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the three limits of classical prime theory-> a contradiction

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  • Roger Bagula
    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-
    Message 1 of 4 , Nov 3, 2003
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      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/
    • 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 2 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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        =CBU1
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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 3 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 4 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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