Численные методы. Ч.1. Исследование функций. Буслов В.А - 32 стр.

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f(x + kh) =
k
X
s=0
C
s
k
∆
s
f(x).
f(x + h) = f(x) + ∆f(x) = (1 + ∆)f(x) ,
f(x + 2h) = (1 + ∆)f(x + h) = (1 + ∆)
2
f(x) ,
. . . ,
f(x + kh) = (1 + ∆)
k
f(x) ,
(1 + ∆)
k
=
k
P
s=0
C
s
k
∆
s
C
s
k
=
k(k−1)...(k−s+1)
s!
=
k!
(k−s)!s!
,
∆
k
f(x) =
k
X
s=0
C
s
k
(−1)
s
f(x + (k −s)h) .
∆ = (1 + ∆) − 1
∆
k
f(x) = [(1 + ∆) − 1]
k
f(x) =
k
X
s=0
C
s
k
(1 + ∆)
k−s
(−1)
s
f(x) =
=
k
X
s=0
C
s
k
(−1)
s
f(x + (k −s)h),
∆
k
f(x) = f(x + kh) − C
1
k
f(x + (k −1)h) + C
2
k
f(x + (k −2)h)+
+ . . . + (−1)
k
f(x).
∆
k
f(x) = (∆x)
k
f
(k)
(x + Θk∆x) ,
0 < Θ < 1 f ∈ C
k
∆f = ∆xf
0
(x+Θ∆x)
[x, x+∆x]
ξ
∆f
∆x
=
f(x+h)−f(x)
∆x
= f
0
(ξ) ξ ∈ [x, x + ∆x] k
∆
k
f(x) = (∆x)
k
f
(k)
(x + Θk∆x) .
∆
k+1
f(x) = ∆(∆
k
f) = ∆[f
(k)
(x + kΘ∆x)]∆
k
x =
= ∆
k
x[f
(k)
(x + ∆x + kΘ∆x) − f
(k)
(x + kΘ∆x)] .

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