2.2 Polylogarithms and Hadamard Products
2.2.2 Hadamard product
Definition 2.4. Let f (z) and g(z) be two functions analytic at 0 with f (z) =∑
n≥0fnzn and g(z) =∑
n≥0gnzn. Then, the Hadamard product of f (z) and g(z) is defined as f (z)⊙ g(z) :=∑
n≥0
fngnzn
Theorem 2.6. Let a and b be two arbitrary complex numbers with neither a,b and a + b is an integer. Then (1− z)a⊙ (1 − z)b is also analytic in a ∆-domain, and admits an infinite expansion
(1− z)a⊙ (1 − z)b ∼∑
k≥0
λ(a,b)k (1− z)k
k! +∑
k≥0
µ(a,b)k (1− z)a+b+1+k
k! ,
where the coefficients λ(a,b)k and µ(a,b)k are given by
λ(a,b)k = Γ(a + b + 1) Γ(a + 1)Γ(b + 1)
(−a)¯k(−b)¯k
(−a − b)¯k , µ(a,b)k = Γ(−a − b − 1) Γ(−a)Γ(−b)
(a + 1)k¯(b + 1)¯k (a + b + 2)¯k . Here, x¯kis defined as x¯k = x(x + 1)· · · (x + k − 1) for k ∈ Z≥0.
Remark 2.5. The case that either a or b is a integer is simple; we have 1. (1− z)a⊙g(z) is a polynomial if a∈Z≥0;
2. (1− z)−a⊙g(z) can be reduced to a derivative of g(z) if a∈Z>0; more precisely, one has
(1− z)−a⊙g(z) = 1
(m− 1)!∂zm−1(zm−1g(z)).
The case that a + b is a integer is more complicated and it can be found in books by Abramowitz and Stegun [1, pp.559-560] and by Whittaker and Waston [11, Section 14.53].
Theorem 2.7. Let f (z) and g(z) be two functions that are analytic in a ∆-domain
∆(R, ϕ). Then, f (z)⊙ g(z) is also analytic in some ∆-domain. Moreover, if f(z) = O((1− z)a) and g(z) = O((1− z)b) for z ∈ ∆(R, ϕ), then f(z) ⊙ g(z) admits an expansion in some ∆-domain as follows:
(1) If a + b + 1 < 0, then
f (z)⊙ g(z) = O((1 − z)a+b+1).
(2) If k < a + b + 1 < k + 1, for some k∈ Z≥−1, then
f (z)⊙ g(z) =
∑k j=0
(−1)j
j! (f ⊙ g)(j)(1)(1− z)j+ O((1− z)a+b+1).
(3) If a + b + 1∈ Z≥0, then
f (z)⊙ g(z) =
∑k j=0
(−1)j
j! (f ⊙ g)(j)(1)(1− z)j+ O((1− z)a+b+1L(z)).
The following corollary is a consequence of Theorem 2.6 and Theorem 2.7.
Corollary 2.2. Let f (z) and g(z) be two functions that are analytic in a ∆-domain
∆(R, ϕ) with singular expansions of type (2.2):
f (z) =
∑m i=0
ci(1− z)αi+ O(|1 − z|A) and g(z) =
∑n j=0
dj(1− z)βj + O(|1 − z|B).
Then, the Hadamard product (f⊙g)(z) is also ∆-analytic and admits the singular ex-pansion of the form:
(f⊙g)(z) =∑
m,n
cmdn(1− z)αm⊙(1 − z)βj + P (1− z) + O(|1 − z|C),
where C := 1 + min(α0+ B, β0+ A) /∈Z≥0and P is a polynomial of degree less than C.
The reason for the polynomial P is that the integral powers of (1− z) do not leave a trace in the asymptotics of coefficients, since their contribution is zero.
In practice, this corollary allows us to establish the following algorithm, called Zigzag Algorithm, which is helpful in the computation of the singular expansions when com-posing function under Hadamard products.
Zigzag-algorithm (Fill, Flajolet and Kapur [4])
Input: two functions f (z) and g(z) that are ∆-analytic and have singular expansions of the form (2.2).
Output: the singular expansion of h(z) := (f⊙g)(z)
Step1. Use singularity analysis to determine the asymptotic expansions of fn = [zn]f (z) and gn= [zn]g(z).
Step2. Compute the asymptotic expansion of hn = [zn]h(z) by multiplying the asymptotic expansions of fnand gn.
Step3. Construct a function H(z) by using singularity analysis in the reverse direc-tion such that the asymptotic expansion of its coefficients is compatible with the asymptotic expansion of hn. By the construction, H(z) is a sum of functions of the form (1− z)αL(z)β, which are all singular at 1.
Step4. Output the singular expansion of (f⊙g) as
(f⊙g)(z) = h(z) = H(z) + P (1 − z) + O(|1 − z|C),
where C can be determined by the previous Corollary and P is a polynomial of degree δ, which is the largest integer less than C. Moreover, P (z) can be determined as follows:
P (z) =
∑δ j=0
(−1)j
j! ∂zj(h(z)− H(z))z=1zj.
Chapter 3
Moments by Singularity Analysis
So far, we have presented the tools which will be used in this thesis. From now on, we return to the analysis of the distributional recurrence:
Xn = Xd Sn+ Xn∗−S
n+ nα, for n≥ 2, X1 = 0. (3.1) In this chapter, we will first give the generating functions of the moments and then compute all moments of the random variable Xn.
3.1 Expected Value -- Proof of Theorem 1.3
It is crucial for us to have an asymptotic expansion of the expectation an=E(Xn), be-cause it is the initial case of mathematical induction to get all higher moments of Xn. As discussed in Chapter 1, such asymptotic expansions were derived by Fill, Flajolet and Kapur. However, there seems to be some imprecisions in their result. Consequently, we will re-derive their result (we closely follow the method in [4].) Now, starting from
the distributional recurrence (3.1), conditioning on the size Snyields, as mentioned in (1.2) and we rewrite it into the form:
pn,j = n
Multiplying the equation by zenn and summing over n≥ 1, we get
∑ Let A(z) and C(z) be the ordinary generating functions of the sequences an and cn, that is
Then the relation (3.3) can be reduced to A(z)⊙C(z/e)−
∫ z 0
A(w)⊙C(w/e)dw
w (3.4)
= (A(z)⊙C(z/e))C(z/e) +∑
n≥1
n− 1 n
cn
ennαzn.
Moreover, C(z) which is known as Cayley function.satisfies the functional equation
C(z) = zeC(z). (3.5)
and it admits the singular expansion at the dominant singularity z = e−1(see [6, Propo-sition 1])
C(z) = 1−√
2(1− ez)1/2− 1
3(1− ez) + O(|1 − ez|3/2). (3.6) By differentiating equation (3.5) on z, we get
C′(z) = eC(z)+ zeC(z)C′(z). (3.7) This yields
zC′(z)− C(z) = zC′(z)C(z).
Finally, by taking the coefficients on both size of the equation, we have ncn− cn=
∑n j=0
jcjcn−j (c0 := 0)
and consequently
ncn− cn=
∑n j=0
n
2cjcn−j. Thus, we have
ncn− cn
n =
∑n j=0
1
2cjcn−j.
Substituting this into (3.4) leads to Then we arrive at
A(z)⊙ C(z/e)− Then by differentiating, we transfer the above integral equation into a linear differential equation
Moreover, from (3.7) and (3.5), we have dC(z/e)
dz =
1
zC(z/e) 1− C(z/e). Substituting this into (3.9) yields
df (z)
Now, we solve this differential equation by the method of variation-of-constants. First, we consider the homogenous part, that is
df (z) where K is a constant. Thus, the solution of the homogenous part is
f (z) = C(z/e) 1− C(z/e)eK.
Letting f (z) = 1−C(z/e)C(z/e) eK(z)and substituting it into (3.10), we obtain that eK(z) =
∫ z
0
∂wt(w)
C(w/e)dw + D,
where D is a constant. Consequently, this gives the solution of equation (3.8), that is f (z) =D C(z/e) Finally, with the initial condition a1 = 0, we obtain
A(z)⊙ C(z/e) = 1 Theorem 3.1. The generating function A(z)⊙ C(z/e) is analytic in some ∆-domain.
Moreover, it admits the following singular expansions at z = 1.
1. For α > 32:
3. For 12 < α < 32:
A(z)⊙ C(z/e) = 2√1πΓ(
α− 12)
(1− z)−α + √1 2πΓ(
α− 12) (α−12 α−1 −76)
(1− z)−α+12 +√42L0(1− z)−12 − 127L0+ O(|1 − z|−α+1).
4. For α = 12:
A(z)⊙ C(z/e) = 2√1π(1− z)−12L(z) + 6√72(1− z)−12 + O(L(z)).
5. For 0 < α < 12:
A(z)⊙ C(z/e) = √42Kα(1− z)−12 +2√1πΓ(
α− 12)
(1− z)−α−127Kα+ O(|1 − z|−α+12).
Proof.
First, we compute the singular expansion of B(z)⊙ C(z/e)2 for each case by the Zigzag-algorithm. Since
C(z/e) = 1−√
2(1− z)1/2− 1
3(1− z) + O(|1 − z|3/2), we obtain
C(z/e)2 = 1− 2√
2(1− z)1/2+4
3(1− z) + O(|1 − z|3/2).
Moreover,
B(z) = ∑
n≥2
nαzn= Γ(1 + α)(1− z)−α−1+ O(|1 − z|−α),
Then, by the method of singularity analysis, we have [zn]C(z/e)2 = −2√
2
Γ(−12)n−32 + O(n−52) and [zn]B(z) = nα and this implies that
[zn]B(z)⊙C(z/e)2 =
√2
πnα−32 + O(nα−52).
Thus, converting back this information to the function by Zigzag-algorithm, we find the singular expansion of B(z)⊙C(z/e)2.
1. For α > 32:
Here, K and K1 are some constants. Then applying differential and integral rules for singularity analysis according to Theorem 2.3 and Theorem 2.4, we obtain the follow-ing.
=
Where L0is the integration constant which can be computed by Theorem (2.4).
4. For α = 12:
Where, L0 is the integration constant which can be computed by Theorem 2.4.
Moreover, setting
Finally, we multiply by 121−C(z/e)C(z/e) which is 1
The end result is then as follows.
1. For α > 32:
= 1
4. For α = 12:
This complete the proof of Theorem 3.1.
Moreover, applying the result in Theorem 2.1 and Theorem 2.2 to Theorem 3.1 gives the excepted value of Xn. This establishes Theorem 1.3.