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In-Depth Information
Since
u
nC1
D
v
and
u
n
D
c, we get the solution
1
1
t
u
nC1
D
u
n
:
(4.8)
Similarly, for any function g that is in the form
g.
u
/
D
˛
C
ˇ
u
;
(4.9)
where ˛ and ˇ are constants, we can solve (
4.3
) directly and get
c
C
˛t
1
ˇt
u
n
C
˛t
1
ˇt
v
D
or
u
nC1
D
:
(4.10)
Moreover, let us also consider the case of
u
0
D
u
2
;
(4.11)
that is
g.
v
/
D
v
2
:
(4.12)
Then (
4.3
) reads
v
t
v
2
D
c;
(4.13)
and we have two possible solutions
1
C
p
1
4t c
2t
v
C
D
(4.14)
and
p
1
1
4t c
2t
v
D
:
(4.15)
If we expand
1
(
4.14
)and(
4.15
)intermsoft ,weget
1
t
c
c
2
t
C
.t
2
/
v
C
D
(4.18)
O
1
What do we mean by “expanding” an expression in terms of something? It means that we derive
a series expansion in terms of a certain parameter. Suppose we are interested in the function
sin.e
h
1/
(4.16)
for small values of h. We can compute the first few terms in a Taylor series, e.g.,
1
2
h
2
sin.e
h
C
O
.h
3
/:
1/
h
C
(4.17)
Here, (
4.16
) is expanded in terms of h.
