Exercício 8. Verifique que J12(x)=raiz2pi x sen x e J-12(x)=raiz2pi x cos x.
Sugestão: Use a série de Jp(x) e a série sen x=sumn=0^infty (-1)^n(2 n+1)! x^2 n+1, cos x=sumn=0^infty (-1)^n(2 n)! x^2 n.

Passo 1

O enunciado pede para mostrar que
Jtfrac12(x)=raiz2pi xsin x e J-tfrac12(x)=raiz2pi xcos x
O que vamos usar: a série geral de Bessel de primeira espécie
Jp(x) =summ=0^infty(-1)^mm!,Gamma(m+p+1)l(x2r)^2m+p
e as séries conhecidas
sin x=sumn=0^infty(-1)^n(2n+1)!,x^2n+1, cos x=sumn=0^infty(-1)^n(2n)!,x^2n

Passo 2

Para p=tfrac12 escrevemos
Jtfrac12(x) =summ=0^infty(-1)^mm!,Gammal(m+tfrac32r) l(x2r)^2m+tfrac12
Como
Gammal(m+tfrac32r) =(2m+1)!2^2m+1,m!,raizpi,
o termo genérico vira
(-1)^mm!cdot2^2m+1m!(2m+1)!,raizpi cdotx^2m+tfrac122^2m+tfrac12 =(-1)^mraiz2raizpi, x^2m+tfrac12(2m+1)!
Portanto
Jtfrac12(x) =raiz2pi summ=0^infty (-1)^m,x^2m+tfrac12(2m+1)!

Passo 3

Fatorando x^-tfrac12 na soma, reconhecemos sin x:
summ=0^infty(-1)^m,x^2m+tfrac12(2m+1)! =x^-tfrac12summ=0^infty(-1)^m,x^2m+1(2m+1)! =x^-tfrac12,sin x
Logo
Jtfrac12(x) =raiz2pi,x^-tfrac12sin x =raiz2pi x,sin x

Passo 4

Para p=-tfrac12 temos
J-tfrac12(x) =summ=0^infty(-1)^mm!,Gammal(m+tfrac12r) l(x2r)^2m-tfrac12
E
Gammal(m+tfrac12r) =(2m)!2^2m,m!,raizpi,
de modo que o termo geral é
(-1)^mm!cdot2^2mm!(2m)!,raizpi cdotx^2m-tfrac122^2m-tfrac12 =(-1)^mraiz2raizpi, x^2m-tfrac12(2m)!
Assim
J-tfrac12(x) =raiz2pi summ=0^infty (-1)^m,x^2m-tfrac12(2m)!

Passo 5

Fatorando x^-tfrac12 de novo, identificamos cos x:
summ=0^infty(-1)^m,x^2m-tfrac12(2m)! =x^-tfrac12summ=0^infty(-1)^m,x^2m(2m)! =x^-tfrac12,cos x
Logo
J-tfrac12(x) =raiz2pi,x^-tfrac12cos x =raiz2pi x,cos x

Resposta

Jtfrac12(x)=raiztfrac2pi xsin x
J-tfrac12(x)=raiztfrac2pi xcos x

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