How do you find the Taylor expansion of e^(-1/x)? Calculus Power Series Constructing a Taylor Series 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) + ... Answer link Related questions How do you find the Taylor series of f(x)=1/x ? How do you find the Taylor series of f(x)=cos(x) ? How do you find the Taylor series of f(x)=e^x ? How do you find the Taylor series of f(x)=ln(x) ? How do you find the Taylor series of f(x)=sin(x) ? How do you use a Taylor series to find the derivative of a function? How do you use a Taylor series to prove Euler's formula? How do you use a Taylor series to solve differential equations? What is the Taylor series of f(x)=arctan(x)? What is the linear approximation of g(x)=sqrt(1+x)^(1/5) at a =0? See all questions in Constructing a Taylor Series Impact of this question 18440 views around the world You can reuse this answer Creative Commons License