Uniqueness of the Taylor series If given two real (or complex) valued function $f$ and $g$, and $f\left(x\right)=g\left(x\right)$ for all $x$, then there Taylor-series are the same. Is it true backward? When given a Taylor-series (which is convergent for all "relevant" $x$) is it define one and only one function? I think it is an equivavelent question if I ask when do the $f^{\left(k\right)}$ derivatives $(k\in\mathbb{N})$ define an $f\in C^{\infty}$ function? (Here $C^{\infty}$ denotes a class of functions which are differentiable infinitely many times.)
What if the Taylor-series is not convergent? Could you give me an example when a function is $f\in C^{\infty}$ but its Taylor-series is not convergent? For this last question is the real moment generating function of a $\xi$ lognormal distributed random variable is a good example?
 A: Representations, such as integrals, differential equations, Taylor series, Dirichlet series, Fourier transforms, etc., each have their own unique properties and theorems. Some allow for uniqueness of representation, some don't. Others will agree with the underlying function on the whole domain, while some will not. Some representations have different kinds of properties on one domain vs another domain.
You have to be very careful with each representation and domain. Take for example,

Given any sequence of complex/real numbers, $a_n$, define $r=1/\limsup \sqrt[n]{|a_n|}$, then the function defined by $f(z)=\sum a_n z^n$, converges absolutely for all $|z|<r$ and diverges for all $|z|>r$. Moreover $f$ is infinitely differentiable on $|z|<r$ and the coefficients are unique (meaning that if we define $g(z)=\sum b_nz^n$ and $g(z)=f(z)$ on all of $|z|<r$, then $a_n=b_n$ for all $n$).

This says nothing about the existence or non-existence of a function represented by the series. For example,
$$\frac{1}{1-z} = 1+z+z^2+z^3+\cdots$$
The left had side is defined for all $z\not=1\in\mathbb{C}$, but the right had side (by the theorem just given) diverges for all $|z|>1$. This means that this Taylor series is useless outside it's radius of convergence and does not equal the underlying function that created the Taylor series. Note that $\frac{1}{1-z}$ is indeed infinitely differentiable on all of $z\not=1\in\mathbb{C}$, and yet has a Taylor series that does not equal it everywhere.
An example of domains, take the strictly different theorems for $\mathbb{R}$ and $\mathbb{C}$ concerning Taylor series.

Suppose $f:B(z_0,r)\to\mathbb{C}$ is a complex differentiable function. Then $f$ is infinitely differentiable and the function $g(z)=\sum \frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$ converges absolutely on $B(z_0,r)$. Moreover, $f(z)=g(z)$ for all $z$ on $B(z_0,r)$.


There exists a function, $f:\mathbb{R}\to\mathbb{R}$, that is infinitely real differentiable on $\mathbb{R}$; such that the function $g(x) = \sum \frac{f^{(k)}(0)}{k!}x^k$ exists and converges on all of $\mathbb{R}$; but $f(x)\not=g(x)$ for all $x\not=0\in\mathbb{R}$. As @lulu pointed out, the function $f(x)=e^{-1/x^2}$ satisfies this property (you take the function to be $0$ at $0$ obviously).

The concepts of complex differentiable and real differentiable are strictly different in the context of Taylor series.
