L´EVY CONSTANTS
ANNA BELOVA AND PETER HAZARD
Abstract. We show that the generating function corresponding to the se- quence of denominators of the best rational approximants of a quadratic ir- rational is a rational function with integer coefficients. Consequently we can compute the L´evy constant of any quadratic irrational explicitly in terms of its partial quotients.
1. Introduction
1.1. Background. The aim of this article is to show the following.
Theorem 1.1. Let θ ∈ R\Q be a quadratic irrational. For each n, let pn/qn
denote the n-th best rational approximant toθ. Then the generating functions F(z) =X
n≥0
pnzn G(z) =X
n≥0
qnzn (1.1)
are both rational with integer coefficients.
In fact, we can compute the generating functions F and G explicitly in terms of the partial convergents of θ. This generalises the case whenθ has pre-periodic continued fraction expansion of period one.
Using our result, we can easily derive the following results concerning the L´evy constantof a quadratic irrational. Recall that, given a real numberθwithnth best rational approximant given by pn/qn for eachn, the L´evy constant of θ, when it exists, is given by the following expression
β(θ) = lim
n→∞
1
nlogqn (1.2)
Paul L´evy showed, following earlier work by A. Ya. Khintchine, that β(θ) = π2
12 log 2 Lebesgue-almost everyθ (1.3) (See [5, 3, 4] for more details.) It was shown by Jager and Liardet [2] that for every quadratic irrational, the L´evy constant exists. As an immediate corollary to Theorem (1.1) above we get a new proof of the following result, which was implicitly contained in [2].
Date: October 8, 2017.
2010Mathematics Subject Classification. Primary: 11J70 11K50 ; Secondary: 05A15 . Key words and phrases. Continued fractions, Generating functions.
This work has been partially supported by CAPES Special Visiting Researcher grant CSF- PVE-S - 88887.117899/2016-00.
1
Theorem 1.2. Let θ∈R\Qbe a quadratic irrational. Letldenote the (eventual) period of the simple continued fraction expansion ofθ. Let Mθ denote the element of PSL(2,Z)corresponding to the simple continued fraction expansion of θ. Then
β(θ) = 1
l log rad(Mθ) (1.4)
where, for every A∈PSL(2,C), we denote by rad(A)the spectral radius of either of the linear transformations corresponding toA.
(The description ofMθwill be given in more detail below.)
1.2. Notation and Terminology. LetN and N0 denote the set of positive and non-negative integers respectively. Denote the set of integers by Z and, for each positive integerl letZldenote the set of integers modulol, i.e. Zl =Z/lZ. LetR and C denote, as usual, the real and complex number fields. Given an arbitrary polynomialη, over either Ror C, we denote the discriminant by discrη.
Givenθ∈R, let⌊θ⌋denote the integer partofθ,i.e., greatest integer less than or equal toθ, and let{θ}=θ− ⌊θ⌋denote thefractional partof θ.
Let Sl denote the symmetric group of size l, i.e., the permutation group on a set of l elements. We will denote the signature of a permutationυ in Sl byǫ(υ).
(Recall, any permutation can be written, non-uniquely, as a product of adjacent transpositions (k k+ 1), and the parity of the total number of such transpositions is the signature.) Such a permutationυcan be expressed as a product of cycles. A single cycle will be expressed as (s1, s2, . . . , sk), for somes1, s2, . . . , sk∈ {1,2, . . . ,l} whereυ(sj) =sj−1for eachj, where addition is taken modl.
We denote the cardinality of a setS by #S.
Acknowledgements. P.H. would like to thank the Mathematics Institute at Up- psala University for their hospitality. We thank Charles Tresser and Edson de Faria for reading an earlier draft of this article.
2. Preliminaries.
2.1. Continued fractions. In this section we recall some basic properties of simple continued fraction expansions. The main aim is to set up notation for the rest of the paper. For more details we recommend that the reader consults [3, 1].
2.1.1. Best rational approximants. Let θ ∈ [0,1]\Q. We note that the discus- sion below can be carried out also for irrational points outside [0,1], with suitable modifications. However, for simplicity we restrict ourselves to the caseθ∈[0,1]\Q.
The simple continued fraction expansion ofθis denoted by θ= [a1, a2, . . .] = 1
a1+ 1
a2+· · · 1
an+· · · (2.1)
where a1, a2, . . . are positive integers called the partial quotients of the continued fraction. Define thenth convergentofθto be
[a1, a2, . . . , an] = 1 a1+
1
a2+· · · 1 an−1+
1
an (2.2)
This is a rational number which we will express as pn/qn, where pn and qn are positive integers having no common factors. The following property is satisfied for
alln∈N
θ−pn
qn
≤p inf
q∈Q:q≤qn
θ−p q
(2.3) For this reasonpn/qn is also called the nth best rational approximant of θ. Iden- tifying C with P(C2)\ {[1 : 0]}, equation (2.2) can be expressed in matrix form as
pn
qn
=
0 1 1 a1
0 1 1 a2
· · ·
0 1 1 an−1
1 an
(2.4) Similarly
pn−1
qn−1
=
0 1 1 a1
0 1 1 a2
· · ·
0 1 1 an−1
0 1
(2.5) Thus combining (2.4) and (2.5) we find
pn−1 pn
qn−1 qn
=
0 1 1 a1
0 1 1 a2
· · ·
0 1 1 an−1
0 1
1 an
(2.6) We therefore inductively get the following recurrence relation
pn−1 pn
qn−1 qn
=
pn−2 pn−1
qn−2 qn−1
0 1 1 an
,
p0 p1
q0 q1
=
0 1 1 a1
(2.7) and, more generally, for any non-negative integerm≤n−2,
pn−1 pn
qn−1 qn
=
pn−m−2 pn−m−1
qn−m−2 qn−m−1
0 1
1 an−m
· · ·
0 1 1 an
(2.8) For each suitablenandmdefine
"
Bn−m−1(m) B(m+1)n−m−1 A(m)n−m−1 A(m+1)n−m−1
#
=
0 1 1 an−m
· · ·
0 1 1 an
(2.9) Observe that theA(m)n−m andBn−m(m) are well-defined since we have the relation
"
Bn−m−1(m) B(m+1)n−m−1 A(m)n−m−1 A(m+1)n−m−1
#
=
0 1 1 an−m
"
Bn−m(m−1) Bn−m(m) A(m−1)n−m A(m)n−m
#
(2.10) This could equally be inferred from the corresponding dual relation
"
Bn−m−1(m) Bn−m−1(m+1) A(m)n−m−1 A(m+1)n−m−1
#
=
"
Bn−m−1(m−1) Bn−m−1(m) A(m−1)n−m−1 A(m)n−m−1
# 0 1 1 an
(2.11) The following are well-known basic properties.
Proposition 2.1. Let θ ∈ [0,1]\Q. Then the best rational approximants pn/qn
satisfy the following
(1) pn−1qn−pnqn−1= (−1)n [Lagrange identity]
(2) pnp−1n = [an, an−1, . . . , a3, a2] (3) qn−1q
n = [an, an−1, . . . , a2, a1] (4) pqnn−1−1z+pz+qnn = [a1, a2, . . . , an,1/z]
Proof. The first item follows by taking the determinant of (2.6). The second and third items follow by taking the transpose of (2.6). The last item follows by apply- ing (2.6) to the vector
z 1
.
To summarize, the numerators and denominators of the convergents satisfy a recursion relation (2.6) which may be (re)stated as
pn =anpn−1+pn−2; p0= 0 p1= 1 (2.12) qn =anqn−1+qn−2; q0= 1 q1=a1 (2.13) and more generally
pn = A(1)n−1pn−1+B(1)n−1pn−2 qn = A(1)n−1qn−1+Bn−1(1) qn−2
= A(2)n−2pn−2+B(2)n−2pn−3 = A(2)n−2qn−2+Bn−2(2) qn−3
... ...
(2.14)
whereA(m)n−m andBn−m(m) are positive integers satisfying the recurrence relations Bn−m(m) =A(m−1)n−m+1 A(m)n−m=Bn−m+1(m−1) +an−m+1A(m−1)n−m+1
Bn(0)= 0 B(1)n−1= 1 =A(0)n (2.15) In fact, we have the following.
Proposition 2.2. Letθ∈[0,1]\Q. Then the coefficientsA(m)n satisfy the following A(m)n−m=an−m+1A(m−1)n−m+1+A(m−2)n−m+2 (2.16) A(m)n−m=anA(m−1)n−m +A(m−2)n−m (2.17) Proof. The first equality is just an application of (2.15). For the second equality,
multiply out the relation (2.11) and apply (2.15)
Proposition 2.3. Let θ ∈ [0,1]\Q. Then the best rational approximants pn/qn
satisfy the following
pn+m+1=pnA(m+1)n +pn−1A(m)n+1 (2.18) (−1)m+1pn−m−1=pnA(m−1)n−m −pn−1A(m)n−m (2.19) qn+m+1 =qnA(m+1)n +qn−1A(m)n+1 (2.20) (−1)m+1qn−m−1=qnA(m−1)n−m −qn−1A(m)n−m (2.21) Proof. The first and third equality follows by applying the expression for Bn−m(m) in (2.15) to (2.14). For the second and fourth equality, equations (2.8), (2.9) and (2.15), imply that
pn−1 pn
qn−1 qn
=
pn−m−2 pn−m−1
qn−m−2 qn−m−1
"
A(m−1)n−m A(m)n−m A(m)n−m−1 A(m+1)n−m−1
#
(2.22) The rightmost matrix factor on the right-hand side has determinant (−1)m+1, as the matrix is the product ofm+ 1 matrices of determinant−1. Thus, applying the inverse of the rightmost matrix to both sides and multiplying out gives the required
equalities.
Proposition 2.4. Let θ∈[0,1]\Q. Then A(m−1)n−m +A(m+1)n−m−1
= (−1)n−m−1(qn−m−1pn−1−pn−m−1qn−1−qn−m−2pn+pn−m−2qn) (2.23)
Proof. Rearranging equation (2.22) from the preceding proposition by applying an appropriate inverse and applying Proposition 2.1(1) to compute the necessary determinant gives
"
A(m−1)n−m A(m)n−m A(m)n−m−1 A(m+1)n−m−1
#
= (−1)n−m−1
qn−m−1 −pn−m−1
−qn−m−2 pn−m−2
pn−1 pn
qn−1 qn
(2.24) Taking the trace of both sides now gives the equation (2.23), as required.
2.1.2. Quadratic Irrationals. Let θ ∈ [0,1]\Q be a quadratic irrational. By this we will mean that θ is an algebraic number with minimal polynomial χ with (strict) degree two. A theorem of Lagrange [3, p. 56] implies that θ has a pre- periodic simple continued fraction expansion. Hence, there exist positive integers a1, a2, . . . , am, . . . , am+lsuch that
θ= [a1, a2, . . . , am, am+1, . . . , am+l] (2.25) We call the minimal suchltheperiod. We call any suchmapreperiodand the least such preperiod theminimal preperiodof the simple continued fraction expansion.
Remark 2.1. Observe that, sincean+l=an whenevern >m, it follows from (2.9) that, for all non-negative integersm andnsatisfying n−m >m, we have
A(m)n−m+l−1=A(m)n−m−1, Bn−m+(m) l−1=Bn−m−1(m) (2.26) Therefore the following quantities are well-defined
A(m)(j) =A(m)m+k, B(j)(m)=Bm(m)+k for any k≥0, m+k=j modl (2.27) Letθ∧m =Tm(θ), whereT denotes the Gauss transformation. ThenTl(θ∧m) = θ∧m. Thusθ∧m is a solution to the equation
θ∧m= 1 am+1+
1
am+2+· · · 1
am+l+θ∧m (2.28) andθcan be expressed as
θ= 1 a1+
1
a2+· · · 1
am+θ∧m (2.29)
These can be written in matrix form as θ∧m
1
=
0 1 1 am+1
0 1
1 am+2
· · ·
0 1 1 am+l
θ∧m 1
(2.30)
and
θ 1
=
0 1 1 a1
0 1 1 a2
· · ·
0 1 1 am
θ∧m 1
(2.31) where we identify matrices with their corresponding linear fractional transforma- tions. Let
M1=
0 1 1 am+1
0 1
1 am+2
· · ·
0 1 1 am+l
(2.32) and
M0=
0 1 1 a1
0 1 1 a2
· · ·
0 1 1 am
(2.33) LetMθ=M0M1M0−1. We call this the element of PSL(2,Z)corresponding to con- tinued fraction expansionofθ. We callM1the element of PSL(2,Z)corresponding to the periodic part of the continued fraction expansion ofθ. Thenθ is a solution
of the fixed point equationMθ(z) =z,θ∧m is a solution of the fixed point equation M1(z) =z, andθ=M0(θ∧m). From equation (2.9), equation (2.15), and applying definition (2.27) we have
M1=
"
B((lm−1)) B((ml))
A((lm−1)) A((lm))
#
=
"
A((lm−2)+1) A((lm−1)+1)
A((lm−1)) A((lm))
#
(2.34) and
M0=
"
B(0)(m−1) B((0)m) A((0)m−1) A((0)m)
#
=
"
A((1)m−2) A((1)m−1) A((0)m−1) A((0)m)
#
(2.35) To M1, or equivalently to θ∧m, there are two degree two polynomials which are naturally associated with it
χθ∧m(z) =z2−
A((lm−2)+1)+A((lm))
z+ (−1)l (2.36)
ωθ∧m(z) =A((lm−1)) z2+
A((lm))−A((lm−2)+1)
z−A((lm−1)+1) (2.37) The first is just the characteristic polynomial of M1, and thus the roots are the eigenvalues ofM1. (Observe that detM1= (−1)l asM1 is a product oflmatrices with determinant−1.) The second is the minimal polynomial ofθ∧m, and thus the roots correspond to the eigenvectors ofM1. Notice that these two polynomials have the same discriminant, and thus the roots are rationally related.
In the same way we may associate a pair of degree two polynomialsχθ and ωθ
to Mθ (as the defining polynomials for eigenvalues and eigenvectors). Since the determinant and trace are conjugacy invariant we find that
χθ(z) =χθ∧m(z) (2.38)
Also observe that the roots of ωθ are the image under M0 of the roots of ωθ∧m. Thus it follows (noticing that
A((1)m−2)−A((0)m−1)z
is the denomiator ofM0−1) that ωθ(z) =
A((1)m−2)−A((0)m−1)z2
ωθ∧m(M0−1(z)) (2.39) 3. Generating functions.
3.1. Generating functions associated to a simple continued fraction ex- pansion. Givenθ∈[0,1]\Q, denote bypn/qnthen-th best rational approximant.
Consider the generating functions F(z) =X
n≥0
pnzn G(z) =X
n≥0
qnzn (3.1)
3.2. Generating functions associated to quadratic irrationals. We now fo- cus on the case whenθ∈[0,1]\Qis a quadratic irrational. Thus, for some positive integersa1, a2, . . . , am+l∈N,
θ= [a1, a2, . . . , am, am+1, . . . , am+l] (3.2) As above, letpn/qn denote the sequence of best rational approximants and denote the corresponding generating functions by
F(z) =X
n≥0
pnzn G(z) =X
n≥0
qnzn (3.3)
The aim of this section is to prove Theorem 1.1. Before doing so, let us set up the following notation. Fork∈N0, define
Ik(z) = X
0≤n≤k
pnzn Jk(z) = X
0≤n≤k
qnzn (3.4)
Form< j≤m+l, define Fj(z) = X
n=jmodl n>m
pnzn Gj(z) = X
n=jmodl n>m
qnzn (3.5) Givenk∈N0, fork=j modl, wherem< j≤m+l, define
F(k)(z) =Fj(z) G(k)(z) =Gj(z) (3.6) First consider the generating functionF(z). Observe that
F(z) =Im(z) + X
m<j≤m+l
F(j)(z) (3.7)
More generally, for anyk∈N0, we may split the generating functionF(z) as
F(z) = X
0≤n≤m+k
pnzn+ X
m<j≤m+l
X
n=j+kmodl n>m+k
pnzn
(3.8)
Fixk ≥1. Form < j ≤m+l, using the expansion (2.14) and periodicity (2.27), we find that,
X
n=j+kmodl n>m+k
pnzn
= X
n=j+kmodl n>m+k
A(k)n−kpn−k+Bn−k(k) pn−k−1
zn (3.9)
=A(k)(j)zk X
n−k=jmodl n−k≥m+1
pn−kzn−k+B(k)(j)zk+1 X
n−k−1=j−1 modl n−k−1≥m
pn−k−1zn−k−1 (3.10)
=A(k)(j)zk X
m=jmodl m≥m+1
pmzm +B(j)(k)zk+1 X
m=j−1 modl m≥m
pmzm (3.11)
However, form< j≤m+l, we have the following equalities X
m=jmodl m≥m+1
pmzm=F(j)(z) (3.12)
and
X
m=j−1 modl m≥m
pmzm=
F(j−1)(z) j6=m+ 1
F(m)(z) +pmzm j=m+ 1 (3.13) Hence, applying (2.27), the equation (3.8) becomes
F(z) (3.14)
=Im+k(z) +B((k)m+1)pmzm+k+1+ X
m<j≤m+l
A(k)(j)zkF(j)(z) +B(j)(k)zk+1F(j−1)(z)
This expression may be rewritten, again using (2.27), as
F(z) (3.15)
=Im+k(z) +B((k)m+1)pmzm+k+1+ X
m<j≤m+l
A(k)(j)zk+B(j+1)(k) zk+1 F(j)(z) Substituting the expression (3.7) in the left-hand side and rearranging therefore gives
X
m<j≤m+l
1−A(k)(j)zk−B(j+1)(k) zk+1
F(j)(z) =Im+k(z)−Im(z) +B((k)m+1)pmzm+k+1 (3.16) For each k ≥ 1, exactly the same argument as for F(z) also yields the following relation forG(z),
X
m<j≤m+l
1−A(k)(j)zk−B(j+1)(k) zk+1
G(j)(z) =Jm+k(z)−Jm(z)+B((k)m+1)qmzm+k+1 (3.17) (In fact, the generating function of any sequence satisfying the recurrence rela- tions (2.12) will satisfy an expression of the above type with appropriately chosen initial conditions.) Let
F(z) =
F(m+1)(z) F(m+2)(z)
... F(m+l)(z)
G(z) =
G(m+1)(z) G(m+2)(z)
... G(m+l)(z)
(3.18)
and
I(z) =
Im+1(z)−Im(z) Im+2(z)−Im(z)
... Im+l(z)−Im(z)
J(z) =
Jm+1(z)−Jm(z) Jm+2(z)−Jm(z)
...
Jm+l(z)−Jm(z)
(3.19)
and, recalling the equations (2.15),
K(z) =
B(1)(m+1)zm+2 B(2)(m+1)zm+3
... B((ml)+1)zm+l+1
=
A(0)(m+2)zm+2 A(1)(m+2)zm+3
... A((lm−1)+2)zm+l+1
(3.20)
Setting
L(z) =
L11(z) L12(z) · · · L1l(z) L21(z) L22(z) ...
... . .. ...
Ll1(z) · · · Lll(z)
(3.21)
where
Lkj(z) = 1−A(k)(m+j)zk−B((k)m+j+1)zk+1 (3.22) the expressions (3.16) and (3.17) respectively become
L(z)F(z) =I(z) +pmK(z) L(z)G(z) =J(z) +qmK(z) (3.23)
Let
M(z) =
1 0 · · · 0
−z−1 z−1 0 · · · 0
0 −z−2 z−2 0 0
... . .. ... . .. ...
... . .. . .. 0
0 · · · 0 −z−l+1 z−l+1
(3.24)
(This matrix encodes certain elementary row moves.) After recalling (cf. equa- tions (2.15)) that we have the initial conditions B(1)(k) = 1, A(1)(k) = ak+1 and the relations
B((k)m+j+1)=A(k−1)(m+j+2) (3.25)
A(k)(m+j)=B(k−1)(m+j+1)+A(1)(m+j)A(k−1)(m+j+1)=A(k−2)(m+j+2)+am+j+1A(k−1)(m+j+1) (3.26) we find that, since am+l =am and am+l+1 =am+1, the matrixN(z) =M(z)L(z) can be expressed in column form as
A1−am+2zA2−z2A3 A2−am+3zA3−z2A4 · · · Al−am+1zA1−z2A2
=
A01+A12+A23 A02+A13+A24 · · · A0l +A1l+1+A2l+2
(3.27) where, throughout, addition in lower indices is taken moduloland
A0
k=Ak =
1 A(1)(m+k) A(2)(m+k)
... A((lm−1)+k)
A1
k=−am+kzAk A2
k =−z2Ak (3.28)
Define thel×l matrix
A= A1 A2 · · · Al
(3.29) We will need the following preliminary result, which follows from a straightforward proof by induction.
Proposition 3.1. Let the matrix A be defined as above. Then detA = 0 if and only if the sequence am+1, am+2, . . . , am+l has period strictly less thanl.
We are now in a position to give a proof of the main theorem. However, before proceeding we recommend that the reader consult Appendix A, which contains the solution to linear equations with polynomial coefficients of a more general type.
Remark 3.1. The above analysis works for any pre-periodm, not just the minimal pre-periodmmin. However, to achieve a rather succinct expression for the generating functions, it will become apparent that it is better to choose, for example, m = mmin+l. (We will draw attention to the two places in the proof below where this is necessary.)
Proof of Theorem 1.1. By the preceding analysis above, the equations (3.23) hold.
We will only consider the equation forF(z), as the case forG(z) is totally analogous.
Applying the matrixM(z) to both sides of the equation in (3.23), and recalling thatN(z) =M(z)L(z), gives
N(z)F(z) =M(z)(I(z) +pmK(z)) (3.30) Recall thatN(z) is expressible as (3.27). Thus, a straightforward application of the matrixN(z), together with (2.14) and (2.15) (recalling thatA(k)(m+l)=A(k)m for each k) gives us the following expressions
M(z) (I(z) +pmK(z)) =pm+1zm+1A1+pmzm+2A2 (3.31) where for these equalities we have used the initial conditions (2.12) . Therefore the equation (3.30) which we must solve becomes
N(z)F(z) =pm+1zm+1A1+pmzm+2A2 (3.32) Claim 1. For each k∈Zl the equation
N(z)E(z) =Ak (3.33)
has solutionEk with entriesEj,k given by
Ej,k(z) =uj,k(z)
v(zl) (3.34)
where, ifκ=κ(j, k) andµ=µ(j, k)are defined as in Theorem A.2, then uj,k(z) =
( z(l−1)−κA((lm−κ−1)+k) +z(l−1)+µ(−1)µA((lm−µ−1)+k−l+µ) j6=k−1
zl−1A((lm−1)+k) j=k−1 (3.35)
and
v(ξ) =1−(−1)m
qmpm+l−1−pmqm+l−1−qm−1pm+l+pm−1qm+l
ξ+ (−1)lξ2 (3.36) Proof of Claim: SinceN(z) is expressible in the form (3.27) and detA6= 0, we may apply Theorem A.2. More precisely, forj = 1,2, . . . ,l, if we take
γj0= 1 γj1=−am+j+1 γj2=−1 (3.37) we take Cj =Aj, and we set t=k, then Theorem A.2 implies that the solution, which we denote byEk(z), exists and has entriesEj,k(z),j= 1,2, . . . ,l, given by
Ej,k(z) =uj,k(z)
v(zl) (3.38)
where v(z) and uj,k(z) are polynomials in the variablez. In fact, uj,k(z) is given by
uj,k(z) =
( z(l−1)−κuκ,0j,k +z(l−1)+µu0,µj,k j6=k−1
zl−1u0,0k−1,k j=k−1 (3.39)
where the coefficientsuκ,0j,k andu0,µj,ksatisfy the recurrence relations (A.37) and (A.38) respectively. Observe that the initial values for this recurrence satisfy
ulk,k−1,0=A(0)(m+k) ulk+1,k−2,0 =A(1)(m+k) (3.40)
Observe that the recurrence relation (A.37), after making the substitutions (3.37) becomes (2.17). Moreover, the initial conditions (3.40) above agree with the initial conditions given by (2.17). Consequently, for allj= 1,2, . . . ,l, we have
uκ,0j,k =A((lm−κ−1)+k) (3.41) Next observe that
u0,k−2,kl−1 = (−1)l−1A(0)(m+k−1) u0,k−3,kl−2 = (−1)l−2A(1)(m+k−2) (3.42) so the recurrence relation (A.38) together with these initial conditions (3.42) agree with the recurrence relation and initial conditions (2.16). Thus, for allj = 1,2, . . . ,l, we have
u0,µj,k = (−1)µA((lm−µ−1)+k−l+µ) (3.43) (Note: here we have used the convention stated in Remark 3.1 above.) From this we get the expression for the numerator ofEj,k in terms ofA((lm−κ−1)+k) andA((lm−µ−1)+k−l+µ)
given by (3.35).
Now consider the denominator ofEj,k. Theorem A.1 tells us that v is of the form
v(ξ) =v0+ξv1+ξ2v2 (3.44) where the coefficients are given by
v0= Y
1≤k≤l
γk0= 1 v2= Y
1≤k≤l
γk2= (−1)l
v1= X
τ∈T+:τ6≡0,2
(−1)ǫ(υτ) Y
1≤k≤l
γkτ(k) (3.45)
The equalities (3.41) and (3.43), and the hypothesis (3.37) together with Proposi- tion A.6, in the cases=l, imply that
v1=u0,1l,l −am+l+1u0,0l,1 −u1,0l,2 =−A((lm−2)+1)−am+l+1A((lm−1)+l+1)−A((lm−2)+l+2) (3.46) (Recall that addition in lower indices is taken in Zl.) Applying the recurrence relation (2.16), after recalling that periodicity implies that A((lm)+l) = A((lm)), we therefore find that
v1=−A((lm−2)+1)−A((lm)) (3.47) By applying Proposition 2.4 we get the equivalent expression
v1=−(−1)m(qmpm+l−1−pmqm+l−1−qm−1pm+l+pm−1qm+l) (3.48) Thus equation (3.36) holds and the claim is shown. //
Linearity then gives us the solution in the particular case (3.32) above. Namely, the above claim in the casek= 1 and 2, followed by summing over allFj, we find that
F(z) =zm+1u(z)
v(zl) (3.49)
wherev is given by (3.36) anduis given by u(z) =pm+1
X
1≤j≤l
uj,1(z) +pmz X
1≤j≤l
uj,2(z) (3.50)
Collecting terms of the same degree gives the expression, using the notation (3.39) used in Claim 1 above,
u(z) =pm+1ul1,1−1,0+ X
2≤j≤l
zj−1
pm+1ulj,1−j,0+pmulj,2−j+1,0
(3.51)
+ X
l+1≤j≤2l−1
zj−1
pm+1u0,j−j,1 l+pmu0,j−j,2 l−1
+z2l−1pmu0,2ll,2−1
(Recall that the lower indices ofuκ,µj,k lie in Zl.) We investigate the four types of expressions in the above sum as follows. By the above (3.41)
pm+1ul1,1−1,0=pm+1A(0)m+k =pm+1 (3.52) For 2≤j≤l, applying (3.41) followed by equality (2.18) we find that
pm+1ulj,1−j,0+pmulj,2−j+1,0=pm+1A(j−1)m+1 +pmA(j−2)m+2 (3.53)
=pm+j (3.54)
Forl+ 1≤j≤2l−1, applying (3.43) followed by equality (2.19) gives us pm+1u0,j−j,1 l+pmu0,j−j,2 l−1= (−1)j−l
pm+1A(2m+1−(2l−j−1)l−j)−pmA(2m+1−(2l−j) l−j)
(3.55)
= (−1)l−1pm+j−2l (3.56)
(Note: here we have also used the convention stated in Remark 3.1 above.) Lastly, equality (3.43) gives
pmu0,l,2l−1=pm(−1)l−1A(0)m+1= (−1)l−1pm (3.57) Together these imply that
F(z) = (3.58)
zm+1
P
1≤j≤lpm+jzj−1+ (−1)l−1P
l+1≤j≤2lpm+j−2lzj−1 1−(−1)m
qmpm+l−1−pmqm+l−1−qm−1pm+l+pm−1qm+l
zl+ (−1)lz2l The same argument also shows,mutatis mutandis, that
G(z) = (3.59)
zm+1
P
1≤j≤lqm+jzj−1+ (−1)l−1P
l+1≤j≤2lqm+j−2lzj−1 1−(−1)m
qmpm+l−1−pmqm+l−1−qm−1pm+l+pm−1qm+l
zl+ (−1)lz2l Since the matrixAis nonsingular, it follows that a similar analysis can be made for an arbitrary right-hand side of equations (3.30). In particular, givenanyinitial conditions to the recurrence relations (2.12) and (2.13).
Corollary 3.1. Take an arbitrary preperiodic sequence aj of positive integers of preperiodl. Letrn be an arbitrary sequence satisfying the recurrence relations given in(2.12)and (2.13). (For any choice of initial conditions.) Then the corresponding generating function is rational.
Corollary 3.2. Letθ∈R\Qbe a quadratic irrational. LetM1 denote the element of PSL(2,Z)corresponding to the periodic part of the continued fraction expansion of θ. Denote the characteristic polynomial by χ. The denominator of the corre- sponding generating functionsF andGhas the form v(zl), where
v(ξ) =ξ2χ 1
ξ
(3.60) In particular,discrv = discrχ.
Proof. Recall thatM1 is given by equation (2.34). Observe that M1 is a product of l matrices of determinant −1. Thus detM1 = (−1)l. Next observe that by equations (3.47), (3.45) and (3.44)
v(ξ) = 1−
A((lm−2)+1)+A((lm))
ξ+ (−1)lξ2 (3.61)
= 1−tr(M1)ξ+ det(M1)ξ2 (3.62)
=ξ2χ 1
ξ
(3.63) Since the discriminant of an arbitrary quadratic polynomial χ is invariant under the involutionχ(ξ)7→ξ2χ(1/ξ), the result follows.
Proof of Theorem 1.2. By the Theorem of Jager and Liardet [2], the L´evy constant β(θ) = limn→∞ 1
nlogqn exists. By the Cauchy-Hadamard formula, and by continu- ity of the logarithm, 1ρ = limn→∞q1/nn , whereρdenotes the radius of convergence of the generating functionG. Thereforeβ(θ) =−logρ. Letuandvbe the functions defined in the proof of Theorem 1.1, for the generating functionG. Observe that the poles ofGmust occur at solutions tov(zl) = 0. Letv± denote the zeroes ofv, where±denotes the sign used in the quadratic formula. Thus
v±= τ±p
τ2−4(−1)l
2(−1)l (3.64)
where τ = trM1 = A(ml−2)+1 +A(ml). Observe that both v− and v+ are real. Also observe that the zero vmin with the smallest modulus (v− in the case when l is even, andv+ in the case when lis odd) is always positive and, in fact, lies in the interval (0,1). Consequentlyv1/minl is a zero ofv(zl), it is also positive real number, and moreover lies in the interval (0,1).
Claim 2. The polynomial u(z)has no zeroes in(0,1).
Proof of Claim: Rearranging the expression for numerator ofG(z) in (3.59) given above we have
u(z) = X
1≤j≤l
zj−1 qm+j+ (−1)lqm+j−lzl
(3.65) Observe that qk > qj for allk > j >0. Thus, for all positive j and allz ∈(0,1), we have the inequality qm+j >|qm+j−lzl| On the interval (0,1) the polynomialu can be expressed as a sum of positive terms, and the claim follows. //
Consequently, vmin1/l is a pole of the generating function G(z). Any other pole of G(z) must lie on the union of circles
n|z|=|v−|1/lo
∪n
|z|=|v+|1/lo
(3.66) Thus ρ = |vmin|1/l. However, by Corollary 3.2 above, it follows that 1/v− and 1/v+ are zeroes of the characteristic polynomial χ. Since vmin is the zero ofv(ξ) of minimal modulus it follows that 1/vminis the zero ofχ(ξ) of maximal modulus, i.e., an eigenvalue of maximal modulus of the matrix Mθ. The result follows.
Appendix A. Matrix Computations.
A.1. The Setup. LetC1,C2, . . . ,Cl be column vectors in Cl and define the l×l matrix
C=
C1 C2 · · · Cl
(A.1)
For 1≤r, s≤l, denote by
• Cr the matrix Cwith therth row removed.
• Crs the matrixCwith therth row andsth column removed1.
• Cs thesth column of the matrixC.
• Crs therth row of thesth column2of the matrixC.
• Crs the columnCswith therth row removed.
For eachs= 1,2, . . . ,landp= 0,1,2, take a polynomialcps(z)∈C[z] and define Cs;p(z) =cps(z)Cs+p ∀s= 1,2, . . . ,l, ∀p= 0,1,2. (A.2) (Thus, for instance,Cr
s;p(z) =cps(z)Cr
s+p, where addition in the lower index is taken modl.) Define
Ns(z) =Cs;0(z) +Cs;1(z) +Cs;2(z) ∀s= 1,2, . . . ,l (A.3) and
N(z) =
N1(z) N2(z) · · · Nl(z)
(A.4) For 1≤r, s≤l, let
Nrs(z) =h Nr
1(z) Nr
2(z) · · · \Nr
s(z) · · · Nrl(z) i
(A.5) where the hat denotes that the column is omitted. (ThusNrs(z) denotes the (r, s)th matrix minor ofN(z).)
A.2. Computation of a Determinant.
Theorem A.1. If
cps(z) =γspzp some γsp∈C, ∀s= 1,2, . . . ,l, ∀p= 0,1,2 (A.6) then
detN(z) = detC×v(zl) (A.7)
wherev is a quadratic polynomial
v(ξ) =v0+v1ξ+v2ξ2 (A.8) with coefficients given by
• v0=Q
1≤k≤lγk0
1i.e., the (r, s)th matrix minor ofC. 2i.e., the (r, s)th entry ofC.
• v1 =v1(γ10, . . .) is a polynomial in γ10, . . ., which is invariant under cyclic permutations in the lower index, but not a symmetric polynomial.
• v2=Q
1≤k≤lγk2
To prove Theorem A.1, we need to set up the following notation and terminology.
(This will be followed by some auxiliary propositions before we start the proof.) We will identify the index set {1,2, . . . ,l} for the collection of columns C1,C2, . . . ,Cl with the cyclic groupZl=Z/lZ. Thus the index set{1,2, . . . ,l}becomes endowed with addition modulo l and the cyclic order in a natural manner. We observe, importantly, that the notion of a closed (oriented) interval makes sense in this setting. Let
T =
τ:Zl→ {0,1,2} (A.9)
We will also representτinT by the corresponding string, over the alphabet{0,1,2}, given by
τ(1)τ(2)· · ·τ(l) (A.10)
As usual, we denote by 0r a (sub)string of 00· · ·0 of lengthrand define 1rand 2r similarly.
Remark A.1. The reason for introducing this notation is the following. Since the determinant is multilinear, equation (A.3)implies that
detN(z) =X
τ∈T
det C1;τ(1) C2;τ(2) · · · Cl;τ(l)
(A.11) We need to compute each summand. For each τ ∈ T, multilinearity of the deter- minant again, together with equation (A.2), implies that
det C1;τ(1) C2;τ(2) · · · Cl;τ(l)
=c(τ) det C1+τ(1) C2+τ(2) · · · Cl+τ(l) (A.12) where
c(τ) = Y
1≤k≤l
cτ(k)k (A.13)
Thus we wish to determine (i) when the determinant on the right-hand side of (A.12) is zero, (ii) if the determinant on the right-hand side of (A.12)is non-zero, calcu- late c(τ).
Forτ∈ T, defineυτ:Zl→Zl by
υτ(k) =k+τ(k) (A.14)
This gives a one-to-one correspondence betweenT and the set U =
υ:Zl→Zl|k≤υ(k)≤k+ 2 modl, ∀k∈Zl (A.15) Givenυ ∈ U, letτυ denote the corresponding element ofT. We say thatτ∈ T is decreasingatk∈Zlif
τ(k+ 1) =τ(k)−1 or τ(k+ 2) =τ(k)−2 (A.16) Ifτ is not decreasing at any point we say it isnon-decreasing. We say that υ∈ U isnon-decreasingif the correspondingτυis non-decreasing. Let
• T+ denote the set of non-decreasingτ∈ T
• U+ denote the set of non-decreasingυ∈ U.
Proposition A.1. For υ∈ U,υ∈ Sl only ifυ∈ U+.
