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Re: [PrimeNumbers] let me introducte myself and a slightly better approx to pi(x) than Li(x)

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  • LALGUDI BALASUNDARAM
    To Compute accurate value of pi(x),please see my book Prime Numbers Some Characteristics ISBN 1-4392-4094-9 pages 19-22. L.J.Balasundaram ... From: pbtoau
    Message 1 of 7 , Jul 18 5:49 AM
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      To Compute accurate value of pi(x),please see my book"Prime Numbers" Some Characteristics ISBN 1-4392-4094-9 pages 19-22.
      L.J.Balasundaram


      --- On Sat, 7/17/10, pbtoau <PbtoAu@...> wrote:

      From: pbtoau <PbtoAu@...>
      Subject: Re: [PrimeNumbers] let me introducte myself and a slightly better approx to pi(x) than Li(x)
      To: primenumbers@yahoogroups.com
      Date: Saturday, July 17, 2010, 11:53 PM







       









      Hi,



      This is probably just a consequence of Li(x) being larger than pi(x) for the first 300 orders of magnitude or so. Without getting into Riemann zeros and the explicit formula the Riemann approximation does much better than Li(x) with a simple adjustment.



      - David



      --- In primenumbers@yahoogroups.com, Matteo Mattsteel Vitturi <mattsteel@...> wrote:

      >

      >

      > Hello Norman.

      > Thank you for your reply: I don't think you give me a

      > stupid answer and I'm not looking for good approx of pi(x).

      > All I

      > said happened by chance and, later, I thought to write to this list.

      > I

      > know a (compicated) formula of pi(x) that takes into account

      > non-trivial zeroes of the zeta-function is common-knowledge, but I'm

      > naively impressed by the fact that when a "simple number" as x^(1/pi) is

      > subtracted to Li(x) then the approximation is improved so much

      > (examining the first two millions integers).

      > Oh, well: I'm using

      > Li(x) = Integral over t from 2 to x of (dt/ln(t))

      > For example,

      > rounding to the greatest integer, we have

      > Li(10)-10^(1/pi) = 4 =

      > pi(10)

      > Li(100)-100^(1/pi) = 25 = pi(100)

      > Li(1000)-1000^(1/pi) =

      > 168 = pi(1000)

      > Li(10000)-10000^(1/pi) = 1227  [that is 2 less than

      > pi(10000)=1229]

      > Li(100000)-100000^(1/pi) = 9592 = pi(100000)

      > Li(1000000)-1000000^(1/pi)

      > = 78563 [that is 65 over pi(1000000)=78498]

      > I know that as few as

      > three evidence is nothing compared to infinity, and this is the reason

      > to ask for your advice.

      > Regards.

      > M.

      >

      >

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