$f(x) = e^{-x^2}$ series representation Let $f(x) = e^{-x^2}$, defined for $x \in \mathbb{R}$. Find a series representation of a function $F:\mathbb{R} \to \mathbb{R}$ such that $F(0)=0$ and $F'(x)=f(x)$ for each $x$.
I know that the answer to this is in some form of a Taylor series, but I don't even know where to start beyond that...
 A: Since this looks like homework, let me do a similar problem instead of the one you ask for. Let $f(x)=\sin(x^2)$. Since the Taylor series expansion (around $x=0$) for $\sin x$ is 
$$\sin x  =\sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!}= x - \frac{x^3}{3!}+\frac{x^5}{5!}-\ldots,$$
we may evaluate this series at $x^2$ to obtain
$$\sin x^2  =\sum_{n=1}^\infty \frac{(x^2)^{2n-1}}{(2n-1)!}= \sum_{n=1}^\infty \frac{x^{2(2n-1)}}{(2n-1)!}= \sum_{n=1}^\infty \frac{x^{4n-2}}{(2n-1)!}=x^2 - \frac{x^6}{3!}+\frac{x^{10}}{5!}-\ldots.$$
If we are looking for a function $F(x)$ such that $F(0)=0$ and $F'(x)=f(x)$, then by the Fundamental Theorem of Calculus we know that $F(x)=\int_0^x f(t) dt$. Thus,
$$F(x)=\int_0^x f(t) dt = \int_0^x \sum_{n=1}^\infty \frac{t^{4n-2}}{(2n-1)!} dt = \sum_{n=1}^\infty \int_0^x \frac{t^{4n-2}}{(2n-1)!} dt = \sum_{n=1}^\infty \frac{x^{4n-1}}{(4n-1)(2n-1)!}.$$
Using the ratio test you can show that the radius of convergence of the series is infinite, so $F(x)$ is defined for every real number $x$ and, by construction, $F(0)=0$ and $F'(x)=f(x)=\sin x^2$.
