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Expansion for the prime counting function

  1. May 3, 2009 #1
    my question is, let us suppose we can find an expansion for the prime number (either exact or approximate)

    [tex] \pi (x) = \sum _{n=0}^{\infty}a_n log(x) [/tex]

    and we have the expression for the logarithmic integral

    [tex] Li (x) = \sum _{n=0}^{\infty}b_n log(x) [/tex]

    where the numbers a(n) and b(n) are known , then my question is , what could one expect about the difference expansion

    [tex] \pi (x) - Li(x) = \sum _{n=0}^{\infty}(a_n - b_n) log(x) [/tex] ??
     
  2. jcsd
  3. May 3, 2009 #2

    CRGreathouse

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    Unless the a_i and b_i are functions of x, no such expansions exist as they would imply that pi(x)/log(x) and Li(x)/log(x) are constant.
     
  4. May 4, 2009 #3
    [tex] \pi (x) = \sum _{n=0}^{\infty}a_n log^{n} (x) [/tex]

    [tex] Li (x) = \sum _{n=0}^{\infty}b_n log^{n} (x) [/tex]

    sorry i made a mistake it should include powers of log(x) and not only logarithm of x , sorry about that.

    [tex] \pi(x) - Li (x) = \sum _{n=0}^{\infty}(a_n - b_n) log^{n} (x) [/tex]
     
    Last edited: May 4, 2009
  5. May 4, 2009 #4

    CRGreathouse

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    About what point are you taking this expansion? I don't see a way to take it about the point at infinity.
     
  6. May 30, 2009 #5
    "the expression for the logarithmic integral"

    While there is an asymptotic expansion for the logarithmic integral of the form you propose it is not convergent. It is only a good representation of li(x) for a certain number of terms depending on x.

    Other series for li(x) that do converge are much trickier. For example Ramanujan's series: http://en.wikipedia.org/wiki/Logarithmic_integral_function#Series_representation.

    Pi(x) - Li(x) = (Difference of Pi(x) - Li(x)) (see wiki on RH for the best estimate on this) + (Error on your expression for Pi(x)) + (Error on your expression for Li(x))

    Note that I added the errors because the sign of the errors is not known. If you could prove the errors are strictly positive you could subtract them.

    Of course everyone wishes that such a nice analytic series exists for Pi(x). Many eminent mathematicians have been looking for such a series for hundreds of years with not much to show for it.
     
    Last edited: May 30, 2009
  7. May 30, 2009 #6

    CRGreathouse

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