CVS2

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Find the Maclaurin series for $\tan^{-1}z\,$ .

$\frac{d}{dz}\tan^{-1}z = \frac{1}{1+z^2} = \sum_{n=0}^\infty (-z^2)^n = \sum_{n=0}^\infty (-1)^n z^{2n}, |z|<1\,$ .

Integrating termwise,

$\int_0^z\sum_{n=0}^\infty(-1)^n\xi^{2n}d\xi = \sum_{n=0}^\infty (-1)^n\int_0^z \xi^{2n}d\xi = \sum_{n=0}^\infty (-1)^n \frac{z^{2n+1}}{2n+1}, |z|<1\,$

Main Page : Complex Variables : Series

CVS2

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