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where c is given and
v
is unknown. We can rewrite this equation in the form
v
D
h.
v
/;
(4.75)
where
h.
v
/
D
c
C
t g.
v
/:
For equations on this form, one can apply a very simple iteration in order to find an
approximation of the solution
v
D
v
satisfying (
4.75
). The iteration reads
v
kC1
D
h.
v
k
/;
(4.76)
where
v
0
is an approximate initial guess.
2
A solution
v
of (
4.75
) satisfies
v
D
h.
v
/:
Since h leaves
v
unchanged, h.
v
/
D
v
, and the value
v
is referred to as a
fixed-
point
of h. Furthermore, since the purpose of iteration (
4.76
) is to compute
v
,(
4.76
)
is referred to as a
fixed-point iteration
.
Let us use (
4.76
) to generate numerical approximations of one specific equation:
x
D
sin.x=10/:
(4.77)
In Fig.
4.7
we have plotted the graph of y
D
x and y
D
sin.x=10/ and we note that
x
D
0 is the only solution of (
4.77
).
y
y
=
x
y
= sin(
x/
10)
−
10
π
10
π
x
D
D
Fig. 4.7
The graph of y
x and y
sin.x=10/
2
It is important to note that (
4.76
) is an iteration to solve (
4.75
), i.e. to solve (
4.73
)at
one
specific
time level. Thus, we have an iterative procedure within each time step.
