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Does there exist a smooth function $f: \mathbb{R} \rightarrow \mathbb{R}$ whose Maclaurin series has a zero radius of convergence? In other words, does there exist a smooth function such that the Maclaurin series $P_n(x):=\sum_{n=0}^{\infty} a_n x^n$ for $f(x)$ satisfies the property that $ \lim_{n \to \infty} P_n(x) = \infty $ for all $x \neq 0$, where $a_n=\frac{f^n(0)}{n!}$ is the $n$-th partial sum of the series? If not why so?

P.S: The function $ f(x) = e^{-1/x^2}$ is not the candidate here, because it has a Maclaurin series that converges to zero for all $ x$, (though not to $f(x)$)

Celestina
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  • See https://math.stackexchange.com/questions/63050/every-power-series-is-the-taylor-series-of-some-c-infty-function for example. – Jens Schwaiger Aug 30 '24 at 05:34

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The result directly follows from Borel's theorem, but one explicit example is the function

$$ f(x) = \sum_{n=0}^\infty e^{-n}\cos(n^2x) $$

as found in Gelbaum's Counterexamples in Analysis, pp68-70. One can check that all derivatives converge for all $x$, and that all series coefficients exceed $1$ for large enough degree.

RDL
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