Information Technology Reference
In-Depth Information
Let us try to do this for the example studied above. We have
f.x/
D
2
C
x
e
x
and we choose x
0
D
3.Since
f
0
.x/
D
1
e
x
;
so
5
e
3
1
e
3
f.x
0
/
f
0
.x
0
/
x
1
D
x
0
D
3
D
2:2096:
Note that
j
f.x
0
/
j D j
f.3/
j
15:086
and
j
f.x
1
/
j D j
f .2:2096/
j
4:902;
so the value of f is significantly reduced. Of course, the steps above can be repeated
to obtain the following algorithm.
Algorithm 4.2
Newton's Method.
Given an initial approximation x
0
and a tolerance ".
k
D
0
while
j
f.x
k
/
j
>"
do
x
kC1
D
x
k
f.x
k
/
f
0
.x
k
/
k
D
k
C
1
end
In Table
4.2
we give values of the number of iterations k, x
k
,andf.x
k
/.Inthe
computations we have used x
0
D
3 and "
D
10
6
.
We note that the convergence is much faster than that of the bisection method.
This fast convergence does not occur just by accident. In fact, Newton's method
usually converges quite rapidly and we will study this closer both by theory and by
examples. The theory is presented in Project
4.8
.
