Derivatives of a function equal to an arbitrary sequence of real numbers at some point I've just come up with this question: for any arbitrary sequence of real numbers $(a_{n})_{n \in \mathbb{N}}$, does there exist a function $f : \mathbb{R} \to \mathbb{R}$ such that the following condition is satisifed:
$$
\exists x_{0} \in \mathbb{R}. \forall k \in \mathbb{N}. \ f^{(k)}(x_0) = a_k
$$
We assume that $\mathbb{N} = \{ 1, 2, 3, \dots \}$. I've given it a thought, but I haven't come up with anything more than a power series $f(x) = \sum_{n=0}^{\infty} a_{n}x^{n}$ and considering $x_{0} = 0$. That would, however, not always work, e.g. when the radius of convergence of the series is equal to $0$.
 A: Fix a smooth function $\varphi $ such that $ \mathbf{1}_{[-\frac{1}{2},\frac{1}{2}]}(x) \leq \varphi(x) \leq \mathbf{1}_{[-1,1]}(x) $ for all $x \in \mathbb{R}$. Also, for any $\varepsilon > 0$, we let
$$
\varphi^{(0)}_{\varepsilon}(x)
= \varphi(x/\varepsilon)
\qquad\text{and}\qquad
\varphi^{(-n)}_{\varepsilon}(x)
= \int_{0}^{x} \frac{(x-t)^{n-1}}{(n-1)!} \varphi^{(0)}_{\varepsilon}(t) \, \mathrm{d}t, \qquad n \geq 1.
$$
We observe that:

*

*$ \frac{\mathrm{d}}{\mathrm{d}x} \varphi^{(-n)}_{\varepsilon}(x) = \varphi^{(-n+1)}_{\varepsilon}(x)$, justifying the choice of the notation.


*For any integer $k$, we have $\varphi^{(k)}_{\varepsilon}(0) = \begin{cases} 1, & k = 0, \\ 0, & k \neq 0. \end{cases}$


*$ \bigl| \varphi^{(-n)}_{\varepsilon}(x) \bigr|
\leq \int_{0}^{\varepsilon} \frac{(|x|-t)^{n-1}}{(n-1)!} \, \mathrm{d}t
\leq \varepsilon \frac{|x|^{n-1}}{(n-1)!} $.
Now let $(a_n)_{n\geq 0}$ be an arbitrary sequence of real numbers. Then we choose $(\varepsilon_n)_{n\geq 0}$ so that $\varepsilon_0$ is an arbitrary positive real number and $\varepsilon_n$ satisfies
$$ |a_n| \biggl( \sup_{\substack{|x| \leq n, |k| \leq n}} \bigl| \varphi^{(-k)}_{\varepsilon_n}(x) \bigr| \biggr) \leq \frac{1}{2^n}, \qquad n \geq 1. $$
(In light of the upper bound of $\bigl| \varphi^{(-n)}_{\varepsilon}(x) \bigr|$ above, we may let $\varepsilon_n = \frac{(n-1)!}{n^{n-1}2^n(|a_n| + 1)}$.) Then by the construction, the function $f$ defined by
$$ f(x) = \sum_{k=0}^{\infty} a_k \varphi_{\varepsilon_k}^{(-k)}(x) $$
is smooth and term-by-term differentiation is valid. (This is because the series $\sum_{k=0}^{\infty} a_k \varphi_{\varepsilon_k}^{(n-k)}(x)$ converges locally uniformly for any $n = 0, 1, \ldots$) Moreover,
$$ f^{(n)}(0)
= \sum_{k=0}^{\infty} a_k \varphi_{\varepsilon_k}^{(n-k)}(0)
= a_n. $$

Addendum. By modifyin this argument, for any sequences $(x_n)$ and $(a_n)$ you can find a smooth function $f$ such that $f^{(n)}(x_n) = a_n$ for all $n \geq 0$.
