I know that the Maclaurin expansion of $e^x$ is $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…$$
But i'm not sure how to find the Maclaurin series here
I tried this
$$
f'_{(0)}=-2xe^{-x^2}=0
$$
And that follows to every derivative that follows, so how can I get a power series out of it?
Best Answer
As Dark mentions in the comments, $$e^y = 1+y+\frac{y^2}{2!}+\frac{y^3}{3!}+\cdots$$ works for any $y\in \Bbb R$. In particular, you can substitute $y=-x^2$ and that should give you the result!
$$e^{-x^2} = 1+\left(-x^2\right)+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!}+\cdots$$
There's no need of using a derivative here though, since you're just simply making a substitution in an identity that holds for any $y$ (here, it is important to notice that the radius of convergence is infinite, we'd need to be more careful if this were not the case).
That's the power of using algebraic expressions. When you say that certain property holds for any $y$ it means that it holds for any expression you put instead of $y$, because that expression represents any number at the very end.