AEM_3e_Chapter_19

AEM_3e_Chapter_19 - 19 19 3. 0, 2, 0, 2, 0 Series and...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
19 19 Series and Residues EXERCISES 19.1 Sequences and Series 3. 0, 2, 0, 2, 0 6. Converges. To see this write the general term as µ 2 5 n 1+ n 2 n i 1+3 n 5 n i . 9. Diverges. To see this write the general term as n µ 1+ 1 n i n . 12. Write z n = µ 1 4 + 1 4 i n in polar form as z n = à 2 4 ! n cos + i à 2 4 ! n sin .Now Re( z n )= à 2 4 ! n cos 0as n →∞ and Im( z n )= à 2 4 ! n sin 0as n →∞ since 2 / 4 < 1. 15. We identify a = 1 and z =1 i . Since | z | = 2 > 1 the series is divergent. 18. We identify a =1 / 2 and z = i . Since | z |− 1 the series is divergent. 21. From lim n →∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 (1 2 i ) n +2 1 (1 2 i ) n +1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1 | 1 2 i | = 1 5 we see that the radius of convergence is R = 5. The circle of convergence is | z 2 i | = 5. 24. From lim n →∞ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 1 ( n +1) 2 (3 + 4 i ) n +1 1 n 2 (3+4 i ) n ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = lim n →∞ µ n n +1 2 1 | 3+4 i | = 1 5 we see that the radius of convergence is R = 5. The circle of convergence is | z +3 i | =5. 27. From lim n →∞ n s ¯ ¯ ¯ ¯ 1 5 2 n ¯ ¯ ¯ ¯ = lim n →∞ 1 25 = 1 25 we see that the radius of convergence is R = 25. The circle of convergence is | z 4 3 i | = 25. 30. (a) The circle of convergence is | z | = 1. Since the series of absolute values X k =1 ¯ ¯ ¯ ¯ z k k 2 ¯ ¯ ¯ ¯ = X k =1 | z | k k 2 = X k =1 1 k 2 292
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
19.2 Taylor Series converges, the given series is absolutely convergent for every z on | z | = 1. Since absolute convergence implies convergence, the given series converges for all z on | z | =1. (b) The circle of convergence is | z | = 1. On the circle, n | z | n →∞ as n →∞ . This implies nz n 6→ 0as n →∞ . Thus by Theorem 19 . 3 the series is divergent for every z on the circle | z | =1. EXERCISES 19.2 Taylor Series 3. DiFerentiating 1 1+2 z =1 2 z +2 2 z 2 2 3 z 3 + ··· gives 2 (1 + 2 z ) 2 = 2+2 · 2 2 z 3 · 2 3 z 2 + ··· .Thu s 1 (1+2 z ) =1 2 · (2 z )+3 · (2 z ) 2 −··· = X k =1 ( 1) k 1 k (2 z ) k 1 where R = 1 2 . 6. Replacing z in e z = X k =0 z k k ! by z 2 and multiplying the result by z gives ze z 2 = X k =0 ( 1) k k ! z 2 k +1 where R = . 9. Replacing z in cos z = X k =0 ( 1) k z 2 k (2 k )! by z/ 2 gives cos z 2 = X k =0 ( 1) k (2 k )! ³ z 2 ´ 2 k where R = . 12. Using the identity cos z = 1 2 (1 + cos2 z ) and the series cos z = X k =0 ( 1) k z 2 k (2 k )! gives cos 2 z = 1 2 + 1 2 X k =0 ( 1) k (2 z ) 2 k (2 k )!
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/03/2011 for the course BIS 511 at Yale.

Page1 / 9

AEM_3e_Chapter_19 - 19 19 3. 0, 2, 0, 2, 0 Series and...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online