Definition 2.4. A partition λ is a sequence of nonnegative integers λ
1
, λ
2
, · · · with λ
1
≥ λ
2
≥ · · · and
∞
X
i=1
λ
i
∞. The length l
λ and the size |λ| of λ are defined as l
λ := max i ∈ N; λ
i
6= 0 and |
λ| :=
∞
X
i=1
λ
i
. We set
λ ⊢ n := λ partition ; |λ| = n
for n ∈ N. An element of λ ⊢ n is called a partition of n.
Remark: we only write the non zero components of a partition. Choose any
σ ∈ S
n
and write it as σ
1
σ
2
· · · σ
l
with σ
i
disjoint cycles of length λ
i
. Since disjoint cycles commute, we can assume that
λ
1
≥ λ
2
· · · ≥ λ
l
. Therefore λ := λ
1
, · · · , λ
l
is a partition of n.
Definition 2.5. We call the partition λ the cycle-type of σ ∈ S
n
.
Definition 2.6. Let λ be a partition of n. We define C
λ
⊂ S
n
to be the set of all elements with cycle type
λ. It follows immediately from 2.5 that Z
n
xσ only depends on the cycle type of σ. If σ, θ ∈ S
n
have different cycle type then Z
n
xσ 6= Z
n
xθ .
Therefore two elements in S
n
can only be conjugated if they have the same cycle type since Z
n
x
is a class function. One can find in [3, chapter 39] or in [12, chapter I.7, p.60] that this condition is sufficient. The cardinality of each C
λ
can be found also in [3, chapter 39].
Lemma 2.7. We have |C
λ
| =
n z
λ
with z
λ
:=
n
Y
r=1
r
c
r
c
r
and c
r
= c
r
λ := i|
λ
i
= r .
2.6 We put lemma 2.7 and 2.5 together and get
Lemma 2.8. Let x ∈ C
p
and s ∈ N
p
be given. Then E
Z
s
n
x
=
X
λ⊢n
1 z
λ p
Y
k=1 l
λ
Y
m=1
1 − x
λ
m
k s
k
. 2.7
Obviously, it is very difficult to calculate E
Z
s
n
x
explicitly. It is much easier to write down the
generating function.
2.3 Generating function of E
Z
s
n
x
We now give the definition of a generating function, some lemmas and apply them to the sequence
E
Z
s
n
x
n∈N
. 1096
Definition 2.9. Let f
n n∈N
be given. The formal power series f t := P
∞ n=0
f
n
t
n
is called the gen- erating function of the sequence f
n n∈N
. If a formal power series f t = P
n∈N
f
n
t
n
is given then f
n
:= f
n
. We will only look at the case f
n
∈ C and f convergent.
Lemma 2.10. Let a
m m∈N
be a sequence of complex numbers. Then X
λ
1 z
λ
a
λ
t
|λ|
= exp X
m=1
1 m
a
m
t
m
with a
λ
:=
l λ
Y
i=1
a
λ
i
. 2.8
If RHS or LHS of 2.8 is absolutely convergent then so is the other. Proof. The proof can be found in [12, chapter I.2, p.16-17] or can be directly verified using the
definitions of z
λ
and the exponential function. In view of 2.7 is natural to use lemma 2.10 with a
m
= Q
p j=1
1− x
m j
s
j
to write down the generating function of the sequence
E
Z
s
n
x
n∈N
. We formulate this a lemma, but with a
m
= Px
m
for P a polynomial. We do this since the calculations are the same as for a
m
= Q
p j=1
1 − x
m j
s
j
.
Lemma 2.11. Let Px =
P
k∈N
p
b
k
x
k
be a polynomial with b
k
∈ C and x
k
:= Q
p j=1
x
k
j
j
. We set for a partition
λ P
λ
x :=
l λ
Y
m=1
P x
λ
m
with x
λ
m
= x
λ
m
1
, · · · , x
λ
m
p
and Ω := ¦
t, x ⊂ C
p+1
; |t| 1, kxk ≤ 1
© .
We then have X
λ
1 z
λ
P
λ
xt
|λ|
= Y
k∈N
p
1 − x
k
t
−b
k
2.9 and both sides of 2.9 are holomorphic on Ω.
We use the principal branch of the logarithm to define z
s
for z ∈ C \ R
−
. Proof. We only prove the case p = 2. The other cases are similar. We use 2.8 with a
m
= Px
m 1
, x
m 2
and get X
λ
1 z
λ l
λ
Y
m=1
Px
λ
m
1
, x
λ
m
2
t
|λ|
= exp X
m=1
1 m
Px
m 1
, x
m 2
t
m
= exp
∞
X
k
1
,k
2
=0
b
k
1
,k
2
∞
X
m=1
t
m
m x
k
1
1
x
k
2
2 m
= exp
∞
X
k
1
,k
2
=0
b
k
1
,k
2
−1Log1 − x
k
1
1
x
k
2
2
t
=
∞
Y
k
1
,k
2
=0
1 − x
k
1
1
x
k
2
2
t
−b
k1,k2
. The exchange of the sums is allowed since there are only finitely many non zero b
k
1
,k
2
. Since we have used the Taylor-expansion of Log1 + z near 0, we have to assume that |t x
k
1
1
x
k
2
2
| 1 if b
k
1
,k
2
6= 0. 1097
We now write down the generating function of E
Z
s
n
x
.
Theorem 2.12. Let s ∈ N
p
and x ∈ C
p
. We set f
s
x, t :=
Y
k∈N
p
1 −
x
k
t
−
s k
−1
k
with
s k
−1
k
:=
p
Y
j=1
s
j
k
j
−1
k
j
. 2.10
Then E
Z
s
n
x
=
f
s
x, t
n
. Proof. This is 2.7 and lemma 2.11 with P
x =
Q
p j=1
1 − x
j s
j
.
Remark: we use the convention E
Z
s
x
:= 1.
Remark: In an earlier draft version, the above proof was based on representation theory. It was at least twice as long and much more difficult. We wish to acknowledge Paul-Olivier Dehaye, who has
suggested the above proof and allowed us to us it.
Corollary 2.12.1. For n ≥ 1, p = s = 1, we have that
E Z
n
x = 1 − x.
2.11 Proof.
1 − x t 1 − t
=
∞
X
n=0
t
n
− x t
∞
X
n=0
t
n
= 1 + 1 − x
∞
X
n=1
t
n
2.4 Asymptotics for kxk
