<s>[INST] What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$? What is the Taylor's expansion of $f(x)=\exp(\frac{1}{2}c^2x^2)$ with a constant $c$.
Note that $f'(x)=\exp(\frac{1}{2}c^2x^2)c^2x$ and $f''(x)=\exp(\frac{1}{2}c^2x^2)(c^2x)^2+\exp(\frac{1}{2}c^2x^2)c^2$. Then $f'(0)=0$ and $f''(0)=c^2$.
Is that
$$
f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+o(x^2)=1+\frac{1}{2}c^2(\frac{1}{2}c^2x^2)^2?
$$
 [/INST] Alright, first to simplify the problem you are asking what the Taylor Series expansion of $\exp(ax^2)$ for a constant $a$. In your case, simply take $a=\frac{1}{2}c^2$. But then
$$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$
$$\Rightarrow \exp(ax^2)=\sum_{n=0}^\infty \frac{(ax^2)^n}{n!}=\sum_{n=0}^\infty \frac{a^n x^{2n}}{n!}$$
Then for $a=\frac{1}{2}c^2$ this becomes
$$\exp(ax^2)=\sum_{n=0}^\infty \frac{c^{2n} x^{2n}}{2^nn!}=1+\frac{c^2x^2}{2}+\cdots$$
Note that this is different than what you wrote in your question as you erronously stated
$$f''(0)x^2=c^2(\frac{1}{2}c^2x^2)^2$$
when it should have been
$$f''(0)x^2=c^2x^2$$
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