(3.3)
Δ
F
=
F
(
J
+1) –
F
(
J
) = 2
B
(
J+
1)
–
4
D
(
J+
1)
3
(3.4)
Δ
F
=
F
(
J–
1)
– F
(
J
)
=
–2
BJ
+ 4
DJ
3