math185f09-hw3sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

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Unformatted text preview: MATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 3 SOLUTIONS For a real-valued function of two real variables, u : R R , we say that u is twice continuously differentiable if all second-order partial derivatives u xx ,u yy ,u xy ,u yx exist and are continuous on R . The set of all twice continuously differentiable functions on R is denoted C 2 ( R ). 1. We mentioned Tauberian theorems in class. Here is an example of an easy one (easy relative to other Tauberian theorems). Let n =0 a n z n be a power series with radius of convergence 1 and suppose lim n na n = 0 . (a) Show that lim m m n =0 n | a n | m = 0 . ( Hint: Problem 4 (a), Problem Set 3 , Math 104 , Spring 2009.) Solution. Let > 0 be given. Since lim n na n = 0, there exists N 1 N such that n | a n | < / 2 whenever n > N 1 . Now by the Archimedean property, there exists N 2 N such that | a 1 | + 2 | a 2 | + + N 1 | a N 1 | N 2 < 2 . Hence | a 1 | + 2 | a 2 | + + N 1 | a N 1 | m < 2 whenever m > N 2 . Now for m > max { N 1 ,N 2 } , m n =0 n | a n | m = N 1 n =0 n | a n | m + m n = N 1 +1 n | a n | m < 2 + m- N 1 m 2 < . Hence lim m m n =0 n | a n | m = 0 . (b) Define a function f by f ( z ) = X n =0 a n z n for all | z | < 1 . Let x be a real variable and suppose the following left limit exists lim x 1- f ( x ) = A. Show that the series n =0 a n converges to A . Date : October 18, 2009 (Version 1.0). 1 Solution. Let x R and 0 x < 1. We have X m n =0 a n- f ( x ) = X m n =0 a n (1- x n )- X n = m +1 a n x n (1- x ) X m n =0 | a n | (1 + x + + x n- 1 ) + X n = m +1 | a n | x n < m (1- x ) m n =0 n | a n | m + X n = m +1 n | a n | x n n . Let > 0. Since lim n na n = 0, there exists M 1 N such that m | a m | < / 2 whenever m > M 1 . By (a), there exists M 2 N such that m n =0 n | a n | m < 2 for m > M 2 . Hence when m > max { M 1 ,M 2 } , X m n =0 a n- f ( x ) < m (1- x ) 2 + 2 m X n = m +1 x n = 2 m (1- x ) + 1 m x m +1 1- x < 2 m (1- x ) + 1 m (1- x ) . Then for 1- x = 1 /m and m > max { M 1 ,M 2 } , we get X m n =0 a n- f 1- 1 m < , which implies that lim m X m n =0 a n = lim m f 1- 1 m . Since lim x 1- f ( x ) = A exists, we must have lim m f 1- 1 m = A and so we get X n =0 a n = A. 2. Recall that C is both a real vector space of dimension 2 and a complex vector space of dimen- sion 1. A function : C C is called R-linear if is a linear transformation of real vector spaces, ie. ( 1 z 1 + 2 z 2 ) = 1 ( z 1 ) + 2 ( z 2 ) for all 1 , 2 R and z 1 ,z 2 C . (2.1) It is called C-linear if is a linear transformation of complex vector spaces, ie....
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This note was uploaded on 02/17/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at University of California, Berkeley.

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math185f09-hw3sol - MATH 185: COMPLEX ANALYSIS FALL 2009/10...

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