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Question Number 35234 by abdo.msup.com last updated on 17/May/18
$${let}\:{f}\left({x}\right)\:={e}^{−{x}^{{n}} } \:\:\:\:\:{with}\:{n}\:{fromN} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by prof Abdo imad last updated on 18/May/18
$${we}\:{have}\:{e}^{−{u}} \:=\sum_{{p}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{p}} {u}^{{p}} }{{p}!}\:\:\:{with}\:{R}=\infty \\ $$$${so}\:{e}^{−{x}^{{n}} } \:\:=\sum_{{p}=\mathrm{0}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{p}} \:{x}^{{np}} }{{p}!}\:\:. \\ $$