PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 137, Number 7, July 2009, Pages 2243–2249
S 0002-9939(09)09787-1
Article electronically published on January 28, 2009
LÉVY CONSTANTS OF TRANSCENDENTAL NUMBERS
CHRISTOPH BAXA
(Communicated by Ken Ono)
√
Abstract. We prove that every γ ≥ log 1+2 5 is the Lévy constant of a transcendental number; i.e., there exists a transcendental number α such that γ =
1
log qm (α), where qm (α) denotes the denominator of the mth converlim m
m→∞
gent of α.
1. Introduction
An irrational number is said to have a Lévy constant β(α) if the limit
1
limm→∞ m
log qm (α) exists and its value is β(α). Here qm (α) denotes the denominator of the mth convergent of the regular continued fraction expansion of α.
Classic results of A.Ya. Khintchine [10] and P. Lévy [11] say that almost every real
number has a Lévy constant and its value √
is β(α) = π 2 /(12 log 2) for almost all α.
Lévy constants satisfy β(α) ≥ log((1 + 5)/2). This follows from [0, 1, . . . , 1] =
m times
Fm−1 /Fm (where Fm denotes the mth Fibonacci number) and therefore
√
1 1 + 5 m+1
qm (α) ≥ Fm ∼ √
as m → ∞.
2
5
H. Jager and P. Liardet [9] proved that every quadratic irrationality has a Lévy
constant. C. √
Faivre [5] showed that every possible value is attained; i.e., for all
γ ≥ log((1 + 5)/2) there is an irrational α such that β(α) = γ. E.P. Golubeva
[6], [7], [8] studied Lévy constants of quadratic irrationalities and their connections
with real quadratic fields and binary quadratic forms. In a recent paper, J. Wu
[16]
proved
√ that the Lévy constants of quadratic irrationalities are dense in
log((1 + 5)/2), +∞ . This implies trivially that the set β(α) | α is algebraic,
√
α ∈
/ Q, β(α) exists is dense in log((1 + 5)/2), +∞ . It is the purpose of the
present paper to prove the following complementary result:
√
Theorem 1. (i) For every γ ≥ log((1 + 5)/2) there exist non-denumerably many,
pairwise not equivalent transcendental
α such that β(α) = γ.
√
(ii) For every γ ≥ log((1 + 5)/2) there exist non-denumerably many, pairwise
not equivalent U2 -numbers α such that β(α) = γ.
Received by the editors August 26, 2008, and, in revised form, September 29, 2008.
2000 Mathematics Subject Classification. Primary 11K50, 11J81.
c
2009
American Mathematical Society
2243
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2244
CHRISTOPH BAXA
Remarks. 1) Two numbers α1 , α2 are called equivalent if their continued fraction expansions have shapes α1 = [b0 , b1 , . . . , bk , a1 , a2 , a3 , . . .] and α2 = [c0 , c1 , . . . , c , a1 ,
a2 , a3 , . . .]. It is easy to see that equivalent numbers have the same Lévy constant
(if one of them has one).
2) Obviously part (ii) of the theorem implies part (i). Nevertheless, we will give
separate proofs of both parts as it is instructive to see how certain parameters have
to be changed when asking for a U2 -number instead of a transcendental number.
2. Continued fractions with prescribed Lévy constant
An important technical tool in our proof will be continuants. Let a1 , . . . , an be
positive integers. The continuant Kn (a1 , . . . , an ) is defined as the determinant
a1
−1
Kn (a1 , . . . , an ) =
1
a2
−1
1
a3
..
.
1
..
.
−1
..
.
.
an
In addition we set K0 := 1 and K−1 := 0.
Lemma 2. (i) Km (a1 , . . . , am ) = Km (am , . . . , a1 ).
(ii) For 0 ≤ m ≤ n we have
Kn (a1 , . . . , an ) = Km (a1 , . . . , am )Kn−m (am+1 , . . . , an )
+ Km−1 (a1 , . . . , am−1 )Kn−m−1 (am+2 , . . . , an ).
(iii) If α = [a0 , a1 , a2 , . . .], then qn (α) = Kn (a1 , . . . , an ).
Proof. Part (i) is trivial. Proofs of parts (ii) and (iii) can be found, e.g., in
O. Perron’s classic textbook [13].
We will drop the index and write K(a1 , . . . , an ) from now on.
Lemma 3. log K(a1 , . . . , an , bm , . . . , b1 ) = log K(a1 , . . . , an ) + log K(b1 , . . . , bm ) +
O(1) with an absolute implied constant.
Proof. Using Lemma 2 we get
log K(a1 , . . . , an , bm , . . . , b1 )
= log K(a1 , . . . , an )K(b1 , . . . , bm ) + K(a1 , . . . , an−1 )K(b1 , . . . , bm−1 )
= log K(a1 , . . . , an ) + log K(b1 , . . . , bm )
+ log 1 +
K(a1 , . . . , an−1 )K(b1 , . . . , bm−1 )
,
K(a1 , . . . , an )K(b1 , . . . , bm )
which implies the assertion.
Lemma 4. If α = [0, a] = [0, a, a, a, . . .], then
log qm (α) = m log[a] + O(1) = m log
a+
√
a2 + 4
+ O(1).
2
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LÉVY CONSTANTS OF TRANSCENDENTAL NUMBERS
2245
Proof. Using induction on m one sees that
1
[a]m+1 − (−[0, a])m+1
qm (α) = √
a2 + 4
for all m ≥ −1, which implies the assertion.
From now on we will use the shorthand notation
√
a + a2 + 4
.
wa = log[a] = log
2
Lemma 5. Let a < b be positive integers and wa < γ < wb . Then there exist nondenumerably many, pairwise not equivalent α = [0, a1 , a2 , a3 , . . .] such that β(α) = γ
and ai ∈ {a, b} for all i ≥ 1.
Proof. Let x ∈ (0, 1) be such that γ = (1 − x)wa + xwb and let (σk )k≥1 be an
increasing sequence of positive integers satisfying min{σ1 x, σ1 (1 − x)} ≥ 1, σk k
as k → ∞ and lim σk = +∞. These conditions imply lim k/(σ1 + · · · + σk ) = 0.
k→∞
Set
k→∞
xσ1 (1−x)σ2 xσ2 α = 0, a(1−x)σ1 , b
,a
,b
,... ,
where the notation aτ means that the partial quotient a is repeated τ times. Let
k
(1 − x)σi + xσi
i=1
Then
<m≤
k+1
(1 − x)σi + xσi .
i=1
xσ1 xσk τa τb
, . . . , a(1−x)σk , b
,a ,b
(a1 , . . . , am ) = a(1−x)σ1 , b
for some τa , τb satisfying either 1 ≤ τa ≤ (1 − x)σk+1 and τb = 0 or τa =
(1 − x)σk+1 and 1 ≤ τb ≤ xσk+1 . Using Lemmata 3 and 4 we get
xσ1 xσk τa τb
log qm (α) = log K a(1−x)σ1 , b
, . . . , a(1−x)σk , b
,a ,b
=
k xσi log K a(1−x)σi + log K b
i=1
(1)
=
k
τb + log K aτa + log K b + O(k)
(1 − x)σi wa + xσi wb + τa wa + τb wb + O(k)
i=1
= (1 − x)(σ1 + · · · + σk )wa + x(σ1 + · · · + σk )wb + O(k).
Analogously we have m = σ1 + · · · + σk + O(k), which, together with (1), implies
the assertion.
3. Transcendence criteria
The transcendence criteria we use follow ideas that originated with E. Maillet
[12] and A. Baker [2], [3]. Recent improvements of their work can be found in
papers by J.L. Davison [4] and B. Adamczewski and Y. Bugeaud [1].
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2246
CHRISTOPH BAXA
Theorem 6 (W.M. Schmidt, [14]; see also [15]). Let α ∈ R be algebraic but neither
rational nor a quadratic irrationality and δ > 0. Then there exist only finitely many
β
∈
R which are rational or quadratic irrationalities such that
|α − β| < H(β)−3−δ .
i
Remark. Here H denotes the classic absolute height; i.e., if p(X) = m
i=0 ai X ∈
Z[X] \ {0} with gcd(a0 , . . . , am ) = 1, p(β) = 0 and deg p is minimal with this
property, then H(β) = max |ai |.
0≤i≤m
Corollary 7. Let α ∈ R have a quasiperiodic but not periodic continued fraction
expansion
α = [0, a1 , . . . , aν1 −1 , aν1 , . . . , aν1 +k1 −1 λ1 , aν2 , . . . , aν2 +k2 −1 λ2 , . . .].
Here aν , . . . , aν+k λ indicates that the partial quotients aν , . . . , aν+k should be ren−1
peated λ times, i.e., νn = ν1 + i=1 λi ki . If α is algebraic, then lim qνi+1 −1 qν−3−δ
i +ki −1
i→∞
< +∞ for every δ > 0.
Proof. For i ≥ 1 we define the quadratic irrationality
βi = [0, a1 , . . ., aν1 −1 , aν1 , . . ., aν1 +k1 −1 λ1 , . . . , aνi−1 , . . ., aνi−1 +ki−1 −1 λi−1 ,
aνi , . . ., aνi +ki −1 ].
For k ≤ νi+1 −1 we have ak (α) = ak (βi ) and we may write pk /qk for pk (α)/qk (α) =
pk (βi )/qk (βi ). Now Li βi2 + Mi βi + Ni = 0 with
Li = qνi −2 qνi +ki −1 − qνi −1 qνi +ki −2
Mi = qνi −1 pνi +ki −2 + pνi −1 qνi +ki −2 − pνi −2 qνi +ki −1 − qνi −2 pνi +ki −1
Ni = pνi −2 pνi +ki −1 − pνi −1 pνi +ki −2
and therefore H(βi ) ≤ max{|Li |, |Mi |, |Ni |} < 2qν2i +ki −1 . Theorem 6 implies
qν−2
> |α − βi | > C(α, δ)H(βi )−3−δ > C(α, δ)2−3−δ qν−6−2δ
i+1 −1
i +ki −1
for a certain C(α, δ) > 0. The corollary follows immediately.
Lemma 8. Let a < b be positive integers and wa < γ < wb . Then there exist nondenumerably many, pairwise not equivalent transcendental α = [0, a1 , a2 , a3 , . . .]
such that β(α) = γ and ai ∈ {a, b} for all i ≥ 1.
Proof. Let x and (σk )k≥1 be as in the proof of Lemma 5. Set
λ1
λ2
xσ1 xσ2 α = 0, a(1−x)σ1 , b
, a(1−x)σ2 , b
,... .
Then β(α) = γ by Lemma 5. Using the notation of Corollary 7 we have
(2) kn = (1−x)σn + xσn and νn −1 =
n−1
λi (1−x)σi + xσi for all n ≥ 1.
i=1
≤ qm (α) ≤ [b] for all m ≥ 0 we get
n
log qνn+1 −1 (α) ≥ −1 +
λi (1 − x)σi + xσi
wa
m−1
Employing that [a]
m
i=1
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LÉVY CONSTANTS OF TRANSCENDENTAL NUMBERS
2247
and
−7/2
log qνn +kn −1 (α) ≥ −
n−1
7 wb λi (1 − x)σi + xσi + (1 − x)σn + xσn wa .
2 wa i=1
Because of Corollary 7, α will be transcendental if the sequence (λn )n≥1 is chosen
such that
n−1
7 wb
7 wb (1 − x)σn + xσn −
lim λn −
−1
λi (1 − x)σi + xσi
n→∞
2 wa
2 wa
i=1
= +∞.
As the sequence (λn )n≥1 can be chosen in non-denumerably many different ways,
the construction yields non-denumerably many, pairwise not equivalent α with
β(α) = γ.
Lemma 9. Keeping the notation of Corollary 7 we have
.
0 < |Ln α2 + Mn α + Nn | < 8qν4n +kn −1 qν−2
n+1 −1
Proof. Let β̄i denote the conjugate of βi . If |β̄i | ≥ 1, it follows from Li β̄i2 + Mi β̄i +
Ni = 0 that
|β̄i |2 ≤ |Li β̄i2 | = |Mi β̄i + Ni | < 2qν2i +ki −1 (|β̄i | + 1) ≤ 4qν2i +ki −1 |β̄i |
and therefore |β̄i | < 4qν2i +ki −1 , which remains true even if |β̄i | < 1. This implies
|α − β̄i | ≤ 1 + |β̄i | < 1 + 4qν2i +ki −1 < 8qν2i +ki −1
and thus
|Li α2 + Mi α + Ni | = |Li | · |α − βi | · |α − β̄i |
< qν2i +ki −1 · qν−2
· 8qν2i +ki −1 = 8qν4i +ki −1 qν−2
.
i+1 −1
i+1 −1
Lemma 10. Let a < b be positive integers and wa < γ < wb . Then there exist
non-denumerably many, pairwise not equivalent U2 -numbers α = [0, a1 , a2 , a3 , . . .]
such that β(α) = γ and ai ∈ {a, b} for all i ≥ 1.
Proof. Let x and (σk )k≥1 be as in the proof of Lemma 5 and α as in the proof of
Lemma 8. Then β(α) = γ by Lemma 5 and (2) holds. We have
H(Ln X 2 + Mn X + Nn ) = max{|Ln |, |Mn |, |Nn |} < 2qν2n +kn −1 ≤ 2[b]2(νn +kn −1) ,
where H denotes the height of a polynomial. Using Lemma 9 we can estimate
0 < |Ln α2 + Mn α + Nn | < 8qν4n +kn −1 qν−2
≤ 8[b]4(νn +kn −1) [a]−2(νn+1 −2)
n+1 −1
−Ψn
= [b]−(2(νn+1 −2)wa −4(νn +kn −1)wb −3 log 2)/wb = 2[b]2(νn +kn −1)
with
Ψn =
2(νn+1 − 2)wa − 4(νn + kn − 1)wb − 3 log 2
.
2(νn + kn − 1)wb + log 2
Replacing νn and kn by the expressions given in (2) we see that lim Ψn = +∞
n→∞
can be achieved by letting the sequence (λn )n≥1 grow sufficiently fast, which proves
the assertion.
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2248
CHRISTOPH BAXA
√
∞
Remark. As log 1+2 5 , +∞ = n=1 (wn , wn+2 ), the assertions of Theorem 1 follow
√
from Lemma 8 and Lemma 10 respectively, at least for every γ > log 1+2 5 . In the
remainder we will briefly indicate how one can take care of the remaining value w1 .
Lemma 11. If α = [a0 , a1 , a2 , . . .] has a Lévy constant, then
1 log[ai , ai+1 , ai+2 , . . .].
m→∞ m
i=1
m
β(α) = lim
Proof. Using well-known identities from the theory of regular continued fractions
one gets
m+1
[ai , ai+1 , . . .] = |qm α − pm |−1 = qm · [am+1 , am+2 , . . .] + qm−1
i=1
(where pm /qm denotes the mth convergent), which implies
m+1
i=1
qm−1 log[ai , ai+1 , . . .] = log qm + log [am+1 , am+2 , . . .] +
qm
= log qm + log [am+1 , am+2 , . . .] + [0, am , . . . , a1 ]
= log qm + log[am+1 , am+2 , . . .] + O(1).
Lemma 12. Let α = [a0 , a1 , a2 , . . .], α = [b0 , b1 , b2 , . . .] and let t(m) denote the
number of i ∈ {1, . . . , m} with ai = bi . If α has a Lévy constant, bi = ai + O(1) as
i → ∞ and t(m) = o(m) as m → ∞, then β(α ) = β(α).
Proof. We have
m
log[bi , bi+1 , . . .] =
i=1
m
i=1
=
m
log[ai , ai+1 , . . .] +
m
log[bi , bi+1 , . . .] − log[ai , ai+1 , . . .]
i=1
m
log[ai , ai+1 , . . .] + O t(m) =
log[ai , ai+1 , . . .] + o(m),
i=1
i=1
which together with Lemma 11 yields the assertion.
λ2
λ4
Lemma 13. Let a < b be positive integers and α = 0, aλ1 , b , aλ3 , b , . . . .
(i) If
7 wb
lim λ1 + · · · + λn −
(λ1 + · · · + λn−1 ) = +∞,
n→∞
2 wa
then α is transcendental.
(ii) If lim λn /(λ1 + · · · + λn−1 ) = +∞, then α is a U2 -number.
n→∞
Proof. This can be proved along the same lines as Lemmata 8 and 10.
Lemma 14. Let a be a positive integer.
(i) There exist non-denumerably many pairwise not equivalent transcendental α
such that β(α) = wa .
(ii) There exist non-denumerably many pairwise not equivalent U2 -numbers α
such that β(α) = wa .
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LÉVY CONSTANTS OF TRANSCENDENTAL NUMBERS
2249
Proof. Set α = 0, aλ1 , a+1, aλ3 , a+1, aλ5 , . . . . By letting the sequence (λ2n+1 )n≥0
grow fast enough we can guarantee that the assumptions of both Lemma 12 and
Lemma 13 are satisfied.
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Department of Mathematics, University of Vienna, Nordbergstraße 15, A-1090,
Wien, Austria
E-mail address: christoph.baxa@univie.ac.at
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