On the exponential sum with
square{free numbers
D. I. Tolev
Introduction Consider the exponential sum
S () =
X
nN
2(n) e(n);
where N is a large integer, e(x) = e2ix and where (n) denotes the Mobius
function. This sum appears naturally in the study of various problems involving squarefree numbers. For example, if r (N ) is the number of representations of N as the sum of square{free numbers, then we have
r (N ) =
Z 1
0
S () e(;N ) d :
In the case 3 Evelyn and Linfoot [5], Mirsky [7] and Brudern and
Perelli [2] used the circle method to obtain asymptotic formulae of the form
6 1
r (N ) = ( ; 1)! 2 S (N ) N ;1 + (N ) ; (N ) = o(N ;1 ) ;
where
Y
Y
S (N ) =
1 ; (1 ;1p2)
1 ; (1 ; p12);1 :
p2 N
p2 jN
-
For 3 the sharpest estimate for (N ) is due to Brudern and Perelli.
They prove ([2], Theorem 1) that if 3 then (N ) = O(N ;3=2+" ) holds,
where here and later " > 0 denotes an arbitrarily small number, which is not
the same in dierent occurences.
Brudern and Perelli also show that the last estimate is best possible up to
the current knowledge about Riemann's zeta function (s). More precisely
1
([2], Theorem 2), if 2 then (N ) = (N ;2+=2;" ), where is the
supremum of the real parts of the zeros of (s). It is however established
([2], Theorem 3) that under the generalized Riemann hypothesis (GRH) for
all Dirichlet L{functions one has (N ) = O(N ;7=4+" ) for 4 and also
3(N ) = O(N 37=28+"). Thus, in the case = 3 they miss the optimal
exponent by 1=14.
One of the key instruments for obtaining these results is the estimate of
the sum S () on the set of minor arcs. Suppose that Q 1 and denote
M(Q) =
[
q
[
qQ a=1
(a;q)=1
a ; Q ; a + Q ; m(Q) = Q ; 1 + Q n M(Q):
q qN q qN
N
N
Theorem 4 of [2] states that if Q N 3=7, then the estimate (2), given below, holds. A similar result was previously obtained by Baker, Brudern and
Harman [1], but under the condition Q N 1=3.
The aim of the present paper is to present the proof of the following:
Theorem. Suppose that
(1)
1 Q N 1=2
and let m(Q) be dened as above. Then for the sum S () we have
(2)
sup jS ()j N 1+" Q;1 :
2m(Q)
It is pointed out by Brudern and Perelli at the end of section 5 of [2] that
from this theorem one can obtain:
Corollary. Suppose that GRH holds. Then we have 3(N ) = O(N 5=4+").
We should mention that the circle method provides a non{trivial estimate for 2(N ) as well (see Brudern et al [3]), but it is weaker then the
estimate 2(N ) = O(N 2=3+"), obtained by elementary methods by Evelyn
and Linfoot [5] (see also Estermann [4] for simpler proof).
Furthermore, Heath{Brown [6] developed the square sieve and applied it
to the related problem of counting square{free twins. Using his method one
can obtain 2(N ) = O(N 7=11+").
To prove our theorem we apply the square sieve as well as some of the
methods used in [2].
2
Acknowledgement The present paper was writen during the authors visit
to the Mathematical Institute of the University of Oxford. The author would
like to thank the Royal Society for nancial support and to the sta of the
Institute for the excellent working conditions.
The author is especially grateful to Professor D.R.Heath{Brown for the
intersting discussions and valuable comments.
Proof of the theorem Let us take arbitrary 2 m(Q). From Dirichlet's
theorem and from the denition of m(Q) it follows that there exist integers
a and q satisfying
Q
a
(3)
(a; q) = 1;
Q<q N
; q qN ;
Q:
It is clear that
X
X
S () =
(d)
e(d2m)
mNd;2
dN 1=2
and therefore, according to Lemma 5.1 of [8], we have
S () log N 1 max1=2 Y (; D);
(4)
2 DN
where
(5)
Y (; D) =
X
D<d2D
min(ND;2 ; jjd2jj;1)
and where jjxjj denotes the dierence from x to the nearest integer.
Brudern and Perelli [2] estimate Y (; D) by three dierent methods and
establish that
(6)
Y (; D) N 1+" Q;1
unless D2 2 NQ ;1; Q4=3 . If Q N 3=7 the later interval is empty, so (6) is
true for all D 2 21 ; N 1=2 . This implies Theorem 4 of [2].
To prove out Theorem we shall estimate Y (; D) using the square sieve.
We can assume that
N D2 Q4=3
(7)
Q
3
because in the other cases the estimate (6) is established in [2].
Let P be a parameter, which we shall specify later. Now we assume only
that
(8)
N P N 1=2 for some constant > 0:
Consider the function
(9)
X k 2
;
(k) = logP P
p
P<p2P
pq
-
; where the summation is taken over primes and p stands for the Legendre
symbol. Obviously (k) 0 for any integer k. Furthermore, we have
(10)
(k) 1 if k = d2 for some d 2 (D; 2D] :
Indeed, in this case
X k P<p2P
pq
-
X
X
p = P<p2P 1 P<p2P 1 ; (dq) ;
p dq
-
where (n) denotes the number of dierent prime factors of n. It is well{
known that (n) 2 log n, so (10) follows from Tchebyshev's prime number
theorem and (8), (9). Using (5) and (10) we get
(11)
Y (; D) X
D2 <k4D2
;
(k) min M; jjkjj;1 ;
where we have put
M = ND;2 :
(12)
We can now expand the function min(M; jjxjj;1) into Fourier series, but
it does not converge very fast and some extra eorts are needed to deal with
the `tail' of this series. To avoid this we take a smooth function G(M; x),
which is periodic with period one, satisfy
(13)
;
min M; jjxjj;1 G(M; x)
4
and which has Fourier expansion
X
(14)
G(M; x) = cn e(nx)
n2
Z
with coecients satisfying
(15)
and
(16)
cn log M
X
jnj>M 1+"
jcnj M ;A
for any arbitrarily large constant A > 0 (the constant in the { symbol
depends on " and A).
To construct such a function we take an innitely dierentiable function
!R (1x), which is supported and positive in the interval (;1; 1) and such that
;1 ;1
;1 ! (x) dx = 1. Then for any > 0 we dene ! (x) = ! ( x) and
consider the convolution
g(M; x; ) =
;
Z 1
;1
;
min M; jjtjj;1 ! (x ; t) dt :
We put G(M; x) = 2g M; x; (10M );1 . It is not dicult to see that the
conditions (13) { (16) hold.
Having in mind (11) { (14) and (16) we nd
(17)
Y (; D) 1 + jZ (; D)j ;
where
X
X
Z (; D) =
(k)
cn e(kn) :
D2 <k4D2
Now we apply (9) to get
2 X
Z (; D) = logP P
D2 <k4D2
where
(18)
; pp0
jnjM 1+"
X
0 2P
P<p;p
(pp0 ;q)=1
k X c e(kn) ;
pp0 jnjM 1+" n
is the Jacobi symbol. We represent the last sum in the form
2
Z (; D) = logP P (E1 + E2) ;
5
where
E1 =
X
X
D2 <k4D2 P<p2P jnjM 1+"
p kq
-
E2 =
X
X
X
D2 <k4D2 p;p0 : (19)
k
pp0
cn e(kn) ;
X
jnjM 1+"
cn e(kn)
and where the summation over p and p0 is taken over primes satisfying
(19)
P < p; p0 2P ; (pp0 ; q) = 1 ; p 6= p0 :
Consider the sum E1. It is clear that
X
X
X
(20)
E1 =
cn
e(kn) + O(N "PD2 ) :
P<p2P D2 <k4D2
p q k60 (mod p)
0<jnjM 1+"
-
According to Lemma 5.1 from [8], for the sum over k we have
(21)
X
D2 <k4D2
k60 (mod p)
e(kn) =
X
D2 <k4D2
e(kn) ;
X
e(kn)
D2 <k4D2
k0 (mod p)
min(D2; jjnjj;1) + min(D2 P ;1; jjpnjj;1) :
Hence from (12), (15), (20) and (21) we get
E1 N "(P E1(1) + E1(2) + PD2) ;
(22)
where
E1(1) =
X
min(D2 ; jjnjj;1) ;
nM 1+"
X
X
(2)
E1 =
nM 1+" P<p2P
min(D2 P ;1; jjpnjj;1)
(the summation is already taken over positive n only).
To estimate E1(1) we apply Lemma 5.4 of [8] and use (1), (3), (7) and (12)
to get
2
E1(1) N 1+" q1 + D12 + M1 + Nq N 1+" Q1 + DN :
(23)
6
Furthermore, we have
E1(2) X
h2PM 1+"
(h) min(D2P ;1 ; jjhjj;1) ;
where (h) is the divisor function. Proceeding as above we nd
2
X
(24) E1(2) N "
min(D2P ;1; jjhjj;1) N 1+"P Q1 + DN :
h2PM 1+"
From (22) { (24) we obtain
(25)
2
E1 N 1+" P Q1 + DN :
Consider now E2. We have
E2 =
X
X
p;p0 : (19) jnjM 1+"
where
(26)
Fn() =
X
D2 <k4D2
cn Fn() ;
k e(nk) :
pp0
We divide E2 into two parts:
(27)
E2 = E2(1) + E2(2) ;
where E2(1) is the contribution of the terms with n 6= 0 and E2(2) comes from
the terms with n = 0.
Using Polya { Vinogradov's inequality (see, for example, Theorem 2.1
of [8] ) we nd
X k p
pp0 log N
F0() =
0
pp
D2 <k4D2
and therefore, according to (12) and (15), we get
(28)
E2(2) N "P 3 :
7
Consider E2(1). Bearing in mind (15) we nd
E2(1) N "
(29)
X
X
p;p0 : (19) nM 1+"
jFn()j :
We shall estimate the sum Fn(), dened by (26). According to (3), we write
in the form
Q:
(30) = aq + ; where (a; q) = 1 ; Q < q N
;
j
j
Q
qN
Write
X k ank G(t) =
pp0 e q :
D2 <kt
Using Abel's formula and (12), (29) and (30) we nd
X k ank Fn() =
(31)
0 e q e(nk )
pp
2
2
D <k4D
Z 4D2
=;
G
(t) dtd e(nt) dt + G(4D2 ) e(N 4D2)
D2
(1 + j jnD2) t2[max
jG(t)j
D2;4D2 ]
N " t2[max
jG(t)j :
D2 ;4D2]
;
Consider the sum G(t). We divide it into O D2 (pp0q);1 complete sums
and at most one incomplete sum modulo pp0 q. However, since (pp0; q) = 1,
for the complete sum we have
pp0 q X
k=1
0
pp
k e ank = X
pp0
q
u=1
q X
uq + vpp0 an(uq + vpp0) v=1
= ppq 0
pp0 X
u=1
pp0
e
q uX
anpp0v = 0 :
e
pp0 v=1
q
Hence we can write G(t) in the form
G(t) =
X
K1 <kK2
8
q
k e ank ;
pp0
q
where
0 K2 ; K1 < pp0 q :
(32)
We proceed in a standard manner to nd
G(t) =
(33)
=
pp0q
X
k=1
1
0
K1 <hK2 pp q
X
X
X
; pp20 q <s pp20 q
s s ;
e s(kpp;0qh) ppk 0 e ank
q
pp0 q
; pp20 q <s 2
where
1 X e ;hs ;
s= 0
pp q K1<hK2 pp0 q
s =
pp0 q X
k=1
k e ank + sk :
pp0
q pp0 q
Using (32) we easily get
;1
s (1 + jsj) :
(34)
Consider s . We have
s =
pp0 X
q X
uq + vpp0 an(uq + vpp0)
e
pp0
q
u=1 v=1
pp0 X
q X
uq
su
(anpp0 + s)v :
e
e
=
0
pp0 v=1
q
u=1 pp
Hence
(35)
s =
( ; q ppqs0 (pp0 )
0
where
(pp0) =
is the Gauss sum.
if anpp0 + s 0 (mod q) ;
otherwise ;
pp0 X
u=1
9
u e u pp0 pp0
0 + s(uqpp+0 qvpp )
p
Using (33) { (35) and the estimate (pp0 ) pp0 (see, for example,
Lemma 1.6 of [8]) we nd
X
p
G(t) q pp0
(36)
0
0<jsj pp2 q
anpp0 +s0 (mod q)
jsj;1 :
From (29), (31) and (36) we obtain
E2(1) N " q P
N" q P
X
X
P<p;p0 2P nM 1+"
X
0<jsj2P 2 q
jsj;1
X
jsj;1
0<jsj2P 2 q
anpp0 +s0 (mod q)
X
2(h) :
h4P 2 M 1+"
ah+s0 (mod q)
Obviously, the inner sum is N "(1 + P 2ND;2 q;1) and using (1) and (3)
we get
2 2
E2(1) N " P q + PDN2 N 1+" P Q1 + DP 2 :
(37)
From (17), (18), (25), (27), (28) and (37) we nd that
1
D2 :
+ DP2 + PN
Y (; D) N 1+" PQ
We choose P = N ";1=2D2. It follows from (1) and (7) that the condition (8)
holds. From (38) we obtain
1
+ N11=2 N 1+" Q;1 :
Y (; D) N 1+" PQ
It remains to apply (4) and the proof of the theorem is complete.
(38)
References
[1] R. C. Baker, J. Brudern, G. Harman, Simultaneous diophantine approximation with square{free numbers, Acta Arith. 63, (1993), 52{60.
10
[2] J. Brudern, A. Perelli, Exponential sums and additive problems involving
square{free numbers, Ann. Squola Norm. Sup. Pisa Cl. Sci. 28, (1999),
no 4, 591{613.
[3] J. Brudern, A. Granville, A.Perelli, R. C. Vaughan, T. D. Wooley, On the
exponential sum over k{free numbers, Philos. Trans. Roy. Soc. London
Ser. A 356, (1998), 739{761.
[4] T. Estermann, On the representations of a number as the sum of two
numbers not divisible by kth powers, J. London Math. Soc. 6, (1931),
37{40.
[5] C. J. A. Evelyn, E. H. Linfoot, On a problem in the additive theory of
numbers, I: Math. Z. 30, (1929), 433{448; II: J. Reine Angew. Math.
164, (1931), 131{140; III: Math. Z. 34, (1932), 637{644; IV: Ann. of
Math. 32, (1931), 261{270; V: Quart. J. Math. 3, (1932), 152{160; VI:
Quart. J. Math. 4, (1933), 309{314.
[6] D. R. Heath{Brown, The square sieve and consequtive square{free numbers, Math. Ann. 266, (1984), no 3, 251{259.
[7] L. Mirsky, On a theorem in the additive theory of nimbers due to Evelyn
and Linfoot, Math. Proc. Cambridge Phil. Soc. 44, (1948), 305{312.
[8] C. D. Pan, C. B. Pan, Goldbach conjecture, Science press, Beijing, 1992.
Department of Mathematics
Plovdiv University \P. Hilendarski"
24 \Tsar Asen" str.
Plovdiv 4000
Bulgaria
Email: [email protected]
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