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Unformatted text preview: Mathematics 185 Intro to Complex Analysis Fall 2009 M. Christ Solutions to Selecta from Problem Set 9 3.2.13 Differentiate the power series (about 0) for (1 z ) 1 to obtain a power series expansion for (1 z ) 2 . What is the radius of convergence of the new series? Solution. (1 z ) 1 = n =0 z n , with radius of convergence 1. The derivative of (1 z ) 1 is (1 z ) 2 . Therefore in D (0 , 1), (1 z ) 2 = X n =1 nz n 1 = X n =0 ( n + 1) z n . The radius of convergence of the new series is still 1. For instance, we can use the root test: ( n +1) 1 /n = e [ln( n +1)] /n , and [ln( n +1)] /n 0, so ( n +1) 1 /n e = 1. Therefore the radius of convergence is 1 = 1 / lim n ( n + 1) 1 /n . 3.2.14 Suppose that the power series n =0 a n z n has radius of convergence R . Let f ( z ) be the sum of this series, for  z  < R . Suppose that  z  < R . Let R be the radius of convergence of the Taylor series for f about z . Show that R  z  R R +  z  . Solution. D ( z ,R  z  ) D (0 ,R ). Indeed, if w is in this disk, then  w  =  ( w z ) + z   w z  +  z  ( R  z  ) +  z  = R. Therefore f is analytic in D ( z ,R  z  ). Therefore its Taylor series expansion about z has radius of convergence R  z  . Now the situation is symmetric, under reversal of the roles of the two centers z and 0. Since R R  z  , 0 D ( z , R ). Therefore the radius of convergence R of the Taylor series about 0 satisfies R R  z  = R  z  . Thus R R +  z  . 3.2.24 Compute the Taylor series of ( z ) = n =1 n z about 2. Solution. Write n z = e z ln( n ) = e 2 ln( n ) e ( z 2) ln( n ) = n 2 X k =0 ( 1) k ln( n ) k k !...
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This note was uploaded on 01/19/2010 for the course MATH 113 taught by Professor Ogus during the Fall '08 term at University of California, Berkeley.
 Fall '08
 OGUS
 Math, Power Series

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