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ασ
≠
ααρ σσμ
=+ =+
a
,
a
,
ασ
,
∈
Z
2
Case 2.
If
, order
where
, and
22
1
1
1
2
2
1
2
2
+
ρμ∈Γ
,
.
ασ
=
ρμ
≠
2
.
If
, then
Similar to Case 1, (21) follows. As
22
0
2
2
2
ασ
≠
ααρσ
=+ =
3
a
,
a
σ μ
+
3
,
, order
where
ασ
,
∈
Z
2
,
ρμ
,
∈Ω
.
2
2
2
3
3
2
33
+
33
0
κ
α ∈Γ
ρμ∈Γ
,
.
Thus, taking finite steps (denoted by
), we obtain
, and
0
κκ
0
ˆ
%
ˆ
%
∫
8 [
π
2
GGk GG ed
(),
⋅
(
⋅−=
)]
()
γ
()
γ
*
⋅
ik
⋅
γ
γ
α
σ
α
σ
2
R
1
ˆ
ˆ
%
∫
*
=
R
G
()
γ
G
() exp{
γ
⋅
ik
⋅
γ γ
}
d
=
LLLLLLLLLL
a
αλ
+
a
βμ
+
2
11
1
1
κ
κ
Q
∏
∏
∫
()
μ
=
{
Q
()
ρ
(
γ
/
aO
l
)}
⋅
⋅
{
(
γ
/
a
l
)}
*
⋅
exp{
− ⋅
ik
γ γ
}
d
=
O
.
l
l
κ
2
l
=
1
([0,2
⋅
a
π
]
l
=
1
ασ
,
∈
Z
2
Therefore, for any
, result (21) is established.
+
, the translation operator
S
is defined to be
For any
vvvZ
=
(, )
∈
2
12
( )( ) ( )
va
, where
a
is a pasitive constant real number.
S
D
x
=−
D
x
va
%
and
%
()
φφ
(), (), ()
xxx
ι
h
h
x
ι∈
J
()
Theorem 3
[7]
.
Let
LR
. As-
sume that conditions in Theorem 1 are satisfied. Then, for any function
2
,
be functions in
2
ι
fx
()
LR
∈
()
, and any integer n, we have
%
7
n
−
1
%
∑
∑∑∑
hh
. (22)
f
,
φφ
( )
x
=
f
,
( )
x
ι
:,
su
nu
,
nu
,
ι
: ,
su
2
2
uZ
∈
ι
= ∞
1
s
uZ
∈
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2.
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∞
Cantorian
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