\[{\operatorname{ber}_{\nu}}^{2}x+{\operatorname{bei}_{\nu}}^{2}x=(\tfrac{1}{2}x%
)^{2\nu}\sum_{k=0}^{\infty}\frac{1}{\Gamma\left(\nu+k+1\right)\Gamma\left(\nu+%
2k+1\right)}\frac{(\frac{1}{4}x^{2})^{2k}}{k!},\]