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x
dP () dP x ()
l+1 − x l = ( l + 1) ( (7.94)
P x)
dx dx l
and

dP x () dP ()
x
l−1
l
x − = lP x () (7.95)
l
dx dx
Adding Eqs. (7.94) and (7.95), we obtain the more symmetric formula:
x
dP () dP ()
x
l+1 − l−1 = 2 ( l + 1) P x( ) (7.96)
dx dx l
Replacing l by l – 1 in Eq. (7.94) and eliminating P ′ x() from Eq.
− l 1
(7.95), we find that:

dP l ()x
2
(1− x ) = lP − l 1 () −x lxP l ()x (7.97)
dx
Differentiating Eq. (7.97) and using Eq. (7.95), we obtain:

d  (1− 2 dP x() + ll( + P x( ) =
l
dx   x ) dx   ) 1 l 0 (7.98a)

which can be written in the equivalent form:

2
dP l ()x dP l ()x
( +
2
(1− x ) − 2x + ll 1 ) ( ) =P x 0 (7.98b)
l
dx 2 dx
which is the ODE for the Legendre polynomial, as previously
pointed out in Section 4.7.1.
5. Next, we want to show that if l ≠ m, we have the orthogonality
between any two elements (with different indices) of the basis;
that is

∫ 1 Px P x dx() m () = 0 (7.99)
l
−1
To show this relation, we multiply Eq. (7.98) on the left by P (x)
m
and integrate to obtain:

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