1 漸化式

Modified Bessel function: $K_{\nu}(z)$配下の関係式を満たす：

$$\frac{2\nu}{z} K_{\nu}(z) = -K_{\nu-1}(z) +K_{\nu+1}(z)$$ (44)

$$2K'_{\nu}(z) = -K_{\nu-1}(z) -K_{\nu+1}(z) \, ; \qquad K'_{\nu}(z) \equiv \dI{z}K_{\nu}(z)$$ (45)

$$K_{-\nu}(z) =K_{\nu}(z) .$$ (46)

1 証明

$$= \int_0^\infty e^{-z\cosh(t)} \cosh(\nu t)dt = \int_0^\infty e^{... ...\left( - \cosh\left[ (\nu-1)t \right] + \cosh\left[ (\nu+1)t \right] \right) dt$$ RHS of Eq.(44)

$$= \int_0^\infty e^{-z\cosh(t)} 2\sinh(\nu t) \sinh(t) dt \, ; \quad \because\, \cosh(x)-\cosh(y) =2 \sinh\frac{x+y}{2}\sinh\frac{x-y}{2}$$

$$= -\frac{2}{z} \int_0^\infty \dI{t} \left( e^{-z\cosh(t)} \right)... ...(\nu t) dt \, ; \quad \because \, (\sinh x)'=\cosh x, \, (\cosh x)'=\sinh x, \,$$

$$= -\frac{2}{z} \left( \left[ e^{-z\cosh(t)} \right]_0^\infty - \i... ...osh(\nu t) dt \right) = \frac{2}{z}\int_0^\infty e^{-z\cosh(t)} \cosh(\nu t) dt$$

$$=$$

$$= 2\dI{z} \int_0^\infty e^{-z\cosh(t)} \cosh(\nu t)dt = -2 \int_0^\infty e^{-z\cosh(t)} \cosh(t) \cosh(\nu t) dt$$ LHS of Eq.(45)

$$= - \left( K_{\nu-1}(z) +K_{\nu+1}(z) \right) =- \int_0^\infty e^... ...[ \left(\nu-1\right)t \right] + \cosh\left[ \left(\nu+1\right)t \right] \right)$$ RHS of Eq.(45)

$$= -2\int_0^\infty e^{-z\cosh(t)} \cosh(\nu t) \cosh(t) dt \, ; \q... ..., \cosh x+\cosh y = 2\cosh\frac{x+y}{2}\cosh\frac{x-y}{2}, \, \cosh(-x)=\cosh x$$

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp