Graphics Programs Reference
In-Depth Information
EXAMPLE 10.5
Use the Fletcher-Reeves method to locate the minimum of
10
x
1
+
3
x
2
−
=
10
x
1
x
2
+
F
(
x
)
2
x
1
00
T
.
Start with
x
0
=
Solution
Since
F
(
x
) isquadratic, we need only two iterations. The gradientof
F
is
20
x
1
−
10
x
2
+
2
∇
F
(
x
)
=
−
10
x
1
+
6
x
2
First iteration:
−
2
0
−
2
0
−
2
s
0
g
0
=−
∇
F
(
x
0
)
=
v
0
=
g
0
=
x
0
+
s
v
0
=
10(2
s
)
2
3(0)
2
f
(
s
)
=
F
(
x
0
+
s
v
0
)
=
+
−
10(
−
2
s
)(0)
+
2(
−
2
s
)
40
s
2
=
−
4
s
f
(
s
)
=
80
s
−
4
=
0
s
=
0
.
05
0
0
05
−
2
0
−
0
.
1
x
1
=
x
0
+
s
v
0
=
+
0
.
=
0
Second iteration:
0
−
−
20(
−
0
.
1)
+
10(0)
−
2
g
1
=−
∇
F
(
x
1
)
=
=
10(
−
0
.
1)
−
6(0)
1
.
0
g
1
·
g
1
1
0
4
.
γ
=
g
0
=
=
0
.
25
g
0
·
0
−
25
−
2
0
−
0
.
5
v
1
=
g
1
+
γ
v
0
=
+
0
.
=
1
.
0
−
1
.
0
s
−
.
−
.
−
.
−
.
0
1
0
5
0
1
0
5
s
x
1
+
s
v
1
=
+
=
0
−
1
.
0
−
s
f
(
s
)
=
F
(
x
1
+
s
v
1
)
5
s
)
2
s
)
2
=
10(
−
0
.
1
−
0
.
+
3(
−
−
10(
−
0
.
1
−
0
.
5
s
)(
−
s
)
+
2(
−
0
.
1
−
0
.
5
s
)
5
s
2
=
0
.
−
s
−
0
.
1
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