Uniform convergence of analytic function on a closed and bounded interval Let $f:[0, a] \to \mathbb{R}$ be an analytic function. That would mean (by my book definition) that for every $c \in [0, a]$ there exist a neighborhood $U_c \subset [0,a]$ such that
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(c) } {n!} (x-c)^n. $$ for all $x \in U_c$. 
A bit later my book says (same function $f$): we can easily see that the power series defined as
$$f(t-x) = f(t) + \sum_{n=1}^{\infty} (-1)^n\frac{f^{(n)}(t) } {n!} x^n$$
converges for all $x$ in an interval that contains (maybe that's a typo, and the book means "is contained in"?) $[0,t]$, thus converges uniformly in $[0,t]$. I think this formula comes from the definition of analyticity on $t$ followed by a change of variables, and that's understable. But how about the convergence? Why from a local one, we get a bigger convergence? And how does that imply the uniform convergence? 
Thanks. 
 A: A useful and general set of results about power series is the following list:
If $ f(z) = \sum_{n=0}^\infty a_n z^n $ is a general power series in $z$, where $z$ and $a_n$ may be complex variables or real variables, then there exists $R\geqslant 0$, called the radius of convergence, such that 


*

*$\displaystyle \frac{1}{R} = \lim_{n \to \infty} r_n $ where each $r_n = \sup\{ \lvert a_k \rvert^{1/k}: k \geqslant n \}   $;  
Here the convention is used that if all the $r_n$ are infinite then $R=0$. Otherwise, the $r_n$ are positive and decreasing, so the limit exists.  If it is equal to zero $R$ is taken to be $\infty$.  
This is also written $\displaystyle \frac{1}{R} = \limsup_{n \to \infty} \lvert a_n\rvert^{1/n} $ 

*The power series converges for all $z$ with $\lvert z \rvert < R$;

*It diverges for all $z$ with $\lvert z \rvert > R$;

*If $r < R$ then the power series converges uniformly on the closed set $\{ z : \lvert z \rvert \leqslant r\}$;

*The power series may differentiated and integrated term by term inside its radius of convergence.


I think then your book is referring to property (4), by saying that if $t$ is real and positive and lies inside the radius of convergence, which is to say $t < R$, then convergence is uniform on $[0,t]$.  Moreover, if the series converges for any $x$ then by property (3) $R\geqslant \lvert x \rvert$.  It follows that if $\lvert t \rvert < \lvert x \rvert $, we also have $\lvert t \rvert < R$ and we can deduce uniform convergence for $\lvert z \rvert \leqslant \lvert t \rvert$.  This explains how convergence at a single point leads to a conclusion about uniform convergence in an interval.
The proofs of (1) to (5) are not too demanding.  For (1) through (3) see the Cauchy-Hadamard theorem.  Property (4) follows because for $\lvert t \rvert <R$ the tail of the series can be dominated by a larger $t'$ with $\lvert t \rvert < \lvert t' \rvert < R$. Property (5) has been covered on math.stack.exchange here and here but you don't need these in your case.
