# §10.27 Connection Formulas

Other solutions of (10.25.1) are $I_{-\nu}\left(z\right)$ and $K_{-\nu}\left(z\right)$.

 10.27.1 $I_{-n}\left(z\right)=I_{n}\left(z\right),$ ⓘ Symbols: $I_{\NVar{\nu}}\left(\NVar{z}\right)$: modified Bessel function of the first kind, $n$: integer and $z$: complex variable A&S Ref: 9.6.6 Referenced by: §10.27, §10.31, §10.44(ii) Permalink: http://dlmf.nist.gov/10.27.E1 Encodings: TeX, pMML, png See also: Annotations for 10.27 and 10
 10.27.2 $I_{-\nu}\left(z\right)=I_{\nu}\left(z\right)+(2/\pi)\sin\left(\nu\pi\right)K_{% \nu}\left(z\right),$
 10.27.3 $K_{-\nu}\left(z\right)=K_{\nu}\left(z\right).$ ⓘ Symbols: $K_{\NVar{\nu}}\left(\NVar{z}\right)$: modified Bessel function of the second kind, $z$: complex variable and $\nu$: complex parameter A&S Ref: 9.6.6 Referenced by: §10.25(iii), §10.30(i), §10.31, §10.47(ii) Permalink: http://dlmf.nist.gov/10.27.E3 Encodings: TeX, pMML, png See also: Annotations for 10.27 and 10
 10.27.4 $K_{\nu}\left(z\right)=\tfrac{1}{2}\pi\frac{I_{-\nu}\left(z\right)-I_{\nu}\left% (z\right)}{\sin\left(\nu\pi\right)}.$

When $\nu$ is an integer limiting values are taken:

 10.27.5 $K_{n}\left(z\right)=\frac{(-1)^{n-1}}{2}\*\left(\left.\frac{\partial I_{\nu}% \left(z\right)}{\partial\nu}\right|_{\nu=n}+\left.\frac{\partial I_{\nu}\left(% z\right)}{\partial\nu}\right|_{\nu=-n}\right),$ $n=0,\pm 1,\pm 2,\ldots$.

In terms of the solutions of (10.2.1),

 10.27.6 $I_{\nu}\left(z\right)=e^{\mp\nu\pi i/2}J_{\nu}\left(ze^{\pm\pi i/2}\right),$ $-\pi\leq\pm\operatorname{ph}z\leq\tfrac{1}{2}\pi$,
 10.27.7 $I_{\nu}\left(z\right)=\tfrac{1}{2}e^{\mp\nu\pi i/2}\left({H^{(1)}_{\nu}}\left(% ze^{\pm\pi i/2}\right)+{H^{(2)}_{\nu}}\left(ze^{\pm\pi i/2}\right)\right),$ $-\pi\leq\pm\operatorname{ph}z\leq\tfrac{1}{2}\pi$.
 10.27.8 $K_{\nu}\left(z\right)=\begin{cases}\tfrac{1}{2}\pi ie^{\nu\pi i/2}{H^{(1)}_{% \nu}}\left(ze^{\pi i/2}\right),&-\pi\leq\operatorname{ph}z\leq\tfrac{1}{2}\pi,% \\ -\tfrac{1}{2}\pi ie^{-\nu\pi i/2}{H^{(2)}_{\nu}}\left(ze^{-\pi i/2}\right),&-% \tfrac{1}{2}\pi\leq\operatorname{ph}z\leq\pi.\end{cases}$
 10.27.9 $\pi iJ_{\nu}\left(z\right)=e^{-\nu\pi i/2}K_{\nu}\left(ze^{-\pi i/2}\right)-e^% {\nu\pi i/2}K_{\nu}\left(ze^{\pi i/2}\right),$ $|\operatorname{ph}z|\leq\tfrac{1}{2}\pi$.
 10.27.10 $-\pi Y_{\nu}\left(z\right)=e^{-\nu\pi i/2}K_{\nu}\left(ze^{-\pi i/2}\right)+e^% {\nu\pi i/2}K_{\nu}\left(ze^{\pi i/2}\right),$ $|\operatorname{ph}z|\leq\tfrac{1}{2}\pi$.
 10.27.11 $Y_{\nu}\left(z\right)=e^{\pm(\nu+1)\pi i/2}I_{\nu}\left(ze^{\mp\pi i/2}\right)% -(2/\pi)e^{\mp\nu\pi i/2}K_{\nu}\left(ze^{\mp\pi i/2}\right),$ $-\tfrac{1}{2}\pi\leq\pm\operatorname{ph}z\leq\pi$.