J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
T
AKAO
K
OMATSU
On inhomogeneous diophantine approximation with
some quasi-periodic expressions, II
Journal de Théorie des Nombres de Bordeaux, tome
11, n
o2 (1999),
p. 331-344
<http://www.numdam.org/item?id=JTNB_1999__11_2_331_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
On
inhomogeneous
Diophantine approximation
with
somequasi-periodic
expressions,
II
par TAKAO KOMATSU
RÉSUMÉ. On s’intéresse aux valeurs de
M(03B8,~)
=lorsque
03B8 est un réel ayant undéveloppement
en fraction continuequasi-périodique.
ABSTRACT. We consider the values
concerning
M(03B8, ~)
=~~
where the continued fraction
expansion
of 03B8 has aquasi-periodic
form. In
particular,
we treat the cases so that eachquasi-periodic
form includes no constant.
Furthermore,
wegive
somegeneral
conditions
satisfying
M (03B8, ~)
= 0.1. INTRODUCTION
Let 0 be irrational
and 0
real. We supposethroughout
that isnever
integral
for anyinteger
q. Define the value of the functionwhich is called
inhomogeneous approximations
constant for thepair
0, 0.
It is convenient to introduce the functionsThen
M(8,cP)
= Several authors have treatedM (0, q5)
orM+(8, Ø)
by
using
their ownalgorithms
(See
[1], [2], [4], [5],
[11]
e.g.),
but it has been difficult to find the exact values ofM (0,
Ø)
forthe concrete
pair
of 8 and~.
Forexample, Cusick,
Rockett and Szusz([2])
obtainwhen 0 =
(1 +
~)/2
=(1;1,1, ... ~.
And author([5])
obtainswhen 0 =
( a2 + 4 - a)/2
=[0;
a,a, ... ~.
However,
it is not easy toapply
these methods to find the value
.Nl(e, ~)
about the differenttypes
of 8. In[6]
author establishes therelationship
between,~l~l (8,
Ø)
and thealgo-rithm of
Nishioka,
Shiokawa and Tamura. If we use thisresult,
we canevaluate the exact value of
M (0,
Ø)
for anypair
of 0and 0
at least when 0 is apositive
real root of thequadratic equation and 0
EQ(8).
Forexample,
are
given
when 0 =( ab(ab +
4) -
ab) / (2a) = [0;
a,b,
a,b, ... .
Furthermore,
in[7]
author is so successfulapplying
theNishioka-Shiokawa-Tamura
algorithm
that the exact value ofM (0, 0)
can be calculated evenif 0 is a Hurwitzian
number,
namely
its continued fractionexpansion
has aquasi-periodic
form. And it is the first time to find a concretepair
of 8and 0
so that = 0. Forexample,
for apositive integer s
is
given.
In this paper we consider the cases so that each
quasi-periodic
form2. NST ALGORITHM
We first introduce the NST
algorithm
(~9~). 8
= a2, - - -] denotes
the continued fraction
expansion
of9,
whereThe k-th
convergent
=ao;
at, ... , ak ~
]
of 6 is thengiven by
therecurrence relations
Denote 0
=0
[
bo; bl, b2, ... , ~
]
be theexpansion of q5
in terms of thesequence
~80, ~1, ... },
whereThen, 0
isrepresented by
where
Dk
=(-1)~‘6oB1...
8k. Now,
thefollowing
theorem isestablished in
(6~.
Theorem 1.
where
Bn =
:E~==1
Remark. It is also known in
[6]
that +~~~
= and + _(1 -
Together
withJ1~1+(B,~)
=M-(C,1 - ~),
one canobtain the value
~l~t(B, ~).
3. THE CASE
,M~B, ~~
= 0Continued fraction
expansions
of the formare called Hurzuitziarc if co is an
integer,
cl, ... , arepositive integers,
...,
Qp(k)
arepolynomials
with rational coefficients which takesis not constant.
Ql(k),
... ,Qp(k)
are said to form aquasi-period.
Theexpansions
where s is a
positive integer
with s >2,
are well-knownexamples
(See
[3],
[8],
[10]
e.g.).
In[7]
fora positive integer s
we haveM( e1/8,
(el/8 -1)/2)
=0,
M(e’/-’, 1/2)
=1/8
andM(eI/8,
1/3)
=0if~=2(mod 3); 1/18
otherwise.Then,
what is the condition such that,Jt~l (9, ~) -
0 holds? It seemsthat a non-constant
polynomial
in aquasi-periodic
part
influences whetherM (0, Ø)
= 0 or not.So,
we consider the cases eachquasi-periodic
formincludes no constant.
where s is a
positive
integer,
orwhere s is an odd
positive integer
with s >3,
is one of the well-knownexamples
(See [10]
e.g.).
In any of two
expansions
of 0 above ak isincreasing
and ak - oo(k
-oo).
So,
one may beapt
toconjecture
thatJl~l(8, ~)
= 0 for almost all of0.
But,
there is a casesatisfying
.M(~,~) 7~
0.Theorem 2.
x B
Proo f .
First,
note that in theexpansion
of ~ Iyielding
It is convenient to see that
and
Hence,
and
yielding
thatM - (0, q5)
=1/4.
Next,
1 - ~ =
(1 - 0)/2
=1/(e1~9
+1)
isexpanded
asand
In a similar manner,
by
’1Contrary
to thisresult,
thereis,
of course, a casesatisfying
M(0,
§)
= 0.Theorem 3.
Remark. It is
interesting
to see that in[7]
, - , B.in
comparison
with Theorem 2 above and this Theorem.Proof. 0
=1/2
isexpanded
asone finds that
We shall show one more case
satisfying
,NI (8,
§)
= 0.Theorem 4.
/ -, "
Proof.
When s - 0(mod 3), ~
=1/3
isexpanded
asas n tends to
infinity.
Hence,
we haveM - (0, 1/ 3)
= 0.1 -
ql
=2/3
isexpanded
asas n tends to
infinity.
Hence,
we haveM+(9,
1/3)
= 0.When 5=1
(mod 3), ~
=1/3
isexpanded
asas n tends to
infinity.
Since for n=1, 2, ...
one finds that
i a.__1I
as n tends to
infinity.
Hence,
we haveM - (0, 1/3)
= 0.as n tends to
infinity.
Since for n =1, 2, ...
1
one finds that
as n tends to
infinity.
Hence,
we have.M+(~ 1/3)
= 0.Therefore,
M(0, 1/3)
= 0. When s - 2(mod 3), ~
=1/3
isexpanded
asas r~ tends to
infinity.
Since for n =1, 2, ...
one finds that
and
4>0
=1/3,
for n =1, 2, ...
as n tends to
infinity.
Since for n =1, 2,
...one finds that
as n tends to
infinity.
Hence,
we haveM+(0,
1/3)
= 0.Therefore,
M (0, 1/ 3)
= 0.Let us calculate
M (0, 0)
whenwhere s is an odd
positive
integer
with s > 3. The situations are a littlebit different from the
previous
results. Notice that an =(2n -
1)s
~ oo,so
On-1 =
1/(an
+8n) -7 0 (rt
=1, 2, ...
-oo).
I
= 1and
qn-, IDn-1 I
= 0 hold for this 8 too. The first result isquite
different from Theorem 2.
and
Since for n =
1, 2, ...
one finds that
and
Since for n =
1, 2, ...
one finds that
Proof. 0
=1/2
isexpanded
asand
as n tends to
infinity. Therefore,
we haveM(~, 1/2) = M±(O, 1/2)
= 0. 0Theorem 7. , ,
-
,-with
s 0, s 1, s - 2
(mod 3),
respectively.
05. SOME CONDITIONS SATISFYING
cp)
= 0We have
already
seen severalexamples
so thatM (0, q5)
= 0 holds.Then,
what is the condition of
M(0,
cp)
= 0? Of course, thefollowing
is clear.Theorem 8.
If
§n -
0 or 1(n
-3f or infinitely
man ypositive
integers
n, then = 0.Proof.
First,
we shall show thatBnlDn-11
[
4 for anypositive
we obtain
On the other
hand,
’--’
Corollary.
When bn =1,
On-, -+
0 i f
and 1(n -
oo).
This is very generous.
So,
we state thefollowing.
Theorem 9.
If
I an -
c and an -+ 00(n
-~oo)
f or infinitely
manypositive
integers
n, thenM (0, §)
= 0.Here,
c is a constant notdepending
npon n.
Remark. In
fact,
an =bn
-~ oo(n
-oo)
holds in allprevious
theorems aboveimplying
JI~! (8,
Ø)
= 0.0
REFERENCES
[1] J. W. S. Cassels, Über limx~+~x|~x + 03B1 2014 y|. Math. Ann. 127 (1954), 288-304.
[2] T. W. Cusick, A. M. Rockett and P. Szüsz, On inhomogeneous Diophantine approximation.
J. Number Theory 48 (1994), 259-283.
[3] C. S. Davis On some simple continued fractions connected with e. J. London Math. Soc.
20 (1945), 194-198.
[4] R. Descombes Sur la répartition des sommets d’une ligne polygonale régulière non fermée. Ann. Sci. École Norm Sup. 73 (1956), 283-355.
[5] T. Komatsu On inhomogeneous continued fraction expansion and inhomogeneous
Dio-phantine approximation. J. Number Theory 62 (1997), 192-212.
[6] T. Komatsu On inhomogeneous Diophantine approximation and the Nishioka-Shiokawa-Tamura algorithm. Acta Arith. 86 (1998), 305-324.
[7] T. Komatsu On inhomogeneous Diophantine approximation with some quasi-periodic
ex-pressions. Acta Math. Hung. 85 (1999), 303-322.
[8] K. R. Matthews and R. F. C. Walters Some properties of the continued fraction expansion
of
(m/n)e1/q.
Proc. Cambridge Philos. Soc. 67 (1970), 67-74.[9] K. Nishioka, I. Shiokawa and J. Tamura Arithmetical properties of a certain power series. J. Number Theory 42 (1992), 61-87.
[10] O. Perron Die Lehre von den Kettenbrüchen. Chelsea reprint of 1929 edition.
[11] V. T. Sós On the theory of Diophantine approximations, II. Acta Math. Acad. Sci. Hung.
9 (1958), 229-241. Takao KOMATSU Faculty of Education Mie University Mie, 514-8507 Japan E-mail : komatsuQedu.mie-u.ac. jp