Mathematical Physics Vol 1

5.10 Examples

287

Solution Differentiating (5.268) by z we obtain

∞ ∑ r = 0

( n + 2 r ) z n + 2 r − 1 2 n + 2 r r ! ( n + r ) ! .

( − 1 ) r

J ′ n ( z )=

(5.269)

If equation (5.268) is now multiplied by n z

, and then added to equation (5.269), we

obtain

∞ ∑ r = 0

( 2 n + 2 r ) z n + 2 r − 1 2 n + 2 r r ! ( n + r ) ! .

n z

( − 1 ) r

J n ( z )+ J ′ n ( z )=

The right hand side of this equation is ∞ ∑ r = 0 ( − 1 ) r 2 ( n + r ) z n + 2 r − 1 2 n + 2 r r ! ( n + r ) ! = ∞ ∑ r = 0

z n + 2 r − 1 2 n + 2 r − 1 r ! ( n + r − 1 ) !

( − 1 ) r

= J n

z ) ,

− 1 (

which proves the first identity. In order to prove the second identity, we will first multiply the equation (5.268) by n z , and then substitute r by r + 1, which yields n z J n ( z )= ∞ ∑ r = 0 ( − 1 ) r nz n + 2 r − 1 2 n + 2 r r ! ( n + r ) ! = (5.270)

∞ ∑ r = − 1

nz n + 2 r + 1 2 · 2 n + 2 r + 1 ( r + 1 ) ! ( n + r + 1 ) ! .

( − 1 ) r + 1

(5.271)

=

If we substitute r by r + 1 in equation (5.269), we obtain

∞ ∑ r = − 1

( n + 2 r + 2 ) z n + 2 r + 1 2 · 2 n + 2 r + 1 ( r + 1 ) ! ( n + r + 1 ) ! .

( − 1 ) r + 1

J ′ n ( z )=

(5.272)

Subtracting equation (5.272) from equation (5.271) yields n z J n ( z ) − J ′ n ( z )= ∞ ∑ r = − 1 ( − 1 ) r + 1 nz n + 2 r + 1

2 · 2 n + 2 r + 1 ( r + 1 ) ! ( n + r + 1 ) ! − ( n + 2 r + 2 ) z n + 2 r + 1 2 · 2 n + 2 r + 1 ( r + 1 ) ! ( n + r + 1 ) ! = nz n + 2 r + 1 2 · 2 n + 2 r + 1 ( r + 1 ) ! ( n + r + 1 ) ! ( n + 2 r + 2 ) z n + 2 r + 1 2 · 2 n + 2 r + 1 ( r + 1 ) ! ( n + r + 1 ) ! − ( − 2 r − 2 ) z n + 2 r + 1 2 · 2 n + 2 r + 1 ( r + 1 ) ! ( n + r + 1 ) ! =

∞ ∑ r = − 1

( − 1 ) r + 1

−

∞ ∑ r = 0

nz n − 1 2 n 0! n ! −

( − 1 ) r + 1

+( − 1 ) 0

=

∞ ∑ r = 0

nz n − 1 2 n 0! n !

( − 1 ) r + 1

( − 1 ) 0

−

∞ ∑ r = 0 ∞ ∑ r = 0

( − 1 ) r + 1

=

z n + 2 r + 1 2 n + 2 r + 1 r ! ( n + r + 1 ) !

( − 1 ) r

= J n + 1 ( z ) ,

=

which was to be proved.