Download Recurrence of incomplete quotients of continued fractions

Recurrence of incomplete quotients of
continued fractions
I.D. Shkredov
1. Statement of main result. The continued fraction corresponding to
a number α is, by denition, the expression α = a0 + a1 + 1 1 = [a0 ; a1 , a2 , . . .].
a2 +...
The number an is called nth incomplete quotient of α, and the rational
number pk /qk = [a0 ; a1 , a2 , . . . , ak ] the k th convergent of the fraction. If α ∈
[0, 1), then we write α = [a1 , a2 , . . .]. The information needed about continued
fractions can be found in [1].
We consider the set ΛM of continued fractions with bounded incomplete
quotients an ≤ M and dene a measure on them in the following manner.
The measure of a set {α | an1 ∈ A1 , . . . , ank ∈ Ak }, Ai ⊆ {1, . . . , M } is equal
Q
to |Ai |/M k , where |Ai | denotes the number of elements in Ai . The following
metric result about the recurrence of incomplete quotients is an immediate
consequence of the theorem on a symbolic ow [2]:
For almost all (with respect to the measure dened above) α = [a1 , a2 , . . .] ∈
ΛM there exists on increasing index sequence {nν }ν∈N such that
a1 = anν +1 , a2 = anν +2 , . . . , akν = anν +kν
(1)
and kν ≥ logM nν . On the other hand, for any ε > 0 the measure of the
α ∈ ΛM for which there exists an increasing sequence {nν }ν∈N satisfying (1)
and the inequality kν ≥ (1 + ε) logM nν is equal to zero.
We now state our main result.
Theorem. 1) For any ε > 0 and for almost all numbers α (with respect
to Lebesgue measure) there exists an increasing index sequence {nν }ν∈N satisfying (1) and kν ≥ (6 ln 2/π 2 − ε) · ln nν .
2) For any δ > 0 the measure of the α for which there exists an increasing
sequence {nν }ν∈N satisfying (1) and kν ≥ (1 + δ)/ ln 2 · ln nν is equal to zero.
2. Numerical recurrence. Let X be a metric space with metric d(·, ·)
and a σ - algebra Φ of measurable sets containing all Borel sets. Moreover,
1
let T be a measurable map of X into itself preserving the measure µ. It is
assumed everywhere below that µ(X) = 1. We consider the usual Hausdor
measure Hs (·) on X , and we say that the measure µ and Hs are congruent
if any µ measurable set is also Hs measurable.
The theorem below was proved by Boshernitzan in [3]. (A similar result
was obtained independently by Moshchevitin in [4]).
Theorem (Boshernitzan). Let X be a metric space, Hs and µ be congruent, let µ(A) = Hs (A) for any µ measurable set A, and let T be a map of
X into itself preserving the measure µ. Then lim inf n→∞ {n1/s · d(T n x, x)} ≤ 1
for almost all x ∈ X .
3. The space of sequences. As is well-known, the Gauss transform
T x = {1/x}, x 6= 0, T 0 = 0 is the left shift in the
continued fraction expanR
1
sion, and it preserves the measure ν(A) = 1/ ln 2 A 1+x
dx, which is equivalent
to Lebesgue measure.
We denote by Pk the following map from the space of innite continued
fraction into the nite continued fractions: Pk ([a1 , a2 , . . .]) = [a1 , . . . , ak ]. The
interval Ik = Ik (a1 , . . . , ak ) of rank k is dened to be the set of numbers α
whose continued fractions expansions start with a1 , . . . , ak . We consider the
following non-Archimedean metric on the space of sequences [a1 , a2 , . . .]:
(
D([a1 , a2 , . . .], [b1 , b2 , . . .]) =
1,
if P1 ([a1 , a2 , . . .]) 6= P1 ([b1 , b2 , . . .]),
ν(Ik (a1 , . . . , ak ))
otherwise,
where k ≥ 1 is the maximal integer such that Pk ([a1 , a2 , . . .]) = Pk ([b1 , b2 , . . .]).
The measures H1 and ν are congruent. It is sucient to show this on
Cn1 ,...,nk (A1 , . . . , Ak ) = {α | an1 ∈ A1 , . . . , ank ∈ Ak }, where Ai ⊆ N. By
covering any such cylinder C with elementary cylinders Ik (a1 , . . . , ak ), we
can verify that ν(C) ≤ H1 (C). Conversely, it follows from the denition of
the metric D(·, ·) that any covering of C can be replaced with a covering by
elementary cylinders. This implies the equality of H1 and ν .
4. Proof of the theorem. 1) Since H1 and ν are congruent, Boshernitzan's theorem implies that for almost all α, and consequently for their
corresponding continued function expansions α = [a1 , a2 , . . .], there exists an
index sequence {nν } such that
nν · D(ω, T nν ω) ≤ 1.
(2)
By the denition of D(·, ·) we have D(ω, T nν ω) = ν(Ik ), where k = kν is the
maximal integer such that a1 = anν +1 , a2 = anν +2 , . . . , akν = anν +kν .
2
2
Let γ = eπ /(12 ln 2) . By a wellknown theorem of Khinchin and Levy [1],
for any ε > 0 and for almost all numbers α there exists an M = M (α) such
that qn < (γ + ε)n , , starting from n = M . Since {kν } increases, n1 can be
assumed large enough that k1 > M . Then [1] ν(Ik ) = O(|Ik |). Consequently,
kν ≥ (6 ln 2/π 2 − ε1 ) · ln nν by (2).
2) Following [1], we denote by Mn (x) the measure of the set of numbers
α in [0, 1] such that a1 = r1 , a2 = r2 , . . . , ak = rk ; zk+n < x, where zn+k =
[an+k+1 , an+k+2 , . . .] and 0 ≤ x ≤ 1. Then for the satisfaction of the conditions
a1 = r1 , a2 = r2 , . . . , ak = rk ; an+k+1 = r1 , an+k+2 = r2 , . . . , an+2k = rk (3)
it is necessary and sucient that zk+n belong to the k thrank interval Ik =
I(r1 , . . . , rk ) = (δ1 , δ2 ), δ1 = pk /qk , δ2 = (pk + pk−1 )/(qk + qk−1 ), of length
1/(qk (qk +qk−1 )). We denoteR the set of numbers satisfying (3) by Pn (r1 , . . . , rk ).
Then µ(Pn (r1 , . . . , rk )) = δδ12 M 0 n (x)dx. As is wellknown ([1], p. 101), the
function sequence M0 0 (x), M1 0 (x), . . . satises Kuz'min's functional equation,
and by [1], Theorem 33, the measure of Pn (r1 , . . . , rk ) is
|
1
ln 2(qk (qk + qk−1 ))
log (1 ±
√
√
|Ik |
−λ n
) ¿ |Ik |2 + |Ik |e−λ n , (4)
pk )| + O(|Ik |e
1 + qk
where λ > 0. Take any δ > 0 and consider the sets PN of those α for which
a1 = aN +1 , a2 = aN +2 , . . . , aK = aN +K ,
T
S
and K = [log2 N 1+δ ]. Then the set E = M ≥1 N ≥M PN is the set of numbers
for which (1) is satised innitely many times. We estimate the measure of
P
P
each PN , µ(PN ) = r1 ,...,rK µ(PN −K√(r1 , . . . , rk )). Since r1 ,...,rK |IK | = 1, we
P
have µ(PN ) ¿ r1 ,...,rK |IK |2 + e−λ N −K by (4).
P
Let Sk = r1 ,...,rk |Ik |2 . We prove by induction that Sk ≤ (1/2)k . This
is obvious for S0 . On the other hand, Sk is the sum of areas of squares
with sides equal to lengths of intervals of rank k . We take any such interval
Ik = (pk /qk , (pk + pk−1 )/(qk + qk−1 )). It can be partitioned into intervals
1
2
1
Ik+1
, Ik+1
, . . . of rank k + 1 with decreasing length. Now |Ik+1
|/|Ik | ≤ 1/2.
P
Consequently, by the induction hypothesis, Sk+1 ≤ 1/2 Ik |Ik |2 = Sk /2 ≤
(1/2)k+1 , which was required. Therefore,
∞
∞
∞
√
X
X
X
e−λ N −(1+δ) log2 N < ∞.
1/N 1+δ < ∞ and
µ(PN ) ≤
N =1
N =1
N =1
Hence the measure of E is zero by BorelCantelli lemma. The theorem is
proved.
3
Bibliography
[1] A.Ya. Khinchin, Continued fractions, 4th ed., Nauka, Moscow 1978;
English transl., Univ. Chicago Press, ChicagoLondon 1964.
[2] F. Spitzer, Principles of random walk, Van Nostrand, PrincetonToronto
London 1964.
[3] M.D. Boshernitzan, Invent. Math. 113 (1993), 617631.
[4] N.G. Moshchevitin, Uspekhi Mat. Nauk 53:1 (1998), 223224; English
transl., Russian Math. Surveys 53 (1998), 219220.
4