ON DENJOY’S CANONICAL CONTINUED FRACTION
EXPANSION
M. IOSIFESCU AND C. KRAAIKAMP
Abstract. Denjoy’s canonical continued fraction (CCF) expansion is studied,
and a number of claims by Denjoy on this expansion are proved. Furthermore,
the ergodic system underlying the CCF expansion is obtained.
1. Introduction
Let x be a real non-integer number with (regular) continued fraction expansion
(1)
1
x = a0 +
a1 +
= [a0 ; a1 , a2 , . . . ],
1
.
a2 + . .
where a0 ∈ Z is such that x − a0 ∈ [0, 1), and an ∈ N for n ≥ 1. As is well-known,
the regular continued fraction (RCF) expansion of x is finite if and only if x ∈ Q.
In this case there are two possible expansions, otherwise the expansion is unique.
Apart from the RCF expansion there are very many other continued fraction
expansions: the continued fraction expansion to the nearest integer, Nakada’s αexpansions, Bosma’s optimal expansion . . . in fact too many to mention (see [6]
and [3] for some background information).
One particular expansion, which attracted no attention whatsoever, and which is
quite different from the continued fraction expansions mentioned above, is Denjoy’s
canonical continued fraction expansion (see [2], or [1], p. 275-6 for the original paper
by Denjoy). In [2], Denjoy stated that every real number x has continued fraction
expansions of the form
(2)
x = [d0 ; d1 , d2 , . . . ],
where d0 ∈ Z is such that x − d0 ≥ 0, and the digits dn are either 0 or 1. Such
a continued fraction expansion of x is called a canonical continued fraction (CCF)
expansion of x. Since
(3)
a+
1
0+
1
b
= a + b,
Denjoy noted that the RCF expansion (1) can be changed into a CCF expansion
(2).
In this note we will prove Denjoy’s claims, and also obtain the ergodic system
underlying Denjoy’s CCF expansion.
Key words and phrases. Hurwitzian numbers, continued fractions, singularization.
2000 Math. Subject Class. 28D05 (11K55).
The first author was supported by the Nederlandse Organisatie voor Wetenschappelijk Onderzoek (NWO).
1
2
M. Iosifescu and C. Kraaikamp
2. Insertions
Denjoy’s remark (3) can be ‘translated’ into two ‘operations’ (called insertions
of type i, where i = 1, 2) on the digits of any continued fraction expansion. They
are based on the following two equations. If a, b ∈ Z, b ≥ 2, and ξ ≥ 0, then
1
1
a+
= a+
1
b+ξ
1+
1
0+
b−1+ξ
and
1
1
a+
= a−1+
.
1
b+ξ
0+
1
1+
b+ξ
In the first case we inserted 1/(1 + 1/(0+)) into a + 1/(b + ξ), while in the second
case 1/(0 + 1/(1+)) was inserted.
Now let x ∈ R \ Z, with RCF expansion (1), and let d0 ∈ Z be such that
x − d0 ≥ 0. Setting k = a0 − d0 , in case k > 0 we can apply the second insertion to
(1) just before a1 . Doing so, we get
x = [a0 − 1; 0, 1, a1 , a2 , . . . ]
as a continued fraction expansion of x. Repeating this procedure k − 1 times we
find
x = [a0 − k; (0, 1)k , a1 , a2 , . . . ],
where (0, 1)k is an abbreviation for the string 0, 1, · · · , 0, 1 of 0’s and 1’s of length
2k. For k = 0 this string is empty, i.e., we have
x = [a0 − 0; (0, 1)0 , a1 , a2 , . . . ] = [d0 ; a1 , a2 , . . . ]
(this would be the case if a0 = d0 ; note that d0 > a0 is impossible, since a0 = bxc).
Next let i ≥ 1 be the first index for which ai > 1. Applying the first insertion
before ai yields
x = [d0 ; (0, 1)k , 1i−1 , 1, 0, ai − 1, ai+1 , . . . ],
where 1k is an abbreviation for the string 1, . . . , 1 consisting of k 1’s, which is
empty if k = 0. Repeating this procedure ai − 2 times we find
x = [d0 ; (0, 1)k , 1i−1 , (1, 0)ai −1 , 1, ai+1 , . . . ].
Note that we would have obtained the same result if the second insertion was used
ai − 1 times ‘behind’ ai .
Repeating this procedure, we find for any d0 ∈ Z with d0 ≤ x the following CCF
expansion of x:
(4)
x = [d0 ; (0, 1)a0 −d0 , (1, 0)a1 −1 , 1, (1, 0)a2 −1 , 1, (1, 0)a3 −1 , 1, . . . ].
In this expansion, never two consecutive digits will both equal 0. It follows from
(4), that if x is irrational, then any CCF expansion of x is infinite and unique once
d0 is given. In case x is rational, any CCF expansion of x is finite. However, with
d0 given, two possible CCF expansions exist in this case.
Note that the first n RCF digits a1 , . . . , an of x yield a1 + · · · + an CCF digits
equal to 1 and a1 + · · · + an − n CCF digits equal to 0. Let Zk be the number of 0’s
among the first k CCF digits d1 , . . . , dk of x, i.e., Zk = #{1 ≤ i ≤ k : di = 0}, and
let Wk be the number of 1’s. Then due to Khintchine’s classical result (see [7]),
that for almost all x (with respect to Lebesgue measure λ):
n
1X
aj = ∞,
lim
n→∞ n
j=1
Denjoy’s canonical continued fraction expansions
3
we deduce that
n
X
Zk
lim
= lim
n→∞
n→∞ Wk
aj
j=1
n
X
−n
= 1
(a.e.)
aj
j=1
So in spite of the fact that the CCF expansion of x ∈ R \ Z always has more 1’s
than 0’s, we see that for almost all x there are asymptotically as many 0’s as 1’s.
We conclude this section by noting that the CCF expansion of an x ∈ Z is
x = [d0 ; (0, 1)x−d0 ]
(40 )
for any d0 ∈ Z with d0 ≤ x.
3. On quadratic irrationalities and Hurwitzian numbers
An old and classical result states that a number x is a quadratic irrationality
(that is, an irrational root of a polynomial of degree 2 with integer coefficients) if
and only if x has an RCF expansion which is eventually periodic, i.e., x is of the
form
(5)
x = [a0 ; a1 , . . . , ap , ap+1 , . . . , ap+` ],
p ≥ 0, ` ≥ 1,
where the bar indicates the period, see [4], [8] or [9] for various classical proofs of
this result. It follows from (4) that x has an eventually periodic RCF expansion of
the form (5) if and only if x has a CCF expansion of the form
x =
[d0 ; (0, 1)a0 −d0 , (1, 0)a1 −1 , 1, . . . , (1, 0)ap −1 , 1,
(1, 0)ap+1 −1 , 1, . . . , (1, 0)ap+` −1 , 1, (1, 0)ap+1 −1 , 1, . . . , (1, 0)ap+` −1 , 1, . . . ],
|
{z
}|
{z
}
period
period
= [d0 ; (0, 1)a0 −d0 , (1, 0)a1 −1 , 1, . . . , (1, 0)ap −1 , 1, (1, 0)ap+1 −1 , 1, . . . , (1, 0)ap+` −1 , 1],
where d0 ∈ Z is such, that x − d0 ≥ 0. Again the bar indicates the period. Thus
we see that x is a quadratic irrationality if and only if the CCF expansion (4) of x
is eventually periodic for every d0 ∈ Z with d0 ≤ x.
A nice generalization of the concept of eventually periodic expansions are the
so-called Hurwitzian numbers. A number x is called Hurwitzian if and only if x has
an RCF expansion of the form
x = [a0 ; a1 , . . . , ap , K1 (k), . . . , K` (k)]∞
k=0 ,
p ≥ 0, ` ≥ 1,
where a0 is an integer, the ai ’s are positive integers, and K1 (k), . . . , K` (k) are
polynomials with rational coefficients which take positive integral values for k =
0, 1, . . . , and at least one of these polynomials is non-constant, see [9]. A well-known
example of a Hurwitzian number is e = [2; 1, 2k + 2, 1]∞
k=0 . Again it is immediate
from (4) that a number x is Hurwitzian if for every d0 ≤ x the CCF expansion of
x is given by
x
= [d0 ; (0, 1)a0 −d0 , (1, 0)a1 −1 , 1, . . . , (1, 0)ap −1 , 1,
(1, 0)K1 (k)−1 , 1, . . . , (1, 0)K` (k)−1 , 1]∞
k=0 .
4
M. Iosifescu and C. Kraaikamp
4. Canonical continued fraction convergents
Let x ∈ R, and let d0 ∈ Z be such that x − d0 ≥ 0. Furthermore, let (4) (or (40 ))
be a CCF expansion of x. Finite truncation yields the sequence of CCF convergents
(Cn )n≥0 of x:
1
Cn := d0 +
d1 +
= [d0 ; d1 , d2 , . . . , dn ], n ≥ 0.
1
.
1
d2 + . . +
dn
The value of Cn is computed using the rules 1/0 = ∞ and 1/∞ = 0. For n ≥ 2
this implies that Cn equals Cn−2 when dn = 0. This means that Cn can equal ∞
(= 1/0). In order to study the CCF convergents Cn of x, we define matrices An ,
Mn , for n ≥ 0 by
µ
¶
µ
¶
1 d0
0 1
A0 :=
, An :=
, Mn := A0 A1 · · · An .
0 1
1 dn
Setting
µ
Mn :=
it follows from Mn = Mn−1 An that
µ
Mn =
rn
sn
pn−1
qn−1
pn
qn
pn
qn
¶
,
¶
,
whence qn ≥ 0, pn and qn are relatively prime, and
p−1 = d0 , p0 = 0, pn = dn pn−1 + pn−2 ,
q−1 = 0,
q0 = 1, qn = dn qn−1 + qn−2 .
These recurrence relations show again that dn = 0, for some n ≥ 1, implies that
pn = pn−2 and qn = qn−2 . In particular, if a0 − d0 > 0, then
p1 = p3 · · · = p2(a0 −d0 )−1 = 1,
p2 = d0 + 1, p4 = d0 + 2, · · · , p2(a0 −d0 ) = a0 ,
and
q1 = q3 = · · · = q2(a0 −d0 )−1 = 0
q2 = q4 = · · · = q2(a0 −d0 ) = 1.
Defining the Möbius-transformations Mn : R∗ → R∗ by
pn−1 t + pn
Mn (t) :=
, n ≥ 1,
qn−1 t + qn
we see by induction that
pn
.
qn
For the RCF expansion matrices similar to An and Mn can be defined. For x ∈ R\Z
with RCF expansion (1), setting
µ
¶
µ
¶
1 a0
0 1
B0 :=
, Bn :=
, Nn := B0 B1 · · · Bn ,
0 1
1 an
Cn = Mn (0) =
one has that
µ
¶
Pn−1 Pn
Nn =
,
Qn−1 Qn
where Qn > 0, Pn and Qn are relatively prime, and
Pn
= [a0 ; a1 , . . . , an ],
Qn
Denjoy’s canonical continued fraction expansions
5
see [6]. Since
µ
and
1 a0
0 1
µµ
¶
µ
=
0 1
1 1
¶µ
1
0
0
1
d0
1
1
0
¶ µµ
0 1
1 0
¶¶ai −1
¶µ
µ
=
0
1
1
1
¶¶a0 −d0
1
0
ai − 1 1
¶
,
we conclude (see also (4)) that
Mk(n) = Nn ,
where k(n) = a0 − d0 + 2(a1 − 1) + 1 + . . . + 2(an − 1) + 1, which implies that the
sequence (Pn /Qn )n≥0 of RCF convergents of x is a subsequence of the sequence
(Cn )n≥0 of CCF convergents of x.
Now let an+1 > 1, for some n ≥ 1. Since dk(n)+2j = 0 for 1 ≤ j ≤ an+1 − 1, we
already saw that Ck(n) = Ck(n)+2j = Pn /Qn for these values of j. This also follows
from the fact that
µµ
¶µ
¶¶j
µ
¶µ
¶
Pn−1 Pn
0 1
0 1
1 0
Mk(n)+2j = Nn
=
1 1
1 0
Qn−1 Qn
j 1
µ
¶
jPn + Pn−1 Pn
=
.
jQn + Qn−1 Qn
What can we say about Ck(n)+2j−1 , for 1 ≤ j ≤ an+1 − 1? Since
µ
¶−1
µ
¶
0 1
Pn jPn + Pn−1
Mk(n)+2j−1 = Mk(n)+2j
=
,
1 0
Qn jQn + Qn−1
we see that Ck(n)+2j−1 is a mediant convergent of x, i.e.,
Ck(n)+2j−1 =
jPn + Pn−1
.
jQn + Qn−1
Thus we see that the collection {Cn : n ≥ −1} consists of the integers d0 , . . . , a0 , of
1/0, and of all RCF and mediant convergents of x. Note that every RCF convergent
Pn /Qn of x appears an+1 times as a CCF convergent of x.
5. The Denjoy map Td
One way of finding the RCF expansion (1) of x is by using the so-called Gaussmap T : [0, 1) → [0, 1), defined by
¹ º
1
1
T (x) := −
, x ∈ (0, 1); T (0) := 0.
x
x
For x ∈ R given by (4) or (40 ), let d0 ∈ Z be such that x−d0 ≥ 0. Setting ξ = x−d0 ,
it is clear that
ξ = [0; d1 , d2 , . . . ].
Similarly to the RCF case, this CCF expansion of ξ can easily be obtained from
a suitable map Td , which we call the Denjoy-map. Let Td : [0, ∞) → [0, ∞) be
defined by
 1

− 1, x ∈ (0, 1],



x



1
Td (x) :=
− 0, x ∈ (1, ∞),


x





0,
x = 0.
6
M. Iosifescu and C. Kraaikamp
Furthermore, setting
½
d1 = d1 (ξ) :=
1, ξ ∈ (0, 1],
0, ξ ∈ (1, ∞),
and in case Tdn−1 (ξ) 6= 0, n ≥ 1:
¡
¢
dn = dn (ξ) := d1 Tdn−1 (ξ) ,
we find in case Tdk (ξ) 6= 0 for k = 0, 1, . . . , n − 1, that
ξ
=
1
1
=
d1 + Td (ξ)
d1 +
1
d2 + Td2 (ξ)
1
=
d1 +
1
.
d2 + . . +
= ··· =
.
1
dn + Tdn (ξ)
There are several algorithms yielding the RCF convergents and mediants, see for
instance [3] or [I], where such algorithms together with the underlying ergodic
systems are described. In [5], for any x ∈ [0, 1), the RCF convergents and mediants
of x are ‘generated’ in the same order—but without the duplication of the RCF
convergents—as in the case of the CCF expansion of x. The underlying map S :
[0, 1] → [0, 1] in [5] is given by
 x

, x ∈ [0, 21 ),

 1−x
S(x) =


 1 − x , x ∈ [ 1 , 1],
2
x
and ν is a σ-finite, infinite S-invariant measure with density g, given by
g(x) =
1
,
x
for x ∈ (0, 1).
Moreover, Ito showed in [5] that the dynamical system ([0, 1), S, ν) is ergodic.
It is easy to find by direct calculation that
 2
 Td (x), x ∈ [0, 12 ),
S(x) =

Td (x), x ∈ [ 12 , 1],
i.e., S can be seen as a jump transformation of Td . Due to this, the ergodic properties of S can easily be carried over to Td . Note that Td2 is used to avoid duplication
of RCF convergents. Of course, since
2(k−1)+1
T (x) = Td
(x),
1
for x ∈ [ k+1
, k1 ), k ∈ N,
which follows from (4) or by direct calculation, the ergodic properties of Td can
also be obtained from the ergodic properties of the RCF expansion.
We have the following result.
Theorem 1. The Denjoy-map Td has a σ-finite, infinite invariant measure µ, with
density f , given by
f (x) =
1
1
1(0,1] (x) +
1(1,∞) (x),
x
1+x
and the dynamical system ([0, ∞), Td , µ) is ergodic.
x ∈ [0, ∞),
Denjoy’s canonical continued fraction expansions
7
Proof. By Theorem 1.1 from [10], to prove that Td is µ-measure preserving, it is
enough to show that
¢
¡
µ Td−1 (A) = µ(A),
for every interval A ⊂ [1, ∞). Let us first assume that A ⊂ [0, 1]. In this case
µ(A) = log ab , and
1
Z a1
Z a+1
¡ −1
¢
dx
dx
µ Td (A) =
+
1
1
x
1
+x
b+1
b
b
= µ(A).
a
Next assume that A ⊂ (1, ∞). In this case
Z b
b+1
dx
µ(A) =
= log
,
1
+
x
a
+1
a
= log
and
¡
¢
µ Td−1 (A) =
Z
1
a+1
1
b+1
dx
b+1
= log
= µ(A).
x
a+1
Let A be a Td -invariant Borel set, i.e., Td−1 (A) = A. In order to show that Td
is ergodic with respect to Lebesgue measure λ (and therefore also invariant with
respect to µ, since λ and µ are equivalent), we should show that either λ(A) = 0
or λ(Ac ) = 0, where Ac = [0, ∞) \ A.
Setting A1 = A ∩ [0, 1], A2 = A ∩ (1, ∞), we have
¡
¢
A1 = Td−1 (A1 ) ∩ [0, 1] ∪ Td−1 (A2 ),
and
A2 = Td−1 (A1 ) ∩ (1, ∞).
Then
S −1 (A1 ) = A1 ,
i.e., A1 is an S-invariant set. Since ([0, 1), S, ν) is a ergodic dynamical system we
see that either ν(A1 ) = 0, or ν([0, 1) \ A1 ) = 0, which implies that either λ(A1 ) = 0,
or λ([0, 1) \ A1 ) = 0. In case λ(A1 ) = 0 we¡ clearly have λ(A¢ 2 ) = 0, hence λ(A) = 0.
In case λ([0, 1) \ A1 ) = 0, it follows from Td−1 ([0, 1) \ A1 ) ∩ (1, ∞) = (1, ∞) \ A2 ,
that λ((1, ∞) \ A2 ) = 0, hence λ([0, ∞) \ A) = 0.
2
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M. Iosifescu and C. Kraaikamp
Romanian Academy, “Gheorghe Mihoc” Centre for Mathematical Statistics, Casa
Academiei Romane, Calea 13 Septembrie nr. 13, RO-76117 Bucharest 5, Romania
E-mail address: [email protected]
Technische Universiteit Delft, ITS (CROSS), Thomas Stieltjes Institute for Mathematics, Mekelweg 4, 2628 CD Delft, the Netherlands
E-mail address: [email protected]