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Primes as roots of same function

  1. Apr 19, 2006 #1
    It,s proven that there can not be any Polynomial that gives all the primes..but could exist a function to its roots are precisely the primes 8or related to them) if we write:

    [tex] f(x)=\prod_{p}(1-xp^{-s})=\sum_{n=0}^{\infty}\frac{\mu(n)}{n^{s}}x^{n} [/tex]

    wher for x=1 you get the classical relationship between the Riemann zeta function and the product of primes as you can see the equation above for x=p^{s} for p prime is 0 the problem is that due to we have to know the function \mu(x) this function is not computable using Abel,s sum formula:

    [tex] ln[f(s)]/s=x\int_{2}^{\infty}dt\frac{\pi(t)}{t(t^{s}-x)} [/tex]

    and although is not perhaps completely correct i think [tex] f(x)=x/Li_{s}(x) [/tex]
     
  2. jcsd
  3. Apr 19, 2006 #2

    Hurkyl

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    I think f(x) doesn't converge at all, except at x = 0. (I was interested in a similar expression once, but I wanted zeroes at all the positive integers)


    So, I suspect you'll need some exponential terms to get something that converges. Maybe something like:

    [tex]
    \prod_{i = 1}^{+\infty} (1 - x / p_i) e^{x / p_i}
    [/tex]

    The general exponential term in an infinite product can be more complicated, but the simplest nontrivial term was good enough for my purposes.
     
  4. Apr 22, 2006 #3
    well using the same trick by Euler..i managed to prove that:

    [tex] \prod_{p}(1-xp^{-s})^{-1}=Li_{s}(x) [/tex]

    where necessarily [tex] |xp^{-s}|<1 [/tex]and LI is the Polylogarithm..if we extend the radius of convergence of Polylogarithm to |x|>1 we could get that p^{s} for p prime is a root of the function 1/Li for a certain s.
     
  5. Apr 22, 2006 #4

    shmoe

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    I don't think you did. Take x=0, the left hand side will be 1, the right hand side will be 0.

    If you expand the terms of the left hand side using geometric series, you get a product over primes of sums 1+(x/p^s)+(x/p^s)^2+... If you expand this product (taking care when this is valid), the power of x that appears with 1/n^s is not in general going to be n like you would need to equal the polylogarithm. It will be the sum of the exponents appearing in the prime decomposition of n.
     
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