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How do I determine the order (big o) of $\omega$ in $e^{-\omega/\epsilon}\leq10^{-9}$ and $e^{-\omega/\epsilon}\leq\epsilon$, where $\epsilon$ is a small parameter.

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Hi, I remember you asking this question yesterday, and I remember telling you that in fact there is no big $O$ of omega but indeed $\omega \in \Omega(\epsilon)$. Did has followed up and you told him that you found this in a book. Why don't you scan or (just write) that part of the book so that we can understand where the problem lies and help you resolve it. – Lord Soth Jul 2 '13 at 13:51
up vote 1 down vote accepted

In the first case, taking logs yields $$-\omega/\epsilon \leq -9 \ln(10)\\ \omega \geq 9 \epsilon \ln(10)\\ \omega = \Omega(\epsilon)$$ and similarly in the second case $\omega \geq \epsilon \ln (\epsilon)$ so $\omega = \Omega(\epsilon\ln (\epsilon))$

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