1.3 Solutions
71
1.80 Legendre’s equation is
(1 − x
2 )
∂
2 P n (x)
∂ x 2 − 2x
∂ P n (x)
∂ x
+ n(n + 1)P n (x) = 0
( 1 )
Put x = cos θ, Eq. (1) then becomes
sin
2
θ
∂
2 P n
∂ cos 2 θ
− 2 cos θ
∂ P n
∂ cos θ
+ n(n + 1)P n = 0
( 2 )
For large n, n(n + 1) → n
2 , and cos θ → 1 for small θ,
sin
2
θ
∂
2 P n
∂ cos 2 θ
− 2
∂ P n
∂ cos θ
+ n
2 P n = 0
( 3 )
Now, Bessel’s equation of zero order is
x
2 d
2 J 0 (x)
dx 2 + x
d
dx
J 0 (x) + x
2 J 0 (x) = 0
( 4 )
Letting x = 2n sin θ/2 = n sin θ, in (4) for small θ , and noting that cos θ →
1, after simple manipulation we get
sin
2
θ
d
2 J 0 (n sin θ)
d cos 2 θ
− 2d
d J 0 (n sin θ)
d cos θ
+ n
2 J 0 (n sin θ ) = 0
( 5 )
Comparing (5) with (3), we conclude that
P n (cos θ ) → J 0 (n sin θ)
1.81 T (x, s) = (1 − 2sx + s
2 )
−1/2
=
p l (x)s
l
(1)
(a) Differentiate (1) with respect to s.
∂ T
∂s
= (x − s)(1 − 2sx + s
2 )
−
3
2
=
(x − s)(1 − 2sx + s
2 )
−1 p l (x)s
l
=
lp l (x)s
l−1
Multiply by (1 − 2sx + s
2 )
(x − s) p l s
l
=
lp l s
l−1 (1 − 2sx + s
2 )
Equate the coefficients of s
l
x p l − p l−1 = (l + 1) p l+1 − 2xlp l + (l − 1) p l−1
or (l + 1) p l+1 = (2l + 1)x p l − lp l−1