[Math] Taylor series of $e^{-z^2}$ around $z_0=0$

Question

I am trying to find the Taylor series of $e^{-z^2}$ around $z_0$=0

I found the general formula for the $n^{th}$ derivative: $f^{(n)}(z)=(-2z)^ne^{-z^2}$

To find the Taylor series, I need to plug in $z_0=0$. However, this will lead to $f^{(n)}(z)=0$, so the Taylor series will be equal to 0.

What goes wrong here?

Answer 1

Hint

It equals to zero only for odd $n$. You can also easily find the Taylor expansion of $e^z$ around $z=0$ by using $${d\over dz}e^z=e^z$$and then substitute $z\to -z^2$