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Re: [PrimeNumbers] the three limits of classical prime theory-> a contradiction

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  • Décio Luiz Gazzoni Filho
    ... Hash: SHA1 Please, man, spare us the laughters (at your cost) and go learn some basic asymptotics. I suggest you look up in particular L Hôspital s rule
    Message 1 of 4 , Nov 3, 2003
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      Please, man, spare us the laughters (at your cost) and go learn some basic
      asymptotics. I suggest you look up in particular L'Hôspital's rule for
      limits.

      Remember: when we get tired of laughing at you, you'll just get moderated and
      then you'll lose your primary source of fun, which is trying to troll us.

      So at least be more intelligent in your attempts.

      Décio

      On Monday 03 November 2003 15:56, Roger Bagula wrote:
      > 1) Limit[Prime[n], n--> Infinity]=n*Log[n] ( Euler, I think)
      > 2) Limit[PrimePi[n],n->Infinity]=n/Log[n] (Hadamard)
      > 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from Ulam,
      > in both books I have by him)
      >
      > How 3) can be made to contradict 2):
      > PrimePi[n]+CompositePi[n]=n
      > PrimePi[n]/CompositePi[n]=PrimePi[n]/(n-PrimePi[n])=1/Log[n]
      > Solving for
      > PrimePi[n]=n/(1+Log[n])
      >
      > I'm puzzled by this which I found this morning.
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      L8tE5TAevtUMoC5SzWpXx0w=
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    • Andy Swallow
      ... n/(log n) is asymptotic to n/(1+log n). Think about it. And 1) is certainly not due to Euler. 1) is an equivalent form of the prime number theorem, and so
      Message 2 of 4 , Nov 3, 2003
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        On Mon, Nov 03, 2003 at 09:56:54AM -0800, Roger Bagula wrote:
        > 1) Limit[Prime[n], n--> Infinity]=n*Log[n] ( Euler, I think)
        > 2) Limit[PrimePi[n],n->Infinity]=n/Log[n] (Hadamard)
        > 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from Ulam,
        > in both books I have by him)
        >
        > How 3) can be made to contradict 2):
        > PrimePi[n]+CompositePi[n]=n
        > PrimePi[n]/CompositePi[n]=PrimePi[n]/(n-PrimePi[n])=1/Log[n]
        > Solving for
        > PrimePi[n]=n/(1+Log[n])
        >
        > I'm puzzled by this which I found this morning.

        n/(log n) is asymptotic to n/(1+log n). Think about it.

        And 1) is certainly not due to Euler. 1) is an equivalent form of the
        prime number theorem, and so wasn't proved until Hadamard (and de la
        Vallee Poussin independently) proved it in 1896. 3) follows painfully
        obviously from 2), and so doesn't need to be quoted from anywhere.

        Before the PNT was proved, it was only known that pi(x) was of order
        x/log x, which is weaker than it being asymptotically x/log x.

        Andy
      • Roger Bagula
        My mistake here was that: PrimePi[n]/CompositePi[n]=(1/Log[n])/(1-1/Log[n])=1/(Log[n]-1) That checks to give: PrimePi[n]=n/Log[n] I had the ratio not the
        Message 3 of 4 , Nov 4, 2003
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          My mistake here was that:
          PrimePi[n]/CompositePi[n]=(1/Log[n])/(1-1/Log[n])=1/(Log[n]-1)
          That checks to
          give:
          PrimePi[n]=n/Log[n]
          I had the ratio not the density.
          I'm sorry.

          Roger Bagula wrote:

          > 1) Limit[Prime[n], n--> Infinity]=n*Log[n] ( Euler, I think)
          > 2) Limit[PrimePi[n],n->Infinity]=n/Log[n] (Hadamard)
          > 3) Limit[PrimePi[n]/CompositePi[n],n-> Infinity]=1/Log[n] ( from
          > Ulam, in both books I have by him)
          >
          > How 3) can be made to contradict 2):
          > PrimePi[n]+CompositePi[n]=n
          > PrimePi[n]/CompositePi[n]=PrimePi[n]/(n-PrimePi[n])=1/Log[n]
          > Solving for
          > PrimePi[n]=n/(1+Log[n])
          >
          > I'm puzzled by this which I found this morning.
          >

          --
          Respectfully, Roger L. Bagula
          tftn@..., 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
          URL : http://home.earthlink.net/~tftn
          URL : http://victorian.fortunecity.com/carmelita/435/
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