5
$\begingroup$

The prime counting function $\pi(x)$ is defined as \begin{equation} \pi(x)=\sum_{p\leq x}1 \end{equation} where $p$ runs over primes.
I have seen many bounds for $\pi(x)$ such as \begin{equation} \frac{x}{\log x}\left(1+\frac{1}{2\log x}\right)<\pi(x)<\frac{x}{\log x}\left(1+\frac{3}{2\log x}\right) \end{equation} \begin{equation} \frac{x}{\log x - 1/2}<\pi(x)<\frac{x}{\log x + 3/2} \end{equation} \begin{equation} \frac{x}{\log x+2}<\pi(x)<\frac{x}{\log x - 4} \end{equation} Till now, what are the best known upper and lower bounds for the prime-counting function? Is there a better bound that $\mathrm{Li}(x)$?

$\endgroup$
12
  • 1
    $\begingroup$ Some better bounds are here $\endgroup$
    – Wojowu
    Oct 10 '20 at 10:10
  • 3
    $\begingroup$ ${\rm Li}(x)$ is neither an upper bound nor a lower bound for $\pi(x)$, so it's not clear to me how you can call it a bound. Maybe what you want is $f(x)$ such that $|f(x)-\pi(x)|<|{\rm Li}(x)-\pi(x)|$ for all $x$, or for all sufficiently large $x$? $\endgroup$ Oct 10 '20 at 11:32
  • 2
    $\begingroup$ $f(x)=\pi(x)$ is the best approximation to $\pi(x)$. A close relative is $f(x)=\pi(x)+1$. $\endgroup$
    – GH from MO
    Oct 10 '20 at 15:03
  • 3
    $\begingroup$ mathguy, do you know the prime number theorem with its error term? That will show that functions of the form $\mathop{\rm Li}(x) \pm C x\exp(-c\sqrt{\log x})$ are upper and lower bounds for $\pi(x)$, and explicit values for the constants $C$ and $c$ can be found. This is nearly the best known bound (the power of $\log x$ inside the exponential has been improved). Certainly no bounds of the form $\mathop{\rm Li}(x) \pm Cx^{1-\delta}$ are known. $\endgroup$ Oct 10 '20 at 19:51
  • 1
    $\begingroup$ @mathguy: In my response below, I added some explicit bounds of the type Greg Martin mentioned. Check them out! $\endgroup$
    – GH from MO
    Oct 10 '20 at 21:18
13
$\begingroup$

The following explicit version of the Prime Number Theorem was proved by Trudgian: $$ |\pi(x)-\mathrm{li}(x)|<x e^{-0.39\sqrt{\ln x}},\qquad x\geq 229.\tag{$\ast$}$$ In fact Trudgian's Theorem 2 is somewhat stronger than $(\ast)$, and with Mathematica it is straightforward to extend the validity of $(\ast)$ to $x\geq 2$.

$\endgroup$

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.