What is the taylor expansion of #e^(-1/x)#?

Question

Isaiah Allen · Accepted Answer

Maclaurin Series The Maclaurin series for #e^x# is given by: #e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...# Replacing #x# with #-1/x#, the Maclaurin series for #e^(-1/x)# is: #e^(-1/x)=sum_(n=0)^oo(-1/x)^n/(n!)=sum_(n=0)^oo(-1)^n/(n!)x^-n# #color(white)(e^(-1/x))=1-1/x+1/((2!)x^2)-1/((3!)x^3)+1/((4!)x^4)+...#

Emma Aldridge · Answer

See description:

The Taylor Series about #x=1# is given by:

#e^(-1/x) = 1/e + (x-1)/e - (x-1)^2/(2e) + ... #

Let # f(x)=e^(-1/x) # The Taylor Series about the pivot point #x=a# is given by: # f(x) = f(a) + f'(a)(x-a) + (f''(a))/(2!)(x-a)^2 + (f^((3))(a))/(3!)(x-a)^3 + ... + (f^((n))(a))/(n!)(x-a)^n + ... # As no pivot point for the Taylor Expansion Series has been provided it would be usual practice to assume that #a=0# which gives us the Maclaurin Series; # f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ... # However, #f(x)# has an essential singularity when #x=0# and so we cannot form the Maclaurin series, (ie the Taylor series pivoted about #x=0#). Technically this is the end of the question - There is no such series. Using the well know series for #e^x# we can expand a series by substituting #x# for #-1/x#. This gives us a power series of increasing negative powers, and is known as a Laurent Series (As Laurent series typically have complex arguments we use #z# by convention rather than #x# where #z in CC#: # e^(-1/z) = 1-1/z+1/(2!z^2)-1/(3!z^3)+1/(4!z^4)+...# We can however form a Taylor Series about another pivot point so lets do so about #x=1#. Firstly, we have: # f(1)=e^(-1) =1/e # We need the first derivative: # f'(x)=e^(-1/x)/x^2 # # :. f'(1)=e^(-1)/1 =1/e # And the second derivative (using quotient rule): # f''(x) = ( (x^2)(e^(-1/x)/x^2)-(e^(-1/x))(2x) ) / (x^2)^2 # # " " = ( e^(-1/x)(1-2x) ) / (x^4) # # :. f''(1) = -1/e # #vdots# And so the Taylor Series about #x=1# is given by: # f(x) = 1/e + 1/e(x-1) + (-1/e)/(2!)(x-1)^2 + (f^((3))(a))/(3!)(x-1)^3 + ... # # " " = 1/e + (x-1)/e - (x-1)^2/(2e) + ... #

Scarlett Allison · Answer

undefined The Taylor expansion of $ e^{-1/x} $ about $ x = 0 $ is:

$$ e^{-1/x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left(\frac{1}{x}\right)^n $$

This is the Maclaurin series for $ e^{-1/x} $, centered at $ x = 0 $.