TITLE: Taylor series of $e^{-z^2}$ around $z_0=0$
QUESTION [0 upvotes]: 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?

REPLY [0 votes]: $$ e^{x} = 1 + x + \dfrac {x^{2}}{2!} +\dfrac {x^{3}}{3!} + \cdots = \sum_{n=1}^{\infty} \; \dfrac{x^{n-1}}{(n -1)!} $$
Now replace the value of $ x $ by $ -z^{2} $ in the above Taylor series representation:
$$ \boxed { e^{-z^{2}} = 1 - z^{2} + \dfrac {z^{4}}{2!} - \dfrac {z^{6}}{3!} + \cdots = \sum_{n=1}^{\infty} \;  (-1)^{n -1} \; \dfrac{z^{2n-2}}{(n -1)!}} $$
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