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where one of the zeros is x
D
p
2. The secant method for this problem reads
x
k
x
k1
f.x
k
/
f.x
k1
/
x
kC1
D
x
k
f.x
k
/
2/
x
k
x
k1
x
k
D
x
k
.x
k
x
k1
x
k
2
x
k
C
x
k1
D
x
k
;
so
x
k
x
k1
C
2
x
k
C
x
k1
x
kC1
D
:
(4.71)
We choose x
0
D
1 and x
1
D
2, and get
x
2
D
1:33333;
x
3
D
1:40000;
x
4
D
1:41463:
This should be compared with
p
x
D
2
1:41421:
Recall also that three iterations of Newton's method gives 1:41422, so again we note
that Newton's method is slightly more accurate than the secant method.
4.5
Fixed-Point Iterations
We started this chapter by considering the ODE
u
0
D
g.
u
/;
u
.0/
D
u
0
:
(4.72)
An implicit scheme for this problem reads
u
nC1
t g.
u
nC1
/
D
u
n
;
(4.73)
where
u
n
u
.t
n
/ and t
n
D
nt for a given time step t > 0. We note that (
4.73
)
is an equation with
u
nC1
as unknown. By setting
v
D
u
nC1
and c
D
u
n
, we obtain
the equation
v
t g.
v
/
D
c;
(4.74)
