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arccsc(z)在0的级数

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hbghlyj posted 2023-6-5 20:47 |Read mode
  1. Series[ArcCsc[z], {z, 0, 3}]
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$$\operatorname{arccsc}z=-\frac{1}{2} \sqrt{-\frac{1}{z^2}} z \log \left(-\frac{4}{z^2}\right)+\frac{1}{4} \sqrt{-\frac{1}{z^2}} z z^2+O\left(z^4\right)$$
这是怎样得到的

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