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Yet another formula for pi(x)

  1. Aug 24, 2005 #1
    Yet another formula for pi(x) (prime number counting function)
    start with wilson's therorem : p is prime iff p divides (p-1)! + 1
    let G(x) be the gamma function
    then p is prime iff sin(pi*(G(x)+1)/x) = 0
    let f be the function x -> sin(pi*(G(x)+1)/x)

    Since f(x) = sin(pi*G(x)/x+pi/x) and because G(x)/x is integer (and even) when x is an integer <> 4,
    for any integer x, non prime and different from 4, f(x) = sin(pi/x)

    Then let h(x) = f(x) / sin(pi/x) = sin(pi*(G(x)+1)/x) / sin(pi/x)

    h(x) = 0 if x is prime
    h(x) = 1 for any non prime integer x >4
    h(4) = -1

    therefore, for x >= 5,

    pi(x) = x-2+sum[k=5..x, h(k)] :smile:

    Questions :
    1) Is there a way to convert this sum into an integral and have a cool expression of pi(x) ?

    2) is there a way to differentiate this sum and then have an expression
    of d pi(x) / dx ?

    3) since sin(pi*x) = pi*x*product[n=1..inf, 1-(x/n)^2)],
    and G(x)=(x-1)*G(x-1),
    is there a way to express h(x) as an infinite product, and then simplify it and simplfy the above expression of pi(x) ?
    Last edited: Aug 24, 2005
  2. jcsd
  3. Aug 24, 2005 #2
    don,t want to discourage you ( i think your formulation is interesting) but there are a lots of function exact (with a triple integral, i myself have obtained several integral forms for the PI(x)...

    A hint to transform a series into an integral you can make use of the equality:

    [tex]\sum_{0}^{\infty}a(n)=\int_{-\infty}^{\infty}dxa(x)w(x) [/tex]

    where a(n) is the general term of the series and w(x) is the Laplace inverse transform of:

    [tex]\frac{1}{1-e^{-s}} [/tex]

    Another approach (exact formula) is using the Poisson,s summation formula:

    [tex]\sum_{n=0}^{\infty}A(n)=\int_0^{\infty}dxA(x)\sum_{n=-\infty}^{\infty}e^{inx} [/tex]

    hope it helps....if you wait a bit you will be able to hear the critics of shmoe and Matt Grime about your approach to Pi(x)..(don,t hope it be positive though..:))
    Last edited: Aug 24, 2005
  4. Aug 24, 2005 #3


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    Segre: you can write a much shorter derivation.

    If I define g by:

    g(n) := \left\{
    1 \quad & n \mbox{ is prime} \\
    0 & n \mbox{ is not prime}

    then it's clear that

    \pi(n) = \sum_{k = 1}^{n} g(k)

    right? I'll leave the rest as an exercise.

    BTW, you might want to check the convergence conditions before doing any sort of transform.
    Last edited: Aug 24, 2005
  5. Aug 25, 2005 #4
    Right, of course, but my "prime predicate" function has a real argument, not integer, and my hope was that i could use some tricks of real function analysis (eg a transform of some kind) to obtain a new expression for pi(x)
  6. Aug 25, 2005 #5

    matt grime

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    you can use transforms, these are well known and have been aobut for 50 years in text books.
  7. Aug 25, 2005 #6


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    Not true: x has to be an integer because it's one of the bounds on your summation. :tongue2:

    And even if you extended the meaning of &Sigma; notation to allow the upper bound to be a noninteger, you could do exactly the same to the summation I posted as well.
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