Численные методы. Ч.2. Решение уравнений. Буслов В.А - 5 стр.

UptoLike

ВУЗ: 

СПбГУ | Санкт-Петербург

Составители: 

Рубрика: 

F (x) = x −
f(x)
f
0
(x)
f ∈ C
2
|F
0
| < 1
F
0
= 1 −
(f
0
)
2
− ff
00
(f
0
)
2
=
ff
00
(f
0
)
2
,
x
∗
α f(x) ≈ a(x −x
∗
)
α
F
0
(x
∗
) =
α−1
α
x
∗
x
j+1
− x
∗
= x
j
− x
∗
−
f(x
j
)
f
0
(x
j
)
x
j+1
− x
∗
(x
j
− x
∗
)
2
=
1
x
j
− x
∗
−
f(x
j
)
f
0
(x
j
)(x
j
− x
∗
)
2
=
1
x
j
− x
∗
−
−
f
0
(x
∗
)(x
j
− x
∗
) +
1
2
f
00
(x
∗
)(x
j
− x
∗
)
2
+ o
¡
(x
j
− x
∗
)
2
¢
(x
j
− x
∗
)
2
[f
0
(x
∗
) + f
00
(x
∗
)(x
j
− x
∗
) + o(x
j
− x
∗
)]
,
lim
j→∞
x
j+1
− x
∗
(x
j
− x
∗
)
2
=
f
00
(x
∗
)
2f
0
(x
∗
)
.
x
∗
|ff
00
|/(f
0
2
) < 1
f(x)
x
j
x
j−1
x
j+1
= x
j
−
f
j
(x
j
− x
j−1
)
f
j
− f
j−1
,
f
j
= f(x
j
)
d =
√
5+1
2
x
∗
= 0
f
j
(x
j
− x
j−1
)
f
j
− f
j−1
=
[f
0
∗
x
j
+
1
2
f
00
∗
x
2
j
+ O(x
3
j
)](x
j
− x
j−1
)
f
0
∗
(x
j
− x
j−1
) +
1
2
f
00
∗
(x
2
j
− x
2
j−1
) + O(x
3
j
− x
3
j−1
)
=
= x
j
1 +
f
00
∗
2f
0
∗
x
j
+ O(x
2
j
)
1 +
f
00
∗
2f
0
∗
(x
j
+ x
j−1
) + O(x
2
j
)
= x
j
·
1 −
f
00
∗
2f
0
∗
x
j−1
+ O(x
2
j
)
¸
.
x
j+1
= x
j
−
f
j
(x
j
− x
j−1
)
f
j
− f
j−1
=
f
00
∗
2f
0
∗
x
j
x
j−1
+ O(x
3
j
) .
x
j+1
= αx
j
x
j−1
α =
f
00
∗
2f
0
∗
x
j+1
= α
c
x
d
j
cd = 1 d
2
−d−1 = 0
d d =
√
5+1
2

Страницы