512 Higher Engineering Mathematics
from which,
a r+2 =
a r
(c+r−1)(c+r)+2(c+r)−k 2 −k
(c+r+1)(c+r +2)
=
a r [(c + r)(c + r + 1) − k(k + 1)]
(c + r + 1)(c + r + 2)
(46)
When c = 0,
a r+2 =
a r [r(r + 1) − k(k + 1)]
(r + 1)(r + 2)
For r = 0,
a 2 =
a 0 [−k(k + 1)]
(1)(2)
For r = 1,
a 3 =
a 1 [(1)(2) − k(k + 1)]
(2)(3)
=
−a 1 [k 2 + k − 2]
3!
=
−a 1 (k − 1)(k + 2)
3!
For r = 2,
a 4 =
a 2 [(2)(3) − k(k + 1)]
(3)(4)
=
−a 2
k
2
+ k − 6
(3)(4)
=
−a 2 (k + 3)(k − 2)
(3)(4)
=
−(k + 3)(k − 2)
(3)(4)
.
a 0 [−k(k + 1)]
(1)(2)
=
a 0 k(k + 1)(k + 3)(k − 2)
4!
For r = 3,
a 5 =
a 3 [(3)(4) − k(k + 1)]
(4)(5)
=
−a 3 [k 2 + k − 12]
(4)(5)
=
−a 3 (k + 4)(k − 3)
(4)(5)
=
−(k + 4)(k − 3)
(4)(5)
.
−a 1 (k − 1)(k + 2)
(2)(3)
=
a 1 (k − 1)(k − 3)(k + 2)(k + 4)
5!
and so on.
Substituting values into equation (43) gives:
y = x
0
a 0 + a 1 x −
a 0 k(k + 1)
2!
x
2
−
a 1 (k − 1)(k + 2)
3!
x
3
+
a 0 k(k + 1)(k − 2)(k + 3)
4!
x
4
+
a 1 (k − 1)(k − 3)(k + 2)(k + 4)
5!
x
5
+ · · ·
i.e. y = a 0
1 −
k(k + 1)
2!
x
2
+
k(k +1)(k − 2)(k + 3)
4!
x
4
−· · ·
+ a 1
x −
(k − 1)(k + 2)
3!
x
3
+
(k − 1)(k − 3)(k + 2)(k + 4)
5!
x
5
−· · ·
(47)
From page 506, it was stated that if two solutions of
the indicial equation differ by an integer, as in this case,
where c = 0 and 1, and if one coefficient is indeterminate, as with when c = 0, then the complete solution is
always given by using this value of c. Using the second
value of c, i.e. c = 1 in this problem, will give a series
which is one of the series in the first solution. (This may
be checked for c = 1 and where a 1 = 0; the result will be
the first part of equation (47) above).
Legendre’s polynomials
(A polynomial is an expression of the form:
f (x) = a + bx + cx 2 + dx 3 + · · ·). When k in equation
(47) above is an integer, say, n, one of the solution series
terminates after a finite number of terms. For example,
if k = 2, then the first series terminates after the term in
x 2 . The resulting polynomial in x, denoted by P n (x), is
called a Legendre polynomial. Constants a 0 and a 1 are
chosen so that y = 1 when x = 1. This is demonstrated
in the following worked problems.
Problem 13. Determine the Legendre polynomial
P 2 (x).
Since in P 2 (x), n =k = 2, then from the first part of
equation (47), i.e. the even powers of x:
y = a 0
1 −
2(3)
2!
x
2
+ 0
= a 0 {1 − 3x
2
}
a 0 is chosen to make y = 1 when x = 1
i.e. 1 = a 0 {1 −3(1) 2 } = −2a 0 , from which, a 0 = −
1
2
from which,
a r+2 =
a r
(c+r−1)(c+r)+2(c+r)−k 2 −k
(c+r+1)(c+r +2)
=
a r [(c + r)(c + r + 1) − k(k + 1)]
(c + r + 1)(c + r + 2)
(46)
When c = 0,
a r+2 =
a r [r(r + 1) − k(k + 1)]
(r + 1)(r + 2)
For r = 0,
a 2 =
a 0 [−k(k + 1)]
(1)(2)
For r = 1,
a 3 =
a 1 [(1)(2) − k(k + 1)]
(2)(3)
=
−a 1 [k 2 + k − 2]
3!
=
−a 1 (k − 1)(k + 2)
3!
For r = 2,
a 4 =
a 2 [(2)(3) − k(k + 1)]
(3)(4)
=
−a 2
k
2
+ k − 6
(3)(4)
=
−a 2 (k + 3)(k − 2)
(3)(4)
=
−(k + 3)(k − 2)
(3)(4)
.
a 0 [−k(k + 1)]
(1)(2)
=
a 0 k(k + 1)(k + 3)(k − 2)
4!
For r = 3,
a 5 =
a 3 [(3)(4) − k(k + 1)]
(4)(5)
=
−a 3 [k 2 + k − 12]
(4)(5)
=
−a 3 (k + 4)(k − 3)
(4)(5)
=
−(k + 4)(k − 3)
(4)(5)
.
−a 1 (k − 1)(k + 2)
(2)(3)
=
a 1 (k − 1)(k − 3)(k + 2)(k + 4)
5!
and so on.
Substituting values into equation (43) gives:
y = x
0
a 0 + a 1 x −
a 0 k(k + 1)
2!
x
2
−
a 1 (k − 1)(k + 2)
3!
x
3
+
a 0 k(k + 1)(k − 2)(k + 3)
4!
x
4
+
a 1 (k − 1)(k − 3)(k + 2)(k + 4)
5!
x
5
+ · · ·
i.e. y = a 0
1 −
k(k + 1)
2!
x
2
+
k(k +1)(k − 2)(k + 3)
4!
x
4
−· · ·
+ a 1
x −
(k − 1)(k + 2)
3!
x
3
+
(k − 1)(k − 3)(k + 2)(k + 4)
5!
x
5
−· · ·
(47)
From page 506, it was stated that if two solutions of
the indicial equation differ by an integer, as in this case,
where c = 0 and 1, and if one coefficient is indeterminate, as with when c = 0, then the complete solution is
always given by using this value of c. Using the second
value of c, i.e. c = 1 in this problem, will give a series
which is one of the series in the first solution. (This may
be checked for c = 1 and where a 1 = 0; the result will be
the first part of equation (47) above).
Legendre’s polynomials
(A polynomial is an expression of the form:
f (x) = a + bx + cx 2 + dx 3 + · · ·). When k in equation
(47) above is an integer, say, n, one of the solution series
terminates after a finite number of terms. For example,
if k = 2, then the first series terminates after the term in
x 2 . The resulting polynomial in x, denoted by P n (x), is
called a Legendre polynomial. Constants a 0 and a 1 are
chosen so that y = 1 when x = 1. This is demonstrated
in the following worked problems.
Problem 13. Determine the Legendre polynomial
P 2 (x).
Since in P 2 (x), n =k = 2, then from the first part of
equation (47), i.e. the even powers of x:
y = a 0
1 −
2(3)
2!
x
2
+ 0
= a 0 {1 − 3x
2
}
a 0 is chosen to make y = 1 when x = 1
i.e. 1 = a 0 {1 −3(1) 2 } = −2a 0 , from which, a 0 = −
1
2