]> $$\begin{array}{rcl}Bi\left(0\right)& =& \frac{1}{3^{\frac{1}{6}}\Gamma \left(\frac{2}{3}\right)}\\ Bi\left(z\right)& =& \sqrt{\frac{z}{3}}\left(I_{\frac{1}{3}}\left(\zeta \right)+I_{-\frac{1}{3}}\left(\zeta \right)\right)& ;& \zeta =\frac{2}{3}{z}^{\frac{2}{3}}\\ Bi\left(-z\right)& =& \sqrt{\frac{z}{3}}\left(J_{-\frac{1}{3}}\left(\zeta \right)-J_{\frac{1}{3}}\left(\zeta \right)\right)\end{array}$$