Question

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

Answer 1

`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) + ...`