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,
Q
N
X
i6=j (z − zi )
.
p(z) =
cj Q
i6=j (zj − zi )
j=1
Written another way, this expresses p(z) as
ℓj (zi ) = δij .
P
cj ℓj (z) where ℓj is a polynomial for which
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
f (1/j) =
(
0,
if j is even
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 − zj ). This obviously requires some care as such an
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:
∞ z2
sin πz Y
=
1− 2 .
πz
n
n=1
This example introduces most of the key ideas. We shall then look more carefully at the
theory of infinite products of numbers, such as
∞ 1
2 Y
=
1− 2
π n=1
4n
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 X 2z
π cot πz = +
z n=1 z 2 − n2
(for z 6= 0, ±1, ±2, . . .).
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 + 21 and x = − N + 21 .
y
x=− N+
1
2
x=N+
y=N
−1
b
0
b
1
N
b
b
N +1
b
1
2
x
y = −N
We first show that on CN there is a bound for | cot πz| which is independent of N, that
is, there is a B such that, for all N = 1, 2, 3, . . .
| cot πz| ≤ B whenever z ∈ CN . Note that
cos πz = cos πx cosh πy − i sin πx sinh πy
sin πz = sin πx cosh πy + i cos πx sinh πy.
Therefore, on x = ± (N + 1/2),
| cos πz|
| sinh πy|
| sinh πy|
=
=p
≤1
| sin πz|
| cosh πy|
1 + sinh2 πy
25
whilst on y = ± N,
p
cos2 πx cosh2 πN + sin2 πx sinh2 πN
| cos πz|
=p
| sin πz|
sin2 πx cosh2 πN + cos2 πx sinh2 πN
p
cos2 πx + sinh2 πN
= p
sin2 πx + sinh2 πN
s
1 + sinh2 πN
1
≤
≤1+
2
sinh 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
Z
π cot πw
1
dw.
IN,z =
2πi CN w 2 − z 2
Note that from the above bounds, and using the fact that if w ∈ CN , then |w 2| ≥ N 2 and
hence |w 2 − z 2 | ≥ N 2 − |z|2 ,
Z
Z
π| cot πw|
1
1
1
|dw| ≤
|dw| = 2
· 2(4N + 1),
|IN,z | ≤
2
2
2
2
2π CN |w − z |
N − |z|2
CN N − |z|
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 CN . The singularities occur where sin πw = 0 and where w 2 − z 2 = 0, that is,
at integers w = −N, −(N − 1), . . . , (N − 1), N and w = ±z.
π cot πw
Let f (w) = 2
. Then each integer is a simple zero of sin πw and thus is a simple
w − z2
pole of f and the Laurent expansion is of the form
f (w) =
c−1
+ c0 + c1 (w − n) + c2 (w − n)2 + · · ·
w−n
Calculating the residue is easy at such a point:
1
π(w − n) cos πw
=
.
w→n (w 2 − z 2 ) sin πw
n2 − z 2
Res(f, n) = lim (w − n)f (w) = lim
w→n
A similar argument shows that the residues at z and −z are
π cot πz
2z
and
π cot(−πz)
−2z
Thus the Residue Theorem gives
IN,z
N
X
1
π cot πz π cot(−πz)
+
+
= Res(f, z) + Res(f, −z) +
Res(f, n) =
2
2z
−2z
n − z2
n=−N
n=−N
N
X
26
Thus, with z fixed, and taking a limit as N → ∞,
lim IN,z
N →∞
∞
X
π cot πz π cot(−πz)
1
=0=
+
+
2
2z
−2z
n − z2
n=−∞
or, using symmetry and pulling out the n = 0 term,
∞
X
1
2
π cot πz
− 2+
.
0=
2
z
z
n − z2
n=1
Therefore, multiplying through by z,
∞
π cot πz −
1 X 2z
=
z
z 2 − n2
n=1
as claimed.
Clearly each term on the right hand side has an antiderivative related to log(z 2 − 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
log(sin πz) − log(πz) = log
or
sin πz
πz
=
∞
X
n=1
log(z 2 − n2 ) = log
∞
Y
(z 2 − n2 )
n=1
∞
sin πz Y 2
=
(z − n2 )
πz
n=1
. . . 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.
y
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 limz→0(π cot πz − z1 ) = 0.
Note also that
−2z
z2
2z
d
2
Log 1 − 2 = n z 2 = 2
.
dz
n
z − n2
1 − n2
Thus, for |z| < 1,
∞
X 2z
d
ℓsin z =
dz
z 2 − n2
n=1
∞
X
d
z2
=
Log 1 − 2
dz
n
n=1
∞
z2
d X
Log 1 − 2 .
=
dz n=1
n
28
Here we could interchange d/dz and
P
since the series of Logs converges uniformly (on
any compact set not containing an integer.) One easy way to prove this is to observe that
|Log(1 − w)| ≤ 2|w| when |w| ≤ 1/2. Therefore
2
2 Log 1 − z ≤ 2|z| ≤ 2 for |z| < 1 and n ≥ 2.
n2 n2
n2
P
2
But ∞
n=2 n2 < ∞ and so we have the result by the Weierstrass M-test (we can clearly forget
a finite number of terms in the series if we want to). Therefore, for |z| < 1,
Log
sin πz
πz
∞
X
z2
=
Log 1 − 2
n
n=1
+C
for some constant C. Substituting z = 0 shows that C = 0. That is,
Log
sin πz
πz
∞
X
N
X
z2
z2
Log 1 − 2 = lim
=
Log 1 − 2 .
N →∞
n
n
n=1
n=1
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
2
.
Exercise: (a) Show that Arg 1 − nz 2 ≤ sin−1 |z|
n2
(b) Use the fact that t ≤ 1.1 sin t for t ∈ [0, 0.25] to show that for |z| < 1 and n ≥ 2,
2
1.1
|z|
−1
<
.
sin
n2
n2
(c) Deduce that for any N,
2
N N
X
X
z 2 π
π
π
1
≤ + 1.1
− 1 < 2.3 < π.
Arg 1 − 2 < + 1.1
2
n
2
n
2
6
n=1
n=2
Pi*x In light of this then,
Log
sin πz
πz
∞
X
z2
=
Log 1 − 2
n
n=1
29
N
X
z2
= lim
Log 1 − 2
N →∞
n
n=1
!
N Y
z2
= lim Log
1− 2
N →∞
n
n=1
Now recalling that exp(Log w) = w for all w 6= 0, and, of course, that exp is continuous,
sin πz
sin πz
= exp Log
πz
πz
!!
N Y
z2
= exp lim Log
1− 2
N →∞
n
n=1
!!
N 2
Y
z
= lim exp Log
1− 2
N →∞
n
n=1
N
Y
z2
= lim
1− 2 .
N →∞
n
n=1
We shall write this as
∞ sin πz Y
z2
=
1− 2
πz
n
n=1
for |z| < 1. We will soon see that this equation holds for all z.
If we substitute z = 1/2 we obtain
1
π
2
so that
∞ ∞
Y
Y
1
(2n − 1)(2n + 1)
=
1− 2 =
4n
(2n)2
n=1
n=1
∞
Y
π
(2n)2
=
2
(2n − 1)(2n + 1)
n=1
or
π
22 42 62
=
···
2
1·3 3·5 5·7
which is traditionally and rather unsatisfactorily written as Wallis’ product
π
22 · 42 · 62 · · ·
= 2 2 2 2 .
2
1 ·3 ·5 ·7 ···
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
∞
Y
pn converges to P if the sequence of partial products PN = p1 p2 · · · pN converges
product
n=1
to a non-zero 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 pn → 1. As we’ll see shortly, this turns out to be correct,
30
so it is common to consider infinite products of the form
Q∞
n=1 (1
+ αn ) where αn 6= −1 for
all but a finite number of n. Clearly we then must have αn → 0 for the product to exist.
∞
Y
pn exists if
Definition: More generally we agree to say that an infinite product
n=1
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
∞ Y
1 2 3
1
= 0· · · ···
1−
n
2 3 4
n=1
1 2
2 3
so that the infinite product exists if and only if limn→∞
But the value is = limn→∞ n1 = 0. Therefore the infinite
n−1
exists and is
n
Q∞
product 1 (1 − n1 ) does
···
non-zero.
not exist.
It diverges to 0. (Ignoring the initial 0, this corresponds to the fact that
N
X
log
n=1
n
= − log n + 1
n+1
diverges to −∞.)
Example: (b) Consider
∞ Y
(−1)n
1
1
1
1 4 3 6
1+
= 1−
1+
1−
··· = · · · ··· .
n+1
2
3
4
2 3 4 5
n=1
Here PN = 1/2 if N is odd and =
N +2
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 =
This need not be the case however.
Q∞
n=1
pn exists then Log P =
Let pn = exp(−iπ/2n ), so that the partial products are given by
PN =
N
Y
n=1
pn = exp(−iπ(1 − 1/2N )).
In this case PN → P = −1, so we have
∞
X
n=1
Log pn = −πi while Log P = πi.
31
P∞
n=1
Log pn .
Lemma 17. If P =
Q∞
pn exists then pn → 1.
n=1
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 6= 0.
With the notation as above then,
P
Pn
=
= 1.
Pn−1
P
lim pn = lim
n→∞
n→∞
Theorem 18. P =
Q∞
n=1
pn exists and is nonzero if and only if
P∞
n=1
Log pn converges.
Proof. As we saw in the concrete example we did earlier, the main issue here is that we
Q
P
might not have Log pn = Log pn . Recall however that for any N
!
N
N
Y
X
Arg
pn =
Arg pn + 2kN π,
some kN ∈ Z,
n=1
n=1
and so
Log
N
Y
pn =
n=1
Thus
PN =
N
Y
pn = exp Log
n=1
N
Y
pn
n=1
One direction is now easy! If
!
N
X
Log pn + 2kN πi.
n=1
= exp
N
X
Log pn + 2kN πi
n=1
P∞
n=1 Log pn
!
= exp
N
X
Log pn
n=1
converges then PN → exp (
P∞
n=1
!
.
Log pn ) as
N → ∞ (since exp is continuous) and so PN is convergent with a non-zero limit.
For the converse1 , assume that P = lim PN exists and is nonzero. Let us first assume that
P 6∈ (−∞, 0). Then Log is continuous at P and hence Log P = Log lim PN = lim Log PN
N
N
exists. As above, for all N,
cN := Log PN =
N
X
Log pn + 2kN πi
n=1
for some integer kN . Since the sequence {cN } converges it is Cauchy, and in particular
cN − cN −1 = Log pN + 2πi(kN − kN −1 ) → 0,
1
Thanks to Joel for googling the fix to this proof!
32
as N → ∞.
But we know that pN → 1, and hence that Log pN → 0. This implies that 2πi(kN −kN −1 ) → 0
too, which can obviously only happen if the sequence {kN } is eventually constant. Thus,
there exists an integer k such that for all sufficiently large N,
N
X
n=1
Log pn = Log PN − 2kπi.
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 pn were real and positive, then P
would be too, so there must exist at least one term, say pm which is not real and positive.
Y
pn converges to a nonzero point which is not on the negative real axis.
In this case P ′ =
n6=m
The previous case would then imply that
X
Log pn converges, and hence that
n6=m
converges too.
∞
X
Log pn
n=1
∞
Y
Remark: It is worth noting that we actually showed that if
pn is non-zero then the sum
n=1
P∞
Log
p
can
only
differ
from
Log
P
by
an
integer
multiple
of 2πi.
n
n=1
The following result provides us with a very simple test.
P∞
P
Log pn converges absolutely if and only if ∞
n=1 (1 − pn ) converges abP
Q∞
solutely. Therefore if (1 − pn ) converges absolutely then n=1 pn converges.
Theorem 19.
n=1
Proof. Let us write pn = 1 + αn . Then, for |αn | < 1, the Taylor series for Log gives
3
4
2
Log(1 + αn ) − αn = − αn + αn − αn + · · · 2
3
4
|αn |2 |αn |3 |αn |4
+
+
+···
≤
2
3
4
|αn |
≤
|αn | + |αn |2 + |αn |3 + · · · .
2
If we further assume |αn | ≤ 1/2 then the geometric series in the last line has sum at most
1/2 and so, for |αn | ≤ 1/2, | Log(1 + αn ) − αn | ≤ 21 |αn | or, equivalently,
3
1
|1 − pn | ≤ Log pn ≤ |1 − pn |.
2
2
The result now follows by the comparison test since clearly for n large enough, we can assume
that |αn | ≤ 1/2.
33
Corollary 20. If
∞
X
n=1
Exercise: Show that
|αn | < ∞ then
∞
X
∞
Y
(1 + αn ) converges.
n=1
|αn | < ∞ if and only if
∞
Y
(1 + |αn |) converges.
n=1
n=1
Q
If either of these above conditions hold then we say that ∞
1 (1+αn ) converges absolutely.
Q
Q∞
It therefore follows that if 1 (1 + αn ) converges absolutely then ∞
1 (1 + αn ) converges.
This is, of course, because the similar result for series is true.
P
To complete the discussion we shall see, by two examples, that the convergence of ∞
1 αn
Q∞
is neither necessary nor sufficient for the convergence of the product 1 (1 + αk ).
P
Example: (a) Let α2n−1 = n−1/2 and α2n = −n−1/2 + n−1 . Then ∞
n=1 αn is divergent
(being the harmonic series). Note that
1
1
1
1
1+ √
1− √ +
= 1 + 3/2
n
n n
n
Q∞
Q∞
so 1 (1 + αn ) = 1 (1 + n−3/2 ) is convergent.
P
1
Example: (b) On the other hand if αn = (−1)n /n 2 then ∞
2 αn is clearly convergent, whilst
Q∞
2 (1 + αn ) is divergent.
Exercise: Prove this last statement! Hint: Prove and use the fact that
1
1
1
(1 + α2n−1 )(1 + α2n ) = 1 − √
1+ √
.
<1−
2n
2n − 1
2n
5.4
Products of functions
Next suppose we are given a sequence of functions {αn : Ω → C}∞
n=1 defined on some
region Ω. As for infinite sums, for each fixed z ∈ Ω we can consider the convergence of the
∞ Y
product
1 + αn (z) . If, for each z ∈ Ω, this infinite product exists then we can define
n=1
the pointwise limit function
∞ Y
P (z) =
1 + αn (z) ,
n=1
z ∈ Ω.
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
N Y
PN (z) =
1 + αn (z)
n=1
is clearly also analytic, but what about P ? How can we guarantee that the infinite product
is analytic?
34
Theorem 21. Let {αn }∞
n=1 be a sequence of analytic functions defined on Ω such that
converges uniformly on each compact subset of Ω. Then
P (z) =
∞
Y
P
|αn |
(1 + αn (z))
n=1
converges uniformly on each compact subset of Ω and P is analytic on Ω.
P
Proof. First observe that if
|αn | converges uniformly on each compact subset of Ω then
P
certainly for any z ∈ Ω,
|αn (z)| converges and hence by the previous results, P (z) =
Q∞
n=1 (1 + αn (z)) exists.
As before, let
PN (z) =
N
Y
(1 + αn (z)),
z∈Ω
n=1
so that
PN +1 (z) − PN (z) = (1 + αN +1 (z))PN (z) − PN (z) = αN +1 (z)PN (z).
For M > N then, using the fact that for t ≥ 0, 1 + t ≤ et ,
−1
X
M
PM (z) − PN (z) = αj+1(z) Pj (z)
j=N
≤
≤
≤
≤
M
−1
X
j=N
M
−1
X
j=N
M
−1
X
j=N
M
−1
X
j=N
P
=e
|αj+1 (z)|
|αj+1 (z)|
|αj+1 (z)|
|αj+1 (z)|
|αj (z)|
M
−1
X
j=N
j
Y
n=1
j
Y
|(1 + αn (z))|
(1 + |αn (z)|)
n=1
j
Y
e|αn (z)|
n=1
∞
Y
e|αn (z)|
n=1
|αj+1(z)|
Let K be a compact subset of Ω and let ǫ > 0. Since
P∞
1
|αj | converges uniformly on
K, it has a continuous limit. In particular, the limit is bounded and so there is a constant
P
C > 0 such that exp ( ∞
1 |αj (z)|) ≤ C for all z ∈ K. Moreover there is a positive integer
P
N0 such that ∞
N |αj (z)| < ǫ/C whenever N > N0 and z ∈ K. Therefore
|PM (z) − PN (z)| < ǫ
35
whenever M, N > N0 and z ∈ K. Thus {PN } forms a Cauchy sequence in C(K) with the
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 {αn }∞
n=1 be a sequence of
P
analytic functions defined on Ω such that
|αn | converges uniformly on each compact subset
Q∞
of Ω. Then, at every point z where P (z) = n=1 (1 + α(z)) is nonzero,
∞
P ′ (z) X αn′ (z)
=
.
P (z)
1
+
α
(z)
n
n=1
Proof. Since P (z) 6= 0, none of the factors 1 + αn (z) vanishes.
The product rule lets us differentiate the partial product PN (z):
PN′ (z)
=
N
X
αn′ (z)
n=1
N
Y
(1 + αk (z)).
k6=n
Thus,
QN
N
N
X
αn′ (z)
PN′ (z) X ′
k6=n (1 + αk (z))
=
αn (z) QN
=
PN (z) n=1
1 + αn (z)
k=1 (1 + αk (z))
n=1
since lots of terms cancel.
Now Pn → P uniformly on each compact set and hence (as the functions are analytic,
see Theorem 12) Pn′ → P ′ uniformly on each compact set.
Thus, if P (z) 6= 0, then (noting perhaps that no Pn (z) is zero either),
Pn′ (z)
P ′(z)
→
Pn (z)
P (z)
as n → ∞. That is
∞
P ′ (z) X αn′ (z)
=
.
P (z)
1
+
α
(z)
n
n=1
Exercise: Investigate the convergence of the series for P ′/P given in the corollary.
Recall that we showed that for |z| < 1
∞ z2
sin πz Y
=
1− 2
πz
n
n=1
We want to extend this identity to all of C. Now the function
see that the product is also analytic everywhere we consider
36
sin πz
πz
P
is analytic everywhere. To
−z 2
|αn | where αn (z) = 2 . We
n
need to show that this converges uniformly on any compact subset of C, or equivalently, that
2
it converges uniformly on any closed disk. Suppose then that |z| ≤ R. Then |αn (z)| ≤ Rn2
P
P
R2
and ∞
<
∞
so
that
|αn | converges uniformly on the disk by the Weierstrass M-test.
2
n=1 n
Finally the product is analytic everywhere by Theorem 21 of this section. Therefore
∞ z2
sin πz Y
1− 2
=
for all z ∈ C.
πz
n
n=1
by the principle of analytic continuation.
Note that the zeros of
sin πz
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 |an | → ∞ as n → ∞. It turns
out that one needs to worry about how fast this happens.
Our first attempt would be
∞ Y
n=1
Unfortunately, this doesn’t work!
z
1−
an
.
Exercise: (Assignment 2) Examine the convergence of
∞ Y
z
1−
n
n=1
37
which has ‘zeros’ at the positive integers.
P
Recall that for this product to converge we need
Log 1 − nz to converge and this is
P −z
essentially
. On the other hand, any expression of the form
n
P (z) =
∞ Y
n=1
z
1−
an
egn (z)
will also have zeros at a1 , a2 , a3 , . . .. A judicious choice of scaling factors fixes the convergence
problem. For example, one could let
P (z) =
∞ Y
n=1
1−
z z/n
e .
n
Why the exponential term. Recall that we want to be able to sum the logarithms of the
terms. Here
z z/n z z
z2
z2
z
z3
z
Log 1 −
e
= Log 1 −
+ = − − 2 − 3 −... + = − 2 − ...
n
n
n
n 2n
3n
n
2n
Indeed, recalling our earlier estimate:
| Log((1 − w)ew )| ≤ |w|2 ,
1
|w| ≤ ,
2
it follows that if |z| ≤ R and n > 2R then
2
Log 1 − z e nz ≤ z 2 ≤ R .
n
n
n2
Thus, by the Weierstrass M-test,
∞
X
Log 1 − z e nz n
n=1
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 E0 (z) = 1 − z and, for k = 1, 2, . . ., let
zk
z2
.
+···+
Ek (z) = (1 − z) exp z +
2
k
The terms in our example then are of the form E1 (z/n).
38
Lemma 23. For |z| ≤ 1,
|1 − Ek (z)| ≤ |z|k+1 .
zk
z2
+···+
so Ek (z) = (1 − z)eqk (z) . The bound is clearly true if
Proof. Let qk (z) = z +
2
k
k = 0, so suppose k ≥ 1. Then
Ek′ (z) = −eqk (z) + (1 − z)qk′ (z)eqk (z) = eqk (z) −1 + (1 − z)(1 + z + · · · + z k−1 = −z k eqk (z) .
This could be expanded as a power series, valid for all z,
q(z)2
′
k
Ek (z) = −z 1 + q(z) +
+ . . . = −z k − z k+1 − HOT.
2
On integrating,
Ek (z) = Ek (0) −
z k+2
z k+1
−
− HOT,
k+1 k+2
or, noting that Ek (0) = 1,
∞
X
z k+1
z k+2
1 − Ek (z) =
bi z i .
+
+ HOT = z k+1
k+1 k+2
i=0
Note that every bi ≥ 0, and that
1 − Ek (1) = 1 =
∞
X
bi .
i=0
Thus, if |z| ≤ 1,
|1 − Ek (z)| ≤ |z|
k+1
∞
X
i=0
i
|bi z | ≤ |z|
k+1
∞
X
i=0
bi = |z|k+1
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 {an } be a sequence of complex numbers such that
an 6= 0 and |an | → ∞ as n → ∞. Then the infinite product
P (z) =
∞
Y
n=1
En−1
z
an
∞ Y
z
=
1−
eqn (z/an )
a
n
n=1
defines an entire function P which has a zero at each an (according to the multiplicity of
occurrence in {an }) and no other zeros in C.
39
In fact we will prove something a little more general. In Version 2, choosing kn = n − 1
always works, and this gives Version 1.
Theorem 25. (Weierstrass Version 2) Let {an } be a sequence of complex numbers such that
an 6= 0 and |an | → ∞ as n → ∞. Let {kn } be a sequence of non-negative integers such that
kn +1
∞ X
r
<∞
|a
|
n
n=1
for every positive r. Then the infinite product
P (z) =
∞
Y
Ekn
n=1
z
an
defines an entire function P which has a zero at each an (according to the multiplicity of
occurrence in {an }) and no other zeros in C.
Proof. Fix r and consider |z| ≤ r. Then
kn +1
∞
∞
X
X
z kn +1 X
r
z
≤
1 − Ekn
≤
< ∞.
an
an
|an |
∞
Hence, by the Weierstrass M-test, the series is uniformly and absolutely convergent on |z| ≤ r
Q
z
(and hence on arbitrary compact subsets) so that, according to Theorem21, ∞
E
k
n
n=1
an
is uniformly convergent on compact sets and is analytic on C, that is, it is entire. 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
∞
X
n=1
converge then the product
z
N
∞
Y
1
|an |λ+1
Eλ
n=1
z
an
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
∞
∞
X
X
1
1
=
aλ+1
n(λ+1)/2
n=1
n=1 n
40
converges if λ > 1, and so the rank of the canonical product here will be 2. The function
Y
∞
∞ Y
z
z
z
z2
E2 √
P (z) =
1− √
=
exp √ +
n
n
n 2n
n=1
n=1
√ √
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)
is also entire, and therefore has an anh(z)
tiderivative, say g0 (see, for example, Corollary 1 of Section 38 of Brown and Churchill).
Proof.
As h never vanishes, the function
Now
d
h′ (z) −g0 (z)
h(z)e−g0 (z) = h′ (z)e−g0 (z) − h(z)
e
=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 a1 , a2 , a3 , . . . (with numbers listed according to the multiplicity of the zero).
Then there exists an entire function g and a sequence {kn } of non-negative integers such
that
g(z) N
f (z) = e
z
∞
Y
Ekn
n=1
z
an
.
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 |an | → ∞ as n → ∞. Therefore there exists a sequence of integers
Q
z
exists and is an entire function with the same
kn such that the product z N ∞
E
n=1 kn an
zeros as f (with the correct multiplicities). Thus
h(z) =
zN
f (z)
n=1 Ekn (z/an )
Q∞
has only removable singularities and has no zeros.
By the lemma then there is an entire function g such that h(z) = exp(g(z)), which
completes the proof.
In the special case where g(z) is a polynomial and the product
Q∞
k=1
Ekn
z
zk
is a canon-
ical product (so that kn is some fixed non-negative integer λ), 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
f (z) = c
N Y
n=1
z
1−
an
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
sin z = z
∞ Y
n=1
z2
1− 2 2
π n
.
This is not the canonical product representation. The rank of the canonical product here is
clearly 1 and so the canonical product is
sin z = z
∞ Y
z z
e πn .
1−
πn
n=1
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). 43