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

Mar 4, 2017

${e}^{- \frac{1}{x}} = 1 - \frac{1}{x} + \frac{1}{2 {x}^{2}} - \frac{1}{6 {x}^{3}} + \frac{1}{24 {x}^{4}} + \ldots$

#### Explanation:

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 $z = - \frac{1}{x}$, 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 + ...

$= 1 - \frac{1}{x} + \frac{1}{2 {x}^{2}} - \frac{1}{6 {x}^{3}} + \frac{1}{24 {x}^{4}} + \ldots$