Environmental Engineering Reference
In-Depth Information
Substituting Eq. (A.23) into Eq. (A.24) yields
1
t
2
+
∞
+
∞
n
=
0
P
n
(
x
)
t
n
n
=
0
nP
n
(
x
)
t
n
−
1
−
2
xt
+
+(
t
−
x
)
=
0
.
Since the coefficients of
t
n
must vanish, we have, for
n
≥
1,
(
n
+
1
)
P
n
+
1
(
x
)
−
(
2
n
+
1
)
xP
n
(
x
)+
nP
n
−
1
(
x
)=
0
.
(A.26)
A repeated application of this recurrence formula will yield the Legendre polyno-
mial
P
N
(
x
)
of any arbitrary degree
N
in terms of
P
0
(
x
)
and
P
1
(
x
)
.
Similarly, substituting Eq. (A.23) into Eq. (A.25) yields
P
n
+
1
(
2
xP
n
(
P
n
−
1
(
x
)
−
x
)+
x
)
−
P
n
(
x
)=
0
.
(A.27)
By taking derivatives of Eq. (A.26) with respect to
x
, we obtain
)
P
n
(
)
+
P
n
+
1
(
xP
n
(
nP
n
−
1
(
(
n
+
1
)
x
)
−
(
2
n
+
1
x
)+
x
x
)=
0
.
This together with Eq. (A.27) leads to, by eliminating
P
n
−
1
(
and
P
n
+
1
(
x
)
x
)
,
P
n
+
1
(
xP
n
(
x
)
−
x
)=(
n
+
1
)
P
n
(
x
)
,
xP
n
(
P
n
−
1
(
x
)
−
x
)=
nP
n
(
x
)
.
Thus, by adding the two equations,
P
n
+
1
(
P
n
−
1
(
x
)
−
x
)=(
2
n
+
1
)
P
n
(
x
)
.
(A.28)
A.6 Associated Legendre Polynomials
The associated Legendre equation is
1
n
x
2
y
x
2
y
−
m
2
2
xy
+
−
(
n
+
1
)
−
=
0
,
(A.29)
1
−
where
m
is a natural number. To express its particular solutions for any natural num-
ber
n
by
P
n
(
x
)
, the particular solutions of Legendre equations, consider the Legendre
equation,
1
x
2
v
−
2
xv
+
−
(
+
)
=
.
n
n
1
v
0
(A.30)
By taking
m
-th derivatives with respect to
x
, we obtain
d
m
d
x
m
1
x
2
v
−
d
x
m
2
xv
+
d
m
d
m
v
d
x
m
=
−
n
(
n
+
1
)
0
.
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