# Can the sequence of derivatives $\{f^{(k)}(0)\}_{k\geq 1}$ be any sequence? [duplicate]

Let $\{a_{k}\}_{k\geq 1}$ be any sequence of real numbers, must there exist a smooth function $f:]-\epsilon,\epsilon[\rightarrow \mathbb{R}$ (for some positive $\epsilon$) such that for every positive integer $k\geq 1$, we have $f^{(k)}(0)=a_k ?$

Thank you a lot.

## marked as duplicate by Andrés E. Caicedo, Lord Shark the Unknown, Community♦Dec 2 '17 at 16:49

• In the case where $\limsup \sqrt[k]{|a_k|} = \epsilon$, we know that $f(x) = \sum\limits_{k=0}^{\infty}a_kx^k$ converges uniformly in $(-\epsilon^{-1},\epsilon^{-1})$ and is infinitely differentiable there, and $f^{(k)}(0) = k!a_k$, so if for your sequence $\limsup\sqrt[k]{\frac{|a_k|}{k!}}<\epsilon$, you get your request – Joshhh Dec 2 '17 at 15:02
• @Joshhh Suure.. I m aware of that,but Thank you so much for your comment – Amr Dec 2 '17 at 15:03

Yes, this is a special case of a theorem of Borel. Given any sequence $(a_n)$ there is a smooth function on $\Bbb R$ whose Maclaurin series is $\sum a_nx^n$.
I outline the proof. There is a smooth function $f:\Bbb R\to\Bbb R$ which equals $1$ on $[-1,1]$ and vanishes outside $[-2,2]$. Then consider $f(x)=\sum_{n=0}^\infty a_n x^n\phi(x/\varepsilon_n)$, where $\varepsilon_n$ is a sequence of positive numbers tending to zero. Then if $\varepsilon_n$ tends to zero rapidly enough, the series for $f$, and its formal derivatives of all orders will converge uniformly, and it will follow that $f$ has the given Maclaurin series.
• Thank you so much, I was thinking along those lines. (by playing with the frequencies $\epsilon_n$). By the way, why do you say "whose maclaurin series is ...".? What if the radius of convergence happens to be zero, then the maclaurin series wont be defined on a non degenerate interval as I requested ? Thanks again – Amr Dec 2 '17 at 15:22
• Aha so maclaurin series is just a name for for what I usually call formal power series (as in the ring of formal power series $\mathbb{R}[[x]]$) – Amr Dec 2 '17 at 15:25