 then the residue is given

by the following form :

Re s ( f , zo ) = 4 × [ −

p ( 3) ( zo ) q ( 4 ) ( zo )

−

3 p ( 2 ) ( zo ) q ( 5 ) ( zo ) × 5 ( q ( 4 ) ( zo )) 2

p ( zo ) q ( 7 ) ( zo ) 1 p (1) ( zo ) q ( 6 ) ( zo ) 1 × − × 35 5 ( q ( 4 ) ( zo )) 2 ( q ( 4 ) ( zo )) 2

p ( zo ) q ( 5) ( zo ) q ( 6 ) ( zo ) 6 ( q ( 5 ) ( zo )) 2 p (1) ( zo ) 2 + × + × 25 25 ( q ( 4 ) ( zo )) 3 ( q ( 4 ) ( zo )) 3 −

…….(11)

( q ( 5 ) ( zo )) 3 p ( zo ) 6 × ] 125 ( q ( 4 ) ( zo )) 4

Proof: If ( z = zo ) is a pole of order (m=4) then by (short – cut - method) we get :

33

Abbas Y.AL-Bayati and Sasan A. Al-Shwani  4 p ( z)   ( z − zo ) q( z)  

1 d3 lim z→ zo 3 3! dz

Re s( f , zo ) =

..….(12)

We expand the analytic function q(z) into Taylor series valid in disk / z − zo / < r =

1 lim z→ zo 3!

= 4 lim z→ zo

    4 ( z − zo ) p( z) d   3  2  ( z − z o ) ( 2) dz (1) q ( zo ) + .....   q ( z o ) + ( z − z o )q ( z o ) + 2!  

…..(13)

d 3  p( z)    dz 3  q( z) 

….(14)

3

Where : U ( z) = q ( 4 ) ( zo ) +

( z − zo ) (5 ) ( z − zo ) 2 ( 6 ) q ( zo ) + q ( zo ) + ..... 5 6×5

….(15)

Then from (14) we get : = 4 lim z→ zo

d 2 U ( z) p (1) ( z) − p ( z) U (1) ( z)    (U ( z)) 2 dz 2  

= 4 lim z→ zo

d U ( z) p ( 2) ( z) − p (1) ( z) U (1) ( z) [ dz (U ( z)) 2

p( z)U ( 2) ( z) + U (1) ( z) p (1) ( z)

−

(U ( z)) 2

= 4 lim z→ zo [ −

(U ( z)) 2

− 3×

p (1) ( z)U ( 2) ( z) (U ( z)) 2

p( z)U (1) ( z)U ( 2) ( z)

− 6×

(U ( z)) 3

……(17) ]

p ( 2 ) ( z) U (1) ( z) p ( 3) ( z) − 3× U ( z) (U ( z)) 2

p( z)U ( 3) ( z)

6×

− 2×

(U (1) ( z)) 2 p( z)

…….(16)

(U ( z)) 3 (U (1) ( z)) 3 p( z) (U ( z)) 4

+ 6×

+

(U (1) ( z)) 2 p (1) ( z)

…..(18)

(U ( z)) 3

]

Where 1 1  ( z − zo )2 q(6) ( zo ) + ... ; limz→zo U ( z) = q(4) ( zo )  U ( z) = q(4) ( zo ) + ( z − zo )q(5) ( zo ) + 5 6×5    1 (5) 2 1 (5) (1) ( 6) (1) U ( z) = q ( zo ) + ( z − zo )q ( zo ) + ................ ; limz→ zo U ( z) = q ( zo )  5 6×5 5     2 (6) 6 1 (6) ( 2) (7) ( 2) ( z − zo )q ( zo ) + ...... ; lim z→ zo U ( z) = q ( zo )  U ( z) = q ( zo ) + 6×5 7× 6× 5 15    6 24 1 ( z − zo )q(8) ( zo ) + .....; limz→zo U (3) ( z) = q(7) ( zo ) U (3) ( z) = q(7) ( zo ) + 7 × 6×5 8× 7 × 6 × 5 35 

34

…..(19)

A New Numerical Procedure to …

Then substituting the above values in (19) in equation (18) we get the desired proof. 3.3) In this approach we can find the residue for (m=5): If ( z = zo ) is a pole of order (m=5) i.e

[q( z ) = 0, q ( z ) = 0, q /

o

o

//

]

( zo ) = 0, q /// ( zo ) = 0, q //// ( zo ) = 0, q ///// ( zo ) ≠ 0 then the

residue is given by the following form : 1 (6) q ( z o ) p ( 3) ( z o ) − 4× 6 Re s ( f , z o ) = 5 × [ (5) q ( zo ) (q (5) ( zo )) 2 p (4) ( zo )

1 1 (7) q ( z o ) p ( 2) ( z ) ( q ( 6) ( z o )) 2 p 2 ( zo ) 21 − 6× + 12 × 6 (5 ) 2 (q ( z o )) (q (5) ( zo )) 3 1 (9 ) 1 (8 ) q ( z o ) p ( zo ) q ( z o ) p (1) ( zo ) 126 56 − − 4× (q ( 5) ( zo )) 2 (q ( 5) ( zo )) 2 1 ( 6) q ( z o ) q (8 ) ( zo ) p ( z o ) 42 + (q ( 5) ( z o )) 3 1 ( 6) 1 q ( zo )q ( 7 ) ( z o ) p (1) ( zo ) ( q ( 7 ) ( z o )) 2 p( zo ) 21 21 + 4× + 6× (q ( 5) ( zo )) 3 (q (5) ( z o )) 3 1 1 (7) q ( zo ) p ( zo ) ( q ( 6) ( zo )) 2 21 − 36 × 6 (q (5) ( z o )) 4 1 1 ( q ( 6) ( zo ))4 p ( zo ) ( q ( 6) ( zo ))3 p (1) ( zo ) − 24 × 6 + 24 × 6 (q (5 ) ( zo ))5 (q (5) ( zo ))4

…(20)

Proof : (By the same way in (m=3)&(m=4)) . 3.4) In this approach we can find the residue for (m=6): If ( z = zo ) is a pole of order (m=6) i.e

[q( z ) = q

( zo ) = q ( 2 ) ( zo ) = q (3) ( zo ) = q ( 4) ( zo ) = q (5) ( zo ) = 0, q ( 6) ( zo ) ≠ 0 then the residue is given by the following form : (1)

o

35

]

Abbas Y.AL-Bayati and Sasan A. Al-Shwani

Re s ( f , zo ) = 6 × [

1 ( 4) p ( zo )q (7 ) ( zo ) p (5) ( zo ) 7 − 5× (q (6) ( zo )) 2 q( 6) ( zo )

1 (3) 1 ( 2) p ( zo )q (8) ( zo ) p ( zo )q (9) ( zo ) 28 84 − 4× − 10 × ( q( 6) ( zo )) 2 (q ( 6) ( zo )) 2

1 ( 2) 1 (1) p ( zo )q(8) ( zo ) p ( zo )q(10) ( zo ) 28 210 − 6× − 5× (q(6) ( zo ))2 (q(6) ( zo ))2 1 1 ( q(7 ) ( zo ))2 p(3) ( zo ) p( zo )q(11) ( zo ) 462 − + 20 × 7 (q( 6) ( zo ))2 (q(6) ( zo ))3 1 (7 ) 1 1 ( × ) p( zo )q( 7) ( zo )q(10) ( zo ) q ( zo )q(8) ( zo ) p( 2) ( zo ) 196 7 210 + 60 × + 10 × (q(6) ( zo ))3 (q(6) ( zo ))3 1 (7 ) 1 1 ( × )q(9) ( zo )q(8) ( zo ) p( zo ) q ( zo )q(8) ( zo ) p(1) ( zo ) 84 196 28 + 8× + 20 × (q( 6) ( zo ))3 (q( 6) ( zo ))3 1 (1) 1 (7 ) )q ( zo )q(9) ( zo ) p(1) ( zo ) ( p ( zo )q(7 ) ( zo )q(9) ( zo ) 196 196 + 32 + 24 × (q( 6) ( zo ))3 (q(6) ( zo ))3 1 (8 ) 1 ( q(7 ) ( zo ))3 p( 2) ( zo ) q ( zo ))2 p(1) ( zo ) 28 + 30 × − 60 × 7 (q(6) ( zo ))3 (q(6) ( zo ))4 (

1 1 1 1 ( q( 7) ( zo ))2 q(9) ( zo ) p( zo ) ( q( 7) ( zo ))2 q(8) ( zo ) p(1) ( zo ) 84 28 − 60 × 7 − 180 × 7 (q( 6) ( zo ))4 (q( 6) ( zo ))4 1 1 1 1 (7 ) ( q(7 ) ( zo ))3 q(8) ( zo ) p( z0 ) q ( zo )( q(8) ( zo ))2 p( zo ) 28 28 7 7 − 90 × + 240 × (q( 6) ( zo ))4 (q( 6) ( zo ))5 1 1  ( q( 7) ( zo ))4 p(1) ( zo ) ( q( 7) ( zo ))5 p( zo )  7 7 + 120 × − 120 ×  (q(6) ( zo ))5 (q(6) ( zo ))6  

Proof: (By the same way in (m=3)&(m=4)) . 3.5) If ( z = zo ) is a pole of order (m=7) i.e

[q( z ) = q o

(1)

...(21)

]

( zo ) = q ( 2 ) ( zo ) = q ( 3) ( zo ) = q ( 4) ( z ) = q ( 5) ( zo ) = q ( 6) ( zo ) = 0, q ( 7 ) ( zo ) ≠ 0

then the residue is given by the following form:

36

A New Numerical Procedure to … 1 p (5) ( zo ) q (8) ( zo ) 8 Re s ( f , z) = 7 × [ (7) − 6× (q (7 ) ( zo ))2 q ( zo ) p (6) ( zo )

−9×

1 (9) 1 (10) q ( zo ) p (3) ( zo ) q ( zo ) 36 120 − 14 × − ( 7) 2 (7 ) (q ( zo )) (q ( zo )) 2

p ( 4) ( zo )

1 (11) 1 (10 ) q ( z o ) p ( 2 ) ( zo ) p (2) ( zo ) q ( zo ) 120 6 − 15 × 330 − × (q ( 7 ) ( zo )) 2 (q ( 7 ) ( zo )) 2 − 6×

1 ( 9) 1 (12 ) q ( zo ) p (1) ( zo ) q ( zo ) 36 792 6 − × − (q ( 7 ) ( z o )) 2 (q ( 7 ) ( z o )) 2

p ( 3) ( zo )

1 1 1 q (13) ( z o ) ( × ) q (8 ) ( zo ) q ( 9 ) ( z o ) p ( 2 ) ( zo ) 1716 + 20 × 8 36 (7) 2 (q ( z o )) (q ( 7 ) ( zo )) 3

p( z o )

1 1 1 ( q (8 ) ( z o )) 2 p ( 4) ( zo ) ( × ) q (8 ) ( zo ) q (9 ) ( zo ) p (3 ) ( zo ) 8 + 30 × + 108 × 8 36 (7) 3 (q ( z o )) (q ( 7 ) ( z o )) 3 1 1 1 1 (8 ) ( × )q (8) ( zo )q (11) ( zo ) p (1) ( zo ) ( × )q ( z o )q (10 ) ( z o ) p ( 2) ( z0 ) 8 330 8 120 + 80 × + 76 × (q ( 7 ) ( zo )) 3 (q ( 7 ) ( zo )) 3 1 1 1 ( × )q (8) ( zo )q (12 ) ( z o ) p( zo ) ( q ( 9) ( z o )) 2 p ( 2) ( zo ) + 12 × 8 792 + 90 × 36 + (q ( 7 ) ( zo )) 3 (q ( 7 ) ( zo )) 3 1 1 1 1 × ) p( z o )q (11) ( zo )q (9 ) ( z o ) ( × )q (8) ( zo )q (10 ) ( zo ) p (1) ( zo ) 36 330 8 120 + 30 × + 8× (q ( 7 ) ( zo )) 3 (q ( 7 ) ( zo )) 3 (

1 1 1 ( 9) q ( z o )) 2 p (1) ( z o ) ( × )q (9 ) ( zo )q (10 ) ( z o ) p (1) ( zo ) 36 120 36 + 8× + 136 × (q ( 7 ) ( zo )) 3 (q ( 7 ) ( zo )) 3 1 (10 ) 1 1 ( ( × )q (8 ) ( z o )q (10 ) ( z o ) p ( 2) ( zo ) q ( z o )) 2 p( zo ) 8 120 120 + 20 × + 56 (q ( 7 ) ( zo )) 3 (q ( 7 ) ( z o )) 3 (

37

Abbas Y.AL-Bayati and Sasan A. Al-Shwani 1 1 1 (9) q ( z o ) p ( 2 ) ( zo ) ( q (8) ( zo )) 3 p ( 3) ( zo ) ( q (8) ( z o )) 2 8 8 36 − 120 × − 540 × (q ( 7 ) ( z o )) 4 (q ( 7 ) ( zo )) 4 1 1 (11) 1 1 (9 ) q ( z o ) p ( zo ) q ( zo ) p (1) ( zo ) ( q (8) ( zo )) 2 ( q (8) ( zo )) 2 8 330 8 36 − 90 × − 24 × (q ( 7 ) ( zo )) 4 (q ( 7 ) ( z o )) 4 1 1 1 ( × × )q (8) ( zo )q ( 9) ( z o )q (10 ) ( zo ) p ( z o ) 8 36 120 − 360 × − (q ( 7 ) ( zo )) 4 1 (9 ) 1 1 q ( z o )) 2 q (8) ( z o ) p (1) ( zo ) ( q ( 9) ( zo )) 2 p ( zo ) 8 − 540 × 36 − 90 × 36 ( 7 ) (q ( 7 ) ( z o )) 4 (q ( zo )) 4 (

1 1 1 (10 ) q ( zo ) p ( z o ) ( q (8) ( zo )) 4 p ( 2 ) ( zo ) ( q (8) ( zo )) 3 8 120 480 + 360 × 8 + × (q ( 7 ) ( zo )) 5 (q ( 7 ) ( z o )) 5 1 1 ( 9) q ( zo ) p (1) ( zo ) ( q (8) ( zo )) 3 36 + 1440 × 8 + (q ( 7 ) ( z o )) 5 1 ( 9) 1 1 1 ( 9) q ( z o )) 2 ( q (8) ( z o )) 2 p( zo ) q ( zo ) p ( zo ) ( q (8) ( zo )) 4 36 8 8 36 1080 × − 1800 × − (q ( 7 ) ( z o )) 5 (q (7 ) ( z o )) 6 (

1 1  ( q (8) ( zo )) 5 p (1) ( zo ) ( q (8 ) ( z o )) 6 p( z o )  720 × 8 + 720 × 8  (q ( 7 ) ( zo )) 6 (q ( 7 ) ( zo )) 7  

Proof:

…..(22)

If ( z = zo ) is a pole of order (m=7) then by (short – cut -method) we

get : Re s( f , zo ) =

1 d6  p ( z)  lim z→ zo 6 ( z − zo ) 7 where q( z)  6! dz 

( f ( z) =

p( z) ) q( z)

…..(23)

We expand the analytic function q(z) into Taylor series valid in disk z − zo < r

=

1 lim z→ zo 6!

   7  ( z z ) p ( z ) − d  o  ...(24)  dz 6  ( z − z o ) 2 ( 2) ( z − zo ) 3 (3) (1) q ( z0 ) + q ( zo ) + ......   q ( zo ) + ( z − z o ) q ( zo ) + 2! 3!   6

38

A New Numerical Procedure to …

= 7 lim z→ zo

d 6  p( z)    dz 6 U ( z) 

…..(25)

where : U ( z) = q ( 7 ) ( zo ) +

( z − zo ) (8 ) ( z − zo ) 2 ( 9 ) ( z − zo ) 3 (10 ) q ( zo ) + q ( zo ) + q ( zo ) + .... …(26) 8 8×9 8 × 9 × 10

Then from (25) we get : = 7 lim z→ zo

d 5 U ( z) p / ( z) − p( z)U (1) ( z)    dz 5  (U ( z)) 2 

= 7 lim z→ zo

p (1) ( z)U (1) ( z) p( z)(U (1) ( z)) 2  d 4  p (2) ( z) p( z)U (2) ( z) 2 2 − − × + ×   dz 4  U ( z) (U ( z)) 2 (U ( z)) 2 (U ( z)) 3 

= 7 lim z→ zo

p ( 2) ( z)U (1) ( z) p( z)U ( 3) ( z) p ( z)U ( 2 ) ( z) d 3  p ( 3) ( z) 3 3 − × − − ×  dz 3  U ( z) (U ( z)) 2 (U ( z)) 2 (U ( z)) 2

+ 6×

p( z)U (1) ( z)U ( 2) ( z) (U ( z)) 3

= 7 lim z→ zo − 6× −

d2 dz 2

+ 12 ×

(U ( z)) 2

(U ( z)) 2

+ 24 ×

p ( z)(U (1) ( z)) 3   (U ( z)) 4 

…..(29)

− 4×

(U ( z)) 3

p (1) ( z)U (3) ( z) (U ( z)) 2

(U ( z)) 3

p ( z)(U ( 2 ) ( z)) 2 (U ( z)) 3

− 36 ×

p (1) ( z)(U (1) ( z)) 3

= 7 lim z→ zo

(U ( z)) 3

− 6×

p ( 2) ( z)(U (1) ( z)) 2

p (1) ( z)U (1) ( z)U ( 2) ( z)

− 24 ×

p (1) ( z)(U (1) ( z)) 2

..(28)

p ( 4) ( z) p (3) ( z)U (1) ( z) − 4× U ( z) (U ( z)) 2

p ( 2) ( z)U ( 2) ( z)

p( z)U ( 4 ) ( z)

6×

[

+ 6×

…..(27)

(U ( z)) 4

+ 8×

p ( z)U (1) ( z)U (3) ( z) (U ( z)) 3

+

p( z)(U (1) ( z)) 2 U ( 2 ) ( z)

+ 24

(U ( z)) 4 p( z)(U (1) ( z)) 4 (U ( z)) 5

…(30) ]

p ( 4) ( z)U (1) ( z) p ( 3) ( z)U ( 2) ( z) p ( 2 ) ( z)U ( 3) ( z) d p ( 5) ( z) [ − 5× − 4 × − 10 × − dz U ( z) (U ( z)) 2 (U ( z)) 2 (U ( z)) 2

39

Abbas Y.AL-Bayati and Sasan A. Al-Shwani 6×

p ( 2 ) ( z)U ( 2) ( z) (U ( z)) 2

+ 30 ×

− 180 ×

p (1) ( z)U ( 4 ) ( z) (U ( z)) 2

U (1) ( z)U ( 2 ) ( z) p ( 2) ( z)

+ 60 ×

+ 20 ×

5×

(U ( z)) 3

U (3) ( z)U ( 2) ( z) p( z) (U ( z)) 3 p (1) ( z)(U ( 2) ( z)) 2

− 60 ×

(U ( z)) 3

(U (1) ( z)) 2 U ( 2 ) ( z) p (1) ( z) (U ( z))

+ 120 ×

4

(U (1) ( z)) 4 p (1) ( z) (U ( z)) 5

p( z)U ( 5) ( z) (U ( z)) 2

(U ( z)) 3

U (1) ( z)U ( 3) ( z) p (1) ( z) (U ( z)) 3

(U (1) ( z)) 3 p ( 2 ) ( z) (U ( z)) 4

− 90 ×

− 120 ×

+ 20 ×

p( z)U (1) ( z)U ( 4) ( z)

+ 10 ×

+ 32 ×

−

− 60 ×

U (1) ( z)(U ( 2 ) ( z)) 2 p( z) (U ( z))

4

(U (1) ( z)) 2 p ( 3) ( z) (U ( z)) 3

+ 8×

U (1) ( z)U ( 2) ( z) p (1) ( z) (U ( z)) 3

+ 24 ×

p (1) ( z)U (1) ( z)U ( 3) ( z) (U ( z)) 3

(U (1) ( z)) 2 U (3) ( z) p( z) (U ( z)) 4

+ 240 ×

(U (1) ( z)) 3 U ( 2 ) ( z) p( z) (U ( z)) 5

(U (1) ( z)) 5 p( z)

…..(31)

(U ( z)) 6

 p ( 6 ) ( z) p ( 5) ( z)U (1) ( z) p ( 4) ( z)U ( 2 ) ( z) p ( 3) ( z)U ( 3) ( z) = 7 lim z→ zo  − 6× − × − × − 14 9 (U ( z)) 2 (U ( z)) 2 (U ( z)) 2  U ( z) 15 ×

U ( 4 ) ( z) p ( 2) ( z)

− 6×

(U ( z)) 2 p (1) ( z)U (5) ( z) (U ( z))

+ 80 × + 90 × + 8×

2

− 6× −

p ( 2 ) ( z)U ( 3) ( z) (U ( z)) 2

p( z)U ( 6) ( z) (U ( z))

U (1) ( z)U ( 3) ( z) p ( 2) ( z) (U ( z)) 3 p ( 2 ) ( z)(U ( 2 ) ( z)) 2 (U ( z)) 3

2

+ 76 ×

+ 30 ×

− 6×

+ 30 ×

p (3) ( z)U ( 2) ( z) (U ( z)) 2

(U (1) ( z)) 2 p ( 4 ) ( z) (U ( z))

3

U (1) ( z)U ( 4 ) ( z) p (1) ( z) (U ( z)) 3

p ( z)U ( 2 ) ( z)U ( 4) ( z) (U ( z)) 3

p (1) ( z)(U ( 2) ( z)) 2 (U ( z)) 3

40

+ 8×

+ 20 ×

+ 108 ×

+ 12 ×

U (1) ( z)U ( 2) ( z) p ( 2 ) ( z) (U ( z)) 3

U (1) ( z)U ( 2) ( z) p (3) ( z) (U ( z)) 3

U (1) ( z)U ( 5) ( z) p( z) (U ( z)) 3

U (1) ( z)U (3) ( z) p (1) ( z) (U ( z)) 3

A New Numerical Procedure to …

+ 136 ×

(2)

( z )U

(3)

( z ) p (1 ) ( z )

(1 )

( z )U

(3)

( z) p (2) ( z)

(U ( z )) 3

− 540 ×

(U

(U

− 24 ×

(1 )

( z )) 2 U

(2)

1440 ×

(U ( z )) (1 )

2

( z )) U

(2)

(U ( z ))

( z) p

(U

(U ( z)) 4 (U (1) ( z)) 4 p ( 2 ) ( z) (U ( z)) 5

+ 480 ×

(U (1) ( z)) 3 U ( 2) ( z) p (1) ( z) (U ( z)) 5 (U (1) ( z)) 4 U ( 2) ( z) p ( z) (U ( z)) 6

(1 )

( z)

4

U (1) ( z)(U ( 2) ( z)) 2 p (1) ( z)

− 1800 ×

_ 120 ×

4

(U ( z)) 4

+ 360 ×

(U

( z) p (2) ( z)

U (1) ( z)U ( 2) ( z)U ( 3) ( z) p( z)

− 540 ×

+ 20 ×

(U ( z )) 3

U

+ 56 ×

360 ×

U

+ 408 × − 90 ×

(3)

( z )) 2 p ( z )

(U ( z )) 3 (1 )

( z )) 3 p ( 3 ) ( z ) (U ( z )) 4

− 90 ×

(U

(1 )

(4)

(U ( z ))

( z) p ( z)

4

− (U (1) ( z)) 2 U ( 3) ( z) p (1) ( z) (U ( z)) 4

(U ( 2 ) ( z)) 2 p( z) (U ( z)) 4

(U (1) ( z)) 4 U (3) ( z) p( z) (U ( z)) 5

+ 1080 × − 720 ×

( z )) 2 U

+

..(32)

(U ( 2 ) ( z)) 2 (U (1) ( z)) 2 p( z) (U ( z)) 5

(U (1) ( z)) 5 p (1) ( z) (U ( z)) 6

Where :

41

+ 720 ×

(U (1) ( z)) 6 p( z) (U ( z)) 7

]

Abbas Y.AL-Bayati and Sasan A. Al-Shwani (z − zo ) (8) (z − zo )2 (9) q (zo ) + q (zo ) + ... 8 9×8 limz→zo U(z) = q(7) (zo )

U(z) = q(7) (zo ) +

1 2 U(1) (z) = q(8) (zo ) + (z −zo)q(9) (zo ) + .......... . 8 9×8 1 limz→zo U(1) (z) = q(8) (zo ) 8 2 (9) 3× 2 (2) U (z) = q (zo ) + (z − zo )q(10) (zo ) + ... 9 ×8 10×9 ×8 1 limz→zo U(2) (z) = q(9) (zo ) 36 3× 2 (10) 4× 3× 2 U(3) (z) = q (zo ) + (z − zo )q(11) (zo ) + ... 10× 9×8 11×10× 9×8 1 limz→zo U(3) (z) = q(10) (zo ) 120 × × 4 3 2 5× 4× 3× 2 U(4) (z) = q(11) (zo ) + (z − zo )q(12) (zo ) + ....... 11×10×9 ×8 12×11×10× 9×8 1 (11) q (zo ) limz→zo U(4) (z) = 330 6 ×5× 4× 3× 2 5× 4× 3× 2 (z − zo )q(13) (zo ) + ... U(5) (z) = q(12) (zo ) + 13×12×11×10×9 ×8 12×11×10× 9×8 1 (12) limz→zo U(5) (z) = q (zo ) 792 7 × 6 ×5× 4× 3× 2 6× 5× 4 ×3× 2 (z − zo )q(14) (zo ) + .... U(6) (z) = q(13) (zo ) + 14×13×12×11×10× 9×8 13×12×11×10× 9×8 1 (13) limz→zo U(6) (z) = q (zo ) 1716

                                 …(33)      

Then substituting the above values in (33) in equation (32) we get the desired proof . 3.6) In the above manner the procedure can be easily extended for any pole of order (m) . 4-Numarical computations : These examples can be found in [1, 4, 6, 7, 8, 10, 11]. Example (4.1) : Evaluate the following integral

Z2 +1

∫ (Z − 2) C

42

4

(Z − 1) 3

dZ

at

C:/Z/ =3 .

A New Numerical Procedure to …

Solution: The function f(Z) has two poles (Z=1 , Z=2) in the region c:/Z / =3 ,and the order of these poles is (m=3) & (m=4), respectively The solution by Cauchy integral formula is as follows : Z2 +1

∫

( Z − 2) 4 ( Z − 1) 3 C:/ Z / =3

dZ = 58πi − 58πi = 0 12

C

cc

Figure 4.1 Then the solution by the new procedure is given as follows :

∫

Z 2 +1

C :/ Z / = 3

dZ = 2πi

(Z − 2) 4 (Z − 1) 3

2

∑ Re s( f , Z ) = 2πi Re s( f , Z i

1

= 1) + 2πi Re s ( f , Z 2 = 2)

i =1

Where p( Z ) = Z 2 + 1

; q( Z ) = ( Z − 2) 4 (Z − 1) 3

q ( 4 ) (2) = 24

P(1)=2

q (3) (1) = 6

p (1) (2) = 4

q (5 ) (2) = 360

p (1) (1) = 2

q ( 4) (1) = −96

p ( 2 ) (2) = 2

q ( 6 ) (2) = 2160

p ( 2) (1) = 2

q (5) (1) = 720

p ( 3) (2) = 0

q ( 7 ) (2) = 5040

p ( 3) (3) = 0

P(2)=5

Table 4.1 Then by using the new procedure for (m=3) we get : 1 2304  Re s ( f ,1) = 3 ×  − 4 + 2.66 + = 28.98 ≅ 29 × 6)  3 ( 36 

Then by using the new procedure for (m=4) we get Re s ( f ,2) = 4 × [− 0.75 − 1.25 − 3 + 22.5 + 9 − 33.75] = −29

Then :

∫

Z2 +1

( Z − 2) 4 ( Z − 1) 3 C:/ Z / =3

dZ = 2πi

2

∑ Re s( f , Z ) = 2πi Re s( f , Z i

1

= 1) + 2πi Re s( f , Z 2 = 2)

i =1

=58 πi − 58πi ≅ 0

Example (4.2): Evaluate the following integral

eZ − Z 6

∫ (Z − 2i)

6

dz at C:/Z/=3 .

c

Solution: the function f(z) has a pole Z=2i of order (m=6) . 43

Abbas Y.AL-Bayati and Sasan A. Al-Shwani

then by Cauchy integral formula we get: eZ − Z 6

∫

C:/ Z / =3

( Z − 2i )

dz =

6

πi 2i (e − 1440i ) . 60

Then by using the new procedure for(m=6) we get ez − Z6

∫

C:/ Z / =3

( z − 2i )

6

dz = 2πi Re s ( f ,2i ) =

πi 2 i (e − 1440i ). 60

Example (4.3) Evaluate the following improper integral ∞ sin( x) dx where m=1,3,5,….. ∫ m 0 x Solution: Now for this type of improper integrations we not can find the result by Cauchy integral formula. The only way to evaluate is the new procedure with general order (m). ∞

If (m=1) then

∫ 0

sin(x) π dx = x 2

CR ∞

∫

If (m=3) then

0 ∞

If (m=5) then

∫

sin(x) x

3

sin( x) x

0

5

dx = =

−π 4

-R -r

sr Sr

r

R

π we shall prove that 48

Figure 4.2 The function f(Z)= ∞

∫

sin(x) x

0

5

R

dx = lim R→∞

∫ 0

e

Z

Z5

has pole (Z=0) of order (m=5),on real line .Then

sin( x) x5

dx

then by Jordan lemma we get:

 −r R sin(x)  lim R→∞ , r → 0 ( + + + ) 5 dx = 0  − R r Sr CR x   

∫ ∫ ∫ ∫

Where

∫ (

sin( x) x5 e iZ

∫Z

5

dx =

e iZ

∫Z

5

dZ on real part (x-axis). Then by lemma we get

dz = 0 ) then

CR

By replacing (x = -x) in the above first term in the LHS then we have :

44

A New Numerical Procedure to …  R sin( x)  e iZ =− lim R→∞ , r →0 2i dx dz = πi Re s ( f ,0) then we find the residue at a 5  r x5  Z Sr  

∫

∫

pole (z=0) of order (m=5) . Using the new procedure to find the residue for order (m=5) Now p(z)= e iZ & q(z)= Z 5 P(0)=1

q(0)=0

q (5) (0) = 120

p (1) (0) = i

q (1) (0) = 0

q ( 6) (0) =0

p ( 2 ) (0) = −1

q ( 2) (0) = 0

p ( 3) (0) = −i

q (3) (0) = 0

p ( 4 ) (0) = 1

q ( 4) (0) = 0 Table 4.2

Then :  1  1 Re s( f ,0) = 5 ×  = 120  24

∞

∫

we get from that

sin( x) x5

0

dx =

π 48

Then by using the new procedure of order (m) only we find general solution. ∞

∫ 0

sin(x) x

m

π  n +1 (−1) dx =  2(m − 1)! 0 

m = 1,3,5,7,9.... & n = 1,2,3,...

if

otherwise

Example (4.4) Evaluate the following improper integral

∞

cos( x) ∫ xm dx

Where m=2,4,6,….

0

Solution: By the same way of example(3) we get the general solution . ∞

∫ 0

cos( x) x

m

π  n (−1) dx =  2(m − 1)! 0 

if

m = 2,4,... & n = 1, 2,3... otherwise

By using the new procedure of order (m) only we get the above solution. 5-conclusions : It is clear that the new procedure for calculating poles of any order (m) is a very effective rule in determining the values of improper integrals while in the most of complex variable methods fail to compute these integrations. 45

Abbas Y.AL-Bayati and Sasan A. Al-Shwani

REFERENCES [1]

Trott, M. , (1997) "Visualization of Riemann Surfaces of Algebraic functions", Mathematica in Education and Research, Vol.(6), no. 4, pp.3-15.

[2]

Churchill, R. , (1974)" Complex Variable and Applications ", New York .

[3]

Rinehart and Winston , (1976) " Elements of Complex Variable " ,Macmilan & Co LTD London.

[4]

Dauglas, N ., (2000)" complex Analysis " .

[5]

Murray, R. , (1981) " Theory and problems of Complex Variable", Ginn & Company, Boston.

[6]

Oliver, K. , (1996) " The Residue theorem and it's Application ". New York .

[7]

Chen, L. (2003), " Introduction to complex analysis ".

[8]

Mike, P. (2005), " Complex Number and Complex Function ". New York .

[9]

Vitally, M. (2005)'' Complex Function theory'' , University of Bristol.

[10]

Sean, M. (2001) " Introduction to Applied mathematics ".

[11]

Risch, R., (1979) "Algebraic Properties of the Elementary Functions of Analysis", American Journal of Mathematics, Vol.(101), pp. 743-759.

46