The answer, when #a=0#, is : #f(x)=sum_{k=0}^inftyx^(2k)/(k!)#
The Taylor series is given by : #f(x)=sum_{k=0}^infty{f^{(k)}(a)}/{k!}(x-a)^k#.
We know that the Taylor series of #e^(x)#, when #a=0#, is :
#f(x)=sum_{k=0}^inftyx^(k)/(k!)#
So now, we just need to replace the #x# of the above series with #(-x)^(2)# (in operations with Taylor series, it is called substitution) :
#f(x)=sum_{k=0}^infty((-x)^2)^(k)/(k!)=sum_{k=0}^infty((-x)^(2k))/(k!)=sum_{k=0}^inftyx^(2k)/(k!)#
If you meant #e^(-(x^(2)))#, it would be :
#f(x) = sum_{k=0}^infty(-x^2)^(k)/(k!)=sum_{k=0}^infty(-1)^(k)*x^(2k)/(k!)#
You got your answer.
