The Prime Number Theorem states $\pi(n)\sim \dfrac{n}{\ln n}$.

Would there be an equally simple expression if Riemann's Hypothesis were proved true?

From Chebyshev Function, would $\pi(n)\sim \dfrac{n}{\ln n} + \sqrt n\ln n$ work?

Addendum : A relevant link : https://mathoverflow.net/questions/70713/error-term-of-the-prime-number-theorem-and-the-riemann-hypothesis

  • $\begingroup$ jstor.org/stable/2005976?seq=1#page_scan_tab_contents $\endgroup$ – Balarka Sen Jan 25 '16 at 11:13
  • 2
    $\begingroup$ I'm not sure if I understand correctly what you mean. $\pi(n)\sim n/\ln n + \sqrt n/\ln n$ is true unconditionally, because $n/\ln n + \sqrt n/\ln n\sim n/\ln n$. $\endgroup$ – Wojowu Jan 26 '16 at 20:26
  • 1
    $\begingroup$ One idea would be to consider the upper bound for the error term $\pi(n)-n/\ln n$. Is that closer to what you mean? If so, this is relevant. $\endgroup$ – Wojowu Jan 26 '16 at 20:31
  • 1
    $\begingroup$ @Wojowu; i think i want to use pi(x) ~ x/logx + x/(logx)^2 + o() from your link, but as stated previously the asymptotes don't work $\endgroup$ – JMP Jan 26 '16 at 21:01
  • 1
    $\begingroup$ if the RH was true, the residual would be $\pi(x) - x/\ln x = \mathcal{o}(x^{1/2+\epsilon})$ for all $\epsilon > 0$. the residual would be $\pi(x) - x/\ln x = \mathcal{O}(x^{\sigma_0} / \ln x)$ if $\zeta(s)$ had a finite number of zeros at $\Re(s) = \sigma_0$ (if the RH was false) $\endgroup$ – reuns Jan 26 '16 at 22:32

Yes, if the RH were proved true, then the error term for $\pi(x)$ in terms of $Li(x)$ would be optimal, namely $$ | \pi(x) - Li(x) | = O(\sqrt{x}\log{x}). $$ But since we can relate $\frac{x}{\log(x)}$ with $Li(x)$, we would also obtain a version with $\frac{x}{\log(x)}$. We have $$ {\rm Li} (x) - {x\over \log x} = O \left( {x\over \log^2 x} \right) \; . $$ Formulated differently, PNT only gives $$ \pi(x)={\rm Li} (x) + O \left(x \mathrm{e}^{-a\sqrt{\log x}}\right) \quad\text{as } x \to \infty $$ for some constant $a>0$, whereas with RH we get even $$ \pi(x) = {\rm Li} (x) + O\left(\sqrt x \log x\right). $$

  • $\begingroup$ What would this "version with $x/\log x$" actually be, though? $\endgroup$ – Wojowu Jan 26 '16 at 20:31
  • $\begingroup$ The series at the end isn't convergent for any $x$. $\endgroup$ – Wojowu Jan 26 '16 at 20:33
  • $\begingroup$ What do you mean take the first terms? First how many? Isn't that an arbitrary decision to make since the terms fo the series get larger and larger? $\endgroup$ – MCT Jan 26 '16 at 20:37
  • $\begingroup$ What is the $a$ in the $\mathcal O(xe^{-a\sqrt{\log x}})$? $\endgroup$ – user153330 Jan 26 '16 at 20:47
  • $\begingroup$ @Soke: The first terms, see the answer here. $\endgroup$ – Dietrich Burde Jan 26 '16 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.