The case of a taylor series expanded around #0# is called a Maclaurin series. The general formula for a Maclaurin series is:
#f(x)=sum_(n=0)^oof^n(0)/(n!)x^n#
To work out a series for our function we can start with a function for #e^x# and then use that to figure out a formula for #e^(-2x)#.
In order to construct the Maclaurin series, we need to figure out the nth derivative of #e^x#. If we take a few derivatives, we can quite quickly see a pattern:
#f(x)=e^x#
#f'(x)=e^x#
#f''(x)=e^x#
In fact, the nth derivative of #e^x# is just #e^x#. We can plug this into the Maclaurin formula:
#e^x=sum_(n=0)^ooe^0/(n!)x^n=sum_(n=0)^oox^n/(n!)=1+x/(1!)+x^2/(2!)+x^3/(3!)...#
Now that we have a taylor series for #e^x#, we can just replace all the #x#'s with #-2x# to get a series for #e^(-2x)#:
#e^(-2x)=sum_(n=0)^oo(-2x)^n/(n!)=sum_(n=0)^oo(-2)^n/(n!)x^n=#
#=1-2/(1!)x+4/(2!)x^2-8/(3!)x^3+16/(4!)x^4...=#
#=1-2x+2x^2-4/3x^3+2/3x^4...#
which is the series we were looking for.
