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Unformatted text preview: Power Series Represent Holomorphic Functions Theorem 25 A power series with center z C and radius of convergence R > represents a holomorphic function in B R ( z ) . The derivatives of this function are obtained by differentiating the original series term by term. All the series thus obtained have the same radius of convergence R . Hence by the first statement, each of them represents a holomorphic function. Corollary 1 A complex function f which is analytic at z C is holomor phic in B R ( z ) , where R > denotes the radius of convergence of the power series representation of f near z . Remark: In the next section, we will show that the opposite is also true (Taylors theorem). Proof of the Theorem: The proof is in three steps: 1. We write the power series representation of f and the derived series by f ( z ) = summationdisplay n =0 a n ( z z ) n , f 1 ( z ) = summationdisplay n =1 na n ( z z ) n 1 , z B R ( z ) . (832) We want to show that f is holomorphic in B R ( z ) and f = f 1 . With termwise addition, we obtain f ( z + z ) f ( z ) z f 1 ( z ) = summationdisplay n =2 a n parenleftbigg ( z + z z ) n ( z z ) n z n ( z z ) n 1 parenrightbigg , (833) for z B R ( z ). 2. With b := z + z z , a := z z , z = b a , we have ( z + z z ) n ( z z ) n z n ( z z ) n 1 = b n a n b a na n 1 , n 2 . (834) We prove by induction that b n a n b a na n 1 = ( b a ) n 2 summationdisplay k =0 ( k + 1) a k b n 2 k bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =: A n , n 2 . (835) 144 n = 2: b 2 a 2 b a 2 a = ( b + a )( b a ) b a 2 a = b a = ( b a ) A 2 (836) m mapsto m + 1: Assume the formula is true for n = m . We write b m +1 a m +1 b a = b m +1 ba m + ba m a m +1 b a (837) = b ( b m a m ) + ( b a ) a m b a = b b m a m b a + a m = b ( ( b a ) A m + ma m 1 ) + a m (838) = ( b a ) ( bA m + ma m 1 ) + ( m + 1) a m . (839) From the definition of A m we obtain bA m + ma m 1 = m 2 summationdisplay k =0 ( k +1) a k b m 1 k + ma m 1 = m 1 summationdisplay k =0 ( k +1) a k b m 1 k = A m +1 , (840) so that b m +1 a m +1 b a ( m + 1) a m = A m +1 . (841) Therefore we have ( z + z z ) n ( z z ) n z n ( z z ) n 1 = z n 2 summationdisplay k =0 ( k +1)( z z ) k ( z + z z ) n 2 k , (842) for n 2. 3. We use this to obtain f ( z + z ) f ( z ) z f 1 ( z ) = summationdisplay n =2 a n z n 2 summationdisplay k =0 ( k +1)( z z ) k ( z + z z ) n 2 k ....
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This note was uploaded on 11/17/2011 for the course MATH 529 taught by Professor Staff during the Spring '08 term at UNC.
 Spring '08
 Staff
 Math, Derivative, Power Series

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