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be hard to use in practice. For scalar equations, the results are fairly easy to use,
but for such equations, we do not really need the help of theoretical results. The
reason is because just by graphing the function, we are able to choose a very good
initial guess that usually leads to fast convergence. For systems of equations with
many unknowns, the situation is much more complex. But for such systems, it is
very hard to check all the conditions of a precise convergence theorem. The most
common computational practice is just to use the method and see what happens. In
preparation for such a practice, we will try to explain (a), (b) and (c) above. Our
explanation is certainly not a full theory, however; we will rely on experiments and
elements of analysis.
We will first look at some examples where the convergence is fast. In fact, we
will show that the convergence is often quadratic, which means that there exists a
constant c such that
j
e
kC1
j
ce
k
;
(4.187)
where e
k
is the error of the kth iterate of Newton's method. Suppose, just for
illustration, that c
D
1 and e
0
D
1=10.Then
j
e
1
j
.1=10/
2
D
1=10
2
;
j
e
2
j
1=10
2
2
D
1=10
4
;
j
e
3
j
1=10
4
2
D
1=10
8
;
so we see that (
4.187
) gives fast convergence.
(a) Consider
f.x/
D
x
2
4:
(4.188)
Use x
0
D
3 and compute x
1
, x
2
, x
3
,andx
4
in Newton's method for the equation
f.x/
D
0:
(4.189)
(b) Define
e
k
D
x
k
x
;
(4.190)
where x
D
2 is a root of f , and consequently e
k
D
x
k
2. Use the numbers
you computed in (a) to compute the ratio
j
e
kC1
j
e
k
c
k
D
(4.191)
for k
D
0; 1; 2; 3. Use the results to argue that
