1000 Solved Problems in Modern Physics

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

)−···