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 $\mathrm{Li}(x)$?
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1$\begingroup$ Some better bounds are here $\endgroup$– WojowuOct 10, 2020 at 10:10
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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 $x$? $\endgroup$– Gerry MyersonOct 10, 2020 at 11:32
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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 MOOct 10, 2020 at 15:03
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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. $\endgroup$– Greg MartinOct 10, 2020 at 19:51
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1$\begingroup$ @mathguy: In my response below, I added some explicit bounds of the type Greg Martin mentioned. Check them out! $\endgroup$– GH from MOOct 10, 2020 at 21:18
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1 Answer
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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 $x\geq 2$.
answered Oct 10, 2020 at 21:13