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Define $f$ as the well known bump function:

$$ f(x)=\begin{cases}\exp\left(\frac{1}{x^2-1}\right) & |x| \lt 1 \\0 & |x|\ge 1\end{cases} $$

Does the series

$$ \sum_{k=0}^{\infty} f^{(k)}(0) $$

converge?

Answers such as this show that the derivatives of any test function cannot be bounded, however they don't give any lower bounds that would prove a series such as the present one divergent. Numerically calculating the first few partial sums shows that the series seems to be growing rapidly but I have no idea how to find a proof.

rapgru
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1 Answers1

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Note that for any analytic function $f(z)=\sum a_nz^n$ on a neighborhood of zero, one cannot have $\sum_{k=0}^{\infty} f^{(k)}(0)$ converge or even $f^{(k)}(0)$ bounded unless the function is entire since $k!a_k=f^{(k)}(0)$ so if $|f^{(k)}(0)| \le M$ one has $|a_k| \le M/k!$ so the power series has infinite radius of convergence.

Trivially here $f$ (which is analytic near $0$ as a composition of analytic functions) has singularities at $\pm 1$ so just by general theory $|f^{(k)}(0)|$ is highly unbounded (for even $k$ as the odd derivatives at zero are $0$ since $f$ even function).

A more interesting question is if $\sum_{k=0}^{\infty} f^{(k)}(0)/k!$ converges (it is Abel summable to zero but convergence is trickier)

Conrad
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  • How would I show that f(x) is analytic on some neighbourhood of zero? The only way I know would be showing that the Taylor series error vanishes, but we don't now much about the derivatives here. – rapgru Oct 19 '24 at 19:55
  • it is a composition of analytic functions $\exp, \frac{1}{x^2-1}$ so it is obviously analytic – Conrad Oct 19 '24 at 20:35