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How fast?

  1. Sep 13, 2006 #1
    hello..following the PNT we know that

    [tex] \frac{\psi(x)}{x}\rightarrow 1 [/tex] and

    [tex] \frac{\pi(x)}{Li(x)}\rightarrow 1 [/tex]

    my question is "how fast" do the expressions:

    [tex] |\frac{\psi(x)}{x}-1|=|f(x)| [/tex] and

    [tex] |\frac{\pi(x)}{Li(x)}-1|=|g(x)| [/tex] tend to 0 ?

    in the sense that for example will the expressions...

    [tex] f(x)x^{1/2} [/tex] and [tex] g(x)x^{1/2} [/tex] tend to 0 or will they tend to infinite?...:rolleyes: :rolleyes: (to give a clearer explanation)
  2. jcsd
  3. Sep 13, 2006 #2


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    You're just asking the Riemann hypothesis now.
  4. Sep 13, 2006 #3


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    Just about any discussion of the prime number theorem will have something to say about the error term, it shouldn't be hard for you to look this up. Something like:

    [tex]\psi(x)=x+O\left(x\exp\left(-C\frac{(\log x)^{3/5}}{(\log\log x)^{1/5}} \right)\right)[/tex]

    for a constant C>0 is known. Much better can be had assuming RH of course, or even larger zero free regions (the above bound comes from a zero free region).

    In the other direction you have:

    [tex]\psi(x)=x+\Omega(\sqrt{x}) [/tex]

    though slightly better is known by a factor of some logloglog(x) I think, you can look it up to check the number of log's.
  5. Sep 13, 2006 #4
    In that case as "Shmoe" posted it would be to see if the limit:

    [tex] \frac{\Omega(\sqrt (x))}{\sqrt(x)}=h(x) [/tex]

    tends to 0 for x-->oo, or if the integral [tex] \int_{0}^{\infty}dxh(x) [/tex] is finite.

    As far as i know i have seen "graphs" of [tex] \psi(x)-x [/tex] and seems (don't know if there is a well math theorem) that has the function [tex] x^{1/2} [/tex] as and "upper" and "lower" limit...( depending on what sign you take when take the square root of x ) i would be interested in knowing if the integral:

    [tex] \int_{c}^{\infty}dx|x^{1/2}(\frac{d\psi}{dx}-1)|^{2} [/tex] exist so it's on an L(c,oo) space.... c=oo or c=0..thanks.
    Last edited: Sep 13, 2006
  6. Sep 13, 2006 #5


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    The RH is equivilent to

    [tex]|\operatorname{Li}(x)-\pi(x)|\le c\sqrt x \ln x[/tex] for some constant c. You're asking if

    [tex]|\operatorname{Li}(x)-\pi(x)|\le c\operatorname{Li}(x)x^{-1/2}\sim c\sqrt x \ln x[/tex] for some constant c.
    Last edited: Sep 13, 2006
  7. Sep 15, 2006 #6
    In fact if we define the "trace" of a certain operator (Hamiltonian ) by:

    [tex] Z=Tr[e^{iuH}]=\sum_{n=-\infty}^{\infty}e^{iuE_{n}} [/tex] (1)

    differentiating V. Mangodlt formula..

    [tex] -\frac{d\psi}{dx}+1-\frac{1}{x^{3}-x}=\sum_{\rho}x^{\rho -1} [/tex]

    If we put...[tex] \rho_{n} = 1/2+iE_{n} [/tex] ,multiplying both sides by [tex] \sqrt (x) [/tex] and letting

    x=exp(u) the V.Mangoldt formula becomes just a "trace"..
    Last edited: Sep 15, 2006
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