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 CN 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 CN 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 w0 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 CN 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 n1 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 2n21 (sum of residues of f ( z ) tan z at poles of f ( z ) ) n n 1 f 2n21 (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