11.7.2 Derive as formas trigonométricas polinomiais de j_{n}(z) e n_{n}(z) .^{27}
\begin{aligned} j_{n}(z)= & \frac{1}{z} \operatorname{sen}\left(z-\frac{n \pi}{2}\right) \sum_{s=0}^{[n / 2]} \frac{(-1)^{s}(n+2 s)!}{(2 s)!(2 z)^{2 s}(n-2 s)!} \\\ & +\frac{1}{z} \cos \left(z-\frac{n \pi}{2}\right) \sum_{s=0}^{[(n-1) / 2]} \frac{(-1)^{s}(n+2 s+1)!}{(2 s+1)!(2 z)^{2 s}(n-2 s-1)!} \\\ n_{n}(z)= & \frac{(-1)^{n+1}}{z} \cos \left(z+\frac{n \pi}{2}\right) \sum_{s=0}^{[n / 2]} \frac{(-1)^{s}(n+2 s)!}{(2 s)!(2 z)^{2 s}(n-2 s)!} \\\ & +\frac{(-1)^{n+1}}{z} \operatorname{sen}\left(z+\frac{n \pi}{2}\right) \sum_{s=0}^{[(n-1) / 2]} \frac{(-1)^{s}(n+2 s+1)!}{(2 s+1)!(2 z)^{2 s+1}(n-2 s-1)!} \end{aligned}

Passo 1

O enunciado pede para obter as chamadas formas trigonométricas polinomiais das funções esféricas de Bessel
\displaystyle j_{n}(z) e \displaystyle n_{n}(z).
O que vamos usar:
1. A representação por derivadas
j_{n}(z) =(-z)^{n}\Bigl(\frac{1}{z}\frac{d}{dz}\Bigr)^{n}\frac{\sin z}{z}, \quad n_{n}(z) =(-1)^{\,n+1}(-z)^{n}\Bigl(\frac{1}{z}\frac{d}{dz}\Bigr)^{n}\frac{\cos z}{z}.
2. A Regra de Leibniz para derivar produtos.
3. As identidades de mudança de fase
\sin(z+\alpha) e \cos(z+\alpha).

Passo 2

Transformamos a definição usando o fato de que
\Bigl(\frac{1}{z}\frac{d}{dz}\Bigr)^{n} f(z) \;=\; z^{-n}\frac{d^{n}}{dz^{n}}\bigl[z^{n}f(z)\bigr].
Assim
j_{n}(z) =(-z)^{n}z^{-n}\frac{d^{n}}{dz^{n}}\bigl[z^{n}\tfrac{\sin z}{z}\bigr] =(-1)^{n}\frac{d^{n}}{dz^{n}}\bigl[z^{\,n-1}\sin z\bigr],
n_{n}(z) =(-1)^{n+1}\frac{d^{n}}{dz^{n}}\bigl[z^{\,n-1}\cos z\bigr].

Passo 3

Aplicamos a Regra de Leibniz a
\displaystyle \frac{d^{n}}{dz^{n}}\bigl[z^{\,n-1}g(z)\bigr],
onde g(z)=\sin z ou g(z)=\cos z.
\frac{d^{n}}{dz^{n}}\bigl[z^{n-1}g(z)\bigr] =\sum_{k=0}^{n}\binom{n}{k} \frac{d^{\,n-k}}{dz^{\,n-k}}\bigl[z^{n-1}\bigr] \;\frac{d^{\,k}}{dz^{\,k}}g(z).
Como
\displaystyle \frac{d^{\,n-k}}{dz^{\,n-k}}z^{n-1} =\frac{(n-1)!}{(n-k-1)!}\,z^{\,n-k-1}
e
\displaystyle \frac{d^{\,k}}{dz^{\,k}}\sin z=\sin\bigl(z+k\pi/2\bigr),
\displaystyle \frac{d^{\,k}}{dz^{\,k}}\cos z=\cos\bigl(z+k\pi/2\bigr),
fica
j_{n}(z) =(-1)^{n} \sum_{k=0}^{n} \binom{n}{k} \frac{(n-1)!}{(n-k-1)!}\, z^{\,n-k-1}\, \sin\bigl(z+k\tfrac{\pi}{2}\bigr),
n_{n}(z) =(-1)^{n+1} \sum_{k=0}^{n} \binom{n}{k} \frac{(n-1)!}{(n-k-1)!}\, z^{\,n-k-1}\, \cos\bigl(z+k\tfrac{\pi}{2}\bigr).

Passo 4

Agora separamos a soma em termos pares ( k=2s) e ímpares ( k=2s+1).
Por exemplo, em j_{n}:
  • Para k=2s, \sin(z+2s\pi/2)=(-1)^{s}\sin z.
  • Para k=2s+1, \sin(z+(2s+1)\pi/2)=(-1)^{s}\cos z.
    Isso gera uma forma preliminar
j_{n}(z) =\frac{1}{z}\Bigl[ \sin z\sum_{s=0}^{\lfloor n/2\rfloor}A_{s}z^{2s} \;-\; \cos z\sum_{s=0}^{\lfloor(n-1)/2\rfloor}B_{s}z^{2s+1} \Bigr],
onde A_{s} e B_{s} são coeficientes em termos de n e s obtidos da combinação
\binom{n}{k}\,(n-1)!/(n-k-1)!.
Simplificando esses coeficientes com manipulação de fatoriais chega-se a
A_{s} =\frac{(-1)^{s}(n+2s)!}{(2s)!\,(n-2s)!\,2^{2s}}, \quad B_{s} =\frac{(-1)^{s}(n+2s+1)!}{(2s+1)!\,(n-2s-1)!\,2^{2s}}.

Passo 5

Para obter a forma final em termos de \sin\bigl(z-\tfrac{n\pi}{2}\bigr) e
\cos\bigl(z-\tfrac{n\pi}{2}\bigr), usamos
\sin z =\sin\Bigl(z-\tfrac{n\pi}{2}\Bigr)\cos\Bigl(\tfrac{n\pi}{2}\Bigr) +\cos\Bigl(z-\tfrac{n\pi}{2}\Bigr)\sin\Bigl(\tfrac{n\pi}{2}\Bigr),
\cos z =\cos\Bigl(z-\tfrac{n\pi}{2}\Bigr)\cos\Bigl(\tfrac{n\pi}{2}\Bigr) -\sin\Bigl(z-\tfrac{n\pi}{2}\Bigr)\sin\Bigl(\tfrac{n\pi}{2}\Bigr),
e os valores \cos(n\pi/2), \sin(n\pi/2)\in\{0,\pm1\}.
Fazendo essa combinação e reorganizando as somas obtemos exatamente
j_{n}(z) =\frac{1}{z}\,\sin\Bigl(z-\tfrac{n\pi}{2}\Bigr)\! \sum_{s=0}^{\lfloor n/2\rfloor} \frac{(-1)^{s}(n+2s)!}{(2s)!\,(n-2s)!\,(2z)^{2s}} \;+\; \frac{1}{z}\,\cos\Bigl(z-\tfrac{n\pi}{2}\Bigr)\! \sum_{s=0}^{\lfloor (n-1)/2\rfloor} \frac{(-1)^{s}(n+2s+1)!}{(2s+1)!\,(n-2s-1)!\,(2z)^{2s}}.

Passo 6

O mesmo procedimento vale para
\displaystyle n_{n}(z) começando com \cos z/z e o fator global
(-1)^{\,n+1}. Através de Leibniz e mudança de fase chega-se a
n_{n}(z) =\frac{(-1)^{n+1}}{z}\,\cos\Bigl(z+\tfrac{n\pi}{2}\Bigr)\! \sum_{s=0}^{\lfloor n/2\rfloor} \frac{(-1)^{s}(n+2s)!}{(2s)!\,(n-2s)!\,(2z)^{2s}} \;+\; \frac{(-1)^{n+1}}{z}\,\sin\Bigl(z+\tfrac{n\pi}{2}\Bigr)\! \sum_{s=0}^{\lfloor (n-1)/2\rfloor} \frac{(-1)^{s}(n+2s+1)!}{(2s+1)!\,(n-2s-1)!\,(2z)^{2s+1}}.

Resposta

\boxed{ \begin{aligned} j_{n}(z)&=\frac{1}{z}\,\sin\Bigl(z-\tfrac{n\pi}{2}\Bigr)\!\n\sum_{s=0}^{\lfloor n/2\rfloor} \frac{(-1)^{s}(n+2s)!}{(2s)!\,(n-2s)!\,(2z)^{2s}} +\frac{1}{z}\,\cos\Bigl(z-\tfrac{n\pi}{2}\Bigr)\! \sum_{s=0}^{\lfloor (n-1)/2\rfloor} \frac{(-1)^{s}(n+2s+1)!}{(2s+1)!\,(n-2s-1)!\,(2z)^{2s}},\\[6pt] n_{n}(z)&=\frac{(-1)^{n+1}}{z}\,\cos\Bigl(z+\tfrac{n\pi}{2}\Bigr)\! \sum_{s=0}^{\lfloor n/2\rfloor} \frac{(-1)^{s}(n+2s)!}{(2s)!\,(n-2s)!\,(2z)^{2s}} +\frac{(-1)^{n+1}}{z}\,\sin\Bigl(z+\tfrac{n\pi}{2}\Bigr)\! \sum_{s=0}^{\lfloor (n-1)/2\rfloor} \frac{(-1)^{s}(n+2s+1)!}{(2s+1)!\,(n-2s-1)!\,(2z)^{2s+1}}. \end{aligned} }]

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