Question

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

Answer 1

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.