0

For what $p \in [0,1]$ would the function $p^n$ be negligible? Specifically; if $p$ is allowed to depend on $n$.

If $p$ is a constant between 0 and 1, (say $p = \frac{1}{2}$), then it clearly is negligible. Judging from this plot, if one chooses $p = 1 - \frac{1}{n}$, the resulting function is not negligible.

What about $p = 1 - \frac{1}{\sqrt{n}}$? Or $p = 1 - \frac{1}{\log{n}}$ ? (or even $p = 1 - \frac{1}{\log\log n}$ etc...?) At which point (between a constant and $1-\frac{1}{n}$) does the function stop being negligible?

Borage
  • 1
  • What is the source of this question? The limit definition is easier to use – kelalaka May 17 '21 at 16:04
  • Prove or disprove: if $0\le\alpha<1$ then $\left(1-n^{-\alpha}\right)^n$ is a negligible function of $n$. Hint: take the $\ln$. – fgrieu May 17 '21 at 16:16
  • @kelalaka The source of the question is: I want to show that, given a sequence of $n$ random bits, if I take the XOR of a random subset of them (where each position is included in the subset with some probability, say $1/\sqrt{n}$) the result is approximately uniformly random. There are a few steps in between that lemma and this question, but eventually I arrived at the fact that for most sequences, the XOR is close to uniformly random, with an error of something like $(1 - 2/\sqrt{n})^{n/4}$. I want to show that that error is negligible. – Borage May 18 '21 at 11:12
  • @fgrieu Thank you for the hint. I'm still a bit stuck, however. I tried starting with the desired inequality $|(1-n^{-\alpha})^n| < n^{-c}$, and taking the logarithm on both sides, but it does not seem to get me anywhere yet. Was this what you had in mind, or were you hinting at another strategy? – Borage May 18 '21 at 11:14
  • My thinking is one the tune of: $$\ln((1-n^{-\alpha})^n)=n\ln(1-n^{-\alpha})=-n(n^{-\alpha}+o(n^{-\alpha}))=-n^{1-\alpha}+o(n^{1-\alpha})$$ which allows to conclude. The then in the proposition can I guess be changed to $\iff$, to get a relatively precise characterization of $p$. There's a ton of justifications to make. – fgrieu May 18 '21 at 12:10

0 Answers0