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
$$
f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+o(x^2)=1+\frac{1}{2}c^2(\frac{1}{2}c^2x^2)^2?
$$
 A: What you have written so far is the definition of a Taylor polynomial of second degree. If you need the whole Taylor series expansion (and let's say around $x_0=0$) you should start from the known Taylor Series of the exponential function:
$$e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}, \; \forall u \in \mathbb{R}$$
And then just substitute in $u=\frac{1}{2} c^2 x^2$
A: 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
$$f''(0)x^2=c^2x^2$$
A: Just use Taylor expansion as if it were $e^X$ where $X = \frac{1}{2}c^2x^2$ hence:
$$e^X = \sum_{k = 0}^{+\infty} \frac{X^k}{k!}$$
$$e^{\frac{1}{2}x^2x^2} = \sum_{k = 0}^{+\infty} \frac{\left(\frac{1}{2}c^2x^2\right)^k}{k!} =\sum_{k = 0}^{+\infty}\frac{1}{2^k k!} (cx)^{2k}$$
The first terms are given by
$$1+\frac{c^2 x^2}{2}+\frac{c^4 x^4}{8}+O\left(x^5\right)$$
