We have two positive and increasing integers $a$ and $b$. We assume that $a$ is a function of $b$ and that we have:

$$a \log a \sim b$$

where $\sim$ is the asymptotic equivalence.

I would like to show that $a = \Theta(\frac{b}{\log b})$.

Thank you.

  • $\begingroup$ Note that the inverse of $x \log(x)$ is $\frac{x}{W(x)}$ where $W$ is the Lambert W function i.e. the inverse of $x e^x$. So you need asymptotics of $W(x)$ for large argument, specifically you want $W(x)=\Theta(\log(x))$. $\endgroup$ – Ian Jul 20 '17 at 15:43
  • $\begingroup$ That is exactly the relation between the $n$-prime $p_n$ and the number $\pi(n)$ of primes less than or equal to $n$. See WIkipedia. $\endgroup$ – lhf Jul 20 '17 at 16:10

$$\frac{b}{\log b}\sim\frac{a\log a}{\log a+\log \log a}\sim a.$$

This is even more precise than $\Theta$.

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  • 1
    $\begingroup$ Don't we need to assume that $a \to \infty$ as $b \to \infty$ ? $\endgroup$ – lhf Jul 20 '17 at 16:13
  • $\begingroup$ Thank you very much for your answer. @lhf $a$ and $b$ are increasing and $a$ is a function of $b$. $\endgroup$ – Dingo13 Jul 20 '17 at 18:24

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