# What is the Taylor series of e^((-x)^2)?

May 12, 2015

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 ${\left(- x\right)}^{2}$ (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 ${e}^{- \left({x}^{2}\right)}$, it would be :

f(x) = sum_{k=0}^infty(-x^2)^(k)/(k!)=sum_{k=0}^infty(-1)^(k)*x^(2k)/(k!)