Graphics Programs Reference
In-Depth Information
EXAMPLE 4.4
Determine the root of
x
3
10
x
2
f
(
x
)
=
−
+
5
=
0thatlies in (0
.
6
,
0
.
8) with Brent's
method.
Solution
Bisection
The starting points are
6
3
6)
2
x
1
=
.
f
1
=
.
−
.
+
=
.
0
6
0
10(0
5
1
616
8
3
8)
2
x
2
=
0
.
8
f
2
=
0
.
−
10(0
.
+
5
=−
0
.
888
Bisectionyields the point
7
3
7)
2
x
3
=
0
.
7
f
3
=
0
.
−
10(0
.
+
5
=
0
.
443
By inspecting the signsof
f
weconcludethat the newbrackets on the root are(
x
3
,
x
2
)
=
(0
.
7
,
0
.
8).
First interpolation cycle
Substituting the above values of
x
and
f
into the numer-
ator of the quotient in Eq. (4.2), we get
num
=
x
3
(
f
1
−
f
2
)(
f
2
−
f
3
+
f
1
)
+
f
2
x
1
(
f
2
−
f
3
)
+
f
1
x
2
(
f
3
−
f
1
)
=
0
.
7(1
.
616
+
0
.
888)(
−
0
.
888
−
0
.
443
+
1
.
616)
−
0
.
888(0
.
6)(
−
0
.
888
−
0
.
443)
+
1
.
616(0
.
8)(0
.
443
−
1
.
616)
=−
0
.
30775
and the denominatorbecomes
den
=
(
f
2
−
f
1
)(
f
3
−
f
1
)(
f
2
−
f
3
)
=
(
−
0
.
888
−
1
.
616)(0
.
443
−
1
.
616)(
−
0
.
888
−
0
.
443)
=−
3
.
9094
Therefore,
f
3
num
den
443
(
−
0
.
30775)
x
=
=
0
.
=
0
.
034 87
(
−
3
.
9094)
and
=
x
3
+
=
.
+
.
=
.
x
x
0
7
0
034 87
0
734 87
Since the result is within the establishedbrackets, we accept it.
Relabel points
As
x
>
x
3
, the points are relabeledas illustratedinFigs. 4.2(b) and
4.3(b):
x
1
←
x
3
=
.
0
7
f
1
←
f
3
=
0
.
443
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