<s>[INST] Series expansion for $e^\frac{x^2}{2}$ $$e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}+ \cdots$$
$$e^{\frac{x^2}{2}}=1+(\frac{x^2}{2})^2+(\frac{x^4}{2\cdot2!})^2+(\frac{x^6}{2\cdot3!})^2+(\frac{x^8}{2\cdot4!})^2+\cdots$$
$=e^{\frac{x^2}{2}}=1+\dfrac{x^4}{8}+\dfrac{x^{8}}{4\cdot2!}-\dfrac{x^{12}}{8\cdot3!}+\dfrac{x^{16}}{16\cdot4!}+\cdots$
Wolfram's answer is $e^{\frac{x^2}{2}}=1+\dfrac{x^2}{2}+\dfrac{x^4}{4\cdot2!}+\dfrac{x^6}{8\cdot3!}+\dfrac{x^8}{16\cdot4!}+\cdots$
So what is wrong with my computation? Is Wolfram correct, I suspect that it cannot be wrong.
 [/INST] $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$
So you just plug in $x^2/2$ for $x$ and this gives you what Wolfram gave you.
</s>