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How can I prove, that $$\log^x{x} > x^{\sqrt{x}}$$ for big $n$ ? I tried to logarithm those expressions, deduct them, somehow estimate the values but no luck.

After few tries, I ended up with expression, where I need to prove that $\log{n}$ grows faster than $$\large{n^{\frac{1}{\sqrt{n}}}}$$, so that is an alternative question.

alkabary
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Nope nope
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1 Answers1

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We have $$\log(x)^x = \exp\left(\log\left(\log(x)^x\right)\right) = \exp\left(x\log(\log(x))\right)$$ and $$x^{\sqrt{x}} = \exp\left(\log\left(x^{\sqrt{x}}\right)\right) = \exp\left(\sqrt{x}\log(x)\right)$$ Now conclude what you want by noticing that $$\sqrt{x}\log(x) \ll x \ll x\log(\log(x))$$

The first part of the inequality is true, since $\log(y) \ll y \implies \log(\sqrt{y}) \ll \sqrt{y} \implies \log(y) \ll \sqrt{y}$

Adhvaitha
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  • I found this also, but how can I prove the last part? When the x^(1/2) is smaller than x outside the logartihm and x is bigger than log(x) inside the logarithm, why can I say, that one is bigger than other? – Nope nope May 18 '15 at 19:15
  • @Nopenope Edited. – Adhvaitha May 18 '15 at 19:26