Let $E \subseteq \R $, $f: E \to \R $.
If $a$ is an interior point of $E$, we say that $f$ is real analytic at $a$ if $\exists r > 0, (a-r , a+r) \subseteq E $ such that $\exists \sum_{ n=0 }^{ \infty }c_n(x - a)^n $ centered at $a$ which has a radius of convergence greater than or equal to $r$, and which converges to $f$ on $(a -r, a+r)$.
If $E$ is an open set, and $f$ is real analytic at every point $a$ of $E$, we say that $f$ is real analytic on $E$.
Let $E \subseteq \R$. We say a function $f: E \to \R$ is once differentiable on $E$ iff it is differentiable.
$\forall k \geq 2 $ we say that $f$ is $k$ times differentiable on $E$, iff $f$ is differentiable and $f' $ is $k-1$ times differentiable.
If $f$ is $k$ times differentiable, we define the $k$th derivative $f^{(k)}: E \to \R $ by the recursive rule:
$$f^{(k)} = \begin{cases} f &k=0 \br f' &k=1\br (f^{(k-1)})' &k \geq 2 \br \end{cases}$$
A function is said to be infinitely differentiable iff it is $k$ times differentiable for every $k \geq 0$.
Let $E \subseteq \R$, let $a$ be an interior point of $E$, and let $f$ be a function which is real analytic at $a$, thus there is an $r > 0$ for which we have the power series expansion
$$f(x) = \sum_{ n=0 }^{ \infty }c_n (x -a)^n $$
for all $x \in (a -r, a +r). $ Then for every $k \geq 0 $, the function $f$ is $k$-times differentiable on $(a-r, a+r) $, and for each $k \geq 0 $ the $k$th derivative is given by
$$\begin{align*} f^{(k)}(x) &= \sum_{ n=0 }^{ \infty } c_{n+k}(n+1)(n+2)…(n+k)(x-a)^n \br &= \sum_{ n=0 }^{ \infty }c_{n+k} \frac{(n+k)! }{ n! }(x-a)^n \end{align*}$$
for all $x \in (a-r, a+r) $.
Let $E$ be an open subset of $\R$, and let $f: E \to \R $ be a real analytic function on $E$. Then $f$ is infinitely differentiable on $E$. Also, all derivatives of $f$ are also real analytic on $E$. Also, all derivatives of $f$ are also real analytic on $E$.
Let $E$ be a subset of $\R$, let $a$ be an interior point $E$, and let $f:E \to \R$ be a function which is real analytic at $a$ and has the power series expansion
$$f(x) = \sum_{ n=0 }^{ \infty }c_n(x -a)^n $$
for all $x \in (a- r, a+r)$ and some $r > 0$. Then for any integer $k \geq 0$, we have
$$f^{(k)}(a) = k!c_k $$
In particular, we have Taylor’s formula
$$f(x) = \sum_{ n=0 }^{ \infty } \frac{ f^{(n)}(a)}{ n! }(x -a)^n $$
for all $x$ in $(a-r, a+r)$.
The power series $\sum_{ n=0 }^{ \infty } \frac{ f^{(n)}(a)}{ n! }(x -a)^n $ is sometimes called the Taylor series of $f$ around $a$. Taylor’s formula thus asserts that if $a$ function is real analytic, then it is equal to its Taylor series.
Let $E$ be a subset of $\R$, let $a$ be an interior point of $E$, and let $f:E \to \R$ be a function which is real analytic at $a$. Suppose that $f$ has two power series expansions
$$f(x) = \sum_{ n=0 }^{ \infty }c_n(x-a)^n $$
and
$$f(x) = \sum_{ n=0 }^{ \infty } d_n(x-a)^n $$
centered at $a$, each with a non-zero radius of convergence. Then $c_n = d_n $, for all $n \geq 0 $.