How can I prove that a Taylor series represents a function $f(x)$ in $\mathbb{R}$ given only the form of the derivatives and the Taylor series of $f$? How can I prove that a Taylor series represents a function $f(x)$ in $\mathbb{R}$ given only the form of the derivatives and the Taylor series of $f$?
$f(x)$ is infinitely differentiable, I know the nth derivation, as well as the form of the Taylor series $T(x)$.
I have to show that the remainder converges to zero.
The form of the function $f(x)$ is not given, so considering the limit of the remainder $R_{n,f,a}(x)=f(x)-T_n(x)$ is not possible.
My idea:
In such a case, does it make sense to consider the limit of the Lagrangian form of the remainder:
Let $x\in \mathbb{R}$ be arbitrary.
There is a $\xi\in (a,x)$ such that: $R_{n,f,a}(x)=\frac{f^{n+1}(\xi)}{(n+ 1)!}(x-a)^{n+1}$
Can one conclude from $\frac{f^{n+1}(\xi)}{(n+1)!}(x-a)^{n+1}\rightarrow 0$ that $T(x)$ the representation of $f(x)$ for all $x\in \mathbb{R}$?
 A: But this just isn't true. It is known that the class of smooth functions (=infinitely differentiable) is not the same as the class of (real) analytic functions.
The function: $$\Bbb R\ni x\mapsto\begin{cases}\exp(-x^{-2})&x\neq0\\0&x=0\end{cases}\in\Bbb R$$Is smooth and its Taylor expansion at zero is identically zero. Yet, this function is not identically zero on any neighbourhood of zero. Thus it is not analytic.
Anyway - you need more information, such as an bound on the remainder terms. For example, having bounded derivatives is a sufficient condition.
Proof that the function's Taylor coefficients are all zero:

Denote this function as $f$. $$\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{1}{h}\exp(-h^{-2})=\lim_{|t|\to\infty}t\exp(-t^2)=0$$
And $f'(x)=2x^{-3}\exp(-x^{-2})=P_1(x^{-1})\exp(-x^{-2})$ for some polynomial $P_1$, on $x\neq0$.
Given that $f^{(n)}(x)$ exists and equals $P_n(x^{-1})\exp(-x^{-2})$ for some polynomial $P_n$, $x\neq0$, and $f^{(n)}(0)=0$, then I claim $f^{(n+1)}$ exists and equals $P_{n+1}(x^{-1})\exp(-x^{-2})$ for $x\neq0$, and is zero at $x=0$.
$$f^{(n+1)}(0)=\lim_{h\to0}\frac{1}{h}P_n(h^{-1})\exp(-h^{-2})=\lim_{|t|\to\infty}tP_n(t)\exp(-t^2)\overset{L.H.}{=}0$$
And $f^{(n+1)}(x)$, $x\neq0$, exists and equals: $$2x^{-3}P_n(x^{-1})\exp(-x^{-2})-x^{-2}P_n(x^{-1})\exp(-x^{-2})=P_{n+1}(x^{-1})\exp(-x^{-2})$$For some new and obviously defined polynomial $P_{n+1}$.
It follows by induction that $f^{(n)}(0)=0$ for all $n$.

