The document discusses recurrence relations for Legendre polynomials. It presents 6 recurrence relations and provides proofs of each one. The relations involve differentiation and combinations of Legendre polynomials of different orders. The proofs use properties of Legendre polynomials and differntiate, combine, and manipulate the recurrence relations to derive new ones.

1. 1.
2.
3.
4.
5.
6.
n
(n 1)Pn  1(x)  (2n 1) x Pn(x) nPn  1(x)
nPn(x)  x P' n (x) - P' n - 1 (x)
(2x 1) n (x)  ' n  1 (x) - P' n  1 (x)
'n (x)x P'n-1 (x)n Pn-1 (x)
2
(1 x )P' n(x)  n[Pn  1(x)  xPn(x)]
2
(1 x )P' n(x)  (n 1)[x Pn(x)  Pn  1(x)]

2. 0
1_
2 -
xt t 2
P x t
0
__________________ n-1
n
3 2
2
We know that
Proo
(1- 2 ) ( )
Diffentiating both side of (i) partually w.
r.t t, w
f
e get
P (x) n t
(1 2 xt t )
(1)
n
n
n
n
x t 




 

 
 
2 ¥ - 1
____(__-_)_ __
 
1 / 2 0
2
(1- 2 ) ( ).
. :
(1- 2 )
n
n
n
x t
xt t P x nt
x
e
t t
i



o 1 2 2
n
x t P x P x t P
( - ){ ( ) ( )
(1 2 ){
  
n
 
2 2 1
    
1 2 3
n
x
  
Eqating the coefficient
P x
     
1 1
1
. :
of t
( ) ... ( ) ...}
( ) 2 ( ) 3 ( ) ... ( ) ...}
both sides,we get,
( 1) ( ) 2 ( ) ( )( 1) ( )
n
n
n n n n
i
P
e
t t
x t P x t
x P x t P x t nP x t
on
n P x nxP x P x n xP x

 
( )
n xP x nP x
  
(2 1) ( ) ( )
n
n n n i e n P

. :( 
1)
 1  1

3. Proof (2):
Differntiatig both sides of (i) partially w.r.t.x we get
'


0
2 3/2


2 -3/2 '
0
_ _ _
_______
(1
2 )
( )
( - )(1- 2 ) ( ) ( ) ....( )
Aga
n
n
n
n
n
n
t
xt t
P x t
or t x t xt t x t t P x ii


 
  

in differntiating (i) partially w.r.t t,we get

2 -3/2 
1

x t xt t n t P x iii
1
0

 
 

1 '
 
t nt P x  x 
t t P x U g iii
0 0
n
( - )(1- 2 )
( ) ....( )
( ) ( ) ( ) [ sin ( )]
Eqating the coefficient of t
n
n
n n
n n
n n
 
'
onbothside we have
nP x  xP x P x
-1
,
( ) ( ) - ( ) n n n

4. n xP x n P x nP x
n n n
 
1 1
' ' '
n xP x n P x n P x nP x
1 1
We have proved that
Proof(3)
(
)
2 1
( ) ( 1) ( ) ( ) [Recurrence relation (1)]
Differtiating w.r.t x,we have
(2 1) ( ) (2 1) ( ) ( 1) ( ) ( ) ..
n n n n
 
   
     
' '
xP x nP x P x
 
1
'
n n n

x P x
n
' '
1 1
...(iv)
From recurrence relation (2),we have
( ) ( ) ( )
Substituting the value of ( ) in (iv) and simplifying we get
n P x P x P x
(2  1) ( )  ( ) 
( )
n n n
 

5. From the recurrence relation (3),we have
(2
' '
1 1
n P x P x P x
  
n n n
 
' '
 
1
Proof (4)
1) ( ) ( ) ( )
From the recurrence relation (2),we have
nP ( x ) xP ( x )
P
n n n

'
1
( )
Eliminating ( ),weget
' '
1

(  1) ( )  ( ) 
( )

Reaplacing n by ( -1),we get
' '
( )  ( ) 
( )
1 1
' '
or ( ) ( ) ( )
 
1 1
n
n n n
n n n
n n n
x
P x
n P x P x xP x
n
nP x P x xP x
P x xP x nP x
 
 

6. ' '
Pn x xPn x nPn x    
1 1
Proof
From the recurrence relatio
(5)
n (4),we h
ave
( ) ( ) ( ) ...(v)
From the recurrence relation (2),we have
' '
xP x P x nP x
( ) - ( ) ( ) ...(vi)
n n 
1
n
Multipying (vi) by x and sub
2 '
x n P x xP x
1
stracting from (v),we get
(1- )
( ) [ ( ) - ( )]
x P
n n 
n 


7. Recurrence relation (1)may e written as
n  xP x  nxP x  n  P x 
nP x
n n n n
n  xP x P x  n P x 
xP x
 
1 1
1 1
Proof (6)
( 1
)
( ) ( ) ( 1) ( ) ( )
or ( 1)[ ( ) - ( )] [ ( ) ( )]
n n n n
 
2 '
2 '
x P x
(1- ) ( ), using recurrence relation (5)
n

x n xP x P x  
( )  (  1)[ ( ) 
( )]
n n n X p
1 (1- )