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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 !?!?
     
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