It appears that the following holds: $$1-e^{-e^{-c}} \sim e^{-c}$$

I do not see why this should hold. How do I prove it? I would be very happy to see multiple ways of proving this, as it is always good to have more tools in one's mathematical toolbox.

Moreover, it seems to me that the ratio between these to value tends to 1 rather quickly (probably exponentially). Is there a way how to say something about the speed of convergence?

  • 3
    $\begingroup$ Why don't you use Taylor series, seeing as the exponential is entire. $\endgroup$
    – mattos
    Jul 2 '19 at 11:25
  • 4
    $\begingroup$ Replace $e^{-c}=x$ where by the statement $x \to 0$ $\endgroup$
    – Yuriy S
    Jul 2 '19 at 11:26
  • $\begingroup$ @Yuriy , performing the suggested subs. we get $1-e^x=x$ which is still hard to do... $\endgroup$
    – NoChance
    Jul 2 '19 at 11:40
  • 1
    $\begingroup$ @NoChance, no, it will be $1-e^{-x} \approx x$ if $x \ll 1$ if I understood the question correctly. It can be proved through any definition of the exponential function. $\endgroup$
    – Yuriy S
    Jul 2 '19 at 11:42
  • $\begingroup$ @YuriyS, thank you for your reply, you are correct. However, I think it is still not very easy to solve. $\endgroup$
    – NoChance
    Jul 2 '19 at 11:44

I'm assuming that this holds as $c\rightarrow +\infty$. In that case, $e^{-c}\rightarrow 0$ and $$\begin{split} 1-e^{-e^{-c}} &= 1-(1-e^{-c}+\mathcal O(e^{-2c}))\\ &= e^{-c}+\mathcal O(e^{-2c})) \end{split}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.