Information Technology Reference
In-Depth Information
f
0
.0/
D
0;
f
00
.0/
D
1;
(4.225)
and
g
0
.0/
D
1;
g
00
.0/
D
0:
(4.226)
Use these two facts and (
4.215
) to explain once more that you expect slow and
fast convergence for (
4.217
)and(
4.218
) respectively.
(t) Compute x
1
, x
2
, x
3
, x
4
in Newton's method for (
4.217
)and(
4.218
)using
x
0
D
1.For(
4.217
) you should also compute e
kC1
=e
k
,andfor(
4.218
) you
should compute
9
e
kC1
=e
k
.
We have now seen one example where the convergence becomes slow and also
one example of very fast convergence. In both cases we can explain these effects by
either graphing or analytically using (
4.215
). Finally, we shall discuss divergence.
In fact, several things can go terribly wrong with Newton's method:
(I) We can generate a sequence
f
x
k
g
where
j
x
k
j!1
as k increases.
(II) We can obtain convergence toward an unwanted solution.
(III) We can generate an infinite loop, reiterating the same values over and over
again without converging toward a solution.
The purpose of the rest of this project is to illustrate these possibilities.
(u) To illustrate (I) above, we consider
f.x/
D
1
x
2
:
(4.227)
Use graphical analysis to see that if x
0
0,then
j
x
1
j
is very large. Furthermore,
if x
0
D
0,thenx
1
becomes infinite.
(v) To illustrate (II), we consider
f.x/
D
.x
C
1/.x
1/.x
2/:
(4.228)
Set x
0
D
0. Since, both x
D
1 and x
D
1 are zeros of f , we would expect
Newton's method to converge toward one of these values. This is however, not
the case. Compute x
1
,graphf , and explain what happens.
(w) To illustrate (III), we consider
f.x/
D
x
x
3
;
(4.229)
which has zeros in
1, 0 and 1. Show that Newton's method for solving (
4.229
)
can be written in the form
9
If
3
j
e
k
C
1
j
c
j
e
k
j
for a constant c not depending on k, we have cubic convergence. If
j
e
k
C
1
j
c
j
e
k
j
, with c<1, we have linear convergence.
