lecture-42 - , . . . . Proof. Similar to the previous...

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Lecture 42: Uniqueness of Series Representations Dan Sloughter Furman University Mathematics 39 May 19, 2004 42.1 Uniqueness of Taylor series Theorem 42.1. If f ( z ) = X n =0 a n ( z - z 0 ) n for all z in an open disk D = { z C : | z - z 0 | < R } , then a n = f ( n ) ( z 0 ) n ! for n = 0 , 1 , 2 , . . . . Proof. Let C be the circle | z - z 0 | = R 1 , where 0 < R 1 < R , and let g n ( z ) = 1 2 πi ( z - z 0 ) n +1 , n = 0 , 1 , 2 , . . . . Then Z C g n ( z ) f ( z ) dz = 1 2 πi Z C f ( z ) ( z - z 0 ) n +1 dz = f ( n ) ( z 0 ) n ! . However, we also have Z C g n ( z ) f ( z ) dz = X m =0 a m · 1 2 πi Z C 1 ( z - z 0 ) n - m +1 dz. 1
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Since Z C 1 ( z - z 0 ) n - m +1 dz = ± 0 , if m 6 = n, 2 πi, if m = n, it follows that f ( n ) ( z 0 ) n ! = Z C g n ( z ) f ( z ) dz = a n . 42.2 Uniqueness of Laurent series Theorem 42.2. If f ( z ) = X n = -∞ c n ( z - z 0 ) n for all z in an open annulus R 0 < | z - z 0 | < R 1 , R 0 0, then c n = 1 2 πi Z C f ( z ) ( z - z 0 ) n +1 dz, where C is any closed contour in the annulus with z 0 in its interior and n = 0 , ± 1 , ± 2
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Unformatted text preview: , . . . . Proof. Similar to the previous proof, let g n ( z ) = 1 2 i ( z-z ) n +1 , n = 0 , 1 , 2 , . . . . Then Z C g n ( z ) f ( z ) dz = 1 2 i Z C f ( z ) ( z-z ) n +1 dz. However, we also have Z C g n ( z ) f ( z ) dz = X m =- a m 1 2 i Z C 1 ( z-z ) n-m +1 dz. Since Z C 1 ( z-z ) n-m +1 dz = , if m 6 = n, 2 i, if m = n, 2 it follows that c n = 1 2 i Z C f ( z ) ( z-z ) n +1 dz. 3...
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This note was uploaded on 11/23/2009 for the course MATHEMATIC Mathematic taught by Professor Dansloughter during the Spring '04 term at Furman.

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lecture-42 - , . . . . Proof. Similar to the previous...

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