Find all $A$ such that $f(x) = \sum_{n \in A} \frac{x^n}{n!}$ is bounded for $x<0$

Asked Mar 19 '19 at 16:20

Active Oct 10 '20 at 19:34

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Let $\mathbb{R}_- = \{ x \in \mathbb{R} \mid x \leq 0 \}$, $\mathbb{N} = \{0,1, 2, ...\}$. Find all subsets $A \subset \mathbb{N}$ such that the function $f : \mathbb{R}_- \to \mathbb{R}$,$$f(x) := \sum_{n \in A} \frac{x^n}{n!}$$ is bounded.

Note that the function $f$ is defined on $\mathbb{R}_-$ and not all of $\mathbb{R}$. Moreover $f$ looks like a lot $x \mapsto e^x$. So one set for which $f$ is bounded is $A = \mathbb{N}$, as well as $A=\{0\}$.

Moreover we need to find an equilibrium in $A$ in the following sense: if there are too many even numbers $f$ will not be bounded, and if there are too many odd numbers $f$ also can't be bounded. So I think there is a $K \in \mathbb{R}$ such that if we want $f$ to be bounded then the number of even numbers in $A$ is at most $K \times$ the number of odd numbers in $A$.

Thank you.

edited Oct 10 '20 at 19:34

Integrand

8,078

asked Mar 19 '19 at 16:20

mouargmouarg

What is $\Bbb {R}_ -$? – little o Mar 19 '19 at 16:24
1

${ a \in \mathbb R : a < 0}$ – GEdgar Mar 19 '19 at 16:25
$\mathbb N, {0}, \mathbb N \setminus {0},\varnothing$. These four. Any others? – GEdgar Mar 19 '19 at 16:26
I'm not sure what the answer is, but don't $A=k\mathbb{N}$ where $k$ is odd not also work? – Keen-ameteur Mar 19 '19 at 17:34
@Keen-amateur let's try $A=3\mathbb{N}$. You get $$\sum_{n=0}^\infty \dfrac{x^{3n}}{(3n)!} = \dfrac{1}{3}\left(e^x+2e^{-x}\cos\left(\dfrac{\sqrt{3}}{2}x\right)\right)$$ which does not converge as $x \to -\infty$. – SlipEternal Mar 20 '19 at 13:15
Vaguely related: https://math.stackexchange.com/q/3728903 – Integrand Oct 10 '20 at 19:35
I think that only subsets like $A_n={k\in\mathbb{N}:k\ge n}$ can be proved to give $f(x)$ bounded at $-\infty$. – Raffaele Oct 10 '20 at 20:26
^ $A_2 = {k \in \mathbb{N} : k \ge 2}$ yields $f(x) = e^x-1-x$ which is not bounded as $x \to -\infty$. Similarly, $A_n$ for $n \ge 2$ doesn't work. – JimmyK4542 Oct 10 '20 at 21:25
What about finite set $A$? – W. Wongcharoenbhorn Oct 10 '20 at 21:52
@W.Wongcharoenbhorn polynomials aren't bounded over $\mathbb{R}_-$. – Integrand Oct 10 '20 at 22:27

Find all $A$ such that $f(x) = \sum_{n \in A} \frac{x^n}{n!}$ is bounded for $x<0$

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