Power series methods of solving ordinary differential equations 513
Hence, P 2 (x)=−
1
2
1 − 3x 2 =
1
2
(3x 2 − 1)
Problem 14. Determine the Legendre polynomial P 3 (x).
Since in P 3 (x), n =k = 3, then from the second part of
equation (47), i.e. the odd powers of x:
y = a 1
x −
(k − 1)(k + 2)
3!
x
3
+
(k − 1)(k − 3)(k + 2)(k + 4)
5!
x
5
− · · ·
i.e. y = a 1
x −
(2)(5)
3!
x
3
+
(2)(0)(5)(7)
5!
x
5
= a 1
x −
5
3
x
3
+ 0
a 1 is chosen to make y = 1 when x = 1.
i.e. 1 = a 1
1 −
5
3
= a 1
−
2
3
from which, a 1 =−
3
2
Hence, P 3 (x) =−
3
2
x−
5
3
x 3
or P 3 (x) =
1
2
(5x 3 − 3x)
Rodrigue’s formula
An alternative method of determining Legendre polynomials is by using Rodrigue’s formula, which states:
P n (x)=
1
2 n n!
d n
x 2 − 1
n
dx n
(48)
This is demonstrated in the following worked problems.
Problem 15. Determine the Legendre polynomial
P 2 (x) using Rodrigue’s formula.
In Rodrigue’s formula, P n (x) =
1
2 n n!
d n
x 2 − 1
n
dx n
and when n =2,
P 2 (x) =
1
2 2 2!
d 2 (x 2 − 1) 2
dx 2
=
1
2 3
d
2
(x
4
− 2x
2
+ 1)
dx 2
d
dx
(x
4
− 2x
2
+ 1)
= 4x
3
− 4x
and
d 2
x 4 − 2x 2 + 1
dx 2
=
d(4x 3 − 4x)
dx
= 12x
2
− 4
Hence, P 2 (x) =
1
2 3
d 2
x 4 −2x 2 +1
dx 2
=
1
8
12x 2 − 4
i.e. P 2 (x) =
1
2
3x 2 − 1
the same as in Problem 13.
Problem 16. Determine the Legendre polynomial
P 3 (x) using Rodrigue’s formula.
In Rodrigue’s formula, P n (x) =
1
2 n n!
d n
x 2 − 1
n
dx n
and
when n = 3,
P 3 (x) =
1
2 3 3!
d 3
x 2 − 1
3
dx 3
=
1
2 3 (6)
d 3
x 2 − 1
x 4 − 2x 2 + 1
dx 3
=
1
(8)(6)
d 3
x 6 − 3x 4 + 3x 2 − 1
dx 3
d
x 6 −3x 4 +3x 2 −1
dx
= 6x
5
− 12x
3
+ 6x
d
6x 5 −12x 3 +6x
dx
= 30x
4
− 36x
2
+ 6
and
d
30x 4 − 36x 2 + 6
dx
= 120x
3
− 72x
Hence, P 3 (x) =
1
(8)(6)
d 3
x 6 − 3x 4 + 3x 2 − 1
dx 3
=
1
(8)(6)
120x 3 − 72x
=
1
8
20x 3 − 12x
i.e. P 3 (x)=
1
2
5x 3 − 3x
the same as in Problem 14.
Now try the following exercise
Exercise 198 Legendre’s equation and
Legendre polynomials
1. Determine the power series solution of
the Legendre equation:
1 − x 2
y − 2x y + k(k + 1)y = 0 when
(a) k = 0 (b) k = 2, up to and including the
Hence, P 2 (x)=−
1
2
1 − 3x 2 =
1
2
(3x 2 − 1)
Problem 14. Determine the Legendre polynomial P 3 (x).
Since in P 3 (x), n =k = 3, then from the second part of
equation (47), i.e. the odd powers of x:
y = a 1
x −
(k − 1)(k + 2)
3!
x
3
+
(k − 1)(k − 3)(k + 2)(k + 4)
5!
x
5
− · · ·
i.e. y = a 1
x −
(2)(5)
3!
x
3
+
(2)(0)(5)(7)
5!
x
5
= a 1
x −
5
3
x
3
+ 0
a 1 is chosen to make y = 1 when x = 1.
i.e. 1 = a 1
1 −
5
3
= a 1
−
2
3
from which, a 1 =−
3
2
Hence, P 3 (x) =−
3
2
x−
5
3
x 3
or P 3 (x) =
1
2
(5x 3 − 3x)
Rodrigue’s formula
An alternative method of determining Legendre polynomials is by using Rodrigue’s formula, which states:
P n (x)=
1
2 n n!
d n
x 2 − 1
n
dx n
(48)
This is demonstrated in the following worked problems.
Problem 15. Determine the Legendre polynomial
P 2 (x) using Rodrigue’s formula.
In Rodrigue’s formula, P n (x) =
1
2 n n!
d n
x 2 − 1
n
dx n
and when n =2,
P 2 (x) =
1
2 2 2!
d 2 (x 2 − 1) 2
dx 2
=
1
2 3
d
2
(x
4
− 2x
2
+ 1)
dx 2
d
dx
(x
4
− 2x
2
+ 1)
= 4x
3
− 4x
and
d 2
x 4 − 2x 2 + 1
dx 2
=
d(4x 3 − 4x)
dx
= 12x
2
− 4
Hence, P 2 (x) =
1
2 3
d 2
x 4 −2x 2 +1
dx 2
=
1
8
12x 2 − 4
i.e. P 2 (x) =
1
2
3x 2 − 1
the same as in Problem 13.
Problem 16. Determine the Legendre polynomial
P 3 (x) using Rodrigue’s formula.
In Rodrigue’s formula, P n (x) =
1
2 n n!
d n
x 2 − 1
n
dx n
and
when n = 3,
P 3 (x) =
1
2 3 3!
d 3
x 2 − 1
3
dx 3
=
1
2 3 (6)
d 3
x 2 − 1
x 4 − 2x 2 + 1
dx 3
=
1
(8)(6)
d 3
x 6 − 3x 4 + 3x 2 − 1
dx 3
d
x 6 −3x 4 +3x 2 −1
dx
= 6x
5
− 12x
3
+ 6x
d
6x 5 −12x 3 +6x
dx
= 30x
4
− 36x
2
+ 6
and
d
30x 4 − 36x 2 + 6
dx
= 120x
3
− 72x
Hence, P 3 (x) =
1
(8)(6)
d 3
x 6 − 3x 4 + 3x 2 − 1
dx 3
=
1
(8)(6)
120x 3 − 72x
=
1
8
20x 3 − 12x
i.e. P 3 (x)=
1
2
5x 3 − 3x
the same as in Problem 14.
Now try the following exercise
Exercise 198 Legendre’s equation and
Legendre polynomials
1. Determine the power series solution of
the Legendre equation:
1 − x 2
y − 2x y + k(k + 1)y = 0 when
(a) k = 0 (b) k = 2, up to and including the