1.3 Solutions
69
H
n − 2ξ H
n + 2n H n = 0
1.76 J n (x) =
k
(−1)
k
x
2
n+2k
k!Γ (n + k + 1)
(a) Differentiate
d
dx
[x
n J n (x)] = J n (x)nx
n−1
+ x
n dJ n (x)
dx
=
k
(−1)
k
x
2
n+2k nx
n−1
k!Γ (n + k + 1)
+
x
n (n + 2k)(−1)
k x
n+2k−1
k!Γ (n + k + 1).2 n+2k
=
k
(−1)
k
x
2
n+2k−1 (n + k)x
n
k!Γ (n + k + 1)
=
(−1)
k
x
2
n+2k−1 x
n
Γ (n + k)
= x
n J n−1 (x)
(b) A similar procedure yields
d
dx
[x
−n J n (x)] = −x
−n J n+1 (x)
1.77 From the result of 1.76(a)
d
dx
[x
n J n (x)] = J n (x)nx
n−1
+ x
n d J n (x)
dx
= x
n J n−1 (x)
Divide through out by x
n
n
x
J n (x) +
dJ n (x)
dx
= J n−1 (x)
Similarly from (b)
−
n
x
J n (x) +
dJ n (x)
dx
= −J n+1 (x)
Add and subtract to get the desired result.
1.78 J n (x) =
∞
k=0
(−1)
k
x
2
n+2k
k!Γ (n + k + 1)
(a) Therefore, with n = 1/2
J 1
2
(x) =
x
2
1/2
Γ
3
2
−
x
2
5/2
1.Γ
5
2
+
x
2
9/2
2!Γ
7
2
− · · ·
Writing Γ
3
2
=
√
π
2
, Γ
5
2
=
3
√ π
4
, Γ
7
2
=
15
8
√
π
J 1
2
(x) =
2
π x
x −
x
3
3!
+
x
5
5!
+ · · ·
=
2
π x
sin x
(b) With n = −1/2
J −
1
2
(x) =
x
2
−1/2
Γ
1
2
−
x
2
3/2
1.Γ
3
2
+
x
2
7/2
2!Γ
5
2
− · · ·