Антоник В.Г. Численные методы: математический анализ и дифференциальные уравнения
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f(x)
[a, b]
I(f) =
b
Z
a
f(x)dx ,
f
i
, i = 1, n f(x)
x
1
, . . . , x
n
[a, b] f
i
= f(x
i
), i = 1, n
b
Z
a
f(x)dx ≈
n
X
i=1
c
i
f(x
i
)
4
= S(f)
S(f)
c
i
x
i
R
n
(f) = I(f) − S(f) .
f(x)
R
n
(f) = 0
S(f)
m
f(x) = x
k
, k = 1, m f(x) =
x
m+1
R
n
(x
k
) = 0, k = 1, m, R
n
(x
m+1
) 6= 0
[a, b]
a = x
0
< x
1
< . . . < x
n−1
< x
n
= b , x
i
− x
i−1
= h , i = 1, n
h =
b − a
n
I(f) =
n
X
i=1
x
i
Z
x
i−1
f(x)dx .
f(x) [x
i−1
, x
i
]
L
0,i
(x)
L
0,i
(x)
x
i−1
f(x
i−1
)
L
0,i
(x) = f(x
i−1
)
x
i
Z
x
i−1
f(x)dx ≈
x
i
Z
x
i−1
L
0,i
dx = hf(x
i−1
) .
S(f) = h
n
X
i=1
f(x
i−1
) .
|R(f)| ≤
h(b − a)
2
max
a≤x≤b
|f
0
(x)|.
L
0,i
(x)
(x
i
, f(x
i
))
S(f) = h
n
X
i=1
f(x
i
) .
x
i
[x
i−1
, x
i
]
x
i
=
x
i−1
+ x
i
2
L
0,i
(x) (x
i
, f(x
i
))
S(f) = h
n
X
i=1
f(x
i
) .
|R(f)| ≤
h
2
(b − a)
24
max
a
≤
x
≤
b
|f
00
(x)|.
1 2
f(x)
L
1,i
(x) f(x
i−1
), f(x
i
)
L
1,i
(x) =
1
h
[(x − x
i−1
)f(x
i
) − ( x − x
i
)f(x
i−1
)] .
x
i
Z
x
i−1
f(x)dx ≈
x
i
Z
x
i−1
L
1,i
(x)dx =
f(x
i−1
) + f(x
i
)
2
h .
S(f) = h
Ã
f(x
0
) + f(x
n
)
2
+
n−1
X
i=1
f(x
i
)
!
.
|R(f)| ≤
h
2
(b − a)
12
max
a≤x≤b
|f
00
(x)|.
[x
i−1
, x
i
] x
i
=
x
i−1
+ x
i
2
L
2,i
(x)
x
i−1
, f(x
i−1
) x
i
, f(x
i
) x
i
, f(x
i
)
x
i
Z
x
i−1
f(x)dx ≈
x
i
Z
x
i−1
L
2,i
(x)dx =
h
6
(f(x
i−1
) + 4 f (x
i
) + f(x
i
)) .
S(f) =
h
6
³
f(x
0
) + f ( x
n
) + 2( f (x
1
) + . . . + f(x
n−1
))+
+4(f(x
1
) + . . . + f(x
n
))
´
.
|R(f)| ≤
h
4
(b − a)
2880
max
a≤x≤b
|f
IV
(x)|.
f(x) = x
2
I =
4
Z
0
f(x)dx
n = 8
a = 0, b = 4
h =
b − a
n
=
1
2
.
f(x)
x
i
f
i
I ≈
1
2
(0 + 0 .25 + 1 + 2.25 + 4 + 6.25 + 9 + 12.25) = 17.5 .
max
0≤x≤4
|f
0
(x)| = max
0≤x≤4
|2x| = 8 .
|R| ≤
4
2 · 2
· 8 = 8 .
I(f) ≈ S(f) =
n
X
i=0
c
i
f(x
i
) .
c
i
, i = 0, n
n
f(x) = x
k
k = 0, n
S(x
k
) = I(x
k
) , k = 0, n .
c
0
, c
1
, . . . , c
n
I(f) =
b
Z
a
f(x)dx ,
n = 1, x
0
= a, x
1
= b
S(f) = c
0
f(x
0
) + c
1
f(x
1
) = c
0
f(a) + c
1
f(b) .
c
0
, c
1
I(x
0
) =
b
Z
a
x
0
dx =
b
Z
a
dx = b − a ,
I(x
1
) =
b
Z
a
x
1
dx =
1
2
b
2
−
1
2
a
2
,
S(x
0
) = c
0
a
0
+ c
1
b
0
= c
0
+ c
1
,
S(x
1
) = c
0
a
1
+ c
1
b
1
= ac
0
+ bc
1
.
c
0
+ c
1
= b − a ,
ac
0
+ bc
1
=
b
2
− a
2
2
.
c
0
= c
1
=
b − a
2
.
S(f) =
b − a
2
f(a) +
b − a
2
f(b) =
f(a) + f(b)
2
(b − a) .
n = 1 ¤
f(x) = (x + 1)
3
I =
4
Z
0
f(x)dx
n
3
2
n
I
f(x) = −(x − 1)
3
I =
3
Z
−1
f(x)dx
n
5
4
n
I
f(x) = x
2
− 1
I =
2
Z
−2
f(x)dx .
n = 8
I
k
=
4
Z
0
x
k
dx , k = 1, 2, 3, 4
h = 1
I =
2
Z
−2
(1 + x + x
2
+ x
3
+ x
4
)dx
h = 1
I
I(f) =
b
Z
a
f(x)dx ,