0
$\begingroup$

Are there numbers $k > 1$ and $c > 1$ such that:

1 ) $\theta(c) \geq c \left( 1-\frac{1}{5 \ln^2(c)} \right) $

2 ) $\frac{c}{1+\frac{1}{\ln^4(c)}} \leq p(\pi(c))$ where $p(n)$ is the $n$-th prime number and $\pi(n)$ is the prime counting function

3 ) $\pi(c) \leq \frac{c}{\ln (c)} (1+\frac{1}{\ln (c)}+\frac{2}{\ln^2 (c)}+\frac{6}{\ln^3 (c)} + \frac{25}{\ln^4 (c)} )$

4 ) $\pi(c) \geq \frac{c}{\ln (c)} (1+\frac{1}{\ln (c)}+\frac{2}{\ln^2 (c)}+\frac{6}{\ln^3 (c)} + \frac{24}{\ln^4 (c)} )$

5 ) $\prod \limits_ {p \leq c} (1+\frac{1}{p^2-1}) \prod \limits_{p \leq c^k} (1+\frac{1}{p}) \leq e^\gamma \ln (c ( 1-\frac{1}{5 \ln^2(c)})+c^k ( 1-\frac{1}{5 k^2 \ln^2(c)}))$?

$\endgroup$
4
  • $\begingroup$ I forgot to mention that $k$ have to be bigger than $1$ $\endgroup$
    – Ahmad
    Dec 25 '16 at 22:04
  • $\begingroup$ Looks like homework. $\endgroup$
    – Fan Zheng
    Dec 25 '16 at 22:48
  • $\begingroup$ not a homework, i think that such numbers could help prove RH partially $\endgroup$
    – Ahmad
    Dec 25 '16 at 22:52
  • 1
    $\begingroup$ Also, the typical error bound coming out of the proof of the PNT is way better than $O(\ln(x)^{-N})$, yet the PNT is still a light-year from the RH. $\endgroup$
    – Fan Zheng
    Dec 25 '16 at 22:59
4
$\begingroup$

Proving such results falls into three parts. First you take an effective version of the prime number theorem, which implies all your desired bound for sufficient. Second you write a computer program (or use computations done by somebody else) to check your inequalities for small $x$. Finaly you bridge the gap between the two ranges by improving the general bounds for large $x$ with more specialized arguments using numerical information on zeros of the Riemann $\zeta$-function.

The first part is easy, since you can just cite the error terms of the effective PNT. As far as I know, the bet results are due to Trudgian ( https://arxiv.org/pdf/1401.2689.pdf ). The difficult part is making the different ranges overlap. Note that as your inequalities are rather weak, and $\pi(x)$ can be computed in sublinear time, you should compute $\pi(x_0)$, use an upper bound sieve to compute some $x_1$, such that your inequality holds in $[x_0, x_1]$. Since $x_1$ will be of magnitude $x_0+\frac{x_0}{\log^C x_0}$, the savings are substantial.

$\endgroup$

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.