5 Infinite Products

5 Inﬁnite products

5.1 Interpolation

A common and classical problem in calculus is to find a function that takes specified values at certain specified points. For example, it is easy to find a polynomial p for which you have specified N values

p(zj)= cj, j =1,...,N.

Indeed you can choose the Lagrange interpolating polynomial,

N (z zi) p(z)= c i6=j − . j (z z ) j=1 Qi6=j j i X − Q Written another way, this expresses p(z) as cjℓj(z) where ℓj is a polynomial for which

ℓj(zi)= δij. P If you wish to find an analytic function that satisfies infinitely many conditions, things are much more complicated. We have already seen, for example, that it is impossible to find an entire function f such that

0, if j is even f(1/j)= (1/j, if j is odd. (It is of course easy to construct a continuous function with these constraints!) A natural question to ask is whether one can ever construct anything like the Lagrange interpolating polynomial with infinitely many terms. This would require us to write some form of infinite product of terms (z z ). This obviously requires some care as such an − j infinite product can easily be zero or unbounded!

One of the amazing truths about analytic functions is that essentially all entire functions can be written as inﬁnite products. This was discovered but not proved, as was so much else, by Euler in about 1750. We shall begin this section by studying one example:

sin πz ∞ z2 = 1 . πz − n2 n=1 Y This example introduces most of the key ideas. We shall then look more carefully at the theory of inﬁnite products of numbers, such as

2 ∞ 1 = 1 π − 4n2 n=1 Y 24 and inﬁnite products of functions. Later we shall apply the ideas to two further examples, the gamma function and the zeta function.

5.2 A nontrivial example

The basis of our example is the following partial fraction series representation of cot πz.

1 ∞ 2z π cot πz = + (for z =0, 1, 2,...). z z2 n2 6 ± ± n=1 X − To prove this we consider the contour CN which is the boundary of the rectangle bounded by the lines y = N, y = N, x = N + 1 and x = N + 1 . − 2 − 2

x = N + 1 x = N + 1 − 2 y = N 2 1 0 1 N N +1 −bb b b b x

y = N −

We ﬁrst show that on C there is a bound for cot πz which is independent of N, that N | | is, there is a B such that, for all N =1, 2, 3,... cot πz B whenever z C . Note that | | ≤ ∈ N

cos πz = cos πx cosh πy i sin πx sinh πy − sin πz = sin πx cosh πy + i cos πx sinh πy.

25 whilst on y = N, ± cos πz cos2 πx cosh2 πN + sin2 πx sinh2 πN | | = sin πz 2 2 2 2 | | psin πx cosh πN + cos πx sinh πN pcos2 πx + sinh2 πN = psin2 πx + sinh2 πN p1 + sinh2 πN 1 1+ ≤ s sinh2 πN ≤ sinh πN 1 1+ < 2. ≤ πN So in fact we can take B = 2. Suppose now that z / Z and make sure that N > z . We will now evaluate ∈ | | 1 π cot πw I = dw. N,z 2πi w2 z2 ZCN − Note that from the above bounds, and using the fact that if w C , then w2 N 2 and ∈ N | | ≥ hence w2 z2 N 2 z 2, | − | ≥ −| | 1 π cot πw 1 1 I | | dw dw = 2(4N + 1), | N,z| ≤ 2π w2 z2 | | ≤ N 2 z 2 | | N 2 z 2 · ZCN | − | ZCN −| | −| | which tends to 0 as N . →∞ On the other hand, for any N, the integral is just the sum of the residues at the singularities inside C . The singularities occur where sin πw = 0 and where w2 z2 = 0, that is, N − at integers w = N, (N 1),..., (N 1), N and w = z. π−cot−πw − − ± Let f(w) = . Then each integer is a simple zero of sin πw and thus is a simple w2 z2 pole of f and the Laurent− expansion is of the form c f(w)= −1 + c + c (w n)+ c (w n)2 + w n 0 1 − 2 − ··· − Calculating the residue is easy at such a point: π(w n) cos πw 1 Res(f, n) = lim (w n)f(w) = lim − = . w→n − w→n (w2 z2) sin πw n2 z2 − − A similar argument shows that the residues at z and z are − π cot πz π cot( πz) and − 2z 2z − Thus the Residue Theorem gives

N π cot πz π cot( πz) N 1 I = Res(f, z) + Res(f, z)+ Res(f, n)= + − + N,z − 2z 2z n2 z2 n=−N n=−N X − X − 26 Thus, with z ﬁxed, and taking a limit as N , →∞ π cot πz π cot( πz) ∞ 1 lim IN,z =0= + − + N→∞ 2z 2z n2 z2 n=−∞ − X − or, using symmetry and pulling out the n = 0 term,

π cot πz 1 ∞ 2 0= + . z − z2 n2 z2 n=1 X − Therefore, multiplying through by z,

1 ∞ 2z π cot πz = − z z2 n2 n=1 X − as claimed. Clearly each term on the right hand side has an antiderivative related to log(z2 n2) and − the left hand side has an antiderivative related to log(sin πz) log(πz). If no-one is looking − you might write that

sin πz ∞ ∞ log(sin πz) log(πz) = log = log(z2 n2) = log (z2 n2) − πz − − n=1 n=1 X Y or sin πz ∞ = (z2 n2) πz − n=1 Y . . . before you realize that you have no idea what any of this means! Our task then is to turn this into something that does make sense to make this rigorous. sin πz Let us begin by considering ℓsin(z) = Log on the punctured disk, 0 < z < 1. πz | | Lemma 16. ℓsin(z) is analytic on the punctured disk, 0 < z < 1, where, as usual, Log is | | the principal branch of log.

Proof. Log f(z) is analytic in any region where the analytic function f(z) is neither zero nor takes a negative real value, since this is the cut for the function Log. First note that sin πz/z can only vanish where sin πz = 0, and none of the zeros is inside the punctured disk. Next suppose that the function takes a negative real value, k, or sin πz = kπz − − where k > 0. Equating real parts gives sin πx cosh πy = kπx or simply sin πx = k′πx − −

27 where k′ > 0.

f(x) = sin πx

1 2 x g(x)= k′πx, k′ > 0 −

From the graphs this can only occur if x = 0 or x > 1. So we can assume that x = 0 | | and equate the imaginary parts to obtain sinh πy = kπy. Since k > 0 the only solution is − y = 0 and therefore the function ℓsin is analytic on the punctured disk.

In fact, the function sin πz/πz clearly has a removable singularity at z = 0 and so we can make ℓsin analytic at z = 0, provided we give sin πz/πz the value 1 there. Consequently ℓsin is analytic on the disk z < 1. Moreover, for 0 < z < 1 we have | | | | d d sin πz d 1 ℓsin z = Log = (Log sin πz Log z Log π)= π cot πz . dz dz πz dz − − − z

From this we deduce that the deriviative of ℓsin at z = 0 must equal lim (π cot πz 1 )=0. z→0 − z Note also that 2 −2z d z n2 2z Log 1 2 = z2 = 2 2 . dz − n 1 2 z n − n − Thus, for z < 1, | | d ∞ 2z ℓsin z = dz z2 n2 n=1 X − ∞ d z2 = Log 1 dz − n2 n=1 X d ∞ z2 = Log 1 . dz − n2 n=1 X

28 Here we could interchange d/dz and since the series of Logs converges uniformly (on any compact set not containing an integer.)P One easy way to prove this is to observe that Log(1 w) 2 w when w 1/2. Therefore | − | ≤ | | | | ≤ z2 2 z 2 2 Log 1 | | for z < 1 and n 2. − n2 ≤ n2 ≤ n2 | | ≥

∞ 2 But 2 < and so we have the result by the Weierstrass M-test (we can clearly forget n=2 n ∞ a ﬁniteP number of terms in the series if we want to). Therefore, for z < 1, | | sin πz ∞ z2 Log = Log 1 + C πz − n2 n=1 X for some constant C. Substituting z = 0 shows that C = 0. That is,

sin πz ∞ z2 N z2 Log = Log 1 = lim Log 1 . πz − n2 N→∞ − n2 n=1 n=1 X X At this point we would like to take the sum inside the logariPi*xthm, but we need to remember that Log a + Log b is not necessarily Log ab. Rather

Log a + Log b = ln a + i Arg a + ln b + i Arg b = ln ab + i(Arg a + Arg b) | | | |

If Arg a + Arg b ( π, π) then everything does work OK. ∈ − 2 z2 −1 |z| Exercise: (a) Show that Arg 1 2 sin 2 . − n ≤ n (b) Use the fact that t 1.1 sin t for t [0, 0.25] to show that for z < 1 and n 2, ≤ ∈ | | ≥ z 2 1.1 sin−1 | | < . n2 n2 (c) Deduce that for any N,

N z2 π N 1 π π2 Arg 1 < +1.1 +1.1 1 < 2.3 < π. − n2 2 n2 ≤ 2 6 − n=1 n=2 X X Pi*x In light of this then,

sin πz ∞ z2 N z2 Log = Log 1 = lim Log 1 πz − n2 N→∞ − n2 n=1 n=1 X X N z2 = lim Log 1 N→∞ − n2 n=1 ! Y

29 Now recalling that exp(Log w)= w for all w = 0, and, of course, that exp is continuous, 6 sin πz sin πz = exp Log πz πz N z2 = exp lim Log 1 N→∞ − n2 n=1 !! Y N z2 = lim exp Log 1 N→∞ − n2 n=1 !! Y N z2 = lim 1 . N→∞ − n2 n=1 Y We shall write this as sin πz ∞ z2 = 1 πz − n2 n=1 Y for z < 1. We will soon see that this equation holds for all z. | | If we substitute z =1/2 we obtain

1 ∞ 1 ∞ (2n 1)(2n + 1) = 1 = − π − 4n2 (2n)2 2 n=1 n=1 Y Y so that π ∞ (2n)2 π 22 42 62 = or = 2 (2n 1)(2n + 1) 2 1 3 3 5 5 7 ··· n=1 Y − · · · which is traditionally and rather unsatisfactorily written as Wallis’ product

π 22 42 62 = · · ··· . 2 12 32 52 72 · · · ··· Our immediate aim is to provide a general theoretical framework for what we have done, to regularize the example and to allow us to treat other examples.

5.3 Products of constants

∞ Definition: If pn n=1 is a sequence of non-zero complex numbers we say that the infinite ∞ { } product p converges to P if the sequence of partial products P = p p p converges n N 1 2 ··· N n=1 to a non-zeroY limit P . If the infinite products converge to zero or infinity then the product is said to diverge. Infinite products which do not converge are said to diverge.

Remark: Intuitively, for the inﬁnite product to converge we’d expect to need that Log pn converges to 0, which means that p 1. As we’ll see shortly, this turns out to be correct, n →

30 so it is common to consider infinite products of the form ∞ (1 + α ) where α = 1 for n=1 n n 6 − all but a finite number of n. Clearly we then must have αQn 0 for the product to exist. → ∞ Definition: More generally we agree to say that an infinite product pn exists if n=1 Y 1. at most a finite number of factors are zero; and

2. the product of the non-vanishing terms exists in the above sense.

Hence a (convergent) infinite product has the value 0 if and only if one or more of its factors is 0. Example: (a) Consider ∞ 1 1 2 3 1 =0 − n · 2 · 3 · 4 ··· n=1 Y so that the infinite product exists if and only if lim 1 2 n−1 exists and is non-zero. n→∞ 2 3 ··· n 1 ∞ 1 But the value is = limn→∞ = 0. Therefore the infinite product (1 ) does not exist. n 1 − n It diverges to 0. (Ignoring the initial 0, this corresponds to the factQ that

N n log = log n +1 n +1 − n=1 X diverges to .) −∞ Example: (b) Consider

∞ ( 1)n 1 1 1 1 4 3 6 1+ − = 1 1+ 1 = . n +1 − 2 3 − 4 ··· 2 · 3 · 4 · 5 ··· n=1 Y N+2 Here PN =1/2 if N is odd and = 2(N+1) if N is even. Clearly then limN→∞ PN =1/2 so that the infinite product exists and equals 1/2. Note that if 0 were prefixed as the first factor in example (b) then the product would still exist and would equal 0. We say that it converges to 0. ∞ ∞ Example: (c) It is tempting to think that if P = n=1 pn exists then Log P = n=1 Log pn. This need not be the case however. Q P Let p = exp( iπ/2n), so that the partial products are given by n − N P = p = exp( iπ(1 1/2N )). N n − − n=1 Y ∞ In this case P P = 1, so we have Log p = πi while Log P = πi. N → − n − n=1 X 31 Lemma 17. If P = ∞ p exists then p 1. n=1 n n → Q Proof. If P exists, but is zero, then we need only look at those terms past the last pn that is zero, so without loss of generality, let’s assume that P = 0. 6 With the notation as above then,

Pn P lim pn = lim = =1. n→∞ n→∞ Pn−1 P

∞ ∞ Theorem 18. P = n=1 pn exists and is nonzero if and only if n=1 Log pn converges. Q P Proof. As we saw in the concrete example we did earlier, the main issue here is that we might not have Log pn = Log pn. Recall however that for any N

Q N P N Arg p = Arg p +2k π, some k Z, n n N N ∈ n=1 ! n=1 Y X and so N N

Log pn = Log pn +2kN πi. n=1 n=1 Y X Thus

N N N N

PN = pn = exp Log pn = exp Log pn +2kN πi = exp Log pn . n=1 n=1 ! n=1 ! n=1 ! Y Y X X One direction is now easy! If ∞ Log p converges then P exp ( ∞ Log p ) as n=1 n N → n=1 n N (since exp is continuous)P and so PN is convergent with a non-zeroP limit. →∞ 1 For the converse , assume that P = lim PN exists and is nonzero. Let us ﬁrst assume that

P ( , 0). Then Log is continuous at P and hence Log P = Log lim PN = lim Log PN 6∈ −∞ N N exists. As above, for all N,

cN := Log PN = Log pn +2kN πi n=1 X for some integer k . Since the sequence c converges it is Cauchy, and in particular N { N }

c c = Log p +2πi(k k ) 0, as N . N − N−1 N N − N−1 → →∞ 1Thanks to Joel for googling the ﬁx to this proof!

32 But we know that p 1, and hence that Log p 0. This implies that 2πi(k k ) 0 N → N → N − N−1 → too, which can obviously only happen if the sequence k is eventually constant. Thus, { N } there exists an integer k such that for all suﬃciently large N,

N Log p = Log P 2kπi. n N − n=1 X Since the right-hand side converges (to Log P 2kπi), the left-hand side converges too. − The remaining situation is if P ( , 0). If every p were real and positive, then P ∈ −∞ n would be too, so there must exist at least one term, say pm which is not real and positive. ′ In this case P = pn converges to a nonzero point which is not on the negative real axis. n6=m Y ∞

The previous case would then imply that Log pn converges, and hence that Log pn n6=m n=1 converges too. X X ∞

Remark: It is worth noting that we actually showed that if pn is non-zero then the sum n=1 ∞ Y n=1 Log pn can only diﬀer from Log P by an integer multiple of 2πi. P The following result provides us with a very simple test.

Theorem 19. ∞ Log p converges absolutely if and only if ∞ (1 p ) converges ab- n=1 n n=1 − n ∞ solutely. ThereforeP if (1 pn) converges absolutely then Ppn converges. − n=1 P Q Proof. Let us write p =1+ α . Then, for α < 1, the Taylor series for Log gives n n | n| α2 α3 α4 Log(1 + α ) α = n + n n + n − n − 2 3 − 4 ··· α 2 α 3 α 4 | n| + | n| + | n| + ≤ 2 3 4 ··· α | n| α + α 2 + α 3 + . ≤ 2 | n| | n| | n| ··· If we further assume α 1/2 then the geometric series in the last line has sum at most | n| ≤ 1/2 and so, for α 1/2, Log(1 + α ) α 1 α or, equivalently, | n| ≤ | n − n| ≤ 2 | n| 1 3 1 p Log p 1 p . 2| − n| ≤ n ≤ 2| − n|

The result now follows by the comparison test since clearly for n large enough, we can assume that α 1/2. | n| ≤

33 ∞ ∞ Corollary 20. If α < then (1 + α ) converges. | n| ∞ n n=1 n=1 X Y ∞ ∞ Exercise: Show that α < if and only if (1 + α ) converges. | n| ∞ | n| n=1 n=1 X Y ∞ If either of these above conditions hold then we say that 1 (1+αn) converges absolutely. ∞ ∞ It therefore follows that if 1 (1 + αn) converges absolutelyQ then 1 (1 + αn) converges. This is, of course, because theQ similar result for series is true. Q ∞ To complete the discussion we shall see, by two examples, that the convergence of 1 αn ∞ is neither necessary nor suﬃcient for the convergence of the product 1 (1 + αk). P −1/2 −1/2 −1 ∞ Example: (a) Let α2n−1 = n and α2n = n + n . ThenQ αn is divergent − n=1 (being the harmonic series). Note that P 1 1 1 1 1+ 1 + =1+ √n − √n n n3/2 ∞ ∞ −3/2 so 1 (1 + αn) = 1 (1 + n ) is convergent. 1 n ∞ Example:Q (b) On theQ other hand if αn =( 1) /n 2 then αn is clearly convergent, whilst − 2 ∞ 2 (1 + αn) is divergent. P Exercise:Q Prove this last statement! Hint: Prove and use the fact that 1 1 1 (1 + α2n−1)(1 + α2n)= 1 1+ < 1 . − √2n 1 √2n − 2n − 5.4 Products of functions

Next suppose we are given a sequence of functions α : Ω C ∞ defined on some { n → }n=1 region Ω. As for infinite sums, for each fixed z Ω we can consider the convergence of the ∞ ∈ product 1+ α (z) . If, for each z Ω, this infinite product exists then we can define n ∈ n=1 the pointwiseY limit function ∞ P (z)= 1+ α (z) , z Ω. n ∈ n=1 Y The harder questions concern the properties of the limit function P . Suppose for example that each function αn is analytic on Ω. In this case each partial product

PN (z)= 1+ αn(z) n=1 Y is clearly also analytic, but what about P ? How can we guarantee that the inﬁnite product is analytic?

34 Theorem 21. Let α ∞ be a sequence of analytic functions deﬁned on Ω such that α { n}n=1 | n| converges uniformly on each compact subset of Ω. Then P ∞

P (z)= (1 + αn(z)) n=1 Y converges uniformly on each compact subset of Ω and P is analytic on Ω.

Proof. First observe that if α converges uniformly on each compact subset of Ω then | n| certainly for any z Ω, Pαn(z) converges and hence by the previous results, P (z) = ∈ | | ∞ n=1(1 + αn(z)) exists. P Q As before, let N P (z)= (1 + α (z)), z Ω N n ∈ n=1 Y so that P (z) P (z)=(1+ α (z))P (z) P (z)= α (z)P (z). N+1 − N N+1 N − N N+1 N For M > N then, using the fact that for t 0, 1+ t et, ≥ ≤ M−1 P (z) P (z) = α (z) P (z) M − N j+1 j j=N X M−1 j α (z) (1 + α (z)) ≤ | j+1 | | n | n=1 jX=N Y M−1 j α (z) (1 + α (z) ) ≤ | j+1 | | n | j=N n=1 X Y M−1 j α (z) e|αn(z)| ≤ | j+1 | j=N n=1 X Y M−1 ∞ α (z) e|αn(z)| ≤ | j+1 | j=N n=1 X Y M−1 = eP |αj (z)| α (z) | j+1 | jX=N Let K be a compact subset of Ω and let ǫ > 0. Since ∞ α converges uniformly on 1 | j| K, it has a continuous limit. In particular, the limit is boundP ed and so there is a constant C > 0 such that exp ( ∞ α (z) ) C for all z K. Moreover there is a positive integer 1 | j | ≤ ∈ ∞ N0 such that αj(zP) < ǫ/C whenever N > N0 and z K. Therefore N | | ∈ P P (z) P (z) < ǫ | M − N | 35 whenever M,N >N and z K. Thus P forms a Cauchy sequence in C(K) with the 0 ∈ { N } supremum norm, and hence it converges uniformly on K. The analyticity of P on Ω now follows from Theorem 13.

Corollary 22. (The Logarithmic Derivative of a Product) Let α ∞ be a sequence of { n}n=1 analytic functions deﬁned on Ω such that α converges uniformly on each compact subset | n| ∞ of Ω. Then, at every point z where P (z)=P n=1(1 + α(z)) is nonzero, P ′(z) Q∞ α′ (z) = n . P (z) 1+ α (z) n=1 n X Proof. Since P (z) = 0, none of the factors 1+ α (z) vanishes. 6 n The product rule lets us diﬀerentiate the partial product PN (z):

N N ′ ′ PN (z)= αn(z) (1 + αk(z)). n=1 X kY6=n Thus, ′ N N N ′ P (z) (1 + αk(z)) α (z) N = α′ (z) k6=n = n P (z) n N 1+ α (z) N n=1 Qk=1(1 + αk(z)) n=1 n X X since lots of terms cancel. Q Now P P uniformly on each compact set and hence (as the functions are analytic, n → see Theorem 12) P ′ P ′ uniformly on each compact set. n → Thus, if P (z) = 0, then (noting perhaps that no P (z) is zero either), 6 n ′ ′ Pn(z) P (z) Pn(z) → P (z) as n . That is →∞ P ′(z) ∞ α′ (z) = n . P (z) 1+ α (z) n=1 n X

Exercise: Investigate the convergence of the series for P ′/P given in the corollary. Recall that we showed that for z < 1 | | sin πz ∞ z2 = 1 πz − n2 n=1 Y C sin πz We want to extend this identity to all of . Now the function πz is analytic everywhere. To z2 see that the product is also analytic everywhere we consider α where α (z)= − . We | n| n n2 P 36 need to show that this converges uniformly on any compact subset of C, or equivalently, that

R2 it converges uniformly on any closed disk. Suppose then that z R. Then α (z) 2 | | ≤ | n | ≤ n ∞ R2 and 2 < so that α converges uniformly on the disk by the Weierstrass M-test. n=1 n ∞ | n| FinallyP the product is analyticP everywhere by Theorem 21 of this section. Therefore sin πz ∞ z2 = 1 for all z C. πz − n2 ∈ n=1 Y by the principle of analytic continuation.

sin πz Note that the zeros of πz occur at the nonzero integers. If you were trying to create a function with zeros at those points you might ﬁrst write down

(z 1)(z + 1)(z 2)(z + 2) − − ···

which of course doesn’t converge! Your second attempt might be

z z z z 1 1+ 1 1+ − 1 1 − 2 2 ··· which is of course what we ended up with. Question: Is an analytic function determined uniquely, up to a scalar multiple, by its zeros (in the same way that a polynomial is)? The answer is no, essentially since there are entire functions which are nowhere zero. In particular, if g is any entire function, then f(z) and eg(z)f(z) have exactly the same zeros with the same multiplicities. Suppose then that we try to construct an entire function f with simple zeros at points a1, a2, a3,... and no other zeros. Obviously the points can’t have a limit point or else f would be identically zero by analytic continuation. Since any bounded sequence must have a limit point, this implies that we only have a hope here if a as n . It turns | n|→∞ → ∞ out that one needs to worry about how fast this happens. Our ﬁrst attempt would be ∞ z 1 . − a n=1 n Y Unfortunately, this doesn’t work! Exercise: (Assignment 2) Examine the convergence of

∞ z 1 − n n=1 Y 37 which has ‘zeros’ at the positive integers. Recall that for this product to converge we need Log 1 z to converge and this is − n −z essentially n . On the other hand, any expressionP of the form

P ∞ z P (z)= 1 egn(z) − a n=1 n Y will also have zeros at a1, a2, a3,.... A judicious choice of scaling factors ﬁxes the convergence problem. For example, one could let

∞ z P (z)= 1 ez/n. − n n=1 Y Why the exponential term. Recall that we want to be able to sum the logarithms of the terms. Here

z z z z z2 z3 z z2 Log 1 ez/n = Log 1 + = . . . + = . . . − n − n n −n − 2n2 − 3n3 − n −2n2 − Indeed, recalling our earlier estimate:

1 Log((1 w)ew) w 2, w , | − |≤| | | | ≤ 2

it follows that if z R and n> 2R then | | ≤ 2 z z z 2 R Log 1 e n . − n ≤ n ≤ n2 Thus, by the Weierstrass M-test,

∞ z z Log 1 e n − n n=1 X

converges uniformly on the compact disk of radius R, and hence on any compact subset of the plane. It follows from Theorem 18 that the product for P (z) converges Exercise: Show that the series converges uniformly on compact sets and hence that the function P deﬁned this way is analytic on C. This gives a clue to a way forward. Deﬁnition: (Weierstrass elementary factors) Let E (z)=1 z and, for k =1, 2,..., let 0 − z2 zk E (z)=(1 z) exp z + + + . k − 2 ··· k

The terms in our example then are of the form E1(z/n).

E′ (z)= eqk(z) + (1 z)q′ (z)eqk(z) = eqk(z) 1+(1 z)(1 + z + + zk−1 = zkeqk(z). k − − k − − ··· − This could be expanded as a power series, valid for all z,

q(z)2 E′ (z)= zk 1+ q(z)+ + . . . = zk zk+1 HOT. k − 2 − − − On integrating, zk+1 zk+2 E (z)= E (0) HOT, k k − k +1 − k +2 −

or, noting that Ek(0) = 1,

zk+1 zk+2 ∞ 1 E (z)= + +HOT = zk+1 b zi. − k k +1 k +2 i i=0 X Note that every b 0, and that i ≥ ∞ 1 E (1)=1= b . − k i i=0 X Thus, if z 1, | | ≤ ∞ ∞ 1 E (z) z k+1 b zi z k+1 b = z k+1 | − k |≤| | | i |≤| | i | | i=0 i=0 X X as required.

The Weierstrass elementary functions are just the things needed to produce analytic functions with the appropriate zeros.

Theorem 24. (Weierstrass Version 1) Let a be a sequence of complex numbers such that { n} a =0 and a as n . Then the inﬁnite product n 6 | n|→∞ →∞ ∞ z ∞ z P (z)= E = 1 eqn(z/an) n−1 a − a n=1 n n=1 n Y Y

deﬁnes an entire function P which has a zero at each an (according to the multiplicity of occurrence in a ) and no other zeros in C. { n} 39 In fact we will prove something a little more general. In Version 2, choosing k = n 1 n − always works, and this gives Version 1.

Theorem 25. (Weierstrass Version 2) Let a be a sequence of complex numbers such that { n} a =0 and a as n . Let k be a sequence of non-negative integers such that n 6 | n|→∞ →∞ { n} ∞ r kn+1 < a ∞ n=1 n X | | for every positive r. Then the inﬁnite product

∞ z P (z)= E kn a n=1 n Y deﬁnes an entire function P which has a zero at each an (according to the multiplicity of occurrence in a ) and no other zeros in C. { n}

Proof. Fix r and consider z r. Then | | ≤ ∞ z ∞ z r kn+1 1 E kn+1 < . − kn a ≤ a ≤ a ∞ n n ∞ n X X X | |

Hence, by the Weierstrass M-test, the series is uniformly and absolutely convergent on z r | | ≤ (and hence on arbitrary compact subsets) so that, according to Theorem21, ∞ E z n=1 kn an is uniformly convergent on compact sets and is analytic on C, that is, it is entire.Q Moreover the zeros of the product are those of the individual factors, which occur at the an with appropriate multiplicities.

In most examples one can pick kn to be a ﬁxed integer. If λ is the smallest integer that makes the series ∞ 1 a λ+1 n=1 n X | | converge then the product ∞ z zN E λ a n=1 n Y associated with the sequence 0,..., 0, a1, a2,... with the 0 occurring N times is known as the canonical product and λ is called the rank of the canonical product.

Example: (a) Let an = √n. Then

∞ 1 ∞ 1 = aλ+1 n(λ+1)/2 n=1 n n=1 X X 40 converges if λ> 1, and so the rank of the canonical product here will be 2. The function

∞ z ∞ z z z2 P (z)= E = 1 exp + 2 √n − √n √n 2n n=1 n=1 Y Y is the canonical product of an analytic function with zeros at 1, √2, √3, 2,... .

(b) There is no canonical product associated with the sequence an = log n since no ﬁxed choice of λ can be made.

We are now in a position to say exactly what a function with zeros at a certain unbounded sequence has to look like. We ﬁrst need a small lemma which is often useful.

Lemma 26. Suppose that h is an entire function which is never zero. The there exists an entire function g such that h(z)= eg(z) for all z.

h′(z) Proof. As h never vanishes, the function is also entire, and therefore has an an- h(z) tiderivative, say g0 (see, for example, Corollary 1 of Section 38 of Brown and Churchill). Now d h′(z) h(z)e−g0(z) = h′(z)e−g0(z) h(z) e−g0(z) =0 dz − h(z) and so h(z) = ceg0(z) for some nonzero constant c. The result follows be letting g(z) = g0(z)+Log c.

Theorem 27. (Weierstrass Factorization Theorem) Let f be a non-zero entire function which has

a zero of order N at z =0 and • zeros at the non-zero numbers a , a , a ,... (with numbers listed according to the mul- • 1 2 3 tiplicity of the zero).

Then there exists an entire function g and a sequence k of non-negative integers such { n} that ∞ z f(z)= eg(z)zN E . kn a n=1 n Y Remark: (i) The factorization is clearly not unique. (ii) You can take N =0 if f doesn’t have a zero at z = 0.

41 Proof. Remember that a as n . Therefore there exists a sequence of integers | n|→∞ →∞ k such that the product zN ∞ E z exists and is an entire function with the same n n=1 kn an zeros as f (with the correct multiplicities).Q Thus

f(z) h(z)= N ∞ z n=1 Ekn (z/an) has only removable singularities and has noQ zeros. By the lemma then there is an entire function g such that h(z) = exp(g(z)), which completes the proof.

∞ z In the special case where g(z) is a polynomial and the product Ek is a canon- k=1 n zk

ical product (so that kn is some ﬁxed non-negative integer λ),Q Laguerre introduced the number p = max(deg g,λ) called the genus of the entire function. Example: (a) Let f(z) = ez. Here we don’t have an inﬁnite product, but deg g = 1, so f 2 has genus 1. Similarly, ez is of genus 2. (b) Let f(z) be a polynomial. This can be written as

N z f(z)= c 1 − a n=1 n Y so the exponential term is trivial, and the Weierstrass elementary functions are of order 0. Thus every polynomial if of genus 0. (c) We saw earlier that ∞ z2 sin z = z 1 . − π2n2 n=1 Y This is not the canonical product representation. The rank of the canonical product here is clearly 1 and so the canonical product is

∞ z z sin z = z 1 e πn . − πn n=1 Y Again the leading exponential term is trivial, so here the genus is 1. Remark: We won’t look further at the genus in this course. It mainly reappears in the Hadamard Factorization Theorem which is a fancier version of Weierstrass’ theorem above. It roughly says that the genus of f is always close to the order of f, which is roughly the

42 smallest µ so that f(z) = O(exp( z µ). Hadamard used his theorem to give an asymptotic | | law for the distribution of prime numbers. Recall that a meromorphic function is one whose singularities are all poles. Meromorphic functions can be thought of as a natural extension of rational functions.

Theorem 28. Every function which is meromorphic (in the whole plane) is the quotient of two entire functions.

Proof. If F (z) is meromorphic we can ﬁnd an entire function g(z) with the poles of F (z) as zeros. The product F (z)g(z) is then an entire function f(z) so that F (z)= f(z)/g(z).