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Question Number 35234 by abdo.msup.com last updated on 17/May/18

$$letf\left(x\right)=e^{-x^n}withnfromN$$$$developpfatintegrserie.$$

Commented by prof Abdo imad last updated on 18/May/18

$$wehave{e}^{-u}=\sum _{p=0}^n\frac{{(-1)}^p{u}^p}{p!}withR=\mathrm{\infty }$$$$so{e}^{-x^n}=\sum _{p=0}^n\frac{{(-1)}^p{x}^{np}}{p!}.$$

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