Annales Academiæ Scientiarum Fennicæ
Mathematica
Volumen 40, 2015, 385–395
MEAN SQUARE ESTIMATE FOR RELATIVELY
SHORT EXPONENTIAL SUMS INVOLVING
FOURIER COEFFICIENTS OF CUSP FORMS
Anne-Maria Ernvall-Hytönen
University of Helsinki, Department of Mathematics and Statistics
P. O. Box 68, FI-00014 University of Helsinki, Finland; [email protected]
Abstract. We estimate the mean square of a short exponential sum involving Fourier coefficients of a cusp form with a linear twist, a smooth weight function, and a relatively short averaging
interval.
1. Introduction
P∞
(κ−1)/2
e(nz) with e(x) = e2πix be a holomorphic cusp
Let f (z) =
n=1 a(n)n
form of weight κ with respect to the full modular group. Estimating exponential
sums involving Fourier coefficients of cusp forms is a classical question. For the so
called long sums, the best possible bound is well-known: Jutila [8] has proved that
X
a(n)e(nα) ≪ M 1/2 ,
1≤n≤M
when α ∈ R. This was an improvement over the classical result by Wilton [12]. By
the Rankin–Selberg mean value theorem [10] this bound is the best possible in the
general case, even though for some values of α it is possible to prove considerably
better bounds. For instance, for rational values of α = hk , the classical bound is
≪ k 2/3 M 1/3+ε .
However, the behavior of short sums, i.e., the sums over an interval [x, y], where
y − x = o(x) is much less known. These sums have been investigated for instance by
the author and Karppinen in [3]. Even though some of the bounds proved there are
sharp, it is likely that in many cases, the actual bounds are much smaller than what
have been proved.
It is generally a very difficult question to prove good bounds for individual sums.
It is much easier to consider the average behavior, namely, to bound expressions of
the type
2
ˆ M +∆
X
a(n)e(αn) dx.
w(x) M
x≤n≤x+y
Mean squares have been used to deduce the average behavior on various types
of objets of interest, including the zeta function and the error term in the divisor
doi:10.5186/aasfm.2015.4019
2010 Mathematics Subject Classification: Primary 11L07, 11F11, 11F30.
Key words: Fourier coefficients, cusp forms, exponential sums, mean square.
The author would like to thank Professor Valentin Blomer for fruitful conversations, and for the
idea of considering this question. Furthermore, the author would also like to thank Esa Vesalainen
for useful conversations during the time working with the problem. Finally, the author would also
like to thank the anonymous referee whose comments made this a much better paper. The work
was funded by the Academy of Finland, grant 138337.
Anne-Maria Ernvall-Hytönen
386
problem (for the latter, see Cramér [1]). For further results on the error term in the
divisor problem, see e.g. [4].
The classical mean square result for long sums can be found in [7], Theorem 1.2:
2
ˆ M X
h a(n)e n dx = c2 (κ)kM 3/2 + O k 2 M 1+ε + O k 3/2 M 5/4+ε .
k 1
n≤x
This result fits very well together with the result we are going to prove in this paper.
A corresponding result also holds for sums involving the divisor function twisted with
a rational parameter.
Questions closely connected to the topic of the current paper√have also been dealt
in [5], where Ivić proves the asymptotic result for α = 0, y ≪ x and ∆ = M, and
in [6], where Jutila proved an asymptotic result for a mean-square involving a sum
√
of values of the divisor function with y ≪ x1/2 and ∆ ≫ M 1/2 . In the case y = x
an exponential sum involving Fourier coefficients of a cusp form was dealt with in
[2]. There the averaging interval depended on the exponential twist similarly as in
the current paper. Very recently, Vesalainen [11] proved a mean square result for
exponential sums of length at most square root twisted with a rational parameter.
This leads to the natural question: what happens when the sum is short but longer
than of square root length, and the sum has an exponential twist.
We give the following answer to the question:
′
Theorem 1. Let 1 > δ > 1/2, T ≍ M δ , 0 < k ≪ M 1/2−ε for some fixed
positive ε′ and let h and k be co-prime. Let ∆ = min(k 2 M 1/2+ε , M), and let w(x) be
an infinitely many times differentiable smooth weight function in R that has support
on the interval [M, M + ∆] satisfying the conditions w(x)(j) ≪ ∆−j for all nonnegative j ∈ Z. Further assume w(x) = 1 for M + ∆′ ≤ x ≤ M + ∆ − ∆′ for some
∆′ ≍ ∆ and w(x) = 0 for x ≤ M and x ≥ M + ∆. Now
2
ˆ M +∆
X
h a(n)e n dx
w(x) k M
x≤n≤x+T
q
2
1+ε
ε 1/2
2
1+ε
ε
1/2
|S|(k M
= S + O(k M
+ ∆M T k) + O
+ ∆M T k) ,
where
2k
S= 2
π
X
2 −3/2
n≤min(M 2 T −2 ,M 2 T −1 ∆−1 )
√
nx
− 2π
k
!
sin2 2π
p
|a(n)| n
ˆ
M +∆
1/2
w(x)x
M
sin
2
p
n(x + T )
2π
k
!
√
n(x + T )
nx π
dx.
+ 2π
−
k
k
4
Furthermore,
(
∆T
S≪
∆M 1/2 k
when k ≫ T M −1/2 ,
otherwise.
The bounds for S given at the end of the theorem are actually the true orders
for the size of that term. Therefore, when the term S dominates, which is when k ≪
T 1/2 M −ε/2 (see Corollary 4), this theorem gives the average order of the exponential
sums, and not just an upper bound for the average.
Mean square estimate for relatively short exponential sums
387
′
Remark 2. We have to assume that 1 ≤ k ≪ M 1/2−ε for some positive ε′ . If we
allow k to be larger, then the error term in the truncated Voronoi type summation
formula would be too large, i.e., we would not get a better average bound than the
one obtained from Jutila’s estimate for long sums [8] using the triangle inequality.
Notice that if we are only interested in the upper bound, we can easily deduce
⋆
a result without the weight function, since by choosing the
function
weight
w (x) to
∆
3∆
be a smooth weight function supported on the interval M − 2 , M + 2 , and such
that w ⋆ (x) = 1 on [M, ∆], we have
2
2
ˆ M +3∆/2
ˆ M +∆ X
X
h h a(n)e n dx.
a(n)e n dx ≤
w ⋆ (x) k k M −∆/2
M
x≤n≤T
x≤n≤x+T
Hence, noticing that the term ∆M ε T 1/2 k is always smaller than the other terms, we
obtain the following corollary:
′
Corollary 3. Let 1 > δ > 1/2, T ≍ M δ , 0 < k ≪ M 1/2−ε for some fixed
positive ε′ and let h and k be co-prime. Let ∆ = min(k 2 M 1/2+ε , M). Then
2
ˆ M +∆ X
h a(n)e n
dx ≪ min ∆T M ε , ∆M 1/2+ε k + k 2 M 1+ε .
k M
x≤n≤x+T
Writing this corollary according to various values of k we obtain the following
bounds:
′
Corollary 4. Let 1 > δ > 1/2, T ≍ M δ , 0 < k ≪ M 1/2−ε for some fixed
positive ε′ and let h and k be co-prime. Let ∆ = min(k 2 M 1/2+ε , M). Then


2
∆M 1/2 k if k ≪ T M −1/2 ,
ˆ M +∆ X

h a(n)e n
if k ≫ T M −1/2 and k ≪ T 1/2 M −ε/2 ,
dx ≪ ∆T

k
M

x≤n≤x+T
∆M ε k 2 if k ≫ T M −1/2 and k ≫ T 1/2 M −ε/2 .
The main advantage in this theorem is the relatively short averaging interval
compared with the length of the sum, when the value of k is small. The averaging
interval is actually similar to the one in [2]. However, in the current paper, the sum
can be much longer than the averaging interval unlike there.
In order to prove Theorem 1, we need some lemmas, which will be presented in
the following section. In the final section, we will have the proof of Theorem 1.
In the following, the ε’s will be positive, not necessarily equal. The constants
implied by symbols ≪, ≫ and O( ) do not depend on M or k, but they do depend
on ε, δ, on the properties of the weight function, etc.
2. Lemmas
The first lemma is partial integration (see e.g. [9, Lemma 6]):
Lemma 5. Let A be a P ≥ 0 times differentiable function which is compactly
supported in a finite interval [a, b]. Assume also that there exist two quantities A0
and A1 such that for any non-negative integer ν ≤ P and for any x ∈ [a, b],
A(ν) (x) ≪ A0 A−ν
1 .
388
Anne-Maria Ernvall-Hytönen
Moreover, let B be a function which is real-valued on [a, b], and regular throughout
the complex domain composed of all points within the distance ̺ from the interval;
and assume that there exists a quantity B1 such that
0 < B1 ≪ |B ′ (x)|
for any point x in the domain. Then we have
P
ˆ b
A1
−P
A(x)e (B(x)) dx ≪ A0 (A1 B1 )
1+
(b − a) .
̺
a
The following lemmas can be proved using partial integration or the previous
lemma. The details are similar to those in [2].
Lemma 6. Let 1 ≤ m, n ≤ M. Then
!!
p
p
ˆ M +∆
nT1 (x)
mT2 (x)
1/2
w(x)x e ± 2
dx
+2
k
k
M
√
√ −P 1−P P/2+1/2
n+ m
≪ kP
∆
M
,
where T1 (x) and T2 (x) are x or x + T (not necessarily but possibly the same).
Lemma 7. Let 1 ≤ m, n ≤ M. Then
!
p
p
ˆ M +∆
nT
(x)
mT
(x)
dx
∓2
w(x)x1/2 e ±2
k
k
M
√
√ −P
≪ k P n − m ∆1−P M P/2+1/2 ,
where T (x) = x or T (x) = x + T .
Lemma 8. Let 1 ≤ m, n ≤ M. Then
p
√ !
ˆ M +∆
m(x
+
T
)
nx
dx
∓2
x1/2 w(x)e ±2
k
k
M
√
√ −P P/2+1/2
m
P 1−P ≪ max k ∆
.
√1 + T x−1 − n M
x∈[M,M +∆]
√
√
When √1+Tmx−1 − n = 0 for some x ∈ [M, M + ∆], Lemma 8 does not give any
information.
3. Proof of Theorem 1
Using a Voronoi-type summation formula (see [7, Theorem 1.2] with x ≍ M and
the choice N = M), we have
X
h
n
a(n)e
k
n≤x
√
√ −1
X
nx π
h̄
−3/4
1/4 1/2
+ O (M ε k)
cos 4π
−
= π 2
x k
a(n)e − n n
k
k
4
n≤M
with hh̄ ≡ 1 mod k. Squaring and integration over the error term gives a total
contribution of at most
∆M ε k 2 ≪ M 1+ε k 2 .
Mean square estimate for relatively short exponential sums
389
We may thus forget it for awhile because the contribution from the cross-terms
between the error term and the other terms can be taken into account using the
Cauchy–Schwarz inequality. Denote
√
√ −1
X
nx π
h̄
−3/4
1/4 1/2
S(x, M1 , M2 ) = π 2
.
cos 4π
−
a(n)e − n n
x k
k
k
4
M <n≤M
1
2
Hence, the expression we need to consider is
ˆ M +∆
w(x) |S(x + T, 0, M) − S(x, 0, M)|2 dx.
M
Let us first split the summation at M ′ :
ˆ M +∆
w(x) |S(x + T, 0, M) − S(x, 0, M)|2 dx
M
ˆ M +∆
2
=
w(x) |S(x + T, 0, M ′ ) − S(x, 0, M ′ )| dx
M
ˆ M +∆
2
+
w(x) |S(x + T, M ′ , M) − S(x, M ′ , M)| dx
M
ˆ M +∆
+O
w(x) |S(x + T, M ′ , M) − S(x, M ′ , M)|
M
′
′
· |S(x + T, 0, M ) − S(x, 0, M )| dx ,
and choose M ′ in the following way:
(
M 2 T −2
M′ =
M 2 T −1 ∆−1
if T ≫ ∆,
if ∆ ≫ T.
Notice that if ∆ = M, then ∆ ≫ T , and if ∆ = M 1/2+ε k 2 , then the condition ∆ ≫ T
is equivalent to M 1/2+ε k 2 ≫ T , so k ≫ T 1/2 M −1/4−ε . Let us concentrate on treating
the first two terms, and then use the Cauchy–Schwarz inequality to estimate the last
term. Let us start with the second term. We have
ˆ M +∆
2
w(x) |S(x + T, M ′ , M) − S(x, M ′ , M)| dx
M
ˆ M +∆
ˆ M +∆
2
2
′
≪
w(x) |S(x + T, M , M)| dx +
w(x) |S(x, M ′ , M)| dx.
M
M
Let us treat the second term as the first one can be treated similarly. We have
X
h̄
2
′
2 −1 1/2
a(n)a(m)e
(m − n) (mn)−3/4
|S(x, M , M)| = (2π ) x k
k
M ′ <m,n≤M
√
√
nx π
mx π
cos 4π
.
−
−
· cos 4π
k
4
k
4
Anne-Maria Ernvall-Hytönen
390
The terms with m = n yield a total contribution
√
ˆ M +∆
X
nx π
2 −1
2 −3/2
1/2
2
dx
−
(2π ) k
|a(n)| n
w(x)x cos 4π
k
4
M
′
M <n≤M
≪ kM
1/2
∆(M ′ )−1/2+ε .
When T ≫ ∆, we have M ′ = M 2 T −2 and hence
kM 1/2 ∆(M ′ )−1/2+ε ≪ kM 1/2 ∆M −1+ε T = k∆T M −1/2+ε ≪ k∆M ε T 1/2 .
When T ≪ ∆, we have M ′ = M 2 T −1 ∆−1 , and hence
kM 1/2 ∆(M ′ )−1/2+ε ≪ kM 1/2 ∆M −1+ε T 1/2 ∆1/2 ≪ k∆M ε T 1/2 .
When m 6= n, we need to estimate integrals
√ ˆ M +∆
√
√
x
1/2
w(x)x e ±2( n − m)
dx
k
M
and
√ √
√
x
w(x)x e ±2( n + m)
dx.
k
M
We only consider the first one as the second one is similar but simpler. We use
Lemma 7 to bound the integral. We obtain
√ ˆ M +∆
√
√
√
√
x
1/2
w(x)x e ±2( n − m)
dx ≪ ∆1−P k P M 1/2+P/2 | n − m|−P .
k
M
ˆ
M +∆
1/2
The sum over these estimates can be made as small as desired if
√
√
√
∆−1 ( n − m)−1 k M ≪ M −ε .
1/2+ε √
This is the case for all except ≍ kM ∆
n values around each n (as long as n is
sufficiently large such that there are values m 6= n in this neighborhood, which is
∆2
the case when n ≫ k2 M
1+ε ). Let us use absolute values in these cases to bound the
integral, and sum over these values:
X
h̄
2 −1
−3/4
(2π ) k
a(n)a(m)(nm)
e (m − n)
k
2
∆
<n≤M
k2 M 1+ε
kM 1/2+ε √
n
|n−m|≪
∆
·
ˆ
M +∆
1/2
w(x)x
M
≪k
X
√
e ±2( n −
√
√ x
m)
dx
k
(nm)ε−3/4 ∆M 1/2 ≪ kM 1/2 ∆
∆2
<n≤M
k2 M 1+ε
kM 1/2+ε √
n
|n−m|≪
∆
X
n
n−3/2+ε
kM 1/2+ε √
n ≪ k 2 M 1+ε .
∆
We have now derived the total contribution from the terms with m, n ≫ M ′ and
the error term in the Voronoi summation formula to be ≪ k 2 M 1+ε + k∆M ε T 1/2 .
Let us now move to considering the part
ˆ M +∆
2
w(x) |S(x + T, 0, M ′ ) − S(x, 0, M ′ )| dx.
M
Mean square estimate for relatively short exponential sums
391
This part is technically somewhat more challenging because it is not sufficient to
just use the triangle inequality, but we need to get some cancellation on the diagonal
terms. Therefore, we will also meet some integrals that are more difficult to consider
than the integrals treated in the first part of the proof.
We have
S(x + T, 0, M ′ )
!
p
X
n(x
+
T
)
k 1/2
π
h̄
= √ (x + T )1/4
−
a(n)e − n n−3/4 cos 4π
k
k
4
π 2
′
n≤M
!
p
n(x + T ) π
k 1/2 1/4 X
h̄
= √ x
−
a(n)e − n n−3/4 cos 4π
k
k
4
π 2
′
n≤M
!
p
X
n(x
+
T
)
h̄
π
k 1/2 −
a(n)e − n n−3/4 cos 4π
− √ (x + T )1/4 − x1/4
k
k
4
π 2
′
n≤M
!
p
n(x + T ) π
k 1/2 1/4 X
h̄
= √ x
−
a(n)e − n n−3/4 cos 4π
k
k
4
π 2
′
n≤M
+ O k 1/2 (M ′ )1/4+ε M −3/4 T .
Thus, we have
S(x + T, 0, M ′ ) − S(x, 0, M ′ )
!
p
1/2
X
n(x
+
T
)
k
π
h̄
= √ x1/4
−
a(n)e − n n−3/4 cos 4π
k
k
4
π 2
n≤M ′
!
√
nx π
− cos 4π
+ O k 1/2 (M ′ )1/4+ε M −3/4 T .
−
k
4
Let us first treat the error term and then concentrate on the main term. Squaring
and integrating over the main term gives the total contribution
∆k(M ′ )1/2+ε M −3/2 T 2 .
If T ≫ ∆, we have M ′ = M 2 T −2 , and hence
∆k(M ′ )1/2+ε M −3/2 T 2 ≪ ∆kM 1+ε T −1 M −3/2 T 2 = kM −1/2+ε ∆T ≪ k∆M ε T 1/2 .
If ∆ ≫ T , we have M ′ = M 2 T −1 ∆−1 , and hence
∆k(M ′ )1/2+ε M −3/2 T 2 ≪ ∆kM 1+ε T −1/2 ∆−1/2 M −3/2 T 2
= kM −1/2+ε ∆1/2 T 3/2 ≪ k∆M ε T 1/2 .
Anne-Maria Ernvall-Hytönen
392
In both cases, the contribution coming from the error term is at most k∆M ε T 1/2 .
Let us now move to the main term:
k 1/2
X
h̄
1/4
w(x) √ x
a(n)e − n n−3/4
k
π
2
M
n≤M ′
!
p
√
!2
n(x + T ) π
nx π
· cos 4π
− cos 4π
−
−
dx
k
4
k
4
ˆ M +∆
h̄
k X
−3/4
a(n)a(m)e (m − n)
(nm)
w(x)x1/2
= 2
2π m,n≤M ′
k
M
!
p
√
!
n(x + T ) π
nx π
· cos 4π
− cos 4π
−
−
k
4
k
4
!
p
√
!
m(x + T ) π
mx π
· cos 4π
− cos 4π
dx.
−
−
k
4
k
4
ˆ
M +∆
Let us first look at the case with n = m. Now
!
p
√
!2
n(x + T ) π
nx π
− cos 4π
−
−
cos 4π
k
4
k
4
!
!
p
p
√
√
n(x
+
T
)
n(x
+
T
)
nx
nx
π
= 4 sin2 2π
sin2 2π
.
− 2π
+ 2π
−
k
k
k
k
4
When n ≪
xk 2
,
T2
we have
p
n(x + T )
−
k
√
√
nx
nT
≪ √ ≪ 1.
k
k x
2
Hence, if MTk2 ≫ 1, and since by our choice of M ′ , we always have M ′ ≫
can split the diagonal sum into two. The first part gives the bound
ˆ M +∆
2k X
2 −3/2
|a(n)| n
w(x)x1/2 sin2
π2
M
2
n≪ M k2
T
M k2
,
T2
p
√ !
n(x + T )
nx
− 2π
2π
k
k
!
p
√
n(x
+
T
)
nx
π
· sin2 2π
dx
+ 2π
−
k
k
4
√
2
X
nT
2 −3/2
1/2
√
∆ ≪ ∆T
|a(n)| n
M
≪k
k M
M k2
n≪
T2
we
Mean square estimate for relatively short exponential sums
393
by partial summation. The second part gives the bound
p
√ !
ˆ M +∆
n(x
+
T
)
2k X
nx
|a(n)|2 n−3/2
w(x)x1/2 sin2 2π
− 2π
2
π
k
k
M
M k2
n≫
T2
!
p
√
n(x
+
T
)
π
nx
dx
+ 2π
−
· sin2 2π
k
k
4
−1/2
X
Mk 2
2 −3/2
1/2
1/2
≪k
≪ ∆T.
|a(n)| n
M ∆ ≪ ∆M k
T2
2
n≫ M k2
T
On the other hand, if
M k2
T2
≪ 1, we have
p
√ !
ˆ M +∆
n(x
+
T
)
nx
2k X
|a(n)|2 n−3/2
− 2π
w(x)x1/2 sin2 2π
2
π n≥1
k
k
M
!
p
√
n(x + T )
nx π
2
dx
+ 2π
−
· sin 2π
k
k
4
X
≪k
|a(n)|2 n−3/2 M 1/2 ∆ ≪ ∆M 1/2 k.
n≥1
We can now move to the terms with n 6= m. We split the cosine product into
exponential terms, and thus, we have to treat integrals of the type
!!
p
p
ˆ M +∆
mT
(x)
nT
(x)
1
2
dx,
±
w(x)x1/2 e ±2
k
k
M
where T1 (x) and T2 (x) are x or x + T , possibly but not necessarily the same. These
integrals have been treated in Lemmas 6–8. Lemma 6 always gives a good enough
bound. Lemma 7 gives a good
bound, namely, the integral can be made
√ enough
√
as small as desired, unless | n − m| ≪ M 1/2+ε k∆−1 . As in the first part of the
√
√
1/2+ε √
n values of m around each n for which | n − m| ≪
proof, there are ≍ kM ∆
M 1/2+ε k∆−1 . The contribution coming from these terms is again
≪ k 2 M 1+ε .
We may now move to the cases when using Lemma 8 fails. Notice first that writing
p
√
n(x + T )
mx
−2
,
fn,m (x) = 2
k
k
we have
√
√
n0
m0
′
− √
= 0,
fn0 ,m0 (x) = √
xk
x + Tk
for some y ∈ [M, M + ∆]. Next we want to show that when n lies
when n0 = m0 y+T
y
1/2+ε √
T
outside the interval Im,c = m 1 + MT+∆ − c, m 1 + M
+ c for c ≍ M ∆ k m,
then
′
′
fn,m
≫ M ε ∆−1
Anne-Maria Ernvall-Hytönen
394
for some ε′ , which is just enough that we can use partial integration Lemma 5 (and
therefore, Lemma 8) to estimate the integral. Write now
T
n=m 1+
− c.
M +∆
Then
q

q
m 1+ T −c
c
T
√ √
1
+
−
′
m
M +∆
M +∆
m
m
f (x) = = √ 

√
√
q
−
1
−
n,m
xk xk
x + Tk
1 + Tx
√ m
T
c
T
≫ √c
− −
≍ M ε ∆−1 .
≍ √
x m xk M + ∆
xmk
T
+ c. Denote Im = m 1 + MT+∆ ,
We can show a similar estimate for n = m 1 + M
∆2
T
. Notice that when m ≪ M 1+ε
, the interval Ic,m \ Im contain at most a
m 1+ M
k2
constant number of terms. Furthermore, byour choice
of M ′ , there is at most one n
for each m satisfying the condition n = m 1 + Ty for some y ∈ [M, M + ∆], since
the length of the interval Im is
T
T
mT ∆
m 1+
−m 1+
≍
.
M
M +∆
M2
T
are greater than m. When
All the integers on the interval m 1 + MT+∆ , m 1 + M
−1
the distance mT M between the largest number on the interval and m is at most
o(1), there cannot be any numbers on the interval. This is the case when m =
o(MT −1 ).
∆2
In particular, when both m = o M 1+ε
and m = o(MT −1 ), the sum is empty.
2
k
2
∆
Therefore, it suffices to estimate the sums m ≫ M 1+ε
and m ≫ MT −1 . We will just
k2
take absolute values of the integrals. The contribution coming from interval Ic,m \ Im
is
X
|a(n)||a(m)|(nm)−3/4 ∆M 1/2 ≪ k 2 M 1+ε .
≪k
2
∆
m≫ 1+ε
M
k2
n∈Im,c \Im
The contribution coming from interval Im is
X
|a(n)||a(m)|(nm)−3/4 ∆M 1/2 ≪ k∆M 1/2
≪k
m≫M T −1
n∈Im
X
m≫M T −1
mε−3/2 ≪ k∆M ε T 1/2 .
We have now derived the contribution from the terms m, n ≪ M ′ to be the contribution from the main term and an error of size at most ≪ k 2 M 1+ε + k∆M ε T 1/2 . Using
the Cauchy–Schwarz inequality completes the proof.
Mean square estimate for relatively short exponential sums
395
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Received 14 November 2013 • Revised received 14 July 2013 • Accepted 26 September 2014