γ
N
=
(−1)
ν
h
ν
N
ν!
∞
Z
−∞
h
f
(ν)
(x + (−1)
ν
uh
N
θ) − f
(ν)
(x)
i
u
ν
K(u)du,
0 < θ < 1.
∞
Z
−∞
|u
ν
K(u)|du < ∞, x ∈ R
1
N → ∞
f
(ν)
(x + (−1)
ν
uh
N
θ) → f
(ν)
(x)
|γ
N
| = o (h
ν
N
) .
T
i
= 0, i =
1, . . . , ν − 1 ♠
b (f
N
(x))
K(u) K(u) ∈ A
ν
, ν ≥ 4.
K(u) ∈ A
2
,
sup
u
K(u) < ∞, K(u) = K(−u)
Z
u
2
K(u)du < ∞.
K(u) ∈ A
4
,
K(u) =
15(3 − 10u
2
+ 7u
4
)/2
5
, |u| ≤ 1,
0, |u| > 1.
K(u) ∈ A
6
,
K(u) =
105(5 − 35u
2
+ 63u
4
− 33u
6
)/2
8
, |u| ≤ 1,
0, |u| > 1.
ρ(u) = {1 − u
2
, |u| ≤ 1; 0, |u| > 1} K(u) ∈ A
ν
,
K(u) = ρ(u)
ν−2
X
j=0
p
j
(0)p
j
(u)(2j + 3)(j + 2)/8(j + 1),
p
j+2
(u) =
j + 3
j + 4
2j + 5
j + 2
up
j+1
(u) − p
j
(u)
,
p
0
(u) = p
0
(0) = 1, p
1
(u) = 2u, p
1
(0) = 0.