On the Seventh Power Moment of $\Delta(x)$

0.19 MB·

by  Jinjiang Li

x
On the Seventh Power Moment of ∆( )
Jinjiang Li
Department of Mathematics,
6
China University of Mining and Technology,
1
0 Beijing 100083,P. R. China
2
Email: [email protected]

n
a
Abstract: Let∆(x)betheerrortermoftheDirichletdivisorproblem. Inthispaper,weestablish
J

1 an asymptotic formula of the seventh-power moment of ∆(x) and prove that
2
T 7(5s (d) 3s (d) s (d))
] ∆7(x)dx= 7;3 − 7;2 − 7;1 T11/4+O(T11/4−δ7+ε)
T Z2 2816π7
N
with δ =1/336, which improves the previous result.
7
.
h Keywords: Dirichlet divisor problem; higher-power moment; asymptotic formula.
t
a
m
1 Introduction and main result
[

1
v Throughout this paper, let d(n) denote the Dirichlet divisor function. In 1838, Dirichlet proved that
5
the error term
1
5 ∆(x):= d(n) xlogx (2γ 1)x, with x>2,
− − −
5
n6x
X
0
. satisfies ∆(x) x1/2. Here γ is Euler’s constant. Since then the determination of the exact order of
1 ≪
0 ∆(x) has been called Dirichlet’s divisor problem. Many writers have sharpened Dirichlet’s bound for
6
∆(x). The latest result is due to Huxley [5], who proved that
1
:
v ∆(x) x131/416(logx)26957/8320.
i ≪
X
r For a survey of the history of this problem, see Kr¨atzel [10].
a
In the opposite direction, Hardy [3] proved that
Ω x1/4(logx)1/4loglogx ,
+
∆(x)=
 Ω (cid:0)x1/4 . (cid:1)
−
The best results in this direction to date are(cid:0) (cid:1)
∆(x)=Ω x1/4(logx)1/4(loglogx)(3+log4)/4exp( c logloglogx)
+
−
(cid:16) p (cid:17)
1
and
∆(x)=Ω x1/4exp(c(loglogx)1/4(logloglogx)−3/4
−
(cid:16) (cid:17)
for some constant c>0, due to Hafner [2] and [1] respectively. It is conjectured that ∆(x) x1/4+ε
≪
is true for every ε > 0. The evidence in support of this conjecture has been given by Tong [14] and
Ivic [6], who proved, respectively, that
T ζ4(3)
∆2(x)dx= 2 T3/2+O(T log5T) (1.1)
6π2ζ(3)
Z2
and
T 35
∆(x)Adx ε T1+A4+ε, for 06A6 (1.2)
| | ≪ 4
Z2
and any ε>0. On the other hand , Vorono¨ı[16] proved that
T
∆(x)dx=T/4+O(T3/4), (1.3)
Z2
whichin conjunction with (1.1) and(1.2) showsthat ∆(x) has a lotof signchangesandcancellations
between the positive and negative portions.
Tsang [15] first studied the third and fourth-power moments of ∆(x). He proved that
T 3c
∆3(x)dx= 1 T7/4+O(T7/4−δ3+ε), (1.4)
28π3
Z2
T 3c
∆4(x)dx= 2 T2+O(T2−δ4+ε), (1.5)
64π4
Z2
where δ =1/14,δ =1/23 and
3 4
c := (αβ(α+β)) 3/2h 9/4 µ(h)d(α2h)d(β2h)d((α+β)2h),
1 − −
| |
α,β,h N
X∈
c2 := (nmkℓ)−3/4d(n)d(m)d(k)d(ℓ),
n,m,k,ℓ N
X∈
√n+√m=√k+√ℓ
and µ(h) is the Mo¨bius function.
In [18], Zhai proved that (1.4) holds for δ = 1/4. Ivi´c and Sargos [8] proved that (1.4) holds
3
for δ = 7/20. Following the approach of Tsang [15], Zhai [18] proved that the equation (1.5) holds
3
for δ = 2/41. This approach used the method of exponential sums. In particular, if the exponent
4
pair conjecture is true, namely, if (ε,1/2+ε) is an exponent pair, then the equation (1.5) holds for
δ = 1/14. Moreover, in [8], Ivi´c and Sargos proved a substantially better result that the equation
4
(1.5) holds for δ =1/12. Later, combining the method of [8] and a deep result of Robert and Sargos
4
[12], Zhai [20] proved that the equation (1.5) holds for δ = 3/28. Recently, Kong [9] proved that
4
δ =1/8.
4
By a unified approach, Zhai [19] proved that the asymptotic formula
T
∆k(x)dx=CkT1+k/4+O T1+k/4−δk+ε (1.6)
Z1 (cid:16) (cid:17)
2
holds for 3 6 k 6 9, where C and 0 < δ < 1 are explicit constants. He gives δ = 1/64,δ =
k k 5 6
35/4742,δ =17/6312,δ =8/9433,δ =13/75216.Theasymptoticformula(1.6)improvedtheresult
7 8 9
of Heath-Brown [4]. When k = 5, the asymptotic formula (1.6) holds for δ = 1/64, which improved
5
an earlier exponent δ =5/816provedin [18] by the approachof Tsang [15]. In [21], Zhang and Zhai
5
improved the previous result of the case k = 5 and proved δ = 3/80. Meanwhile, Wang [17] studied
5
the case k =6 and proved δ =3/248, which improved the result of Zhai [19], i.e. δ =35/4742.
6 6
The aimofthis paperistoimprovethe valueofδ =17/6312,whichisachievedbyZhai[19]. The
7
main result is the following
Theorem 1.1 We have
T 7(5s (d) 3s (d) s (d))
∆7(x)dx= 7;3 − 7;2 − 7;1 T11/4+O(T11/4−δ7+ε)
2816π7
Z2
with δ =1/336, where
7
d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q)
s (d)= ,
7;3 (nmkℓrsq)3/4
n,m,k,Xℓ,r,s,q∈N∗
√n+√m+√k+√ℓ=√r+√s+√q
d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q)
s (d)= ,
7;2 (nmkℓrsq)3/4
n,m,k,Xℓ,r,s,q∈N∗
√n+√m+√k+√ℓ+√r=√s+√q
d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q)
s (d)= .
7;1 (nmkℓrsq)3/4
√n+√mn+,m√,kk,+Xℓ,√r,ℓs+,q√∈rN+∗√s=√q
Notations. Throughout this paper, x denotes the distance from x to the nearest integer, i.e.,
k k
x = min x n. [x] denotes the integer part of x; n N means N < n 6 2N; n N means
k k n Z| − | ∼ ≍
C N 6 n∈6 C N with positive constants C ,C satisfying C < C . ε always denotes an arbitrary
1 2 1 2 1 2
smallpositiveconstantwhichmaynotbethesameatdifferentoccurances. Weshallusetheestimates
d(n) nε. Suppose f :N R is any function satisfying f(n) nε, k >2 is a fixed integer. Define
≪ → ≪
f(n )f(n ) f(n )
s (f):= 1 2 ··· k , 16ℓ<k. (1.7)
k;ℓ (n n n )3/4
n1,···,nℓ,nXℓ+1,···,nk∈N∗ 1 2··· k
√n1+···+√nℓ=√nℓ+1+···+√nk
Weshalluses (f)todenotebothoftheseries(1.7)anditsvalue. Supposey >1isalargeparameter,
k;ℓ
and we define
f(n )f(n ) f(n )
s (f;y):= 1 2 ··· k , 16ℓ<k.
k;ℓ (n n n )3/4
1 2 k
n1,···,nℓ,nXℓ+1,···,nk6y ···
√n1+···+√nℓ=√nℓ+1+···+√nk
2 Preliminary Lemmas
Lemma 2.1 Suppose k >3,(i ,i , ,i ) 0,1 k 1 such that
1 2 k 1 −
··· − ∈{ }
√n1+( 1)i1√n2+( 1)i2√n3+ +( 1)ik−1√nk =0.
− − ··· − 6
3
Then we have
√n1+( 1)i1√n2+( 1)i2√n3+ +( 1)ik−1√nk max(n1,n2, ,nk)−(2k−2−2−1).
| − − ··· − |≫ ···
Proof. See Lemma 2.2 of [19].
Lemma 2.2 If g(x) and h(x) are continuous real-valued functions of x and g(x) is monotonic, then
b v
g(x)h(x)dx max g(x) max h(x)dx .
Za ≪(cid:18)a6x6b| |(cid:19)(cid:18)a6u<v6b(cid:12)Zu (cid:12)(cid:19)
(cid:12) (cid:12)
Proof. See Lemma 1 of [15]. (cid:12)(cid:12) (cid:12)(cid:12)
Lemma 2.3 Suppose A,B R,A=0. Then
∈ 6
2T
cos(A√t+B)dt T1/2 A 1.
−
≪ | |
ZT
Proof. It follows from Lemma 2.2 easily.
Lemma 2.4 Suppose K >10,α,β R,2K 1/2 6 α K1/2 and 0<δ <1/2. Then we have
−
∈ | |≪
# k K : β+α√t <δ Kδ+K1/2+ε.
{ ∼ k k }≪
Proof. See Lemma 4 of [20].
Lemma 2.5 Let a,δ be real numbers, 0 < δ < a/4, and let k be a positive integer. There exists a
function ϕ(y) which is k times continuously differentiable and such that
ϕ(y)=1, for y 6a δ,
| | −

 0<ϕ(y)<1, for a−δ <|y|<a+δ,
ϕ(y)=0, for y >a+δ,
and its Fourier transform  | |
+
∞
Φ(x)= e( xy)ϕ(y)dy
−
Z−∞
satisfies the inequality
k
1 1 k
Φ(x) 6min 2a, , .
| | π|x| π|x| (cid:18)2π|x|δ(cid:19) !
Proof. See [11] or [13].
Lemma 2.6 Let d(n) denote the divisor function. Then we have
s (d) s (d;y) y 1/2+ε, 16ℓ<k.
k;ℓ k;ℓ −
| − |≪
Proof. See Lemma 3.1 of [19].
4
Lemma 2.7 Suppose k >3;(i ,i , ,i ) 0,1 k 1,(i ,i , ,i )=(0,0, ,0);
1 2 k 1 − 1 2 k 1
··· − ∈{ } ··· − 6 ···
1<N ,N , ,N ;0<∆ E1/2,E =max(N ,N , ,N ). Let
1 2 k 1 2 k
··· ≪ ···
A =A(N ,N , ,N ;i ,i , ,i ;∆)
1 2 k 1 2 k
··· ···
denote the number of solutions of the inequality
√n1+( 1)i1√n2+( 1)i2√n3+ +( 1)ik−1√nk <∆
| − − ··· − |
with n N ,j =1,2, ,k. Then
j j
∼ ···
A ∆E 1/2N N N +E 1N N N .
− 1 2 k − 1 2 k
≪ ··· ···
Proof. See Lemma 2.4 of [19].
Lemma 2.8 Suppose 1 6 N 6 M 6 K 6 L,1 6 R 6 S 6 Q,L Q and 0 < ∆ Q1/2. Let
≍ ≪
A (N,M,K,L,R,S,Q;∆) denote the number of solutions of the inequality
1
0< √n+√m+√k+√ℓ √r √s √q <∆ (2.1)
| − − − |
with n N,m M,k K,ℓ L,r R,s S,q Q. Then
∼ ∼ ∼ ∼ ∼ ∼ ∼
A (N,M,K,L,R,S,Q;∆) ∆Q1/2NMKLRS+NMKRSL1/2+ε.
1
≪
In particular, if ∆Q1/2 1, then
≫
A (N,M,K,L,R,S,Q;∆) ∆Q1/2NMKLRS.
1
≪
Proof. If (n,m,k,ℓ.r,s,q) satisfies (2.1), then
(√n+√m+√k √r √s)+√ℓ=√q+θ∆
− −
for some 0< θ <1. Thus, we have
| |
2ℓ1/2(√n+√m+√k √r √s)+(√n+√m+√k √r √s)2+ℓ=q+u
− − − −
with u = 2q1/2θ∆+θ2∆2 62q1/2∆+∆2 ∆Q1/2. Then we have
| | | | ≪
q =2ℓ1/2(√n+√m+√k √r √s)+(√n+√m+√k √r √s)2+ℓ u
− − − − −
with u 6C∆Q1/2 for some absolute positive constant C >0. Hence the quantity of
| |
A (N,M,K,L,R,S,Q;∆) does not exceed the number of solutions of
1
2ℓ1/2(√n+√m+√k √r √s)+(√n+√m+√k √r √s)2+ℓ q <C∆Q1/2 (2.2)
− − − − −
(cid:12) (cid:12)
(cid:12) (cid:12)
with(cid:12)n N,m M,k K,ℓ L,r R,s S,q Q. (cid:12)
∼ ∼ ∼ ∼ ∼ ∼ ∼
If∆Q1/2 1,thenforfixed(n,m,k,ℓ,r,s),thenumberofqforwhich(2.2)holdsis 1+∆Q1/2
≫ ≪ ≪
∆Q1/2. Hence
A (N,M,K,L,R,S,Q;∆) ∆Q1/2NMKLRS.
1
≪
5
Now suppose ∆Q1/2 61/4C. Then for fixed (n,m,k,ℓ,r,s), there is atmostone q such that(2.2)
holds. If such q exists, then we have
2ℓ1/2(√n+√m+√k √r √s)+(√n+√m+√k √r √s)2 <C∆Q1/2. (2.3)
− − − −
(cid:13) (cid:13)
We sha(cid:13)ll use Lemma 2.4 to bound the number of solutions of (2.3) with α(cid:13)= 2(√n+√m+√k
(cid:13) (cid:13) −
√r √s),β = (√n+√m+√k √r √s)2. Let D denote the number of solutions of (2.3) with
1
− − −
α >2L 1/2, and D the number of solutions of (2.3) with α <2L 1/2. By Lemma 2.4, we get
− 2 −
| | | |
D (∆Q1/2L+L1/2+ε)NMKRS
1
≪
∆Q1/2NMKLRS+NMKRSL1/2+ε.
≪
Now we estimate D . From α < 2L 1/2, we can get K R. If √n+√m+√k = √r+√s, then
2 −
| | ≍
from (2.2), we get ℓ=q. This contradicts to the fact that √n+√m+√k+√ℓ √r √s√q >0.
| − − |
Therefore,wehave√n+√m+√k =√r+√s.ByLemma2.1,wehave √n+√m+√k √r+√s
6 | − |≫
S 15/2 for any such (n,m,k,r,s). By a splitting argumentand Lemma 2.7, there exists a δ satisfying
−
S 15/2 δ L 1/2, which holds
− −
≪ ≪
D log2S 1
2
≪ ·
δ<√n+√m+X√k √r+√s62δ
| − |
log2S (δS1/2NMKR+NMKR)
≪ ·
L 1/2S1/2+εNMKR+NMKRSε
−
≪
NMKRSε,
≪
which can be absorbed into the estimate of D . This completes the proof of Lemma 2.8.
1
Lemma 2.9 Suppose 1 6 N 6 M 6 K 6 L,1 6 R 6 S 6 Q,L Q and 0 < ∆ Q1/2. Let
≍ ≪
A (N,M,K,L,R,S,Q;∆) denote the number of solutions of the inequality
2,
±
0< √n+√m+√k+√ℓ+√r √s √q <∆
| ± − |
with n N,m M,k K,ℓ L,r R,s S,q Q. Then
∼ ∼ ∼ ∼ ∼ ∼ ∼
A (N,M,K,L,R,S,Q;∆) ∆Q1/2NMKLRS+NMKRSL1/2+ε.
2,
± ≪
In particular, if ∆Q1/2 1, then
≫
A (N,M,K,L,R,S,Q;∆) ∆Q1/2NMKLRS.
2,
± ≪
Proof. The proof of Lemma 2.9 is similar to that of Lemma 2.8, so we omit the details.
Lemma 2.10 Suppose N >2 (j =1,2,3,4,5,6,7) are real numbers, ∆>0. let
j
A (N .N ,N ,N ,N ,N ,N ;∆) denote the number of solutions of the inequality
, 1 2 3 4 5 6 7
±±
0< √n +√n +√n +√n √n √n √n <∆
1 2 3 4 5 6 7
| ± ± − |
with n N ,(j =1,2,3,4,5,6,7),n N . Then we have
j j j ∗
∼ ∈
7
A (N .N ,N ,N ,N ,N ,N ;∆) ∆1/7N13/14+N5/7 Nε.
±,± 1 2 3 4 5 6 7 ≪ j j j
jY=1(cid:16) (cid:17)
6
Proof. Taking a = 6∆/5,δ = ∆/5 in Lemma 2.5, there exists a function ϕ (y), which is ℓ =
1
[7log(N N N N N N N )] times continuously differentiable such that
1 2 3 4 5 6 7
ϕ (y)=1, if y 6∆,
1
| |

 0<ϕ1(y)<1, for ∆<|y|<7∆/5,
ϕ (y)=0, for y >7∆/5.
1
Let  | |
+
∞
Φ (x)= e( xy)ϕ (y)dy,
1 1
−
Z−∞
then it satisfies
ℓ
12∆ 1 1 5ℓ
Φ (x) 6min , , , (2.4)
1
| | 5 π|x| π|x|(cid:18)2π|x|∆(cid:19) !
and
+
∞
ϕ (y)= e(xy)Φ (x)dx. (2.5)
1 1
Z−∞
Set
R = ϕ (√n +√n +√n +√n √n √n √n ).
, 1 1 2 3 4 5 6 7
±± ± ± −
j=1n,2jX,∼3,N4,j5,6,7
By the definition of ϕ (y), we get
1
A (N .N ,N ,N ,N ,N ,N ;∆)6R . (2.6)
, 1 2 3 4 5 6 7 ,
±± ±±
We estimate R first. By (2.5), we have
,
−−
R = ϕ (√n +√n +√n +√n √n √n √n )
, 1 1 2 3 4 5 6 7
−− − − −
j=1n,2jX,∼3,N4,j5,6,7
+
∞
= e(x(√n +√n +√n +√n √n √n √n ))Φ (x)dx.
1 2 3 4 5 6 7 1
− − −
j=1n,2jX,∼3,N4,j5,6,7Z−∞
Let S(x;N):= e(x√n), we have
n N
∼
P
+
∞
R = S(x;N )S(x;N )S(x;N )S(x;N )S(x;N )S(x;N )S(x;N )Φ (x)dx,
, 1 2 3 4 5 6 7 1
−−
Z−∞
if we notice that S(x;N)= e( x√n). Applying Ho¨lders inequality, we get
−
n N
∼
P
7 + 1/7
R 6 ∞ S(x;N )7 Φ (x)dx . (2.7)
, j 1
−− | | | |
jY=1(cid:18)Z−∞ (cid:19)
Let
+
T(N):= ∞ S(x;N)7 Φ (x)dx.
1
| | | |
Z0
7
It is sufficient to estimate T(N), where N =N for some j 1,2,3,4,5,6,7 .Let
j
∈{ }
100ℓ2
K := , ℓ=[7log(N N N N N N N )]; K :=N1/2.
1 2 3 4 5 6 7 0
∆
Using the trivial estimate S(x;N) N and the estimate
≪
ℓ
1 5ℓ
Φ (x) 6 ,
1
| | π x 2π x∆
| |(cid:18) | | (cid:19)
we have
ℓ
∞ S(x;N)7 Φ (x)dx N7 ∞ Φ (x)dx N7 5ℓ ∞ 1 dx
| | | 1 | ≪ | 1 | ≪ 2π∆ xℓ+1
ZK ZK (cid:18) (cid:19) ZK
N75ℓ ℓ ℓ N75ℓ 1
≪ ℓ 2πK∆ ≪ ℓ ℓℓ
(cid:18) (cid:19)
N7(N N )7log5
1 7
··· 1. (2.8)
≪ (N1···N7)7log7(N1···N7)7loglog(N1···N7) ≪
For the mean square of S(x;N), we have
K0 K0
S(x;N)2dx = e(x(√n √m))dx
| | −
Z0 Z0 n Nm N
X∼ X∼
K0
= e(x(√n √m))dx
−
n Nm NZ0
X∼ X∼
K0
= + e(x(√n √m))dx.
−
(cid:26)n Nm N n Nm N(cid:27)Z0
X∼n=mX∼ X∼n=mX∼
6
If we use the trivial estimate S(x;N) N, then
≪
K0
e(x(√n √m))dx N3/2.
− ≪
n Nm NZ0
X∼n=mX∼
For the case n=m, we have
6
K0
e(x(√n √m))dx
−
n Nm NZ0
X∼n=mX∼
6
1 1
N1/2
≪ √n √m ≪ n m
n Nm N | − | n Nm N | − |
X∼n=mX∼ X∼n=mX∼
6 6
1
N1/2 N3/2logN.
≪ r ≪
n N16r N
X∼ X≪
Therefore, we have
K0
S(x;N)2dx N3/2logN. (2.9)
| | ≪
Z0
8
If we notice Φ (x) ∆ from (2.4) and the trivial estimate S(x;N) N, then
1
| |≪ ≪
K0
S(x;N)7 Φ (x)dx
1
| | | |
Z0
K0
∆N5 S(x;N)2dx
≪ | |
Z0
∆N13/2logN. (2.10)
≪
If K 6K , then from (2.8) and (2.10) we get
0
T(N) ∆N13/2logN. (2.11)
≪
Now suppose K <K. By a splitting argument, we have
0
K 2U
S(x;N)7 Φ (x)dx ∆logK max S(x;N)7dx. (2.12)
1
ZK0 | | | | ≪ ×K06U6KZU | |
From Lemma 2.1, we can get ∆ 1 max(N ,N ,N ,N ,N ,N ,N )63/2 and thus logK ℓ2. On
− 1 2 3 4 5 6 7
≪ ≪
the other hand, we have
2U 2U
S(x;N)7dx max S(x;N)2 S(x;N)5dx
ZU | | ≪ U6x62U| | ×ZU | |
N2(N9/2+UN3)Nε
≪
(N13/2+UN5)Nε, (2.13)
≪
which is derived from equation (2.18) of Zhang and Zhai [21].
From (2.12) and (2.13) and noticing ∆K ℓ2, we get
≪
K
S(x;N)7 Φ (x)dx ∆logK (N13/2+UN5)Nε
1
| | | | ≪ ×
ZK0
(∆N13/2+N5)Nε,
≪
which combining (2.8) and (2.10) gives
T(N)= ∞ S(x;N)7 Φ (x)dx (∆N13/2+N5)Nε. (2.14)
1
| | | | ≪
Z0
From (2.6), (2.7) and (2.14), we get the result of Lemma 2.10 for the case “ , ”. By noting the
− −
propertiesofconjugation,theestimatesofothercasesareexactlythesameasthatofthecase“ , ”.
− −
This completes the proof of Lemma 2.10.
3 Proof of Theorem
In this section, we shall prove the theorem. We begin with the following truncated form of the
Vorono¨ı’s formula ([7], equation (2.25) ), i.e.
1 d(n)
∆(x)= x1/4cos(4π√nx π/4)+O(x1/2+εN−1/2), (3.1)
√2π n3/4 −
n6N
X
9
for 16N x. Set ∆(x):=R +R , where
1 2
≪
1 d(n)
R :=R (x)= x1/4cos(4π√nx π/4), R :=R (x)=∆(x) R .
1 1 √2π n3/4 − 2 2 − 1
n6y
X
Take y =T1/4. By the elementary estimate (a+b)7 a7 ba6+ b7, we have
− ≪| | | |
T T T T
∆7(x)dx= R7(x)dx+O R 6 R dx+ R 7dx .
Z1 Z1 1 Z1 | 1| | 2| Z1 | 2| !
By a splitting argument, it is sufficient to prove the result in the interval [T,2T]. We will divide the
process of the proof of the theorem into two parts.
Proposition 3.1 For fixed T >10,N =T,y =T1/4, we have
2T 7(5s (d) 3s (d) s (d))
R17dx= 7;3 −28176;2π7 − 7;1 T11/4+O(T11/4−1/336+ε). (3.2)
ZT
Proof. Let
d(n)d(m)d(k)d(ℓ)d(r)d(s)d(q), if n,m,k,ℓ,r,s,q6y,
(nmkℓrsq)3/4
g :=g(n,m,k,ℓ,r,s,q)=
 0, otherwise.
According to the elementary formula
1
cosa cosa cosa = cos a +( 1)i1a + +( 1)ih−1a ,
1 2··· h 2h 1 1 − 2 ··· − h
− (i1,i2,···,ihX−1)∈{0,1}h−1 (cid:0) (cid:1)
we can write
R7 =S (x)+S (x)+S (x)+S (x)+S (x)+S (x)+S (x),
1 1 2 3 4 5 6 7
where
35 π
S (x) := cos gx7/4,
1
64 4
n,m,k,ℓ,r,s,q6y
X
√n+√m+√k+√ℓ=√r+√s+√q
35
S (x) := gx7/4
2
64
n,m,k,ℓ,r,s,q6y
X
√n+√m+√k+√ℓ=√r+√s+√q
6
π
cos 4π √n+√m+√k+√ℓ √r √s √q √x ,
× − − − − 4
21 (cid:16)3π(cid:16) (cid:17) (cid:17)
S (x) := cos gx7/4,
3
64 4
n,m,k,ℓ,r,s,q6y
X
√n+√m+√k+√ℓ+√r=√s+√q
21
S (x) := gx7/4
4
64
n,m,k,ℓ,r,s,q6y
X
√n+√m+√k+√ℓ+√r=√s+√q
6
3π
cos 4π √n+√m+√k+√ℓ+√r √s √q √x ,
× − − − 4
(cid:18) (cid:16) (cid:17) (cid:19)
10