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For $\epsilon>0$, why the following holds with $x$ larger enough $$O(x^{-\epsilon/2}\log x)=O(2^{-\log^{1/2}x})$$

Zoe
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1 Answers1

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All you need is $\log(x) = o(x^c)$ for any $c > 0$.

Taking logs of the question, this gives $-c \log(x)$ on the left where we can choose $c = \epsilon/4$ and $-d \log^{1/2}(x)$ on the right for some $d > 0$ and this makes it clear since $\log(x) \to \infty$.

marty cohen
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  • I took logs of the left, i got $-\epsilon/2\log x +\log\log x$. How to get the term $-c\log(x)$, $c=\epsilon/4$? – Zoe Feb 02 '20 at 15:02
  • The log log doesn't matter since it's small. And to get the c log x use the result in my first line. – marty cohen Feb 02 '20 at 19:05
  • The right hand side is $-\epsilon/2\log x$ (if we ignore the term $\log\log x$), and the left hand side we have $-d\log^{1/2} (x)$ for some $d>0$. Is it suffices to conclude that $O(x^{-\epsilon/2}\log x)=O(2^{-\log^{1/2}x})$ with $\log(x)\rightarrow \infty$? Sorry, I can't understand why we choose $c=\epsilon/4$. – Zoe Feb 03 '20 at 04:17