Transcript Lect11PoleZeroExpansions

ECE 6382
Pole and Product Expansions,
Series Summation
D. R. Wilton
ECE Dept.
8/24/10
Pole Expansion of
Meromorphic Functions
Mittag -Leffler1 Theorem :
 f ( z ) has simple poles at z  an , n  1,
, where 0  a1  a2  a3 
 f ( z ) has residue bn at z  an , n  1,

f ( z )  M , independent of N , on circles of CN of radius RN that enclose
N poles, not passing through any of them and such that RN 

N 
Then
 bn
bn 
f ( z )  f (0)   
 
an 
n 1  z  an

Note that a pole at the origin is not allowed!
1Historical
note: It is often claimed that friction between Mittag-Leffler and Alfred
Nobel resulted in there being no Nobel Prize in mathematics. However, it seems this
is not likely the case; see, for example, www.snopes.com/science/nobel.asp
Proof of Mittag-Leffler Theorem
For z  aN , z  an , consider the sequence of contour integrals
1
f ( w) dw
 f (0) f ( z )
IN 



2 i CN w( w  z )

z
z

bn
 N

 n 1 an (an  z )
But we also have for w  RN ei on C N
IN
1
f ( w) dw

2 i CN w( w  z )
1

2
2

0
aN 2
aN 1
f ( RN ei ) RN d
RN ( RN  z )
1 2 M RN


 0
N 
2 RN ( RN  z )
a2
C4 C2
CN
Hence, taking the limit in the first equation above,

 bn
zbn
bn 
f ( z )  f (0)  
 f (0)   
 
an 
n 1 an ( an  z )
n 1  ( z  an )

Im w
a1
w0
w z
Re w
Extended Form of the
Mittag-Leffler Theorem
Extended Mittag -Leffler Theorem :
 f ( z ) has simple poles at z  an , n  1,
, where 0  a1  a2  a3 
 f ( z ) has residue bn at z  an , n  1,
 f ( z )  M z , independent of N , on circles of CN of radius RN that encloses
p
N poles, not passing through any of them and such that RN 

N 
Then considerations of the integral
1
f ( w) dw
IN 
2 i CN w p 1 ( w  z )
lead to the expansion
f ( z )  f (0)  zf (0) 
z p f ( p ) (0)   bn z p 1 / anp 1 

 

p!
z

a
n 1 
n

Example: Pole Expansion of cot z
cos z
has poles z  n , n  0, 1, 2, with residues
sin z
( z  n ) cos z
( z  n ) cos z
lim
 lim
1
n
z  n
z

n

sin z
sin  z  n  1
cot z 
Then since a pole at z  0 is not allowed, consider
1
cot z 
z
for which the singularity at the origin has been removed and which has
the (finite) limit
1

 z cos z  sin z 
lim  cot z    lim 

z 0
z 0
z
z sin z



 
z2  
z3  
 1 1 3
   z  O ( z5 )
 z 1   z   

2!  
3!  
 2 6
 lim  

lim
 0
2
4
3
z 0 
z

0

z  O (z )

z 
z z  


3!




Example: Pole Expansion of cot z (cont.)
To check that cot z 
RN 
(2 N  1)
so that C N threads between the poles at z  n , and that
2
cot z 
2

cos z
is bounded on C N , note that z  RN ei where
sin z
cos  RN e
sin  RN e
i
i


2

cos  RN  cos   i sin   
2
sin  RN  cos   i sin   
cos  RN cos   cosh  RN sin    i sin  RN cos   sinh  RN sin  
sin  RN cos   cosh  RN sin    i cos  RN cos   sinh  RN sin  
2
cos 2  RN cos   cosh 2  RN sin    sin 2  RN cos   sinh 2  RN sin  

sin 2  RN cos   cosh 2  RN sin    cos 2  RN cos   sinh 2  RN sin  
cosh 2  RN sin    sin 2  RN cos  

cosh 2  RN sin    cos 2  RN cos  
where we've used sinh 2 x  cosh 2 x  1 in both numerator and denominator.
Example: Pole Expansion of cot z (cont.)
Now if we plot the expression
cosh 2  RN sin    sin 2  RN cos  
cot z 
cosh 2  RN sin    cos 2  RN cos  
2
for 0     / 2 , for various values of N , it appears that cot z  1.88823
2
so that cot z  M  1.88823 , independent of N .
|cot z|^2 on circular arcs R_n =(N-1/2)pi/2
1.4
1.2
|cot z|^2
1
N=1
N=2
0.8
N=3
0.6
N=4
0.4
N=8
0.2
0
0
0.5
1
Theta (radians)
1.5
2
• Actually, it isn’t necessary
that the paths CN be
circular; indeed it is
simpler in this case to
estimate the maximum
value on a sequence of
square paths of
increasing size that pass
between the poles
Example: Pole Expansion of cot z (cont.)
Alternatively, we could consider the expression on the contours shown below :
2
2
cos( x  iy )
cos x cosh y  i sin x sinh y
cos 2 x cosh 2 y  sin 2 x sinh 2 y
cot z 


sin( x  iy )
sin x cosh y  i cos x sinh y
sin 2 x cosh 2 y  cos 2 x sinh 2 y
2
 tanh 2 y  1,
x    2 N  1  2

  cos 2 x cosh 2 y  sin 2 x cosh 2 y
2
2

coth
y

coth
 2, y    2 N  1  2
 sin 2 x sinh 2 y  cos 2 x sinh 2 y

It is easy to show that I N 
1
f ( w) dw

 0 on these CN also.
N 

2 i CN w( w  z )
cot z  coth
C3
C2

2
cot z  coth
C1

2
3
2
coth (x) ―
cot z  coth
5
2

2

2
Example: Pole Expansion of cot z (cont.)
1
Collecting our results, we now have (1) f ( z )  cot z  , where
z
1
1
1
2
on CN , cot z   cot z   M 

M

, and
1
z
z
N



 2
where (2) f ( z ) has poles at an  n , n  1, 2, with (3) residues bn  1
and (4) f (0)  0. Hence

 bn
1
bn    bn
b 
f ( z )  cot z   f (0)   
  
 n
z
an  n 1  ( z  an ) an 
n 1  ( z  an )

1
1   
1
1 
 0  




n  n 1  ( z  n ) n 
n 1  ( z  n )
or rearranging and combining the two series term - by - term,

1  
2z

cot z     2

z n 1  z  n 2 2 
Ques. : Can we alternatively consider expanding f ( z )  z cot z ?
Other Pole Expansions
1
1 
2z

  2
sin z
z n 1 z  n 2 2
1


cos z
 tan z 
 2n  1 

2
n  0   2n  1  
 z2







2
2z
  2n  1  
2

 z
2



1
2z
 cot z    2
z n 1 z  n 2 2
n 0
2
1
1 
2z

  2
sinh z
z n 1 z  n 2 2
1


cosh z
 tanh z 
 2n  1 

2
n  0   2n  1  
 z2







2
2z
  2n  1  
2

 z
2



1
2z
 coth z    2
z n 1 z  n 2 2
n 0
2
• The Mittag-Leffler theorem generalizes the partial fraction representation of a
rational function to meromorphic functions
Infinite Product Expansion of
Entire Functions
Weierstrass's Factor Theorem :
 f ( z ) is an entire function
 f ( z ) has simple zeros at z  an , n  1,

, where 0  a1  a2  a3 
f ( z )
 M , independent of N , on circles CN of radius RN that do not
f ( z)
pass through zeros of f ( z ) and such that RN 

N 
Then
f ( z )  f (0) e
z f  (0) 
f (0)

z  an
1


e

an 
n 1 
z
 There exists a generalization to multiple order zeros (Schaum's, p. 267)
Product Expansion Formula
Proof :
At a simple zero, f ( z ) must have the form f ( z )   z  an  g ( z ) ,
where g ( z ) is analytic and non - vanishing at z  an . Hence the logarithmic
derivative,
f ( z ) d
d
d
 ln f ( z ) 
ln  z  an   ln g ( z ) 
f ( z ) dz
dz
dz
1
g ( z )

,
 z  an  g ( z )
has a simple pole at z  an with residue 1 ! By the Mittag - Leffler
f ( z ) f (0)   1
1 
theorem,

 
  , which, on integrating, yields
f ( z)
f (0) n 1  z  an an 
z
f ( z )
f ( z)
f (0)    z  an  z 
 .
 ln
0 f ( z ) dz  ln f ( z )  ln f (0)  ln f (0)  z f (0)  
 an
an 
n 1 
Upon exponentiating, we obtain the desired result,
f ( z )  f (0) e
z f  (0)
f (0)

 
z 
z 
 ln 1 an  an 
e n1
 f (0) e
z f  (0) 
f (0)
e
n 1
 z 
ln 1 
 an 
e
z
an
 f (0) e
z f  (0) 
f (0)

z
1



an
n 1 
 an
e .

z
Useful Product Expansions

z2 
 sin z  z 1  2 2 
n 
n 1 

z2 
 sinh z  z  1  2 2 
n 
n 1 


4z2
 cos z   1 

2
2 

n 1
  2n  1  


4z2
 cosh z   1 

2
2 

n 1
  2n  1  



 tan z  z 
n 1

1  n2z 2
2



2
4
z
1 
2 2 
2
n

1
 






 tanh z  z 
n 1

1  n2z 2
2



2
4
z
1 
2 2 
2
n

1
 



• Product expansions generalize for entire functions the factorization of the
numerator and denominator polynomials of a rational function into products
of their roots
The Argument Principle
Assume f ( z ) has a pole or zero of order ( multiplicity ) M n at z  an ; we write
f ( z )   z  an 
Mn
g ( z) ,
where g ( z ) is analytic and non - vanishing at z  an . Note M n  0 for
poles and M n  0 for zeros. Hence the logarithmic derivative,
d
f ( z )
d
d
ln f ( z ) 
 M n ln  z  an   ln g ( z )
dz
f ( z)
dz
dz
Mn
g ( z )


g ( z ) ( 0)
 z  an 
has a simple pole at z  an with residue M n . Therefore,
1
2 i

C
d
ln f ( z ) dz   M n
dz
n
where the sum is over the poles and zeros of f ( z ) enclosed by C .
The Argument Principle (cont.)

1
2 i

C
d
ln f ( z ) dz  N  P
dz
 N  number of zeros,
where 
encircled by C, counting multiplicities.
P

number
of
poles

Since the integrand is an exact differential, we also have
1
2 i

C
endpoint of C
z
d
1
ln f ( z ) dz 
ln f ( z )
dz
2 i
beginning point of C
C
3
endpoint of C
1
arg f ( z )
endpoint of C

ln f ( z ) beginning point of C 
2 i
2 beginning point of C
0

1
 change in arg f ( z ) as z goes around C
2
a4
4
a3
2
a2
z  a1
This is the result from which the "argument principle" gets its name.
1
Summation of Series
The residue theorem is also frequently used to sum series. Some of the
most important results are obtained from integrals of the form
   dz over
C
the contour C shown below and are summarized as follows :


f (n)  (sum of residues of  f ( z ) cot  z at poles of f ( z ) )
n 

  1
n
f (n)  (sum of residues of  f ( z ) csc  z at poles of f ( z ) )
n 

 f  2n21   (sum of residues of
 f ( z ) tan  z at poles of f ( z ) )
n 



n

1
   f 2n21  (sum of residues of  f ( z ) sec yz at poles of f ( z ) )
n 
C
where all poles of f ( z ) are used.
x
… -3
-2 -1 0
1
2
3
…
Summation of Series, cont’d
Example :

1
.

2
n  n  1
Evaluate the sum
From


f (n)  (sum of residues of  f ( z ) cot  z at poles of f ( z ) )
n 
1
1

 f ( z ) has poles at z  i,  i
2
n  1 (n  i )(n  i )
 cot(i )
 Res  f (i ) cot( i )  lim ( z i)
z  i
( z  i) ( z i)
with f (n) 


 cos(i )
 cosh( )
 coth( )


for both residues
2i sin( i )
2sinh( )
2

1
  coth( )

2
n  n  1