Bài giảng: Phương pháp tính pdf
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TR
:
:
: 17201
TRÌNH :
DÙNG CHO SV NGÀNH :
- 2008
11.1. 2
CNTT
: 3
60
45
15
0
0
0
Sinh
TS
LT
TH/Xemina
BT
KT
10
8
2
0
1.1
1
1.2
2
1.3
2
1
2
1
1
1
.
15
10
4
1
2.1
1
2.2
1
2.3
2
1
2
1
2.5.
2
1
2
1
3.
12
9
3
0
3.1
2
3.2
2
1
3.3
2
1
3
1
4.
12
8
3
1
4.1
4
1
4.2
4
2
5.
11
7
3
1
1
5.
3
2
5.-Kutta
3
1
TS
LT
TH/Xemina
BT
KT
60
42
15
3
-
-
-
-
-
1
:
1.1.
1.
.
, .
.
.
Aa
(
).
,
a.
a
Aa
:
Aa
a
(1.1)
a
.
a
a
.
a
(1.1)
a
A = a
a
(1.2)
( 1.1)
:
a -
a A
a +
a
(1.3)
2.
:
a
Aa
A
Aa
(
).
.
:
a
=
a
a
( 1.4)
.
a
=
a
a
( 1.5)
2
(1.4) (1.5)
.
a
( 1.4)
a
,
a
( 1.5)
a
.
Do ( 1.5) nên ( 1.2) :
A= a ( 1
a
) (1.6)
a
a
.
3.
,
.
:
= 10
a
= 0,05
= 2
b
= 0,05m.
.
, :
a
10
05,0
= 0,5% <
b
=
2
05,0
= 2,5%
1.2.
1.
,
.
2,74 3
, 0,0207 .
2.
:
A =
10
s
s
a
(1.7)
:
a
s
0 9,
65,807 :
65,807 = 6.10
1
+ 5.10
0
+ 8.10
-1
+ 0.10
-2
+ 7.10
-3
( 1.7)
:
1
= 6,
o
= 5,
-1
= 8,
-2
= 0,
-3
= 7
a
.
s
.
a
0,5 .10
s
s
,
a
> 0,5 .10
s
s
.
3
.
: Cho a = 65,827
a
6, 5, 8, 2 ,
7, 4 .
a
= 0,0067 6, 5, 8,
2, 7, 4 .
s
s
.
3.
a
.
.
(1.2)
( 1.6) .
: .
không
. ,
v v
.
1.3.
1.
.
,
.
.
,
:
,
5
, ,
iên
5
, tiên < 5
.
: 62,8274
(
) 62,827;
62,83;
(
a) 62,8.
2.
.
4
a
.
aa '
.
a
:
aa '
( 1.8)
,
.
.
:
- - a + a - A
:
Aa '
aa '
+
Aa
a
a
a
+
(1.9)
a
.
3.
: = (
2
- 1 )
10
.
(Newton)
:
(
2
- 1)
10
= 3363 - 2378
2
( 1.10)
:
2
= 1,41421356
(1.10)
2
(
1.1):
Bng 1.1
2
1,4
0,0001048576
33,8
1,41
0,00013422659
10,02
1,414
0,00014791200
0,508
1,41421
0,00014866399
0,00862
1,414213563
0,00014867678
0,0001472
.
(1.10) ,
(1.10) .
5
1.4.
1.
.
:
u = f( x,y) (1.11)
, .
:
, ,
, y, u
Dx, dy,
, y, u
x
y
u
, y, u.
(1.1) ta
:
x
x
;
y
y
(1.12)
u
u
u
2. Sai
= x + y
Ta suy ra:
u
x
+
y
( 1.12)
:
u
x
y
:
x+y
x
y
(1.13)
:
u
u
.
:
(
)
(
) .
. = x-
.
:
u
=
u
u
=
yx
yx
yx
.
.
6
3.
= xy
du = ydx + xdy
u
y
x
+
x
y
y
x
+
x
y
u
=
y
x
+
x
y
:
u
=
u
u
=
xy
xy
yx
=
x
x
y
y
:
xy
=
x
+
y
( 1.14)
: (
)
(
) . :
(x
n
)
= n
x ;
4.
:
:
x/y
=
x
+
y
( 1.16)
5.
:
Cho : u = f( x
1
, x
2
, ,x
n
)
u
=
x
i
n
n
f
1
i
( 1.17)
u
(1.4)
: (
) (
)
:
V=
6
1
3
= 3,7 0,05
= 3,14.
.
, theo (1.14) (1.15)
:
v
=
+ 3
d
7
d
= 0,05/3,7 =0,0135
Suy ra:
V
= 0,0005 + 3.0,0135 = 0,04
: V=
6
1
3
= 26,5 cm
3
V
= 26,5 .0,04 = 1,06 1,1cm
3
V= 26,5 1,1 cm
3
1.5.
1.
.
.
.
ng,
.
.
.
2.
a) :
A =
3
1
1
-
3
2
1
+
3
3
1
-
3
4
1
+
3
5
1
-
3
6
1
.
6 .
. .
:
3
1
1
=
1
1
= 1,000
1
= 0
3
2
1
=
8
1
= 0,125
2
= 0
8
3
3
1
=
27
1
= 0,037
3
= 4.
4
10
3
4
1
=
64
1
= 0,016
4
= 4.
4
10
3
5
1
=
125
1
= 0,008
5
= 0
3
6
1
=
216
1
= 0,005
6
= 4.
4
10
a =1,000 - 0,125 + 0,037 - 0,016 + 0,008 - 0,005 = 0,899
aA
=
1
1
1
3
-
125,0
2
1
3
+
037,0
3
1
3
-
016,0
4
1
3
+
008,0
5
1
3
-
005,0
6
1
3
aA
1
1
1
3
+
125,0
2
1
3
+
037,0
3
1
3
+
016,0
4
1
3
+
008,0
5
1
3
+
005,0
6
1
3
1
+
2
+
3
+
4
+
5
+
6
= 9.
4
10
a = 0,899 9.
4
10
:
= 0,899
9.
4
10
( 1.18 )
b) `
B =
3
1
1
-
3
2
1
+
3
3
1
-
1
1
n
3
1
n
5.
3
10
.
.
, ta
.
, n s:
9
n
B
=
3
1
1
-
3
2
1
1
1
n
3
1
n
n
B
.
n
BB
,
5.10
-3
.
:
n
BB
=
333
1
1
2
1
1
1
nnn
(
),
= 6 :
3
3
6
10.3
334
1
7
1
BB
6
B
=
(xem 1.18):
6
B
= A = 0,899
4
10.9
.899,0
:
B - 0,889 = B -
6
B
+ A - 0,899
899,0899,0
6
ABBB
343
10.410.910.3899,0
B
,0
899
4.
3
10
:
,
.
1.6. 1
1.
(
)
.
.
.
10
.
.
,
,
.
2.
y
1i
=qy
i
, ( 1.19 )
y
0
.
i
i
(
),
i
y
i
~
. :
y
y
i
i
( 1. 20 )
i+1
y
~
i + 1
:
y
~
i + 1
= q
y
~
i
= > 0
( 1.21)
(1.19)
:
y
~
i + 1
- y
i+1
=
q
yy
ii
q
y
~
i + 1
- y
i+1
=
q
(
)
~
yy
ii
:
y
i
~
2
= q
y
i
~
1
y
i 2
= q
y
i 2
:
y
i
~
2
-
y
i 2
= q(
y
i
~
1
-
y
i 1
)
= q
2
(
y
i
~
-
y
i
)
:
11
y
ni
~
-
y
ni
= q
n
(
y
i
~
-
y
i
)
yy
nini
~
=
q
n
yy
ii
~
`
,
y
y
1
~
=
+
yy
nini
~
=
q
n
;
1.
q
1
q
n
yy
nini
~
(
).
.
2.
q
1 -
q
n
q
n
,
yy
nini
~
khi n
,
,
,
.
1.
: a = 21
o
o
2.
:
a = 13267 ;
a
= 0,1%
b = 2,32 ;
b
= 0,7%
3.
:
a = 0,39410;
a
= 0,25 .10
-2
12
b = 38,2543 ;
b
= 0,25 .10
-2
4.
sau:
a = 1,8921 ;
a
= 0,1.10
-2
b = 22,351;
b
= 0,1.
5. (
)
:
a) 2,1514; b)0,16152;
c)0,01204; d) - 0,0015281.
6.
:
a) u = ln ( x + y
2
) ; x = 0,97 ; y = 1,132
b) u = (x + y
2
)/z ; x = 3,28; y= 0,932 ; z= 1,132.
7. :
S =
11
1
+
12
1
+
13
1
+
14
1
+
15
1
+
16
1
+
17
1
8. : e = 1 +
!1
1
+
!2
1
+ +
!
1
n
+
10
-4
1.
a
= 0,13.10
-4
;
b
= 0,28.10
-3
2.
a
= 0,13.10
2
;
b
= 0,16.10
-1
3. a) 2; b) 4.
4. a) 3; b)1.
5. a)2,15; = 0,14.10
-2
; = 0,65.10
-3
b) 0,162; = 0,48.10
-3
; = 0,3.10
2
c) 0,0120; = 0,4.10
-4
; = 0,33.10
-2
d) -0,00153; = 0,19.10
-5
; = 125. 10
-2
6. a) u = 0,81;
u
= 0,27. 10
-2
;
u
= 0,33. 10
-2
b) u = 3,665;
u
= 0,7. 10
-2
;
u
= 0,20. 10
-2
7. S = 0,511.
13
8. e = 2,7183 0,0001.
14
:
2.1.
1.
:
f(x) = 0 (2.1)
:
.
Ngh
(2.1) (2.1)
:
f() = 0 (2.2)
2.
:
y= f(x) (2.3)
(2-1).
= 0 = .
:
0 = f() (2.4)
2-1
(2.1)
(2.1)
g(x) = h(x) (2.5)
2 (2-2)
y = g(x),
y = h(x) (2.6)
= :
g() = h() (2.7)
M
x
y
2.2
x
y
M
f
g
15
2
(2.6)
(2.5),
(2.1).
3.
(2.1)
(2.1)
.
2 trên.
:
2.1 -
2
(a<b) sao cho f(a) (b)
f(a).f(b) < 0 (2.8)
(x)
[a, b] [a, b]
(2.1).
(2 - 3). = f(x)
x
,
,
,
b.
(2.1)
[a, b].
2-3
4. (
)
2.1 - [a, b]
(2.1)
.
:
y
a
B
b
x
A
16
2.2 - [a, b] (x)
,
(a) (b) ,
(2.8) [a, b]
(2.1).
( 2 - 4).
= f(x)
[a, b].
[a, b]
(2.1).
(x)
. ta
:
2.3 - [a, b]
(x)
,
(x)
(a), f(b)
[a, b]
(2.1)
2-4
(2.1)
= f(x)
2.3.
5.
f(x) = x
3
- x - 1 = 0 (2.9)
.
:
(x).
,
y
a
b
x
A
B
17
2
- 1 = 0 =
3
1
x
-
-1/
3
1/
3
+
+
0
-
0
+
f(x)
-
M
m
+
: M = f (-
3
1
) = -
33
1
+
3
1
- 1 <0
(h. 2-5),
(2.9)
,
.
f(1) = 1
3
- 1 - 1 < 0
f(2) = 2
3
- 2 - 1 > 0
[1, 2]
(2.9)
2-5
trong [1, 2].
,
(2. 9)
,
[1, 2].
y
x
31
31
18
2.2.
1.
(2.1)
[a, b].
x
[a, b]
< b - a.
.
[a, b],
= (a + b)/2.
[a, c] hay [c, b].
(c). (c) = 0
ngh
.
(c) 0. (c)
f(a) . (c) (a)
[a, c]. (c) (a) ng phân ly
[c, b].
[ a, b]
[a, c] hay [c, b], [a
1
, b
1
],
[a, b] [a, b]
:
b
1
- a
1
=
2
1
(b - a).
[a
1,
, b
1
]
, [a
2
, b
2
], [a
1
, b
1
]
[a,
b] [a
1,
, b
1
] :
b
2
- a
2
=
2
1
(b
1
- a
1
) =
2
2
1
(b - a)
, [a
n
, b
n
], [a, b] 1/2
n
[a, b] :
a
n
b
n
; b
n
- a
n
=
n
ab
2
)(
n
, :
19
nnn
aba
n
ab
2
)(
(2.10)
n
, :
nnn
abb
n
ab
2
)(
(2.11)
, a
n
hay b
n
.
Khi n
n
, b
n
.
.
:
. :
.
2.
(2.9)
[1, 2].
:
[1, 2] (1) = 1 - 1 - 1 < 0
f(2) = 2
3
- 2 - 1 > 0
[1, 2]
3/2.
f
2
3
=
2
3
2
-
2
3
- 1 > 0 (1).
[1, 3/2].
[1, 3/2],
5/4.
(5/4) < 0,
f(1).
[5/4, 3/2].
[5/4, 3/2],
11/8.
(11/8) > 0,
(5/4).
[5/4, 11/8].
[5/4, 11/8],
21/16.
(21/16) < 0,
(5/4).
[21/16, 11/8].
[21/16, 11/8],
43/32.
(42/32) > 0,
(21/16).
[21/16, 42/32].
20
21/16 = 1,3125 hay 43/32 =
1,34375 1/2
5
= 1/32 =
0,03125.
5
[1,2] 2 - 1 = 1, (
(2.10) (2.11)).
3.
1)
(x) = 0
2) .
3) [a, b]
21
4)
= (a+b)/2, (c)
f(c)f(a)< 0
Thay b=c
Thay a=c
= b - a
e <
:
a
b
-
a
<
-
b
<
S
S
22
2. 3.
1.
(2.1)
[a,b];
(2.1)
:
X = (x) (2.12)
(2.1)
0
[a,b]
x
n
:
x
n
= (x
n-1
), n = 1,2 (2.13)
x
0
[a,b] (2.14)
,
.
2.
:
2.2 -
n
khi n
(2.13) (2.14)
.
n
.
n
.
n
. .
:
2.4 -
(2.13)(2.14)
1) [a,b] (2.1)
(2.12):
2)
n
(2.13) (2.14) [a,b]:
3) (x) :
| q <1, a<x<b (2.15)
.
