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

1 Answer
Eddie
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) + ...#

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