What is the Maclaurin series for $e^(-x)$ ?

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Answer 1 · 2017-11-03T14:01:37

$e^(-x) = 1 -x +(x^2)/(2!) - (x^3)/(3!) + x^4/(4!) - ...$

We can start with the well known series for $e^(x)$

$e^x = 1 +x +(x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + ...$

If we replace $x$ in the above series by $-x$ we get:

$e^(-x) = 1 +(-x) +((-x)^2)/(2!) + ((-x)^3)/(3!) + ((-x)^4)/(4!) + ...$
$\ \ \ \ \ \ = 1 -x +(x^2)/(2!) - (x^3)/(3!) + x^4/(4!) - ...$

What is the Maclaurin series for e^(-x)e^(-x)?