Page 42 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

Notes
(n 2)(n 3)
C = C
4 n (2n 3) 2
(n 3)(n 2)(n 1)n
= C 0
1.2.4.(2n 3)(2n 1)

(n 4)(n 5)
C = C 4
6 6(2n 5)
(n 5)(n 4)(n 3)(n 2)(n 1)n
= C 0
2.4.6(2n 5)(2n 3)(2n 1)
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Substituting these values of C’s in equation (B) we have for s = n

( n n 1) x 2 ( n n 1)(n 2)(n 3) n 4
n
y = C 0 x x x ... ...(N)
2(2n 1) 2.4 (2n 1)(2n 3)
= y (say)
1
For the second value of s n 1, we have from equation (K)

( n r 1)( n r )
C = C r 2
r (2n r 1)( r )
(n r 1)(n r )
or C = C r 2 ...(O)
r r (2n r 1)
Putting the values of r 2, 4, 6, ... in equation (O)

(n 1)(n 2)
C = C 0
2 2(2n 3)

(n 3)(n 4)
C = C 2
4 4(2n 5)
(n 4)(n 3)(n 2)(n 1)
= C 0
2.4(2n 3)(2n 5)
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Substituting these values of C’s in equation (B) we have for s n 1

y = C x n 1 C x n 3 C x n 5 ....
0 2 4
(n 1)(n 2) n 3 (n 1)(n 2)(n 3)(n 4) n 5
n
1
= C 0 x x x ...
2.(2n 3) 2.4(2n 3)(2n 5)
= y ...(P)
2

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