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QUESTION. Given reals $0 < \epsilon, \delta < 1$, is it always possible to find $m, n \in \mathbb{N}$ such that $$\begin{cases} \qquad \,\,\,\, \,(1-\delta^m)^n < \epsilon \\ 1-(1-(\frac{\delta}2)^m)^n <\epsilon \,\,? \end{cases}$$

This might not be so hard but I wish to know.

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    $\begingroup$ Not at research level, I'm afraid. $\endgroup$
    – Fan Zheng
    Commented Nov 27, 2016 at 3:57
  • $\begingroup$ One way is to solve n in terms of m, and show that if m is large enough then the interval always contains an integer. $\endgroup$
    – Fan Zheng
    Commented Nov 27, 2016 at 4:09
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    $\begingroup$ Here's a way of looking at it: for large $m$, the expression $\frac{\log(1 - \delta^m)}{\log(1 - (\delta/2)^m)}$ is more or less on the order of $2^m$ by looking at Taylor series for the numerator and denominator, so for all sufficiently large $m$ we have $(1.5)^m < \frac{\log(1 - \delta^m)}{\log(1 - (\delta/2)^m)}$. Choose such $m$ so large that $\frac{\log(\epsilon/2)}{\log(1-\epsilon)} < (1.5)^m$, and take $n$ to be the ceiling of $\frac{\log(\epsilon/2)}{\log(1 - \delta^m)}$. Then we have $(1 - \delta^m)^n < \epsilon/2$ and $1 - \epsilon < (1 - (\delta/2)^m)^n$. $\endgroup$
    – Todd Trimble
    Commented Nov 27, 2016 at 7:14
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    $\begingroup$ Putting @ToddTrimble's comment another way, the first line is approximately $e^{-n\delta^m}<\epsilon$, while the second line is approximately $e^{-n(\delta/2)^m}>1-\epsilon$. This can clearly be done simultaneously: first choose $m$ large enough so that the ratio between the two exponents is gigantic (you need to find an $a$ and an $m$ such that $e^a/e^{a/2^m}<\epsilon/(1-\epsilon)$). Then find an $n$ such that $n\delta^m=a$. $\endgroup$
    – Anthony Quas
    Commented Nov 27, 2016 at 8:16

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