\[\operatorname{ber}_{\nu}x+i\operatorname{bei}_{\nu}x=\sum_{k=0}^{\infty}\frac{%
e^{(3\nu+k)\pi i/4}x^{k}J_{\nu+k}\left(x\right)}{2^{k/2}k!}=\sum_{k=0}^{\infty%
}\frac{e^{(3\nu+3k)\pi i/4}x^{k}I_{\nu+k}\left(x\right)}{2^{k/2}k!}.\]