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Since $|e^{ix}|=1$ and $$f_n(x)=\sum_{k=0}^{n}\frac{(ix)^k}{k!}$$ approximates $e^{ix}$ well near the origin, we can know that $|f_n(x)|\approx 1>0$ for small $x$. My question is that whether there is a constant $C$ such that $$\min_{x\in\mathbb{R}} |f_n(x)|>C$$ holds for all $n$ (or at least for any $n\in S$, where $S$ has positive density in $\mathbb{N}$)?

My current attempt is to use the fact that $$\mathrm{Re} [f_n(x)] = \sum_{k=0\\k\ \mathrm{even}}^{n}(-1)^{k/2}\frac{x^k}{k!}$$ and $$\mathrm{Im}[f_n(x)] = \sum_{k=0\\k\ \mathrm{odd}}^{n}(-1)^{(k-1)/2}\frac{x^k}{k!}$$ are coprime in $\mathbb{Q}[x]$, which enables us to apply the Bezout's identity to deduce a lower bound for $|f_n(x)|$. However, this method can only prove that $|f_n(x)|>e^{-O(n\log(n))}$, which is not a uniform lower bound.

Are there any tricks to improve this method, or are there any method more analytic rather than using this fact in number theory?

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    My guess would be that $|f_n(x)|$ does not have a uniform positive lower bound. The asymptotic location of the zeros of $e_n(z)=\sum_{k=0}^n \frac{z^k}{k!}$ are known (see for example https://math.stackexchange.com/a/109605/42969), and that indicates that many $e_n$ have zeros on or close to the imaginary axis. – Martin R Aug 26 '21 at 17:25
  • Can you add your proof of $|f_n(x)|>e^{-O(n\log(n))}$ to the question? – Martin R Aug 28 '21 at 09:07

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