10.4.59

Questa รจ la mia formula:

Ai $\mathcal(z)\sim\frac{1}{2}\pi^{-\frac{1}{2}}\mathcal{z}^{-\frac{1}{4}}\mathcal{e}^{-\zeta}\sum_{0}^{\infty}(-1)^{\mathcal{k}}\mathcal{c}_{\mathcal{k}}\zeta^{\mathcal{k}} \quad (|arg \ \mathcal{-z}| < \pi)$

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