Find the power series representation of $e^{-x^2}$ 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?
 A: 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.
A: That is easier than what you think.
General expansion:
$$e^A = 1 + A + \frac{1}{2}A^2 + \frac{1}{6}A^3 + \cdots$$
So if $A = -x^2$ you get
$$e^{-x^2} = 1 - x^2 + \frac{1}{2}(-x^2)^2 + \frac{1}{6}(-x^2)^3 + \cdots$$
So
$$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$
To find the general series expansion, you just have to remember the series expansion of $e^A$:
$$e^A = \sum_{k = 0}^{+\infty} \frac{A^k}{k!}$$
So again: if $A = -x^2$ you gain finally
$$\boxed{e^{-x^2} = \sum_{k = 0}^{+\infty} \frac{(-x^2)^k}{k!} = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k!} x^{2k}}$$
A: You're mistaken in saying "And that follows to every derivative that follows".
The product rule will come into play as you keep differentiating, giving terms that don't vanish. Here are a few steps:
$f(x)=e^{-x^2}$, so $f(0)=1$.
$f'(x) = -2xe^{-x^2}$, so $f'(0) = 0$.
$f''(x) = -2x[-2xe^{-x^2}] -2[e^{-x^2}] = (4x^2-2)e^{-x^2}$, so $f''(0)=-2$.
And so on. I think you will find that the odd-numbered derivatives ($f'(0), f^{(3)}(0),f^{(5)}(0),\ldots$) vanish , but the even-numbered ones ($f(0), f''(0),f^{(4)}(0),\ldots$) do not.
