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Lecture20 - Power Series Represent Holomorphic Functions...

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Power Series Represent Holomorphic Functions Theorem 25 A power series with center z 0 C and radius of convergence R > 0 represents a holomorphic function in B R ( z 0 ) . 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 0 C is holomor- phic in B R ( z 0 ) , where R > 0 denotes the radius of convergence of the power series representation of f near z 0 . Remark: In the next section, we will show that the opposite is also true (Taylor’s 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 0 ) n , f 1 ( z ) = summationdisplay n =1 na n ( z z 0 ) n 1 , z B R ( z 0 ) . (832) We want to show that f is holomorphic in B R ( z 0 ) and f = f 1 . With termwise addition, we obtain f ( z + Δ z ) f ( z ) Δ z f 1 ( z ) = summationdisplay n =2 a n parenleftbigg ( z + Δ z z 0 ) n ( z z 0 ) n Δ z n ( z z 0 ) n 1 parenrightbigg , (833) for z B R ( z 0 ). 2. With b := z + Δ z z 0 , a := z z 0 , Δ z = b a , we have ( z + Δ z z 0 ) n ( z z 0 ) n Δ z n ( z z 0 ) 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
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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 + Δ z z 0 ) n ( z z 0 ) n Δ z n ( z z 0 ) n 1 = Δ z n 2 summationdisplay k =0 ( k +1)( z z 0 ) k ( z z z 0 ) n 2 k , (842) for n 2. 3. We use this to obtain f ( z + Δ z ) f ( z ) Δ z f 1 ( z ) = summationdisplay n =2 a n Δ z n 2 summationdisplay k =0 ( k +1)( z z 0 ) k ( z z z 0 ) n 2 k . (843) For | z | ≤ R 0 and | z + Δ z | ≤ R 0 , R 0 < R , we estimate vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle n 2 summationdisplay k =0 ( k + 1)( z z 0 ) k ( z + Δ z z 0 ) n 2 k vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle n 2 summationdisplay k =0 ( k +1) R n 2 0 = n ( n 1) 2 R n 2 0 .
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