# §10.31 Power Series

For $I_{\nu}\left(z\right)$ see (10.25.2) and (10.27.1). When $\nu$ is not an integer the corresponding expansion for $K_{\nu}\left(z\right)$ is obtained from (10.25.2) and (10.27.4).

When $n=0,1,2,\dotsc$,

 10.31.1 $K_{n}\left(z\right)=\tfrac{1}{2}(\tfrac{1}{2}z)^{-n}\sum_{k=0}^{n-1}\frac{(n-k% -1)!}{k!}(-\tfrac{1}{4}z^{2})^{k}+(-1)^{n+1}\ln\left(\tfrac{1}{2}z\right)I_{n}% \left(z\right)+(-1)^{n}\tfrac{1}{2}(\tfrac{1}{2}z)^{n}\sum_{k=0}^{\infty}\left% (\psi\left(k+1\right)+\psi\left(n+k+1\right)\right)\frac{(\tfrac{1}{4}z^{2})^{% k}}{k!(n+k)!},$

where $\psi\left(x\right)=\Gamma'\left(x\right)/\Gamma\left(x\right)$5.2(i)). In particular,

 10.31.2 $K_{0}\left(z\right)=-\left(\ln\left(\tfrac{1}{2}z\right)+\gamma\right)I_{0}% \left(z\right)+\frac{\tfrac{1}{4}z^{2}}{(1!)^{2}}+(1+\tfrac{1}{2})\frac{(% \tfrac{1}{4}z^{2})^{2}}{(2!)^{2}}+(1+\tfrac{1}{2}+\tfrac{1}{3})\frac{(\tfrac{1% }{4}z^{2})^{3}}{(3!)^{2}}+\dotsi.$

For negative values of $n$ use (10.27.3).

 10.31.3 $I_{\nu}\left(z\right)I_{\mu}\left(z\right)=(\tfrac{1}{2}z)^{\nu+\mu}\sum_{k=0}% ^{\infty}\frac{(\nu+\mu+k+1)_{k}(\tfrac{1}{4}z^{2})^{k}}{k!\Gamma\left(\nu+k+1% \right)\Gamma\left(\mu+k+1\right)}.$