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This question already has an answer here:

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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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 $\endgroup$ – Joshhh Dec 2 '17 at 15:02
  • $\begingroup$ @Joshhh Suure.. I m aware of that,but Thank you so much for your comment $\endgroup$ – Amr Dec 2 '17 at 15:03
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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.

For a more general result, see Theorem 1.2.6 in Volume 1 of Hormander's The Analysis of Linear Partial Differential Operators.

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  • $\begingroup$ 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 $\endgroup$ – Amr Dec 2 '17 at 15:22
  • $\begingroup$ Can you also outline why the series for f and its formal derivatives will converge uniformly ? $\endgroup$ – Amr Dec 2 '17 at 15:24
  • $\begingroup$ The Maclaurin series of a function, needn't be convergent, and even if it is, need not converge to the given function. $\endgroup$ – Lord Shark the Unknown Dec 2 '17 at 15:24
  • $\begingroup$ 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 $\mathbb{R}[[x]]$) $\endgroup$ – Amr Dec 2 '17 at 15:25
  • $\begingroup$ @Amr For more details see en.wikipedia.org/wiki/Borel%27s_lemma $\endgroup$ – Lord Shark the Unknown Dec 2 '17 at 15:27

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