Page 92 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 4: Laguerre Polynomials

n
For fixed values of r, the coefficient of t on the R.H.S. is L r ( ), obtained by putting Notes
x
r s 1 n or s n r 1.
n
Total Coefficient of t is obtained by summing over all allowed values of r.
Since s n r 1 and r 0

Therefore n r 1 0 or r (n 1).
n 1
n
x
Coefficient of t on the R.H.S. = L r ( )
r 0
n
Therefore equating coefficient of t , on both sides of (i), we have
n 1
L n ( ) L r ( ).
x
x
r 0
Illustrative Examples

Example 1: Prove that L (0) 1.
n
Solution: We have

tx
1 (1 t )
n
x
t L n ( ) = e
n 0 1 t
Putting x 0, we have
1 1
t L (0) = (1 t )
n n (1 t )
n 0
= 1 t t 2 t n

= t n
n 0
L n (0) = 1

Example 2: Expand x 3 x 2 3x 2 in a series of Laguerre polynomials.

x
Solution: We know that L n ( ) is a polynomial of degree n. Since x 3 x 2 3x 2 is a polynomial
of degree 3, we may write
3
x
x 3 x 2 3x 2 = C L ( ) ...(i)
r r
r 0
x
Putting values of L 0 ( ), L 1 ( ), L 2 ( ) and ( ) from section 4.3, we have
L
x
x
x
3
x 3 x 2 3x 2 = c c (1 x ) c . 1 2 4 x 2 c 3 6 18x 9x 2 x 3
0 1 2
2! 3!
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