3-4 The oblique frame of reference of the ellipsoid-of-revolution 147
Solution (the third step).
R
2 (A) + R
2
1 (A) = A 1
1 − E
2
1 + E
2
1 − cos
2 I cos
2 A
+
+E
4
1 − cos
2 I cos
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
1 − cos
2 I cos
2 A
3 + 3 cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O 1
E
8
.
(3.122)
End of Solution (the third step).
Solution (the fourth step).
dS
dA
1
A 1
√
1 − E 2
= (1 + x)
+1/2 = 1 +
1
2
x −
1 · 1
2 · 4
x
2 +
1 · 1 · 3
2 · 4 · 6
x
3
−
1 · 1 · 3 · 5
2 · 4 · 6 · 8
x
4 + O +
x
5
, (3.123)
subject to
x := E
2
1 − cos
2 I sin
2 A
+ E
4
1 − cos
2 I sin
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
1 − cos
2 I sin
2 A
2 + 3 cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O +
E
8
;
(3.124)
dS(B = 0) = A 1
1 − E 2
1 +
1
2
E
2
1 − cos
2 I sin
2 A
+
+E
4
3
8
1 − cos
2 I sin
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
5
16
1 − cos
2 I sin
2 A
3 +
5
4
cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O +
E
8
dA .
(3.125)
End of Solution (the fourth step).
Note that all series (3.120), (3.121), (3.122), and (3.125) are uniformly convergent. Accordingly, we
can interchange integration and summation within (3.117) when we substitute (3.125) as a series
expansion. An alternative useful expansion of S(A) in terms of powers of ∆A is provided by the
following formulae:
S (A 0 + ∆A) = S (A 0 ) + S 1 (A 0 ) ∆A + S 2 (A 0 ) (∆A)
2 + O S
(∆A)
3
,
(3.126)
S 1 (A) =
1
1!
dS
dA
= A 1
1 − E 2
1 +
1
2
E
2
1 − cos
2 I sin
2 A
+
+E
4
3
8
1 − cos
2 I sin
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
5
16
1 − cos
2 I sin
2 A
3 +
5
4
cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O
E
8
,
(3.127)
S 2 (A) =
1
2!
d
2 S
dA
2 = A 1
1 − E 2
− E
2 cos
2 I sin A cos A−
−E
4
3
4
1 − cos
2 I sin
2 A
cos
2 I sin A cos A − 2 cos
2 I sin A cos
3 A + cos
4 I sin
3 A cos A
−
−E
6
15
8
1 − cos
2 I sin
2 A
2 cos
2 I sin A cos A +
5
2
cos
2 I sin
3 A cos A−
−
5
2
1 − cos
2 I sin
2 A
cos
4 I sin A cos
3 A +
5
4
1 − cos
2 I sin
2 A
cos
4 I sin
3 A cos A
+ O
E
8
.
(3.128)
Solution (the third step).
R
2 (A) + R
2
1 (A) = A 1
1 − E
2
1 + E
2
1 − cos
2 I cos
2 A
+
+E
4
1 − cos
2 I cos
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
1 − cos
2 I cos
2 A
3 + 3 cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O 1
E
8
.
(3.122)
End of Solution (the third step).
Solution (the fourth step).
dS
dA
1
A 1
√
1 − E 2
= (1 + x)
+1/2 = 1 +
1
2
x −
1 · 1
2 · 4
x
2 +
1 · 1 · 3
2 · 4 · 6
x
3
−
1 · 1 · 3 · 5
2 · 4 · 6 · 8
x
4 + O +
x
5
, (3.123)
subject to
x := E
2
1 − cos
2 I sin
2 A
+ E
4
1 − cos
2 I sin
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
1 − cos
2 I sin
2 A
2 + 3 cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O +
E
8
;
(3.124)
dS(B = 0) = A 1
1 − E 2
1 +
1
2
E
2
1 − cos
2 I sin
2 A
+
+E
4
3
8
1 − cos
2 I sin
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
5
16
1 − cos
2 I sin
2 A
3 +
5
4
cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O +
E
8
dA .
(3.125)
End of Solution (the fourth step).
Note that all series (3.120), (3.121), (3.122), and (3.125) are uniformly convergent. Accordingly, we
can interchange integration and summation within (3.117) when we substitute (3.125) as a series
expansion. An alternative useful expansion of S(A) in terms of powers of ∆A is provided by the
following formulae:
S (A 0 + ∆A) = S (A 0 ) + S 1 (A 0 ) ∆A + S 2 (A 0 ) (∆A)
2 + O S
(∆A)
3
,
(3.126)
S 1 (A) =
1
1!
dS
dA
= A 1
1 − E 2
1 +
1
2
E
2
1 − cos
2 I sin
2 A
+
+E
4
3
8
1 − cos
2 I sin
2 A
2 + cos
4 I sin
2 A cos
2 A
+
+E
6
5
16
1 − cos
2 I sin
2 A
3 +
5
4
cos
4 I
1 − cos
2 I sin
2 A
sin
2 A cos
2 A
+ O
E
8
,
(3.127)
S 2 (A) =
1
2!
d
2 S
dA
2 = A 1
1 − E 2
− E
2 cos
2 I sin A cos A−
−E
4
3
4
1 − cos
2 I sin
2 A
cos
2 I sin A cos A − 2 cos
2 I sin A cos
3 A + cos
4 I sin
3 A cos A
−
−E
6
15
8
1 − cos
2 I sin
2 A
2 cos
2 I sin A cos A +
5
2
cos
2 I sin
3 A cos A−
−
5
2
1 − cos
2 I sin
2 A
cos
4 I sin A cos
3 A +
5
4
1 − cos
2 I sin
2 A
cos
4 I sin
3 A cos A
+ O
E
8
.
(3.128)
