Subsections

6-4

次の式を示せ。

$\displaystyle e^{iz\sin\varphi } = \sum_{n=-\infty}^{+\infty}J_{{n}}(z)e^{+in\varphi }$ (48)

6-4解答

複素Fourier級数:

$\displaystyle f(x)=f(x+n\lambda)
=
\sum_{n=-\infty}^{+\infty}
c_n
e^{+in(2\pi/\...
...
\quad
c_n
=
\frac{1}{\lambda}
\int_0^\lambda
f(x)e^{-in (2\pi/\lambda) x} dx.
$

$\displaystyle \Longrightarrow\quad
f(\varphi )
=e^{iz\sin\varphi }
=e^{iz\sin\l...
...0^{2\pi}
e^{iz\sin\varphi -in \varphi } d\varphi
=J_{{n}}(z),
\quad
\because\,$   Eq.(46)$\displaystyle $

% latex2html id marker 3374
$\displaystyle \therefore\,
e^{iz\sin\varphi }
=
\sum_{n=-\infty}^{+\infty}J_{{n}}(z)e^{+in\varphi }
$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp