Loading ...
Sorry, an error occurred while loading the content.

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

Expand Messages
  • 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 1 of 4 , Nov 3, 2003
    • 0 Attachment
      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 2 of 4 , Nov 4, 2003
      • 0 Attachment
        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/
      Your message has been successfully submitted and would be delivered to recipients shortly.