Acta Mathematica Academiae Paedagogicae Nyı́regyháziensis
28 (2012), 1–11
www.emis.de/journals
ISSN 1786-0091
MEAN VALUE OF HARDY SUMS OVER SHORT
INTERVALS
WEIXIA LIU
Abstract. The main purpose of this paper is to study the mean value
properties of certain Hardy sums over short intervals by using the mean
value theorems of the Dirichlet L-functions, and to give two interesting
asymptotic formulae.
1. Introduction and the main result
For a positive integer q and an arbitrary integer h, the Dedekind sum S(h, q)
is defined by
q X
a
ah
S(h, q) =
,
q
q
a=1
where
x − [x] − 21 , if x is not an integer;
0,
if x is an integer,
[x] is the largest integer less than or equal to x. The various properties of
S(h, q) were investigated by many authors. Maybe the most famous property
of the Dedekind sums is the reciprocity formula (see references [1, 3, 7])
((x)) =
h2 + q 2 + 1 1
−
12hq
4
valid for all (h, q) = 1, h > 0, q > 0. A three term version of (1.1) was
discovered by H. Rademacher [6]. In this paper we study the Hardy sums
(1.1)
S(h, q) + S(q, h) =
H(h, k) =
k−1
X
(−1)j+1+[hj/k] .
j=1
Since the sums H(h, k) can be expressed explicitly in terms of Dedekind sums,
they are connected with Dedekind sums. Some arithmetical properties of
2010 Mathematics Subject Classification. 11F20.
Key words and phrases. Hardy sums, mean value, asymptotic formula.
Supported by Technology Project of Minjiang University (YKY:1014).
1
2
WEIXIA LIU
H(h, k) can be found in B. C. Berndt [2] and R. Sitaramachandra Rao [8].
In [10] an asymptotic formula for the 2m-th power mean of these sums was
proved, namely
p−1
2
−m
X
)
6 ln p
2m
2m ζ (2m)(1 − 4
2m−1
|H(h, p)| = p
+O p
exp
,
−m )
ζ(4m)(1
+
4
ln
ln
p
h=1
here p is an odd prime, m is a positive integer and ζ(s) is the Riemann zeta
function.
What about the mean value of H(h, p) when h runs through an interval
[1, λp], 0 < λ < 1? It seems difficult to obtain an asymptotic formula even in
case like λ = 31 , 14 . Here we study the better behaved mean value:
X X
X X
H(2ab̄, p),
H(2ab̄, p),
a<p/3 b<p/3
a<p/3 b<p/4
where b̄ is defined by the equation bb̄ ≡ 1(modp). The factor 2 is necessary
since H(d, p) = 0 for an odd number d (see Lemma 1).
Theorem 1. Let p ≥ 5 be a prime. We have
(I)
X X
1
H(2ab, p) = p2 + O(p1+ε ).
5
a≤p/3 b≤p/3
(II)
X X
a≤p/3 b≤p/4
H(2ab, p) =
27 2
p + O(p1+ε ).
320
There are also five other types of Hardy sums, all of them closely connected
with Dedekind sums [4]. Since their relation to L-functions is quite similar to
that used here (see Lemma 1), analogues of our theorem can be obtained for
these sums.
2. Some lemmas
To prove the theorem, we need the following lemmas.
Lemma 1. Let p be a prime and h be a positive integer with (h, p) = 1. We
have
X

2
16p
χ(h) L(1, χχ02 ) , if 2|h;

 − π2 (p−1)
χ mod p
H(h, p) =
χ(−1)=−1


0,
if 2 - h,
where χ02 denotes the principal character modulo 2 and the sum is over the odd
character (modp).
Proof. See [10, Lemma 3].
MEAN VALUE OF HARDY SUMS OVER SHORT INTERVALS
3
Lemma 2. Let p ≥ 5 be a prime. For any non-principal character χ modulo
p, we have
X
3τ (χ)
L(1, χχ03 )
χ(a) =
2πi
a≤p/3
and
X
τ (χ)
[2 + χ(2) − χ(4)] L(1, χ),
2πi
a≤p/4
P
where χ03 is the principal character modulo 3, and τ (χ) = pa=1 χ(a)e2πia/p is
the Gauss sum.
χ(a) =
Proof. These identities can be easily deduced from the Fourier expansion for
primitive character sums(see [5])
τ (χ) P+∞ χ(n) sin(2πnλ)
X
,
if χ(−1) = 1;
π Pn=1
n
χ(a) = τ (χ)
+∞ χ(n)(1−cos(2πnλ))
, if χ(−1) = −1.
n=1
πi
n
a≤λp
See also reference [9].
Lemma 3. Let χ02P
and χ03 be the principal
P characters modulo 2 and 3, respectively. If r1 (n) = d|n χ02 (d), r2 (n) = d|n χ02 (d)χ03 ( nd ), then for any integer
m ≥ 0, we have the identity
∞
X
r1 (n)r2 (2m n)
π4
=
.
2
n
40
n=1
Proof. Noting that r1 (n), r2 (n) are multiplicative functions, we can write
! ∞
∞
∞
j
m+j
X
X
X r1 (n)r2 (n)
r
(2
)r
(2
)
r1 (n)r2 (2m n)
1
2
m
(2.1)
=
r
(2
)
+
2
n2
4j
n2
n=1
n=1
j=1
=
1+
∞
X
j=1
1
4j
!
2-n
∞
X
n=1
2-n
r1 (n)r2 (n)
n2
4 X r1 (n)r2 (n)
.
3 n=1
n2
∞
=
2-n
For the summation
∞
X
r1 (n)r2 (n)
n2
n=1
2-n
∞
X
r1 (n)r2 (n)
n=1
2-n
n2
=
Y
p≥3
, we can write
r1 (p)r2 (p) r1 (p2 )r2 (p2 )
1+
+
+ ...
p2
p4
4
WEIXIA LIU
by using the Euler product formula. Note that
r1 (pα ) = 1 + α
and
α
r2 (p ) =
1,
if p = 3,
α + 1, if p =
6 3,
for any positive integer α and an odd prime p. Hence
(2.2)
∞
X
r1 (n)r2 (n)
n=1
2-n
n2
=
Y
2
3
22 32
(α + 1)2
= 1 + 2 + 4 + ...
1 + 2 + 4 + ··· +
+ ...
3
3
p
p
p2α
p≥5
−4 81 Y
1
1
27 4
3π 4
=
1− 2
1− 4 =
ζ (2)/ζ(4) =
,
64 p≥5
p
p
100
160
π2
6
where we used that ζ(2) =
we have
and ζ(4) =
π4
.
90
∞
X
r1 (n)r2 (2m n)
n2
n=1
Therefore, from (2.1) and (2.2),
=
π4
.
40
This proves Lemma 3.
Lemma 4. For any integer m ≥ 0, we have the asymptotic formula
X
2
χ(2m ) L(1, χχ02 ) L(1, χχ03 )L(1, χ) =
χ mod p
χ(−1)=−1
Proof. For convenience, we put
X
A(y, χ) =
χ(n)r1 (n),
N <n≤y
π4
(p − 1) + Om (pε ).
m
90 · 2
B(y, χ) =
X
χ(n)r2 (n),
N <n≤y
where N is a parameter with p ≤ N ≤ p4 and r1 , r2 were defined in Lemma
3. Then from Abel’s identity, we have
∞
X
X χ(n)r1 (n) Z ∞ A(y, χ)
χ(n)r1 (n)
0
L(1, χχ2 )L(1, χ) =
=
+
dy;
2
n
n
y
N
n=1
1≤n≤N
L(1, χχ02 )L(1, χχ03 )
=
∞
X
χ(n)r2 (n)
n=1
n
X χ(n)r1 (n) Z ∞ B(y, χ)
=
+
dy.
2
n
y
N
1≤n≤N
MEAN VALUE OF HARDY SUMS OVER SHORT INTERVALS
5
Hence, we can write
X
2
(2.3)
χ(2m ) L(1, χχ02 ) L(1, χχ03 )L(1, χ) =
χ mod p
χ(−1)=−1
!
X χ(n1 )r1 (n1 ) Z ∞ A(y, χ)
=
χ(2m )
+
dy ×
n1
y2
N
1≤n
≤N
χ(−1)=−1
1
!
X χ(n2 )r1 (n2 ) Z ∞ B(y, χ)
×
+
dy
n2
y2
N
1≤n2 ≤N
!
!
X
X χ(n1 )r1 (n1 )
X χ(n2 )r1 (n2 )
=
χ(2m )
n1
n2
1≤n1 ≤N
1≤n2 ≤N
χ(−1)=−1
!
X
X χ(n1 )r1 (n1 ) Z ∞ B(y, χ) χ(2m )
+
dy
2
n
y
1
N
1≤n1 ≤N
χ(−1)=−1
!
X χ(n2 )r1 (n2 ) Z ∞ A(y, χ) X
χ(2m )
+
dy
n2
y2
N
1≤n2 ≤N
χ(−1)=−1
Z ∞
Z ∞
X
A(y, χ)
B(y, χ)
m
χ(2 )
+
dy
dy
y2
y2
N
N
X
χ(−1)=−1
:= M1 + M2 + M3 + M4 .
Now, we calculate each term in the expression (2.3).
(i) From the orthogonality of Dirichlet characters we can write
(2.4)
M1 =
X
χ(−1)=−1
X χ(n1 )r1 (n1 )
χ(2m )
n1
1≤n ≤N
1
!
X χ(n2 )r1 (n2 )
n2
1≤n ≤N
!
2
X r1 (n1 )r2 (n2 ) X
1 X
(1 − χ(−1))χ(2m n1 n2 )
=
2 1≤n ≤N 1≤n ≤N
n1 n2
χ mod p
1
2

=

X
X r1 (n1 )r2 (n2 ) X
X r1 (n1 )r2 (n2 ) 
p−1
0
0
0
0
,

−

2 1≤n ≤N 1≤n ≤N
n1 n2
n
n
1 2
1≤n ≤N 1≤n ≤N
1
2
2m n1 ≡n2 ( mod p)
1
2
2m n1 ≡−n2 ( mod p)
P
0
where
denotes the summation over all 1 ≤ n1 ≤ N such that
1≤n1 ≤N
(n1 , p) = 1.
For convenience, we split the sum over n1 or n2 into following cases:
6
WEIXIA LIU
a.) p/2m ≤ n1 ≤ N, p ≤ n2 ≤ N ;
b.) 1 ≤ n1 ≤ p/2m − 1, p ≤ n2 ≤ N ;
c.) p/2m ≤ n1 ≤ N, 1 ≤ n2 ≤ p − 1;
d.) 1 ≤ n1 ≤ p/2m − 1, 1 ≤ n2 ≤ p − 1.
Then we have
X
X r1 (n1 )r2 (n2 )
p−1
0
2
p/2m ≤n1 ≤N
2m n1 ≡n2 (
0
p≤n2 ≤N
mod p)
X
p
n1 n2
X
1≤s1 ≤2m N /p 1≤s2 ≤N/p
X
X
p
1≤s1 ≤2m N /p 1≤s2 ≤N/p
X
pε
X
p−1 p−1
X
X r1 (s1 p + `1 )r2 (s2 p + `2 )
(s1 p + `1 )(s2 p + `2 )
` =1 ` =1
1
X
0
X
0
1≤n1 ≤p/2m −1 p≤n2 ≤N
2m n1 ≡n2 ( mod p)
2
`1 ≡`2 ( mod p)
p−1
X
[(s1 p + `1 )(s2 p + `1 )]ε
(s1 p + `1 )(s2 p + `1 )
` =1
1≤s1 ≤2m N /p 1≤s2 ≤N/p
and
p−1
2
1
1
m pε
s1 s2
r1 (n1 )r2 (n2 )
p
n1 n2
X
X
(rpn1 )ε−1 pε .
1≤n1 ≤p/2m −1 1≤r≤N/p
Moreover,
p−1
2
X
0
X
p/2m ≤n1 ≤N 1≤n2 ≤p
2m n1 ≡n2 ( mod p)
0
r1 (n1 )r2 (n2 )
pε ,
n1 n2
where we have used the estimate r1 (n) nε , and r2 (n) nε .
For the case 1 ≤ n1 ≤ p/2m − 1, 1 ≤ n2 ≤ p − 1, the solution of the
congruence 2m n1 ≡ n2 (modp) is 2m n1 = n2 . Hence,
X
X
X
p−1
r1 (n1 )r2 (2m n1 )
p−1
0
0 r1 (n1 )r2 (n2 )
= m+1
2
n1 n2
2
n12
m
m
1≤n ≤p−1
1≤n1 ≤p/2 −1
1≤n1 ≤p/2 −1
2
2m n1 ≡n2 ( mod p)
p − 1 X r1 (n)r2 (2m n)
+ Om (pε ).
m+1
2
2
n
n=1
∞
=
Now, from Lemma 3, we can immediately get
π4
p − 1 X 0 X 0 r1 (n1 )r2 (n2 )
(2.5)
=
(p − 1) + Om (pε ).
2 1≤n ≤N 1≤n ≤N
n1 n2
90 · 2m
1
2
2m n1 ≡n2 ( mod p)
MEAN VALUE OF HARDY SUMS OVER SHORT INTERVALS
Similarly, we can also get
(2.6)
p − 1 X 0 X 0 r1 (n1 )r2 (n2 )
=
2 1≤n ≤N 1≤n ≤N
n1 n2
1
2
2m n1 ≡−n2 ( mod p)
=
p − 1 X 0 X 0 r1 (n1 )r2 (n2 ) φ(p) X 0 X 0 r1 (n1 )r2 (n2 )
+
2 1≤n ≤N 1≤n ≤N
n1 n2
2 1≤n ≤N 1≤n ≤N
n1 n2
1
2
1
2m n1 +n2 =p
p
X
X
2m r1 ( p−n
)r2 (m)
0
2m
+p
n(p
−
n)
1≤n≤p−1
1≤n ≤N
2
m
X
X
(n(p − n))ε
+
n
1≤n≤p−1
1≤n ≤N
2
m p + p
ε
ε
2
2m n1 +n2 =`p,`≥2
N X
N
X
nε `ε
N +n2
p
2
p
nε2 (`p − n2 )ε
`n2 − n22 /p
m pε .
2
`n2
n2 =1 `=1
2≤n2 ≤
X
2≤n2 ≤
2
X 2m r1 ( `p−n
)r2 (n2 )
2m
(`p − n2 )n2
N +n
Then from (2.4), (2.5) and (2.6), we have
(2.7)
M1 =
π4
p + Om (pε ).
m+4
5·2
(ii) Note the partition identity
B(y, χ) =
+
X
√
X
√
n≤ y
−
√
−

+
X
χ(n)χ02 (n)
n≤y/m
χ(n)χ02 (n)
X
√
n≤ y
X
χ(m)χ03 (m) −

χ(n)χ02 (n) 
X
√
n≤ N
X
√
n≤ y

χ(n)χ02 (n) 
X
√
m≤ N
m≤N/n
n≤ N

χ(m)χ03 (m)
m≤y/n
χ(m)χ03 (m)
m≤ y
X
X
χ(n)χ02 (n)
χ(m)χ03 (m)

χ(n)χ03 (n)
X
√
n≤ N

χ(n)χ03 (n) .
X
n≤N/m
χ(n)χ02 (n)
7
8
WEIXIA LIU
Applying Cauchy inequality, the estimate
2
2
X X
X X
χ(n) =
χ(n)
χ6=χ0 N <n≤M
χ6=χ0 N <n≤M ≤N +p
X
X
= (p − 1)
χ0 (n) − 2
(p − 1)2
χ0 (n) ≤
4
N <n≤M ≤N +p
N <n≤M ≤N +p
and noting that the identity
X
X
χ(n)χ02 (n) =
µ(d)χ(d)
N <n≤M
d|2
(2.8)
√ X
|B(y, χ)|2 y
+
+
√
y
X
√
n≤ y χ mod p
χ(−1)=−1
χ mod p
χ(−1)=−1
X
X
√
m≤ y χ mod p
χ(−1)=−1
X
X
√
m≤ y χ mod p
χ(−1)=−1
Similarly,
χ(n),
N/d<n≤M/d
we have
X
X
2
X
0
χ(m)χ2 (m)
m≤y/n
2
X
0
χ(n)χ
(n)
2
n≤y/m
2 2
X
X
0
0
×
yp2+ε .
χ(n)χ
(n)
χ(n)χ
(n)
2
3
√
√
n≤ y
n≤ y
X
|A(y, χ)|2 yp2+ε .
χ mod p
χ(−1)=−1
Then from Cauchy inequality and (2.8) we can write
!
X
X χ(n1 )r1 (n1 ) Z ∞ B(y, χ) (2.9)
χ(2m )
dy
M2 =
n1
y2
N
1≤n1 ≤N
χ(−1)=−1


Z ∞
X
X
1 
nε−1
|B(y, χ)| dy
1
2
y
N
1≤n ≤N
1
Z
N
∞
p
3
+ε
2
√
χ(−1)=−1
y
3
p 2 +ε
dy 1 −ε .
y2
N2
(iii) Similarly to (ii), we can also get
ε
N
3
(2.10)
M3 p 2 +ε
N 2 −ε
1
.
MEAN VALUE OF HARDY SUMS OVER SHORT INTERVALS
9
(iv) By the same argument as in (ii), we can write
(2.11)
Z
X
M4 =
m
χ(2 )
χ(−1)=−1
Z ∞Z ∞
≤
N
Z
N
∞
1
y2
N
Z
∞
∞
N
N
1
y2z2
Z
A(y, χ)
dy
y2
X

∞
N
B(y, χ)
dy
y2
|A(y, χ)||B(z, χ)|dydz
χ(−1)=−1
∞
Z
X
 12 
 21
X
1 
|B(y, χ)|2  dydz
|A(y, χ)|2  
2
z
N
χ(−1)=−1
χ(−1)=−1
2
p1+ε
p2+ε
.
3 dy
N
y2
Now, taking N = p3 , combining (2.3)-(2.11), we obtain the asymptotic formula
X
2
π4
χ(2m ) L(1, χχ02 ) L(1, χχ03 )L(1, χ) =
p + Om (pε ). m+4
5
·
2
χ mod p
χ(−1)=−1
Lemma 5. Let χ02 and χ03 be the principal characters modulo 2 and 3, respectively. Then we have
X
2 2
π4
(p − 1) + O(pε ).
χ(2) L(1, χχ02 ) L(1, χχ03 ) =
180
χ mod p
χ(−1)=−1
Proof. From the method of proving Lemma 3, we can easily get this Lemma.
3. Proof of the Theorem
In this section we complete the proof of our theorem.
Proof. i) From Lemma 1 and Lemma 2 we have
XX
H(2ab, p)
a< p3 b< p3
=−
=
16p
2
π (p − 1)
36p
4
π (p − 1)
X
2 X
χ(a)
χ(b)
χ(2) L(1, χχ02 )
X
χ mod p
χ(−1)=−1
X
χ mod p
χ(−1)=−1
a≤ p3
b≤ p3
2 2
χ(2) |τ (χ)|2 L(1, χχ02 ) L(1, χχ03 ) ,
10
WEIXIA LIU
using τ (χ)τ (χ) = p if χ(−1) = −1. Thus, applying Lemma 4, we obtain
XX
1
H(2ab, p) = p2 + O(p1+ε ).
5
p
p
a< 3 b< 3
This complete the proof of i) in the theorem.
ii) Lemma 1 and Lemma 2 imply
XX
H(2ab, p)
a< p3 b< p4
=−
16p
2
π (p − 1)
X
2 X
χ(a)
χ(b)
χ(2) L(1, χχ02 )
X
χ mod p
χ(−1)=−1
a≤ p3
b≤ p4
X
2
24p
χ(2) χ(4)
= 4
χ(2) 1 +
−
|τ (χ)|2 L(1, χχ02 ) L(1, χχ03 )L(1, χ)
π (p − 1)χ mod p
2
2
χ(−1)=−1
=−
2
12p
4
π (p − 1)
X
2
χ(2) [2χ(2) + χ(4) − χ(8)] L(1, χχ02 ) L(1, χχ03 )L(1, χ)
χ mod p
χ(−1)=−1
where using τ (χ)τ (χ) = p if χ(−1) = −1. Thus, applying Lemma 4, we obtain
XX
27 2
H(2ab, p) =
p + O(p1+ε ).
320
p
p
a< 3 b< 4
This complete the proof of ii) in the theorem.
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MEAN VALUE OF HARDY SUMS OVER SHORT INTERVALS
Received May 2, 2011.
Weixia Liu,
Straits Institute of Minjiang University,
Fuzhou 350108,
P. R. China.
E-mail address: liuweixia0201@126.com
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