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It's well-known that exponential growth eventually overtakes polynomial growth (link, link).

So for any non-negative integer $d$ and positive $\epsilon$, there exists $t^* \ge 0$ for which

$$ 1 + t + \frac{t^2}{2!} + \ldots + \frac{t^d}{d!} \le e^{\epsilon t} $$

for all $t \ge t^*$.

In other words, it takes $t^*$ seconds for the exponential to catch up with the polynomial.

I'd like to know if there's a closed-form expression for $t^*$ in terms of $d$ and $\epsilon$. Something like

$$ t^* = d/\epsilon. $$

Thoughts?

Community
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ConvexMartian
  • 347

2 Answers2

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In general there is no such closed-form expression, but closed-form bounds are possible.

EDIT: Of course if $\epsilon \ge 1$, the inequality is true for all $t \ge 0$, so consider the case $0 < \epsilon < 1$. I'll denote the left side as $M_d(t)$. We can consider this as $e^{t} \mathbb P[X \le d]$, where $X$ is a random variable having Poisson distribution with parameter $t$. Now for any $\mu > 0$, $$\mathbb P[X \le d] = \mathbb P\left[e^{-\mu X} \ge e^{-\mu d}\right] < e^{\mu d} \mathbb E\left[ e^{-\mu X}\right] = e^{\mu d} e^{-t + t \exp(-\mu)} $$ Thus it suffices to have $ \mu d + t \exp(-\mu) \le \epsilon t $, i.e. $\exp(-\mu) < \epsilon$ with $$ t \ge \dfrac{\mu d}{\epsilon-\exp(-\mu)}$$ It's not quite optimal, but you could e.g. take $\mu = -\ln(\epsilon/2)$, and then the bound is $$ t \ge \dfrac{2 d \ln(2/\epsilon)}{\epsilon} $$

Robert Israel
  • 448,999
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It is enough to take $t^* = e(d+1)!/\epsilon^{d+1}$.

Proof:

The LHS of the desired inequality is at most $t^d/0! + t^d/1! + \cdots t^d/d! \le et^d$ (assuming $t\ge 1$ which is fine).

The RHS of the desired inequality is at least $(\epsilon t)^0/0! + \cdots + (\epsilon t)^{d+1}/(d+1)! \ge (\epsilon t)^{d+1}/(d+1)!$.

Therefore, it is enough if we ensure that $et^d \le (\epsilon t)^{d+1}/(d+1)!$. But this is equivalent to $t\ge e(d+1)!/\epsilon^{d+1}$.

With more effort one could presumably improve the dependence of $t^*$ on $d$ and $\epsilon$.

Gabriel Nivasch
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