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Unformatted text preview: Math 185. Sample Answers to Problem Set #9 Page 214 8. Let X n =0 a n ( z z ) n be the Taylor series for f ( x ) about z = z , and let R be its radius of convergence. The assumption that f ( n ) ( z ) = 0 for n m implies that a = a 1 = = a m = 0, so f ( z ) = X n = m +1 a n ( z z ) n = ( z z ) m +1 X n =0 a n + m +1 ( z z ) n for all z with  z z  < R . (Here the last equality holds because the factor ( z z ) m +1 is just a constant as far as the summation is concernedsee Exercise 7 on page 181.) Therefore g ( z ) = X n =0 a n + m +1 ( z z ) n for all z satisfying 0 <  z z  < R . This is also true when z = z since g ( z ) = a m +1 = f ( m +1) ( z ) ( m + 1)! . Therefore g ( z ) has a Taylor series at z , so it is analytic there. 10. For all N Z + and all z in the given annular domain, let N ( z ) = S 2 ( z ) N 1 X n =0 b n ( z z ) n , so that, as with (3) on page 207 (since 1 / ( z z ) n is also continuous on C ), Z C g ( z ) S 2 ( z ) dz = N 1 X n =0 b n Z C g ( z )( z z ) n dz + Z C g ( z ) N ( z ) dz . Next, we need to show that the given power series is uniformly convergent on C (and this is the main difficulty in this part of the problem). Assume that our annular domain is R 1 <  z z  < R 2 . Then the power series T ( w ) = X n =1 b n w n 1 2 is convergent on the set  w  < 1 /R 1 since T (1 / ( z z )) = S 2 ( z ), term for term. If the contour...
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This note was uploaded on 09/01/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.
 Fall '07
 Lim
 Taylor Series

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