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
Search WWH ::




Custom Search