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Table 22.17
Series expansion
q
(
b
)
q
(
b
)=
N
r
=1
q
r
b
r
(22.108)
E
2
1
cos
B
0
1
−
q
1
=
Q
(
B
0
)=
(22.109)
E
2
sin
2
B
0
1
−
2!
Q
(
B
0
)=
sin
B
0
1+
E
2
1
3sin
2
B
0
+
E
4
−
2+3sin
2
B
0
−
q
2
=
1
(22.110)
2cos
2
B
0
1
− E
2
sin
2
B
0
2
q
3
=
3!
Q
(
B
0
)=
1
6cos
2
B
0
(1
−
E
2
sin
2
B
0
)
2
∗
∗
1+
sin
2
B
0
−
E
2
−
1+5sin
2
B
0
+2sin
4
B
0
−
E
4
2
9sin
6
B
0
10 sin
2
B
0
+11sin
4
B
0
−
(22.111)
−
E
6
sin
2
B
0
6
13 sin
2
B
0
+9sin
4
B
0
−
−
q
4
=
4!
Q
IV
(
B
0
)=
sin
B
0
24 cos
4
B
0
(1
−E
2
sin
2
B
0
)
4
∗
[5+sin
2
B
0
+
E
2
−
5sin
4
B
0
+
E
4
18 sin
2
B
0
−
∗
1
−
20
17 sin
6
B
0
(22.112)
63 sin
2
B
0
+96sin
4
B
0
−
−
+
E
6
−
27 sin
8
B
0
+
24 + 104 sin
2
B
0
−
159
sin
4
B
0
+82sin
6
B
0
−
E
8
sin
2
B
0
−
65 sin
4
B
0
+27sin
6
B
0
]
24 + 68 sin
2
B
0
−
q
5
=
5!
Q
V
(
B
0
)=
1
120 cos
5
B
0
(1
−E
2
sin
2
B
0
)
5
∗
E
2
1+22sin
2
B
0
+93sin
4
B
0
+4sin
6
B
0
−
[5+18
sin
2
B
0
−
∗
E
4
−
86 sin
8
B
0
−
20 + 136 sin
2
B
0
−
338 sin
4
B
0
+68sin
6
B
0
−
(22.113)
E
6
24
220 sin
10
B
0
−
380 sin
2
B
0
+ 1226 sin
4
B
0
−
1588 sin
6
B
0
+ 1178 sin
8
B
0
−
−
E
8
sin
2
B
0
240
81 sin
10
B
0
1100 sin
2
B
0
+ 1936 sin
4
B
0
−
1633 sin
6
B
0
+ 518 sin
8
B
0
−
−
E
10
sin
4
B
0
(120
420
sin
2
B
0
+ 541
sin
4
B
0
−
298
sin
6
B
0
+
sin
8
B
0
)]
−
−
Table 22.18
Harmonic map, “
Northing
”, first boundary value problem,
b
representation
y
(
b,l
)=
y
0
+
y
10
b
+
y
20
b
2
+
y
02
l
2
+
y
30
b
3
+
y
12
bl
2
+
y
40
b
4
+
y
22
b
2
l
2
+
y
04
l
4
+
y
50
b
5
+
y
32
b
3
l
2
+
y
14
bl
4
+
O
b
6
,l
6
(22.114)
22-35 The Symmetry Condition for Eastern Harmonic Function
Let us assume that the “
Eastern function
”
x
=
ξ
(
q,l
) fulfills the symmetry condition
x
=
ξ
(
q,l
)=
ξ
(
−
q,l
)
(22.115)
Such a constraint produces a symmetry “around the
Reference Parallel B
0
or Q
0
”: Points which
are situated South or North of the
Reference Parallel B
0
or Q
0
(
−q
verse +
q
)havethesame
harmonic map. If we compare such a symmetry condition with the general from Eq.(
22.78
)of
the harmonic function of Table
22.11
,weareledtothe
postulate
{
α
1
=0
,β
2
=0
,α
3
=0
,
β
4
=0
,α
5
=0
,etc
.
}
. In consequence we have determined all
Eastern harmonic coe
cients
{
. Accordingly we are led to “
Easting
” (Rechtswert) of type Eq.(
22.134
)ofTable
22.21
,
“
l
-representation” (Fig.
22.3
).
α
}
,
{
β
}
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