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f.x
0
;y
0
/
C
x
@f
@x
.x
0
;y
0
/
C
y
@f
@y
.x
0
;y
0
/
D
0;
(4.160)
g.x
0
;y
0
/
C
x
@g
@x
.x
0
;y
0
/
C
y
@g
@y
.x
0
;y
0
/
D
0:
Do you understand that (
4.160
) is a linear system of equations with the unknowns
x and y ? To do that, keep in mind that x
0
and y
0
are known. It means that
f.x
0
;y
0
/ is just a known number. Similarly, every term in (
4.160
) evaluated at the
point .x
0
;y
0
/ is known. So the only unknowns are x and y . To simplify the
notation we let f
0
D
f.x
0
;y
0
/, and so on. Then, system (
4.160
) can be rewritten
in the form
7
@f
0
@x
!
x
y
D
f
g
0
:
@f
0
@y
(4.161)
@g
0
@x
@g
0
@y
Let us assume that the matrix
A
D
@f
0
!
@f
0
@y
@x
(4.162)
@g
0
@x
@g
0
@y
is nonsingular. Then
D
@f
0
!
1
f
0
g
0
x
y
;
@f
0
@y
@x
(4.163)
@g
0
@x
@g
0
@y
and we can define
@f
0
!
1
f
g
0
x
y
1
D
x
y
0
C
x
y
D
x
y
0
:
@f
0
@y
@x
(4.164)
@g
0
@x
@g
0
@y
As in the scalar case, this argument can be repeated and we obtain Newton's method
for the 2
2 system (
4.156
),
x
kC1
y
kC1
@f
k
!
1
f
k
g
k
D
x
y
k
:
@f
k
@y
@x
(4.165)
@g
k
@x
@g
k
@y
Here,
@f
k
@x
@f
@x
.x
k
;y
k
/;
f
k
D
f.x
k
;y
k
/;
D
andsoon.
@f
0
@x
D
@f
@x
.x
0
;y
0
/, and so on.
7
Note that
