article01_3. 282KB Jun 04 2011 12:06:50 AM

Journal de Th´eorie des Nombres
de Bordeaux 19 (2007), 1–25

Sign changes of error terms related to
arithmetical functions
par Paulo J. ALMEIDA

P
6
´sume
´. Soit H(x) = n≤x φ(n)
Re
e par une conn − π 2 x. Motiv´
jecture de Erd¨os, Lau a d´evelopp´e une nouvelle m´ethode et il a
d´emontr´e que #{n ≤ T : H(n)H(n + 1) < 0} ≫ T. Nous consiP
d´erons des fonctions arithm´etiques f (n) = d|n bdd dont l’addition
P
peut ˆetre exprim´ee comme n≤x f (n) = αx + P (log(x)) + E(x).


P
Ici P (x) est un polynˆ

ome, E(x) = − n≤y(x) bnn ψ nx + o(1) avec
ψ(x) = x − ⌊x⌋ − 1/2. Nous g´en´eralisons la m´ethode de Lau et
d´emontrons des r´esultats sur le nombre de changements de signe
pour ces termes d’erreur.
P
6
Abstract. Let H(x) = n≤x φ(n)
n − π 2 x. Motivated by a conjecture of Erd¨
os, Lau developed a new method and proved that
#{n ≤ T : H(n)H(n + 1) < 0} ≫ T. We consider arithmetical
P
functions f (n) = d|n bdd whose summation can be expressed as
P
+ E(x), where P (x) is a polynomial,
n≤x f (n) = αx + P (log(x))


P
E(x) = − n≤y(x) bnn ψ nx + o(1) and ψ(x) = x − ⌊x⌋ − 1/2. We
generalize Lau’s method and prove results about the number of
sign changes for these error terms.

1. Introduction
We say that an arithmetical function f (x) has a sign change on integers
at x = n, if f (n)f (n + 1) < 0. The number of sign changes on integers of
f (x) on the interval [1, T ] is defined as
Nf (T ) = #{n ≤ T, n integer : f (n)f (n + 1) < 0}.
We also define zf (T ) = #{n ≤ T, n integer : f (n) = 0}. Throughout this
work, ψ(x) = x − ⌊x⌋ − 1/2 and f (n) will be an arithmetical function such
Manuscrit re¸cu le 8 janvier 2006.
This work was funded by Funda¸ca
˜o para a Ciˆ
encia e a Tecnologia grant number
SFRH/BD/4691/2001, support from my advisor and from the department of mathematics of
University of Georgia.

Paulo J. Almeida

2

that

f (n) =

X bd
d|n

d

for some sequence of real numbers bn .

The motivation for our work was a paper by Y.-K. Lau [5], where he
proves that the error term, H(x), given by
X φ(n)
6
= 2 x + H(x)
n
π

n≤x

has a positive proportion of sign changes on integers solving a conjecture

stated by P. Erd¨os in 1967.
An important tool that Lau used to prove his theorem, was that the
error term H(x) can be expressed as


X µ(n)  x 
1
+O
ψ
(1)
H(x) = −
, (S. Chowla [3])
n
n
log20 x
n≤ x
log5 x

We generalize Lau’s result in the following way
Theorem 1.1. Suppose H(x) is a function that can be expressed as



X bn  x 
1
(2)
H(x) = −
ψ
+O
,
n
n
k(x)
n≤y(x)

where each bn is a real number and
1

(i) y(x) increasing, x 4 ≪ y(x) ≪
(3)

X

n≤x

x
(log x)5+

D
2

, for some D > 0, and

b4n ≪ x logD x;

(ii) k(x) is an increasing function, satisfying limx→∞ k(x) = ∞.
(iii) H(x) = H(⌊x⌋) − α{x} + θ(x), where α 6= 0 and θ(x) = o(1).

Let ≺ ∈ { 0 and
αH(n) ≺ 0. In particular,
(1) #{n ≤ T : αH(n) > 0} ≫ T ;

(2) if #{n ≤ T : αH(n) < 0} ≫ T , then NH (T ) ≫ T or zH (T ) ≫ T .

We consider arithmetical functions f (n) for which, the error term of the
summation function satisfies the conditions of Theorem 1.1. A first class is
described in the following result

Sign changes of error terms

3

Theorem 1.2. Let f (n) be an arithmetical function and suppose the sequence bn satisfies condition (3) and


X
x
(4)
bn = Bx + O
logA x
n≤x
for some B real, D > 0 and A > 6 +

D
2,

respectively. Let α =

P∞

bn
n=1 n2 ,

X b

X
B log 2πx γb
n
f (n) − αx +
− B log x and H(x) =
+ .
x→∞

n
2
2

γb = lim

n≤x

n≤x

If α 6= 0, then Theorem 1.1 is valid for the error term H(x). Moreover, if
f (n) is a rational function, then, except when α = 0, or B = 0 and α is
rational, we have
NH (T ) ≫ T

if and only if

#{n ≤ T : αH(n) < 0} ≫ T.

Notice that this class of arithmetical functions is closed for addition, i.e.,

if f (n) and g(n) are members of the class then also is (f + g)(n). In the
case considered by Lau, it was known that H(x) has a positive proportion
of negative values (Y.-F. S. P´etermann [6]), so the second part of Theorem
n
1.2 generalizes Lau’s result. Another example is f (n) = φ(n)
.
Using a result of U. Balakrishnan and Y.-F. S. P´etermann [2] we are able
to apply Theorem 1.1 to more general arithmetical functions:
Theorem 1.3. Let f (n) be an arithmetical function and suppose the sequence bn satisfies condition (3) and
(5)

∞
X
bn
= ζ β (s)g(s)
ns

n=1

for some β real, D > 0, and a function g(s) with a Dirichlet series expansion absolutely convergent for σ > 1−λ, for some λ > 0. Let α = ζ β (2)g(2)

and
( P
f (n) − αx,
if β < 0,
P⌊β⌋
Pn≤x
H(x) =
β−j
if β > 0,
n≤x f (n) − αx −
j=0 Bj (log x)
where the constants Bj are well defined. If α 6= 0, then Theorem 1.1 is
valid for the error term H(x).
Theorem 1.3 is valid for the following examples, where r 6= 0 is real:






σ(n) r
φ(n) r
φ(n) r
,
,
.
n
n
σ(n)

Paulo J. Almeida

4

2. Main Lemma
The main tool used by Y.-K. Lau was his Main Lemma, where he proved
P
6
that if H(x) = n≤x φ(n)
n − π 2 x then
2
Z 2T Z t+h
H(u) du
dt ≪ T h,
t

T

for sufficiently large T and any 1 ≤ h ≪ log4 T . Lau’s argument depends
essentially on the formula (1). In this section, we obtain a generalization
of Lau’s Main Lemma.
Main Lemma. Suppose H(x) is a function that can be expressed as (2)
and satisfies conditions (i)
and (ii) of Theorem 1.1. Then, for all large T
and h ≤ min log T, k 2 (T ) , we have
2
Z 2T Z t+h
3
H(u) du
dt ≪ T h 2 .
(6)
T

t

For any positive integer N , define
X bd  x 
(7)
HN (x) = −
ψ
.
d
d
d≤N

The Main Lemma will follow from the next result.
Lemma 2.1. Assume the conditions of the Main Lemma and take D > 0
satisfying condition (i). Let E = 4 + D
2 , then
(a) For any δ > 0, large T , any Y ≪ T and N ≤ y(T ), we have
Z T +Y
Y
Y
+ y(T + Y ) (log T )E ;
(H(u) − HN (u))2 du ≪ 1−δ + 2
k (T )
N
T


(b) For all large T , N ≤ y(T ) and 1 ≤ h ≤ min log T, k 2 (T ) , we have
2
Z 2T Z t+h
3
HN (u) du
dt ≪ T h 2 + N 3 (log N )E .
T

t

Now we prove the Main Lemma:
1

Proof. Take N = T 4 and δ > 0 small. Cauchy’s inequality gives us
Z t+h
Z t+h
2
2
H(u) du ≤ 2
HN (u) du
t

t

+2

Z

t

t+h

(H(u) − HN (u)) du

2

.

Sign changes of error terms

5

1

Since N = T 4 then, for sufficiently large T , N 3 logE N ≪ T . So, using
part (b) of Lemma 2.1 we have
2
Z 2T Z t+h
3
3
HN (u) du
dt ≪ T h 2 + N 3 logE N ≪ T h 2 .
t

T

Using Cauchy’s inequality and interchanging the integrals,
2
Z 2T Z t+h
(H(u) − HN (u)) du dt
t

T

2T

≤h

Z

2T +h

≤h

Z

Z

≪h

Z

min(u,2T )

max(u−h,T )

T

2T +h

Z

2

2

T





(H(u) − HN (u)) du dt

t

T

≤ h2

t+h

2

!

(H(u) − HN (u)) dt du

(H(u) − HN (u))2 du

T +h T +h
+ 2
+ y(2T + h) (log T )E
k (T )
N 1−δ
3



≪ T + Th ≪ Th2
since y(2T + h) ≪

T
(log T )E+1

Z

2T

T

and h ≤ min(log T, k 2 (T )). Hence
2
Z t+h
3
dt ≪ T h 2 .
H(u) du
t



3. Step I
In this section, we will prove part (a) of Lemma 2.1. Using expression
(2) and Cauchy’s inequality, we obtain
2



Z T +Y
Z T +Y
y(u)


X
u 
bm
Y

du+O
ψ
.
(H(u) − HN (u))2du ≤ 2
m
m
k 2 (T )
T
T
m=N +1


−1
−1
Let η(T, m, n) = max T, y (m), y (n) , then
2

Z T +Y
y(T +Y )
y(u)
 u  u
X bm  u 
X bm bn Z T +Y


du = 2
ψ
ψ
ψ
du.
2
m
m
mn η(T,m,n) m
n
T
m=N +1

m,n=N +1

Paulo J. Almeida

6

The Fourier series of ψ(u) = u − ⌊u⌋ − 12 , when u is not an integer, is
given by
∞
1 X sin(2πku)
ψ(u) = −
,
π
k

(8)

k=1

so we obtain
2
π2

y(T +Y )





Z
∞
ku
bm bn X 1 T +Y
lu
sin 2π
sin 2π
du.
mn
kl η(T,m,n)
m
n

X

m,n=N +1

k,l=1

Now, the integral above is equal to






Z
l
l
k
k
1 T +Y
cos 2πu
+
− cos 2πu
−
du.
2 η(T,m,n)
m n
m n
For the first term we get



Z T +Y
l
k
+
du ≪
cos 2πu
m n
η(T,m,n)
If

k
m

k
m

= nl , then
T +Y

Z



cos 2πu

η(T,m,n)



l
k
−
m n



1

.
+ nl

du ≤ Y,

otherwise
Z

T +Y



cos 2πu

η(T,m,n)



l
k
−
m n



1
.
du ≪  k
 − l
m
n

Part (a) of Lemma 2.1 will now follow from the next three lemmas.
Lemma 3.1. Let E = 4 +
X

m,n≤X

|bm bn |

D
2 as
∞
X

k,l=1

kn6=lm

in Lemma 2.1. Then
1
≪ X (log X)E .
kl | kn − lm|

Lemma 3.2. If D > 0 satisfies condition (3), then
X

m,n≤X

|bm bn |

∞
X

k,l=1

D
1
≪ X (log X)1+ 2 .
kl ( kn + lm)

Lemma 3.3. For any δ > 0,
X

N 0.
2
n
n
N

n≤N

n>N

Proof. Follows from Cauchy’s inequality, partial summations and the fact
that, for any ǫ > 0, τ (n) = O(nǫ ).

Remark. If H(x) can be expressed in the form (2), then


X |bn |
D
1
(9)
|H(x)| ≤
+O
≪ (log x)1+ 4 .
n
k(x)
n≤y(x)

Proof of Lemma 3.2 : Since the arithmetical mean is greater or equal to
the geometrical mean, we have
∞
∞
X
X
X
X
1
1
√
|bm bn |
|bm bn |
≤2
kl(kn + lm)
kl knlm
m,n≤X
m,n≤X
k,l=1
k,l=1
 X |b | 2
√m
≪
m
m≤X

 X  X b2 
M
≪
1
M
m≤X

M ≤X
D

≪ X(log X)1+ 2 .


Proof of Lemma 3.3 : For the second sum, take d = (m, n), m = dα and
n = dβ. Since kn = lm, then α|k and β|l. Taking k = αγ, we also have
l = βγ. As
∞
∞
X
1 X 1
π 2 (m, n)2
1
=
=
.
(10)
kl
αβ
γ2
6 mn
k,l=1

kn=lm

γ=1

Paulo J. Almeida

8

Then,
∞
|bm bn | X 1
π2
=
mn
kl
6

X

N