The Taylor series expansion of $e^{x}$ at $x=0$ is given by:
$$
e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}
$$
So, by substituting $-x^{2}$ for $x$, we get the Taylor series expansion of $e^{-x^{2}}$ at $x=0$:
$$
e^{-x^{2}}=\sum_{n=0}^{\infty}
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