Therefore, by taking derivative on both sides of equation 31 with respect to q, Z
q
n; q, v = nλ
n−1
q, vξq, vλ
q
q, v + λ
n
q, vξ
q
q, v
+nA
n−1
q, vrq, vA
q
q, v + A
n
q, vr
q
q, v.
We are only interested in the above quantity at q = 1, at which ξ1, v = Z0; 1, v and r1, v = 0.
This shows that, by dividing nλ
n
1, v, we obtain
Z
q
n; 1, v
nλ
n
1, v
= λ
q
1, v λ1, v
Z0; 1, v +
ξ
q
1, v
n +
A1, v λ1, v
n
r
q
1, v
n .
As n → ∞, obviously ξ
q
1, vn → 0. Since the matrix Aq, v is primitive A1, v
λ1, v
n
→ D1, v, as n → ∞,
for some matrix D1, v, according to Theorem 8.5.1 in [HJ86]. Hence
A1, v λ1, v
n
r
q
1, v
n → 0, as n → ∞.
Consequently, Z
q
n; 1, v
nλ
n
1, v
→ λ
q
1, v λ1, v
Z0; 1, v, as n → ∞.
Since Z
q
n; 1, v is a vector of Z
q
Γn, k; 1, v for all Γ ∈ G , we have for any noncrossing partition
Γ of C
k
, Z
q
Γn, k; 1, v
nλ
n
1, vZΓ0, k; 1, v
→ λ
q
1, v λ1, v
. as n → ∞. If Γ is the partition consisting of k isolated vertices, then Γn, k is the cylinder graph P
n
× C
k
and Γ0, k is the cycle C
k
with only type 2 edges. Therefore, by Lemma 2 and the fact that ZC
k
; 1, v = 1 + v
2 k
, Lemma 3 is proved.
Now by dominated convergence theorem, Theorem 3 follows easily from the next lemma, details are omitted.
Lemma 4.
Z
q
P
n
× C
k
; 1, v
n1 + v
2 n+1k
1 + v
1 nk
≤ n + 1k
nk .
Proof of Lemma 4: Let the edge subset A = A
1
∪ A
2
, where the set A
1
, A
2
consists of type 1 and type 2 edges in A respectively. By the definition of the multivariate Tutte polynomial,
ZG
nk
; q, v =
X
A⊆E
q
kA
v
|A
1
| 1
v
|A|−|A
1
| 2
, where 0 ≤ |A
1
| ≤ nk and 0 ≤ |A
2
| ≤ n + 1k. As such, by taking derivative on both sides of the above,
Z
q
G
nk
; q, v =
X
A⊆E
kAq
kA−1
v
|A
1
| 1
v
|A|−|A
1
| 2
, 130
then Z
q
G
nk
; 1, v =
|E|
X
m=0
X
|A|=m
kAv
|A
1
| 1
v
m−|A
1
| 2
. 32
Since the number of components in a graph is bounded by the number of vertices the graph has, for the cylinder graph P
n
× C
k
, 1 ≤ kA ≤ n + 1k.
Then from equation 32 and the fact that |E| = 2n + 1k, Z
q
P
n
× C
k
; 1, v ≤ n + 1k
2n+1k
X
m=0
X
|A|=m
v
|A
1
| 1
v
m−|A
1
| 2
= n + 1k
2n+1k
X
m=0 nk
X
i=0
X
|A
1
|=i,|A|=m
1
v
i 1
v
m−i 2
= n + 1k
2n+1k
X
m=0 nk
X
i=0
nk i
n + 1k m − i
v
i 1
v
m−i 2
= n + 1k1 + v
1 nk
1 + v
2 n+1k
, which completes the proof of Lemma 4.
Theorem 3 shows that to calculate the asymptotic value of EL
M S T
G
nk
n for a specific value of k, one only needs to find λ
q
1, v. The calculation of λ
