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)
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