Does there exist a function $f$ that is a lower bound of the prime number function $\pi$ with $f \sim \pi$?

  • 5
    $\begingroup$ Just to be a little puckish, I will note that $$f(x)=\pi(x)-1$$ satisfies your requirements :) $\endgroup$ – Zev Chonoles Aug 23 '11 at 18:10
  • $\begingroup$ Yep, thanks. My wording should be more precise... $\endgroup$ – user14947 Aug 23 '11 at 18:36

This segment of the Wikipedia article mentions a few such bounds. In all cases, the bounds hold from a certain explicit $N$ on. They can be easily be made unconditional without changing the asymptotic behaviour by suitably redefining the functions for $x \lt N$.

For instance, the article mentions the bounds $$\frac{x}{\ln x+2}<\pi(x)<\frac{x}{\ln x-4},$$ valid for $x \ge 55$, as well as much stronger bounds by Pierre Dusart. Since $\pi(x)$ is asymptotically $x/\ln x$, the two bounds above have the right asymptotic properties. Tweaking the lower bound so that it is valid below $55$ is easy. The crudest method is to use the function which is $-1$ for $x \lt 55$, and $x/(\ln x +2)$ for $x \ge 55$. Asymptotic behaviour is unaffected.


I found an more 'elegant' lower bound:

$$ \frac{n}{\log\,n} - 2 \leq \pi(n), \; n \geq 2 $$ Primality Testing in Polynomial Time

  • $\begingroup$ There is a proof on page 46: springerlink.com/content/p9rft7x5nkbwjkq6 $\endgroup$ – user14947 Sep 24 '11 at 1:41
  • $\begingroup$ Sorry, was thinking of $\text{li}(x)$, that bounces back and forth from below $\pi(x)$ to above. $\endgroup$ – André Nicolas Sep 24 '11 at 1:49
  • $\begingroup$ Although, I am not sure what is meant by $\log$ in the paper. The author distinguishes between $\ln$ (natural logarithm) and $\log$ (base 2 or base 10, maybe?)... $\endgroup$ – user14947 Sep 24 '11 at 1:57
  • $\begingroup$ On second or third thought, I am again very surprised at the result, and doubtful. Maybe there is confusion with estimates of $p_n$. I do not trust Wolfram Mathworld assertions, they are often wrong. $\endgroup$ – André Nicolas Sep 24 '11 at 6:18
  • $\begingroup$ mmh, I don't think so, Martin Dietzfelbinger seems like a reliable source. $\endgroup$ – user14947 Sep 24 '11 at 12:20

Your Answer

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