$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$

We start from the following really strange

**Theorem 1.**
Let $\{a_m\}$ with $m=0,1,\ldots $ be an arbitrarily sequence. Then there exists $f\in C^\infty(\bR)$ such that
\begin{equation}
f^{(m)}(0)=a_m\qquad \forall m=0,1,\ldots
\label{eq-1.2.1}
\end{equation}

*Proof (optional reading).*
Let $\varphi\in C_0^\infty ([-2,2])$ (which means that $\varphi\in C^\infty (\bR)$ and $\varphi(x)=0$ as $x\notin [-2,2]$). Assume that $\varphi(x)=1$ as $x\in [-1,1]$.Such functions exist.

Let us consider series \begin{equation} f(x):= \sum_{n=0}^\infty \varphi (b_n x)\, a_n \frac{x^n}{n!}. \label{eq-1.2.2} \end{equation} We claim that for some sequence $b_n\to +\infty $ this series converges (with all its derivatives) to a function $f$ satisfying (\ref{eq-1.2.1}).

Observe that for given $x\ne 0$ only a finite number of the terms in (\ref{eq-1.2.1}) differ from $0$ and therefore series (\ref{eq-1.2.1}) indeed converges to function $f$ which is $C^\infty$ may be except $0$. Clearly, as $x=0$ this series also converges to $a_0$.

On the other hand, assume that $b_{n+1}\ge 2 b_n $ for all $n$. Then all terms with $b_nx\ge 1$ vanish and terms with $n\ge 2$ do not exceed $|a_n |b_n^{-n+1} /n!\times |x|$ and if we assume in addition that $b_n\ge |a_n|$ for all $n$ then $|f(x)-a_0 \varphi (b_0 x)|\le C|x|$ and (\ref{eq-1.2.1}) holds for $m=0$.

Consider now $f^{(m)}(x)$ for $m\ge 1$: \begin{multline*} f^{(m)}(x)= \sum_{0\le k\le m} \frac{m!}{k!(m-k)!} \sum_{n=k}^\infty\varphi^{(m-k)} (b_nx) a_n b_n^{m-k} (b_nx) \frac{x^{n-k}}{(n-k)!} \end{multline*} and similarly to the previous arguments sum of the terms with $n\ge m+1$ does not exceed $C_m|x|$ (prove it by yourself as a bonus problem!).

Therefore $|f^{(m)}(x)-g_m^{(m))}(x)|\le C|x|$ where $g_m(x)$ is defined by the same sum (\ref{eq-1.2.1}) restricted to $n\le m$. Consider $|x|\le \frac{1}{2}b_m^{-1}$; then $\varphi(b_nx)=1$ as $n\le m$ and and then $g_m^{(m))}(x)= a_m$ which implies (\ref{eq-1.2.1}).

So let $f(x)$ be some function satisfying (\ref{eq-1.2.1}). Consider its Taylor series: \begin{equation} \sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!} = \sum_{n=0}^\infty a_n\frac{x^n}{n!}. \label{eq-1.2.3} \end{equation}

*Question.* Does this series necessarily converge?

The answer to the first question is simple: (\ref{eq-1.2.3}) converges for $|x|<\varepsilon$ if $|a_m|\le C\varepsilon^{-m}m!$ for all $m$. If the latter is not fulfilled, this series is not converging as either $x=\varepsilon$ or $x=-\varepsilon$.

*Question.* Does this series necessarily converge to $f(x)$?

The answer is also simple: (\ref{eq-1.2.3}) converges for $|x|<\varepsilon$ to $f(x)$ iff $f(z)$ is an analytic function in the disk $D(0,\varepsilon)={z:|z|<\varepsilon}$.

**Example.** $f(x)=\exp(-1/x^2)$ (as $x\ne 0$; $f(0)=0$) is flat at $0$: $f^{(m)}(0)=0$ for all $m$.

*Question.* If (\ref{eq-1.2.3}) does not converge for $|x|<\varepsilon$ to $f(x)$, does it make any sense?

The answer is "Yes" \begin{equation} f(x)\sim \sum_{n=0}^\infty a_n\frac{x^n}{n!} \label{eq-1.2.4} \end{equation} but in what sense? The more detailed answer is:

For each $N$ and $m\le N$ \begin{equation} |\bigl(f(x)- S_N(x)\bigr)^{(m)}|\le C_{N}|x|^{N-m}\qquad \text{as }\ |x|\le c \label{eq-1.2.5} \end{equation} where \begin{equation} S_N(x):=\sum_{n=0}^{N-1}a_n\frac{x^n}{n!} \label{eq-1.2.6} \end{equation} is a partial sum.

So, Theorem 1 could be reformulated as

**Theorem 2.**
Let $\{a_m\}$ with $m=0,1,\ldots $ be an arbitrarily sequence. Then there exists $f\in C^\infty(\bR)$ such that (\ref{eq-1.2.4}) holds in the sense of(\ref{eq-1.2.5})-(\ref{eq-1.2.6}).

Consider two functions: $f(x)$ and $g(x)$ as $x\to x^*$.

Here $x^*$ may be a real number $x^*=\bar{x}$, or $x^*= \bar{x}\pm 0$ (the limit from the righ or left), or $x^*=\pm \infty$ (infinity of sign $+$ or $-$), or just $x^*=\infty$ (just infinity, so $x\to \infty\iff |x|\to +\infty$).

**Definition 1.**
We say that

- $f$ is
*$O$-large of*$g$, $f=O(g)$ if $|f|\le C|g|$ (i.e. $\bigl|\frac{f}{g}\bigr|\le C$) as $x\to x^*$ - $f$ is
*$O$-small of*$g$*, $f=o(g)$ if $\lim \frac{f}{g}=0 as $x\to x^*$. - $f$ is
*asymptotically equivalent to*$g$, $f\sim g$ if $\lim \frac{f}{g}=1 as $x\to x^*$. - $f$ is
*of magnitude of*$g$, $f\asymp g$ if $f=O(g)$ and $g=O(f)$ as $x\to x^*$.

Obviously $f\sim g$ iff $f-g=o(f)$. Also $f\asymp g$ iff $C^{-1}\le \bigl|\frac{f}{g}\bigr|\le C$.

One of our main tasks would be for given albeit difficult to calculate $f$ find $g$ which is easy to calculate such that $g\sim f$ (as $x\to x^*$). We call $g$ *the principal part of the asymptotics* (or *the main part of the asymptotics*) and $f-g$ *the remainder*.

If we prove that $f-g=O(h)$ or $f-g=o(h)$ with some $h=h(x)$ we call it *remainder estimate*. Clearly we would like to have as small as possible remainder estimate (then we call asymptotics *sharp*).

Even better to find $f$ in the form of *asymptotic series* or *asymptotic decomposition*
\begin{equation}
f(x)\sim
\sum_{n=0}^\infty g_n(x)
\label{eq-1.2.7}
\end{equation}
in the sense adapted to the problem we study.