Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Questions about the prime counting function

  1. Feb 9, 2012 #1
    greetings . i have a couple of questions about the prime counting function .
    when [itex]\pi _{0}(x)[/itex] changes by 1, then it's logical to assume that it should happen at a prime argument . meaning :

    [tex]\lim_{\xi\rightarrow 0}\pi _{0}(x+\xi )-\pi _{0}(x-\xi )=1 [/tex]

    implies that x is a prime .
    is this a true assumption ?

    according to the literature, we can expand [itex] \pi_{0}(x)[/itex] using the riemann R function .

    [tex] R(x) = 1+\sum_{k=1}^{\infty}\frac{(ln x)^{k}}{k!k\zeta (k+1)}[/tex]

    [tex] \pi_{0}(x)= R(x)-\sum_{\rho} R(x^{\rho })-\frac{1}{lnx}-\frac{1}{\pi}tan^{-1}\left( \frac{\pi}{lnx}\right) = \sum_{k=1}^{\infty}\frac{(lnx)^{k}[1-\sum \rho^{k} ]}{k!k\zeta(k+1)}-\frac{1}{lnx}-\frac{1}{\pi}tan^{-1}\left( \frac{\pi}{lnx}\right) [/tex][itex] \rho [/itex] being the nontrivial zeros of the zeta function .
    is this correct !?!? i mean , is the expansion correct !?!?
    Last edited: Feb 9, 2012
  2. jcsd
  3. Feb 9, 2012 #2
    i think i got my answer .

    [tex] \Pi _{0}(x)=\sum_{n=1}^{\infty}\frac{1}{n}\pi_{0}(x^{1/n})[/tex]

    and [itex] \Pi_{0}(x)[/itex] does change by 1 at primes .

    now i am intrigued by the terms [itex] \sum \rho^{k} [/itex] , it seems to me we don't need to know [itex] \rho [/itex] themselves to evaluate [itex]\Pi_{0}(x) [/itex] , we just need to evaluate [itex] \sum \rho^{k} [/itex] . is this correct !?!?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook