Page 58 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 2: Legendre’s Polynomials

Notes
2n
So comparing the coefficients of h on both sides we have
1 2
P n 2 ( )dx = ...(G)
x
2n 1
1
1 0 for m n
x
x
Thus P m ( )P n ( )dx = 2
=
for m n
1 2n 1
From the properties of Legendre’s polynomials we can prove certain results.
2.6 Expansion of a f(x) in terms of Legendre’s Polynomials

Since P 0 ( ), ( ),P 2 ( )... a set Legendre polynomials are orthogonal in the range of , ( 1,1),
x
x
x
x
P
1
x
any function ( )f x can be expressed in terms an expansion series involving P n ( ) i.e.
x
f ( ) = C P ( ) for x in the range 1 x 1 ...(i)
x
n n
n 0
M ultiplying equation (i) by P m ( ) and integrating over the limit 1 to 1, we have
x
1 1
x
x
x
f ( )P m ( )dx = C n P m ( )P n ( )dx ...(ii)
x
1 n 0 1
Now
1 0 if m n
P ( ) P ( )dx = ...(iii)
x
x
m n 2
if m n
1 2n 1
Substituting in (i) we have
1
2
f ( )P m ( )dx = C m ...(iv)
x
x
2m 1
1
Example: Expand ( )f x in the form
C P ( ),
x
r r
r 0
Where
0 1 x 0
f ( ) = ...(i)
x
1, 0 x 1
We know

x
x
f ( ) = C P ( ) ...(ii)
r r
r 0

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