\[\frac{{\partial}^{2q}}{{\partial x_{1}}^{2q}}\Psi_{K}\left(\boldsymbol{{0}}%
\right)=\frac{2}{K+2}\Gamma\left(\frac{2q+1}{K+2}\right)\exp\left(i\frac{\pi}{%
2}\left(\frac{2q+1}{K+2}+2q\right)\right),\]