Information Technology Reference
In-Depth Information
4.6.4
A Nonlinear Example
Before we return to the nonlinear system (
4.143
), we will consider an example of a
nonlinear algebraic system just to test method (
4.165
). The system
e
x
e
y
D
0;
ln.1
C
x
C
y/
D
0;
(4.166)
can be solved analytically. It follows from the first equation that x
D
y and thus the
second equation is reduced to
ln.1
C
2x/
D
0:
Note that ln.
z
/
D
0
)
z
D
1, thus we have
1
C
2x
D
1;
which gives x
D
0 and so y
D
0. Since we have the exact solution, we can use
this example to check the properties of Newton's method applied to this system. By
defining
f.x;y/
D
e
x
e
y
and
g.x; y/
D
ln.1
C
x
C
y/;
we can specify Newton's method (
4.165
)as
e
x
k
!
1
e
x
k
e
y
k
ln.1
C
x
k
C
y
k
/
x
kC1
y
kC1
D
x
y
k
(4.167)
e
y
k
1
1Cx
k
Cy
k
1
1Cx
k
Cy
k
Recall that, in general, if ad
bc
ยค
0,
ab
cd
1
D
d
b
ca
;
1
ad
bc
(4.168)
D
2
so the inverse needed in (
4.167
) is readily computed. By choosing x
0
D
y
0
,
we get the following results:
