A refinement of a classic theorem on continued fractions

0.11 MB·English

by  P. Bengoechea

A refinement of a classic theorem on continued
fractions
Paloma Bengoechea
3
January 28, 2013
1
0
2

n
Abstract
a
J

5 We express the set of transformations occuring in two different continued
2 fraction algorithms as subspaces of PGL(2,Z) defined by certain simplelinear

inequalities. As a consequence, we improve a Hurwitz’s classic theorem on
]
T continued fractions giving, for γ PGL(2,Z), a bound depending only on γ
N ∈
for the index of the term from which the continued fractions of two irrational
.
h numbers related by γ start being identical.
t
a
m
Different algorithms of Diophantine approximation give different expansions of
[
a real number x in continued fraction. In the first section we express the set of

1 transformations appearing in the algorithm of classic positive continued fraction for
v
an irrational number x as a subspace of PGL(2,Z) defined by linear inequalities
4
4 on the action on and x. In the second section we give a similar description for
9 ∞
the transformations appearing in a different algorithm of continued fraction playing
5
. an important role in [B]. This proves a special case of a more general reduction
1
0 conjecture formulated by Zagier.
3
1 In the third section, we improve Hurwitz’s classic theorem which says that the
: classic continued fractions of two irrational numbers are the same after a finite
v
i number of steps if and only if the numbers are PGL(2,Z)-equivalent. In Hurwitz’s
X
theorem the index from which the continued fractions start being identical depends
r
a
on the irrational numbers themselves and hence it is not bounded. We give, for
γ PGL(2,Z), a bound depending only on γ for the index of the term from which
∈
the continued fractions of two irrational numbers on the orbit of γ coincide.
1 Classic continued fraction
We denote by Γ the group PGL(2,Z) andby ε and T the transformations that corre-
a b ax+b
spond to the inversion and the translation for the usual action x :=
(cid:18)c d(cid:19) cx+d
1
on the projective line:
0 1 1 1
ε = , T = .
(cid:18)1 0(cid:19) (cid:18)0 1(cid:19)
The positive continued fraction of a real number x,
1
x = n + (n Z, n 1 i 1),
0 1 0 ∈ i ≥ ∀ ≥
n +
1 1
n +
2 ...
also denoted by x = [n ,n ,...], is given by the algorithm
0 1
1
x = x, n = [x ], x = = εT−ni(x ) (i 0). (1.1)
0 i i i+1 i
x n ≥
i i
−
Clearly each x is the image of x by a matrix γ = γ Γ, given explicitly by
i i i,x
∈
0 1 0 1
γ = γ := Id, γ = γ := (i 1) (1.2)
0 0,x i i,x (cid:18)1 n (cid:19)···(cid:18)1 n (cid:19) ≥
i−1 0
− −
and recursively by
γ = Id, γ = εT−niγ (i 0). (1.3)
0 i+1 i
≥
Akeyideaistoreplacethesequence (γ ,γ ,γ ,...)ofelementsinΓbytheunordered
1 2 3
set Γ(x) = γ ,γ ,γ ,... Γ.
1 2 3
{ } ⊂
p
i
The i-th convergent of x is denoted by = [n ,...,n ]. The integers p and q
0 i i i
q
i
satisfy the recurrence
p = 0 p = 1, p = n p +p (i 0),
−2 −1 i i i−1 i−2
≥
q = 1, q = 0, q = n q +q (i 0),
−2 −1 i i i−1 i−2
≥
the equation
p q p q = ( 1)i (1.4)
i+1 i i i+1
− −
and the inequalities
(1) q q 0 for all i 0 and q > q > 0 for all i 2,
i i−1 i i−1
≥ ≥ ≥ ≥
(2) p p for all i 2 and p > p for all i 3,
i i−1 i i−1
| | ≥ | | ≥ | | | | ≥
1 1+√5 i
(3) q for all i 0,
i
≥ √5(cid:16) 2 (cid:17) ≥
1 1+√5 i
(4) p for all i 0.
i
| | ≥ √5(cid:16) 2 (cid:17) ≥
2
It is well known that any rational number p/q satisfying
p 1
x <
(cid:12)q − (cid:12) 2q2
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
is a convergent of x.
The numbers δ (i 1) defined by
i
≥ −
δ = ( 1)i(p q x) (1.5)
i i−1 i−1
− −
satisfy the recurrence
δ
i−1
δ = x, δ = 1, δ = n δ +δ with n =
−1 0 i+1 i i i−1 i
− (cid:20) δ (cid:21)
i
and the inequalities 1 = δ > δ > ... 0. If x is rational, then x = p /q for some
0 1 i i i
≥
i and the recurrence stops with δ = 0; if x is irrational, the δ are all positive and
i+1 i
converge to 0 with exponential rapidity. With these notations, one has
p p x δ
γ−1 = i−1 i−2 , γ = i−1 . (1.6)
i (cid:18)q q (cid:19) i(cid:18)1(cid:19) (cid:18) δ (cid:19)
i−1 i−2 i
The algorithm (1.1) is a reduction algorithm in the sense that the norms of the
x
vectors γ decrease to zero.
i(cid:18)1(cid:19)
r s r s
Lemma 1.1 If and are two rational numbers such that t,u > 0, x
t u t ≤ ≤ u
r s
and ru st = 1, then or is a convergent of x.
− ± t u
r s
Proof. Let x be an element in [ , ]. If one of the inequalities
t u
r 1 s 1
x < or x <
(cid:12)t − (cid:12) 2t2 (cid:12)u − (cid:12) 2u2
(cid:12) (cid:12) (cid:12) (cid:12)
r s (cid:12) (cid:12) (cid:12) (cid:12)
is satisfied, then or is a convergent of x. If not, then
t u
1 r s r s 1 1
= = x + x + ,
tu t − u t − u − ≥ 2t2 2u2
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) r s
which can only happen if t = u = 1. Then s = r+1, so [ , ] = [r,r+1] and either
t u
s p r p
0 0
= x = or = [x] = .
u q t q
0 0
(cid:3)
3
Theorem 1.2 For all x R irrational, the set Γ(x) equals W (W W ), where
1 2
∈ − ∪
W = γ Γ 1 γ( ) 0, γ(x) > 1
{ ∈ | − ≤ ∞ ≤ }
and
W = γ W γ( ) = 0, det(γ) = 1 , W = γ W γ( ) = 1, det(γ) = 1 .
1 2
{ ∈ | ∞ } { ∈ | ∞ − − }
The sets W and W have respectively exactly one and at most one element.
1 2
Proof. One easily checks that
1 1+n
0 1 − 0 if n 2,
W = − , W =  (cid:26)(cid:18) 1 n (cid:19)(cid:27) 1 ≥
1 (cid:26)(cid:18)1 1 n (cid:19)(cid:27) 2 − 0
− − 0  if n = 1.
1
∅

This proves the second statement. It is also easy to see that Γ(x) W (W W ).
1 2
⊆ − ∪
δ q
i−1 i−2
Indeed γ (x) = > 1 and γ ( ) = for i 1 with inequalities (1) imply
i i
δ ∞ −q ≥
i i−1
γ W (W W ).
i 1 2
∈ − ∪
We therefore have to show that W Γ(x) W W . Let γ Γ satisfy
1 2
⊆ ∪ ∪ ∈
1 γ( ) < 0, γ(x) > 1.
− ≤ ∞
Denote
r s u s
γ−1 = , γ = − .
(cid:18) t u (cid:19) (cid:18) t r (cid:19)
−
The conditions γ( ) < 0 and γ(x) > 0 imply that u and t have the same sign, as
∞ r s r s r s r s s r
well as ux s and tx+r, so x , ( , denotes [ , ] if or [ , ]
− − ∈ |t u| |t u| t u t ≤ u u t
r s
otherwise). By Lemma 1.1, or is a convergent of x.
t u
r
If is a convergent of x, then γ and γ−1 are
t
p p +kp q +kq p kp
γ−1 = i i−1 i , γ = i−1 i − i−1 − i (1.7)
(cid:18)q q +kq (cid:19) (cid:18) q p (cid:19)
i i−1 i i i
−
or
p p kp q kq p +kp
γ−1 = i − i−1 − i , γ = − i−1 − i i−1 i (1.8)
(cid:18)q q kq (cid:19) (cid:18) q p (cid:19)
i i−1 i i i
− − −
q
with k Z and i 0. In the case (1.7), 1 γ( ) = i−1 k < 0 if and
∈ ≥ − ≤ ∞ − q −
i
only if k = 0 and i 1, or k = 1 and i = 0, because q q . Moreover
i−1 i
≥ ≤
δ δ
i i
γ(x) = k > 1 if and only if k n 1 because = n . Hence
i+1 i+1
δ − ≤ − (cid:20)δ (cid:21)
i+1 i+1
4
1 γ( ) < 0 and γ(x) > 1 if and only if k = 0 and i 1 or n 2, k = 1 and
1
− ≤ ∞ ≥ ≥
i = 0. In the first case
q p
γ = i−1 − i−1 = γ (i 1);
(cid:18) q p (cid:19) i+1 ≥
i i
−
in the second case
1 1+n
γ = − 0 W .
(cid:18) 1 n (cid:19) ∈ 2
0
−
q
i−1
In the case (1.8), if 1 γ( ) = + k < 0, then k < 0 and γ(x) =
− ≤ ∞ q
i
δ
i
+k < 0, so γ W.
−δ 6∈
i+1
s
If is a convergent of x, then γ and γ−1 are:
u
p +kp p q p
γ−1 = i−1 i i , γ = i − i (1.9)
(cid:18)q +kq q (cid:19) (cid:18) q kq p +kp (cid:19)
i−1 i i i−1 i i−1 i
− −
or
p kp p q p
γ−1 = − i−1 − i i , γ = i − i (1.10)
(cid:18) q kq q (cid:19) (cid:18)q +kq p kp (cid:19)
i−1 i i i−1 i i−1 i
− − − −
δ
with k Z and i 0. In the case (1.9), γ(x) = i+1 > 1 if and only if
∈ ≥ δ kδ
i i+1
−
δ q
i i
k = n because = n . If k = n , then 1 γ( ) = < 0. So
i+1 i+1 i+1
(cid:20)δ (cid:21) − ≤ ∞ −q
i+1 i+1
q p
γ = i − i = γ (i 0).
(cid:18) q p (cid:19) i+2 ≥
i+1 i+1
−
q
i
In the case (1.10), if 1 γ( ) = < 0, then k < 0 and γ(x) =
− ≤ ∞ q +kq
i−1 i
δ
i+1
< 0, so γ W.
δ +kδ 6∈
i i+1
−
0 1
Let γ = be the unique element of W that satisfies γ( ) = 0 and
(cid:18)1 u(cid:19) ∞
1
det(γ) = 1. We have 1 < γ(x) = if and only if 0 < x + u < 1, namely
− x+u
u = [x], and γ = γ Γ(x).
1
− ∈
(cid:3)
Remark 1.3 For x Q, Theorem 1.2 is still true if the value is allowed for γ(x),
∈ ∞
with γ W.
∈
5
2 The slow continued fraction
We consider the slower version of the classic algorithm of reduction (1.1)
x +1 = T(x ) if x 0,
i i i
≤

 1
x = x, x =  1 = T−1ε(x ) if 0 < x 1, (2.1)
0 i+1  i i
 x − ≤
 i
 xi −1 = T−1(xi) if xi > 1.


With this algorithm the expansion of x in continued fraction is
1
x = 1 1+
± ±···± 1
1+ +1+
|n0| ··· 1
| {z } n1 1+···+1+...
| {z } n2
| {z }
where n ,n ,n ,..., are given in (1.1) and each equals the sign of n .
0 1 2 0
± | |
Each x is the image of x by a matrix γ′ = γ′ Γ given recursively by
i i i,x ∈
Tγ′ if x 0,
i i ≤

γ′ = Id, γ′ =  T−1εγ′ if 0 < x 1, (2.2)
0 i+1  i i ≤

T−1γ′ if x > 1.
 i i



Inasimilarwayasintheprevioussection, wewillreplacethesequence(γ′,γ′,γ′,...)
1 2 3
of elements in Γ by the unordered set Γ(x)′ = γ′,γ′,γ′,... Γ, given explicitly
{ 1 2 3 } ⊂
by
q +kq p kp
Γ(x)′ = i−2 i−1 − i−2 − i−1 , 1 k n . (2.3)
(cid:26)(cid:18) q p (cid:19) ≤ ≤ i(cid:27)
− i−1 i−1 i≥1
Theorem 2.1 For all x R irrational, the set Γ(x)′ equals W′ W′, where
∈ − 1
W′ = γ Γ γ( ) 1, γ(x) > 0
{ ∈ | ∞ ≤ − }
and
W′ = γ W′ γ( ) = 1, det(γ) = 1 .
1 { ∈ | ∞ − }
The set W′ has exactly one element.
1
Proof. One easily proves the second statement checking that
1 n
W′ = − 0 .
1 (cid:26)(cid:18) 1 n +1(cid:19)(cid:27)
0
−
6
δ
It is also easy to see that Γ(x)′ W′ W′. Indeed, γ′(x) = i−1 k > 0 because
⊆ − 1 i δ −
i
δ q
i−1 = n and γ′( ) = i−2 k and inequalities (1) imply γ′ W′ W′.
(cid:20) δ (cid:21) i i ∞ −q − i ∈ − 1
i i−1
We therefore have to see that W′ Γ(x)′ W′. Let γ Γ satisfy
⊆ ∪ 1 ∈
γ( ) 1, γ(x) > 0.
∞ ≤ −
As for Theorem 1.2, the conditions γ( ) < 0 and γ(x) > 0 imply that γ−1( ) or
∞ ∞
γ−1(0) is a convergent of x. Hence γ is equal to
q +kq p kp q kq p +kp
i−1 i − i−1 − i or − i−1 − i i−1 i ,
(cid:18) q p (cid:19) (cid:18) q p (cid:19)
i i i i
− −
or
q p q p
i − i or i − i
(cid:18) q kq p +kp (cid:19) (cid:18)q +kq p kp (cid:19)
i−1 i i−1 i i−1 i i−1 i
− − − −
with k Z and i 0.
∈ ≥
q
i−1
In the first case we have, when n 2, γ( ) = k 1 if and only
1
≥ ∞ − q − ≤ −
i
if k 1; when n = 1, γ( ) 1 if and only if k 0 and k = 0 for i = 1,
1
≥ ∞ ≤ − ≥ 6 6
δ
i
because q q . Moreover, γ(x) = k > 0 if and only if k n because
i−1 i i+1
≤ δ − ≤
i+1
δ
i
= n . So γ( ) 1 and γ(x) > 0 if and only if
i+1
(cid:20)δ (cid:21) ∞ ≤ −
i+1
q +kq p kp
γ = i−1 i − i−1 − i (1 k n )
(cid:18) q p (cid:19) ≤ ≤ i+1
i i
−
or, when n = 1,
1
1 n
γ = − 0 W′.
(cid:18) 1 n +1(cid:19) ∈ 1
0
−
q δ
i−1 i
In the second case, if γ( ) = + k 1, then k < 0 and γ(x) = +
∞ q ≤ − −δ
i i+1
k < 0, so γ W′.
6∈
q q
i i−1
In the third case we have γ( ) = 1 if and only if 0 < +
∞ −q +kq ≤ − q
i−1 i i
k 1, if and only if k = 0 and i 1, or k = 1 and i = 0. Moreover, γ(x) =
≤ ≥
δ
i+1
> 0 if and only if k n . So γ( ) 1 and γ(x) > 0 if and only if
i+1
δ kδ ≤ ∞ ≤ −
i i+1
−
q p
γ = i − i (i 1),
(cid:18) q p (cid:19) ≥
i−1 i−1
−
or
1 n
γ = − 0 W′.
(cid:18) 1 1+n (cid:19) ∈ 1
0
−
7
q δ
i i+1
Inthelastcase, ifγ( ) = 1, thenk < 0andγ(x) = < 0,
∞ q +kq ≤ − δ +kδ
i−1 i i i+1
so γ W′. −
6∈
(cid:3)
Remark 2.2 Theorem 2.1 proves a special case of a more general reduction con-
jecture formulated by Zagier. For a Fuchsian group G, consider a partition of P1(R)
into a finite set of intervals I , where α is a generator of G, together with a map
α
ρ : P1(R) P1(R) such that ρ = α. The map ρ can be viewed as the map
→ |Iα
x
ρ¯ : R2 R2 defined by (0,0) (0,0) and (x,y) = (0,0) α if (x : y) I .
→ 7→ 6 7→ (cid:18)y(cid:19) ∈ α
Such an algorithm is a reduction algorithm if ρ¯(x,y) (x,y) for all (x,y) R2
k k ≤ k k ∈
with equality only if (x,y) = (0,0) or α is parabolic and (x : y) ∂I . Zagier
α
∈
conjectured that, if ∆ is the diagonal x = y of R2, the map ρ¯ possesses a global
attractor set D = ∞ ρ¯r(R2 ∆) with a finite rectangular structure, on which
∩r=0 −
ρ¯ is essentially bijective and such that every point (x,y) of R2 ∆ is mapped to
−
D after finitely many iterations of ρ¯. Katok and Ugarcovici ([KU]) proved it for
the special two-parameter family of maps ρ (suggested by Zagier) defined by the
a,b
transformations of SL(2,Z)
x+1 if x < a
1
ρ (x) =  if a x < b
a,b
 −x ≤
x 1 if x b.
− ≥


3 Main theorem
For the classic continued fraction defined in Section 1, the following theorem is well
known:
Theorem 3.1 (Hurwitz) Twoirrationalnumbers x = [n ,n ,...]andy = [m ,m ,...]
0 1 0 1
are Γ-equivalent if and only if there exists s,t 0 such that n = m for all i 0.
s+i t+i
≥ ≥
The proof of this classic theorem can be found in [HW]. As we said in the
introduction, if we consider x and y belonging to a fixed irrational orbit of Γ, the
indexes s and t in Theorem 3.1 are not bounded. In the next theorem, we show that
there is a bound for s and t in terms of the matrix in Γ relating x and y.
A
Definition 3.2 For α = Q with B > 0 and (A,B) = 1, we define
B ∈
log(√5min( A ,B))
M(α) := | | +2r +3,
log((1+√5)/2)
where r is the number of convergents of α.
8
Theorem 3.3 Let γ be an element of Γ. There exists N = N(γ) such that for all
x R Q, if we write x = [n ,n ,...] and γ(x) = [m ,m ,...], then there exists
0 1 0 1
∈ −
0 s,t N with n = m for all i 0. We can take N = 3 if γ( ) = and
s+i t+i
≤ ≤ ≥ ∞ ∞
N = max(M(γ( )),M(γ−1( ))) otherwise.
∞ ∞
Proof. Denote by (x ) and (x′) the respective series defined in (1.1) for x
i i≥0 i i≥0
and γ(x). Because x = γ (x) and x′ = γ (γ(x)), if there exists s,t 0 such
s s,x t t,γ(x) ≥
that γ = γ γ, then x = x′. In this case, n = m and so γ = γ γ
s,x t,γ(x) s t s t s+1,x t+1,γ(x)
(by (1.3)). Again x = x′ and n = m . By induction we would have
s+1 t+1 s+1 t+1
γ = γ γ and n = m for all i 0.
s+i,x t+i,γ(x) s+i t+i
≥
To prove that such s,t smaller than N exist, we may bound by N the cardinal
number of the sets γ Γ(γ(x))γ and γ Γ(x)γ−1 .
{ i,x 6∈ }i≥0 i,γ(x) 6∈ i≥0
(cid:8) (cid:9)
1 b
If γ( ) = , then γ = ± with b Z. One easily sees from Theorem 1.2
∞ ∞ (cid:18) 0 1(cid:19) ∈
that
1 0
Γ(x+b)Tb = Γ(x), γ Γ( x) − if γ ( ) 0,1 ,
i,x ∈ − (cid:18) 0 1(cid:19) i,x ∞ 6∈ { }
which is the case for all i 3. So
≥
γ Γ(γ(x))γ < 3, γ Γ(x)γ−1 < 3.
i,x i≥0 i,γ(x) i≥0
|{ 6∈ }| | 6∈ |
(cid:8) (cid:9)
Suppose γ( ) = ; we will prove the inequality
∞ 6 ∞
γ γ−1 Γ(γ(x)) M(γ−1( )).
i,x 6∈ i≥0 ≤ ∞
(cid:12)(cid:8) (cid:9)(cid:12)
Applying this inequali(cid:12)ty to γ−1 instead of(cid:12)γ and γ(x) instead of x, we will obtain
γ γ Γ(x) M(γ( ))
i,γ(x) 6∈ i≥0 ≤ ∞
(cid:12)(cid:8) (cid:9)(cid:12)
and thus the theorem wi(cid:12)ll be proved. (cid:12)
p
Suppose there exists i 0 such that γ γ−1 Γ(γ(x)). We put γ−1( ) =
i,x
≥ 6∈ ∞ q
with (p,q) = 1. The description of Γ(γ(x)) given by Theorem 1.2 and the fact that
γ γ−1(γ(x)) = γ (x) > 1 reduce the supposition above to one of the following
i,x i,x
conditions:
p
(i) γ 0,
i,x
q ≥
(cid:16) (cid:17)
p
(ii) γ 1.
i,x
q ≤ −
(cid:16) (cid:17)
For the case (i) we have
p q p p q
i−2 i−2
γ = − 0.
i,x
q q p+p q ≥
(cid:16) (cid:17) i−1 i−1
−
9
p p p p p
i−2 i−1 i−2 i−1
This implies , and hence (Lemma 1.1) or is a convergent
q ∈ |q q | q q
i−2 i−1 i−2 i−1
p
of .
q
For the case (ii) we have
p q p p q
i−2 i−2
γ = − 1.
i,x
q q p+p q ≤ −
(cid:16) (cid:17) i−1 i−1
−
p p p p p p p
i−1 i−2 i−1 i−1 i−2 i−1
This implies − , and hence (Lemma 1.1) − or is
q ∈ |q q q | q q q
i−1 i−2 i−1 i−1 i−2 i−1
p − −
a convergent of .
q
p p p
i−1 i−2
If − is a convergent of , then we deduce from inequalities (1) and (2)
q q q
i−1 i−2
−
and the recurrence satisfied by the convergents that
p (n 1)p +p = p p p (i 4)
i−3 i−1 i−2 i−3 i−1 i−2
| | ≤ | − | | − | ≤ | | ≥
q (n 1)q +q = q q q (i 2).
i−3 i−1 i−2 i−3 i−1 i−2
≤ − − ≤ ≥
We deduce from inequalities (3) and (4)
log(√5 p ) log(√5q)
i | | +3, i +3 (i 4).
≤ log((1+√5)/2) ≤ log((1+√5)/2) ≥
Finally we obtain
γ γ−1 Γ(γ(x)) M(γ−1( )).
i,x 6∈ i≥0 ≤ ∞
(cid:12)(cid:8) (cid:9)(cid:12)
(cid:12) (cid:12)
(cid:3)
References
[B] Bengoechea, P.: From continued fractions and quadratic functions to modular
forms, In preparation.
[HW] Hardy, G.H.; Wright, E.M.: An Introduction to the Theory of Numbers, 4th
ed., Oxford: Clarendon Press (1965).
[KU] Katok, S.; Ugarcovici, I.: Structure of attractors for (a,b)-continued fraction
transformations, Journal of Modern Dynamics, 4 (2010), 637 - 691.
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