COMPLEX ANALYSIS–Spring 2014 Homework 5. NOTE: Exercises 7 and 8 won’t be graded. They are there for your amusement only. If you are not amused, ignore them. 1. Let {pn } be the sequence of primes (p1 = 2, p2 = 3, p3 = 5, . . .) Prove: ∞ Y

∞ X 1 1 = 1 z n 1 − z p n=1 n=1

for all z ∈ C with Rez > 1. 2. Textbook, Chapter 5, # 6 (p. 154). Prove Wallis’ product formula 2·2 2m · 2m π = · 4 · 43 · 5 · · · ··· . 2 1·3 (2m − 1) · (2m + 1) Solution. The formula follows at once putting z = 1/2 into the product formula for sin πz, namely  ∞  Y z2 sin πz = πz 1− 2 . n n=1 3. Textbook, Chapter 5, # 7 (p. 154-55). Establish the following properties of infinite products. P∞ 2 (a) Show that for all n, then the n=1 |an | < ∞ and an 6== −1 P Q∞if ∞ product n=1 (1 + an ) converges∗ if and only if n=1 an converges. (b) P Find an example of Q a sequence of complex numbers {an } such that an converges but (1 + an ) diverges. Q P (c) Also find an example such that (1 + an ) converges but (1 + an ) diverges. Solution. P∞ 2 i. Assume n=1 |an | < ∞ and an 6= 0 for all n ∈ N . Then limn→∞ an = 0 and there is N0 Q ∈ N such that |an | < 1/2 for all n≥ N . Let us notice first that 0 n (1+an ) converges if and only P∞ if n=N0 log(1 + an ) converges, where log(1 + z) for |z| < 1 is the analytic determination of log(1+z) given by the power series log(1+z) = z −z 2 /2+z 3 /3∓· · · . The proof that the convergence ∗ I use convergence for an infinite product in the standard sense; the limit of the partial products is a non-zero complex number. If the patial products converge to 0, the product is said to diverge.

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of the log series implies that of the product is essentially done in the textbook on page 141. Here it is done again. We have NY 0 +N

(1 + an ) = e

PN0 +m n=N0

log(1+an )

n=N0

and if B = limm→∞ α=

∞ Y

PN0 +m n=N0

log(1 + an ) exists, then

(1 + an ) = lim

NY 0 +m

m→∞

n=N0

(1 + an ) = eB 6= 0.

n=N0

Conversely, assume that ∞ Y

α=

NY 0 +m

(1 + an ) = lim

m→∞

n=N0

(1 + an )

n=N0

exists and α 6= 0. Let Log be an analytic determination of the logarithm valid in some neighborhood U of α. Then NY 0 +m

(1 + an ) = e

PN0 +m n=N0

log(1+an )

∈U

n=N0

for m large enough and we have ! N +m NY 0 +m 0 X Log (1 + an ) = log(1 + an ) + 2πikm , n=N0

n=N0

for some km ∈ Z for all m large enough. The function Log is analytic, thus for m large enough ! ! NY N0 +m+1 0 +m Y (1 + an ) − Log (1 + an ) < π. Log n=N0

n=N0

This means that for m large enough we will have | log(1 + aN0 +m+1 ) + 2πi(km+1 − km )| < π. However an → 0 and since log(1 + an ) is the principal determination of log, we have log(1 + an ) → 0, so that if km+1 6= km we have | log(1+aN0 +m+1 )+2πi(km+1 −km )| ≥ 2π|km+1 −km |−| log(1+aN0 +m+1 )| > π for m large enough, contradicting the previous inequality. It follows that km = k is constant for m large enough and we see that ∞ X log(1 + an ) = Log(α) + 2πik. n=N0

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Having established this, we notice next that f (z) = [log(1 + z) − z]/z 2 is analytic for |z| < 1 thus there is C ≥ 0 such that |f (z)| ≤ C for |z| ≤ 1/2; i.e., | log(1 + z) − z| ≤ c|z|2 for |z| ≤ 1/2. It follows that if N0 ≤ n ≤ m, then m m X X X log(1 + ak ) − mak ≤ C |ak |2 . k=n

k=n

k=n

P∞ 2 Since P n=1 |an | < ∞, it is clear that the sequence of partial sums of n=N0 ∞ log(1+an )P is a Cauchy sequence if and only the sequence of partial sums of n=N0 ∞an is a Cauchy sequence. Putting it all together, we are done. P Q ii. We need to find {an } such an converges while (1 + an ) P that 2 diverges. By part a, |an |√ must diverge so that an obvious n−1 first choice is an = (−1) / n for n ∈ N. With this choice the product diverges to 0. But there is an even more obvious choice, namely √  1/ √ k + 1, if n = 2k − 1, k ∈ N, an = −1/ k + 1, if n = 2k, k ∈ N, Then, if n = 2k,         1 1 1 1 1 1 √ √ √ √ √ √ Pn = 1+ 1− 1− 1+ 1− ··· 1 + k+1 k+1 2 2 3 3 k+1 Y j−1 1 = = , j k + 1 j=2 thus limk→∞ P2k = 0 proving (since the partial products form a decreasing sequence) that the product diverges to 0. iii. For a sequence P∞ {an } to have the required properties, it is necessary that n=1 |an |2 be divergent. One can try a barely converging product, one in which 1 + an is above 1, followed by 1 + an+1 below 1, or vice-versa. The simplest example of a sequence with P |a |2 < ∞ and an → 0 (for convergence of a product n n √ one must have the factors converging to 1!) is, perhaps, {1/ n}. So consider the sequence {an } such that the corresponding sequence {1 + an } is  −1  −1  −1 1 1 1 1 1 1 1+ √ , 1+ √ , 1+ √ , 1+ √ , 1+ √ ,...; 1+ √ , 2 2 3 3 4 4 that is, the sequence 1 √ , 2

−

1 √ , 1+ 2

1 √ , 3 3

−

1 √ , 1+ 3

1 √ , 4

−

1 √ , 1+ 4

It is clear that this sequence does the trick. The partial products are all either 1 or 1 + √1n , depending on whether we stop at an P even or odd index, so the product converges to 1. The series n an provides a nice and not totally obvious example for calculus 2 teachers of an alternating series that diverges because at first glance it would seem that the sequence of absolute values of the terms decreases. But if we add two consecutive terms we have 1 1 1 1 √ − √ =√ √ ∼ , n n 1+ n n(1 + n) and divergence follows. 4. Textbook, Chapter 5, # 8 (p. 155). Prove that for every z ∈ C, z 6= ±π, ±2π, ±3π, . . ., the product below converges, and cos(z/2) cos(z/4) cos(z/8) · · · =

∞ Y

cos(z/2 ∗ k) =

k=1

sin z . z

Comments: I added z 6= ±π, ±2π, ±3π, . . . because if (and only if) z is a non-zero multiple of π there is a factor on the left that is 0 and technically the product diverges (to 0). The equality is, however, trivially true in this case. Once one is past the zero factor, if there is one, the product converges uniformly on compact sets because |cos(z/2k ) − 1| ≤ c|z|2 /4k . For k ∈ N let

Solution.

fk (z) =

k Y

cos(z/2j ),

j=1

gk (z) =

k Y

sin(z/2j ).

j=1

Using the hint, fk (z)gk (z) = 2−k

k Y

sin(z/2j−1 ) = 2−k

j=1

gk (z) sin z . sin(z/2k )

One of the beauties of working with analytic functions is that you can cancel not identically zero functions and equalities holding for all z before continue holding for all z, including those z at which the canceled function is 0. Thus sin z sin z z/2k fk (z) == 2−k = · . sin(z/2k ) z sin(z/2k ) z/2k = 1, we get k→∞ sin(z/2k )

Since lim

∞ Y k=1

cos(z/2k ) = lim

k→∞

k Y

cos(z/2j ) = lim fk (z) = k→∞

j=1

4

sin z . z

5. Textbook, Chapter 5, # 12 (p. 155). Suppose f is entire and never vanishes, and that none of the higher derivatives of f ever vanish. Prove that if f is also of finite order, then f (z) = eaz+b for some constants a, b. Solution. I find the meaning of “higher derivatives,” somewhat unclear. If we understand it to mean simply “derivatives,” then the exercise is almost trivial. So I will interpret it as meaning that there is M ∈ N such that f (m)(z) 6= 0 for all z ∈ C if m ≥ M . In fact, it will suffice to assume f (m) (z) 6= 0 for all z ∈ C for some m ∈ N. By Hadamard’s factorization theorem, because f never vanishes, f (z) = eP (z) for some polynomial P of degree k, where k ≤ ρ < k + 1, and ρ is the order of f . Claim: f m (z) = (Qm (z) + (P 0 (z))m ) eP( z) , where Qm is a polynomial of degree < m(q − 1). The claim is easily verified by induction, so I assume it established. Since P 0 is a polynomial of degree k − 1, it follows that Qm (z) + (P 0 (z))m is a polynomial of degree m(k − 1) and the only way it will never be 0 (assuming it is not the zero polynomial) is if m(k − 1) = 0; i.e., P is of degree 1. 6. Textbook, Chapter 5, # 16 (p. 156). Suppose that Qn (z) =

Nn X

cnk z k

k=1

are given polynomials for n = 1, 2, . . .. Suppose also that we are given a sequence of complex numbers {an } without limit points. Prove that there exists a meromorphic function f (z) whose only poles are at {an }, and so that for each n the difference   1 f 9z) − Qn z − an is holomorphic near an . In other words, f has prescribed poles and a prescribed principal part at each of these poles. This result is due to Mittag-Leffler. Solution. This was a surprisingly hard exercise, in comparison with other exercises in the text, I will start with a simple estimate. Let 0 < |δ| < 1. Then 1 − (1 − δ)n ≤ 2n |δ| for n = 1, 2, . . .. The proof is, of course, trivial, 1 − (1 − δ)n = |δ| [1 + (1 − δ) + · · · (1 − d)n−1 ] ≤ 2n |δ| since |1 − δ| ≤ 2 and 1 + 2 + · · · 2n−1 ≤ 2n .

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Now let Qn be one of the polynomials of the statement; assume |an | ≥ 1 (I’m not sure this is needed, but it does no harm). Let Nn be the degree of Qn . Let Ek be the canonical factor defined on page 145 of the textbook. Set  
1 1 − Ekn (z/an )Nn Fn (z) = Qn z − an where kn ∈ N is to be determined. A thing to notice is Fn is meromorphic with a single pole at an and its principal part is Qn (1/(z − an )). In fact, that it is meromorphic with a single pole at an is clear. For the principal part statement notice that no matter what kn ∈ N is, Ekn (z/an )Nn has a zero of order Nn at an so that we can write it in the form (z − an )Nn hn (z) where hn is holomorphic, and  
cnNn cn1 n + ··· + + c0 Fn (z) = 1 − (z − an )Nn hn (z) , (z − an )Nn z − an from which (I hope) the result follows at once. Assume now |z| < |an |/2. By Lemma 4.2 of the textbook, kn +1 z 2e ≤ kn +1 . |1 − Ekn (z/an )| ≤ 2e an 2 Notice that 2e/2kn +1 ≤ 1 if kn ≥ 2. Applying now our first estimate with δ = δn = 1 − Ekn (z/an ) and n replaced by Nn , we get 1 − EkNnn = 1 − (1 − δn )Nn ≤ 2Nn |δn | ≤ 2Nn −kn e. Letting Mn = max|z| 0, 0 ≤ y ≤ 1. Prove: Z y ∞ X 1 y n+α−1 ds = , sα−2 log (1) 1−s n(n + α − 1) 0 n=1 and (can be otained at once from (1) Z 1 ∞ X (1 − y)n+α−1 1 (2) . (1 − s)α−2 log ds = s n(n + α − 1) y n=1 (b) Specialize to the case α = 1 to get Z y 1 1 (3) log ds = 1−s 0 s Z 1 1 1 (4) log ds = 1 − s s y

∞ X yn , n2 n=1 ∞ X (1 − y)n . n2 n=1

(c) Show that  Z 1 Z 1 1 1 1 1 1 1 log ds = − log log + log ds 1 − s s 1 − y y 1 − s s y y and use this in (4) to get  Z 1 ∞ 1 1 X (1 − y)n 1 1 (5) ds = log log + log . 1−s s 1−y y n=1 n2 y (d) From (3) and in (5) obtain that Z 1 ∞ 1 1 1 X y n + (1 − y)n 1 log ds = log log + (6) . 1−s 1−y y n=1 n2 0 s (e) Set α = 1, y = 1 in (1), comppare to (6) and conclude that (7)

∞ ∞ X 1 1 1 X y n + (1 − y)n = log log + n2 1−y y n=1 n2 n=1

for all y ∈ [0, 1]. 1 1 (Since limy→0+ log 1−y log y1 = limy→1− log 1−y log y1 = 0, the formula makes sense if y = 0 or 1. Of course, it is trivial in these cases.) 7

(f) Prove: (8)

∞ ∞ X X 1 21−n 2 = (log 2) + . n2 n2 n=1 n=1

(g) Euler calculated log 2 with good precision using log 2 =

∞ X 1 , n n2 n=1

P∞ and used (log 2)2 = 0.480453. To calculate n=1 1/n2 to a precision of 6 exact decimals requires adding a million terms; impossible in Euler’s day and age. But getting that precision from the series on the right in (8) only requires adding 17 terms; quite an improvement. So Euler gets, up to 6 correct decimals (9)

∞ X 1 = 1.644934 2 n n=1

For the last part of this exercise, stare at 1.644934 and, as Euler did, say: “Obviously, π 2 over 6.”

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8.

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