Subsections

6-1

以下の関係式を示せ。

$\displaystyle \frac{2n}{z} J_{{n}}(z)$ $\displaystyle = J_{{n-1}}(z)+J_{{n+1}}(z)$ (40)
$\displaystyle 2 \dI{z}J_{{n}}(z)$ $\displaystyle = J_{{n-1}}(z)-J_{{n+1}}(z)$ (41)

6-1解答

RHS of Eq.(40) $\displaystyle =\frac{1}{\pi} \int_0^\pi \left\{ \cos\left(\left[n-1\right]\varp...
...{\pi} \int_0^\pi \cos\left(n\varphi -z\sin\varphi \right)\cos\varphi \,d\varphi$    
$\displaystyle \quad n\varphi -$ % latex2html id marker 3294
$\displaystyle z\sin\varphi = \xi \to d\xi =nd\varph...
...{c\vert ccc} \varphi & 0 & \to & \pi \\ \hline \xi & 0 & \to & n\pi \end{array}$    
  $\displaystyle = \frac{2}{\pi} \int_0^\pi \cos \left(n\varphi -z \sin\varphi \ri...
...hi -z \sin\varphi \right)\,d\varphi -\frac{2}{z\pi}\int_0^{n\pi} \cos\xi \,d\xi$    
  $\displaystyle =\frac{2n}{z}J_{{n}}(z)+0 =$   LHS of Eq.(40)    

LHS of Eq.(41) $\displaystyle = \frac{2}{\pi} \int_0^\pi \left\{ -\sin\left(n\varphi -z\sin\varphi \right) \right\} \left(-\sin\varphi \right)\,d\varphi$    
  $\displaystyle = \frac{1}{\pi} \int_0^\pi \left\{ \cos\left(\left[n-1\right]\var...
...right]\varphi -z\sin\varphi \right) \right\}d\varphi =J_{{n-1}}(z)-J_{{n+1}}(z)$    
  $\displaystyle =$   RHS of Eq.(41)    

$\displaystyle \cos A+\cos B
=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
;\quad
\cos A-\cos B
=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp