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Let $f(t) = e^{-|t|}$, $t \in \mathbb{R}$. The function $f$ has the everywhere convergent power series expansion

$$f(t) = \sum_{n=0}^\infty \frac{(-1)^n|t|^n}{n!}.$$

I would like to know whether there exists some $C > 0$ independent of $t$ so that for all integers $M \ge 0$, $$\Big|\sum_{n=0}^M \frac{(-1)^n|t|^n}{n!} \Big| \le C.$$

If we were to instead consider the power series expansion, for, say, $e^t$, then we do not expect there to be a uniform constant, because $e^t$ becomes too large as $t \to \infty$. However, since $f$ is decaying at both $\pm \infty$, I suspect such a $C$ can be found, and one will need to rely somehow on the cancellation of the positive and negative terms in the series.

Hints or solutions are greatly appreciated!

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  • $\begingroup$ Even for some fixed (nonzero) M, do you think there exists many polynomials that are bounded on the whole real line? $\endgroup$ – Did Mar 1 '18 at 15:31
  • $\begingroup$ How about $C=1$? Isn't it true that $$\Big|\sum_{n=0}^M \frac{(-1)^n|t|^n}{n!} \Big| \leq \sum_{n=0}^M \Big|\frac{(-1)^n|t|^n}{n!}\Big|$$ $$= \sum_{n=0}^M \frac{|t|^n}{n!}$$ $$\leq \sum_{n=0}^{\infty} \frac{|t|^n}{n!}$$ $$=e^{-|t|}$$ $$\leq1$$ Am I missing something? $\endgroup$ – MPW Mar 1 '18 at 15:33
  • $\begingroup$ @MPW Yes - the sum you wrote down is $e^{|t|}$, not $e^{-|t|}$. $\endgroup$ – user296602 Mar 1 '18 at 15:42
  • $\begingroup$ Ah, indeed, my bad. $\endgroup$ – MPW Mar 1 '18 at 15:46
  • $\begingroup$ That is not a power series expansion. $\endgroup$ – zhw. Mar 1 '18 at 16:15
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For any $M>0$ the term $\sum_{n=0}^M \frac{(-1)^n|t|^n}{n!}$ is a polynomial in $|t|$ which is unbounded as $|t| \to \infty$.

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Hint. Note that by letting $t=M\in\mathbb{N}^+$, we have that $$\lim_{M\to +\infty}\Big|\sum_{n=0}^M \frac{(-1)^n M^n}{n!} \Big|=+\infty$$

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