How do you find the Taylor expansion of $e^{- \frac{1}{x}}$ $e^(-1/x)$ ?

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Answer 1 · 2017-03-04T01:21:36

$e^(- 1/x) =1- 1/x+ 1/(2x^2)- 1/(6x^3) + 1/(24 x^4) + ...$

If you mean about $x = 0$ , in which case it is really called a Maclaurin series, you may know the exponential Maclaurin series:

$e^z= sum_{k=0}^{\infty } z^{k}/( k!)$

With $z = - 1/x$ , we substitute:

$e^(- 1/x)= sum_{k=0}^{\infty } (- 1/x)^{k}/( k!)$

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

$=1- 1/x+ 1/(2x^2)- 1/(6x^3) + 1/(24 x^4) + ...$

How do you find the Taylor expansion of e^(-1/x)e−1xe^(-1/x)?