# 17 if z 2 then z 2 5 n 4 5 n apply the weierstrass m

• Homework Help
• 186
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 87 - 90 out of 186 pages.

17. If | z | ≤ 2, then z + 2 5 n 4 5 n . Apply the Weierstrass M -test with M n = ( 4 5 ) n . Since M n is convergent (a geometric series with r < 1), it follows that the series X n =0 z + 2 5 n converges uniformly for all | z | ≤ 2. 21. If 2 . 01 ≤ | z - 2 | ≤ 2 . 9, then ( z - 2) n 3 n 2 . 9 3 n = A n , and 2 n ( z - 2) n 2 2 . 01 n = B n . Apply the Weierstrass M -test with M n = ( A n + B n ). Since M n is convergent (two geo- metric series with ratios < 1), it follows that the series X n =0 ( z - 2) n 3 n + 2 n ( z - 2) n converges uniformly in the annular region 2 . 01 ≤ | z - 2 | ≤ 2 . 9. 25. (a) For z - 1 2 < 1 6 , we have | z | = z - 1 2 + 1 2 z - 1 2 + 1 2 < 2 3 . Since X n =0 2 3 n is convergent, we conclude from the Weierstrass M-test that X n =0 z n con- verges uniformly on z - 1 2 < 1 6 . (b) We claim that X n =0 z n does not converge uniformly on | z - 1 2 | < 1 2 : We have that, for | z - 1 2 | < 1 2 , | z | = z - 1 2 + 1 2 z - 1 2 + 1 2 < 1 , and we know that the series X n =0 z n converges pointwise to s ( z ) = 1 1 - z in | z | < 1. Its n th partial sum is s n ( z ) = 1 - z n +1 1 - z . Take z = x to be a real number with | x | < 1. Then | s ( x ) - s n ( z ) | = x n +1 1 - x = x n +1 1 - x → ∞ as x 1 .
Section 4.1 Sequences and Series of Functions 87 Therefore, the maximum difference M n between each partial sum and the series sum is unbounded on | z | < 1, so the sequence of partial sums, and, hence, the series, does not converge uniformly on | z | < 1. 29. (a) Let δ > 1 be a positive real number. To show that the series ζ ( z ) = X n =1 1 n z (principal branch of n z ) converges uniformly on the half-plane H δ = { z : Re z δ > 1 } , we will apply the Weier- strass M -test. For all z H δ , we have | n z | = e ( x + iy ) ln n = e x ln n = n x > n δ . So 1 n z 1 n δ = M n . Since M n = 1 n δ is a convergent series (because δ > 1), it follows from the Weierstrass M -test that that n =1 1 n z converges uniformly in H δ . (b) Each term of the series n =1 1 n z is analytic in H = { z : Re z > 1 } (in fact, each term is entire). To conclude that the series is analytic in H , it is enough by Corollary 2 to show that the series converges uniformly on any closed disk contained in H . If S is a closed disk contained in H , S is clearly disjoint from the imaginary axis. Let H δ ( δ > 0) be a half-plane containing S . By part (a), the series converges uniformly on H δ , consequently, the series converges uniformly on S . By Corollary 2, the series is analytic in H . (Note the subtilty in the proof. We did not show that the series converges uniformly on H . In fact, the series does not converge uniformly in H .) (c) To compute ζ 0 ( z ), according to Corollary 2, we can differentiate the series term-by- term. Write 1 n z = 1 e z ln n = e - z ln n . Using properties of the exponential function, we have d dz 1 n z = d dz e - z ln n = - ln ne - z ln n = - ln n 1 n z . So, for all z H , ζ 0 ( z ) = - X n =1 ln n n z . 33. To show that f n { f n } converges uniformly on Ω, it is enough to show that max z Ω | f n ( z ) - f m ( z ) | can be made arbitrarily small by choosing m and n large. In other words, given ε > 0, we must show that there is a positive integer N such that if m , n N , then max z Ω | f n ( z ) - f m ( z ) | < ε.
88 Chapter 4
• • • 