1000 Solved Problems in Modern Physics

(Romina) #1

1.3 Solutions 69


H

′′
n−^2 ξH


n+^2 nHn=^0

1.76 Jn(x)=



k

(−1)k

(x
2

)n+ 2 k

k!Γ(n+k+1)

(a) Differentiate
d
dx

[xnJn(x)]=Jn(x)nxn−^1 +xn
dJn(x)
dx

=


k

(−1)k

(x
2

)n+ 2 k
nxn−^1
k!Γ(n+k+1)

+

∑xn(n+ 2 k)(−1)kxn+^2 k−^1
k!Γ(n+k+1). 2 n+^2 k

=


k

(−1)k

(x
2

)n+ 2 k− 1
(n+k)xn
k!Γ(n+k+1)

=

∑(−1)k
(x
2

)n+ 2 k− 1
xn
Γ(n+k)

=xnJn− 1 (x)

(b) A similar procedure yields
d
dx

[x−nJn(x)]=−x−nJn+ 1 (x)

1.77 From the result of 1.76(a)
d
dx


[xnJn(x)]=Jn(x)nxn−^1 +xn

dJn(x)
dx

=xnJn− 1 (x)
Divide through out byxn
n
x

Jn(x)+

dJn(x)
dx

=Jn− 1 (x)
Similarly from (b)


n
x

Jn(x)+

dJn(x)
dx

=−Jn+ 1 (x)
Add and subtract to get the desired result.

1.78 Jn(x)=


∑∞

k= 0

(−1)k

(x
2

)n+ 2 k

k!Γ(n+k+1)

(a) Therefore, withn= 1 / 2

J 12 (x)=

(x
2

) 1 / 2

Γ

( 3

2

)−

(x
2

) 5 / 2

1 .Γ

( 5

2

)+

(x
2

) 9 / 2

2!Γ

( 7

2

)−···

WritingΓ

( 3

2

)

=

√π
2 ,Γ

( 5

2

)

=^3

√π
4 ,Γ

( 7

2

)

=^158


π

J 12 (x)=


2

πx

[

x−

x^3
3!

+

x^5
5!

+···

]

=


2

πx

sinx

(b) Withn=− 1 / 2

J− 12 (x)=

(x
2

)− 1 / 2

Γ

( 1

2

) −

(x
2

) 3 / 2

1 .Γ

( 3

2

)+

(x
2

) 7 / 2

2!Γ

( 5

2

)−···
Free download pdf