Finding an order of $n$ such that a product is exponentially small in $n$

Question

I am looking for a function $f$ s.t. if $t=O(f(n))$, where $O$ stands for Big-O notation, then:

$$\frac{t}{n}\times\frac{t-1}{n}\times\cdots \times \frac{2}{n}\sim e^{-an} \tag{1}$$

i.e. when the LHS of $(1)$ is exponentially small in $n$?

Say $t=n/2$, so $t=O(n)$, then the LHS of $(1)\leq \left(\frac{1}{2}\right)^n$ so it is exponentially small in $n$. But I was wondering if I can do better than that like what if $t=O(\sqrt{n})$ or something like that in which case I don't know if the LHS of $(1)$ would still be exponentially small?

Can you explain what you mean by the "$=$" sign in expression $(1)$? Do you mean equivalence? — Bcpicao, Nov 07 '20 at 20:12
HINT: take the $\log$ and try to think about comparing the continuous equivalent of these sequences. — Bcpicao, Nov 07 '20 at 20:25
@Bcpicao Sorry didn't understand what you mean by continuous equivalent of these sequences. May I know which sequences? — trickymaverick, Nov 07 '20 at 20:27

score 2 · Accepted Answer · answered Nov 07 '20 at 20:33

You want $$\frac{f(n)!}{n^{f(n)-1}}\sim e^{-an}.$$ Since $$f(n)!\sim \sqrt{2\pi f(n)}\left(\frac{f(n)}e\right)^{f(n)},$$ so we want $$n\sqrt{2\pi f(n)}\left(\frac{f(n)}{en}\right)^{f(n)}\sim e^{-an}.$$ Taking logs and writing $f(n)=f$ for simplicity, we have that $$\log n+\frac{\sqrt{2\pi}}2+\frac{\log f}2+f\cdot(\log f-\log n-1)+an$$ is asymptotically constant. Writing $g=f/n$ (since $f$ should be a bit smaller than $n$), $$\frac{\sqrt{2\pi}}2+\frac{\log g+3\log n}2+gn(\log g-1)+an\to c$$ for some constant $c$. The asymptotically largest terms here are $gn\log g$ and $an$, so we want $$-g\log g\sim a,$$ so $g$ should be (around) a constant. So, $f$ should be asymptotically some constant times $n$.

I was gonna use functional analysis and integrals but indeed Stirling already did the job for us :) Nice one (+1) — Bcpicao, Nov 07 '20 at 20:36
@manifolded Isn't the LHS exactly $\frac{f(n)!}{n^{f(n)-1}}$? — Carl Schildkraut, Nov 07 '20 at 21:51

score 1 · Answer 2 · answered Nov 08 '20 at 05:45

Starting from @Carl Schildkraut'answer, you want to solve for $f(n)$ the equation

$$\frac{[f(n)]!}{n^{f(n)-1}}= e^{-an}$$ To simplify notations, let $f(n)=y$ and $k=\frac{1}{n}e^{-a n}$ to make $$y!=k\, n^y\tag 1$$

Looking at this question of mine, you will find a superb approximation proposed by @robjohn who is a very eminent user in this forum.

Adapted to $(1)$, it will give $$y=e\,n\, \exp\Big[W\left(\frac{1}{2 e n}\log \left(\frac{k^2}{2 \pi n}\right)\right) \Big]-\frac 12$$ where $W(.)$ is Lambert function.

Back to the original notations, this will give $$f(n)=e\, n\,\exp\Big[W\left(-\frac{a}{e}-\frac{3 \log (n)+\log (2 \pi )}{2 e n}\right) \Big]-\frac 12$$

Finding an order of $n$ such that a product is exponentially small in $n$

2 Answers2