On Inhomogeneous Continued Fraction Expansions and Inhomogeneous Diophantine Approximation

Journal of Number Theory  NT2060 journal of number theory 62, 192212 (1997) article no. NT972060

On Inhomogeneous Continued Fraction Expansions and Inhomogeneous Diophantine Approximation Takao Komatsu Centre for Number Theory Research, Macquarie University, New South Wales 2109, Australia Communicated by Alan C. Woods Received August 17, 1995; revised March 4, 1996

We consider a new algorithm yielding the values M(%, ,)=lim inf |q|   |q|&q%&,&. Specifically, we compute M(%, 1a), M(%, 12a), M(%, 1- a 2 +4), and M(%, 12) when %=(- a 2 +4&a)2=[0, a, a, ...].

 1997 Academic Press

1. INTRODUCTION In this paper we compute the inhomogeneous approximation constant M(%, ,)=lim inf |q| &q%&,& |q|  

for the pair of irrationals % and ,, subject to q%&, not being integral for any integer q. In order to deal with this function it is convenient to introduce the functions M+(%, ,)=lim inf q &q%&,& q  +

and M&(%, ,)=lim inf q &q%+,&=lim inf |q| &q%&,&. q  +

q  &

Then M(%, ,)=min(M+(%, ,), M&(%, ,)). This notation is introduced by Cusick et al. [4]. M+(%, ,) has been treated by several authors, for example Cassels [3], Descombes [5], and Sos [10], by using algorithms for inhomogeneous diophantine approximation in which , is expressed in terms of the continued fraction expansion of %.

192 0022-314X97 25.00 Copyright  1997 by Academic Press All rights of reproduction in any form reserved.

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INHOMOGENEOUS DIOPHANTINE APPROXIMATION

193

There does not seem to have been any significant progress on this problem since [10]. However, recently Cusick et al. have suggested a new and simpler method. They use and cite the previous work of Cassels, Descombes, and Sos, but the links between the expansion they actually use and the cited results are not at all clear. In this paper we use a different algorithm for the inhomogeneous continued fraction expansion to obtain the values M(%, ,). In particular we consider the general case (a+- a 2 +4)2 =[a, a, . . .], which includes the example a=1 of [4].

2. INHOMOGENEOUS CONTINUED FRACTION EXPANSIONS We first exhibit several alternative ways to express , in terms of the continued fraction expansion of %. As usual, %=[a 0 , a 1 , a 2 , . . .] denotes the continued fraction expansion of %, where %=a 0 +% 0 , 1% n&1 =a n +% n ,

a 0 =w%x , a n =w1% n&1 x

(n=1, 2, . . .).

The kth convergent p k q k =[a 0 , a 1 , ..., a k ] of % is then given by the recurrence relations p k =a k p k&1 + p k&2 (k=0, 1, . . .),

p &2 =0,

p &1 =1,

q k =a k q k&1 +q k&2

q &2 =1,

q &1 =0.

(k=0, 1, . . .),

Let , be real, with 0<,<1 without loss of generality. We also set D k = q k %& p k . Then (&1) k D k >0, q k&1 D k &q k D k&1 =(&1) k, and D k+1 = a k+1 D k +D k&1 . We assume that n%&, is never integral for any integer n. The first two algorithms for representing , in terms of D k are well-known. As usual, [ ] denotes the fractional part. v Algorithm 1 (Sos [10]). (1-1) c 1 =r.

If [(r&1) %]<,<[r%] for an integer r with 2ra 1 +1, set

(1-2) Otherwise, namely, if [(a 1 +1) %]<,<[%], set c 1 =a 1 +1 and c 2 =0. Then next determine c 3 . (2-1) Other than in case (1-2), if [(c 1 q 0 +(r&1) q 1 ) %]>,> [(c 1 q 0 +rq 1 ) %] for an integer with 1ra 2 , set c 2 =r. (2-2) Again other than in case (1-2), if [(c 1 q 0 +a 2 q 1 ) %]>,> [((c 1 &1) q 0 ) %], set c 2 =a 2 and c 3 =0; and next determine c 4 .

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TAKAO KOMATSU

(k-1)

Suppose k is odd. If [(c 1 q 0 +c 2 q 1 + } } } +(r&1) q k&1 ) %] <,<[(c 1 q 0 +c 2 q 1 + } } } +rq k&1 ) %]

for an integer r with 1ra 3 ; set c k =r. Next, determine c k+1 . (k-2) If [(c 1 q 0 +c 2 q 1 + } } } +a k q k&1 ) %]<, <[(c 1 q 0 +c 2 q 1 + } } } +(c k&1 &1) q k&2 ) %], set c k =a k and c k+1 =0. Then determine c k+2 . Finally, if k is even the inequalities above are reversed, becoming } } } >,> } } } . Then, , is represented by 

,= : c k+1 D k , k=0

where the c k are integers with 2c 1 a 1 +1 and 0c k a k (k2). Examples. Let %=(- 5&1)2=[0, 1, 1, ...] and ,=12 (or ,=1- 5). (We use these % and , in the examples of every algorithm below.)  1 +1=2D 0 +D 2 + : (D 3k+1 +D 3k+2 ). 2 k=1

1 -5



+1=2D 0 + : (D 4k&2 +D 4k&1 +D 4k ). k=1

Remark. As shown in [4], there are alternative patterns to express the c k , though each c k is decided uniquely in the above method. For example, if we allow c 1 =1, the decision c 1 =a 1 +1 and c 2 =k is the same as c 1 =1 and c 2 =k+1 (k=0, 1, ..., a 2 &1) because (a 1 +1) q 0 +kq 1 =q 0 +(k+1) q 1 . In fact, this algorithm may produce the expression for ,+1 instead of ,, as above when c 1 =a 1 +1. With this alteration we have 12=D 0 +  k=0 (D 3k+1 +D 3k+2 ) and 1- 5=D 0 +D 1 +  k=1 (D 4k&2 +D 4k&1 +D 4k ). v Algorithm 2 (Descombes [5], cf. Cassels [3]). Let \ n =x n %& ,& y n , where the sequence [(x n , y n )] is characterized by the following: Put r n+1 =w1% n &\ n D n x (n0). Set x 0 =1 and y 0 =w%&,x. Go to step (A 0 ) below.

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INHOMOGENEOUS DIOPHANTINE APPROXIMATION

(A n )

(1)

195

If r n+1 {a n+1 , put

x n+1 =x n +r n+1 q n +q n&1

and

y n+1 = y n +r n+1 p n + p n&1 ,

and

y n+1 = y n + p n+1 ,

then go to (A n+1 ). (2) If r n+1 =a n+1 , put x n+1 =x n +q n+1

then go to (B n+1 ). (B n ) Put x n+1 =x n &q n and y n+1 = y n & p n , then go to (A n+1 ). Then , is represented by ,= lim (x n %& y n ). n

Indeed, (x n+1 %& y n+1 )&(x n %& y n ) =\ n+1 &\ n r n+1 D n +D n&1 , = D n+1 , &D n ,

{

if if if

(A n )(1); (A n )(2); (B n ).

Examples (cf. [4, Lemma 3]). When %=(- 5&1)2 and ,=12, we have n

x 0 =1,

x 1 =2,

n&1

x 3n = : q 3k ,

x 3n\1 = : q 3k

k=0 n

y 0 =0,

y 1 =1,

(n1),

k=0 n&1

y 3n = : p 3k ,

y 3n\1 = : p 3k

k=0

(n1).

k=0

Therefore, 1 = lim 2 n

\

n

n

+



: q 3k %& : p 3k = : D 3k . k=0

k=0

k=0

When %=(- 5&1)2 and ,=1- 5, we have x 0 =x 2 =1, x 1 =x 3 =2, y 0 = y 2 =0, y 1 = y 3 =1, and for n1, n&1

x 4n =x 4n+2 =q 0 + : q 4k+3 , k=0 n&1

y 4n = y 4n+2 = p 0 + : p 4k+3 , k=0

n&1

x 4n+1 =x 4n+3 =q 0 + : q 4k+3 +q 4n+1 , k=0 n&1

y 4n+1 = y 4n+3 = p 0 + : p 4k+3 + p 4n+1 . k=0

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TAKAO KOMATSU

Therefore, 1 -5

= lim n

\\

n

+ \

n

q 0 + : q 4k+3 %& p 0 + : p 4k+3 k=0

k=0

++



=D 0 + : D 4k+3 . k=0

The following two algorithms were originally used to display arithmetical k m properties of the Mahler function   k=1  1mk%+, x y . v Algorithm 3 (Nishioka et al. [9], cf. Ito [6]). Expand , in terms of the sequence [% 0 , % 1 , . . .] by setting ,=b 0 &, 0 ,

b 0 =W,X ,

, n&1 % n&1 =b n &, n ,

b n =W, n&1 % n&1 X

(n=1, 2, . . .).

Then , is represented by ,=b 0 &b 1 % 0 +b 2 % 0 % 1 & } } } +(&1) k b k % 0 % 1 } } } % k&1 &(&1) k % 0 % 1 } } } % k&1 , k 



=b 0 & : (&1) k b k+1 % 0 % 1 } } } % k =b 0 & : b k+1 D k . k=0

k=0

Examples.  1 =1& D 0 +D 1 +2D 2 + : (2D 3k +D 3k+1 +D 3k+2 ) , 2 k=1

\ 1 =1& D +D +2D + : (2D \ -5

+



0

1

2

4k&1

+

+2D 4k +D 4k+1 +D 4k+2 ) .

k=1

v Algorithm 4 (Borwein and Borwein [1]).

The algorithm is given

by # 0 =,,

# n&1 % n&1 =d n +# n ,

d n =w# n&1 % n&1 x

(n1).

Then ,=# 0 =d 1 % 0 +d 2 % 0 % 1 + } } } +d k+1 % 0 % 1 } } } % k +# k+1 % 0 % 1 } } } % k 



= : d k+1 % 0 % 1 } } } % k = : (&1) k d k+1 D k . k=0

k=0

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INHOMOGENEOUS DIOPHANTINE APPROXIMATION

197

Examples.  1 = : (&D 6k&5 +D 6k&2 ), 2 k=1

1 -5



=& : D 4k&3 . k=1

The final algorithm which is the one we mainly use in this paper, is related to the smallest value of &q%&,& for 0q
For each positive integer n, define an integer ! n by &! n %&,&= min &q%&,&. 0q
From [8, Theorem 2] (cf. [2, Theorem 7.1]) ! n is given by !n =

{

(&1) n q n&1(w &q n ,x+t) qn , qn

=

where t=&1, 0, or 1 when n is odd; t=0, 1, or 2 when n is even. Since , satisfies lim n   &! n %&,&=0, , is represented by ,= lim [! n %]. n

Examples (cf. [4, Lemma 3]). Because n&1

! 3n&2 =! 3n&1 =! 3n = : q 3k

(n1)

k=0

for ,=12 and n&1

! 4n =! 4n+1 =! 4n+2 =! 4n+3 =q 0 + : q 4k+3

(n1)

k=0

for ,=1- 5, we have  1 = : D 3k 2 k=0

and

1 -5



=D 0 + : D 4k+3 . k=0

3. THE VALUE M+(%, ,) How to calculate M+(%, ,) by using the first two algorithms is already well known.

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TAKAO KOMATSU

v Algorithm 1. M+(%, ,)=lim inf min(Q n &Q n %&,&, Q$n &Q$n %&,&), n  +

where Q n = n&1 k=0 c k+1q k , and Q$n =Q n &q n&1 if c n+1 {0; Q$n =Q n &q n if c n+1 =0. v Algorithm 2 [3, Proposition 7.] M+(%, ,)=lim inf min(x n | \ n |, x$n | \$n | ), n  +

where x$n =x n +q n&1 , y$n =y n + p n&1 , and \$n =x$n %&,& y$n . But these two algorithms are themselves not simple enough to readily yield the actual value of M+(%, ,). Algorithms 3 and 4 have not been used for the present purpose but it is not difficult to guess how they might be. We will use Algorithm 5. From here on we shall consider just the case %=(- a 2 +4&a)2= [0, a, a, . . .]; its nth convergent is pn =[0, a, ..., a ] qn n

where p n+1 =ap n + p n&1 ,

p &1 =1,

p 0 =0;

q n+1 =aq n +q n&1 ,

q &1 =0,

q 0 =1.

For convenience set :=1% and denote by ; its conjugate, namely, ;=&%. Then q n = p n+1 =

: n+1 : n+1 &; n+1 : n+1 &; n+1 t . = :&; - a 2 +4 - a 2 +4

Note that :+;=a, :&;=- a 2 +4, and :;=&1. From D n+1D n =&% we have D n =(&1) n % n+1 (n=0, 1, 2, . . .). 4. ALGORITHM 5 We obtain some values of M+(%, ,) by using Algorithm 5. Theorem 1. M+(%, ,)=lim inf ! n &! n %&,&. n  +

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INHOMOGENEOUS DIOPHANTINE APPROXIMATION

Proof.

199

It suffices to show that min(! n&1 &! n&1 %&,&, ! n &! n %&,&)q &q%&,&

for any integer q with q n&1 q
pn

\q &%+ & n

[ &q n ,] 1 . < qn q n+1 +q n %

Thus, &} n %&,&=[} n %&,]= >

1 [ &q n ,] pn &} n &% + qn qn qn

\

+

1 1 , & q n q n+1 +q n %

&\ n %&,&=1&[ \ n %&,] =\ n

pn

\ q &%+ & n

[ &q n ,] 1 . < qn q n+1 +q n %

When a2, clearly q n+1 +q n %>2q n . When a=1, from q n+1 q 3 3 5&- 5 =2&%  = > qn q2 2 2

(n1)

we again have q n+1 +q n %>2q n . Therefore, &! n %&,&=&\ n %&,&<

1 1 < =&q n&1 %& p n&1 &. q n+1 +q n % q n +q n&1 %

If q%&,>( p n q+w &q n ,x)q n , then 0<

1 q n&1 p [ &q n ,] . &q n &% < & qn qn q n q n(q n+1 +q n %)

\

+

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200

TAKAO KOMATSU

Thus, &} n %&,&=

[ &q n ,] 1 q n&1 pn , &} n &% < & qn qn q n q n(q n+1 +q n %)

&\ n %&,&=

1 [ &q n ,] q n&1 pn . & +\ n &% > qn qn qn q n(q n+1 +q n %)

\

+

\

+

Therefore, &q n&1 %& p n&1 &&&! n %&,&>

q n&1 1 1 + & =0. q n +q n&1 % q n(q n+1 +q n %) q n

We may set q=! n &kq n&1 +lq n , where 0
\ +

M+ %, Lemma 1. : 2n &; 2n #Z : 2 &; 2 Proof.

: 2n&1 &; 2n&1 #1 (mod a). :&;

These claims are easy to check.

Lemma 2. Proof.

and

When a2, ! 2n&1 =! 2n =q 2n&1 a (n1).

From Lemma 1 we have

\

: 2n &; 2n q &q 2n&1 =& 2 =& 2n&1 . a : &; 2 a



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201

INHOMOGENEOUS DIOPHANTINE APPROXIMATION

Since for an integer q p 2n&1 q+w &q 2n&1 ax&1 q 2n&1 1 p 2n&1 q+w &q 2n&1 ax p 2n&1 q+w &q 2n&1 ax+1
{

= {

= {

=

and } 2n&1 &q 2n&2(w &q 2n&1 ax&1) = q 2n&1 q 2n&1

{ q = { a

2n&2

+

=

q 2n&2 1 q 2n&2 1 q 2n&1 = + = + . q 2n&1 a q 2n&1 a q 2n&1

= {

=

Since (\ 2n&1 +} 2n&1 )

\

1 p 2n&1 2 q 2n&2 1 &% < + < , q 2n&1 a q 2n&1 q 2n q 2n&1

+ \

+

we obtain

{

} 2n&1 %&

1 1 } 2n&1 p 2n&1 +w &q 2n&1 ax&1 =} 2n&1 %& & a a q 2n&1

=

>

\ 2n&1 p 2n&1 +w &q 2n&1 ax 1 & \ 2n&1 %& q 2n&1 a

\

{

=1& \ 2n&1 %&

1 >0. a

=

Therefore, ! 2n&1 =\ 2n&1 =q 2n&1 a. Similarly, from Lemma 1 we have q 2n 1 : 2n+1 &; 2n+1 q 2n 1 = & =& & 1& . a a :&; a a

\  \ &



\ +

Since, for 0q
\

q %&

p 2n q q 1 1 1 = < < , q 2n q 2n :q 2n +q 2n&1 q 2n aq 2n +q 2n&1 aq 2n

+

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+

202

TAKAO KOMATSU

we have 1 p 2n q+w &q 2n ax+1 p 2n q+w &q 2n ax+2 p 2n q+w &q 2n ax
{ = 1 q q q q q q = { q \& a +a+= = {& a + aq = ={ aq = = aq 2n&1

2n

2n

2n&1

2n

2n&1

2n

2n&1

2n

and } 2n q 2n&1 w &q 2n ax = q 2n q 2n

{ = 1 q q = { q \& a &1+a+= q q = & { aq q = 1 q =1& 1& \ a+ q . 2n&1

2n

2n

2n&1 2n

2n&1 2n

2n&1 2n

Since, for a2,

\

(} 2n +\ 2n ) %&

p 2n 2 w &q 2n ax+1 2 >0 + , q 2n q 2n a

+

we obtain

{

} 2n %&

1 1 } 2n p 2n +w &q 2n ax =} 2n %& & a a q 2n

=

>

\ 2n p 2n +w &q 2n ax+1 1 & \ 2n %& q 2n a

\

{

=1& \ 2n %&

1 >0. a

=

Therefore, ! 2n =\ 2n =q 2n&1 a.

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+

2n

203

INHOMOGENEOUS DIOPHANTINE APPROXIMATION

Proof of Theorem 2. It is known that in the case a=1 M+(%, 1)=lim inf q &q%&= q

1 -5

.

Let a2. From Lemma 2 ! 2n =

q 2n&1 n&1 = : q 2k =! 2n&1 . a k=0

Indeed, 



: D 2k = : % 2k+1 = k=0

k=0

% 1 = . 1&% 2 a

Since

"

! 2n %&

n&1 1 1 % 2n %(1&% 2n ) 1 , = : D 2k & = & = a a 1&% 2 a a k=0

" "

" "

"

we finally obtain 1 1 1 : 2n &; 2n % 2n } . = = lim ! 2n ! 2n %& = lim 2 2 a n a n   a - a +4 a a - a 2 +4

\ +

M+ %,

"

"

Theorem 3. 1

1

\ a+ =a - a +4 .

M %, Proof.

2

2

This result is trivial when a=1 or 2. Let a3. We show that 1 1 = 1& a a

2

1

\ + \ + - a +4 ,

M& %,

2

which yields the claim. By an argument similar to the proof of Theorem 2 we have ! 2n&1 =\ 2n&1 =

a&1 q 2n&1 a

and

! 2n =} 2n =q 2n &

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q 2n&1 . a

204

TAKAO KOMATSU

Note that M&(%, ,)=lim inf n  + ! n &! n %+,& as in the Corollary to Theorem 1. Using that q 2n&1 a= n&1 k=0 q 2k ,

"

! 2n&1 %+

n&1 1 1 = (a&1) : D 2k + a a k=0

" " " %(1&% ) 1 1 = (a&1) + = 1& % " 1&% a" \ a+ 2n

2n

2

and

"

! 2n %+

n&1 1 1 = D 2n & : D 2k + a a k=0

" " = % "

2n+1

&

"

%(1&% 2n ) 1 % 2n 2n+1 . + =% + 1&% 2 a a

"

Hence, we have

"

! 2n&1 ! 2n&1 %+

2

1 1 = 1& a a

: 2n &; 2n

" \ + - a +4 % 1 1  1& \ a+ - a +4

2n

2

2

(n  )

2

and

"

! 2n ! 2n %+

1 1 : 2n &; 2n = : 2n+1 &; 2n+1 & a a - a 2 +4

"

% 2n a

\ +\ + 1 % : 1 1 1 (n  ).  1& + & = 2& a a a + \ a + - a +4 - a +4 \ 2

2

% 2n+1 +

2

2

Therefore, we obtain 1 1 } min = a - a 2 +4

\ +

M& %,

1 a

\\ + 1&

2

, 2&

1 1 = 1& a2 a

2

5. KHINCHIN'S RESULTS The homogeneous approximation constant *(%) is given by *(%) :=lim inf q &q%&= lim * n = q

n

1

+ \ + - a +4 .

1 - a 2 +4

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2

205

INHOMOGENEOUS DIOPHANTINE APPROXIMATION

for %=[0, a, a, . . .], where * n =q n &q n %&. Khinchin's results show the following concerning the relationship between the homogeneous constant *(%) and the inhomogeneous constant M(%) :=sup , M(%, ,). Lemma 3 [7, Theorem]. When a=1 or when a is even 1 a . M(%)= - 1&4(*(%)) 2 = 4 4- a 2 +4 When a3 is odd, 14 - a 2 +4, 12 - a 2 +4, M(%) a a+1 , a+2 4 - a 2 +4

{

if if

3 n - 2 * n ; - 2 * n 3 n - 8 * n ;

if

3 n - 8 * n ,

where 3 n =2 &q n ,&. This raises the question of whether there exists a real , satisfying M(%, ,)=a4 - a 2 +4. The readily guessed case ,=12a does not satisfy this. Actually, for a1 we have 1

1

\ 2a+ =4a - a +4 ,

M %,

2

\

M %,

2

1

1

, = - a +4 + (a +4) - a +4 2

2

2

and for a=1, 2, 4,

\

M %,

-a a . = 2 4 - a 2 +4

+

When a3 is odd we obtain 1 1 , = 2 4 - a 2 +4

\ +

M %,

because then q 2n&1 is odd and q 2n even (n1) entails 3 2n&1 =1- 8 * n and 3 2n =0- 2 * n . For a=1, these are Lemma 6, Lemma 7, and Theorem 3 in [4].

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TAKAO KOMATSU

Theorem 4. 1 1 1 =M+ %, = 2 2a 2a 4a - a 2 +4

\ +

\ +

(1)

M %,

(2)

M %,

(3)

M %,

-a -a a =M\ %, = 2 2 4 - a 2 +4

(a=1, 2, 4),

(4)

M %,

1 1 = 2 4 - a 2 +4

(a is odd 3).

\

1 2

- a +4

+

\

=M\ %,

\ + \ +

Proof.

\

1 2

- a +4

+

=

(a1), 1

(a1),

2

(a +4) - a 2 +4

+

(4) When a3 is odd, [q 2n&1 2]=12 and [q 2n 2]=0. Since

p 2n&1 q+w &q 2n&1 2x 1 p 2n&1 q+w &q 2n&1 2x+1
{

= {

=

and \ 2n&1 q 2n&2 &q 2n&2(w &q 2n&1 2x+1) q 2n&2 q 2n&2 & = = =1& . q 2n&1 q 2n&1 2 2q 2n&1 2q 2n&1

{

= {

=

But (} 2n&1 +\ 2n&1 )( p 2n&1 q 2n&1 &%)>0 implies ! 2n&1 =} 2n&1 =q 2n&2 2. Therefore,

"

! 2n&1 ! 2n&1 %&

1 1 : 2n&1 &; 2n&1 % 2n&1  = 2 2 2 2 - a +4 4 - a 2 +4

"

Since 1 p q+w &q 2n 2x+1 p 2n q+w &q 2n 2x
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(n  ).

INHOMOGENEOUS DIOPHANTINE APPROXIMATION

207

therefore q 2n&1 1 } 2n &q 2n&1 w &q 2n 2x = = & = q 2n q 2n 2 2

{

= {

=

and q 2n&1 q 2n&1 q 2n&1 1 &q 2n&1(w &q 2n 2x+1) \ 2n + = = & = + . q 2n q 2n 2 q 2n q 2n 2

{

= {

=

Hence (} 2n +\ 2n )(%& p 2n q 2n )<1q 2n implies ! 2n =} 2n =q 2n 2. Similarly, lim n   ! 2n &! 2n %&12&=14 - a 2 +4. (1)

Note that

{

q 2n&1 =0 2a

=

or

1 ; 2a

1

q 2n

{ 2a = =2a

or

a+1 . 2a

When [q 2n 2a]=12a and [q 2n&1 2a]=0, we have ! 2n =\ 2n =q 2n&1 2a, which yields

"

! 2n&1 ! 2n&1 %&

1 1 : 2n &; 2n % 2n  = 2 2 2a 2a 2a - a +4 4a - a 2 +4

"

(n  ).

When [q 2n&1 2a]=0 and [q 2n&2 2a]=12a, we have ! 2n&1 =\ 2n&1 =q 2n&1 2a, which also yields lim n   ! 2n&1 &! 2n&1 %&12a&=14a 2 - a 2 +4. In the other cases

"

lim ! n ! n %&

n

1 1 .  2 2a 4a - a 2 +4

"

Similarly, we obtain 1 1 = 1& 2a 2a

\ + \

M& %,

2

1

+ - a +4 , 2

proving (1). 6. PROOF OF THEOREM 4 (2) We provide just a sketch of the outline of the proof in the cases (2) and (3) because the methods are quite similar albeit that the calculations are more complicated.

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208

TAKAO KOMATSU

When %=(- a 2 +4&a)2=[0, a, a, ..., ], the continued fraction expansion of 1 - a 2 +4 is given by ,=

1 2

- a +4

_

= 0, a,

1 1 1 1 a, 2a, a, 2a, a, . . . = 0, a, a, 2a . 2 2 2 2

& _

&

(One may prefer to write ,=[0, a, (a&1)2, 1, 1, (a&1)2, 2a] instead, when a is odd. But the following argument holds for both the odd and even cases.) Denote the n th convergent of , by p$n q$n . By induction we have for n1 p$2n&1 = p 2n&1 ,

p$2n = p 2n 2;

q$2n&1 =q 2n&1 +q 2n&3 ,

q$2n =(q 2n +q 2n&2 )2.

Using q 2n q 2n&2 &q 22n&1 =1, q 4n =q 22n +q 22n&1 , and q 4n&1 =q 2n&1 q 2n + q 2n&2 q 2n&1 , we obtain a series of inequalities q 0 q 1 &1 p$2 q 1 q 2 q 2 q 3 &1 p$4 q 3 q 4 < < < < < < }}} q2 q$2 q4 q6 q$4 q8 <

1 p$2n&2 q 2n&3 q 2n&2 q 2n&2 q 2n&1 &1 p$2n < < < < }}} < q$2n&2 q 4n&4 q 4n&2 q$2n - a 2 +4

< }}} <

<

q 22n q 4n+1

=

p$2n+1 q 22n&1 +1 q 22n&2 p$2n&1 < < = < }}} q$2n+1 q 4n&1 q 4n&3 q$2n&1

q 24 p$5 q 23 +1 q 22 p$3 p 3 q 20 p$1 = < < = < < = , q 9 q$5 q7 q 5 q$3 q 3 q 1 q$1

yielding for n1 q 4n

\ - a +4  q \ - a +4 =q q  - a +4 |=q q  - a +4 |=q 2

2n

=q 2n&1 q 2n = : (&1) k q 2k&1 , k=1 2n&1

4n&2 2

4n&1 2

4n&3 2

2n&2

q 2n&1 &1= : (&1) k&1 q 2k&1 &1, k=1 2n&1

2 2n&1

+1= : (&1) k&1 q 2k , k=1 2n&1

2 2n&2

= : (&1) k&1 q 2k&2 . k=1

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209

INHOMOGENEOUS DIOPHANTINE APPROXIMATION

By comparing [} 4n&3 %&,] and 1&[ \ 4n&3 %&,], &q 4n&4 w &q 4n&3 - a 2 +4x q 4n&3 q 4n&3

{ q = {

! 4n&3 =

=

4n&4

q 22n&2

q 4n&3

=q

4n&3

=q 22n&2 .

Similarly, we have &q 4n&2 w &q 4n&1 - a 2 +4x q 4n&1 q 4n&1

{ = q (q +1) = (a=1), { q = q =q &q (w &q - a +4x+1) = { =q q q &q = { q = q =q +q &q q (w&q - a +4x+1) &q q = q = { = { q q q w&q - a +4x = { =q q &q q q = { q = q =q .

! 4n&1 =

2 2n&1

4n&2

2 2n&2

4n&1

4n&1

2

! 4n&1

4n&2

4n&1

4n&1

4n&1

2 2n&2

4n&2

2 2n&2

4n&1

4n&1

(a2),

4n&2

4n&1

2

! 4n

4n&1

4n

4n&1

2n&1

q 2n

4n

4n

4n

=q

4n

=q 22n ,

2

! 4n&2

4n&3

4n&2

4n&2

4n&2

4n&3

2n&2

2n&1

4n&2

2 2n&2

4n&2

It follows that

"

1

" 1 : &; = : (&1) D & \ - a +4 + " - a +4 " 1 : +; +2 1&(&% ) = "% 1&(&% ) &- a +4 " a +4

q 22n&2 q 22n&2 %& 2n&1

- a 2 +4 2n&1

2

2n&1

k&1

2k&2

2

4n&2

4n&2

2

=

2

k=1

2 2n&1 2

2

1 : 4n&2 +; 4n&2 +2 % 4n&2  2 2 2 a +4 - a +4 (a +4) - a 2 +4

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(n  ).

210

TAKAO KOMATSU

Similarly,

"

q 22n q 22n %&

1 - a 2 +4

"

=

: 4n+2 +; 4n+2 +2 % 4n+2 a 2 +4 - a 2 +4



1

(n  )

2

(a +4) - a 2 +4

and

"

(q 22n +q 4n&1 &q 4n&2 ) (q 22n +q 4n&1 &q 4n&2 ) %& =

\

1 - a 2 +4

: 4n&2 +; 4n&2 +2 : 4n &; 4n &: 4n&1 +; 4n&1 + a 2 +4 - a 2 +4

\

_ % 4n +% 4n&1 &

% 4n&2 - a 2 +4

+



"

+

a 2 +3 (a 2 +4) - a 2 +3

(n  ).

In like manner we get M&(%, ,)=1(a 2 +4) - a 2 +4, which yields the result.

7. PROOF OF THEOREM 4 (3) The claim is trivial for a=1 or a=4. Denote the n th convergent of ,=1- 2=[0, 1, 2, 2, . . .] by p"n q"n . Then by induction p"n = p n =q n&1 and q"n =q n & p n =q n &q n&1 . A series of inequalities q 21 &q 20 p"2 q 22 &q 21 &1 q 23 &q 22 p"4 q 24 &q 23 &1 < < < < < < }}} q2 q"2 q4 q6 q"4 q8 <

1 p"2n q 22n &q 22n&1 &1 q 22n+1 &q 22n p"2n+2 < < < < }}} < q"2n q 4n q 4n+2 q"2n+2 -2

< }}} <

<

2q 22n p"2n+1 (q 2n &q 2n&1 ) 2 2q 22n&2 p"2n&1 = < < = < }}} q 4n+1 q"2n+1 q 4n&1 q 4n&3 q"2n&1

2q 24 p"5 (q 4 &q 2 ) 2 2q 22 p"3 (q 2 &q 1 ) 2 2q 20 p"1 = < < = < < = q 9 q"5 q7 q 5 q"3 q3 q 1 q"1

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211

INHOMOGENEOUS DIOPHANTINE APPROXIMATION

shows that for n1 q 4n

2n&1

\- 2 q \ - 2  =q &q =q +2 : (&1) q  - 2 | =2q =2 : (&1) q , q  - 2 | =(q &q ) =q &2 : (&1) q

=q 22n &q 22n&1 &1=q 4n +2 : (&1) k q 2k &1, k=0

2n&2

4n&2

2 2n&1

2 2n&2

k&1

4n&2

q 2k ,

k=0

2n&2

4n&3

2 2n&2

k

2k

k=0

2n

4n&1

2

2n

k

2n&1

4n

2k&1

.

k=1

By comparing [} 4n&3 %&1- 2] and 1&[ \ 4n&3 %&1- 2] we have ! 4n&3 =} 4n&3 =

{

&q 4n&4 w &q 4n&3 - 2x q 4n&3 q 4n&3

=

{

q 4n&4 } 2q 22n&2 q 4n&3 =2q 22n&2 . q 4n&3

1

" \

=

=

Hence,

"

! 4n&3 ! 4n&3 %&

-2

=2

=

: 2n&1 &; 2n&1 - 2 2 +4

2

+"

2%

1 1&(&% 2 ) 2n&1 & 1&(&% 2 ) -2

1 : 4n&2 +; 4n&2 +2 % 4n&2  4 -2 4 -2

"

(n  ).

Similarly, we can obtain ! 4n&1 =(q 2n &q 2n&1 ) 2, and

! 4n =q 22n &q 22n&1 ,

! 4n&2 =q 22n&1 &q 22n&2 +1,

yielding

"!

4n&1

%&

1

= ! - 2" "

4n

%&

1

= ! - 2" "

4n&2

%&

1

1

 - 2" 4 - 2

(n  ).

In like manner we get M&(%, 1- 2)=14 - 2, which yields the claim.

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212

TAKAO KOMATSU

8. OPEN PROBLEMS RELATED TO KHINCHIN'S RESULTS What is the value of , satisfying M(%, ,)=a4 - a 2 +4 for even a6, if there is any? Is there a real , satisfying M(%, ,)>14 - a 2 +4 for odd a3? It is easy to guess that there are , so that M(%, ,)=, 24 - a 2 +4. What is the condition on , to make this so?

REFERENCES 1. J. M. Borwein and P. B. Borwein, On the generating function of the integer part: [n:+#], J. Number Theory 43 (1993), 293318. 2. D. Bowman, Approximation of wn:+sx and the zero of [n:+s], J. Number Theory 50 (1995), 128144. 3. J. W. S. Cassels, Uber  x  + x |x+:& y|, Math. Ann. 127 (1954), 288304. 4. T. W. Cusick, A. M. Rockett, and P. Szusz, On inhomogeneous Diophantine approximation, J. Number Theory 48 (1994), 259283. 5. R. Descombes, Sur la repartition des sommets d'une ligne polygonale reguliere non fermee, Ann. Sci. Ecole Norm. Sup. 73 (1956), 283355. 6. Sh. Ito and H. Tachii, A Diophantine algorithm and a reduction theory of ternary forms, Tokyo J. Math. 16 (1993), 261289. 7. A. Ya. Khinchin, On the problem of Tchebychef, Izv. Akad. Nauk. SSSR 10 (1946), 281294 [in Russian]. 8. T. Komatsu, The fractional part of n%+, and Beatty sequences, J. Theorie des Nombres de Bordeaux 7 (1995), 387406. 9. K. Nishioka, I. Shiokawa, and J. Tamura, Arithmetical properties of a certain power series, J. Number Theory 42 (1992), 6187. 10. V. T. Sos, On the theory of Diophantine approximations, II, Acta Math. Acad. Sci. Hungar. 9 (1958), 229241.

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