Please check my error this series expansion

Question

I have beeen able to derive the series for $e^{-x^2}$ and want to find the series representation for $e^{-\frac{x^2}{2}}$:

$$e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}+ \cdots$$ $$e^{\frac{x^2}{2}}=1-\frac{x^2}{2}+\frac{x^4}{2\cdot2!}-\frac{x^6}{2\cdot3!}+\frac{x^8}{2\cdot4!}+\cdots$$

The coefficient seems wrong, for example the third term is $\dfrac{x^4}{4}$, while Wolfram gives me the coefficient as $\dfrac{x^4}{8}$.

Answer 1

\begin{align} \textbf{wrong: } & & & e^{-x^2/2}=1-\frac{x^2}{2}+\frac{x^4}{2\cdot2!}-\frac{x^6}{2\cdot3!}+\frac{x^8}{2\cdot4!}+\cdots \\[10pt] \textbf{right: } & & & e^{-x^2/2}=1-\frac{x^2}{2^1\cdot1!}+\frac{x^4}{2^2\cdot2!}-\frac{x^6}{2^3\cdot3!}+\frac{x^8}{2^4\cdot4!}+\cdots \end{align} This is because $\left( \dfrac {x^2} 2 \right)^4 = \dfrac{x^8}{2^4}, $ etc.

Answer 2

The general term for $e^x$ is $\dfrac{x^n}{n!}$.

For $e^{-x^2/2}$, this becomes $\dfrac{(-x^2/2)^n}{n!} =\dfrac{(-1)^nx^{2n}}{2^nn!} $.