Page 84 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 4: Laguerre Polynomials

Notes
(k r )
a r 1 = a r
(k r 1) 2
for k = 0

r
a 1 = a ...(iv)
r 2 r
(r 1)
Putting r = 0, 1, 2, ..., in (iv), we have

a 1 = a 0 ( 1) a 0
1

1 2 ( 1)
a 2 = a 1 ( 1) a 0
2 2 (2!) 2
2 3 ( 1)( 2)
a 3 = a 2 ( 1) a 0 etc.
3 2 (3!) 2

Hence a r = (1) r ( 1)( 2) ( r 1) a 0
2
( !)
r
From (ii), we have

r r
y = a x a 0 a x a x 2 a x r
2
r
1
r 0
( 1) 2 ( 1)( 2) 3
= a 0 1 x 2 x 2 x
(2!) (3!)
( 1) ( r 1) r
r
( 1) x ...(v)
r
( !) 2
If = n

n ( n n 1) 2 2 (n n 1) (n r 1)
y = a 0 1 2 . x 2 x ( 1) 2
1 (2!) ( !)
r
n
= a ( 1) r ( n n 1) (n r 1) x r
0 ( !) 2
r
r 0
n
= a ( 1) r ! n x r
0 (n r )!( !) 2
r
r 0
Laguerre Polynomials

The standard solution of Laguerre equation for which a 0 1 is called the Laguerre polynomial
x
of order n and is denoted by L n ( ).

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