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Re: Primes in the interval (Problem)

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  • djbroadhurst
    ... The discussion after Theorem 3.1 [Mertens] in Riesel s book http://www.amazon.com/Numbers-Computer-Factorization-Progress-Mathematics/dp/0817637435
    Message 1 of 3 , Dec 4, 2010
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      --- In primenumbers@yahoogroups.com,
      <chitatel2000@...> wrote:

      > Determine the largest value of the error

      I have already explained why your error is unbounded:

      > It is interesting to see how Sergey fooled himself into
      > making his false conjecture. His mistake was to believe
      > that we may rely on the sieve of Eratosthenes to give
      > the "right" constant in the prime number theorem. In fact,
      > we should not: there is a mismatch by the celebrated factor
      > 2*exp(-Euler) > 1

      The discussion after Theorem 3.1 [Mertens] in Riesel's book
      http://www.amazon.com/Numbers-Computer-Factorization-Progress-Mathematics/dp/0817637435
      explains this rather clearly.

      David
    • djbroadhurst
      ... Off-list, Sergey asked me for an example of a large error. Let p[n] = 10^1000 + 453. Then p[n-1] = 10^1000 - 1769 is the previous prime. Sergey defines Q
      Message 2 of 3 , Dec 6, 2010
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        --- In primenumbers@yahoogroups.com,
        "djbroadhurst" <d.broadhurst@...> wrote:

        > I have already explained why your error is unbounded

        Off-list, Sergey asked me for an example of a large error.

        Let p[n] = 10^1000 + 453.
        Then p[n-1] = 10^1000 - 1769 is the previous prime.
        Sergey defines Q as the number of primes
        between p[n-1]^2 and p[n]^2.
        Using the prime number theorem, we may estimate
        Q =~ (p[n]^2 - p[n-1]^2)/log(p[n]^2) =~ 9.65*10^999

        Sergey defines his "error" E by
        E = (p[n]^2 - p[n-1]^2)*prod(k=1,n,1-1/p[k]) - Q

        Using Mertens' theorem
        http://mathworld.wolfram.com/MertensTheorem.html
        we may easily estimate
        E =~ 2*exp(-Euler)*Q - Q =~ 1.19*10^999
        whose integer part has 1000 decimal digits.

        Clearly the error is unbounded, since it is
        of the same order as the n-th prime, p[n].

        I remark that Sergei has already advertised his "blogspot" in 9
        messages to this list, while making no significant contribution
        to our own discussions. I hope that we shall not receive a 10th
        advertisement for a site that seems to make no attempt to learn
        from careful correction of its obvious failures.

        David Broadhurst
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