Graphics Programs Reference
In-Depth Information
EXAMPLE 4.11
Aroot of the equation
P
3
(
x
)
x
3
0
x
2
.
Findamore accurate valueofthis root by one application of Laguerre's iterative
formula.
=
−
4
.
−
4
.
48
x
+
26
.
1is approximately
x
=
3
−
i
Solution
Use the given estimate of the root as the starting value. Thus
x
2
x
3
=
−
=
−
=
−
x
3
i
8
6
i
18
26
i
Substituting these values in
P
3
(
x
) and its derivatives, we get
x
3
0
x
2
P
3
(
x
)
=
−
4
.
−
4
.
48
x
+
26
.
1
=
(18
−
26
i
)
−
4
.
0(8
−
6
i
)
−
4
.
48(3
−
i
)
+
26
.
1
=−
1
.
34
+
2
.
48
i
P
3
(
x
)
0
x
2
=
3
.
−
8
.
0
x
−
4
.
48
=
3
.
0(8
−
6
i
)
−
8
.
0(3
−
i
)
−
4
.
48
=−
4
.
48
−
10
.
0
i
P
3
(
x
)
=
6
.
0
x
−
8
.
0
=
6
.
0(3
−
i
)
−
8
.
0
=
10
.
0
−
6
.
0
i
Equations(4.14) thenyield
P
3
(
x
)
P
3
(
x
)
−
4
.
48
−
10
.
0
i
G
(
x
)
=
=
=−
2
.
36557
+
3
.
08462
i
−
1
.
34
+
2
.
48
i
P
3
(
x
)
P
3
(
x
)
10
.
0
−
6
.
0
i
G
2
(
x
)
08462
i
)
2
H
(
x
)
=
−
=
(
−
2
.
36557
+
3
.
−
−
1
.
34
+
2
.
48
i
=
0
.
35995
−
12
.
48452
i
The term under the square root sign of the denominatorinEq. (4.16) becomes
(
n
1)
nH
(
x
)
G
2
(
x
)
F
(
x
)
=
−
−
2
3(0
08462
i
)
2
=
.
35995
−
12
.
48452
i
)
−
(
−
2
.
36557
+
3
.
√
5
=
.
67822
−
45
.
71946
i
=
5
.
08670
−
4
.
49402
i
Nowwe must find which sign in Eq. (4.16) produces the larger magnitude of the
denominator:
|
G
(
x
)
+
F
(
x
)
| = |
(
−
2
.
36557
+
3
.
08462
i
)
+
(5
.
08670
−
4
.
49402
i
)
|
= |
2
.
72113
−
1
.
40940
i
| =
3
.
06448
|
G
(
x
)
−
F
(
x
)
| = |
(
−
2
.
36557
+
3
.
08462
i
)
−
(5
.
08670
−
4
.
49402
i
)
|
= |−
7
.
45227
+
7
.
57864
i
| =
10
.
62884
Using the minussign, weobtain fromEq. (4.16) the following improved approxi-
mation for the root
n
3
r
=
x
−
=
(3
−
i
)
−
G
(
x
)
−
F
(
x
)
−
7
.
45227
+
7
.
57864
i
=
3
.
19790
−
0
.
79875
i
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