Trabalho de Propriedades de Bessel
1. A função de Bessel de primeiro tipo de ordem
v
é dada por:
Jv
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+v
22n+vn !
(
n+v
)
!
Mostre que:
Jv−1
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+v−1
22n+v−1n!
(
n+v−1
)
!
Mostre que:
d
(
xvJv
)
dx =xvJv−1
RESPOSTA:
Dada a função de Bessel tal que:
Jv
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+v
22n+vn !
(
n+v
)
!
Para
v=k=0,1,2,3 , …
Jk
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+k
22n+kn !
(
n+k
)
!
De forma particular, aplica-se os números:
k=0→ J 0
(
x
)
=∑
n=0
∞
(
−1
)
nx2n
22nn!
(
n
)
!
k=1→ J1
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+1
22n+1n !
(
n+1
)
!
k=2→ J2
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+2
22n+2n!
(
n+2
)
!
Assim, pode-se afirmar que:
k=
(
v−1
)
→ J
(
v−1
)
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+
(
v−1
)
22n+
(
v−1
)
n !
(
n+
(
v−1
)
)
!→ J v−1
(
x
)
=∑
n=0
∞
(
−1
)
nx2n+v−1
22n+v−1n !
(
n+v−1
)
!
Ainda, multiplicando
Jv
por
xv
:
xvJv
(
x
)
=xv∑
n=0
∞
(
−1
)
nx2n+v
22n+vn !
(
n+v
)
!=∑
n=0
∞
(
−1
)
nx2n+2v
22n+vn!
(
n+v
)
!
A derivada de
xvJv
(
x
)
com relação à
x
:
d
(
xvJv
(
x
)
)
dx =d
dx
[
∑
n=0
∞
(
−1
)
nx2n+2v
22n+vn !
(
n+v
)
!
]
=∑
n=0
∞d
dx
[
(
−1
)
nx2n+2v
22n+vn !
(
n+v
)
!
]
Fazendo a derivada dentro do somatório, tem-se:
d
(
xvJv
(
x
)
)
dx =∑
n=0
∞d
dx
[
(
−1
)
nx2n+2v
22n+vn !
(
n+v
)
!
]
=∑
n=0
∞
(
−1
)
n
22n+vn !
(
n+v
)
!
[
d
dx
(
x2n+2v
)
]
=∑
n=0
∞
(
−1
)
n
22n+vn !
(
n+v
)
!
[
(
n+v
)
x2n+2v−1
]
=∑
n=0
∞
(
−1
)
n
(
n+v
)
x2n+v−1xv
22n+v−1n !
(
n+v
) (
n+v−1
)
!
Logo:
