Geology Reference
In-Depth Information
using the expression (3.268) for D 2 f , and recognising the identity
2 f
r ∂y +
2 f
∂y
f i
2 =
Df y .
(3.287)
For the term following (3.286), we have
h
3
1
3!
2
k 2
2
r +
f
∂y
h
h 2
2 f
r 2 +
2 f
r ∂y +
2 f
∂y
1
24
2
k 2
2
2 hk 2
k 2
=
r +
2
∂y
O h 4 r = r i
h 3
48
r +
2 f
r 2 +
2 f
r ∂y +
2 f
∂y
f
∂y
2 f
f 2
=
+
2
O h 4 r = r i
h 3
48
3 f
r 3 +
3 f
r 2
3 f
r ∂y
3 f
∂y
3 f
3 f 2
f 3
=
∂y +
2 +
+
3
48 D 3 f + O h 4 r = r i
h 3
=
(3.288)
upon substitution from (3.270). From (3.284), with substitution from (3.283),
(3.286) and (3.288), we find that
hf r i
hf
8 D 2 f
2 f y Df
h 2
2 Df
h 3
h
2 ,y i
k 2
2
k 3
=
+
+
=
+
+
+
6 Df Df y r = r i +
48 D 3 f
h 4
O h 5 .
3 f y D 2 f
+
+
+
(3.289)
In turn, this expression for k 3 can be used to develop the expansion,
h
2
h f
k 3 f
1
2!
k 3
∂y
f ( r i
+
h ,y i
+
k 3 )
=
f ( r i ,y i )
+
r +
∂y +
r +
f
h
3
1
3!
k 3
∂y
+
r +
f
+···
(3.290)
Search WWH ::




Custom Search