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

1 Answer
Mar 4, 2017

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

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 = - 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) + ...