Geography Reference
In-Depth Information
f
(0) = 0 =
c
→
(23.24)
f
(
V
)=
23
16
tanhV
+
1
7
1
48
tanh
3
V
80
tanh
5
V
+
O
7
(
tanhV
)
16
V
−
−
(23.25)
Box 23.11 (The integral
1+
tanh
2
xdx
).
√
1+
x
2
=1+
1
1
2
·
1
4
x
4
+
1
1
4
·
3
6
·
2
x
2
x
6
+0
8
,
−
2
·
∀|
x
|
<
1
(23.26)
1+
tanh
2
x
=1+
1
1
8
tanh
4
x
+
1
2
tanh
2
x
16
tanh
6
x
+
O
8
(
tanhx
)
,
−
∀|
tanh
x
|
<
1
(23.27)
tanh
2
xdx
=
−tanhx
+
x
(23.28)
tanh
4
xdx
=
3
tanh
3
x
+
tanh
2
xdx
=
1
1
3
tanh
3
x
−
−
−
tanhx
+
x
(23.29)
tanh
6
xdx
=
−
5
tanh
5
x
+
tanh
4
xdx
=
−
1
1
1
5
tanh
5
x −
3
tanh
3
x
−
tanhx
+
x
(23.30)
1+
tanh
2
xdx
=
x
1+
1
tanhx
1
+
1
8
+
1
1
8
+
1
2
−
−
2
−
(23.31)
16
16
3
tanh
3
x
1
1
1
16
1
80
tanh
5
x
+
O
7
(
tanhx
)
8
−
−
2
:=
2
onto the
Box 23.12 (Equiareal mapping of the rotational symmetric hyperboloid
M
H
circular cylinder
C
1
).
f
=cosh
V
√
cosh 2
V
=
1+
tanh
2
h
Λ
1
Λ
2
(
V
)=1
∀
V
↔
1
− tanh
2
h
(23.32)
f
(
V
)=
(1
tanh
2
V
)
−
1
1+
tanh
2
V
1
/
2
dV
−
(23.33)
f
(0) = 0 =
c
→
(23.34)
f
(
V
)=
85
69
16
tanhV
45
23
48
tanh
3
V
80
tanh
5
V
+
O
7
(
tanhV
)
16
V
−
−
−
(23.35)
tanh
2
x
−
1
1+
tanh
2
xdx
).
Box 23.13 (The integral
1
−
1
x
2
−
1
√
1+
x
2
=1+
3
2
x
2
+
11
8
x
4
+
23
16
x
6
+
O
8
∀|
−
x
|
<
1
(23.36)
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