Unformatted text preview: Chapter 7 Power Series
It is a pain to think about convergence but sometimes you really have to. Sinai Robins 7.1 Sequences and Completeness As in the real case (and there will be no surprises in this chapter of the nature `real versus complex'), a (complex) sequence is a function from the positive (sometimes the nonnegative) integers to the complex numbers. Its values are usually denoted by an (as opposed to, say, a(n)) and we commonly denote the sequence by (an ) , (an )n1 , or simply (an ). The notion of convergence of a sequence n=1 is based on the following sibling of Definition 2.1. Definition 7.1. Suppose (an ) is a sequence and a C such that for all > 0, there is an integer N such that for all n N , we have an  a < . Then the sequence (an ) is convergent and a is its limit, in symbols lim an = a .
n If no such a exists then the sequence (an ) is divergent. Example 7.2. limn
in n Example 7.3. The sequence (an = in ) diverges: Given a C, choose = 1/2. We consider two cases: If Re a 0, then for any N , choose n N such that an = 1. (This is always possible since a4k+2 = i4k+2 = 1 for any k 0.) Then a  an  = a + 1 1 > 1 . 2 = 0: Given > 0, choose N > 1/. Then for any n N , n n n i  0 = i = i = 1 1 < . n n n n N If Re a < 0, then for any N , choose n N such that an = 1. (This is always possible since a4k = i4k = 1 for any k > 0.) Then a  an  = a  1 1 > 68 1 . 2 CHAPTER 7. POWER SERIES The following limit laws are the relatives of the identities stated in Lemma 2.4. Lemma 7.4. Let (an ) and (bn ) be convergent sequences and c C. (a) lim an + c lim bn = lim (an + c bn ) .
n n n n 69 (b) lim an lim bn = lim (an bn ) .
n n (c) limn an = lim n limn bn In the quotient law we have to make sure we do not divide by zero. Moreover, if f is continuous at a then lim f (an ) = f (a) if lim an = a ,
n n an bn . where we require that an be in the domain of f . The most important property of the real number system is that we can, in many cases, determine that a sequence converges without knowing the value of the limit. In this sense we can use the sequence to define a real number. Definition 7.5. A Cauchy sequence is a sequence (an ) such that
n lim an+1  an  = 0. We say a metric space X (which for us means Z, Q, R, or C) is complete if, for any Cauchy sequence (an ) in X, there is some a X such that limn an = a. In other words, completeness means Cauchy sequences are guaranteed to converge. For example, the rational numbers are not complete: we can take a Cauchy sequence of rational numbers getting arbitrarily close to 2, which is not a rational number. However, each of Z, R, and C is complete. It is the completeness of the reals that allows us to know sequences converge without knowing their limits. We will assume that the reals are complete as an axiom. There are many equivalent ways of formulating the completeness property for the reals, including: Axiom (Monotone Sequence Property). Any bounded monotone sequence converges. Remember that a sequence is monotone if it is either nondecreasing (xn+1 xn ) or nonincreasing (xn+1 xn ). Example 7.6. If 0 r < 1 then limn rn = 0: First, the sequence converges because it is decreasing and bounded below by 0. If the limit is L then, using the laws of limits, we get L = limn rn = limn rn+1 = r limn rn = rL. From L = rL we get (1  r)L = 0, so L = 0 since 1  r = 0 The following is a consequence of the monotone sequence property, although it is often listed as a separate axiom: Theorem 7.7 (Archimedean Property). If x is any real number than there is an integer N which is greater than x. CHAPTER 7. POWER SERIES 70 This essentially says that there are no infinities in the reals. Notice that this was already used in Example 7.2. For a proof see Exercise 4. It is interesting to see that the Archimedean principle underlies the construction of an infinite decimal expansion for any real number, while the monotone sequence property shows that any such infinite decimal expansion actually converges to a real number. We close this discussion of limits with a pair of standard limits. The first of these can be established by calculus methods (like L'Hospital's rule, by treating n as the variable); both of them can be proved by more elementary considerations. Lemma 7.8. (a) Exponentials beat polynomials: for any polynomial p(n) and any b R with b > 1, limn p(n) = 0. bn (b) Factorials beat exponentials: for any a R, limn Note this lemma also works for a, b C.
an n! = 0. 7.2 Series A series is a sequence (an ) whose members are of the form an = n bk an = n bk ); here (or k=1 k=0 (bk ) is the sequence of terms of the series. The an = n bk (or an = n bk ) are the partial k=0 k=1 sums of the series. If we wanted to be lazy we would for convergence of a series simply refer to convergence of the partial sums of the series, after all, we just defined series through sequences. However, there are some convergence features which take on special appearances for series, so we should mention them here explicitly. For starters, a series converges to the limit (or sum) a by definition if n lim an = lim bk = a .
n n k=1 To express this in terms of Definition 7.1, for any > 0 we have to find an N such that for all nN n bk  a < . k=1 In the case of a convergent series, we usually express its limit as a = bk or a = k1 bk . k=1 Example 7.9. Occasionally we can find the limit of a sequence by manipulating the partial sums: n 1 1 1 = lim  k(k + 1) n k k+1 k1 k=1 1 1 1 1 1 1 1 = lim 1 +  +  + +  n 2 2 3 3 4 n n+1 1 1 1 1 1 1 1 = lim 1  +  +  + +  n 2 2 3 3 4 n n+1 1 = lim 1  = 1. n n+1 A series where most of the terms cancel like this is called a telescoping series. CHAPTER 7. POWER SERIES 71 Most of the time we need to use the completeness property to check convergence of a series, and it is fortunate that the monotone sequence property has a very convenient translation into the language of series of real numbers. The partial sums of a series form a nondecreasing sequence if the terms of the series are nonnegative, and this observation immediately yields: Lemma 7.10. If bk are nonnegative real numbers then bk converges if and only if the partial k=1 sums are bounded. If bk are nonnegative real numbers and the partial sums the series bk are unbounded of k=1 then the partial sums "converge" to infinity, so we can write bk = . Using this terminology, k=1 we can rephrase Lemma 7.10 to say: bk converges in the reals if and only if it is finite. k=1 We have already used the simple fact that convergence of a sequence (an ) is equivalent to the convergence of (a ), and both of these sequences have the same limit. If an is the nth partial n1 sum of the series k1 bk then an = an1 + bn . From this we conclude: Lemma 7.11. If k1 bk converges then limn bn = 0. A common mistake is to try to use the converse of this result, but the converse is false: 1 Example 7.12. The harmonic series k1 k diverges (even though the limit of the general term is 0): If we assume the series converges, say to L, then we have 1 1 1 L = 1 + + + + ... 2 3 4 1 1 1 1 1 = 1 + + + ... + + + + ... 3 5 2 4 6 1 1 1 1 1 1 > + + + ... + + + + ... 2 4 6 2 4 6 1 1 1 1 1 1 1 1 = 1 + + + + ... + 1 + + + + ... 2 2 3 4 2 2 3 4 1 1 = L + L = L. 2 2 1 1 Here the inequality comes from k > k+1 applied to each term in the first sum in parentheses. But now we have L > L, which is impossible. There is notion of convergence that's special to series: we say that one k1 ck converges absolutely if k1 ck  < . Be careful: We are defining the phrase "converges absolutely," but this definition does not say anything about convergence of the series k1 ck ; we need a proof: Theorem 7.13. If a series converges absolutely then it converges. Proof. First consider the case when the terms ck are real. Define c+ to be ck if ck 0, or 0 if k + + ck < 0. Then ck 0 and k1 ck k1 ck  < so k1 c+ converges; let P be its limit. k Similarly, define c to be ck if ck 0, or 0 if ck > 0. Then c 0 and k1 c k1 ck  < k k k so k1 c converges; let N be its limit. Since ck = c+  c we see that k1 ck converges to k k k P  N. In case ck is complex, write ck = ak +ibk where ak and bk are real. Then k1 ak  k1 ck  < and k1 bk  k1 ck  < . By what we just proved, both k1 ak and k1 bk converge to real numbers, say, A and B. But then k1 ck converges to A + iB. CHAPTER 7. POWER SERIES 72 Another common mistake is to try to use the converse of this result, but the converse is false: k+1 Example 7.14. The alternating harmonic series k1 (1) converges, but not absolutely: This k series does not converge absolutely, according to the previous example. To see that it does converge, rewrite it as follows: (1)k+1 1 1 1 1 1 = 1  +  +  + ... k 2 3 4 5 6 k1 1 1 1 1 1 = 1 +  +  + ... 2 3 4 5 6 (Technically, there is a small detail to be checked here, since we are effectively ignoring half the partial sums of the original series. See Exercise 13.) The reader can verify the inequality 2k(2k1) k(k + 1) for k > 1, so the general term satisfies 1 1 1 1  = , 2k  1 2k 2k(2k  1) k(k + 1) so the series converges by comparison with the telescoping series of Example 7.9. For the rest of this book we shall be concerned almost exclusively with series which converge absolutely. Hence checking convergence of a series is usually a matter of verifying that a series of nonnegative reals is finite. We have already used the technique of comparing a series to a series which is known to converge; this is often called a "comparison test." Some variants of the comparison test will appear when we look at power series. One handy test is the following: Lemma 7.15 (Integral Test). Suppose f is a nonincreasing, positive function defined on [1, ). Then f (t) dt f (k) f (1) + f (t) dt
1 k=1 1 This is immediate from a picture: the integral of f (t) on the interval [k, k + 1] is bounded between f (k) and f (k + 1). Adding the pieces gives the inequalities above for the N th partial sum versus the integrals from 1 to N and from 1 to N + 1, and the inequality persists in the limit. 1 Example 7.16. k1 kp converges if p > 1 and diverges if p 1. 7.3 Sequences and Series of Functions The fun starts when one studies sequences (fn ) of functions fn . We say that such a sequence converges at z0 if the sequence (of complex numbers) (fn (z0 )) converges. If a sequence of functions, (fn ), converges at all z in some subset G C then we say that (fn ) converges pointwise on G. So far nothing new; but this notion of convergence does not really catch the spirit of the function as a whole. Definition 7.17. Suppose (fn ) and f are functions defined on G C. If for all > 0 there is an N such that for all z G and for all n N we have then (fn ) converges uniformly in G to f . fn (z)  f (z) < CHAPTER 7. POWER SERIES 73 What's the big deal about uniform versus pointwise convergence? It is easiest to describe the difference with the use of quantifiers, namely denoting "for all" and denoting "there is." Pointwise convergence on G means ( > 0) ( z G) ( N : n N fn (z)  f (z) < ) , whereas uniform convergence on G translates into ( > 0) ( N : ( z G) n N fn (z)  f (z) < ) . No big dealwe only exchanged two of the quantifiers. In the first case, N may well depend on z, in the second case we need to find an N which works for all z G. And this can make all the difference . . . The first example illustrating this difference says in essence that if we have a sequence of functions (fn ) which converges uniformly on G then for all z0 G
n zz0 lim lim fn (z) = lim lim fn (z) .
zz0 n We will need similar interchanges of limits constantly. Proposition 7.18. Suppose (fn ) is a sequence of continuous functions on the region G converging uniformly to f on G. Then f is continuous on G. Proof. Let z0 G; we will prove that f is continuous at z0 . By uniform convergence, given > 0, there is an N such that for all z G and all n N fn (z)  f (z) < . 3 Now we make use of the continuity of the fn 's. This means that given (the same) > 0, there is a > 0 such that whenever z  z0  < we have fn (z)  fn (z0 ) < . 3 All that's left is putting those two inequalities together: by the triangle inequality f (z)  f (z0 ) = f (z)  fn (z) + fn (z)  fn (z0 ) + fn (z0 )  f (z0 ) < , that is, f is continuous at z0 . Once we know the above result about continuity, we can ask about integration of series of functions. The next theorem should come as no surprise, however, its consequences (which we will only see in the next chapter) are wide ranging. Proposition 7.19. Suppose fn are continuous on the smooth curve and converge uniformly on to f . Then lim fn = f .
n f (z)  fn (z) + fn (z)  fn (z0 ) + fn (z0 )  f (z0 ) CHAPTER 7. POWER SERIES Proof. By Proposition 4.4(d), we can estimate fn  f = fn  f max fn (z)  f (z) length() . z 74 But fn f uniformly on , and we can make maxz fn (z)  f (z) as small as we like. Since uniform convergence is often of critical importance, we give two practical tests: one arguing for uniformity and the other against. They are formulated for sequences that converge to 0. If a sequence gn converges to a function g then we can usually apply these tests to fn = g  gn , which does converge to 0. Lemma 7.20. If fn is a sequence of functions and Mn is a sequence of constants so that Mn converges to 0 and fn (z) Mn for all z in the set G fn converges uniformly to 0 on G. For example, z n  rn if z is in the closed disk Dr (0), and rn 0 if r < 1, so z n 0 uniformly in Dr (0) if r < 1. Lemma 7.21. If fn is a sequence of functions which converges uniformly to 0 on a set G and zn is any sequence in G then the sequence fn (zn ) converges to 0. This is most often used to prove nonuniform convergence. For example, let fn (z) = z n and let G be the open unit disk D1 (0). Then z < 1 if z is in G, so zn 0, and so z n 0. However, let 1 zn = exp( n ). Then zn is in G but fn (zn ) = e1 so fn (zn ) does not converge to 0. Therefore z n does not converge uniformly to 0 on D1 (0). All of these notions for sequences of functions go verbatim for series of functions. Here we also have a notion of absolute convergence (which can be combined with uniform convergence). There is an important result about series of functions, often called the Weierstra M test. Proposition 7.22. Suppose (fk ) are continuous on the region G, fk (z) Mk for all z G, and k1 Mk converges. Then k1 fk converges absolutely and uniformly in G. Proof. For each fixed z we have k1 fk (z) k1 Mk < , so k1 fk (z) converges; call the limit f (z). This defines a function f on G. To see that fn converges uniformly to f , suppose > 0. Since k1 Mk converges there is N so that
k>n Mk = k=1 Mk  n k=1 Mk < for all n > N . Then, for any z in G, if n N then n f (z)  fk (z) = fn (z) fn (z) Mk < k=1 k>n k>n k>n and this satisfies the definition of uniform convergence. We end this section by noting that everything we've developed here could have been done in greater generality  for instance, for functions from Rn or Cn to Rm or Cm . CHAPTER 7. POWER SERIES 75 7.4 Region of Convergence For the remainder of this chapter (indeed, these lecture notes) we concentrate on some very special series of functions. Definition 7.23. A power series centered at z0 is a series of functions of the form ck (z  z0 )k .
k0 The fundamental example of a power series is the geometric series, for which all ck = 1. k Lemma 7.24. The geometric series k0 z converges absolutely for z < 1 to the function 1/(1  z). The convergence is uniform on any set of the form { z C : z r } for any r < 1. Proof. Fix an r < 1, and let D = { z C : z r }. We will use Proposition 7.22 with fk (z) = z k and Mk = . Hence the uniform convergence on D of the geometric series will follow if we can rk show that k0 rk converges. But this is straightforward: the partial sums of this series can be written as n 1  rn+1 rk = 1 + r + + rn1 + rn = , 1r
k=0 whose limit as n exists because r < 1. Hence, by Proposition 7.22, the geometric series converges absolutely and uniformly on any set of the form {z C : z r} with r < 1. Since r can be chosen arbitrarily close to 1, we have absolute convergence for z < 1. It remains to show that for those z the limit function is 1/(1  z), which follows by k0 z k = lim n n k=0 z k = lim 1  z n+1 1 = . n 1  z 1z By comparing a general power series to a geometric series we can give a complete description of its region of convergence. Theorem 7.25. Any power series k0 ck (z  z0 )k has a radius of convergence R. By this we mean that R is a nonnegative real number, or , satisfying the following. (a) If r < R then k0 ck (z  z0 )k converges absolutely and uniformly on the closed disk Dr (z0 ) of radius r centered at z0 . (b) If z  z0  > R then the sequence of terms ck (z  z0 )k is unbounded, so k0 ck (z  z0 )k does not converge. The open disk DR (z0 ) in which the power series converges absolutely is the region of convergence. (If R = then DR (z0 ) is the entire complex plane, and if R = 0 then DR (z0 ) is the empty set.) By way of Proposition 7.18, this theorem immediately implies the following. Corollary 7.26. Suppose the power series k0 ck (z  z0 )k has radius of convergence R. Then the series represents a function which is continuous on DR (z0 ). While we're at it, we might as well state what Proposition 7.19 implies for power series. CHAPTER 7. POWER SERIES 76 Corollary 7.27. Suppose the power series k0 ck (z  z0 )k has radius of convergence R and is a smooth curve in DR (z0 ). Then k ck (z  z0 ) dz = ck (z  z0 )k dz . k0 k0 In particular, if is closed then Proof of Theorem 7.25. Define C to be the set of positive real numbers for which the series k0 ck tk converges, and define D to be the set of positive real numbers for which it diverges. Clearly every positive real number is in either C or D, and these sets are disjoint. First we establish three facts about these sets. () If t C and r < t then r C, and k0 ck (z  z0 )k converges absolutely and uniformly on Dr (z0 ). To prove this, note that k0 ck tk converges so ck tk 0 as k . In particular, this sequence is bounded, so ck  tk M for some constant M . Now if z Dr (z0 ) we have k c  r k and ck (z  z0 ) k k0 k0 ck (z  z0 )k dz = 0. ck  rk = k0 ck  tk r k t k0 M r k t =M r k
k0 t = M < . 1  r/t At the last step we recognized the geometric series, which converges since 0 r < t, and so 0 r/t < 1. This shows that r C, and uniform and absolute convergence on Dr (z0 ) follows from the Weierstra M test. () If t D and r > t then r D, and k0 ck (z  z0 )k diverges on the complement of Dr (z0 )  that is, for z  z0  r. To prove this, assume that ck rk is bounded, so ck  rk M for some constant M . But now exactly the same argument as in (), but interchanging r and t, shows that k k0 ck t converges, contradicting the assumption that t is in D. ( ) There is an extended real number R, satisfying 0 R , so that 0 < r < R implies r C and R < r < implies r D. Notice that R = 0 works if C is empty, and R = works if D is empty; so we assume neither is empty and we start with a0 in C and b0 in D. It is immediate from () or () that a0 < b0 . We shall define sequences an in C and bn in D which "zero in" on R. First, let m0 be the midpoint of the segment [a0 , b0 ], so m0 = (a0 + b0 )/2. If m0 lies in C then we define a1 = m0 and b1 = b0 ; but if m0 lies in D then we define a1 = a0 and b1 = m0 . Note that, in either case, we have a0 a1 < b1 b0 , a1 is in C, and b1 is in D. Moreover, a1 and b1 are closer together than a0 and b0 ; in fact, b1  a1 = (b0  a0 )/2. We repeat this procedure to define a2 and b2 within the interval [a1 , b1 ], and so on. Summarizing, we have an an+1 an < bn bn bn+1 an C bn D bn  an = (b0  a0 )/2n The sequences an and bn are monotone and bounded (by a0 and b0 ) so they have limits, and these limits are the same since limn (bn  an ) = limn (b0  a0 )/2n = 0. We define R to be this limit. If 0 < r < R then r < an for all sufficiently large n, since an converges to R, so r is in C by (). CHAPTER 7. POWER SERIES 77 On the other hand, if R < r then bn < r for all sufficiently large n, so r is in D by (). Thus R verifies the statement ( ). To prove Theorem 7.25, first assume r < R and choose t so that r < t < R. Then t C by ( ), so part (a) of 7.25 follows from (). Similarly, if r = z  z0  > R then choose t so that R < t < r. Then t D by ( ), so part (b) of 7.25 follows from (). It is worth mentioning the following corollary, which reduces the calculation of the radius of convergence to examining the limiting behavior of the terms of the series. Corollary 7.28. ck  rk 0 for 0 r < R but ck  rk is unbounded for r > R. Warning: Neither Theorem 7.25 nor Corollary 7.28 says anything about convergence on the circle z  z0  = R . Example 7.29. Consider the function f (z) = exp(z). You may recall from calculus that the real k defined, realvalued function ex has an expansion as the power series k0 x . In fact, a similar k! k expression holds for the complexdefined, complexvalued f (z). Let g(z) = k0 x . Then k! g (z) = Thus, g(z) has the correct derivative. The question still remains whether f (z) = g(z) or not. To see that f (z) = g(z), first note that 1 1 = = exp(z) = f (z). f (z) exp(z) Thus, the function f (z)g(z) has 0 derivative: d [f (z)g(z)] = f (z)g(z) + f (z)g (z) = f (z)g(z) + f (z)g(z) = 0. dz
g(z) This means that f (z) = f (z)g(z) = c for some constant c C. Evaluating at z = 0, we see c = 1, so g(z) = f (z) as desired. We use the Ratio Test to determine the radius of convergence. Since k+1 x k! x x = (k + 1)! xk k + 1 = k + 1 0 d xk xk1 xk d zk = = = = g(z). dz k! dz k! (k  1)! k!
k0 k0 k1 k0 as k , the power series converges absolutely for all x. The radius of convergence is R = . The region of convergence is all of C, the "disk of radius infinity" about the origin (the center of the series). There are many operations we may perform on series. We may add constants and polynomials to power series. We may rearrange the terms of a series in the case that the series converges absolutely. That absolute convergence is both necessary and sufficient for rearrangement is left as an exercise. Thus, we may add two power series together on a common region of convergence and rearrange their sum to collect coefficients of the same degree together, as the next example demonstrates. We have seen that we may differentiate and integrate power series. We may also multiply series by constants, or multiply power series by polynomials. In fact, we may multiply power series together on their common region of convergence. We leave it as an exercise to determine a formula for multiplying power series together. CHAPTER 7. POWER SERIES 78 Example 7.30. We can use the power series expansion for exp(z) to find a power series expansion of the trigonometric functions. For instance, consider f (z) = sin(z). Then f (z) = sin z = = = = = = = 1 iz e  eiz 2i 1 (iz)k (iz)k  2i k! k! k0 k0 1 1 (iz)k  (1)k (iz)k 2i k!
k0 1 2i k0,k odd i2l+2 z 2l+1
l0 1 2i2l+1 z 2l+1 2i (2l + 1)!
l0 2ik z k k! (2l + 1)! z 2l+1 (1)l+1
l0 (2l + 1)! = z z3 z5 z7 +  + .... 3! 5! 7! Note that we are allowed to rearrange the terms of the two added sums because the corresponding series have infinite radius of convergence. Exercises
1. For each of the sequences, prove convergence/divergence. If the sequence converges, find the limit. (a) an = ein/4 . (b)
(1)n n .
2 (c) cos n.
in (d) 2  2n2 +1 . 1 (e) sin n . 2. Determine whether each of the following series converges or diverges. n 1+i (a) k1 3 n (b) k1 n i n 1+2i (c) k1 5 CHAPTER 7. POWER SERIES (d) 1 k1 n3 +in 79 3. Show that the limit of a convergent sequence is unique. 4. Derive the Archimedean Property from the monotone sequence property. 5. Prove: (a) lim an = a
n n = n (b) lim an = 0 6. Prove Lemma 7.8. lim an  = a. lim an  = 0. n 7. Prove: (cn ) converges if and only if (Re cn ) and (Im cn ) converge. 8. Prove Lemma 7.4. 9. Prove that Z is complete. 10. Use the fact that R is complete to prove that C is complete. 11. Suppose an bn cn for all n and limn an = L = limn cn . Prove that limn bn = L. State and prove a similar theorem for series. 12. Find sup Re e2it : t Q \ Z . 13. Suppose that the terms cn converge to zero, and show that cn converges if and only if n=0 (c2k + c2k+1 ) converges. Moreover, if the two series converge then they have the same k=0 limit. Also, give an example where cn does not converge to 0 and one series diverges while the other converges. 14. Prove that the series k1 bk converges if and only if limn bk = 0 . k=n 15. (a) Show that k1 21 = 1. One way to do this is to write 21 as a difference of powers of k k 2 so that you get a telescoping series. 1 (b) Show that k1 k2k diverges. (Hint: compare the general term to 2k .) +1 (c) Show that k1 k3k converges. (Hint: compare the general term to k12 .) +1 16. Discuss the convergence of k0 z k for z = 1. 17. Prove Lemma 7.20. 18. Prove Lemma 7.21. 19. Discuss pointwise and uniform convergence for the following sequences (a) (nz n ) . n (b) zn for n > 0. 1 (c) 1+nz , defined on {z C : Re z 0}. CHAPTER 7. POWER SERIES 20. Let fn (x) = n2 xenx . 80 (a) Show that limn fn (x) = 0 for all x 0. Treat x = 0 as a special case; for x > 0 you can use L'Hospital's rulebut remember that n is the variable, not x. 1 (b) Find limn 0 fn (x) dx. (Hint: the answer is not 0.) (c) Why doesn't your answer to part (b) violate Proposition 7.19? 21. Prove that absolute convergence is a sufficeint and necessary condition to be able to arbitrarily rearrange the terms of a series without changing the sum. 22. Derive a formula for the product of two power series. 23. Find a power series (and determine its radius of convergence) of the following functions. (a) (b) (c)
1 1+4z . 1 3 z . 2 z2 (4z)2 for z < 4 24. Find a power series representation about the origin of each of the following functions. (a) cos z (b) cos(z 2 ) (c) z 2 sin z (d) (sin z)2 25. (a) Suppose that the sequence ck is bounded and show that the radius of convergence of k k0 ck (z  z0 ) is at least 1. 26. Find the power series centered at 1 for the following functions, and compute their radius of convergence: (a)
1 z. (b) Suppose that the sequence ck does not converge to 0 and show that the radius of con vergence of k0 ck (z  z0 )k is at most 1. (b) Log z. 27. Use the Weierstra M test to show that each of the following series converges uniformly on the given domain: (a) zk
k1 k0 k2 on D1 (0). 1 (b) on {z : z 2}. zk (c) k0 zk on Dr (0), where 0 r < 1. zk + 1 CHAPTER 7. POWER SERIES
1 28. Suppose L = limk ck 1/k exists. Show that L is the radius of convergence of (Use the natural interpretations if L = 0 or L = .) 81 k0 ck (z  z0 )k . 29. Find the radius of convergence for each of the following series. 2 (a) ak z k , a C. (b) k0 k1 k0 k0 k n z k , n Z. z k! . (c) (d) (e) (f) (g) (1)k k . zk k0 k0 k1 z k(k+1) . kk cos(k)z k . 4k (z  2)k . 30. Find a function in "closed form" (i.e. not a power series) representing each of the following series. (a) (b) (c) z 2k k2 k1 k0 k! k(z  1)k1 k(k  1)z k 1 31. Define the functions fn (t) = n et/n for n > 0 and 0 t < . (a) Show that the maximum of fn (t) is 1 n. (b) Show that fn (t) converges uniformly to 0 as n . (c) Show that 0 fn (t) dt does not converge to 0 as n (d) Why doesn't this contradict the theorem that "the integral of a uniform limit is the limit of the integrals"? ...
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