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# Continued Fraction Expansions of Two Close Irrational Numbers

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## Continued Fraction Expansions of Two Close Irrational Numbers

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### Abstract

For an irrational numberxin(0,1), we consider its approximation by continued fraction expansion and by decimal expansion. Comparing the two approximations,kn(x)is deﬁned by the number of digits of continued fraction expansion ofxthat are determined by those of then-th decimal approximation ofx, for anyn≥1. G. Lochs proved in 1964 that the ratio knn(x) converges to a constant 6 log 2 log 10

π2 asn→∞ (see ). The observable point is that we can prove the theorem without using the decimal expansion.

Focusing on this fact, we drop the base notation and take NxNx as an approximation ofxfor any integer N. This approximation converges tox asN increases like then-th decimal expansion approximation.

Therefore we deﬁne a quantityKN(x)in a similar sense, comparing the continued fraction approximation with NxNx. In this thesis, we restate the theorem of Lochs in terms ofKN(x)and logN and prove the renewed statement. Furthermore we restate and reprove several results about the distribution of KlogN(Nx) given by C. Faivre whenN=10n.

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### Contents

I Introduction . . . 1

II Preliminaries . . . 3

2.1 Basic Properties of Continued Fractions . . . 3

2.2 Ergodic Theory and Lévy Constant . . . 5

2.3 A Central Limit Theorem for the sequencekn(x) . . . 11

2.4 Transfer Operators and Measure of Exceptional Set . . . 12

III N-approximation version of Theorem of Lochs . . . 16

IV N-approximation version of Central Limit Theorem . . . 21

V N-approximation version of Conditional Probability of Exceptional Set . . . 24

References . . . 29

Acknowledgements . . . 30

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### I Introduction

For a real numberx, consider an expression

x=a0+ 1 a1+a 1

2+a3+···1

fora0,a1,a2,··· ∈N. Such expression is called the continued fraction ofx, denoted by[a0;a1,a2,···]. Fork≥0, we write pqk

k = [0;a1,a2,···,ak], which is called thek-th convergent of[a0;a1,a2,···]. Each aiis called the thei-th digit of the continued fraction.

If there is a positive integernsuch thatx= [a0;a1,···,an], we call it a ﬁnite continued fraction. If not, we call it an inﬁnite continued fraction. It is well known that a rational number has a ﬁnite continued fraction expansion and an irrational number has an inﬁnite continued fraction expansion that converges to itself. Hence for an irrational number x∈[0,1], we have its inﬁnite continued fraction expansion x= [0;a1,a2,···]. Then pqk

k converges toxask→∞, so we obtain an approximation ofx. On the other hand, we have another approximation ofx, the decimal expansion ofx. More precisely, consider ann-th decimal approximationdn(x)anden(x)ofx, which are given as

dn(x) =10nx

10nx and en(x) =10nx+1 10nx .

Sincedn(x)anden(x)are rational numbers, they have continued fraction expansions. Denoting thei-th digit ofxbyai(x), write the continued fraction expansion ofdn(x)anden(x)as:

dn(x) = [0;a1(dn(x)),a2(dn(x)),a3(dn(x)),...,ak(dn(x)),...] en(x) = [0;a1(en(x)),a2(en(x)),a3(en(x)),...,ak(en(x)),...] Now deﬁnekn(x)as

kn(x) =max{k≥0|ai(dn(x)) =ai(en(x)) for all 0≤i≤k}

and denoteai=ai(dn(x))=ai(en(x))for 0i≤kn(x). Then

dn(x) = [0;a1,a2,a3,...,akn(x),dkn(x)+1] and

en(x) = [0;a1,a2,a3,...,akn(x),ekn(x)+1], (1) where1/dkn(x)+1 =1/ekn(x)+1by the deﬁnition ofkn(x). Now, consider a set

I(a1,a2,·,an) ={y∈(0,1):a1(y) =a1,a2(y) =a2,···,an(y) =an}

fora1,···,anN. It is well-known thatI(a1,a2,·,an)forms an interval in(0,1). Then we have

x∈[dn(x),en(x)]⊂I(a1,a2,...,akn(x)), (2) which leads us to conclude that

x= [0;a1,a2,a3,...,akn(x),...].

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Therefore the ﬁrstkn(x)digits of continued fraction expansion ofxare exactly those ofdn(x)anden(x). It is natural to ask what kind of relation betweenkn(x)andnwould have. One easy observation is that as nincreases,kn(x)must increase without an upper bound so that limnkn(x) =∞. Furthermore, one can ask how fastkn(x)grows asngoes to inﬁnity. In other words, we want to see the asymptotic behavior of knn(x) asn→∞. Gustav Lochs proved the following results in 1964.

Theorem 1(Theorem of Lochs). For almost all x,

nlim

kn(x)

n = 6 log 2 log 10

π2 0.9703 with respect to the Lebesgue measure.

One observation is that using decimal notation does not play a crucial role in his proof(see pp105- 114 of ). The only effect of decimal notation is the appearance of log 10 in the fraction. Hence it is easy to guess that we can prove the theorem by using other bases and it will cause a mere change.

Furthermore, we may drop the base notation and use even a more elementary way of approximation with a large numberNas follows:

For a real numberxand a positive integerN, dN(x) =Nx

Nx and eN(x) =Nx+1 Nx .

We call this theN-approximation ofx. Note that theN-approximation ofxdoes not mean approximating xwith baseN. It is replacing 10nby an integerN. Also, note that theN-approximation converges tox asNgoes to inﬁnity.

SincedN(x)andeN(x)are rational numbers, each has a ﬁnite continued fraction, dN(x) = [0;a1(dN(x)),a2(dN(x)),a3(dN(x)),···]

and

eN(x) = [0;a1(eN(x)),a2(eN(x)),a3(eN(x)),···],

which are calledN-expansions ofx. FordN(x)andeN(x)we deﬁneKN(x)in the same sense. To avoid confusion, we useKN(x)instead ofkN(x). Note thatkn(x) =K10n(x).

This thesis’s main goal is to study the new version of Lochs’ Theorem, written in terms ofKN(x)and logNand observe the difference between the two approximations ofx. Furthermore, we introduce more advanced results about the distribution of kn(nx), which are given by C. Faivre in next section. Stating theirN-approximation version by writing them in terms of KlogNN(x) and proving them is another main goal of this thesis.

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### II Preliminaries

In this section, we introduce more advanced results about the distribution of knn(x). The theorems are the results of some dynamical properties of continued fraction. We introduce several well-known theorems and concepts related to the dynamics of continued fractions, which will be useful in later sections. Before that, we need to see the basic properties of continued fractions that are used frequently.

2.1 Basic Properties of Continued Fractions

Recall from Section 1 that for a real numberxwith its continued fraction expansion x=a0+ 1

a1+a 1

2+a+31+···

= [a0;a1,a2,···].

Fork≥0, we have itsk-th convergent pqk

k = [a0;a1,a2,···]. And for any(a1,a2,···,an)Nn, we have a setI=I(a1,a2,···,an), deﬁned by

I(a1,···,an) ={y∈(0,1)|a1(y) =a1,a2(y) =a2,···,an(y) =an}.

We are already informed that the setIforms an interval in(0,1). In this section, we prove some useful properties of pk,qk ﬁrst, by following the proofs in . Using these, we will prove that the setI is an interval in(0,1), ﬁnding its precise form.

Proposition 1. For k≥2,

pk=akpk1+pk2 and qk=akqk1+qk2. (3) Proof. We prove this by induction onk. Fork=2,

p2

q2 =a0+ 1 a1+a12, hence it is a simple calculation to see that (3) is true fork=2.

Now, assume that (3) holds for k <n and consider the case for k= n. For a continued fraction [a0;a1,a2,...], write

pk

qk = [a1;a2,...,ak].

Then we obtain

pn

qn =a0+ 1

pn−1 qn−1

, which gives

pn=a0pn1+qn1 qn=pn1

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Then by the inductive hypothesis, we get

pn1=anpn2+pn3 and

qn1=anqn2+qn3. Combined with above equation, we obtain

pn=a0(anpn2+pn3) + (anqn2+qn3)

=an(a0pn2+qn2) + (a0pn3+qn3)

=anpn1+pn2

and

qn=pn1=anpn2+pn3

=anqn1+qn2, which complete the proof.

Proposition 2. For any k≥2,

qkpk1−pkqk1= (1)k. (4) Proof. From Proposition 1, we have

pkqk1=akpk1qk1+pk2qk1

qkpk1=akqk1pk1+qk2pk1,

which are obtained by multiplying the ﬁrst and the second formulae in (3) byqk1andpk1, respectively.

Subtracting the second one by the ﬁrst one, we have

qkpk1−pkqk1=(qk1pk2−pk1qk2) for anyk≥2. Hence

qkpk1−pkqk1=(qk1pk2−pk1qk2)

= (1)2(qk2pk3−pk2qk3) ...

= (1)k, proving the proposition.

Now we are ready to prove thatI(a1,a2,···,an)forms an interval in(0,1). The interval is called a fundamental interval.

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Proposition 3. For any a1,···,anN, the set I(a1,a2,···,an)forms an interval in(0,1). More pre- cisely,

I(a1,···,an) =

⎧⎪

⎪⎩

pn+pn−1

qn+qn−1,qpnn

for odd n pn

qn,pqnn++qpn−n−11

for even n

(5) where pqk

k = [0;a1,a2,···,an], and its length is qn(qn+1qn−1).

Proof. Deﬁne a function f(z) = [0;a1,a2,···,an+z]on z∈[0,1]. Then observe thatI(a1,···,an) = f((0,1))and f is continuous so thatI(a1,···,an)forms an interval. Ifnis odd, one can easily see that f is decreasing and increasing ifnis even. Thus we obtain

I(a1,a2,···,an) =

⎧⎨

[f(1),f(0)] for odd n [f(0),f(1)] for even n. Note that Proposition 1 gives

f(1) = [0;a1,···,an+1] = pn+pn1

qn+qn1

and f(0) = [0;a1,···,an] = pn

qn. It remains to show that the length ofI is given by q 1

n(qn+qn−1). By Proposition 2,

| pn+pn1

qn+qn1 −pn

qn |=|qnpn1−pnqn1| qn(qn+qn1)

= 1

qn(qn+qn1), completing the proof.

2.2 Ergodic Theory and Lévy Constant

For an irrational numberx, we know that its partial quotient pqn

n of continued fraction converges tox asn→∞. Measuring how fast the convergence is, one can consider how fast|x−pqnn |converges to 0.

For an irrational numberx, we can writex= [0;a1,a2,...,an1,rn]with rn= [an;an+1,an+2,...].

Note that

rn=an+ 1 rn+1,

which implies thatrn−an<1. By Proposition 1 and Proposition 2, one can easily see that

|x−pn

qn |< 1

(qn1rn+qn2)(qn1an+qn2) < 1 qn2. Thus the growth of qn measures the accuracy of the approximation ofx by pqn

n. In fact, the following inequality holds forn≥1,

qn≥Gn2,

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whereGis the golden ratio(see ). Therefore it is natural to consider the following limit for an irrational numberx

β(x) = lim

n

logqn(x)

n ,

when the limit exists and ﬁnite. In that case, the limit is called Lévy constant (p. 513, ). P. Lévy proved that Lévy constant exists and has its value as 12 log 2π2 for almoast allx∈[0,1], which is stated as following (see ), and G. Lochs proved his theorem by using the theorem of Lévy.

Theorem 2(Theorem of Lévy). For almost all x∈[0,1],

nlim

logqn(x)

n = π2

12 log 2

This theorem is proved by applying a dynamical concept called Ergodic System. In this thesis, we introduce the very basics of Ergodic theory. For more details about Ergodic Theory, see .

Deﬁnition 1. Suppose that(X,B,μ)is a probability space. The measure space(X,B,μ)together with a measurable map T :X →X is called a measure-preserving system if T is measure-preserving with respect to μ, i.e. μ(T1(A)) =μ(A) for any A∈B. The map T is said to be ergodic if T1(A) =A for A∈Bimplies that μ(A) =0orμ(A) =1. A measure-preserving system(X,B,μ,T)is called an ergodic system if T is ergodic.

Consider a setX= [0,1]and the Borelσ-algebraBonX. LetT be the Gauss transformation onX, given byT(x) =1x1x. Note that the map gives a shift map on the continued fraction expansion,

T[0;a1,a2,···] = [0;a2,a3,···], thus

Tk[0;a1,a2,···] = [0;ak+1,ak+2,···].

Fork≥0, it is well known thatT preserves that the Gauss measureμ, the measure with densitylog 21 1+1x. In fact, for an interval[a,b]inX, we have

T1[a,b] =

k=1

1 b+k, 1

a+k

. Thus

μ(T1[a,b]) = 1 log 2

k=1

1

a+k b+k1

1 1+xdx

= 1

log 2

### ∑

k=1

log

1+ 1 a+k

log

1+ 1 b+k

= 1

log 2 lim

n[log(1+a+n)log(1+a)log(1+b+n) +log(1+b)]

= 1

log 2[log(1+b)log(1+a)]

= 1

log 2

b

a

1

1+xdx=μ[a,b].

Therefore, the Gauss mapT is measure-preserving with respect toμ. Hence(X,B,μ,T)is a dynamical system. In the following theorem, we show thatT is ergodic so that(X,B,μ,T)is an ergodic system.

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Theorem 3. Let X = (0,1),Bbe the Borelσ-algebra on X, andμ be the Gauss measure onB. Let T be the Gauss map on X. Then(X,B,μ,T)is an ergodic system.

Proof. It sufﬁces to show that the Gauss mapT is ergodic, which is equivalent to show that ifA∈B withT1(A) =A, then μ(A) =0 or μ(A) =1. For any(a1,a2,···,an)Nn, consider a fundamental intervalI(a1,a2,···,an). LetAbe an measurable subset in(0,1)withT1(A) =A. Note that it is enough to show that for anyA∈B,

μ(Tn(A)∩I(a1,a2,···,an))μ(A)μ(I(a1,a2,···,an)), (6) where fgmeans there exists some absolute constantsCandDsuch that

C f ≤g≤D f.

In fact, recall from Proposition 3 that m(I(a1,a2,···,an)) = qn(qn+1qn−1), where pqn

n = [0;a1,a2,···,an],

pn−1

qn−1 = [0;a1,a2,···,an1] andm is the Lebesgue measure. Then it is easy to deduce that for anyn m(I(a1,a2,···,an)) = qn(qn+1qn−1)<2n−12 from Proposition 1. Then we have

nlimm(I(a1,a2,···,an)) =0,

where the convergence is uniform. Hence the collection of subsets {I(a1,···,an)|(a1,···,an)Nn} generatesB. Therefore (6) is equivalent to say thatμ(A∩B) =μ(A)μ(B)for anyB∈B. Then, taking B= (0,1)−A, the claim provides that 0=μ(A)μ((0,1)−A), which is,μ(A) =0 orμ(A) =1.

Now it remains to prove (6). Note that it is enough to show this for an arbitrary intervalA= [d,e] (0,1). By Proposition 1,

u∈I(a1,···,an) ⇐⇒ u= [0;a1,···,an+Tn(u)] = pn+pn1Tn(u) qn+qn1Tn(u) and by proposition 3,

u∈I(a1,···,an)∩T1(A) ⇐⇒ u is between pn+pn1d

qn+qn1d and pn+pn1e qn+qn1e. Thus we have

m(I(a1,···,an)∩T1(A)) =| pn+pn1d

qn+qn1d −pn+pn1e qn+qn1e |

= (e−d) |pnqn1−pn1qn| (qn+qn1e)(qn+qn1d)

= (e−d) 1

(qn+qn1e)(qn+qn1d). Thus

m(I(a1,···,an)∩Tn(A)) =m(A)m(I(a1,···,an)) qn(qn+qn1) (qn+qn1e)(qn+qn1d) m(A)m(I(a1,···,an)).

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Noticing that there are following inequalities between the Lebesgue measuremand the Gauss measure μ:

m(B)

2 log 2 ≤μ(B) m(B)

log 2 (B∈B), we have proved thatμ(I(a1,···,an)∩Tn(A))μ(A)μ(I(a1,···,an)).

Ergodicity has useful applications on various ﬁelds. In particular, the following well known theorem called Birkhoff’s Ergoodic Theorem plays a crucial part in proving Theorem of Lévy.

Theorem 4(Birkhoff’s Erogodic Theorem). Let(X,B,T,μ)be an ergodic system and f∈L1(μ). Then for almost all x,

nlim

1 n

n1 i

### ∑

=0

f(Tix) = 1 μ(X)

f dμ. Now we can prove Theorem of Lévy.

Proof of Theorem of Lévy. Recall that the Gauss mapT is a shift map on continued fraction, T[0;a1,a2,...] = [0;a2,a3,...].

Hence forx= [0;a1,a2,...]withai1,

Tkx= [0;ak+1,ak+2,...].

So we can write

x= pn1Tnx+pn

qn1Tnx+qn

(8) and

Tnx= qnx−pn

qn1x−pn1. (9)

By plugging (9) into the numerator of (8), we have

x= −pn1qnx+pnqn1x (qn1Tnx+qn)(qn1x+qn)

= (1)n1 qn1Tnx+qn

1 qn1x+qn, which implies

qn|qn1x−pn1|= qn

qn1Tnx+qn. Since

1> qn

qn1Tnx+qn > qn

2qn =1 2, we have

1>qn|qn1x−pn1|>1 2. From (9), we can deduce that

xT xT2x...Tnx= (1)n+1(pnx−qn) =|qnx−qn|, (10)

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that is,

1>qn·xT xT2x...Tnx> 1 2. Taking log on each side of the inequalities, we have

|logqn+n

### ∑

1

i=0

logTix|<log 2 (11) Applying Birkhoff’s Ergodic Theorem on (11), we can deduce that

nlim

logqn

n =lim

n

1 n

n1 i

### ∑

=0

log(Tix)

= 1

0

logxdμ(x)

= π2

12 log 2,which will be denoted byβ,

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proving Theorem of Lévy.

Combining the concept of Lévy constants, C. Faivre proved the following results (see Theorem 2 of ):

Theorem 5. If x has a Lévy constantβ(x)and its partial quotient an(x)satisﬁes an(x) =o(αn)for all α>1, then we have

nlim

kn(x)

n =log 10 2β(x). Note that Theorem of Lochs can be induced from Theorem 5.

The equality (10) has another application on showing the result about the convergence of continued fraction expansion of an irrational number.

Proposition 4. For each n≥1and x∈[0,1], we have|logx−logqpn

n |≤ 2n−21 . Proof. We split the case into two:

Case 1:nis even For evenn, we have the following from Proposition 2 and (10):

pn1qn−pnqn1=1

(qn1x−pn1) =qn−1T1nx+qn

From (8)

x= pn1Tnx+pn

qn1Tnx+qn

= pn1Tnx

qn1Tnx+qn+ pn

qn1Tnx+qn

= pn1Tnx

qn1Tnx+qn−pn(qn1x−pn1).

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Thus

(1+pnqn1)x=pn1

Tnx

qn1Tnx+qn+pn

pn1qnx=pn1

Tnx

qn1Tnx+qn+pn

qn

pn

x= Tnx

pn(qn1Tnx+qn)+1. Taking log on both sides, we obtain

logx−logpn

qn =log

1+ Tnx

pn(qn1Tnx+qn)

. By the inequality log(1+x)≤xforx>−1,

logx−logpn

qn Tnx

pn(qn1Tnx+qn) 1

pnqn 1 2n2. The last inequality comes from the elementary property of continued fractions,

pn=anpn1+pn22pn222pn4≥ ··· ≥2n−21 and

qn=anqn1+qn22qn222qn4≥ ··· ≥2n−21, so thatpnqn2n2.

Case 2:nis odd For oddn, we have:

pn1qn−pnqn1=1

qn1x−pn1=qn−1T1nx+qn. Thus

x= pn1Tnx

qn1Tnx+qn+pnqn1x−pnqn

(1−pnqn1)x=pn1pn

Tnx

pn(qn1Tnx+qn)1

−pn1qnx=pn1pn

Tnx

pn(qn1Tnx+qn)1

. Taking log on both sides,

logx−logpn

qn =log

1 Tnx

pn(qn1Tnx+qn)

. By the inequality log(1−x)≥xforx>0,

logx−logpn

qn Tnx pn(qn1Tnx+qn) logpn

qn logx≤ − Tnx

pn(qn1Tnx+qn) Tnx

pn(qn1Tnx+qn) 1 2n2. Thus we have shown that for everyn≥1,|logx−logqpn

n |≤ 2n−12. This proposition will be useful in later section.

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2.3 A Central Limit Theorem for the sequencekn(x)

C. Faivre proved that a central limit theorem is valid for the sequence (kn(x)) in  by using Theorem of Lochs, Theorem of Lévy, and a dynamical results of digits of continued fraction.

Note that we can see the digits of continued fraction expansion of an irrational number as a sequence (an(x))n0. Considering each digitan(x)as a random variable, the sequence (an(x))n0is a sequence of random variables, in other words, a stochastic process. In particular,(an(x))n0also satisﬁes a good conditions, called a stationary process with uniform-mixing property (see p. 166 of ). See the below deﬁnitions.

Deﬁnition 2(Uniform-Mixing). A stationary process(Xn)n1 is called uniform-mixing orφ-mixing if there exist a sequence(φn)n1such tat

φn0as n→and

– for all A∈σ(X1,X2,···,Xt), B∈σ(Xt+n,···)and n≥1,|P(A∩B)−P(A)P(B)|≤φnP(A) whereσ(X1,X2,···,Xt)is theσ-algebra generated by X1,X2,···,Xt.

More precisely, the stationary process (an) satisﬁes more powerful mixing inequality (see p. 169 of )

|P(A∩B)−P(A)P(B)|≤CrnP(A)P(B),

for some positive constantsCand 0<r<1. Such uniform-mixing stationary process satisﬁes following theorem, which is proved in pp. 187-190 in .

Theorem 6. Let(Xn)n1be aφ-mixing stationary process withn1

√φn<. Consider a stationary process(Yn)n1satisfying the following:

(i). E(Yn)= 0 and Var(Yn) <

(ii). Ynis of the from Yn= f(Xn,Xn+1,···)

(iii).n1βn<, whereβn=Y1E(Y1|X1,X2,···,Xn)2. Let Sn=Y1+Y2+···+Ynfor all n≥1. Then for anyλ>0,

lim sup

n P

1maxin|Si|≥λ√ n

16K λ2 with K=supn1E(Snn2)<.

Applying Theorem 6 to the uniform-mixing stationary process(an(x))of digits of continued fraction expansions ofx, C. Faivre proved the following central limit theorem forkn(x):

Theorem 7. For any real x,

nlimm{x∈[0,1]:kn(x)−αn σ√

n ≤z}= 1

2π

z

exp(−t2/2)dt, (13) for some constantσ >0 =6 log 2 log 10

π2 , and m is the Lebesgue measure.

(20)

2.4 Transfer Operators and Measure of Exceptional Set Forε>0, consider a set

E(ε) ={|kn(x)

n −α |≥ε}, whereα= 6 log 2 log 10

π2 . Note that Theorem of Lochs says that the set E(ε)has measure zero for almost all ε andx, hence we call it an exceptional set in this thesis. The last theorem is given by C. Faivre, providing an upper bound of the measure of the exceptional set for anyε>0.

Theorem 8. For allε >0, there exist constants C>0and0 <1such that P(|kn(x)

n −α|≥ε)≤Cλlogn for any integer N.

The theorem will be proved by showing the followings:

lim sup

n

1 nlogP

kn(x)

n ≤α−ε

≤θ1(ε) (0 ) and

lim sup

n

1 nlogP

kn(x)

n ≥α+ε

≤θ2(ε) whereθ1andθ2are deﬁned by

θ1(ε) = inf

0<t<1/2

1

t+1(−t+ (α−ε)logλ(22t))<0 and

θ2(ε) = inf

η>0(η+ (α+ε)logλ(2+2η))<0.

Note that we can derive Theorem of Lochs as a corollary of this theorem with Borel-Cantelli lemma (see ), saying that this result is more powerful than Theorem of Lochs. The upper boundλ in the statement is the dominant eigenvalue of a transfer operator. The appearance of eigenvalue of trans- fer operator comes from the following proposition, which is about its connection with the measure of exceptional set.

Proposition 5. (i). For each a>0, there exists a constant C such that E

1 qn2a

≤Cλn(2+2a) (14)

(ii). For each t<12, there exist a constant C such that

E(qn2t)≤Cλn(22t). (15) .

참조

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