26481
354
1
1/ 4
1/ 4
354
331776 357904
26481
1/ 4
1/ 4
357904
3
3
3
3
3
6
36
2
28
2
3 2
a
a
b b a
b
b a
a1
a0
a2
a0
a1
a
1 a0
a0
a2
a0
3 1 i 6
y
y
m
m
y m
n
y
m
n 1
m
x
1 x 2
x
1 x2
1
x
1 1 x
n
1 1 1
2
a0 a1
a2
an 1
an
b0
b1
bn 2
bn 1
b0
b1
b2
bn
1
bn
f( )
The remainder
1
2
f ( )
2!
f ( )
f ( )
1!
2!
f ( )
f ( )
f ( )
1!
2!
3!
f ( )
1!
f ( )
f ( )
2!
3!
f ( )
2!
1 5 6 7 9
1 4 2 9
1 4 2 9 0
2
3 5
7
10
2
4 2
1 7
14
7
14
4
2
remainder
f (2 )
f ( 3)
2!
Synthetic division of a polynomial by a quadratic expression.
Example 14.
a0
a1
b0
a2
b1
b0
an 1
bn 2
bn x
an
bn 1
bn 2
b0
b1
b2
bn
bn
x 2
1
1x
2
1
1
2
2
1
2
1 2
3
2
7
2
3
2
7
2
1
2
1
2
2
2
2
2
2
2
2
2
2
2
2
3
4
2
2
2
2
2
3
4
y
P
y = (x – 1)2
(1, 0)
O
x = 1.8
x
y=
lo
g
x
y=
x3
y
x = 0.7
x
O
y=
–x
–
1
x
4
x 2
100
x
1 x
1 x
a0
b0
2
a1
1
2 b1
2
12 2
1
2
1
2
1 1
2 2
1
1
2 2
1
1
2 2
1
1
2 2
1
1
2 2
1
1
2 2
1
1
2 2
1
1
2 2
1
1
2 2
1
1
2 2
1
x
1 1 2 2
1
1
2 2
3
3
1
1
2 2
1
1
2 2
3
3
2
y
A[xo, f (xo)]
x3
O
x2
x1
r (x)
xo
x
B[x1, f (x1)]
f ( x1 )
x1
f ( x0 )
x0
x1
f ( x1 )
x0
f ( x) 0
x
f
when f = f ( x ), f
x0
f0
f0
1
1
1
0 = f ( x0 )
We can also write [2a] as:1
m1 = x 2 = x 0
m1 = x 2 =
x f
f
0 1
1
x1 f0
f0
(2a )
gives the first approximation (2b)
x
f
x2
f2
1
1
f
2
2 1849
.
4 0159
.
x
f
1
1
x3
f3
f
3
2 18548
.
4 0.019
x
f
1
1
x0
f0
1
317798
.
f
0
x
f
1
1
x2
f2
f
2
0.68533
2.69785
x
f
1
1
x3
f3
f
3
0.55327
2.38154
x0 f 1
f1
x
f
x1 f 0
f0
1
1
x0
f0
f
0
x
f
x2
f2
1
1
f
2
x2 f 1
f1
x1 f 2
f2
(0.8506) (0.2136) 0.9 ( 0.2196)
0.2136 ( 0.2196)
0.36812
0.4332
x
f
1
1
x3
f3
f
3
x1 f 3
0.8506 0.2136 0.9 ( 0.00697)
f2
0.2136 ( 0.00697)
x3 f 1
f1
0187972
.
0.22057
x
f
1
1
x4
f4
f
4
x4 f 1
f1
x1 f 4
f4
0.8523 (0.2136) (0.9) ( 0.00166)
0.2136 ( 0.00166)
0183545
.
0.21526
x
f
1
1
3
0.23136
x0
f0
f
0
2
0.59794
( 0.59794)
x
f
1
1
x2
f2
f
2
3 2.721
0.23136 0.0171
x
f
1
1
x3
f3
f
3
3 2.74020
0.23136 0.00039
1
x
f
x0
f0
1
1
f
0
1
( 9)
9
10
(2.9) (1)
1 (
x
f
1
1
x2
x f
2 1
f1
f2
x1 f 2
f2
3 ( 0.711)
0.711)
f (1) = 2
f (2) = 39
x
f
1
1
x0
f0
f
0
0
0
3
y =f (x)
y
B
f (b)
A
O
f (a)
C
m
c
a
b
x
y
x
f (a )
f (b)
a
b
f (a )
a
b
f (b)
a
f (a )
x1
f ( x1 )
x0
f ( x0 )
x1
f1
x0
f
f0 0
x2
f ( x2 )
x1
f ( x1 )
x2
f2
x1
f
f1 1
mk = x k
= xk
1
xk
fk
1
xk
fk
1
1
f k 1 , k = 1, 2,
x1
f1
x0
f
f0 0
2 1
2
7
5
5
3 2
x2
f2
x1
f1 14
. 2
1056
.
3
( 0.6)
0.6
4
.
056
4.056
x3
f3
x2
f2 14
. ( 0.3436) 15562
.
( 1056
. )
0.3436 1056
.
x1
f1
x0
f0 1 0
2.17798 1
( xk
f ( xk )
xk 1 )
f ( xk
1)
f ( xk 1 )
( ek ek 1 )
f (m ek ) f (m ek 1)
1
2 1
2
1
2
1
2
1
1 (e
2 k
ek 1 )
f ( m)
f ( m)
1 1
f ( m) ( ek
f ( m ek )
ek 1 )
1
1
f ( m ek 1 )
f (m)
1 (e
2 k
ek 1)
1 (e
2 k
ek 1)
f (m)
f (m)
f (m)
f (m)
1
1
1e
2 k
1
1
f ( m)
f ( m)
1 (e
2 k
ek 1 )
f (m)
1
2 f (m)
f (m)
f (m)
1/ p
1 |e |
k
k 1
1
2 1 |e |
k
k 1
1/ p
1
2(
k 1)
1/ p
| f (m)|
| f (m)|
| f (m)|
| f (m)|
1
p
1
5
2
xk 1 xk
f ( xk 1 ) f ( xk )
ek 1
f ( m ek 1 )
ek
f ( m ek )
ek f (m ek 1 ) ek 1 f (m ek )
f (m ek 1 ) f (m ek )
f (m)
1
2 f (m)
ek
1
=
1e e
2 k k
1
f (m)
f (m)
h = x
h = x
3
=
2
1
1
x1 = 0.5,
x0 = 0.5,
=
h2
= 1,
h1
2
=1
2
2
2 f0
2
2 f1
(
2
2
2
g =
c =
Since g2
2
2
(
2 f0
=2
2
2 ) f2
= 2.5
f 2 = 0.75
2 f1
0, we get
2
g22
g2
2 f2
4
2 f 2 c2
=
12
.
= 1616286
24.25
2.5
= + ( )
= 0.191857,
= 0.047777
= = 0.808143
=
h3
= 1.61628,
h2
=
=
(
2
3 f3
+(
=1+
+
) = 0.616286
= 2.559263
+ ) = 0.680961.
Since < 0, we get
=
g32
g3
4
3 f 3c3
0.058889
2.559263
6.469625
=
= 0.011541
= + ( )
= 0.201184,
= 0.002223
= = 0.009327
=
h4
= 0.0115441,
h3
=
=
(
+ (
+
=1+
) = 0.988459
= 0.044908
+ ) = 0.000120.
2
4 f4
g42
g4
4
4 f 4 c4
0.004395
0.044908
0.002016
x5 = 201640
xk
1
= xk
( xk
xk 1 )
k 1
k = 1, 2,
1
D
1
D
D = ( xk
= ( xk
= ( xk
1
1
xk )2 ( xk
xk ) ( xk
xk 1 ) ( xk
2
2
xk ) ( xk
xk ) ( xk
xk 2 ) ( xk
1
1
2
xk )2 ( xk
xk xk
xk 2 )
2
1
xk )
xk )
xk
1
= xk
a12
a1
a12 4a0a2
2a 0
2a 0
a1
a12
4a 0 a 2
2a 2
, where a1
Dv
a12
4a 0 a 2 = D v
4a0 a2
a12
4a0 a2 a1
a12
4a 0 a 2
a
a
a
= f 2 = 3, D = ( x2 x1 ) ( x2 x0 ) ( x1 x0 ) = 0.25
1 [( f
f 1 ) ( x2 x0 ) 2 ( f 2 f 0 ) ( x2 x1 ) 2 ] = 2.5
1 =
2
D
1 [( f
f 1) ( x2 x0 ) ( f 2 f 0) ( x2 x1 )] = 15
.
0 =
2
D
2
x2
2a2
a1
a12
4a 0 a 2
1
6
2.5
24.25
a
a
a
= f 3 = 0.047777, D = ( x3 x1 ) ( x3 x2 ) ( x2 x1 ) = 0124512
.
1
2
2
[( f 3 f 2 ) ( x2 x1 )
( f 3 f 1) ( x3 x2 ) ] = 5138588
.
1 =
D
1 [( f
f 2 ) ( x3 x1 ) ( f 3 f 1) ( x3 x2 )] = 1691854
.
.
0 =
3
D
2
2a 2
a1
a12
a
a
a
4a 0 a 2
0.095554
5138588
.
26.081760
= f 4 = 0.002228, D = ( x4 x2 ) ( x4 x3 ) ( x3 x2 ) = 0.006020
1 [( f
f 2 ) ( x4 x2 ) 2 ( f 4 f 2 ) ( x4 x3 ) 2 ] = 4.871483
1 =
3
D
1 [( f
f 3) ( x4 x2 ) ( f 4 f 2 ) ( x4 x3 )] = 1393112
.
.
0 =
4
D
2
2a 2
a1
a12
4a 0 a 2
0.004456
4.871483
23.718931
x5 = 0.201640
2
2.1
2.13
2.136
4
4.2
4.26
4.264
2
2.7
2.71
2.714
1
19
.
191
.
1912
.
1.9129
p = p h, q = q k k = 0,1,
where h = b c b c
c
c (c
b )
b (c
b ) bc
k=
c
c (c
b )
k 1
k 1
k
k
n n 3
2
n 2
n 3
n 1 n 2
n 1
n 1 n 1
n 1
2
n 2
n 3
n 1
n 1
n n 2
n 1
XXX
XXX
XXX
lim
k
lim
k
lim lim k
k
h = k = c12
bc b ) b3c0 b2 c1
c0 (c2 b2 )
b2 (c2 b2 )
c12 c0 (c2
3 1
2
b
b
n
n 1
cn 1h cn 2 k = 0
cn 2 h cn 3 k = 0
bn = 2; bn
1
= 2; cn
1
= 68; cn
= 40; cn
2
3
= 7
2 68h 40k = 0 2 40h 7k = 0
h
7
2
40
2
k
40
2
68
2
1
40
68
7
40
k
h
1
80 136
1600 476
14 80
h
66
66 33 56 14
1124
562
1124
281
p = p
q = q
1
0
1
0
k
1
56
1124
h = 2 0.0580 = 19420
.
,
k = 2 0.0457 = 19543
.
,
x2
19420
.
x
1.9543
b
b
n
n 1
cn 1h cn 2 k = 0
cn 2 h cn 3 k = 0
62h 39 k 50 = 0
39h 8k 62 = 0
h
39
8
50
k
62
39
62
50
62
1
62 39
39
8
h = k = 1
2018
1894
1025
2018
1894
1025
1025
.
2136
. h 22.82 k = 0 7190
16.09 22.82h 4.06k = 0
c22
b4 c1 b3c2
c1 (c3 b3 )
b3 (c3 b3 ) b4 c2
c22 c1 (c3 b3 )
p1 = p0 h = 0.6667
.
q1 = q0 k = 10
0119
.
143
. h 2.8k = 0
019
.
2.8h k = 0
0.651
6.22
0.6410
6.22
Bk
Bk
1
1
Bk
Bk
1
1
Note:
2 (1)( 5) = 10
Note:
50 = 2(1) (25)
B B B B 1/ 4
1
0
2
0
1/ 4
625
626
1/ 4
Note: 32 = ( 2)(1)( 16)
256 = ( 2)(8)(16)
Note: 0 = ( 2)(1)(0)
49152 = ( 2)
Note: 4 = ( 2)(1)( 2)
0 = 2(0)(2) = 0
Note:
8 = ( 2)(1)(4)
32 = ( 2)(4)(4)
B B 1
1
16
0
2
2
B3
B1
2
2
0 = ( 2)(0)
24 = ( 2)(3)( 4)
48 = ( 2)(1)(24)
288 = ( 2)(9)(16)
576 = ( 2)(1)(288)
16896 = ( 2)(33)(256)
132096 = ( 2)(1)(66048)
67239936 = ( 2)(513)(65536)
B3
B2
( 2)(1)(39.5)
y
y = f (x)
A
f (a)
C
O
m
c
a
x
f (a )
f (a )
f (a )
f (a )
that is, m
a
f ( a)
and x
f ( a)
a
f ( a)
respectively
f (a)
f (m)
ek f (m)
f (m)
ek f (m)
1 e 2 f (m)
2! k
1 e 2 f (m)
2! k
ek [ f ( m)
ek f (m)
e
2
k
] [ f (m) ek f (m)
f ( m) ek f (m)
ek2
2
f
ek2
2
f (m)
]
(m)
f (m)
ek2
2
f (m)
f ( m)
f ( m)
2 f ( m)
f (m)
2 f (m)
ek
1
f ( m)
= 1 ek2
2
f ( m)
x k
1
= xk
f ( xk )
f ( xk )
f ( xk )
f ( xk )
x x4
10
xk
4 x k3
3x x4 10
4 x k3 1
1
3(15
. ) 4 10
4(15
. )3 1
f ( xk )
f ( xk )
x k3 3x k 5
3( x k2 1)
2 x k3
3( x k2
5
1)
2 x03
3( x02
16 5
5
21
3
4
1
9
(
)
1)
2 x13
3( x12
5
2 (2.3333) 3 5
1)
3{(2.3333) 2 1}
2 x23
3( x22
5
2 (2.2806) 3 5
1)
3{(2.2806) 2 1}
2 (2.2790) 3 5
3{(2.2790) 2 1}
2
f ( xk )
f ( xk )
x k 2 sin x k
1 2 cos x k
2 (sin x k x k cos x k )
1 2 cos x k
sin x k x k cos x k
0.5 cos x k
1
1
2
x
x
1
1
x
2
x
1 1
x
k 1
xk
f ( xk )
, k
f ( xk )
0,1, 2
f ( x)
f ( x)
2e
1
x
x
2e
1
2
1
x
1
x
(x
1
2)
2
(x
1) 2
0.689753, 0.689753 x1 = 0.737984,
x2 = 0.699338,
x3 =
x4 =
0.68975 is the required root
correct to 5 places
3x k cos x k
3 sin x k
1
x k sin x k cos x k
3 sin x k
1
0.5 (0.5) cos(0.5) 1
3 sin (0.5)
2.11729533
3.479425539
First approximation: x1 = 0.608518649
0.608518649
sin (0.608518649) cos(0.608518649) 1
3 sin (0.608518649)
2.16835703
357165265
.
Second approximation: x2 = 0.607101878
0.607101878 sin (0.607101878) cos(0.607101878) 1
3 sin (0.607101878)
2.16765013 35748962
.
1 (cosh x k ) (cos x k )
(cosh x k ) (sin x k ) (sinh x k )(cos x k )
1 (cosh x0 ) (cos x0 )
(cosh x0 ) (sin x0) (sinh x0 )(cos x0)
Let f ( x )
log x
cos x , since f (13
. )
1 sin x
. )
and f (14
x
so that f ( x )
0.00513 0
.
01665
0
log x k cos x k
1 sin ( x )
k
xk
log ( x0 ) cos( x0 )
1 sin ( x )
0
x0
(13
. ) log (13
. ) (13
. ) (cos13
. c)
1 (13
. ) (sin (13
. c)
0.00667
2.2526
c
(13029
.
) log (13029
.
) (13029
.
) cos (13029
.
)
c
1 (13029
.
) sin (13029
.
)
0.0001444
199606
.
f ( xk )
f ( xk )
f ( x0 )
f( )
f ( )
f ( x0 )
( sin
cos )
cos
x1 =
1
2.8233.
(2.8233) (sin (2.8233)) cos(2.8233)
(2.8233) (cos(2.8233))
x2 = 2.7986
(2.7986) (sin (2.7986)) (cos(2.7986))
(2.7986) (cos(2.7986))
x3 = 2.7984
(2.7984) (sin (2.7984)) (cos(2.7984))
(2.7984) (cos(2.7984))
x4 = 2.7984
cos x
cos x
1
when | x | 1
(x
2
1)
2
when | x |
i
0; x1
1
cos1 c
sin 1
f ( xi ) c
f ( xi )
1
2
x
2
x 3 6
2
x 3 x
2
6
0.3 x
e
6
f ( xk )
f ( xk )
Let
f ( x)
x2
f ( x)
2x
4 cos x ,
4 sin x;
f ( x0 )
f ( x0 )
x02
2 x0
4 sin x0
4 cos x0
( 19
. )2
2 ( 19
. )
4 sin ( 19
. )
4 cos( 19
. )
3.61 3.78
38
. 1293
.
x12
2 x1
4 sin x1
( 193
. )2
4 cos x1
2 ( 193
. )
x2
4 sin ( 193
. )
4 cos ( 193
. )
0.0198
5.266
19338
.
1
2
1
2
1
x
1
2
x
1
( x 1)
8
2
x 2 e 0.3 x
x
2
6
f ( x)
f ( x)
g ( xk )
g ( xk )
{ f ( x) 2 } f ( x) f ( x)
{ f ( x )}2
(f
k
fk fk
)2 f k f
k
f ( xn )
f ( xn )
1
p
f ( x)
f ( x0 )
f ( x0 )
f ( x) f ( x0 ) f ( x0 )
x3
14045
.
0.0139
15.7909
14036
.
x1
5.0938
24.3125
12905
.
12887
.
12.8676
xn
xn 1 N
xn 2
xn ( 2
1
Nxn )
Square root of a number
xn2 N
2 xn
1 xn
2
xn
1
N
xn
1 x
2 n
N
xn
The p th root of a number N
x
xnp N
pxnp 1
( p 1) xnp N
p xnp 1
n 1
1 ( p 1) x
n
p
N
xnp 1
The reciprocal p th root of N
fk
fk
x
N
xk p
p xk ( p
xk
[p 1
p
k 1
1)
N x kp ]
12 8 1 xn
2
N
xn
.
31416
15.
.
1 1.7972 31416
2 17972
.
31416
.
1 1772626
.
2 1772626
.
31416
.
1 17724559
.
2 17724559
.
1 15
.
2
1 xn
2
N
xn
12
N x
1 35
. 12
2 35
. 12
1 3.464
2 3.464 12
1 3.4641
2 3.4641 1 x0
2
0
12
4 8 9
8 8
N x
8
1 x
x 2
8
1 2.85
2 2.85 1 xn
2
0
n
0
m
xk
1
( xk ) 2
2
xk 1
1
2
1
2
1
2
(0.02661096) 2
0.473389
(0.0083721) 2
0.02425308
x
x
x
7
8
9
0.517744 0.517757 0.517782,
3
2
2
1
3
sin x
3
2
3 1
2
xn
( xn 1 ), n 1
lim n
1 4
Near 0.3,
Near 2.1,
d 1 ex = 1 ex
4
dx 4
d [log 4 x ] = 1
dx
x
1
4
1 1
15
30
x2
(2!) 2
x3
(3!) 2
x4
(4!) 2
3
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