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Chapter 4
Other Things & Some Applications
4.1 Unimodality, Log-concavity & Real-rootedness
(4.1) Denition
Let A = (a0 , a1 , . . . , an ) be a sequence of positive real numbers. The
sequence A is said to be:
unimodal if a0 ≤ a1 ≤ · · · ≤ ad ≥ · · · ≥ an for some d ∈ N0 ;
log-concave if a2i+1 ≥ ai ai+2 for all 0 < i < n;
strictly log-concave if a2i+1 > ai ai+1 for all 0 < i < n.
(4.2) Remark
a2i+1 ≥ ai ai+2
(4.3) Example
⇔
log ai+1 ≥
log ai + log ai+2
2
A = (1, 7, 10, 3, 2) is unimodal but not log-concave (32 6≥ 2 · 10).
A = (2, 5, 12, 15, 10, 4, 1) is unimodal & log-concave & strictly log-concave.
(4.4) Proposition
Proof.
♦
If A ∈ (R+)n+1 is log-concave, then it is unimodal.
If the sequence is
not
ar−1 > ar < ar+1
unimodal, then there exists 0 < r < n such that
⇒
a2r < ar−1 ar+1
⇒
A is
The contrapositive of this statement is the result.
not
log-concave.
If the zeros of a polynomial p(x) (of degree n > 1) are all real and
negative, then the zeros of p0(x) are all real and negative.
(4.5) Lemma
Proof.
Suppose
p(x) = c0 + c1 x + . . . + cn xn = a(x + r1 )m1 · · · (x + rk )mk ,
ri , ci , a ∈ R, mi ∈ N, ri > 0
Suppose also that 0 < ri < · · · < rk . The polynomial p0 (x) has degree n − 1. Since
(x + ri )mi is a factor of p(x), we know (x + ri )mi −1 is a factor of p0 (x). From this
−r1 , . . . , −rk are zeros of p0 (x) of multiplicities m1 −1, . . . , mk −1, respectively. Now, the
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CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
function p(x) is continuously dierentiable of R. If 1 ≤ i < k then p(−ri ) = p(−ri+1 ) = 0.
By Rolle's theorem, there exists ri∗ ∈ ] − ri+1 , −ri [ such that p0 (ri∗ ) = 0. Thus, there are
k − 1 distinct numbers r1∗ , . . . , rk∗ , none equal to {−r1 , . . . , −rk }, that are zeros of p0 (x).
We have accounted for
(m1 − 1) + · · · + (mk − 1) + (k − 1) = m1 + · · · + mk − k + k − 1 = n − 1
zeros of p0 (x). Since deg(p0 (x)) = n − 1, all roots are real and negative.
(4.6) Lemma
Let
f (x, y) = c0 xn + c1 xn−1 y + · · · + cn y n
be a polynomial whose zeros ( xy ) are real. Let
g(x, y) =
∂i ∂j
f (x, y)
∂xi ∂y j
If g(x, y) 6= 0, then all zeros of g(x, y) are real.
Proof.
Suppose f (x, y) has k distinct roots λ1 , . . . , λk . Then
f (x, y) = a(x + λ1 y)m1 · · · (x + λk y)mk
for a ∈ R, mi ∈ N. By repeated use of the previous lemma, all non-zero partial derivatives
of f (x, y) will have real roots.
Let p(x) = c0 + c1x + . . . + cnxn be a polynomial whose zeros are all
real and negative. Then the sequence (c0, c1, . . . , cn) is strictly log-concave.
(4.7) Theorem
Since the zeros of p(x) are real and negative, the sequence of coecients is strictly
positive (expand a(x + r1 )m1 · · · (x + rk )mk ). If p(x) = c0 + c1 x + . . . + cn xn , let
Proof.
f (x, y) = c0 xn + c1 xn−1 y + . . . + cn y n
Partially dierentiate f (x, y):
∂ m ∂ n−m−2
f (x, y)
∂xm ∂y n−m−2
We have
∂m
f (x, y) = (n)m c0 xn−m + (n − 1)m c1 xn−m−1 y + · · · + (m)m cn−m−2 y n−m−2
∂xm
Now do
∂ n−m−2
∂y n−m−2
∂m
f (x, y)
∂xm
=
X
ci (n − i)m (i)m xn−m−i y 2−(n−m−i) ≡ g(x, y)
i
In the sum, i ∈ {n − m − 2, n − m − 1, n − m}. Thus
1
cn−m
cn−m−2
g(x, y) = (m + 1)!(n − m − 1)!
(n − m)y 2 + cn−m−1 2xy +
(m + 2)x2
2
m+1
n−m−1
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CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
Let cj =
n
j
pj for all j , then
1 g(x, y) = n! pn−m y 2 + 2pn−m−1 xy + pn−m−2 x2
2
According to the previous lemma, this polynomial has real roots, since it is nonzero.
Thus the discriminant is ≥ 0, i.e.
(2pn−m−1 )2 − 4(pn−m pn−m−2 ) ≥ 0
⇒
p2n−m−1 ≥ pn−m pn−m−2
⇒
c2n−m−1 ≥
m+2 n−m
m+1 n−m−1 cn−m cn−m−2
> cn−m cn−m−2
Thus the sequence is strictly log-concave.
(4.8) Remark
{constants of neg. real rooted polynom.} ⊂ {strictly log-concave} ⊂ {log-concave}
⊂ {unimodal}
(4.9) Example
Is {
n
i
}ni=0 strictly log-concave? Well
n
n
n n
p(x) =
+
x + ··· +
x = (1 + x)n
0
1
n
which has the single negative root x = −1. By the last theorem, the answer is yes.
(4.10) Proposition
is log-concave.
Proof.
♦
The sequence of Stirling numbers of the 2nd kind (S(n, 1), . . . , S(n, n))
The generating function of these numbers is
Bn (x) =
n
X
S(n, k)xk
k=1
We know
S(n, k) = S(n − 1, k − 1) + kS(n − 1, k)
By dierentiating, we have
0
Bn (x) = xBn−1 (x) + xBn−1
(x),
i.e.
B1 (x) = x,
B2 (x) = x + x2 ,
(4.1)
n > 0, B0 (x) = 1
B3 (x) = x + 3x2 + x3 ,
···
Multiply each side of (4.1) by ex
0
ex Bn (x) = xex Bn−1 (x) + xex Bn−1
(x)
d x
=x
(e Bn−1 (x))
dx
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CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
Let fn (x) = ex Bn (x). Then
0
fn (x) = xfn−1
(x)
We claim that all zeros of fn (x), of which there are n, are real, distinct and negative
(except for x = 0). This is true for n = 1, n = 2. Assume true for f1 (x), . . . , fn−1 (x).
since Bn−1 (x) has n − 1 distinct zeros, then (ex Bn−1 (x))0 has n − 2 distinct zeros that
lie between the zeros of Bn−1 (x). Multiplying by x gives another zero at x = 0. Finally,
since ex Bn−1 (x) → 0 as x → 0, there is another zero of (ex Bn−1 (x))0 in ]−∞, r[ where r
is the smallest root of Bn−1 (x).
Thus fn (x) has n distinct real and negative (except at x = 0) roots/zeros. Hence, the
sequence of coecients is log-concave.
4.2 Transfer Matrix Method-Theory
(4.11) Denition A directed graph (digraph) D consists of vertices V , edges
E and an orientation map φ, D = (V, E, φ).
o
• v2
v1
•
• v7
V = {v1 , . . . , v7 }
v4 •
v3
o • o •v6
/
E = {{v1 , v2 }, {v1 , v3 }, · · · , {v7 , v7 }}
Given {a, b} ∈ E,
φ(a, b) = (int(a, b), n(a, b)),
e.g. φ(v1 , v2 ) = (v2 , v1 )
v5 •
For every edge e = {v, v 0 } we dene int(e) = v and n(e) = v 0 . A walk Γ in D of length
n is a sequence of edges (e1 , . . . , en ) such that n(ei ) = int(ei+1 ) for all 1 ≤ i ≤ n. If
n(en ) = int(e1 ) then the walk is called closed.
(4.12) Denition
Let w : E → R be a weight function on the edges of the digraph
D = (V, E, φ). For any walk Γ = (e1 , . . . , en ) we dene
w(Γ) = w(e1 )w(e2 ) · · · w(en )
Dene
Aij (n) =
X
w(Γ)
Γ
where the sum is over all walks Γ of lengths n in D with int(Γ) = vi and n(Γ) = vj .
Dene
Aij (0) = δij
The adjencency matrix of D is the matrix
A = [Aij ],
Aij =
X
e∈E
int(e) = vi
n(e) = vj
84
w(e)
CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
(4.13) Example
o
C
v2
o
•
77
77
77
77
v3
•
w(edge) = b
?
A11 = w(loop) = a;
A12 = b;
A13 = b
A21 = b;
A22 = a;
A23 = 0;
A31 = 0;
A32 = b;


a b b


A =  b a 0
0 b a
A33 = a
(4.14) Theorem (4.2.1.)
Proof.
w(loop) = a
•77v1
♦
For n ∈ N, the (i, j) entry of An is Aij (n).
The (i, j) entry of An is
(An )ij =
X
Aii1 Ai1 i2 · · · Ain−1 j
where the sum is over all possible sequences (i1 , . . . , in−1 ). The sum will equal to zero if
there is no walk of length n from vi → vj . If an entry in the sum is non-zero, then the
product Aii1 Ai1 i2 · · · Ain−1 j equals the weight of the corresponding walk.
(4.15) Denition
From this,
(An )ij = Aij (n)
Dene the generating function
Fij (D, λ) =
+∞
X
Aij (n)λn
n=0
(All information about weighted walk from vi to vj .)
For a matrix T , let (T : j, i) be the matrix obtained by removing row j and column i of
T.
(4.16) Theorem (4.2.2.)
If
Fij (D, λ) =
+∞
X
Aij (n)λn
n=0
then
Fij (D, λ) =
In other words
(−1)i+j det(I − λA : j, i)
det(I − λA)
Fij (D, λ) = ((I − λA)−1 )ij
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CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
(4.17) Remark
deg Fij (D, λ) < n0 = multiplicity of the eigenvalue 0 of A
Proof.
Obvious using simple linear algebra. Fij (D, λ) is the (i, j) entry of the matrix
+∞
X
λn An = (I − λA)−1
n=0
and then use expression for adj(I − λA)
If
cA (λ) = det(A − λI)
= (−1)p (αp−n0 λn0 + · · · + α1 λp−1 + λp )
is the characteristic polynomial of A, then
det(I − λA) = αp−n0 λp−n0 + · · · + α1 λ + 1
where n0 is the multiplicity of eigenvalue 0 of A. So,
deg(det(I − λA)) = p − n0 ,
deg(det(I − λA : j, i)) ≤ p − 1
This gives
deg(det(I − λA : j, i)/ det(I − λA)) ≤ p − 1 − (p − n0 ) = n0 − 1 < n0
(4.18) Example
Let
f (n) = #{(a1 , . . . , an ) : ai ∈ {1, 2, 3}, (ai , ai+1 ) ∈
/ {(1, 1), (2, 3)}}
Then
f (1) = |{(1), (2), (3)}| = 3
f (2) = |{(1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (3, 2), (3, 3)}| = 7
f (3) = . . . = 16
Let D be the directed graph on {1, 2, 3} with (i~,j) as an edge if j is allowed to follow i
in the sequence.
•1
C
2
•
o
[
•3
For any edge e, let w(e) = 1. Adjacency matrix


0 1 1


A = 1 1 0
1 1 1
86
?
CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
7
}|
{
z
The sequence (1, 2, 2, 1, 3, 1, 3) is represented by the walk
(1, 2)(2, 2)(2, 1)(1, 3)(3, 1)(1, 3)
|
{z
}
6 edges
The quantity we want is thus
f (n) = A11 (n − 1) + A12 (n − 1) + A13 (n − 1) + · · ·
3
X
=
Aij (n − 1) =
i,j=1
(An−1 )ij
i,j=1
Now

(I − λA)−1 =
3
X
(1 − λ)2
λ(1 − λ)
λ

1


λ2
λ(1 − λ) 1 − λ − λ2

1 − 2λ − λ2 + λ3
2
λ
λ(1 + λ) 1 − λ − λ
and, e.g. the 1,1-entry is equal to
X
X
(1 − λ)2
=
A11 (n)λn =
f11 (n + 1)λn
2
3
1 − 2λ − λ + λ
F11 (D, λ) =
n≥0
n≥0
More precisely
3
X
i,j=1
+∞
X
1 + λ − λ2
Fij (D, λ) =
=
f (n + 1)λn
1 − 2λ − λ2 + λ3
n=0
and we have obtained the generating function of f (n + 1). The o.g.f. of f (n) is thus
+∞
X
n
f (n)λ = 1 +
n=0
+∞
X
f (n + 1)λn+1 = 1 +
n=0
λ(1 + λ − λ2 )
1−λ
=
2
3
1 − 2λ − λ + λ
1 − 2λ − λ2 + λ3
♦
If our attention is restricted to closed walks in D of length n, then let
CD (n) = A11 (n) + · · · + App (n) = Tr(An )
(4.19) Theorem (4.2.3.)
Let Q(λ) = det(I − λA). Then
+∞
X
n=1
Proof.
CD (n)λn =
−λQ0 (λ)
Q(λ)
Let w1 , . . . , wq be the non-zero eigenvalues of A. Then
CD (n) = Tr(An ) = w1n + · · · + wqn
So
+∞
X
n
CD (n)λ =
n=1
+∞
X
(w1n + · · · + wqn )λn =
n=1
wq λ
w1 λ
+ ··· +
1 − w1 λ
1 − wq λ
Since Q(λ) = (1 − w1 λ) · · · (1 − wq λ), it is easy to see that
+∞
X
n=1
CD (n)λn =
wq λ
w1 λ
−λQ0 (λ)
+ ··· +
=
1 − w1 λ
1 − wq λ
Q(λ)
87
CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
4.3 Transfer Matrix Method Applications
(4.20) Example
Let f (n) bet the number of sequences (x1 , . . . , xn ) of length n that
have elements in {a, b} and such that
(1) (xi , xi+1 ) 6= (a, a)
(2) x1 = a, xn = b.
∀1 ≤ i < n
Let g(n) be the total number of sequences that satisfy (1). Let h(n) be the total number
of sequences that satisfy (1) and (xn , x1 ) 6= (a, a).
Invoke the transfer matrix method:
D digraph:
a
•
o
/ b
•
Adjacency matrix:
"
A=
w(edge) = 1,
A11 A12
#
=
A21 A22
Note
"
I − λA =
"
0 1
#
1 1
−λ
1
(v1 , v2 ) = (a, b)
#
−λ 1 − λ
We have that
f (n) = A12 (n − 1)
Using theorem (4.16) (i.e. 4.2.2.)
F12 (D, λ) =
+∞
X
A12 (n)λn =
n=0
i.e.
+∞
X
λ
(−1)3 (−λ)
=
2
1−λ−λ
1 − λ − λ2
f (n + 1)λn =
n=0
Thus
+∞
X
f (n)λn = λ
n=0
λ
1 − λ − λ2
+1=
λ
1 − λ − λ2
λ2 + 1 − λ − λ2
1−λ
=
1 − λ − λ2
1 − λ − λ2
For g(n) two ways to do this. First way:
g(n) = A11 (n − 1) + A12 (n − 1) + A21 (n − 1) + A22 (n − 1)
Since deg |I − λA| = 2 (quadratic in λ), we know
+∞
X
(A11 (n) + A12 (n) + A21 (n) + A22 (n))λn =
n=0
88
P (λ)
1 − λ − λ2
CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
where deg(P ) < 2. Let P (λ) = a + bλ. So
P (λ)
= (a + bλ)(1 + (λ + λ2 ) + (λ + λ2 )2 + · · · )
1 − λ − λ2
= a + λ(a + b) + O λ2
= 2 + 3λ + O λ2
(get 2 and 3 by counting paths,
see original sum)
giving a = 2 and b = 1. Hence
+∞
X
g(n + 1)λn =
n=0
i.e.
+∞
X
g(n)λn =
n=0
Second way:
We have
(I − λA)
−1
2+λ
1 − λ − λ2
1+λ
1 − λ − λ2
"
#
1−λ λ
1
=
1 − λ − λ2
λ
1
So
+∞
X
(A11 (n) + A12 (n) + A21 (n) + A22 (n))λn =
n=0
=
1
((1 − λ) + (λ) + (λ) + (1))
1 − λ − λ2
2+λ
1 − λ − λ2
as before, giving
+∞
X
g(n)λn =
n=0
1+λ
1 − λ − λ2
For h(n):
h(n) = A11 (n) + A22 (n)
(since sequences of length n satisfying (1) and (xn , x1 ) 6= (a, a) are isomorphic to sequences of length n+1 satisfying (1) and x1 = xn check the function that sets xn+1 = x1
[inverse: removes xn+1 ], it's a bijection because of (1)). Using theorem (4.19) (i.e. 4.2.3.):
CD (n) = A11 (n) + A22 (n) = h(n)
thus
+∞
X
n=1
CD (n)λn =
−λ(1 − λ − λ2 )0
λ + 2λ2
=
1 − λ − λ2
1 − λ − λ2
♦
89
CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
(4.21) Example
Dene
g(n) = #{(a1 , . . . , an ) : ai ∈ {1, 2, 3}, (ai , ai+1 ) ∈
/ {(1, 1), (2, 3)}, 1 ≤ i ≤ n}
where an+1 = a1 . Transfer (adjacency) matrix


0 1 1


A = 1 1 0
1 1 1
We have
g(n) = A11 (n) + A22 (n) + A33 (n) = CD (n)
and
+∞
X
g(n)λn =
n=0
−λQ0 (λ
2λ + 2λ2 − 3λ3
=
= ...
Q(λ)
1 − 2λ − λ2 + λ3
since
Q(λ) = det(I − λA) = 1 − 2λ − λ2 + λ3
(4.22) Example
♦
Let
f (n) = #{(a1 , . . . , an ) : ai ∈ {1, 2, 3},
(ai , ai+1 ) 6= (1, 2),
1≤i<n
(ai , ai+1 , ai+2 ) ∈
/ {(2, 1, 3), (2, 2, 2), (2, 3, 1), (3, 1, 3)},
1 ≤ i < n − 1}}
To apply TMM, look at the digraph on V = {(1, 1), (1, 2), . . . , (3, 3)}. Insert a directed
edge from (a, b) → (b, c) if we are allowed the sequence (a, b, c)

/
•(1,1)
•(1,3)
•
(1,2)
O

(2,2)
O (2,1)•?
o
•
??
??
??
??
?_ ?
O
??
??
??
?
(3,1)
90
•
(3,2)
•
o
/
?
•(2,3) o

•(3,3)
?
CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
Using the order {11, 12, 13, 21, 22, 23, 31, 32, 33}, the transfer matrix becomes
11

12
13
1
21
22
23
31
32
33

1






 1

A = 




 1



1
1
1
1
1
1
1
1
1
1



1







1





1
Now
Q(λ) = det(I − λA)
and
f (n) =
9
X
Aij (n − 2)
i,j=1
Then
+∞
X
f (n)λn =
n=0
P (λ)
Q(λ)
where deg P < 8 and deg Q = 8.
♦
4.4 Transfer Matrix Method Application to Ising Model
Consider the 1-dimensional Ising model (with periodic boundary conditions).
[ Fig. Circle with spins . . . , sn , s1 , s2 , s3 , . . .,
si ∈ {−1, +1} (like magnets) ]
There are 2n possible spin congurations,
s = (s1 , . . . , sn )
For any such conguration we associate an energy, given by the Hamiltonian
H(s) = J(s1 s2 + s2 s3 + · · · + sn s1 ) + h(s1 + · · · + sn )
|
{z
} |
{z
}
internal energy
due to external eld
J and h are parameters of this `system', showing how much weight is given to a unit of
each type of energy.
In statistical mechanics we are interested in systems that represent `something', solving
them and comparing experimental and theoretical results. The central object of study is
91
CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
the partition function,
X
Zn (J, h, x) =
xH(s)
all possible
s=(s1 ,...,sn )
(In physics: x = e−βc , β =
1
T .)
From this we calculate the
free energy
1
log Zn (J, h, x)
n→+∞ n
f (J, h, x) = lim
Every spin conguration can be represented as a walk of length n on the directed graph
D(V, E, φ) where V = {v1 , v2 } = {+1, −1},
/
−1
•
o
+1
•
O
(si , si+1 ) ∈ {(+1, +1), (+1, −1), (−1, +1), (−1, −1)}, 4 edges. To the edge (si , si+1 ) we
associate the weight
h
w(si , si+1 ) = xJ(si si+1 )+ 2 (si +si+1 )
Then
Zn (J, h, x) =
X
=
X
=
X
xH(s) =
all s
X
xJ(s1 s2 +s2 s3 +···+sn s1 )+h(s1 +···+sn )
s
J(s1 s2 )+ h
(s +s2 ) J(s2 s3 )+ h
(s +s3 )
2 1
2 2
x
x
h
· · · xJ(sn s1 )+ 2 (sn +s1 )
s
w(s1 , s2 )w(s2 , s3 ) · · · w(sn , s1 )
s
= Tr(An )
where A is the transfer matrix of D:
#
"
# "
A11 A12
xJ+h x−J
A=
=
A21 A22
x−J xJ−h
By theorem (4.19) (i.e. 4.2.3.) we have
+∞
X
Zn (J, h, x)λn =
n=1
where
−λQ0 (λ)
Q(λ)
Q(λ) = det(I − λA) = 1 − λ(xJ+h + xJ−h ) + λ2 x2J − x−2J
Thus
Zn (J, h, x) = [λn ]
λ(xJ+h + xJ−h ) − 2λ2 x2J
1 − λ(xJ+h + xJ−h ) + λ2 x2J − x−2J
In order to calculate the fre energy, we must look at the eigenvalues of A
xJ+h − µ
x−J
0 = det(A − µI) = −J
J−h
x
x
−µ = µ2 − 2µxJ cosh(h log x) + 2 sinh(2J log x)
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CHAPTER 4. OTHER THINGS & SOME APPLICATIONS
giving
µ=x
J
q
2
−4J
cosh(h log x) ± sinh (h log x) + x
Let µ⊕ be the larger of the eigenvalues and µ be the smaller of the eigenvalues. By
diagonalizing A, we have
Tr(An ) = (µ⊕ )n + (µ )n
Thus
1
1
log Zn (J, h, x) = lim
log((µ⊕ )n + (µ )n )
n→+∞ n
n
1
log(µ⊕ )n = log(µ⊕ )
= lim
n→+∞ n
f (J, h, x) = lim
n→+∞
so
q
J
2
−4J
f (J, h, x) = log x cosh(h log x) + sinh (h log x) + x
is the free energy of the 1-d Ising model.
93