In this section, we treat generating functions as analytic functions. In the sequel, we will often encounter generating functions with a singularity at ζ and (local) behavior
f (z) =
1 − z
ζ
−α
log 1 1 − zζ
!β
, where α and β are arbitrary complex numbers.
For convenience, we use a transformation to let the singularity z = ζ be on the unit circle |z| = 1. More precisely, note that by the scaling rule, we have
g(z) ≡ f (zζ) =
1 −zζ
ζ
−α
log 1 1 −zζζ
!β
= (1 − z)−α
log 1 1 − z
β
.
So, g(z) has a singularity at z = 1, and in this way, we can get fn = [zn]f (z) as follows,
fn= [zn]f (z) = ζ−n[zn]f (zζ) = ζ−n[zn]g(z) = ζ−ngn,
where gn = [zn]g(z). Since all generating functions can be brought in this form, we only need to discuss generating functions with a singularity at z = 1 and (local) behavior
(1 − z)−α
log 1 1 − z
β
,
where α and β are arbitrary complex numbers.
First, we consider the special case (1 − z)−r, where r ∈ Z≥1. Then, (1 − z)−r =
∞
X
n=0
n + r − 1 n
zn.
Consequently, the coefficients are given by [zn] (1 − z)−r =n + r − 1
n
= (n + r − 1) (n + r − 2) · · · (n + 1) (r − 1)!
= nr−1 (r − 1)!
1 + O 1 n
.
Now, for (1 − z)−α, where α ∈ C, we expect a similar result:
[zn] (1 − z)−α =n + α − 1 α − 1
= cnα−1
1 + O 1 n
, (2.1)
where c is a constant which will turn out to be related to the Gamma function Γ(α) which is defined as follows
Γ(α) :=
Z ∞ 0
e−ttα−1dt
for <(α) > 0. Moreover, we recall the following integral representation 1
Γ(α) = 1 2πi
I
C
(−t)−αe−tdt, (2.2)
where the contour C comes from ∞ + i, goes around 0 in counterclockwise direction, and then goes back to ∞ − i (see [12, p. 745]).
The formula (2.1) is made precise in the following theorem.
Theorem 2.3. Given a function f (z) = (1 − z)−αwithα ∈ C \ Z≤0, we have for the n-th coefficient
fn= [zn]f (z) ∼ nα−1
Γ(α) 1 +X
k≥1
ek
nk
! , whereekis a polynomial inα of degree 2k.
Proof. We first prove the case where k = 1. By Cauchy’s coefficient formula, we have fn = 1
2πi Z
C
(1 − z)−α dz zn+1 , where we choose the contour C as follows:
C
R R
1
0 n 1 0 1 1
Figure 2.1: Our contour C is the curve on the left and we separate it into two parts, the center one and the right one.
Next, we break C into two parts eC and H, where C =e z
z = Reiθ, R > 1, arcsin(1/nR) ≤ θ ≤ 2π − arcsin(1/nR) and H =
z
z = 1 +eiθ
n , π/2 ≤ θ ≤ 3π/2
∪ (
z
1 ≤ <(z) ≤ r
R2− 1
n2, I(z) = ±1 n
) .
Then, we consider the integral over eC and H separately.
C : Since |z| = R, ze −nis bounded by R−n. Thus, the integral over eC is exponentially small. More precisely,
1 2πi
Z
Ce
(1 − z)−α dz zn+1
≤ cR · 1
Rn+1 = O
1 Rn
, where c = sup|z|=R|(1 − z)−α|.
H : We are going to break H into 3 parts: or not. Then, the integral of the part with <(t) > log2n is at most
= O integral exponentially small; on the other hand, if <(α) < 0, by integration by parts which is exponentially small, too. The same estimate holds for the corresponding part of ¯H−.
Next, for t with <(t) ≤ log2n, we use the following asymptotic expansion for 1 + nt−n−1
. After plugging this in, we can add the part with <(t) > log2n since this part is again
expo-nentially small. Overall, we can let the contour become eH = eHo∪ eH+∪ eH−, where
He+=t
t = ω + i, ω ≥ 0 ; He−=t
t = ω − i, ω ≥ 0 ; Heo =
t
t = eiθ, θ ∈ [π 2,3π
2 ]
. Then,
Z
He
(−t)−αe−tdt
1 + O log4n n
= 1
Γ(α)
1 + O log4n n
, where the last line follows from (2.2).
Finally, by putting every thing together, we obtain fn = nα−1
Γ(α)
1 + O log4n n
.
This concludes the proof of k = 1. For the general case, one only needs to use more terms in (2.5). This then gives
fn ∼ nα−1
Γ(α) 1 +X
k≥1
ek nk
! .
Theorem 2.4. Given a function f (z) = (1 − z)−α 1zlog1−z1 β
with α ∈ C \ Z≤0, β ∈ C, we have for the n-th coefficient
fn = [zn]f (z) ∼ nα−1
Γ(α)(log n)β 1 +X
k≥1
dk logkn
! ,
wheredkis a polynomial inα of degree 2k.
Proof. Again by Cauchy’s coefficient formula, fn = 1
2πi Z
C
(1 − z)−α 1
z log 1 1 − z
β
dz zn+1,
where the contour C = eC ∪ H is as in Theorem 2.3:
As in the proof of Theorem 2.3, we consider the integral over eC and H separately.
C : Since |z| = R, ze −nis bounded by R−n. Thus, the integral over eC is exponentially
with ¯H = ¯Ho∪ ¯H+∪ ¯H−, where where c is a suitable constant.
Since this integral is exponentially small, we only need to consider the part where
<(t) ≤ log2n.
For this part, we use the following asymptotic expansion for 1 + nt−n−1−β
Moreover, we have (again for <(t) ≤ log2n)
Now, as in the proof of Theorem 2.3, plugging this in and adding the tail <(t) >
log2n which is exponentially small, shows that the integral (2.6) is asymptotic to
Heo =
t
t = eiθ, θ ∈ [π 2,3π
2 ]
. This proves the claimed result.
Definition 2.5. A domain is a ∆-domain at 1 if it can be written as
∆(R, φ) = {z | |z| < R, z 6= 1, | arg(z − 1)| > φ},
whereR > 1 and 0 < φ < π/2. Moreover, if a function is analytic in some ∆-domain, the function is called a∆-analytic function.
Theorem 2.6. Let α , β ∈ R and f (z) be a function that is analytic in ∆ := ∆ (R, φ).
If
f (z) = O (1 − z)−α
log 1 1 − z
β! , wherez ∈ ∆ and approaching 1, then
fn= [zn]f (z) = O
nα−1(log n)β
. Similarly, if
f (z) = o (1 − z)−α
log 1 1 − z
β! , wherez ∈ ∆-domain and approaching 1, then
fn= [zn]f (z) = o(nα−1(log n)β) . Proof. By Cauchy’s coefficient formula, we have
fn = 1 2πi
Z
C
f (z) dz zn+1,
and C is a closed contour in the unit disc. Since f (z) is not analytic at z = 1, we change the contour C into a union of following 4 parts:
γ1 = {z
|z − 1| = 1n, | arg (z − 1) | ≥ θ } (inner circle) γ2 = {z
|z − 1| ≥ n1, |z| ≤ r, arg (z − 1) = θ } (top line segment) γ3 = {z | |z| = r, | arg (z − 1) | ≥ θ } (outer circle) γ4 = {z
|z − 1| ≥ n1, |z| ≤ r, arg (z − 1) = −θ } (bottom line segment), where 1 < r < R, and φ < θ < π2, so that our contour C lies entirely inside our
∆-domain. We let fn[1] , fn[2] , fn[3] , fn[4] be the integral along γ1, γ2, γ3, γ4, i.e., fn[i] = 1
2πi Z
γi
f (z) dz zn+1. Then, we have
fn= 1 2πi
Z
C
f (z) dz
zn+1 = 1
2πi fn[1]+ fn[2]+ fn[3] + fn[4] . So now, we will discuss the integrals separately.
1. Inner circle (γ1):
The line integral of fn[1] will be at most the length of γ1 times the maximum of
|f (z)|,
fn[1] ≤ |γ1| · max|f (z)|
z ∈ γ1 . Since f (z) is O
(1 − z)−α log1−z1 β
, there is a constant c > 0 such that
|f (z)| ≤ c · |1 − z|−α
log1−z1
β. Hence, for the contour γ1where |z − 1| = 1/n and | arg (z − 1) | ≥ θ,
fn[1]
≤ |γ1| · max|f (z)|
z ∈ γ1
≤ 2π1 n · c
1 − z
−α|log 1 − z|β
= O 1 n
· nα
log |1 − z| + iφ
β
= O 1 n
· nα log2|1 − z| + φ2β/2
= O nα−1 · log2n + φ2β/2
,
where φ ∈ [0, 2π). Next, observe for large n proof of Theorem 2.3, the integral over t ∈ log2n, ∞ is exponentially small,
so we can change the range to 1 ≤ t ≤ log2n. Then, obviously The proof of the small-o part is similarly.
Definition 2.7. A log-power function at 1 is a finite sum of the form σ(z) =
whereα1 < · · · < αrand eachck(z) is a polynomial. A log-power function at a finite set of pointsZ = {ζ1, · · · , ζm}, is a finite sum
Σ(z) =
m
X
j=1
σj
z ζj
,
where eachσj is a log-power function at 1.
Theorem 2.8. If a function f (z) is analytic on a ζ ·∆ domain and there exist log-power functionsσ and τ such that
f (z) = σ(z/ζ) + O (τ (z/ζ)) asz → ζ in ζ · ∆, thenfn= [zn]f (z) will be asymptotic to
fn= ζ−nσn+ O ζ−nτn , whereσn = [zn]σ(z) and τn = [zn]τ (z).
Proof. Let g(z) = f (ζz). Then, g(z) is ∆-analytic, with a singularity at 1 and g(z) = σ(z) + O (τ (z)) as z → 1. Now, since τ (z) is a log-power function, so
gn= [zn]g(z) = [zn]σ(z) + [zn]O (τ (z))
= [zn]σ(z) + O ([zn]τ (z))
= σn+ O(τn).
Finally, since fn = ζ−ngn, we get
fn = ζ−ngn = ζ−n(σn+ O(τn)) = ζ−nσn+ O ζ−nτn .
According to the latter result, we know how to find fnwhen f (z) has one singu-larity at ζ. Similarly, fn can be found if f (z) has finitely many singularities (see [12, p. 398] for a proof).
Theorem 2.9. Let f (z) be analytic in |z| < ρ with a finite number of singularities ζ1, · · · , ζk on the circle|z| = ρ. Suppose there exists a ∆-domain such that f (z) is analytic in the domain
D =
k
\
i=1
(ζi· ∆) ,
Moreover, we havek log-power functions σ1, · · · , σk, and a log-power functionτ (z) = (1 − z)−α(log1−z1 )β such that
f (z) = σi(z/ζi) + O (τ (z/ζi)) as z → ζi in D.
Then, the coefficientsfn = [zn]f (z) satisfies
fn=
k
X
i=1
ζi−n(σi)n+ O ρ−nnα−1(log n)β ,
where(σi)n= [zn]σi.
The last result is quite powerful and has many applications (for some of them see Chapter 3). However, it can be only applied if f (z) is analytic in a domain which is larger than |z| < ρ. Sometimes, however, we only know that a function is analytic on
|z| < ρ and analytic extension is either hard to prove or not possible (some examples for this will be given in Section 3.4). Then, singularity analysis cannot be applied and we need other methods. We are going to introduce two such methods in the next section.