Questa รจ la formula 10.4.21
Bi' $(-z)=(z/\sqrt{3})[J_{-2/3}(\zeta)+J_{2/3}(\zeta)] = \frac {1}{2} i(z/\sqrt{3})[e^{-x 1/6}H^{1}_{2/3}(\zeta)-e^{x 1/6}H^{2}_{2/3}(\zeta)]$
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