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We will derive Newton's method for this system along the lines used in the scalar
case. But before we do that, we observe that the first equation of (
4.145
) implies that
x
D
˛
t y
3
(4.146)
and thus the second equation reads
y
t .˛
t y
3
/
3
ˇ
D
0:
(4.147)
Here we note that (
4.147
) is a scalar equation. We can therefore solve (
4.147
)bythe
techniques derived above. The resulting approximation y
y
can then be used in
(
4.146
) to compute an approximation x
x
.
For the present simple 2
2 system, this is a possible approach. But keep in mind
that we are preparing for really big systems. And for such systems, this substitu-
tion technique is not always workable and leads to horrible complications and an
extremely high likelihood of errors. Therefore, we should solve (
4.145
) as a system.
4.6.3
Newton's Method
Our aim is now to derive Newton's method for solving the system
f.x;y/
D
0;
g.x; y/
D
0:
(4.148)
But let us start by briefly recalling the scalar derivation, where we consider the scalar
equation
p.x/
D
0;
(4.149)
and we assume that x
0
is an approximation of x
that solves (
4.149
). By a Taylor
series expansion, we have
p.x
0
C
h/
D
p.x
0
/
C
hp
0
.x
0
/
C
.h
2
/:
(4.150)
O
Since we want h to be such that p.x
0
C
h/
0, we can define h to be the solution
of the linear equation
p.x
0
/
C
hp
0
.x
0
/
D
0;
(4.151)
and thus
h
D
p.x
0
/=p
0
.x
0
/:
(4.152)
