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hw9 - Mathematics 185 Intro to Complex Analysis Fall 2009 M...

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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 0 = 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 0 | < R . Let ˜ R be the radius of convergence of the Taylor series for f about z 0 . Show that R - | z 0 | ≤ ˜ R R + | z 0 | . Solution. D ( z 0 , R - | z 0 | ) D (0 , R ). Indeed, if w is in this disk, then | w | = | ( w - z 0 ) + z 0 | ≤ | w - z 0 | + | z 0 | ≤ ( R - | z 0 | ) + | z 0 | = R. Therefore f is analytic in D ( z 0 , R - | z 0 | ). Therefore its Taylor series expansion about z 0 has radius of convergence R - | z 0 | . Now the situation is symmetric, under reversal of the roles of the two centers z 0 and 0. Since ˜ R R - | z 0 | , 0 D ( z 0 , ˜ R ). Therefore the radius of convergence R of the Taylor series about 0 satisfies R ˜ R - | z 0 - 0 | = ˜ R - | z 0 | . Thus ˜ R R + | z 0 | . 3.2.24 Compute the Taylor series of ζ ( 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 ! ( z - 2) k . Now if we plug each of these series into the definition of ζ ( z ) and boldly interchange the two infinite summation signs, we get: ζ ( z ) = X k =0 ( - 1) k k ! X n =1 n - 2 ln( n ) k ( z - 2) k .
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