Under what conditions does $f$ equal its Taylor series about $a$ on the closed interval $[a,b]$. I am interested in the necessary (but perhaps not sufficient) conditions in which $f$ equals it's Taylor series about $a$ on the entire closed interval $[a,b]$. More specifically, I'm more interested in continuity/differentiability conditions. 
I first thought that perhaps f has to be infinitely differentiable on $[a,b]$ for f's Taylor series about $a$ to converge on $[a,b]$, but after a while, I realized this is not the case. For example, take $f(x)=\arcsin(x)$. This function is not differentiable on $[0,1]$ (not diffrentiable at $x=1$) but it's Taylor series about $x=0$ converges to $\arcsin(x)$ on $[0,1]$.
I realize that what I'm interested in is somewhat related to analytic functions, but it is not exactly the same. Analytic functions are locally convergent on an open interval. 
I think that $f$ has to be infinitely differentiable on $[a,b)$ but not sure how to prove this. It's pretty easy to see it must be so at $x=a$ but I don't know if it does have to be so on the open interval $a,b)$. 
 A: It is not sufficient that $f\in C^\infty$.  For example, let $f$ be the function defined as 
$$f(x)=\begin{cases}e^{-1/x^2}&x\ne0\\\\0&,x=0\end{cases}$$ 
It is easy to show that $f\in C^\infty$.  But for each $n$, $f^{(n)}(0)=0$.  And hence, the Taylor series for $f$ is $0$.  So, $f(x)$ cannot be represented by its Taylor series.

Taylor's Formula for a function that is differentiable $k$ times at $x=a$ is given by
$$f(x)=\sum_{n=0}^k \frac{f^{(n)}(a)}{n!}(x-a)^n+h_k(x)(x-a)^k$$
where $\lim_{x\to a}h_k(x)=0$.
If $f\in C^\infty$ in a neighborhood of $a$, and $\lim_{k\to\infty}h_k(x)(x-a)^k\to 0$ in that neighborhood of $a$, then $f(x)$ can be represented by its Taylor series.  We say then that $f$ is real analytic around $a$.

Note that for the function $f(x)=e^{-1/x^2}$ for $x\ne0$ and $f(0)=0$, $h_k(x)x^k=e^{-1/x^2}$ and $h_k(x)x^k$ does not approach $0$ as $k\to \infty$.  Here in fact, the remainder term is $f(x)$ itself.
A: If the Taylor series of $f$ about $a$ converges on $[a,b]$, then the radius of convergence of that series is at least $r=b-a$, and the sum of that series (call it $g(z)$) is analytic in the open disc of radius $r$ about $a$.  If the series converges to $f$ on $[a,b)$, that implies $f=g$ on $[a,b)$.  Actually $f$ should be defined in some open interval around $a$ if we want it to be differentiable at $a$, but there's no requirement for $f$ to agree with $g$ on $(a-\epsilon, a)$. 
In addition, we want the series to converge (perhaps only conditionally) to $f(b)$ at $b$.  Abel's theorem implies that $f$ is continuous from the left at $b$, but there's not much else we can say there: $g$ need not be analytic at $b$. 
