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Chapter 4 Power Series and Laurent Series

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Chapter 4 Power Series and Laurent Series Section 4.1 Sequences and Series of Complex Numbers 73 Solutions to Exercises 4.1 1. We have in π e 2 1 0 ≤|an| = = → 0asn →∞. n n So an → 0 by the squeeze theorem or by applying (2), Sec. 4.1. 5. We have cosh in = cos n (see (25), Sec. 1.6). Hence cosh in cos n 1 |an| = = ≤ → 0asn →∞. n2 n2 n2 So an → 0 by the squeeze theorem. 9. (a) Suppose that {an} is a convergent sequence and let bn = an+1. We want to show that {bn} is a convergent sequence and that limn→∞ an = limn→∞ bn. Let L denote the limit of the sequence {an}. By definition of convergence, given >0, we can find N ≥ 0, such that for all n ≥ N, |an − L| <. But n ≥ N implies that n +1≥ N. So for all n ≥ N, |an+1 − L| <. But an+1 = bn, so for all n ≥ N, |bn − L| <, which implies that {bn} is a convergent sequence with same limit as {an}. (b) Define a = i and a = 3 . Given that {a } is convergent, then by part (a), we have 1 n+1 2+an n limn→∞ an = limn→∞ an+1. Let L = limn→∞ an. Then 3 3 L = lim an = lim an+1 = lim = . n→∞ n→∞ n→∞ 2+an 2+L Solving for L,wefind 3 L = ⇒ L(2 + L)=3; 2+L ⇒ L2 +2L − 3=0, (L + 3)(L − 1)=0,L= −3orL =1. Because the limit of a convergent sequence is unique, we have to decide whether L =1orL = −3. Note that |a1| = 1. If we can show that Re an ≥ 0, then Re L = limn→∞ Re an ≥ 0, and this would eliminate the value L = −3. Let’s prove by induction that Re an ≥ 0. The statement is clearly true if n = 1, because a1 = i and so Re i =0≥ 0. Now suppose that Re an ≥ 0 and let’s prove that Re an+1 ≥ 0. For this purpose, we compute 3 2+an ! Re an+1 =Re =3Re 2+an (2 + an) · (2 + an) 3 2 = · Re (2 + Re an − i Im an)=3α (2 + 2 Re an) ≥ 0, (2 + an) · (2 + an) 2 1 1 where α = = 2 . This completes the proof by induction and shows that L =6 −3, (2+an )·2+an |2+an | so L =1. 74 Chapter 4 Power Series and Laurent Series ∞ 3 − i 13. The series X is a constant multiple of a convergent geometric series and so it is (1 + i)n n=3 convergent. To find its sum proceed as follows: ∞ 3 − i 1 1 X =(3− i) + + ··· (1 + i)n (1 + i)3 (1 + i)4 n=3 1 1 1 =(3− i) 1+ + + ··· (1 + i)3 1+i (1 + i)2 ∞ 3 − i 1 3 − i 1 = X = · , (1 + i)3 (1 + i)n (1 + i)3 1 − r n=0 where r = 1 , |r| = √1 < 1. So the sum is 1+i 2 3 − i 1 3 − i −1 − 3i 3 i S = · = = = − + . (1 + i)3 − 1 (1 + i)2 i (1 + i)2 2 2 1 1+i ∞ 1 1 17. The series X is absolutely convergent by comparison to the series 2 P 2 : (n + i)((n − 1) + i) n n=2 1 |(n + i)((n − 1) + i)| = (n + i) (n − 1) + i ≥ n(n − 1) ≥ n2 if n ≥ 2; 2 so 1 2 ≤ . (n + i)((n − 1) + i) n2 To sum the series, use partial fractions and the terms will telescope, as follows: 1 1 1 = − . (n + i)((n − 1) + i) (n − 1) + i) n + i So the Nth partial sum is N 1 1 s = X + N (n − 1) + i) n + i n=2 1 1 1 1 1 1 = − + − + ···+ − (2 − 1) + i) 2+i (3 − 1) + i) 3+i (N − 1) + i) N + i 1 1 1 = − → − 0, 2+i N + i 1+i →∞ 1 1−i as N . So the sum is S = 1+i = 2 . ∞ 1+3i n √ 21. The series X is a geometric series with z = 1+3i . Since |z| = 1 10 < 1, the series 4 4 4 n=0 converges to 1 1 4 4 1 2 = = = = (1 + i). 1 − z − 1+3i 3 − 3i 3 1 − i 3 1 4 25. Write (1+2in)n 1 n = +2i . nn n Section 4.1 Sequences and Series of Complex Numbers 75 1 | |≥| | It is now clear that the series is a geometric series with z = n +2i. Since z Im z = 2, it follows that the series diverges. 29. For n ≥ 2, we have n −n cos(in) cosh n e + e 1 3 3 3 = = = (e−n +n + e−n −n) ≤ e−n /2. en3 en3 2en3 2 ∞ 3 The series is absolutely convergent by comparison with the convergent series X e−n /2. n=1 33. The series converges as long as z < 1 ⇒|z| < 2. 2 37. The series converges as long as 1 < 1 ⇒ 1 < |2 − 10z| 2 − 10z 1 1 ⇒ 1 < | − z| 10 5 1 ⇒ 10 < | − z|. 5 1 Thus the series converges outside the closed disk of radius 10 and center at 5 . The limit in this region is (use the geometric series minus the first term, which is 1) 1 2 − 10z 1 − 1= − 1= . − 1 − − 1 2−10z 1 10z 1 10z i 41. If the nth partial sum of the series is sn = n , then the series converges and its limit is i lim sn = lim =0. n→∞ n→∞ n An example of such a series is ∞ i i 1+X − . n +1 n n=1 45. The test of convergence in Theorem 12 is really about series with positive real terms. For a proof, see any calculus book and use the same proof for real series. 49. To establish the addition formula for cosines, we will manipulate the series for cosine, using several properties of absolutely convergent series and Cauchy products. We have ∞ ∞ z2n z2n cos z = X(−1)n 1 ; cos z = X(−1)n 2 ; 1 (2n)! 2 (2n)! n=0 n=0 n z2k z2(n−k) cos z cos z = X(−1)k 1 (−1)n−k 2 1 2 (2k)! (2(n − k))! k=0 (−1)n n (2n)! = X z2kz2(n−k) (2n)! (2k)!(2(n − k))! 1 2 k=0 n n (−1) 2n − = X z2kz2(n k). (2n)! 2k 1 2 k=0 76 Chapter 4 Power Series and Laurent Series Similarly, ∞ ∞ z2n+1 z2n+1 sin z = X(−1)n 1 ; sin z = X(−1)n 2 ; 1 (2n + 1)! 2 (2n + 1)! n=0 n=0 n z2k+1 z2(n−k)+1) sin z sin z = X(−1)k 1 (−1)n−k 2 1 2 (2k + 1)! (2(n − k) + 1)! k=0 (−1)n n (2(n + 1))! = X z2k+1z2(n+1)−(2k+1) (2(n + 1))! (2k)!(2(n +1)− (2k + 1)))! 1 2 k=0 n n (−1) 2(n +1) − = X z2k+1z2(n+1) (2k+1). (2(n + 1))! 2k +1 1 2 k=0 Hence ∞ n n (−1) 2n − cos z cos z − sin z sin z = X X z2kz2(n k) 1 2 1 2 (2n)! 2k 1 2 n=0 k=0 ∞ n n (−1) 2(n +1) − − X X z2k+1z2(n+1) (2k+1) (2(n + 1))! 2k +1 1 2 n=0 k=0 ∞ n n (−1) 2n − =1+X X z2kz2(n k) (2n)! 2k 1 2 n=1 k=0 ∞ n−1 n−1 (−1) 2n − − X X z2k+1z2n (2k+1), (2n)! 2k +1 1 2 n=1 k=0 where in the first series on the right, we wrote the first term separately, and in the second series on the right, we shifted the index by changing n to n − 1. Combining the two series, we see that the last displayed sum equals ∞ n n n−1 (−1) ( 2n − 2n − ) 1+X X z2kz2(n k) + X z2k+1z2n (2k+1) . (2n)! 2k 1 2 2k +1 1 2 n=1 k=0 k=0 But the two inner sums in k add up to 2n 2n X zkz2n−k =(z + z )n , k 1 2 1 2 k=0 because the first sum adds the even-indexed terms and the second sum adds the odd-indexed terms. Hence ∞ (−1)n cos z cos z − sin z sin z =1+X (z + z )n 1 2 1 2 (2n)! 1 2 n=1 ∞ (−1)n = X (z + z )n = cos(z + z ), (2n)! 1 2 1 2 n=0 as desired. Note: While this was a good exercise in Cauchy products, it is not the most efficient way to prove the addition formula for the cosine. For a more elegant proof, see Example 2(b), Sec. 4.6. Section 4.2 Sequences and Series of Functions 77 Solutions to Exercises 4.2 sin nx 1. The sequence of functions f (x)= converges uniformly on the interval 0 ≤ x ≤ π. To see n n this, let Mn = max |0 − fn(x)| = max |fn(x)|, where the maximum is taken over all x in [0,π]. Then ≤ 1 → →∞ Mn = n . Since Mn 0, as n , it follows that fn converges uniformly to f =0on[0,π]. In fact, we fact uniform convergence on the entire real line. 5. (a) and (b) First, let us determine the pointwise limit of the sequence of functions fn(x)= nx .Forx in the interval 0 ≤ x ≤ 1, we have n2x2 − x +1 nx nx x f (x)= = = → 0, as n →∞.
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