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In-Depth Information
By Taylor-series expansions we have
.x
2
/;
ln.1
C
x/
D
x
C
(3.42)
O
so
ln
1
C
D
O
10
4
=N
4
;
10
2
N
2
10
2
N
2
C
(3.43)
and thus
N ln
1
C
10
2
N
2
10
2
N
(3.44)
for large values of N .By(
3.41
)wenowhave
1
C
N
10
2
N
2
e
10
N
:
(3.45)
Again, by a Taylor expansion,
e
x
D
1
C
x
C
.x
2
/;
(3.46)
O
so for large values of N , it follows that
e
10
2
=N
1
C
10
2
N
;
(3.47)
and thus
1
C
N
10
2
N
2
10
2
N
1
C
:
(3.48)
From (
3.40
), we now find that
r
0
D
1
C
10
2
=N
2
N
1
r
0
1
C
10
2
=N
1
r
0
D
10
2
r
N
N
r
0
D
10t r
0
;
since t
D
10=N . We see that
r
N
r
0
r
0
10t;
